University College Dublin
Mathematics, Course
MATH40060 (formerly called MATH4105) Differential Geometry url:- http://mathsci.ucd.ie/courses/math40060
September 23, 2009
Dr. J. Brendan Quigley
comments to:- Dr J.Brendan.Quigley Department of Mathematics University College Dublin, Belfield ph. 716–2584, 716–2580; fax 716–1196 email [email protected]. prepared using LATEX running under Redhat Linux drawings prepared using gnuplot ii
c jbquig-UCD September 23, 2009 Contents
0 Organization of math40060 semester1 yr 09-10 1
1 geometry and SO(3,R) 3 1.1 Euclidean three space R3 ...... 3 1.1.1 notation and technicalities ...... 3 1.1.2 geometric concepts; distance, angle, volume, orientation ...... 4 1.1.3 Cauchy-Schwarz inequality (CSI) ...... 4 1.1.4 justifying our definition of distance §§(1.1) ...... 5 1.1.5 justifying our definition of angle, (formula(1.2)) ...... 6 1.1.6 justifying the definition of signed volume ...... 6 1.1.7 positively oriented orthonormal basis ...... 8 1.1.8 special orthogonal change of basis matrix ...... 8 1.2 projection, reflection, rotation ...... 9 1.2.1 line and plane,vector resolution ...... 9 1.2.2 Five linear mappings and their matrices ...... 9 1.2.3 Five linear mappings by similarity transform ...... 12 1.2.4 characteristic polynomial, traces, determinant ...... 14 1.3 Lie group SO(3,R) and Lie algebra so(3,R) ...... 15 1.3.1 the Lie group SO(3,R) ...... 15 1.3.2 equivalence of special orthogonal and rotation ...... 17 1.3.3 rotation continuously varying in time ...... 19 1.3.4 the Lie algebra so(3,R) ...... 20 1.3.5 SO(3,R) and so(3,R), standard form, eigenvalues and eigenvectors ...... 21 1.4 exponential matrix ...... 21 1.4.1 exp(0) ...... 23 1.4.2 exp(A)exp(B) ...... 23 1.4.3 powers and exponentials of similar matrices ...... 24 d 1.4.4 exp(tA) ...... 24 dt 1.4.5 Convergence of the exponential matrix power series ...... 25 1.5 exponential of angular velocity, exp(tDn), ∈ so(3,R) ...... 25 1.6 Euler angles ...... 26 1.6.1 formulae for rotation ...... 27 1.7 topology of SO(3,R) ...... 28 1.7.1 the Quaternionic sphere as a Lie group ...... 28 1.7.2 the Lie group SU(2,C) ...... 28 1.7.3 the Lie algebra su(2,C) ...... 29 1.7.4 su(2,C) as an inner product space ...... 29 1.7.5 mapping F from SU(2,C) to SO(3,R) ...... 30 1.7.6 SU(2,C) is a double cover of SO(3,R)...... 30 1.8 problem set ...... 31
iii iv CONTENTS
2 curves 35 2.1 plane curves ...... 35 2.1.1 parametrization of regular curves ...... 35 2.1.2 re parametrization ...... 36 2.1.3 velocity, speed, acceleration ...... 37 2.1.4 arc-length ...... 38 2.1.5 arc-length parametrization ...... 39 2.2 curvature ...... 42 2.2.1 technicalities ...... 42 2.2.2 the Serret-Frenet frame ...... 43 2.2.3 classical definition of curvature ...... 43 2.2.4 curvature as angular speed ...... 45 2.2.5 a.l.p. of the osculating circle ...... 46 2.2.6 formulae for t,n,κ in R2 ...... 47 2.2.7 involute and evolute ...... 48 2.3 curves in R3 ...... 49 2.3.1 curves in R3, basic definitions ...... 50 2.3.2 Serret-Frenet formulae ...... 52 2.3.3 3-space formulae for t,n,b,κ,τ ...... 54 2.3.4 curves in R3 with constant curvature and torsion ...... 58 2.4 problem set ...... 61
3 surfaces 65 3.1 surfaces in R3 ...... 65 3.2 surface presentation ...... 65 3.2.1 presentation as a level set or contour ...... 65 3.2.2 graphical presentation ...... 66 3.2.3 parametric presentation ...... 66 3.3 first example, the sphere ...... 67 3.4 saddle and monkey saddle ...... 69 3.5 surface of revolution ...... 69 3.5.1 the sphere revisited ...... 70 3.5.2 the torus as surface of revolution ...... 70 3.5.3 hyperboloids and cone as surfaces of revolution ...... 70 3.6 ruled surfaces ...... 72 3.6.1 cylinder as ruled surface ...... 72 3.6.2 cone as ruled surface ...... 73 3.6.3 single sheeted hyperboloid as ruled surface ...... 73 3.6.4 saddle surface is doubly ruled ...... 73 3.6.5 the right helicoid or screw surface is ruled ...... 74 3.6.6 the tangent developable helicoid ruled surface ...... 74 3.7 charts, affine linear approximate ...... 74 3.7.1 chart, surface element ...... 74 3.7.2 affine linear approximate ...... 75 3.7.3 the tangent plane ...... 75 3.7.4 normal vector ...... 76 3.8 Gauss and Weingarten maps ...... 79 3.9 self adjointness of L ...... 80 3.10 eigenvalues and eigenvectors of a self adjoint mapping ...... 81 3.11 invariants of the Weingarten mapping ...... 81 3.12 geometric meaning of Weingarten mapping ...... 82 3.13 problem set ...... 83
c jbquig-UCD September 23, 2009 CONTENTS v
4 bilinear forms 85 4.1 definition of bilinear forms I, II and III ...... 85 4.2 matrix representation of the three fundamental forms ...... 85 4.2.1 the matrix of a form ...... 86 4.2.2 g,h,e matrices representing I,II,III ...... 86 4.2.3 the matrix of the Weingarten mapping ...... 88 4.2.4 Examples, computing g,h,e,L ...... 88 4.3 explicit presentation and curvature ...... 93 4.4 surface as a graph over the tangent plane at a point ...... 95 4.5 problem set ...... 98
I Answers to problem sets of part I 101
5 Answers to questions in Chapter1 103 5.1 question and answer ...... 103 5.1.1 answer ...... 103 5.1.2 answer ...... 103 5.1.3 answer ...... 103 5.1.4 answer ...... 104 5.1.5 answer ...... 104 5.2 question and answer ...... 104 5.2.1 answer ...... 104 5.2.2 answer ...... 104 5.2.3 answer ...... 105 5.2.4 answer ...... 105 5.2.5 answer ...... 105 5.3 question and answer ...... 107 5.3.1 answer ...... 107 5.4 question and answer ...... 107 5.4.1 answer ...... 107 5.4.2 answer ...... 107 5.4.3 answer ...... 108 5.5 question and answer ...... 108 5.5.1 answer ...... 108 5.5.2 answer ...... 108 5.5.3 answer ...... 109 5.5.4 answer ...... 109 5.6 question and answer ...... 109 5.6.1 answer ...... 109 5.6.2 answer ...... 109 5.6.3 answer ...... 109 5.6.4 answer ...... 110 5.6.5 answer ...... 110 5.6.6 answer ...... 110 5.7 question and answer ...... 110 5.7.1 answer ...... 111 5.8 question and answer ...... 111 5.8.1 ?? ...... 111 5.8.2 answer ...... 112 5.8.3 answer ...... 112 5.8.4 answer ...... 112 5.9 question and answer ...... 112 5.9.1 answer ...... 113
September 23, 2009 c jbquig-UCD vi CONTENTS
5.9.2 answer ...... 113 5.9.3 answer ...... 113 5.9.4 answer ...... 113 5.10 question and answer ...... 113 5.10.1 answer ...... 114 5.10.2 answer ...... 114 5.10.3 answer ...... 114 5.10.4 answer ...... 115 5.10.5 answer ...... 115 5.11 question and answer ...... 115 5.11.1 answer ...... 116 5.11.2 answer ...... 116 5.11.3 answer ...... 117 5.11.4 answer ...... 117 5.11.5 answer ...... 117
6 Answers to questions in chapter2 119 6.1 question and answer ...... 119 6.1.1 logarithmic spiral ...... 119 6.1.2 archimedean spiral ...... 121 6.1.3 catenary ...... 121 6.1.4 cycloid ...... 122 6.1.5 cardioid ...... 123 6.1.6 tractoid ...... 123 6.2 question and answer ...... 124 6.2.1 t and n ...... 125 6.2.2 κ ...... 125 6.2.3 evolute parametrization, e(t) ...... 125 6.2.4 involute parametrization i(t) ...... 125 6.2.5 comparison Γ,E(Γ),I(Γ) ...... 125 6.3 question and answer ...... 125 6.3.1 drawing of curves ...... 126 6.3.2 the catenary ...... 126 6.3.3 tractrix ...... 126 6.3.4 involute of catenary ...... 127 6.3.5 evolute of the tractrix ...... 127 6.4 question and answer ...... 127 6.4.1 deltoid parametrization ...... 127 6.4.2 drawing ...... 127 6.4.3 t, n and κ ...... 128 6.4.4 evolute of the deltoid ∆ ...... 128 6.4.5 involute of the deltoid curve ...... 129 6.5 question and answer, helix ...... 130 6.5.1 answer ...... 130 6.5.2 answer ...... 130 6.5.3 ...... 130 6.5.4 ...... 131 6.5.5 ...... 131 6.6 question and answer, conical helix ...... 131 6.6.1 answer ...... 132 6.6.2 answer ...... 132 6.6.3 answer ...... 132 6.7 question and answer, Viviani’s curve ...... 133 6.7.1 ...... 133
c jbquig-UCD September 23, 2009 CONTENTS vii
6.7.2 answer ...... 133 6.7.3 answer ...... 133 6.7.4 answer ...... 134 6.7.5 answer ...... 134
7 Answers to questions in chapter3 137
8 Answers to questions in chapter(4) 139 8.1 question and answer ...... 139 8.1.1 answer ...... 139 8.2 question and answer ...... 139 8.2.1 answer ...... 139 8.3 question and answer ...... 140 8.3.1 answer ...... 140 8.4 question and answer ...... 140 8.4.1 answer ...... 140 8.5 question and answer ...... 140 8.5.1 answer ...... 141 8.6 question and answer ...... 141 8.6.1 answer ...... 141 8.7 question and answer ...... 141 8.7.1 answer ...... 141 8.8 question and answer ...... 141 8.8.1 answer ...... 142 8.9 question and answer ...... 142 8.9.1 answer ...... 142 8.10 question and answer ...... 142 8.10.1 answer ...... 142 8.11 question and answer ...... 143 8.11.1 answer ...... 143 8.12 question and answer ...... 143 8.12.1 answer ...... 143 8.13 question and answer ...... 143 8.13.1 answer ...... 143
September 23, 2009 c jbquig-UCD viii CONTENTS
c jbquig-UCD September 23, 2009 List of Figures
2.1 (i) circle (ii) cubic parabola (iii) cycloid ...... 36 2.2 kArchimedean spiral ...... 40 2.3 (i) helix(3 turns) (ii) intersection cylinder and screw surfaces ...... 50 2.4 Viviani’s curve, insersection of sphere and cylinder ...... 52
3.1 two parametrizations of each of the saddle and monkeysaddle ...... 68 3.2 sphere and torus surfaces ...... 69 3.3 single and double sheeted hyperboloids, cone, hyperbolic system ...... 71 3.4 cylinder ruled and rotated, saddle doubly ruled ...... 73 3.5 the single sheeted hyperbola as ruled surface ...... 74 3.6 right and tangent developable helicoid ruled surfaces ...... 75
4.1 (i) elliptic (ii) hyperbolic (iii) parabolic (iv) umbilic ...... 96
6.1 (i) conical helix curce (ii) cone and screw surfaces intersect ...... 120
ix x LIST OF FIGURES
c jbquig-UCD September 23, 2009 Chapter 0
Organization of math40060 semester1 yr 09-10
Times
day time venue lecturer topic Tue 02 : 00 − 02 : 50 SR − math − SB JBQ Di f f erentialGeometry Wed 04 : 00 − 04 : 50 TR − math − SB JBQ Di f f erentialGeometry Thu 011 : 00 − 011 : 50 SR − math − SB JBQ Di f f erentialGeometry
JBQ=J.Brendan Quigley SB=Science Building Belfield, EB=Engineering Building Belfield
Lecturer J.Brendan Quigley room:- 35, Science Building ph:- 716-25848, 716-2500(sec) email:- [email protected]
Books
see bibliography at end
urls (universal resource locators)
http://maths.ucd.ie/courses/math40060
1 2 CHAPTER 0. ORGANIZATION OF MATH40060 SEMESTER1 YR 09-10
......
c jbquig-UCD September 23, 2009 Chapter 1 geometry and SO(3,R)
1.1 Euclidean three space R3 1.1.1 notation and technicalities We use standard notation for R3 its points and inner product. We write
x x1 3 x = y = x2 ∈ R z x3
We use columns and superscripts. We have three standard unit basic vectors in R3
1 0 0 x i = 0 , j = 1 , k = 0 , x = y = xi + yj + zk 0 0 1 z
Let a d g p = b , q = e and r = h c f i We have the standard inner product scalar
a d < p,q > = < b , e > = ad + be + c f ∈ R c f
The row vector qt = (d,e, f ) is the transpose of the column vector q. We have a useful expression for the inner product in terms of the transpose.
a t < p,q > = q p = (d,e, f ) b c
The norm of a vector in R3 is a scalar. p ||p|| = < p,p >1/2 = a2 + b2 + c2
In R3 we have the cross-product vector a d i j k b f − ce
p × q = b × e = a b c = cd − a f
c f d e f ae − bd
3 4 CHAPTER 1. GEOMETRY AND SO(3,R)
The triple product of vectors is the scalar
a p g a d g < p × q,r > = det(p,q,r) = < b × q , h > = det b e h ∈ R c r i c f i
So far < p,q >∈ R, ||p||, p×q ∈ R3 and < p×q,r > are mere number formulae ; but fundamental geometric concepts can be expressed (and even clarified) in terms of these formulae. These geometric concepts are, angle, distance, volume and orientation.
1.1.2 geometric concepts; distance, angle, volume, orientation remark To do geometry in R3 we need three concepts, distance, angle, and signed volume. We define these in terms of the the technical devices developed in §§(1.1.1). One might reasonably object on two grounds. Are these definitions valid? Do they describe the real world. We will address these objections in §§(1.1.3)–§§(1.1.6).
In what follows you will see that distance and angle are generated by the inner product alone. The inner product satisfies the famous CSI inequality. The CSI is used to prove that our “distance” is a proper distance and our “angle” a proper angle and that both describe the real world.
The inner product can be used to define positive volume but does not suffice to define signed volume, i.e to incorporate orientation into volume. We need the scalar triple product which contains the cross product which is an extra ingredient. Orientation, in the real world, is FRHR versus FLHR.
We define d(p,q) the distance in R3 from p to q.
q d(p,q) = ||q − p|| = < q − p,q − p >1/2 = (d − a)2 + (e − b)2 + ( f − c)2 (1.1)
We define the angle θ, 0 ≤ θ < π, between vectors p,q, by
< p,q > θ = arccos , 0 ≤ θ ≤ π (1.2) ||p|| · ||q||
We define signed volume of the ordered triple (u,v,w) of vectors to be
< u × v,w > = det(u,v,w) ∈ R (1.3)
The (unsigned ) volume of the triple is
|< u × v,w >| = |det(u,v,w)| ≥ 0 (1.4)
+1 if < u × v,w > > 0 If volume is not zero the orientation of the triple is defined and is −1 if < u × v,w > < 0
1.1.3 Cauchy-Schwarz inequality (CSI) Here is a useful equation, it is in fact a concealed form of, cos2 +sin2 = 1. lemma 1 < p,q >2 + ||p × q||2 = ||p||2 · ||q||2 (1.5)
c jbquig-UCD September 23, 2009 1.1. EUCLIDEAN THREE SPACE R3 5
proof
< p,q >2 + ||p × q||2 a d i j k 2 2 = < b , e > + || a b c ||
c f d e f h i = (ad + be + c f )2 + (b f − ce)2 + (cd − a f )2 + (ae − bd)2 = (a2d2 + b2e2 + c2 f 2 + 2adbe + 2adc f + 2bec f ) + (b2 f 2 − 2b f ce + c2e2) + (c2d2 − 2cda f + a2 f 2) + (a2e2 − 2aebd + b2d2) = (a2d2 + b2e2 + c2 f 2) + (b2 f 2 + c2e2) + (c2d2 + a2 f 2) + (a2e2 + b2d2) = (a2 + b2 + c2)(d2 + e2 + f 2) = ||p||2||q||2
Immediately we obtain the famous Cauchy-Schwarz inequality (CSI) in R3. lemma 2
< p,q >2 ≤ ||p||2||q||2, ∀p,q ∈ R3, ( equality iff p,q are l.d.) (1.6)
1.1.4 justifying our definition of distance §§(1.1) A valid distance function d : R3 × R3 → R (p,q) 7→ d(p,q) must satisfy 3 conditions. positive definiteness d(p,q) ≥ 0, = 0, iff p = q symmetry d(p,q) = d(q,p). triangle law d(p,r) ≤ d(p,q) + d(q,r) We must check validity of d, see definition (1.1) Positive definiteness holds since, it is easily checked (for any v ∈ R3) that ||v|| ≥ 0, = 0 iff v = 0. Symmetry holds since || − v|| = ||v|| for any v ∈ R3. Triangular law is a little harder to check but writing v = q − p and w = r − q this law follows from
||v + w|| ≤ ||v|| + ||w||
But
||v + w|| ≤ ||v|| + ||w|| ⇔ (||v + w||)2 ≤ (||v|| + ||w||)2 ⇔ < v + w,v + w > ≤ ||v||2 + 2||v|| · ||w|| + ||w||2 ⇔ < v,v > +2 < v,w > + < w,w > ≤ < v,v > +2||v|| · ||w||+ < w,w > ⇔ < v,w > ≤ ||v|| · ||w||
September 23, 2009 c jbquig-UCD 6 CHAPTER 1. GEOMETRY AND SO(3,R)
which follows from the CSI written in the form.
−||v|| · ||w|| ≤ < v,w > ≤ ||v|| · ||w||
remark Note that
d(p,q)2 = < q − p,q − p > = (d − a)2 + (e − b)2 + ( f − c)2 (1.7)
This is 3 dimensional version of Pythagoras theorem (make a little sketch). The square on the hypotenuse of the right quadrangle is equal to the sum of the squares on the other three sides. remark I n the real world equation (1.7) holds being Pythagoras’s theorem applied twice. Thus our dis- tance function (1.1) is that of the real world.
1.1.5 justifying our definition of angle, (formula(1.2)) The CSI tells us that ||p|| · ||q|| ≤ < p,q > ≤ ||p|| · ||q|| (1.8) and that < p,q > −1 ≤ ≤ 1 (1.9) ||p|| · ||q|| The fraction in equation(1.9) lies between ±1: it is a genuine cosine! This proves that our definition(1.2) of angle is valid. remark F rom (1.2) we obtain the often used
< p,q > = ||p|| · ||q|| · cosθ (1.10) remark Consider the triangle ∆0pq. The (real world) cosine rule of trigonometry says
d(p,q)2 = d(0,p)2 + d(0,q)2 − 2d(0,p)d(0,q)cos(6 p0q) ⇔ < q − p,q − p > − < p,p >2 − < q,q)2 = −2||p||||q||cos(6 p0q) ⇔ −2 < p,q > = −2||p||||q||cos(6 p0q) < p,q > ⇔ 6 p0q = arccos ||p||||q|| Which proves that our definition of angle(1.2) is that of the real world.
1.1.6 justifying the definition of signed volume §§() remark What criteria should a genuine “signed volume” satisfy? Does det(−,−,−) satisfy these criteria, i.e. is this a genuine signed volume? Does det(−,−,−) correspond to the commonly accepted value of signed volume, i.e. does det(−,−,−) represent the real world? To answer these questions. (i) We clarify the usual definition of ordinary signed volume osv(−,−,−). (ii) From osv(−,−,−) we extract the criteria which a genuine signed volume should have.
c jbquig-UCD September 23, 2009 1.1. EUCLIDEAN THREE SPACE R3 7
(iii) We prove that det(−,−,−) satisfies these criteria.
(iv) We prove, see theorem(1) that there is only one genuine signed volume function on R3.
(v) Thus det(−,−,−) = osv(−,−,−). remark common definition of signed volume The ordinary person understands volume as a positive quantity. Any three vectors u,v and w generate a parallelepiped P = P(u,v,w) ⊂ R3. P has eight vertices 0,u,u + v,v,w,u + w,u + v + w,v + w, you may sketch P. The ordinary volume of P can be computed using lengths and angles, a gruesome computation. This volume can be signed, ±1 according as u,v and w follow FRHR or FLHR. This (hopefully) clarifies the ordinary concept of signed volume osv(u,v,w). remark criteria satisfied by a signed volume The reader should check (draw diagrams) that osv is linear in all three vector variables. The reader should check that osv is skew symmetric, i.e. that switching two vectors reverses FRHR and so negates osv. The reader should check that osv(i,j,k) = 1, i.e. that the standard cube has ordinary signed volume equal to +1. We conclude that a genuine signed volume function K should be a skew-symmetric trilinear form
K : R3 × R3 × R3 → R (u,v,w) 7→ K(u,v,w) with K(i,j,k) = 1. By trilinear we mean that K(u,v,w) is linear in each vector variable u,v and w. By skew-symmetric we mean that
K(v,u,w) = −K(u,v,w) switch u,v K(w,v,u) = −K(u,v,w) switch u,w K(u,w,v) = −K(u,v,w) switch v,w
remark det(−,−,−) satisfies the criteria It is trivial to check that
< u × v,w > = det(u,v,w) is linear in all three vector variables and is skew-symmetric, use column operations on the determinant. It is trivial that det(i,jk) = 1. Thus det(−,−,−) is a genuine signed volume.
theorem 1 There exists exactly one skew symmetric trilinear form K on R3 with K(i,j,k) = 1. proof Since det(−,−,−) is such a form we need only prove uniqueness. Let K be such a form. Each of p,q,r is a linear combination of {i,j,k}. By trilinearity K(p,q,r) is completely determine by just 27 values of K where each entry is one of i,j,k. By skew symmetry K = 0 if two values are equal. Thus K depends on only 6 values where the entries are permutations of i,j,k. But by repeated switching any permutation can be reduced to the trivial (i,j,k). Thus K depends on one value K(i,j,k) which must be 1. Thus K is unique.
September 23, 2009 c jbquig-UCD 8 CHAPTER 1. GEOMETRY AND SO(3,R)
1.1.7 positively oriented orthonormal basis Let {l,m,n} is a basis for R3. We say that {l,m,n} is an orthonormal basis iff the vectors are mutually orthogonal < m,n > = < n,l > = < l,m > = 0 and each is of length 1 (normal) ||l|| = ||m|| = ||n|| = 1. We say that {l,m,n} is positively oriented iff
< l × m,n > = det(l,m,n) > 0
It is trivial to check that i,j,k is a positively oriented orthonormal basis (PONB). ex. If {l,m,n} is a PONB then det(l,m,n) = 1. ex. If l ∈ R3 we can find m 6= 0 with < l,m >= 0. l m l × m Then , , , is a PONB for R3. ||l|| ||m|| ||l|| · ||m|| If, in addition, ||l|| = ||m|| = 1 then {l,m,l × m} is a PONB.
1.1.8 special orthogonal change of basis matrix Let {l,m,n} be a PONB, see §§(1.1.7) for R3. Let E be the 3 × 3 matrix
E = (l,m,n) (1.11)
Then we have the matrix equation E−1 = Et (1.12) Indeed
Et E = (l,m,n)t (l,m,n) lt = mt (l,m,n) nt lt l lt m lt n = mt l mt m mt n nt l nt m nt n < l,l > < l,m > < l,n > = < m,l > < m,m > < m,n > < n,l > < n,m > < n,n > 1 0 0 = 0 1 0 0 0 1 = I remark Later we will refer to any matrix E with E−1 = Et and with detE > 0 as special orthogonal. Such matrices will play an important role in this course.
c jbquig-UCD September 23, 2009 1.2. PROJECTION, REFLECTION, ROTATION 9
remark The matrix equation E = IE can be written
(l,m,n) = (i,j,k)E (1.13)
We can interpret this as the statement that changing from the standard PONB to a general PONB can be accomplished by means of a special orthogonal change of basis (COB) matrix E. remark Note that
Ei = l ⇔ Et l = i , Ej = m ⇔ Et m = j , Ek = n ⇔ Et n = k (1.14)
1.2 projection, reflection, rotation 1.2.1 line and plane,vector resolution 3 2 2 Let n = ai+bj+ck ∈ R be a unit vector, ||n|| = 1 ⇔ a +b = 1. Let L = Ln denote the line through the origin 0 in the direction n. Let P = Pn denote the plane through 0 with normal vector n. x x = λa L = {λn |λ ∈ R } = y y = λb , λ ∈ R (1.15)
z z = λc x P = {x | < x,n >= 0 } = y ax + by + cz = 0 (1.16)
z Here we see the parametric equation of L and the implicit equation of P.
Let x ∈ R. Then
xL =< x,n > n ∈ L , xP = x− < x,n > n ∈ P , xL ⊥ xP and x = xL + xP (1.17)
We have resolved x into its components in L (⊥ P) and P. xL is called the orthogonal projection of x into the line L.xP is called the orthogonal projection of x into the plane P. Consider the vector x − 2 < x,n > n = x− < x,n > n + − < x,n > n
We can think of this vector as resolving x as above and then building a new vector but with the component perpendicular to the plane P reversed. We have found the reflection of x in the plane P.
1.2.2 Five linear mappings and their matrices We will study 5 linear mappings associated with the unit vector n = ai + bj + ck ∈ R3. We will compute the 3 × 3 matrix (w.r.t. the standard basis {i,j,k}) of each.
3 3 (i) A = An : R → R being orthogonal projection onto the line L = Ln. 3 3 (ii) B = Bn : R → R being orthogonal projection onto the plane P = Pn. 3 3 (iii) C = Cn : R → R being reflection in the plane P = Pn. 3 3 (iv) D = Dn : R → R being orthogonal projection into the plane P followed by rotation by π/2 about the normal vector n.
3 3 3 (v) E = En,θ : R → R being rotation of R by angle θ about the axis n of rotation.
September 23, 2009 c jbquig-UCD 10 CHAPTER 1. GEOMETRY AND SO(3,R)
We have A : R3 → R3 x 7→ < x,n > n = n < x,n >= nnt x Thus a a2 ab ac t A = nn = b (a,b,c) = ba b2 bc (1.18) c ca cb c2 Arguing similarly 1 − a2 −ab −ac t B = I − nn = −ba 1 − b2 −bc (1.19) −ca −cb 1 − c2 Again 1 − 2a2 −2ab −2ac t C = I − 2nn = −2ba 1 − 2b2 −2bc (1.20) −2ca −2cb 1 − 2c2
Let x ∈ R3. We claim that Dx = n × x. First n × x ⊥ n which proves that n × x ∈ P. Second n × x ⊥ Bx = (x− < x,n > n) since n × x ⊥ x and n × x ⊥ n and Bx is a linear combination of x and n. Thus indeed Dx = nx. We introduce the sensible notation D = [n×]. We must find the matrix of D. But
bz − cy 0 −c b x Dx = [n×]x = n × x = cx − az = c 0 −a y ay − bx −b a 0 z
Summing up 0 −c b D = [n×] = c 0 −a (1.21) −b a 0
Consider the simplest case of rotation S = Sk,θ by angle θ and with axis k. S is completely determined by its action on the PONB {i,j,k}. A simple sketch shows that
Si = cosθi + sinθj , Sj = −sinθi + cosθj , Sk = k (1.22)
Thus the matrix of this simplest rotation is
cosθ −sinθ 0 S = Sk,θ = sinθ cosθ 0 (1.23) 0 0 1
With this preparation we are ready to tackle the general rotation by angle θ with axial vector n. The three vectors {Bx,Dx,Ax} are not quite a PONB for R3 but have useful properties, (exercises, see exercise sheet §§1.8) question(5))
(i) Ax is axial, i.e. a scalar multiple of n.
(ii) Bx,Dx,Ax are mutually orthogonal.
(iii) ||Bx|| = ||Dx||
(iv) The triple Bx,Dx,n is positively oriented, det(Bx,Dx,n) > 0.
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Arguing as for the simplest case, see equation(1.22)
S(Bx) = cosθBx + sinθDx , S(Dx) = −sinθBx + cosθDx , S(Ax) = Ax (1.24)
We can now compute
Sx = S(Ax + Bx), since x = Ax + Bx = S(Ax) + S(Bx)), by linearity of S = Ax + S(Bx), since Ax is a multiple of n and Sn = n = Ax + cos(θ)Bx + sin(θ)Dx compare S(i) in equation(1.22)
We have an expression for the rotation S = Sn,θ
S = A + cos(θ)B + sin(θ)D = nnt + cos(θ)(I − nnt ) + sin(θ)[n×] (1.25)
And we can compute the matrix.
a2 ab ac 1 − a2 −ab −ac 0 −c b S = ba b2 bc + cos(θ) −ba 1 − b2 −bc + sin(θ) c 0 −a (1.26) ca cb c2 −ca −cb 1 − c2 −b a 0 remark ambiguity in definition of rotation This remark might be skipped on a first reading. There are two rotations about axis n by angle θ; the expression is ambiguous. Appealing to geometric intuition we always mean that rotation which conforms to FRHR. But this is vague. Here is the clarification
By S the rotation about n by θ we mean that S has the property
sin(θ)det(x,Sx,n) ≥ 0 for all x ∈ R3 (1.27)
Let us check that we have made the right choice in our definitions of D and S above.
First
sin(π/2)det(x,Dx,n) = 1 · det(x,n × x,n) = det(n,x,n × x) = < n × x,n × x > = ||n × x||2 ≥ 0
Second
sin(θ)det(x,Sx,n) = sin(θ)det(x,Ax + cos(θ)Bx + sin(θ)Cx,n) h i = sin(θ) det(x,Ax,n) + cos(θ)det(x,Bx,n) + sin(θ)det(x,Cx,n) h i = sin(θ) det(x,< n,x > n,n) + cos(θ)det(x,x− < n,x > n,n) + sin(θ)det(x,Cx,n) h i = sin(θ) det(x,< n,x > n,n) + cos(θ)det(x,x,n) + cos(θ) < n,x > det(x,n,n) + sin(θ)det(x,Cx,n) h i = sin(θ) 0 + 0 + 0 + sin(θ)det(x,n × x,n) = sin2(θ)det(x,n × x,n) = sin2(θ)det(n,x,n × x) = sin2(θ) < n × x,n × x > = sin2(θ)||n × x||2 ≥ 0
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We will be mainly interested in rotation but to a lesser degree in reflection. In due course we will develop 4 different constructions for the matrix of rotation S.
The following formulae will prove very useful, the proofs are left as exercises. All become obvious when expressed in geometric language. ex. see question sheet §§(1.8) question(8) (i) AA = A , BB = B , CC = I , DD = −B (ii) AB = BA = AD = DA = 0 , BD = DB = D (iii) At = A , Bt = B , Ct = C−1 = C , Dt = −D (iv) S−1 = St
ex. see question sheet §§(1.8) question(2) Let n be the unit vector in the direction i + j + k. (i) Compute A,B,C,D (ii) Prove that B “is” the branch of mechanical drawing known as isometric projection. (iii) Compute rotation S when θ = 2π/3.
1.2.3 Five linear mappings by similarity transform The standard versions of the five linear mappings in §§(1.2.2) are by definition found by putting n = k. These standard mappings are
0 0 0 t ortho-projection onto z-axis A = kk = 0 0 0 0 0 1 1 0 0 t ortho-projection onto xy-plane B = I − kk = 0 1 0 0 0 0 1 0 0 t reflection in xy-plane C = I − 2kk = 0 1 0 0 0 −1 rotation by π/2 about z-axis 0 −1 0 with D = [k×] = 1 0 0 ortho-projection onto xy-plane 0 0 0 cosθ −sinθ 0 t t rotation by θ about z-axis Sθ = kk + cos(θ)(1 − kk ) + sinθ[k×] = sinθ cosθ 0 0 0 1
Let n ∈ R3 be an arbitrary unit vector. Recall from §§(1.1.8) that we can concoct a PONB (l,m,n) with n as third entry. The general linear mappings An,Bn,Cn,Dn,Sn,θ become standard when referred to this new basis and so their matrices w.r.t. to the basis {l,mn} are the simple matrices A,B,C,D,Sθ. We want the matrices of these linear mappings but referred to the standard basis {i,j,k}. Let E be the 3 ×3 change of basis matrix
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from the standard PONB {i,j,k} to the new PONB {l,m,n}. As discussed in §§(1.1.7) the COB matrix E is special orthogonal with (easily computed) inverse E−1 = Et .
{l,m,n} = {i,j,k}E
By standard linear algebra the matrices w.r.t. one basis are obtained from those w.r.t. the other basis by similarity transform, using the COB matrix E as follows
−1 t An = EAE = EAE (1.28) −1 t Bn = EBE = EBE (1.29) −1 t Cn = ECE = ECE (1.30) −1 t Dn = EDE = EDE (1.31) −1 t Sn,θ = ESθE = ESθE (1.32) (1.33) remark H ere is an expanded version of the rotation formula(1.32). Suppose
d g a l = e , m = h , n = b f i c
Then d g a cosθ −sinθ 0 d e f Sn,θ = e h b sinθ cosθ 0 g h i (1.34) f i c 0 0 1 a b c
3 0 0 03 00 00 003 remark H ere is what you need to know about change of basis in R . Two bases {e1,e2,e } and {e1,e2,e } are related by a 3 × 3 COB matrix 00 00 00 0 0 0 {e1,e2,e3} = {e1,e2,e3}E There are 5 abstract linear algebra entities which concern us
(i) A scalar α ∈ R.
(ii) A vector v ∈ R3.
(iii) A scalar valued linear mapping f : R3 → R
(iv) A vector valued linear mapping K : R3 → R3
(v) A bi-linear form B : R3 × R3 → R.
With respect to the first basis these abstract entities can be represented by concrete blocks of numbers
(i) α by itself a 1 × 1 matrix.
(ii) v by v0 a column vector or 3 × 1 matrix.
(iii) f by f 0 a row vector or 1 × 3 matrix. n o 0 0 0i (iv) K by K a 3 × 3 matrix, K = K j 1 ≤ i, j ≤ 3 . n o 0 0 0 (v) B by B a 3 × 3 matrix B = Bi, j 1 ≤ i, j ≤ 3 .
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With respect to the second basis these representations become different concrete blocks of numbers α,v00,f00,K00,B00.
The punchline is the five rules of change
α = α, scalars do not change v00 = E−1v0 f00 = f0E K00 = E−1K0E, change by similarity B00 = Et B0E, change by conjugacy
If the COB matrix is special orthogonal (which it often will be in this course) then E−1 = Et and life becomes easier or more confusing according to your point of view.
Note that a 3×3 matrices may represent a linear mapping in which case entries are sub and super-scripted or it may represent a bi-linear form in which case entries are double sub-scripted. In the first case change is by similarity in the second by conjugacy. A matrix may appear in a third guise, as a change of basis matrices.
1.2.4 characteristic polynomial, traces, determinant As you will know already the characteristic polynomial of the real 3 × 3 matrix
a b c K = d e f g h i is 3 2 p = pK(λ) = det(λI − K) = λ + Aλ + Bλ +C = (λ − α)(λ − β)(λ − γ) Here α,β,γ are zeros of p called eigenvalues of K and the coefficients are given by
−A = tr K = d + h + c = α + β + γ
e f a c a b +B = tr K = + + = βγ + γα + αβ h i g i d e −C = detK = αβγ
Since the matrix K is assumed real p is a real polynomial, i.e. has real coefficients. Thus, by the following well known argument if λ is a zero of p then the complex conjugate λ is also a zero.
p(λ) = p(λ) = 0 = 0
Thus eigenvalues of a real matrix are either real or come in complex conjugate pairs (1.35) The characteristic polynomial is invariant under similarity transformation
p(EKE−1) = det(λI − EKE−1) = det(EλIE−1 − EKE−1) = det(E(λI − K)E−1) = det(E) · det(λI − K) · det(E−1) = det(E) · det(λI − K) · (det(E))−1 = det(λI − K) = p(E)
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and thus the traces, determinant and eigenvalues are also invariant.
From formulae(1.28) the characteristic polynomial, eigenvalues, traces and determinant are the same (inde- pendent of n) for all An,Bn,Cn,Dn,S(n,θ) and are readily computed from the standard versions
p tr tr det α,β,γ A λ3 − λ2 1 0 0 1,0,0 B λ3 − 2λ2 + λ 2 1 0 1,1,0 (1.36) C λ3 − λ2 − λ + 1 1 −1 −1 1,1,−1 D λ3 + λ 0 1 0 0,±i S λ3 − (1 + 2cos(θ))λ2 + (1 + 2cos(θ))λ − 1 1 + 2cos(θ) 1 + 2cos(θ) 1 1,e±iθ The reader should familiarize himherself with the eigenspace of each real eigenvector for each of A,B,C,D,S. For example n is an eigenvector associated with the eigenvalue λ = 1 of rotation Sn,θ, the associated eigenspace is the line Ln; i.e every vector on the axis of rotation is fixed. For a reflection, Cnx = 1 · x for all x ∈ Pn and Cn = −1 · n; vectors in the plane of reflection are fixed but the normal vector is reversed.
1.3 Lie group SO(3,R) and Lie algebra so(3,R) 1.3.1 the Lie group SO(3,R) Recall the definition of a group.A group is a set G with an associative operation · : G × G 3 ((g,h)) 7→ g · h = gh ∈ G there is a unique identity element e ∈ G, inverses exist: see algebra course for details and examples. The reader is most likely familiar with finite groups. In this course the underlying set G will not be finite; the typical group element g will vary through a differential manifold G and the multiplication will be continuous, in fact infinitely differentiable. Such a group is called a Lie group. We do not pursue these technicalities but rather study one very important Lie group SO(3,R).
3 3 Mat(3,3,R) = A : R → R A is linear (1.37) Thus Mat(3,3,R) is the set of all 3 × 3 real matrices; this is not a group A−1 does not exist if detA = 0.
GL(3,R) = {A ∈ Mat(3,3,R) | detA 6= 0 } (1.38) GL(3,R) is a Lie group called the general linear group in three dimensions. It consists of all invertible 3 ×3 matrix. So far we have not mentioned geometry in the present context.
A ∈ Mat(3,3,R) carries an ordered triple of vectors (p,q,r) to the ordered triple (Ap,Aq,Ar) with signed volume det(Ap,Aq,Ar) = det A(p,q,r) = det(A)det(pq,r) Thus det(A) is the factor by which signed volume is scaled by the linear mapping A. In particular A preserves orientation (FRHR) iff det(A) > 0. Ordinary (positive) volume is scaled by the factor |det(A)|.
GL+(3,R) = {A ∈ Mat(3,3,R) | detA > 0 } (1.39) GL+(3,R) ⊂ GL(3,R) is a Lie subgroup and consists of orientation preserving linear mappings A : R3 → R3.
SL(3,R) = {A ∈ Mat(3,3,R) | detA = 1 } (1.40) SL(3,R) ⊂ GL+(3,R) ⊂ GL(3,R) is a Lie subgroup and consists of all volume and orientation preserving linear mappings A : R3 → R3.
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A ∈ GL(3,R) is said to be orthogonal iff A preserves distance and angle which is the same as preserving the inner product. The set of all such linear mappings forms a Lie group called the orthogonal group and is de- noted O(3,R). It is easy but not completely trivial to prove (ex) equivalence of the following nine definitions.
A ∈ Mat(3,3,R)is said to be orthogonal iff A preserves both the lengths of vectors and the angles between vectors. iff A preserves the inner product <,> iff < Ax,Ay > = < x,y >, ∀x,y iff (Ax)t (Ay) = xt y ∀x,y iff xt (At A)y) = xt y ∀x,y 1 0 0 t iff A A = I = 0 1 0 0 0 1
iff the columns {A1,A2,A3} of A are orthonormal 1 0 0 t iff AA = I = 0 1 0 0 0 1 iff the rows (transposed) {(A1)t ,(A2)t ,(A3)t } of A are orthonormal iff A−1 = At
We have t −1 O(3,R) = A ∈ Mat(3,3,R) A = A (1.41) but the geometric idea underpinning this is that A preserves the inner product and thus distances and angles but not necessarily orientation of space.
Finally if A ∈ Mat(3,3,R) preserves distance angle and orientation A is said to be special orthogonal (or a rigid transformation of space) and the set of all such linear mappings forms the special orthogonal Lie group
\ t SO(3,R) = O(3,R) SL(3,R) = A ∈ Mat(3,3,R) A = A and detA > 0 (1.42) ex. see question sheet §§1.8 question(??)
(i) If A ∈ SO(3,R) then det(A) = 1.
(ii) If A ∈ O(3,R) then det(A) = ±1.
(iii) Each rotation is special orthogonal.
(iv) Each reflection is orthogonal but not special.
(v) If A ∈ O(3,R) then A = BC where B is a reflection and C is special orthogonal.
(vi) Let p,q be unit vectors. Prove that the composition of reflections BpBq is a rotation Sn,θ. Find the axis n and the angle θ of rotation. This problem describes double reflection as at an old fashioned dressing table with swinging side mirrors. remark SO(3,R) is a compact hyper surface (or sub-manifold) of dimension 3 in the nine dimension vector space Mat(3,3,R) ≡ R9. K ∈ SO(3,R) is given by one axial vector and one angle: n lies on the 2 dimensional unit sphere S2(1) ⊂ R3 (see §(1.6)) and θ lies on the unit circle S1(1) ⊂ R2. There are “3 continuous degrees
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of freedom” on SO(3,R) and this explains (roughly) why it is a manifold of dimension 3. The simplest compact manifold of dimension 3 is the sphere S3(1) ⊂ R4, often called the unit quaternions. It turns out that SO(3,R) is not the sphere but that S3(1) [the Lie group of unit quaternions also known as SU(2,C)] is the double cover of SO(3,R). This situation gives rise to the phenomenon of the spin of the electron and contributes to the form of the periodic table and so of matter itself.
1.3.2 equivalence of special orthogonal and rotation remark It is easy to prove that every rotation is special orthogonal. The opposite is also true but harder to prove, i.e. if a linear mapping preserves distance, angle and orientation in R3 then it must be a rotation. See theorem(2). It is not easy to prove that the composition of several rotations is a single rotation but this follows immediately from theorem(2).
theorem 2
K ∈ SO(3,R) ⇔ K is a rotation proof The proof falls into two parts. In the first (easier) part, we prove that if K is a rotation, then K ∈ SO(3,R). The second part is more challenging, we prove that if K ∈ SO(3,R) then K is a rotation.
First part of proof Assume that K = Sn,θ is a rotation, then K−1 −1 = E SEt , E and S as in equation(1.32) −1 = (Et )−1 S E−1 since (ABC)−1 = C−1B−1A−1 t t −1 = E S Et since Et = E−1, used twice and S = S t = (Et )t S Et since (Et )t = E t = E SEt since (ABC)t = Ct Bt At = Kt .
Also
det(K) t = det(E Sθ E ) with E and Sθ as in §§(??) t = det(E)det(Sθ)det(E ) = det(E)(cos2 θ + sin2 θ)det(E) = det(E)2 > 0
Thus K−1 = Kt and det(k) > 0, i.e. K ∈ O(3,R) T SL(3,R) = SO(3,R). second part of proof Assume that K ∈ SO(3,R). By (1.35) K has either (i) three real eigenvalues or (ii) one real and a complex conjugate pair of eigenvalues. Thus K has at least one real eigenvalue α: let n be the associated eigenvector. Then
K ∈ SO(3,R)
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⇒ < Kn,Kn > = < n,n > ⇒ < αn,αn > = < n,n > ⇒ α2 < n,n > = < n,n > ⇒ α2 = 1 Thus α = ±1. Similarly any real eigenvalue equals ±1.
Denote the other two eigenvalues by β,γ. Note that αβγ = detK = 1. In case(i) above w.l.o.g. α = β = γ = 1 or α = 1, β = γ = −1. In case (ii) above, w.l.o.g. α = ±1 and γ = β, then 1 = detK = αβγ = αββ = α|β|2: thus α = +1. In all cases α = 1 and K n = n (1.43) Let l,m be chosen so that {l,m,n} is a PONB for R3. Kl = al + bm + cn, for some a,b and c ∈ R (1.44) Km = dl + em + f n, for some d,e and f ∈ R (1.45) Kn = n (1.46) Since {l,m,n} is a PONB < l,n > = 0 , < m,n > = 0 , < l,l > = 1 , < m,m > = 1 Since K preserves < , > < Kl,Kn > = 0 , < Km,Kn > = 0 , < Kl,Kl > = 1 , < Km,Km > = 1 The latter can be expressed as four equations in a,b,c,d,e, f . (i) c = 0, (ii) f = 0, (iii) a2 + b2 = 1, (iv) d2 + e2 = 1 (1.47) For example we prove (1.47)(iii), 1 = < Kl,Kl > = < al + bm,dl + em >, since c = 0 by (i) = a2 < l,l > +ab < l,m > +ba < m,l > +b2 < m,m > = a2 · 1 + ab · 0 + ba · 0 + b2 · 1 = a2 + b2
From equations(1.47)(iii), a = cosφ and b = sinφ for some φ, 0 ≤ φ < 2π and From equations(1.47)(iv), d = cosη and e = sinη for some η, 0 ≤ η < 2π. Equations(1.44) tell us that the matrix of K w.r.t. the PONB {l,m,n} is a b c cos(φ)(cosη) 0 d e f = sin(φ) sin(η) 0 (1.48) 0 0 1 0 0 1 Now we use the fact that K ∈ SO(3,R) ⊂ SL(3,R), thus detK = 1. 1 = det(K) = cosφsinη − sinφcosη = sin(η − φ) (1.49) Thus η − φ = π/2 ⇔ η = φ + π/2. The matrix of K w.r.t. the PONB {l,m,n} is cos(φ) −sin(φ) 0 sin(φ) cos(φ) 0 (1.50) 0 0 1 From §§(1.2.3) K is a rotation, in fact by angle φ about the axis n.
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1.3.3 rotation continuously varying in time remark An element A = (l,m,n) ∈ SO(3,R) can be viewed either as a rotation of space or as a PONB or orthonormal frame. When such A is varied with time one obtains a path {A(t) |t ∈ R } through the Lie group manifold SO(3,R) ⊂ R9. A0(t) is the velocity of this path at time t; this is a velocity of rotation also called an angular velocity. If A(α) = I ∈ SO(3,R), i.e. the path passes through the identity at time α then A0(α) is a tangent vector to the path and lies in the tangent space at I to the Lie group manifold. This tangent space is a vector space denoted so(3,R). We will see that it is endowed with a special type of multiplication [ , ] called the Lie product. In fact (so(3,R),[ , ] is a Lie algebra.
Let A : [a,b] 3 t 7→ A(t) ∈ O(3,R) be a C 1 path or curve (see chap2) through SO(3,R) ⊂ Mat(3,3,R) = R9 such that A(α) = I where a < α < b. Then A(t)At (t) = I for all t ∈ (a,b). (1.51) Differentiating this relationship we obtain
A0(t)At (t) + A(t)A0t = 0 for all t ∈ (a,b). (1.52)
Putting t = α and using A(α) = I and writing B = A0(α) we obtain B + Bt = 0 ⇔ Bt = −B. From (1.55) below
B + Bt = 0 ⇔ < Bv,u > + < v,Bu >, ∀ u,v ∈ R3 (1.53)
We say that the matrix B is skew symmetric or equivalently the linear mapping B is skew adjoint. The matrix B is of the form 0 −c b B = c 0 −a , (1.54) −b a 0 proof of (1.53)
B + Bt = 0 (1.55) ⇔ (B + Bt )v = 0, ∀v ∈ R3 (1.56) ⇔ < (B + Bt )v,u > = 0, ∀v,u ∈ R3 (1.57) ⇔ < (Bv,u > + < Bt v,u > = 0, ∀v,u ∈ R3 (1.58) ⇔ ut Bv + ut (Bt v) = 0, ∀v,u ∈ R3 (1.59) ⇔ ut Bv + (ut Bt )v) = 0, ∀v,u ∈ R3 (1.60) ⇔ ut Bv + (Bu)t v) = 0, ∀v,u ∈ R3 (1.61) ⇔ < Bv,u > + < v,Bu > = 0, ∀v,u ∈ R3 (1.62)
eg. Let n ∈ R3 be a fixed unit vector. Define a path A in SO(3,R) as follows
A : (−∞,∞) 3 t 7→ A(t) = Sn,t ∈ SO(3,R)
We will compute B = A0(0).
B
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= A0(0)
d = Sn,t dt t=0
d t t = n n + cos(t)(I − n n) + sin(t)[n×] dt t=0 = 0 − sin(0)(I − nt n) + cos(0)[n×] = 0 − 0(I − nt n) + 1 · [n×] = [n×] 0 −c b a = c 0 −a if n = b b a 0 c
[n×] is the angular velocity at time t = 0 of space rotating at unit angular speed about the axis n. remark If n is a unit vector the skew symmetric matrix [n×] can represent, according to context, two quite different geometric entities. 3 3 (i) [n×] = Dn : R → R projection into the plane Pn followed by a π/2 rotation.
(ii) [n×] is angular angular velocity of the frame {i,j,k} spinning along the path Sn,t at time t = 0. remark We differentiated a time varying rotation and obtained an angular velocity. Soon we will reverse the process and obtain a rotation by exponentiation of an angular velocity. In general pass from Lie group to Lie Algebra by differentiation and back again by exponentiation.
1.3.4 the Lie algebra so(3,R) definition Lie Algebra A pair (V,[ , ]), where V is a vector space over the real number field and
[ , ] : V ×V 3 (A,B) 7→ [A,B] ∈ V is an operation called Lie multiplication , is said to be a Lie Algebra over the real number field iff [ , ] satisfies three properties (i) bi-linearity, [αA + βB,γC + δD] = αγ[A,C] + αδ[A,D] + βγ[B,C] + βδ[B,D] for all A,B,C,D ∈ V and α,β,γ,δ ∈ R (ii) anti-commutativity, [A,B] = −[B,A] for all A,B ∈ V (iii) cyclic associativity, [A,[B,C]] + [B,[C,A]] + [C,[A,B]] = 0 for all A,B,C ∈ V definition (so(3,R),[ , ]) Define the set t so(3,R) = B ∈ Mat(3,3,R) B + B = 0
Clearly so(3,R) ⊂ Mat(3,3,R) = R9 is a vector subspace. Define the Lie product operation
[ , ] : so(3,R) × so(3,R) 3 (A,B) 7→ [A,B] = AB − BA ∈ so(3,R)
Then (so(3,R),[ , ]) is a Lie algebra called the lie algebra of the Lie group SO(3,R).
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Now is a good time to do the following exercises from quastion sheet §(1.8. (i) Prove that (so(3,R),[ , ]) is indeed a Lie algebra. See exercise(6) (ii) Prove that (R3,×) is a Lie algebra. See exercise(6) (iii) Prove that (so(3,R),[ , ]) is isomorphic as a Lie algebra to (R3,×). See exercise(6) remark In this course we will make considerable use of angular velocities but not of the Lie product.
1.3.5 SO(3,R) and so(3,R), standard form, eigenvalues and eigenvectors For both a special orthogonal matrix A ∈ SO(3,R) and a skew symmetric matrix B ∈ so(3,R) we wish to find standard form, characteristic polynomial, traces, determinants, eigenvalues and eigenvectors. But this work has already been done since (i) A is a rotation and (ii) B = [~n×] for some~n ∈ R3. Here are the details.
3 Let A ∈ SO(3,R). A is a rotation, thus A = Sn,θ for some θ ∈ R and unit vector n ∈ R . Eigenvalues are 1,e±iθ. n is the only eigenvector (except when θ is 0 or π) and is associated with eigenvalue 1. tr (A) = 3 2 tr (A) = 1 + 2cos(θ), det(A) = 1 and pA(λ) = λ − (1 + 2cosθ)λ + (1 + 2cosθ)λ − 1. Let E = (l,m,n) where vectors l,m have been chosen so that (l,m,n) is a PONB then
cosθ −sinθ 0 t A = Sn,θ = E sinθ cosθ 0 E (1.63) 0 0 1
Let B ∈ so(3,R). B is an angular velocity and B = [~n×] for some~n ∈ R3. Write~n = dn where d = ||n|| (the angular speed) and n is the associated unit (axial) vector. Then B = [dn×] = d[n×] = d Dn. By §§(1.36) the eigenvalues of B are 0,±di. n is the only eigenvector and is associated with eigenvalue 0. tr B = 0, tr B = 2 3 2 d ,detB = 0 and pB(λ) = λ + D . Let E = (l,m,n) where vectors l,m have been chosen so that (l,m,n) is a PONB then 0 −d 0 t B = d Dn = E d 0 0 E (1.64) 0 0 1
1.4 exponential matrix definition the exponential matrix n o i Let A = A j 1 ≤ i, j ≤ N be an N × N matrix with complex entries.
∞ 1 1 1 1 1 exp(A) = ∑ Ak = I + A + A2 + A3 + A3 + ··· + Ak + ··· k=0 k! 2! 3! 3! k! is called the exponential matrix of A. remark as with any power series, one must check convergence, see §§1.4.5 below. It is onerous to compute a high power Ak of an N × N matrix, especially if N is large; it is even more
September 23, 2009 c jbquig-UCD 22 CHAPTER 1. GEOMETRY AND SO(3,R)
onerous to compute exp(A), a sum of an infinity of such terms. However exp(A) can be readily computed, by diagonalization of A, or if that fails by reduction of A to Jordan form. eg. computation of exp(tD)
Recall 0 −1 0 1 0 0 D = 1 0 0 and B = 0 1 0 0 0 0 0 0 0 It is easy to check that n D2 = −B BD = DB = D and B = B, ∀n ∈ Z+ Thus exp(tD) = ∞ 1 ∑ (tD)n n=0 n! = ∞ 1 ∞ 1 ∑ (tD)2m + ∑ (tD)2m+1 m=0 2m! m=0 2m + 1! = ∞ 1 m ∞ 1 m ∑ (tD)2 + ∑ (tD)2 (tD) m=0 2m! m=0 2m + 1! = ∞ 1 ∞ 1 ∑ (−t2B)m + ∑ (−t2B)mtD m=0 2m! m=0 2m + 1! = ∞ 2m ∞ 2m+1 t m t m ∑ (−1)m B + ∑ (−1)m B D m=0 2m! m=0 2m + 1! = ∞ t2m ∞ t2m+1 ∑ (−1)m B + ∑ (−1)m BD m=0 2m! m=0 2m + 1! = ∞ t2m ∞ t2m+1 ∑ (−1)m B + ∑ (−1)m D m=0 2m! m=0 2m + 1! = " # " # ∞ t2m ∞ t2m+1 ∑ (−1)m B + ∑ (−1)2m+1 D m=0 2m! m=0 2m + 1! = 1 0 0 0 −1 0 cos(t) 0 1 0 + sin(t) 1 0 0 0 0 0 0 0 0 = cost −sint 0 sint cost 0 0 0 1
= Sn,t Summary 0 −t 0 cos(t) −sin(t) 0 exp t 0 0 = sin(t) cos(t) 0 (1.65) 0 0 0 0 0 1
c jbquig-UCD September 23, 2009 1.4. EXPONENTIAL MATRIX 23
1.4.1 exp(0) Let 0 denote the N ×N matrix with zero entries and I the N ×N identity matrix. It follows from the definition that exp(0) = I (1.66)
1.4.2 exp(A)exp(B) Let A and B be N × N matrices then
(A + B)2 = (A + B)(A + B) = A2 + AB + BA + B2 thus (A + B)2 = A2 + 2AB + B2 , if AB = BA i.e. if A and B commute. Similarly (exercise)
(A + B)3 = A3 + 3A2B + 3AB2 + B3 , if AB = BA
In fact (exercise) one may prove the Binomial Theorem for commuting matrices.
theorem 3 Binomial Let A and B be N × N matrices which commute, AB = BA. Let n ≥ 0 be an integer. Then
n n (A + B)n = ∑ AkBn−k k=0 k
The binomial theorem can be put to use to prove a fundamental (but expected) property of exponential ma- trices.
theorem 4 Let A and B be N × N matrices which commute, AB = BA, then
exp(A)exp(B) = exp(A + B) proof
exp(A)exp(B) ! ! ∞ 1 ∞ 1 = ∑ Ak · ∑ Bl k=0 k! l=0 l! ∞ ∞ 1 1 = ∑ ∑ AkBl k=0 l=0 k! l! ∞ 1 = AkBl ∑ ∑ k!l! n=0 k,l ≥ 0 k + l = n ∞ n 1 = ∑ ∑ AkBn−k n=0 k=0 k!n − k! ∞ 1 n n! = ∑ ∑ AkBn−k n=0 n! k=0 k!n − k!
September 23, 2009 c jbquig-UCD 24 CHAPTER 1. GEOMETRY AND SO(3,R)
∞ 1 n n = ∑ ∑ AkBn−k n=0 n! k=0 k ∞ 1 = ∑ (A + B)n n=0 n! = exp(A + B)
As a consequence of this theorem we have, for any N × N matrix A
(exp(A))−1 = exp(−A) (1.67)
Indeed, since A and − A commute
exp(A)exp(−A) = exp(A − A) = exp(0) = I
1.4.3 powers and exponentials of similar matrices If A and K are N × N matrices, K being non-degenerate (K−1 exists ⇔ det(K) 6= 0) then
K−1BK and B are said to be similar matrices. Note that
(K−1BK)2 = K−1BKK−1BK = K−1BIBK = K−1BBK = K−1B2K
In general (easy exercise)
(K−1BK)n = K−1BnK for any integer, n ≥ 0 (1.68)
Next ! ∞ 1 ∞ 1 ∞ 1 exp(K−1BK) = ∑ (K−1BK)n = ∑ (K−1BnK) = K−1 ∑ Bn K = K−1 (exp(B))K n=0 n! k=0 n! k=0 n! summary
exp(K−1BK) = K−1 exp(B)K (K−1BK)n = K−1BnK, for any integer, n ≥ 0 (1.69)
d 1.4.4 exp(tA) dt Let A be an N × N constant matrix and t ∈ R, as expected
d exp(tA) = Aexp(tA) (1.70) dt proof d exp(tA) dt ! d ∞ 1 = ∑ tkAk dt k=0 k!
c jbquig-UCD September 23, 2009 1.5. EXPONENTIAL OF ANGULAR VELOCITY, EXP(TDN), ∈ SO(3,R) 25
∞ d 1 = ∑ tkAk k=0 dt k! ∞ 1 = ∑ ktk−1Ak k=0 k! ∞ 1 = ∑ A tk−1Ak−1 k=1 k − 1! ! ∞ 1 = A ∑ tk−1Ak−1 k=1 k − 1! ! ∞ 1 = A ∑ tlAl l=0 l! = Aexp(tA)
1.4.5 Convergence of the exponential matrix power series Finally, the question of convergence . The definition of exp(A) above is equivalent to the following definition of N2 matrix entries ∞ i i 1 k (exp(tA)) j = A , 1 ≤ i, j ≤ N (1.71) ∑ j k=0 k! One must check that each power series is convergent. It is easy to show (exercise) that
i k k A ≤ M , 1 ≤ i, j ≤ N , k ≥ 0 j where n o i M = N(max A j 1 ≤ i, j ≤ N ) It suffices then to prove convergence of ∞ Mk ∑ k=0 k! For large k, M/k < 2; after a few details the question reduces to convergence of the Geometric progression.
∞ 1k ∑ k=0 2 an easy exercise.
1.5 exponential of angular velocity, exp(tDn), ∈ so(3,R)
Consider the most general angular velocity with unit angular speed
t Dn = [n×] ∈ so(3,R) = E DE (1.72)
Here E = {l,m,n} is a PONB. We are now ready to compute exp(tDn) in two different ways. First
t t t exp(tDn) = exp(E tDE ) = E exp(tD)E = E St E = Sn,t see equation(1.28). (1.73)
Second, we will write for short B = Bn and D = Dn. Recall from example(??) that
2 2 B = B , D = −B , BD = DB = D , exp(tD) = St (1.74)
September 23, 2009 c jbquig-UCD 26 CHAPTER 1. GEOMETRY AND SO(3,R)
Then
exp(tDn) = exp(tD) ∞ tn = ∑ Dn n=0 n! ∞ t2k ∞ t2l+1 = D0 + ∑ D2k + ∑ D2l+1 k=1 2k! l=0 2l + 1! ∞ 2k ∞ 2l+1 t k t l = Id + ∑ D2 + ∑ D2 D k=1 2k! l=0 2l + 1! ∞ t2k ∞ t2l+1 = Id + ∑ −Bk + ∑ −BlD k=1 2k! l=0 2l + 1! ∞ t2k ∞ t2l+1 = Id + ∑ (−1)k Bk + ∑ (−1)l BlD k=1 2k! l=0 2l + 1! ∞ t2k ∞ t2l+1 = Id + ∑ (−1)k B + ∑ (−1)l BD k=1 2k! l=0 2l + 1! ∞ t2k ∞ t2l+1 = (A + B) + ∑ (−1)k B + ∑ (−1)l D k=1 2k! l=0 2l + 1! ( ) ( ) ∞ t2k ∞ t2l+1 = A + ∑ (−1)k B + ∑ (−1)l D k=0 2k! l=0 2l + 1! = A + {cos(t)}B + {sin(t)}D
= An + cos(t)Bn + sin(t)Dn
= Sn,t We have proven, in two different ways
t t exp(tD) = Sn,t ⇔ exp(t[n×]) = nn + cos(t)(Id − nn ) + sin(t)[n×] (1.75)
1.6 Euler angles
There is a fourth approach to rotations by Euler angles. We define the unit sphere, a surface in R3. n o 2 3 S = S (1) = n ∈ R ||n|| = 1 (1.76) The spherical polar parametrization of S2(1) is well known x(θ,φ) sinφcosθ θ h : [−π,π] × [0,π] 3 7→ x(θ,φ) = y(θ,φ) = sinφsinθ ∈ S2(1) ⊂ R3 (1.77) φ z(θ,φ) cosφ
The angles θ and φ are known as the longitude and co-latitude of the point x = h(θ,φ) ∈ S2(1).
Let sinβcosα 2 p = sinβsinα ∈ S (1) cosβ there are many rotations K ∈ SO(3,R) which carry the north pole k to p; we pick a special simple one
Kp = Sk,α ◦ Sj,β (1.78)
c jbquig-UCD September 23, 2009 1.6. EULER ANGLES 27
Then, indeed
sinβ sinβcosα Kp(k) = Sk,α(Sj,β(k)) = Sk,α( 0 ) = sinβsinα = p (1.79) cosβ cosβ
There are 3 very simple types of rotation, those with axial vectors i,j or k. Thus Kp is composed of two 2 simple rotations. You can visualize Kp = Sk,α ◦ Sj,β as follows. First rotate the sphere S (1) about the y-axis by the angle β; the North Pole (co-latitude φ = 0) travels down the Greenwich meredian (longitude θ = 0) until it reaches the point p0 with co-latitude φ = β and of course with longidude θ = 0. Next rotate S2(1) about the z-axis by angle α. The point p0 travels along the circle of constant co-latidude φ = β until it reaches the point p with longitude θ = α.
Let K ∈ SO(3,R) be an arbitrary rotation. Let α and β be the longitude and co-latitude of p = K(k), the point to which the North pole k is carried by K. We will define a third angle γ and prove that
K = Sk,α ◦ Sj,β ◦ Sk,γ (1.80)
The triple α,β,γ are known as Euler angles. Our claim is that any rotation can be constructed as
(i) A rotation about the z-axis (by an angle γ), followed by
(ii) A rotation about the y-axis (by an angle β), followed by
(iii) A rotation about the z-axis (by an angle α)
−1 It remains to justify this claim and to find the third angle γ. Consider the rotation (Kp) ◦ K. This mapping −1 −1 −1 carries k to p and back to k, i.e. (Kp ◦ K)(k) = Kp (K(k)) = Kp (p) = k and is a simple rotation Sk,γ by some angle γ about the z-axis.
−1 Kp ◦ K = Sk,γ ⇔ K = Kp ◦ Sk,γ ⇔ K = Sk,α ◦ Sj,β ◦ Sk,γ (1.81)
We sum up. Let K ∈ SO(3,R) let α and β be the longitude and co-latitude of K(k) ∈ S2(1). Let tr (Sk,− ◦ Sj,−α ◦ K) − 1 γ = arccos β ⇔ tr (K−1 ◦ K) = tr (S ) = 1 + 2cos(γ) (1.82) 2 p k,γ
Then cosα −sinα 0 sinβ 0 cosβ cosγ −sinγ 0 K = Sk,α ◦Sj,β ◦Sk,γ = sinα cosα 0 0 1 0 sinγ cosγ 0 (1.83) 0 0 1 cosβ 0 −sinβ 0 0 1
1.6.1 formulae for rotation One could say that the end product so far of the current chapter is:- four formulae for the rotation matrix developed from four very different points of view.
Sn,t = nnt + cos(t)(Id − nnt ) + sin(t)[n×] t = E St E = exp(t[n×])
= Sk,α ◦ Sj,β ◦ Sk,γ
There is a fifth way to generate a rotation. The composite of two reflections is a rotation; see question sheet §§(1.8) question(7)
September 23, 2009 c jbquig-UCD 28 CHAPTER 1. GEOMETRY AND SO(3,R)
1.7 topology of SO(3,R)
We can use two of our rotation formulae to construct two continuous surjective mappings from simple topo- logical objects onto SO(3,R).
2 1 F : S (1) × S 3 (n,exp(iθ)) 7→ Sn,θ ∈ SO(3,R)
1 1 1 G : S × S × S 3 (exp(iα),exp(iβ),exp(iγ)) 7→ Sk,α ◦ Sj,β ◦ Sk,γ ∈ SO(3,R) Frustratingly both F and G fail to be injective in quite complicated ways. We might conclude that SO(3,R) is a homogeneous compact manifold of dimension 3 but details of its topology remain obscure. We need a new approach. The unit 3-dimensional sphere S3(1) ⊂ R4 is a simple three dimensional compact manifold and as the ‘unit quaternions’ it has a multiplication operation giving it a Lie group structure. Might it possibly be true that S3(1) is isomorphic as a Lie group to SO(3,R)? This is not true but there is a two to one mapping
Φ : S3(1) → SO(3,R) called a double covering of SO(3,R) by the sphere. This observation neatly describes the topology of SO(3,R) The underlying manifold of SO(3,R) is the 3-sphere with diametrically opposed points identi- fied and can also be view as the space of real lines passing through 0 in R4, a space denoted P3(R) and called real projective three space. There is a Lie group SU(2,C) of 2 × 2 complex matrices which is isomorphic to S3(1) and which is easier to work with. SU(2,C) is called the special unitary group1 or, in physics, the spin group. All this can be expressed in one line.
SO(3,R) ≡ SU(2,C)/{±Id} ≡ S3(1)/± ≡ P3(R)
1.7.1 the Quaternionic sphere as a Lie group Let q ∈ R4 = Q write
1 0 0 0 a 0 1 0 0 b 1 = , I = , J = , K = , q = 0 0 1 0 c 0 0 0 1 d
The general vector q may be written q = a1 + bI + cJ + dK Hamilton showed that there was a ’multiplication’ in R4 esentially given by the rules I2 = J2 = K2 = −1, IJ = −JI = K, JK = −vbKJ = I, KI = −IK = J: the Quaternions were born (on Broom bridge in Dublin). From the modern perspective (Q,+,·) is a noncommutative associative algebra. We are only interested in a small part of Q, S3(1) ⊂ Q, called the quaternionic sphere. One may check that (S3(1),·), the unit sphere in 4-space endowed with quaternionic multipication, is a Lie group.
1.7.2 the Lie group SU(2,C) In the end we wish to prove that S3(1) is a double cover of SO(3,R). It is easier to do this if we represent (S3(1),·) as a group of 2 × 2 complex matrices (SU(2,C),×) called the special unitary group. Here are the details of this representation.
We define the Hermitian inner product on the complex vector space C2
< , > : C2 × C2 → C (u,v) 7→ vt u
1 goggle for ’u2’ get ’Bono and the pop group’, google for su2 get wikipedia on the Lie group
c jbquig-UCD September 23, 2009 1.7. TOPOLOGY OF SO(3,R) 29
Thus a c < , > = ca + db b d The matrix A ∈ Mat(2,2,C) is said to be unitary iff
t t < Au,Av >=< u,v > for all u,v ∈ C2 ⇔ A A = Id ⇔ A−1 = A
A is said to be special iff det(A) = 1. Finally we define the special unitary group n o −1 t SU(2,C) = A ∈ Mat(2,2,C) A = A and det(A) = 1 It is left as an exercise to check that a + ib c + id A ∈ SU(2,C) iff A = where det(A) = a2 + b2 + c2 + d2 = 1 −c + id a − ib
Here a,b,c,d are real. We see that topologically SU(2,C is the 3-sphere and we can write A = a1 + bI + cJ + dK where 1,I,J,K are 2 × 2 matrices 1 0 i 0 0 1 0 i 1 = , I = , J = , K = , 0 1 0 −i −1 0 i 0
This (after a few checks) proves that SU(2,C) is the unit quaternionic group.
1.7.3 the Lie algebra su(2,C) In §§(1.3.3) we derived the Lie algebra so(3,R) from the Lie group SO(3,R). We will follow a similar pro- ceedure here to derive the Lie algebra su(2,C) from the Lie group SU(2,C). We will be interested in su(2,C because, it is an inner product space and as such, is isomorphic to the standard inner product space (R3,<>).
Let {A(t) ∈ SU(2,C) |t ∈ R } be a path in the special unitary group passing through Id when t = 0. Differ- t entiating (compare with SO(3,R)) the relation A(t)A (t) = Id at t = 0 and writing B = A0(0) we obtain the relation t B + B = 0 and tr (B) = 02 Thus we define the special unitary Lie algebra n o t su(2,C) = B ∈ Mat(2,2,C) B + B = 0 and tr B = 0 It is left as an exercise to check that ip q + ir B ∈ su(2,C) ⇔ B = p,q,r ∈ R −q + ir −ip
1.7.4 su(2,C) as an inner product space Next we will show that su(2,C) is a real 3-dimensional vector space with an inner product << >>. We will show that this inner product space (su(2,C),<< >>) is just standard Euclidean 3-space (R3,< >) in disguise.
The inner product on su(2,C) is
<< , >> : (su(2,C),su(2,C)) → C 1 t (A,B) 7→ << A,B >>= 2 B A = pl + qm + rn 2 Differentiation of the equation detA(t) ≡ 1 at t = 0 and using A(0) = Id yields tr B = 0: a non trivial exercise.
September 23, 2009 c jbquig-UCD 30 CHAPTER 1. GEOMETRY AND SO(3,R)
where we have taken ip q + ir il m + in A = and B = −q + ir −ip −m + in −il
The following mapping Ω is an isomorphism of inner product spaces.
Ω : (R3,< , >) → (su(2,C),<< , >>) p ip q + ir v = q 7→ Ω(v) = −q + ir −ip r
In particular Ω preserves inner products, i.e.
< A,B > = << Ω(A),Ω(B) >> for all A,B ∈ R3
1.7.5 mapping F from SU(2,C) to SO(3,R) Recall that our goal in this §(1.7) is to prove that SU(2,C) is a double cover of SO(3,R). We are now ready to find a ’two to one’ mapping from F : SU(2,C) → SO(3,R).
A 2 × 2 matrix A ∈ SU(2,C) acts naturally on C2. We Given K ∈ SU(2,C) define the mapping FK from su(2,C) to su(2,C by
t FK : su(2,C) 3 A 7→ KAK ∈ su(2,C)
One can prove that t (i) A ∈ su(2,C) ⇒ FK(A) ∈ su(2,C), i.e. that FK(A) = −FK(A) and tr (FK(A)) = 0.
(ii) FK preserves the inner product << >> on su(2,C).
(iii) given K,L ∈ SU(2,C) then FK◦L = FK ◦ FL Thus we have a group homomorphism
F : SU(2,C) 3 K 7→ FK ∈ SO(3,R)
1.7.6 SU(2,C) is a double cover of SO(3,R). One can check that
F = S and F ! = S eiθ/2 0 k,θ cos(φ/2) −sin(φ/2) j,φ 0 e−iθ/2 sin(φ/2) cos(φ/2)
By Euler angle theory Sk,θ and Sj,φ generate all of SO(3,R). We deduce that F is surjective.
F is not injective. Instead exactly two distinct elements of su(2,C) are sent by F to each element of SO(3,R).
Conclusion (some details need clarification) the topology of the special orthogonal Lie (rotation) group is
SO(3,R) ≡ SU(2,C)/{±Id} ≡ S3(1)/± ≡ P3(R)
c jbquig-UCD September 23, 2009 1.8. PROBLEM SET 31
1.8 problem set algebra 1. (i) If A ∈ SO(3,R) then det(A) = 1. (ii) If A ∈ O(3,R) then det(A) = ±1. (iii) Each reflection is orthogonal but not special. (iv) Every orthogonal matrix is either a rotation or is the product of a rotation and a reflection. (v) Give a 3× orthogonal matrix A which is not special and not a reflection. Prove your assertion. Express A as the product of a rotation and a reflection.
2. Let n be the unit vector in the direction i + j + k. Let θ ∈ R.
(i) Compute Ai,Bi,Ci,Di,Si,θ.
(ii) Compute Aj,Bj,Cj,Dj,Sj,θ.
(iii) Compute Ak,Bk,Ck,Dk,Sk,θ.
(iv) That was to warm up. Now compute An,Bn,Cn,Dn.
(v) Compute the matrix of rotation Sn,2π/3. Compute it by four different methods (when we finish Chapter 1).
3. Prove that, for a 3 × 3 reflection matrix
At = A = A−1
4. (i) Find the matrices of orthogonal projection Bk,Bi,Bj; in mechanical drawing these are called plan, frontal and side elevation. 3 (ii) Find n ∈ R , ||n|| = 1 so that Bn is isometric projection (this is the unique orthogonal projection in which all 3 principle directions (EW,NS,UD) are scaled equally). Compute the matrix B = Bn. (iii) (See expanded version of (iii) in next question, added sept 2009) The eight vertices of the standard cube C = [−1,1]×[−1,1]×[−1,1] of volume 8 are ±i±j±k = ±1 ±1 . Compute B(±i ± j ± k). Express all 8 as linear combinations of l,m where {l,m,n} ±1 is a PONB. Actually draw the isometric projection of the standard cube in the 2-D image plane Pn.
5. Let ||n|| = 1, n ∈ R3. Let P = {x | < x,n >= 0 } ⊂ R3. Let B : R3 → R3 denote orthogonal projection into the plane P. Choose h,v ∈ P (many choices are possible) so that {h,v,n} is a P.O.N.Basis for R3. Then {h,v} is an O.N. Basis for the plane P. Any vector in P can be written as a linear combination Xh +Yv. Denote by B the 2 × 3 matrix which represents B : R3− > P w.r.t. bases {i,j,k} and {h,v} for R3 and R2 respectively. B is the most useful matrix representing O.P into the plane P.
September 23, 2009 c jbquig-UCD 32 CHAPTER 1. GEOMETRY AND SO(3,R)
t h (i) Prove that B = . vt (ii) Argue that h = v × n with v = B(k)/||k|| is the most apt choice of the basis of the picture plane P. (iii) Compute h,v and the 2 × 3 matrix B for isometric projection. (iv) Now easily do the last part of the preceeding question.
3 6. Let n ∈ R with ||n|| = 1. Write A = An, B = Bn, D = Dn. Prove the following facts which were used during lectures to justify the formula
Sn,θ = A + cos(θ)B + sin(θ)D
Given any x ∈ R3 (i) Ax is axial, i.e. a scalar multiple of n. (ii) < Bx,Dx > = < Ax,Bx > = < Ax,Dx > = 0, i.e. the three vectors Ax,Bx and Dx are mutually orthogonal. (iii) ||Bx|| = ||Dx||, i.e. the two vectors Bx and Dx have the same length. (iv) The triple Bx,Dx,n is positively oriented, det(Bx,Dx,n) > 0. (NB this part has been corrected: in a previous erroneous version, Ax was in the place of n.)
7. (i) Find skew symmetric matrices I,J,K which represent unit angular velocity about the i,j,k axes. (ii) Prove that (R3,×) is a Lie Algebra. (iii) Prove that (so(3,R),[ , ]) is a Lie Algebra. You must check i. closure of so(3,R) under the operation [ , ] ii. bilinearity of the operation [ , ] iii. anticommutativity of the operation [ , ] iv. cyclic (Lie) associativity of the operation [ , ] (iv) Prove that the mapping
F : R3 3 ai + bj + ck 7→ aI + bJ + cK ∈ so(3,R)
is a vector space isomorphism. (v) Prove that F (u × v) = [F (u),F (v)] for all u,v ∈ R3. (vi) Conclude that (R3,×) and (so(3,R),[ , ]) are isomorphic as Lie algebras.
8. Let A,B : R3 → R3 be reflections in the plane < x,n >= 0 and < x,m >= 0 respectively. Prove that BA is a rotation. Find the angle and axis of rotation. (Hint, try reflections in lines in R2 to start with.)
3 9. Let n ∈ R with n = 1 and θ ∈ R. For short write A,B,C,D,S for An,Bn,Cn,Dn,Sn,θ respectively. Prove
c jbquig-UCD September 23, 2009 1.8. PROBLEM SET 33
(i) AA = A , BB = B , CC = I , DD = −B (ii) AB = BA = AD = DA = 0 , BD = DB = D (iii) At = A , Bt = B , Ct = C−1 = C , Dt = −D (iv) S−1 = St
10. Let a ∈ R. Let A be the basic k × k Jordan block matrix given by
a i = j i A j = 1 i = j − 1 0 otherwise
For example when k = 5 a 1 0 0 0 0 a 1 0 0 A = 0 0 a 1 0 0 0 0 a 1 0 0 0 0 a (i) For k = 5 prove that a4 4a3 6a2 4a 1 0 a4 4a3 6a2 4a 4 A = 0 0 a4 4a3 6a2 0 0 0 a4 4a3 0 0 0 0 a4 (ii) For k = 5 prove that
1 t t2/2 t3/6 t4/24 0 1 t t2/2 t3/6 at exptA = e 0 0 1 t t2/2 0 0 0 1 t 0 0 0 0 1
(iii) Compute An then exp(tA) for general n ∈ N and k = 5. (iv) Compute A2,A3,A4 then An then exp(tA) for general k
11. Let 0 −1 0 S = 0 0 −1 1 0 0 (i) Prove that K is a rotation.
(ii) By (i) K is rotation about the axis n, (||n|| = 1) by angle θ, i.e. K = Sn,θ. Find n and θ. (iii) Express K in the form
cosφ −sinφ 0 −1 −1 K = E ◦ Sφ ◦ E = E sinφ cosφ 0 E 0 0 1
Find φ and give E and E−1 in full detail.
September 23, 2009 c jbquig-UCD 34 CHAPTER 1. GEOMETRY AND SO(3,R)
(iv) Express K in the form exp(J). Give the matrix J in full detail. (v) Find (Euler) angles α,β,γ such that
K
= Sk,α ◦ Sj,β ◦ Sk,γ cosγ −sinγ 0 = sinγ cosγ 0 · 0 0 1 sinβ 0 cosβ 0 1 0 · cosβ 0 −sinβ cosα −sinα 0 sinα cosα 0 0 0 1
[Footnote In this question an angle 0 ≤ θ < 2π should be specified by giving both cosine and sin as fractions. For example 3 −4 3 −4 θ = arccos = arcsin ⇔ cos(θ) = and sin(θ) = 5 5 5 5 ]
12. Let −12/325 −316/325 3/13 S = 309/325 12/325 4/13 −4/13 3/13 12/13 (i) Prove that K is a rotation.
(ii) By (i) K is rotation about the axis n, (||n|| = 1) by angle θ, i.e. K = Sn,θ. Find n and θ. (iii) Express K in the form
cosφ −sinφ 0 −1 −1 K = E ◦ Sφ ◦ E = E sinφ cosφ 0 E 0 0 1
Find φ and give E and E−1 in full detail. (iv) Express K in the form K = exp(J). Give the matrix J in full detail. (v) Find (Euler) angles α,β,γ such that
cosγ −sinγ 0 sinβ 0 cosβ cosα −sinα 0 K = Sk,α ◦Sj,β ◦Sk,γ = sinγ cosγ 0 0 1 0 sinα cosα 0 0 0 1 cosβ 0 −sinβ 0 0 1
[Footnote In this question an angle 0 ≤ θ < 2π should be specified by giving both cosine and sin as fractions. For example
θ = arccos(3/5) = arcsin(−4/5 ⇔ cos(θ) = 3/5 and sin(θ) = −4/5
]
c jbquig-UCD September 23, 2009 Chapter 2 curves
2.1 plane curves 2.1.1 parametrization of regular curves For the present we will study curves in the plane R2. Later we will study curves in R3. definition curve A curve Γ ⊂ R2 is a subset of the plane of the form
{c(t) |a ≤ t ≤ b } ⊂ R2 where c : [a,b] 3 t 7→ c(t) ∈ R2 is a continuous vector valued function on a closed interval. The function c is said to be a parametrization of Γ. The parametrization is said to be regular iff c is everywhere continuously differentiable and c0(t) 6= 0 for all t ∈ [a,b]. A curve Γ which admits a regular parametrization is said to be a regular curve.
We will demand that curves be regular and so exclude pathological examples from consideration. The re- quirement that c be continuous forces the curve to be unbroken. The requirement that c is differentiable ensures well defined velocity vector and tangent vectors to the curve. The requirement that c0 is continuous means that there the curve is smooth, i.e. the tangent vector does not suddenly change direction. The require- ment that c0(t) 6= 0 ⇔ ||c0(t)|| 6= 0 forces c to be locally injective, (the term immersion is used); this does not force c to be injective, i.e. a curve may cross itself. notation We write x(t) dc x0(t) c(t) = x(t) = , c0(t) = (t) = y(t) dt y0(t) eg. circle [figure(2.1]) The circle (of radius a > 0, center 0) is a subset of R2 x 2 2 2 C = Ca = x = x + y = a y Here is a regular parametrization of C which is therefore a regular curve
x(t) acos(t) c : [0,2π] 3 t 7→ c(t) = = ∈ R2 y(t) asin(t)
35 36 CHAPTER 2. CURVES
Figure 2.1: (i) circle (ii) cubic parabola (iii) cycloid
Here x0(t) −sin(t) c0(t) = = ; c0(t) 6= 0, since ||c0(t)|| = 1 ∀t. y0(t) cos(t) In passing I mention that
(i) x2 + y2 = 1 is said to be the implicit equation of the curve C. p (ii) y = x2 + y2 is said to be the explicit equation of the curve C.
x(t) = acos(t) (iii) , 0 ≤ t ≤ 2π is said to be a parametric equation equation of the curve C. y(t) = asin(t)
(iv) z(t) = aexp(it) = acos(t)+iasin(t), 0 ≤ t ≤ 2π is alternatively a complex parametrization of the curve C.
eg. irregular curve [figure(2.1)]
2 x(t) t Γ = c(t) = = t ∈ R y(t) t3 is a curve but is not a regular curve. c0 exists for all t but is zero at one point t = 0. The given parametrization is irregular. There are other parameterizations but it can be proven that none is regular. eg. cycloid curve [figure(2.1)] x(t) t sin(t) Γ = c(t) = = − t ∈ R y(t) 1 cos(t) is the cycloid curve. It is the curve traced out by the valve on a wheel of radius 1 on a bicycle moving at speed 1. This is not a regular curve. c0 exists for all t but is periodically zero ( at points t = 2nπ, n ∈ Z). The given parametrization (and all others) are irregular. Restricting parametrization to the domain 0 < t < 2π) yields a regular (but uninteresting) curve. See also http://en.wikipedia.org/wiki/Cycloid and http://mathsci.ucd.ie/courses/math40060/maple
2.1.2 re parametrization Let c : [a,b] 3 t 7→ c(t) ∈ Γ ⊂ R2
c jbquig-UCD September 23, 2009 2.1. PLANE CURVES 37
be a regular parametrization of the curve Γ. Let
α : [a,b] 3 t 7→ α(t) = t ∈ [a,b] dα be C 1 with (t) = α0(t) > 0 for all t ∈ [a,b]. Then α is strictly monotone increasing and bijective with dt inverse mapping β : [a,b] 3 t 7→ β(t) = t ∈ [a,b] β is also C 1 with β0(t) > 0,∀t ∈ [a,b]. Of course
α ◦ β = Id : [a,b] → [a,b] → [a,b] and β ◦ α = Id : [a,b] → [a,b] → [a,b]
We can use α to produce from c a new regular parametrization e of Γ: we can use β to convert the parametriza- tion e back to the original parametrization c as follows.
e = c ◦ α : [a,b] → [a,b] → Γ ∈ R2 , e(t) = c(α(t)) = c(t)
c = e ◦ β : [a,b] → [a,b] → Γ ∈ R2 , c(t) = e(β(t)) = e(t) Here e is said to be a re-parametrization and c and e are said to be equivalent parametrizations of the curve Γ. A regular curve has many equivalent parametrizations but amongst these one is canonical and important, parametrization by arc length and is discussed in detail in §§(2.1.4). eg. re-parametrization of circle by arc length Define s α : [0,2πa] 3 s 7→ t = α(s) = ∈ [0,2π] a with inverse function β : [0,2π] 3 t 7→ s = β(t) = at ∈ [0,2πa] We reparameterize the circle C of example(2.1.1)
acos(s/a) γ = c ◦ α : [0,2πa] 3 s 7→ γ(s) = ∈ C ⊂ R2 (2.1) asin(s/a)
We can recover the original parametrization in the form c = γ ◦ β. Although we have not yet studied the concept of ’arc-length’ the idea is intuitive for the circle: it is intuitively clear that the parameter s is arc length. It is customary notation to use ’s’ for arc length and ’γ’ for arc-length parametrization.
2.1.3 velocity, speed, acceleration Let c : [a,b] → Γ ⊂ R2 t 7→ c(t) be a regular parametrized curve.
dc x˙(t) = c˙(t) = c0(t) = dt y˙(t) is the velocity of c at time t. dc q || || = ||c˙(t)|| = x˙(t)2 + y˙(t)2 dt is the speed of c at time t. d2c x¨(t) = c¨(t) = c00(t) = dt2 y¨(t) is the acceleration of c at time t.
September 23, 2009 c jbquig-UCD 38 CHAPTER 2. CURVES
Let e = c ◦ α (2.2) be a reparameterization as in §§2.1.2 Differentiate 2.2 and use the chain rule
de dc dα (t) = (α(t)) · (t) ⇔ e0 = α0 · c0 (2.3) dt dt dt Take norms in 2.3, use α0 > 0 ⇒ |α0 | = α0
de dc dα || (t)|| = || (α(t))|| · (t) ⇔ ||e0|| = α0 · ||c0|| (2.4) dt dt dt Differentiate 2.3, use the product rule and the chain rule
d2e d2c dα 2 dc d2α (t) = (α(t)) · (t) + (α(t)) · (t) ⇔ e00 = α02 · c00 + α00 · c0 (2.5) dt2 dt2 dt dt dt2
2.1.4 arc-length Let c : [a,b] → Γ ⊂ R2 be a regular parametrization of the curve Γ. c(a) is the start point and c(b) is the end point of Γ. We will define L, the total arc length of Γ. Moreover if γ(t) ∈ Γ we will define the arc length s(t) measured along the curve from the start c(a) to c(t).
Let c, α, e = c ◦ α and Γ be as in §§2.1.3. Let p = c(a) be the start point and q = c(t) be an arbitrary point on Γ. Define the arc-length distance s(t) from p = c(a) to q = c(t) (or distance measured along the curve) as
Z t s(t) = ||c(˙p)||dp (2.6) a
This is a reasonable definition since ’distance equals speed by time’. However to be consistent it must be independent of parametrization of the curve. We must prove that if t = α(t) then
Z t Z t ||c˙||(p)dp = ||e˙||(p)dp) (2.7) a a But Z t ||c˙||(p)dp a Z α(t) = ||c˙(α(t))||dα(p) α(a) Z t = ||c˙(α(p))|| · α0(p)dp) a Z t = ||c˙(α(p))|| · ||α0(p)||dp), since α0 > 0 a Z t = ||c˙(α(p)) · α0(p)||dp) a Z t = ||e˙||(p)dp) a Thus proves 2.7 and justifies our definition (2.6). It is no surprise that the rate of change of arc length distance with time equals speed; indeed apply the fundamental theorem of calculus to 2.6 to obtain
s0(t) = ||c˙(t)||, i.e. the speed (2.8)
c jbquig-UCD September 23, 2009 2.1. PLANE CURVES 39
Define the total arc length L of Γ to be the distance from start to end
Z b Z b L = ||c˙(t)||dt = ||e˙(t)||dt (2.9) a a
Of course L is independent of parametrization
2.1.5 arc-length parametrization Given a distance s, 0 ≤ s ≤ L we will show that there is a unique point γ(s) ∈ Γ whose distance from c(a) measured along the curve is s. We will thus obtain a special parametrization
γ : [0,L] 3 s 7→ γ(s) ∈ Γ ⊂ R2 called arc-length parametrization. Now for the details.
Let c : [a,b] 3 t 7→ c(t) ∈ Γ ⊂ Rn be a regular parametrization of the curve Γ whose total arc length is L. Consider the mapping from the interval [a,b] to the interval [0,L].
Z t α : [a,b] 3 t 7→ α(t) = s(t) = ||c˙|| ∈ [0,L] a Since c is C 1 and s0 = ||c˙|| > 0 then α is C 2 and ⇑. Since α(a) = 0 and α(b) = L, by standard calculus, α is a C 2 bijection and an C 2 inverse function
β : [0,L] 3 s 7→ t = β(s) ∈ [a,b] exists. The re-parametrization
γ = c ◦ β : [0,L] 3 s 7→ γ(s) = c(β(s)) ∈ Γ ⊂ Rn (2.10) is known as the parametrization by arc-length of the curve Γ. We have
γ = c ◦ β : [0,L] → Γ ⊂ Rn ⇔ c = γ ◦ α : [a,b] → Γ ⊂ Rn (2.11)
From the chain rule and the fundamental theorem of calculus
||c0|| = ||γ0|| · α0 = ||γ0|| · s0 = ||γ|| · ||c0|| (2.12)
Thus ||γ0|| = 1 ⇔ < γ˙,γ˙ >= 1 (2.13) The speed of arc-length parametrization is 1. γ(s) is the unique point on Γ whose (arc-length) distance from the start point, γ(0), is s. Differentiate 2.13
< γ¨,γ˙ > + < γ˙,γ¨ >= 0 ⇔ < γ˙,γ¨ >= 0
For arc length parametrization, speed is one, acceleration is perpendicular to velocity.
< γ˙,γ˙ >= 1 and < γ¨,γ˙ >= 0 (2.14)
Equations (2.3) and (2.5) yield
c0 = s0γ0 and c00 = (s0)2γ00 + s00γ0 (2.15)
Which give c0 s0c00 − s00c0 γ0 = and γ00 = (2.16) s0 s03
September 23, 2009 c jbquig-UCD 40 CHAPTER 2. CURVES
Figure 2.2: kArchimedean spiral eg. circle in R2 Let 1 6= a ∈ R x(t) acos(t) c : [0,2π] 3 t 7→ c(t) = = ∈ Γ ⊂ R2 y(t) asin(t) is a regular parametrization of the circle curve x 2 2 2 Γ = x + y = a y centered at the origin and with radius a > 0.
acost −asint −acost c(t) = , c0(t) = , ||c0(t)|| = a , c00(t) = asint acost −asint
The speed is constant. c0 ⊥ c00 even though this is not arc-length parametrization (since ||c0|| = a 6= 1). Total length is Z 2π Z 2π L = ||c0|| = adt = 2πa 0 0 Take α : [0,2π] → [0,L] = [0,2πa] Z t Z t , where s(t) = ||c0|| = adt = at t 7→ α(t) = s(t) 0 0 The inverse function β is s β : [0,2πa] 3 s 7→ β(s) = t = ∈ [0,2π] a Arc length parametrization is
γ = c ◦ β : [0,2πa] → Γ ⊂ R2 cos(s/a) , i.e. γ(s) = a s 7→ γ(s) = c(β(s)) = c(s/a) sin(s/a)
eg. Archemedian spiral, see figure2.2 The Archimedean spiral Γ is the curve traced out by the end of a taut string as it is unwound from a spool. We will first obtain a natural parametrization e(t) of Γ. From this we will compute the arc length parametrization γ(s). The Archimedean spiral is an example of an involute curve, see §§(??).
Arc length parametrization of the unit circle C (for notational reasons we use c and t,not the usual γ and s) is.
cost c : [0,2π] 3 t 7→ c(t) = ∈ C ⊂ R2 sint
c jbquig-UCD September 23, 2009 2.1. PLANE CURVES 41
From this will build a parametrization e of the Archimedean spiral Γ. Start at the point i ∈ C. The unit tangent vector at the point i to the circle C is j. Let S(t) be the 2 × 2 matrix of ACW rotation of R2 by angle t, thus
cost −sint S(t) = sint cost
The general point on the circle C is
cost −sint 1 cost c(t) = S(t)i = = sint cost 0 sint
The unit tangent vector to the circle C at the point c(t) is
cost −sint 0 −sint t(t) = S(t)j = = sint cost 1 cost
To the point c(t) ∈ C add a vector −tt(t) in the reverse tangential direction with length the arc length t from c(0) to c(t) measured along the circle C. We obtain a point e(t) on the Archimedean spiral Γ. Our initial parametrization of the Γ is
cost −sint e : [0,2π] 3 t 7→ d(t) = S(t)i − tS(t)j = −t ∈ Γ sint cost
There is a convenient complex version with 1,i and exp(it) playing the role of i,j and S(t) respectively.
z(t) = exp(it) · 1 −t · i · exp(it) = exp(it) − it exp(it)
Next we compute the total arc length L of (one turn of) the Archimedean spiral Γ
L Z 2π = ||e˙(t)||dt 0 Z 2π d cos(t) +t sin(t) = || ||dt 0 dt sin(t) −t cos(t)
Z 2π d = (exp(it) − it exp(it)) dt, avoid onerous trigonometry, use complex methods 0 dt Z 2π = |iexp(it) − iexp(it) +t exp(it))| dt 0 Z 2π = |t exp(it))| dt 0 Z 2π = t |exp(it))| dt 0 Z 2π = t · 1dt 0 = 2π2
Next we compute the arc length function
α : [0,2π] 3 t 7→ α(t) = s(t) ∈ [0,L] = [0,2π2] where Z t Z t t2 α(t) = s(t) = |e˙(p)| dp = pdp = 0 0 2 The inverse function of α : [0,2π] → [0,2π2] is √ β : [0,2π2] 3 s 7→ β(s) = 2s ∈ [0,2π]
September 23, 2009 c jbquig-UCD 42 CHAPTER 2. CURVES
Finally arc-length parametrization of the Archimedean spiral curve is
γ = d ◦ β : [0,2π2] → [0,2π] → Γ ⊂ R2 s 7→ γ(s) = d(β(s)) thus √ √ √ cos( 2s) + 2ssin( 2s) √ √ γ(s) = √ √ √ = 1 − i 2s exp(i 2s), 0 ≤ s ≤ 2π2 sin( 2s) − 2scos( 2s) See also
http://en.wikipedia.org/wiki/Archimedean spiral and http://mathsci.ucd.ie/courses/math40060/maple
2.2 curvature
The intuitive approach to curvature of a curve Γ ⊂ R2 goes as follows. Begin with the circle. A small circle is very curved but a big circle is only slightly curved. It makes sense to quantify curvature κ by κ = ±1/r where ± indicates that the circle is traversed ACW or CW. A line is a degenerate circle of radius r = ∞ and curvature κ = 1/∞ = 0. Curvature is the same at all points on a circle. On a general curve Γ the curvature differs from point to point. The curvature at the point p ∈ Γ was defined (historically) as being the curvature of the circle which best fitted Γ at the point p; this circle is called the osculating circle. A more modern approach is that the the curvature at p ∈ Γ is the angular speed with which the tangent-normal frame (known as the Serret-Frenet frame)) at p turns as p moves along Γ with unit speed. These two definitions will be proven to be equivalent.
2.2.1 technicalities remark matrix of rotation Let S(t) be the matrix of positive (a.c.w.) rotation of R2 by the angle t. It is easy to check the following formulae which will prove useful in discussion of the osculating circle.
cos(t) −sin(t) S(t) = , S0(t) = S(π/2)S(t) , S00(t) = S(π)S(t) = −S(t) sin(t) cos(t) remark arc-length parametrization of the circle In example(2.1.2) we met a.c.w. arc-length parametrization γ of the circle with center 0, radius a > 0,curvature κ = 1/a. Write q = γ(0) then q = ai = i/κ. Equation(2.1) may be written
s 1 γ(s) = aS( ) = 0 + S(κ(s − 0))(q − 0) (2.17) a 0
Arc length parameterization γ of the the general circle C with center p and curvature κ and with γ(s0) = q ∈ C is
γ(s) = p + S(κ(s − s0))(q − p) (2.18) This formula will prove useful when we construct the equation of the osculating circle.
c jbquig-UCD September 23, 2009 2.2. CURVATURE 43
remark computing κ on the circle Let γ be arc-length parameterization γ of a circle with curvature κ, then
1 cos(κs) −sin(κs) −cos(κs) γ(s) = , γ0(s) = , γ(s) = κ κ sin(κs) cos(κs) −sin(κs)
From this we get a formula for curvature of a circle
κ = det(γ0,γ00) (2.19)
2.2.2 the Serret-Frenet frame Let p ∈ Γ where Γ is a regular curve. There is well defined unit tangent vector t to Γ at the point p. Let n be a unit vector orthogonal to t. Unfortunately n(p) is defined only up to ±. To avoid ambiguity let n be that unit vector such that passage from t to n is a.c.w. by the angle π/2. Thus
n = S(π/2)t or equivalently < t,n >= 0 and det(t,n) = 0
The pair (t,n) is known as the Seret-Frenet frame at the point p on the curve Γ. x(s) If γ(s) = is a.l.p. of the curve Γ the Serret-Frenet frame at the point γ(s) ∈ Γ is y(s)
x0(s) −y0(s) (t(s),n(s)) = γ0(s),S(π/2)γ0(s) = ( , ) y0(s) x0(s)
x(t) If c(t) = is regular parametrization of Γ the Serret-Frenet frame at the point c(t) ∈ Γ is y(t)
(t(t),n(t)) 1 1 = c0(t), S(π/2)c0(t) ||c(t)|| ||c(t)|| 1 x0(t) 1 −y0(t) = ( 0 , 0 ) px0(t)2 + y0(t)2 y (t) px0(t)2 + y0(t)2 x (t)
For arc length parameterization γ00 ⊥ γ0 = t so γ00 = hn where the signifance of the scalar h demands expla- nation, see §§(2.2.3).
2.2.3 classical definition of curvature 2 Let γ be arc length parametrization of the C curve Γ. Let s0 be a fixed value of s the osculating circle at the point γ(s0) ∈ Γ is that circle which approximates the curve Γ to the second order near the point γ(s0). More precisely the osculating circle has arc length parameterization c with
0 0 00 00 c(s0) = γ(s0) , c (s0) = γ (s0) , c (s0) = γ (s0) (2.20)
Now we define curvature κ(s0) of the general curve Γ to be that of the osculating circle. From (2.19) and (2.20) we obtain 0 00 0 00 κ(s0) = det(c (s0),c (s0) = det(γ (s0,γ (s0)) To sum up we have the classical definition of curvature, in terms of arc-length parametrization, of the C 2 curve Γ at the point γ(s0). κ = det(γ0,γ00) (2.21)
September 23, 2009 c jbquig-UCD 44 CHAPTER 2. CURVES
We take up a point left hanging at the end of §§(2.2.2). For arc length parametrization < γ00,γ0 >= 0, Recall that there γ00 = hn for some scalar h. But h = h.1 = h.det(t,n) = det(t,hn) = det(γ0,γ00) = κ Thus γ00 = κn (2.22) eg. curvature of the Archimedean Spiral Γ We will carry out this computation using a.l.p. of Γ as found in example(2.1.5). Computations involving a.l.p γ(s) can be onerous so we will eventually find formulae for curvature κ in terms of an arbitrary regular parametrization c(t). Before we begin, here is a slick (work saving) observation. Let a c u = , v = , w = a + ib , z = c + id b d Then < u,v > = Re(wz) and det(u,v) = Im(wz) Referring to the notation of example(2.1.5) the curvature is κ(s) = det(γ0(s),γ00(s)), too much like hard work = Im(z0(s),z00(s)), the complex version is less work
From example(2.1.5) √ √ z(s) = (1 − i 2s)exp(i 2s)
z0(s) i √ √ i √ = 0 − √ · exp(i 2s) + (1 − i 2s) · √ exp(i 2s) 2s 2s √ = exp(i 2s) and z00(s) √ 0 = exp(i 2s) √ iexp(i 2s) = √ 2s Finally κ(s) = Im z0(s)z00(s) √ ! √ iexp(i 2s) = Im exp(−i 2s) · √ 2s i = Im √ 2s 1 = √ 2s
c jbquig-UCD September 23, 2009 2.2. CURVATURE 45
2.2.4 curvature as angular speed remark theory of SO(2,R) and so(2,R) In chapter(1) we studied the Lie group SO(3,R) and its associated Lie algebra so(3,R) in some detail. The comparable theory in the Euclidean plane (R2,<,>) with Lie group SO(2,R) and Lie algebra so(2,R) is trivial. Here are the details.
A 2 × 2 real matrix A is said to be special orthogonal iff it preserves the inner product and orientation in R2. It is easy to check that this is so iff A−1 = At and detA = 1. A rotation in R2 is a matrix of the form
cos(t) −sin(t) A = sin(t) cos(t)
It is trivial to check that A is special orthogonal iff A is a rotation. We define the Lie group
−1 t SO(2,R) = A ∈ Mat(2 × 2,R) A = A and det(A) = 1
Next let A(t) be a path in SO(2,R) with A(α) = Id ∈ SO(2,R). Differentiating the relation At (t)A(t) = Id at t = α and writing B = A0(α) we obtain
Bt + B = 0 ⇔ Bt = −B
Thus B is skew symmetric and it is trivial to prove that
0 −κ B = , for some κ ∈ R κ 0
That is all we need to know about SO(2,R) and so(2,R).
Let γ : [a,b] 3 s 7→ γ(s) ∈ Γ ⊂ R2 be C 2 arc length parametrization of the C 2 curve Γ ∈ R2 and
(t(γ(s)),n(γ(s))) = (t(s),n(s)) = (t,n)(s) denote the Serret-Frenet frame at the point γ(s) ∈ Γ. The following is a C 1 regular path in SO(2,R).
(t,n) : [a,b] → SO(2,R) ⊂ R4 s 7→ (t(s),n(s))
Let s0 ∈ [a,b] be a fixed value of s. This path passes through the point (Lie group element) (t(s0),n(s0)) (for −1 short write (t0,n0)) when s = s0. Multiplying by the inverse Lie group element (t0,n0) we obtain a new path in SO(2,R) which passes through Id the identity element when s = s0.
−1 4 (t0,n0) (t,n) : [a,b] → SO(2,R) ⊂ R s 7→ (t0,n0)(t(s),n(s))
Differentiating at s = s0 we obtain a skew symmetric element of
0 −κ B = 0 κ0 0 the Lie algebra of angular velocities so(2,R). −1 0 0 0 −κ0 (t(s0),n(s0)) (t (s0),n (s0)) = κ0 0
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We have denoted the single scalar (which determines all matrix entries) by κ0 = κ(s0) and we call it the curvature at the point γ(s0 ∈ Γ. Rearranging the latter equation we obtain 0 0 0 κ(s0) (t (s0),n (s0)) = (t(s0),n(s0)) κ(s0) 0
Changing from s0 to general s and suppressing mention of the variable we have the famous Serret-Frenet equations. 0 κ t0 = κn (t0,n0) = (t,n) ⇔ (2.23) κ 0 n0 = −κt
We now define the curvature κ(s) at each point γ(s) ∈ Γ by equation(2.23).
The second Serret-Frenet equation tells us that
t0 = κn ⇔ (γ0)0 = κn
Comparison with (2.22) shows that this definition of curvature (as angular speed of rotation of the S.F. frame) is the same as the classical definition (curvature of the osculating circle).
2.2.5 a.l.p. of the osculating circle Given κ ∈ R, q ∈ R3 and a unit vectorn there is a unique circle C with curvature κ and passing throug q with normal n there. The center of that circle is n p = q + (2.24) k and arc length parameterization c with c(s0) = q is h ni n c(s) = q + + S(κ(s − s )) − (2.25) k 0 κ
Using this formula we can find arc length parametrization of the osculating circle at γ(s0) ∈ Γ
γ00(s ) −γ00(s ) c(s) = γ(s ) + 0 + S κ(s )(s − s ) 0 , where κ = det(γ0(s ),γ00(s )) (2.26) 0 κ2 0 0 κ2 0 0
0 0 00 00 ex. With c as in equation(2.26) prove that c(s0) = γ(s0) , c (s0) = γ (s0) and c (s0) = γ (s0).
The center of the osculating circle at an arbitrary point γ(s) ∈ Γ is
n(s) γ00(s) e(s) = γ(s) + = γ(s) + , where κ(s) = det(γ0(s),γ00(s)) (2.27) κ(s) κ2(s)
As γ(s) moves along the curve Γ the center e(s) of the osculating circle traces out a new curve Evlt(Γ) called the evolute of the original curve Γ eg. Find the evolute curve of the Archimedean spiral Γ. Before we begin let run through our previous treatments of the Archimedean spiral. We started in exam- ple(2.1.2) with a.l.p. of the unit circle C
cos(s) c : [0,2π] 3 s 7→ c(s) = ∈ C ⊂ R2 sin(s)
c jbquig-UCD September 23, 2009 2.2. CURVATURE 47
In example(2.1.5) we defined Γ as the involute curve I(C) of C and found a natural parameterization
cos(t) +t sin(t) d : [0,2π] 3 t 7→ d(t) = ∈ Γ ⊂ R2 sin(t) −t cos(t)
Also in example(2.1.5)we replaced this parameterization of γ by a.l.p. √ √ √ cos( 2s) + 2ssin( 2s) √ √ γ : [0,2π2] 3 s 7→ γ(s) = √ √ √ = (1 − i 2s)exp(i 2s) ∈ Γ ⊂ R2 sin( 2s) − 2scos( 2s)
In example(2.2.3) we computed γ0(s), γ00(s) and κ(s) which were
√ i √ 1 γ0(s) = exp(i 2s) , γ00(s) = √ exp(i 2s) , κ(s) = √ 2s 2s
After this review we are ready to find the evolute curve E(Γ)of the Archimedean spiral Γ.
e(s) γ00(s) = z(s) + κ2(s) √ √ i √ √ 2 = (1 − i 2s)exp(i 2s) + √ exp(i 2s) 2s 2s √ √ √ √ = (1 − i 2s)exp(i 2s) + i 2sexp(i 2s) √ = exp(i 2s) here then is parametrization of the evolute curve E(Γ). √ e : [0,2π2] 3 t 7→ e(t) = exp(i 2t) ∈ E(Γ) ⊂ R2
This is a non standard parametrization of the original circle C. Thus
E(I(C)) = E(Γ) = C. (2.28)
To sum up:- the involute of the unit circle is the Archimedian spiral the evolute of the Archimedean spiral is the unit circle.
We will see in §§?? that equation(2.28) holds for a all C 2 regular curves.
2.2.6 formulae for t,n,κ in R2 In §2.2.4 we find formulae for t,n and κ in terms of arc length parameterization γ(s)
γ00 t = γ0 , n = , κ = det(γ0,γ00) (2.29) det(γ0,γ00)
Changing to the parameterization c(t) we have the (often more usable) formulae.
c0 c00 < c0,c0 > −c0 < c0,c00 > det(c0,c00) t = , n = , κ = (2.30) < c0,c0 >1/2 < c0,c0 >1/2 det(c0,c00) < c0,c0 >3/2
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or writing c(t) = x(t)i + y(t)j the very convenient formulae
1 x0 1 −y0 x0y00 − y0x00 t = 0 , n = 0 , κ = (2.31) px02 + y02 y px02 + y02 x (x02 + y02 + z02)3/2
In addition we have a variant of the Serret-Frenet equations
dt 0 0 1/2 d 0 −κ = < c ,c > κn (t,n) = < c0,c0 >1/2 (t,n) ⇔ dt (2.32) κ 0 dn dt = − < c0,c0 >1/2 κt dt proof
2.2.7 involute and evolute A C 2 regular curve Γ generate two new curves I(Γ) and E(Γ) called the involute and the evolute respectively. The involute is the curve traced out by unwinding a taut string laid along the original curve. The evolute is the curve traced out by the center of the osculating circle as the point of contact moves along the original curve. If γ : [a,b] 3 s 7→ γ(s) ∈ Γ ⊂ R2 is arc-length parametrization of Γ we a have parameterization i of the the involute I(Γ)
i : [a,b] 3 s 7→ i(s) = γ(s) − s ∗ t(s) ∈ I(Γ) ⊂ R2 and a parametrization e of the evolute E(Γ)
n(s) e : [a,b] 3 s 7→ e(s) = γ(s) + ∈ E(Γ) ⊂ R2 κ(s) where t(s),n(s) and κ(s) are unit tangent and normal vectors and curvature of the original curve Γ. Whereas the parameter s is arc-length on Γ it is most likely not arc length on either of the two new curves. remark on the involute We compute i0(s), i00(s) and det(i0(s),i00(s)) for future use
i0(s) h i0 = γ(s) − st(s) = γ0(s) − t(s) − st0(s) = t(s) − t(s) − sκ(s)n(s), by the Serret-Frenet formulae = −sκ(s)n(s)
Writing (t(s),n(s)) for the Serret-Frenet frame at i(s) on I(Γ) this tells us that t(s) = −sgn(κ(s))n(s) and so also n(s) = −sgn(κ(s))t(s). Next
i00(s) = [−sκ(s)n(s)]0 = −κ(s)n(s) − sκ0(s)n(s) − sκ(s)n0(s) = −κ(s)n(s) − sκ0(s)n(s) − sκ2(s)t(s), by the Serret-Frenet formulae
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Next
det(i0(s),i00(s)) = det −sκ(s)n(s) , −κ(s)n(s) − sκ0(s)n(s) − sκ2(s)t(s), = det −sκ(s)n(s) , −sκ2(s)t(s),, by elementary column operations = −s2κ3(s)det(t(s),n(s),), by more elementary column operations = −s2κ3(s) · 1 = s2κ3(s)
With these basic computations done we are ready to find the center of p(s) of the osculating circle at the point i(s) on the involute I(Γ), (i.e. the evolute of the involute).
p(s) n(s) = i(s) + , where κ,n are curvature and unit normal of I(Γ) κ(s) −sgn(κ(s))t(s) = i(s) + , κ(s) h i −sgn(κ(s))t(s) = γ(s) − st(s) + det(i0(s),i00(s))/ < i0(s),i0(s) >3/2 h i −sgnκ(s)t(s) = γ(s) − st(s) + −s2κ3(s)/s3 |κ|3 (s) h i −sgn(κ(s))t(s) = γ(s) − st(s) + −sgn(κ(s))/s h i = γ(s) − st(s) + st(s) = γ(s)
We have proven that
E(I(Γ) = Γ (2.33)
The evolute of the involute of a C 2 regular curve is the original curve itself.
2.3 curves in R3
Our goal is to study curves in three space. We quickly skim over the concepts of velocity, acceleration, speed, and arc-length parameterization which involve no ideas which are not already present in planar curves. We introduce the cylindrical helix, an important example.
To study curvature we introduce the Serret-Frenet frame which consists of a positively oriented orthonormal triple (t,n,b) at each point on a curve Γ ⊂ R3. These three vectors are called the tangent, normal and binormal respectively. We obtain a path through SO(3,R) whose angular velocity B = (t˙,n˙,b˙) lies in so(3,R). Surprisingly B yields only two scalar entries (you might expect three), These scalars are κ and τ called curvature and torsion respectively.
September 23, 2009 c jbquig-UCD 50 CHAPTER 2. CURVES
Figure 2.3: (i) helix(3 turns) (ii) intersection cylinder and screw surfaces
2.3.1 curves in R3, basic definitions As in §(2.1.1), which dealt with plane curves, we define the concept of a curve Γ lying in Euclidean three space R3. We define regular parametrization
x(t) 3 c : [a,b] 3 t 7→ c(t) = y(t) ∈ Γ ⊂ R z(t)
As in §§(2.1.3), we define velocity and acceleration vectors and a speed scalar.
x0(t) x00(t) 0 00 0 q c (t) = y0(t) , c (t) = y00(t) , ||c (t)|| = x0(t)2 + y0(t)2 + z0(t)2 z0(t) z00(t)
As in §§(2.1.5), we define (parametrization independent) total arc length
Z b Z b q L = ||c0(t)||dt = x0(t)2 + y0(t)2 + z0(t)2 dt a a and an arc-length α function with inverse β Z t q α : [a,b] 3 t 7→ α(t) = s(t) = x0(p)2 + y0(p)2 + z0(p)2 dp ∈ [0,L] a and arclength parametrization γ(s) = c ◦ β : [0,L] → γ ⊂ R3. eg. cylindrical helix in R3 Let a,B ∈ R and put b = B/2π. Consider the helix curve Γ which turns once around the z − axis with radius a while ascending to height B = 2πb.
x(t) acos(t) 3 c : [0,2π] 3 t 7→ c(t) = y(t) = asin(t) ∈ Γ ⊂ R (2.34) z(t) bt
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is a regular parameterization of this helix curve which can be viewed as the intersection of a cylinder surface and a screw surface, see figure(2.3). x x 2 2 2 \ −1 y Γ = y x + y = a y z = b tan x z z
acost −asint −acost 0 0 p 00 c(t) = asint , c (t) = acost , ||c (t)|| = a2 + b2 , c (t) = −asint bt b 0 √ The speed is constant. c0 ⊥ c00 even though this is not arc-length parameterization (unless ||c0|| = a2 + b2 = 1). Total length is Z 2π p L = ||c0|| = 2π a2 + b2 0 Take Z t p h p i α : [0,2π] 3 t 7→ α(t) = s(t) = ||c0|| = a2 + b2t ∈ [0,L] = 0,2π a2 + b2 0 whose inverse is h p i s β : 0,2π a2 + b2 3 s 7→ β(s) = t = √ ∈ [0,2π] a2 + b2 Arc length parametrization is s acos √ a2 + b2 s h p i s 3 γ : 0,2π a2 + b2 3 s 7→ γ(s) = c(β(s)) = c √ = asin √ ∈ Γ ⊂ R 2 2 a2 + b2 a + b s b√ a2 + b2 We define the involute curve of the helix as in the case of R2. We use the same formula to parametrize it.
e(t) = c(t) − s(t)t(t) x(t) x0(t) Z t q 1 = y(t) − x0(p)2 + y0(p)2 + z0(p)2 dp y0(t) p 0 2 0 2 0 2 z(t) 0 x (p) + y (p) + z (p) z0(t) acos(t) asin(t) q 1 = asin(t) − (a2 + b2)t √ −acos(t) 2 2 bt a + b b cos(t) −t sin(t) = a sin(t) + cos(t) 0 The evolute of the cylindrical helix lies entirely in the xy plane and is an Archimedean Spiral curve.
We will see in chapter(??) that during traversal of the helix the tangent line sweeps out a surface (which the screw surface as you might expect) which is flat called the tangential helicoid surface. See the evolute curve in figure(??) and the tangential helicoid surface in figure (??). eg. conical helix in R3 The conical helix is that curve Γ ⊂ R3 which lies on the cone surface z2 = x2 +y2 and turns once around the z − axis while ascending to height 2π.
x(t) t cos(t) 3 c : [0,2π] 3 t 7→ c(t) = y(t) = t sin(t) ∈ Γ ⊂ R (2.35) z(t) t
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Figure 2.4: Viviani’s curve, insersection of sphere and cylinder is a regular parameterization of Γ which can be viewed as the intersection of a conical surface K and a screw surface S, see figure(6.1). x x \ 2 2 2 \ −1 y Γ = K S = y x + y = z y z = b tan x z z eg. Viviani’s curve in R3 Let S and C be the spherical surface and cylindrical surfaces x x 2 2 2 2 2 S = y x + y + z = 4 and C = y (x − 1) + y = 1
z z
Γ = S ∩C ⊂ R3 is known as Viviani’s curve.
x(t) cos(2t) + 1 3 c : [0,2π] 3 t 7→ c(t) = y(t) = sin(2t) ∈ Γ ⊂ R (2.36) z(t) 2sin(t) is a regular parameterization Γ, see figure(2.4,2.4).
2.3.2 Serret-Frenet formulae Let γ : [0,L] → Γ ⊂ R3 be a regular C 2 curve parameterized by arc length where in addition, γ00(s) 6= 0,∀s ∈ [0,L]. (2.37)
For each s ∈ [0,L] define the unit tangent vector t(s) at γ(s), the same way we did in R2, i.e.
t(s) = γ0(s) (2.38)
Next we would like to define the unit normal vector n(s) but the situation is different from R2: mere geometry does not yield a unique unit vector perpendicular to t(s), instead there is a complete circle of unit vectors
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each perpendicular to t(s). We used the curve γ to define t and we use it again to define a distinguished unit normal vector; define γ00(s) n(s) = (2.39) ||γ00(s)|| This works because (i) < γ00(s),γ0(s) >= 0 and (ii) γ00(s) 6= 0 by assumption(2.37).
Next define the binormal b(s) to be the unique unit vector which is perpendicular to both t(s) and n(s) and such that the orthonormal triple (t,n,b) satisfies F.R.H.R, i.e. det(t(s),n(s),b(s)) > 0. In fact here is an explicit expression for b γ0(s) × γ00(s) b(s) = t × n = (2.40) ||γ00(s)|| The positively oriented orthonormal triple γ00(s) γ00(s) t, n, b (s) = t(s), n(s), b (s) = γ0(s), , γ0(s) × ||γ00(s)|| ||γ00(s)|| is called the Serret-Frenet frame at the point γ(s) of the curve Γ.
Thus the original path γ (parameterized by arc length) has generated a new path in the Lie group SO(3,R).
(t,n,b) : [0,L] → SO(3,R) γ00(s) γ0(s) × γ00(s) (2.41) s 7→ (t,n,b)(s) = γ0(s) , , ||γ00(s)|| ||γ00(s)||
1 Let s0 ∈ [0,L] be fixed but arbitrary. Then (t,n,b) : [0,L] → SO(3,R) is a C path passing through the point 0 0 0 0 (t,n,b)(s0) ∈ SO(3,R) when s = s0 with velocity vector (t ,n ,b )(s0) = (t,n,b) (s0). Now −1 (t,n,b) (s0)(t,n,b) : [0,L] → SO(3,R) −1 s 7→ (t,n,b) (s)(s0)(t,n,b)(s) −1 is a path in SO(3,R) passing through the identity Id when s = s0 (here (t,n,b) (s0) is a constant matrix in SO(3,R)) with velocity vector, 0 −κ(s0) σ(s0) −1 0 (t,n,b) (s0)(t,n,b) (s0) = κ(s0) 0 −τ(s0) ∈ so(3,R) −σ(s0) τ(s0) 0 We have used the fact, that from the preceding section, this velocity vector is skew symmetric i.e. lies in the special orthogonal Lie algebra so(3,R). Multiplying across by the Lie group element (t,n,b)(s0) and replacing s0 by general s we obtain the differential equation satisfied by (t,n,b)(s) 0 − (s) (s) d κ σ (t,n,b) = (t,n,b)(s) κ(s) 0 −τ(s) , ∀s ∈ [0,L] ds −σ(s) τ(s) 0 From this we see that t0 = κn − σb. But t0 = (γ0)0 = γ00 = ||γ00||n. From this it follows that σ = 0, ∀s and that κ = ||γ00||. The differential equation becomes
dt = κ(s)n(s) ds 0 −κ 0 d dn (t,n,b) = (t,n,b)(s) κ 0 −τ (s) ⇔ = −κ(s)t(s) + τ(s)b(s) ds ds 0 τ 0 db = −τ(s)n(s) ds (2.42) The expanded form here is known as the Serret-Frenet formulae The vectors t,n and b are known as tangent, normal and binormal respectively: the scalars κ and τ are known as curvature and torsion respectively.
September 23, 2009 c jbquig-UCD 54 CHAPTER 2. CURVES
2.3.3 3-space formulae for t,n,b,κ,τ
We collect formulae for the tangent,normal, and binormal vectors and also for the curvature and torsion scalars. These are given both for arc length parametrization γ(s) and arbitrary parameterization c(t).
c0 t = γ0 = < c0,c0 >1/2
γ00 (c0 × c00) × c0 n = = < γ00,γ00 >1/2 ||c0 × c00|| · ||c0||
γ0 × γ00 c0 × c00 (2.43) b = = < γ00,γ00 >1/2 ||c0 × c00||
||c0 × c00|| κ = < γ00,γ00 >1/2 = ||c0||3
< γ0 × γ00,γ000 > < c0 × c00,c000 > τ = = < γ00,γ00 > < c0 × c00,c0 × c00 > proof γ(s) column Formulae for t,n are the very definition of these vectors. The formula for b follows since b = t × n. The formula for κ is the norm of γ00 = t0 = κn. A little effort is required to find a formula for the torsion scalar τ. γ000 = (κn)0 = κ0n + κn0 = κ0n + κ(−κt + τb) = −κt + κ0n + κτb. Taking the inner product with b we obtain < γ000,b >= κτ. Substituting for b and κ we obtain our formula for τ. proof c(t) column Let γ(s) be arc length parameterization of Γ. Then c(t) = γ(s(t)) so that c0 = γ0s0. We have our formula for t = γ0 = c0/s0 = c0/ < c0,c0 >1/2. Next c00 = γ00(s02)+γ0s00 = κn(s0)2 +ts00 Take the cross product with c0 = s0t, obtain c0 ×c00 = κ(s0)3t×n = κ(s0)3b and take norms to get our formula for κ: also since c0 ×c00 is seen to be a scalar multiple of the unit vector b we get our formula for b. Next n = b×t yields our formula for n. Next c000 = (s00t + s02κn)0 = s000t + s00t0 + 2s0s00κn + s02κ0n + s02κn0 000 00 dt 0 0 00 02 0 02 dn 0 000 00 0 0 00 02 0 03 = s t + s ds s + 2s s κn + s κ n + s κ ds s = s t + s s κn + 2s s κn + s κ n + s κ(−κt + τb) = [s000 − s03κ2]t + [3s00s0κ + s02κ0]n + [s03κτ]b. Take the inner product with b and obtain < c000,c,×c00 > /||c0 × c00|| = s03κτ = ||c0 × c00||τ, from which follows our formula for τ.
If we are not using arc-length parametrization γ(t) but rather an abitrary parametrization c(t) we have a variant of the Serret Frenet equations. This variant follows from (2.42) using the chain rule
d ds d d = =< c0(t),c0(t) > . dt dt ds ds
0 − 0 d κ (t,n,b) = < c0(t),c0(t) >1/2 (t,n,b)(t) κ 0 −τ (t) (2.44) dt 0 τ 0
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⇔
dt = < c0(t),c0(t) >1/2 κ(t)n(t) dt dn = − < c0(t),c0(t) >1/2 κ(t)t(t) + < c0(t),c0(t) >1/2 τ(t)b(t) dt db = − < c0(t),c0(t) >1/2 τ(t)n(t) dt (2.45) eg. helix Serret-Frenet frame, curvature, torsion In example(2.3.1) we met a regular parametrization c of the cylindrical helix curve Γ ⊂ R3: position, velocity, acceleration and speed were
acost −asint −acost 0 00 0 p c(t) = asint , c (t) = acost , c (t) = −asint ., ||c (t)|| = a2 + b2 bt b 0
Using formulae(2.43) we compute t,n and b, the tangent, normal and binormal vectors at the point c(t) ∈ Γ.
First we compute −asin(t) c0(t) 1 t(t) = = √ acos(t) (2.46) 0 0 1/2 2 2 < c (t),c (t) > a + b b Next it is convenient to compute b; for that we will need c0 × c00 and ||c0 × c00||. i j k absin(t) bsin(t) 0 00 c (t) × c (t) = −asin(t) acos(t) b = −abcos(t) = a −bcos(t) (2.47)
−acos(t) −asin(t) 0 a2 a and q p ||c0 × c00|| = a b2(sin2(t) + cos2(t)) + a2 = a a2 + b2 (2.48) With these items we can construct
b c0(t) × c00(t) = ||c0(t) × c00(t)|| −asin(t) −acos(t) 1 = √ acos(t) × −asin(t) 2 2 a a + b b 0 i j k 1 = √ −asin(t) acos(t) b 2 2 a + b −cos(t) −sin(t) 0 bsin(t) 1 = √ −bcos(t) 2 2 a + b a
Next we will compute n: to do that we could use formula(2.43) but it is simpler (since we already have t and b) to use n = b × t. Then
n(t) bsin(t) −asin(t) 1 1 = b × t = √ −bcos(t) × √ acos(t) 2 2 2 2 a + b a a + b b
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bsin(t) −asin(t) 1 = −bcos(t) × acos(t) a2 + b2 a b i j k 1 = bsin(t) −bcos(t) a a2 + b2 −asin(t) acos(t) b −(a2 + b2)cos(t) 1 = −(a2 + b2)sin(t) a2 + b2 0 cos(t) = − sin(t) 0 Putting it all together so far, we have the Serret-Frenet frame for the helix −asin(t) cos(t) bsin(t) 1 1 t(t), n(t),b(t) = √ acos(t) , − sin(t) , √ −bcos(t) 2 2 2 2 a + b b 0 a + b a
(2.49) Next we will compute the curvature κ(t) and the torsion τ(t) at the point c(t) on the helix curve Γ. √ ||c0(t) × c00(t)|| a a2 + b2 a κ(t) = 0 3 = √ 3 = 2 2 (2.50) ||c (t)|| a2 + b2 a + b To compute the torsion τ(t) we need −acos(t) sin(t) d c000t) = (c00(t))0 = −asin(t) = a −cos(t) dt 0 0
Thus τ(t)
= < c0(t) × c00(t),c000(t) > < c0(t) × c00(t),c0(t) × c00(t) >
= < c0(t) × c00(t),c000(t) > ||c0(t) × c00(t)||2
= bsin(t) sin(t) 1 < a −bcos(t) , a −cos(t) > a2(a2 + b2) a 0
= a2b a2(a2 + b2)
= b a2 + b2
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Again we summarize
a b κ(t) = , τ(t) = (2.51) a2 + b2 a2 + b2 The physical interpretation of all this is as follows. As a point passes along a curve Γ moving as dictated by the parametrization c the S-F-frame rotates with angular velocity, [see(2.44)]
0 − (t) 0 d κ (t,n,b) = < c0(t),c0(t) >1/2 κ(t) 0 −τ(t) ∈ so(3,R) dt 0 τ(t) 0
This is interpreted as a vector angular velocity (the eigenvector)
τ(t) 0 0 1/2 < c (t),c (t) > 0 κ(t) but here referred to the instantaneous S-F-frame. The angular velocity of the S-F-frame is thus
τ 0 0 1/2 0 0 1/2 Ω = < c ,c > (t,n,b) 0 = < c ,c > (τt + 0n + κb) (2.52) κ
(called the Darboux vector) and angular speed of rotation of the S-F-frame is p ||Ω|| = < c0,c0 >1/2 κ2 + τ2 (2.53)
For the current example, using (2.52) we obtain
Ω −asin(t) bsin(t) p b 1 a 1 = a2 + b2 √ √ acos(t) + √ √ −bcos(t) 2 2 2 2 2 2 2 2 a + b a + b b a + b a + b a −asin(t) bsin(t) 1 = √ b acos(t) + a −bcos(t) 2 2 a + b b a 0 1 = √ 0 2 2 a + b a2 + b2 0 p = a2 + b2 0 1 and from (2.53) s p a 2 a 2 p ||Ω|| = a2 + b2 √ + √ = a2 + b2 (2.54) a2 + b2 a2 + b2 remark arc-length parametrization of the helix For the helix we have proven that
a b κ = , τ = a2 + b2 a2 + b2
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From this it follows that 1 κ τ a2 + b2 = and thus a = , b = κ2 + τ2 κ2 + τ2 κ2 + τ2 a.l.p. of the helix curve may be written
s κ p acos √ 2 + 2 s a2 + b2 2 2 cos κ τ κ + τ s κ p γ(s) = asin √ = sin κ2 + τ2 s (2.55) a2 + b2 κ2 + τ2 τ p s · κ2 + τ2 s b · √ κ2 + τ2 a2 + b2
2.3.4 curves in R3 with constant curvature and torsion It is a simple exercise to prove that in R2, a curve has constant curvature κ iff it is a circle. In this section we investigate curves in R3 with constant curvature κ and torsion τ. Let
γ : [0,L] 3 s 7→ γ(s) ∈ Γ ⊂ R3 be such a curve, parameterized by arc length. The Serret-Frenet matrix is the constant 3 × 3
0 −κ 0 F = κ 0 −τ ∈ so(3,R), κ > 0, τ ∈ R (2.56) 0 τ 0
The Serret-Frenet frame (t,n,b) obeys the matrix differential equation
(t,n,b)0(s) = (t,n,b)(s)F (2.57) the solution of which is (t,n,b)(s) = H exp(sF) (2.58) since exp(sF)0 = F exp(sF) = exp(sF)F , F being constant. Here H ∈ SO(3,R) is short for (t,n,b)(0) the initial Serret-Frenet frame. In particular −1 H t(s) = exp(sF)1, the first column of exp(sF) Next Z s Z s γ(s) − γ(0) = γ0(p)dp = t(p)dp (2.59) 0 0 so that Z s Z s Z s −1 −1 −1 H (γ(s) − γ(0)) = H ( t(p)dp) = H t(p)dp) = exp(pF)1 dp (2.60) 0 0 0 But (see §§1.34 where A ∈ so(3,R) is diagonalized and §§1.4.3 which treats exponentiation of similar matri- ces) F = EDE−1 so that exp(pF) = E exp(pD)E−1 (2.61) √ where taking rho = κ2 + τ2
0 −ρ 0 0 −κ τ 1 D = ρ 0 0 ∈ so(3,R) and E = ρ 0 0 ∈ SO(3,R) (2.62) ρ 0 0 0 0 τ κ
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Thus
H−1(γ(s) − γ(0))
= Z s −1 E exp(pD)E1 dp 0
= Z s t E exp(pD)dp E1 0
= cosρp −sinρp 0 0 Z s 1 E sinρp cosρp 0 dp −κ ρ 0 0 0 1 τ
= p=s sinρp cosρp 0 0 1 1 E −cosρp sinρp 0 −κ ρ ρ 0 0 ρp p=0 τ
= sinρs cosρs 0 0 1 0 0 1 E −cosρs sinρs 0 − −1 0 0 −κ ρ2 0 0 ρs 0 0 0 τ
= −κcosρs −κ 1 E −κsinρs − 0 ρ2 τρs 0
= κ cos(ρs) 2 ρ 1 −1 0 0 κ κ ESk,π sin(ρs) − 0 , recall Sk,π = 0 −1 0 ρ2 ρ2 τ 0 0 0 1 (ρs) ρ2
We may write this result κ p cos( κ2 + τ2 s) κ2 + τ2 κ p γ(s) = v + S sin( κ2 + τ2 s) (2.63) κ2 + τ2 τ p ( κ2 + τ2s) κ2 + τ2 κ where v = γ(0) − HES i is a constant vector (the details of v don’t matter) κ2 + τ2 k,π and S = HESk,π is a rotation (the details of S don’t matter).
A transformation of R3 consisting of a translation combined with a rotation is called a Euclidean motion. We have proven (see formula(2.55)) the theorem theorem 5 Let Γ ∈ R3 be a C 3 regular curve with constant curvature κ > 0 and constant torsion τ then, up to Euclidean
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motion, Γ is the standard helix with this curvature and torsion, i.e. with arc length parametrization √ κcos( κ2 + τ2 s) 1 √ γ(s) = κsin( κ2 + τ2 s) κ2 + τ2 √ τ · ( κ2 + τ2 s)
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2.4 problem set
Atempt at least problems marked ♠. arc-length parameterization in R2 1. For each regular parameterized curve curve. Make a rough sketch. Prove that s(t) is as given. Repa- cost rameterize by arc length. (The simple spiral c(t) = t should be taken as a starting point sint when drawing spirals. Unfortunately arc length parameterization of the simple spiral in very very messy: and so was not included in this list.) cost √ (i) Logarithmic spiral, c(t) = exp(t) , −∞ < t < ∞, s(t) = t exp(t). sint cost −sin(t) (ii) Archimedean spiral♠, c(t) = t + , 0 < t < ∞, s(t) = t2/2. sint cos(t) t t (iii) Catenary♠, c(t) = = , 0 ≤ t < ∞, s(t) = sinh(t) cosht (expt + exp−t)/2 t −sint (iv) Cycloid♠, c(t) = + , 0 ≤ t < ∞, s(t) = 4 − 4cos(t/2) 1 −cost (v) Cardioid, z(t) = 2exp(it) − exp(2it), 0 ≤ t < ∞, s(t) = 8 − 8cos(t/2) 0 1 1 (vi) Tractoid c(t) = + , 0 ≤ t ≤ ∞, s(t) = log(cosht) t cosht sinht
Serret-Frenet frame, curvature, involute, evolute 2. ♠(cycloid) The cycloid Γ is the curve traced out by a point on the rim of a rolling wheel, such as the valve on a bicycle wheel and has a parameterization. t 0 t − sin(t) c(t) = + S(−t) = or z(t) = t + i − iexp(−it) 1 −1 1 − cos(t)
(i) Compute t(t),n(t) the Serret-Frenet frame at the point c(t) ∈ Γ. (ii) Compute κ(t) the curvature at c(t).(ans 1/4sin(t/2)) t + sin(t) (iii) Parametrize the evolute curve E(Γ). (ans ) cos(t) − 1 t + sin(t) (iv) Parametrize the involute curve I(Γ). (ans ) 3 + cos(t) (v) Prove that both I(Γ) and E(Γ) are cycloids and make a single rough (sketch) showing all 3 cycloids Γ,I(Γ) and E(Γ).
3. ♠(catenary-tractrix)♠ The catenary is the curve formed by a clothesline or telephone wire, one parameterization is exp(t) + exp(−t) c(t) = cosh(t) = 2 The tractrix is the curve follows by a resisting dog on a lease as its owner walks a straight line, one parametrization is t − tanh(t) d(t) = sech (t)
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(i) Show both curves in a single sketch. (ii) Compute (t(t),n(t)) and κ(t) at a point c(t) on the catenary. (iii) Compute (t(t),n(t)) and κ(t) at a point c(t) on the tractrix. (iv) Prove that the involute of the catenary is the tractrix. (v) Prove that the evolute of the tractrix is the catenary.
4. ♠(deltoid) When a circle of radius 1 rolls round inside a circle of radius 3 the initial point of contact traces out a deltoid curve ∆. (i) Prove that the following is a parametrization of the deltoid.
2cos(t) + cos(t) c(t) = or z(t) = 2exp(it) + exp(−it) 2sin(t) − sin(t)
(ii) Sketch the deltoid curve inside the circle of radius 3. (iii) Compute (t(t),n(t)) and κ(t) at a point c(t) on ∆. (iv) Prove that the evolute E(∆) is also a Deltoid 3 times the size of the original. (v) Prove that the involute I(∆) is also a Deltoid one third the size of the original
5. Let a,b,α > 0. Let Γ be the cylindrical helix curve which turns 1 radian, anticlockwise while ascending height b, on the cylinder C,x2 + y2 = a2, in time period 1/α of a second, starting from position ai ∈ C when t = 0. A parametrization of Γ is
x(t) acos(αt) c(t) = y(t) = asin(αt) z(t) bαt
(i) Sketch the curve. (ii) Compute c0,c00,c000,s0(t). √ (iii) Prove that s(t) = tα a2 + b2. Find the arc length paramaterization γ(s) of the curve c. (iv) Compute t(t), n(t), b(t), κ(t),τ(t). (v) Prove that κ and τ are constants and express both in terms of a and b. (vi) Find the eigenvector v(t) (with associated eigenvalue 0) of the Serret-Frenet matrix (i.e.the in- stantaneous axis of rotation of the S-F-frame). Express v(t) in terms of the moving S-F-frame. Express v(t) in terms of the fixed standard basis (i,j,k); hence prove that v(t) is a constant vector.
6. Let Γ be the conical helix curve with parametrization
x(t) t cos(t) c(t) = y(t) = t sin(t) z(t) t
(i) Find the equation of a conical surface K and a screw surface S such that Γ = K TS.
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(ii) Sketch Γ = K TS ⊂ R3. (iii) Compute c0,c00,c000,s0(t). (iv) Compute t(t), n(t), b(t), κ(t),τ(t).
7. Let S ⊂ R3 be the spherical surface of radius 2, x2 +y2 +z2 = 4. Let C ⊂ R3 be the cylindrical surface (x − 1)2 + y2 = 1 and Γ = S ∩C be the curve of intersection, known as Viviani’s curve. (i) Prove that the following is a parametrization of Γ.
x(t) cos(2t) + 1 c = y(t) = sin(2t) z(t) 2sin(t)
(ii) Roughly sketch Γ ⊂ S ∩C ⊂ R3 showing the curve and both surfaces. (iii) Compute c0, c00, c000 and < c0(t),c0(t) >. (iv) Compute c0 × c00, < c0 × c00,c0 × c00 >, and < c0 × c00,c000 >. (v) Compute t(t), n(t), b(t), κ(t),τ(t).
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c jbquig-UCD September 23, 2009 Chapter 3 surfaces
The Blurb on the topic TODO.
3.1 surfaces in R3 p One meets surfaces in advanced calculus. For example the sphere as z = a2 − x2 − y2 or as x2 +y2 +z2 = a2 or when using spherical polar coordinates as x = asinφcosθ, y = asinφsinθ,z = acosφ. remark A surface is a 2 dimensional differentiable Σ manifold embedded in R3. It is not useful, at this point, to explain the general concepts alluded to here. We will use the following serviceable definition. definition surface in R3 A subset Σ ⊂ R3 is said to be a surface iff given p ∈ Σ there is a C ∞ injective mapping φ : U ⊂ R2 → R3 carrying a neighbourhood (nhd.) U of 0 in R2 bijectively onto a nhd. φ(U) of p in Σ. Thus Σ ⊂ R3 is a surface iff close to each point it resembles R2.
3.2 surface presentation
There are three ways to present a surface Σ. • as a level set f −1(d), d ∈ R or contour of a C ∞ function f : U ⊂ R3 → R of an open subset of R3 to the real line. This method uses the implicit equation of Σ.
• as the graph Gr(g) of a C ∞ function g : V ⊂ R2 → R of an open subset of R2 to the real line. This method uses the explicit equation of Σ.
• as the image h(W) = Im(h) of a C ∞ function h : W ⊂ R2 → R3 of a open subset of R2 to 3-space. This method uses the parametric equation of Σ. We will use this method most of the time.
3.2.1 presentation as a level set or contour Let U ⊂ R3 be an open subset. Let
x f : U 3 y 7→ w = f (x,y,z) ∈ R z
65 66 CHAPTER 3. SURFACES
be a C ∞ real valued function defined over U ⊂ R3. Let d ∈ Im( f ) ⊂ R. Then x −1 3 Σ = f (d) = y ∈ U f (x,y,z) = d ⊂ R
z is a surface called the level set or contour on which f has constant value d.
x 3 f (x,y,z) = d, y ∈ U ⊂ R z is said to be the implicit equation of Σ.
3.2.2 graphical presentation Let V ⊂ R2 be an open subset. Let
x g : V 3 7→ z = g(x,y) ∈ R y be a C ∞ real valued function defined over V ⊂ R2. Then x x x x 3 Σ = Gr(g) = y z = g(x,y), ∈ V = y ∈ V ⊂ R y y z g(x,y) is a surface called the graph of g.
x z = g(x,y), ∈ V ⊂ R2 y is said to be the explicit equation of Σ.
3.2.3 parametric presentation Let W ⊂ R2 be an open subset. Let
h1(r,s) r h : W 3 7→ h(r,s) = h2(r,s) ∈ R3 s h3(r,s) be a C ∞ vector valued function defined over W ⊂ R3. Then
1 x(r,s) h (r,s) r r 3 Σ = Im(h) = h(r,s) ∈ W = y(r,s) = h2(r,s) ∈ W ⊂ R s s z(r,s) h3(r,s) is a surface called the image set of the parametric function h. W is called the domain of parameterization, r and s are called parameters
x(r,s) = h1(r,s) r y(r,s) = h2(r,s) , ∈ W s z(r,s) = h3(r,s) are said to be the parametric equations of Σ.
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3.3 first example, the sphere
The sphere, with center the origin and radius a > 0 is
S = S2(a) = {x |||x|| = a } ⊂ R3
Let f be the function x 3 2 2 2 f : R 3 x = y 7→ f (x,y,z) = x + y + z ∈ R z
Then S is the level surface f −1(a2) and the implicit equation of S is f (x,y,z) = a2
Let D ⊂ R2 be the disc, an open set
x 2 2 2 2 D = x + y ≤ a ⊂ R y
Let g be the function x p g : D 3 7→ g(x,y) = a2 − x2 + y2 ∈ R y
Then S = Gr(g), the graph of g. The explicit equation of S is z = g(x,y). In fact Gr(g) is only part of S, namely the top (open) hemisphere.
Let θ Q = (0,π) × (−π,π) = − π < θ < π and 0 < φ < π φ a rectangle in R2. Let h be the spherical polar function or transformation
asinφcosθ φ h : Q 3 7→ h(φ,θ) = asinφsinθ ∈ R3 θ acosφ
Then S = Im(h), the image of h in R3. Parametric equations of S are
x(φ,θ) = asinφcosθ y(φ,θ) = asinφsinθ , −π < θ < π, 0 < φ < π. z(φ,θ) = acosφ
In fact Im(g) is only part of S, namely S ∩ (R3 − H). where H ⊂ R3 is the hemiplane on which longitude θ = ±π: H = {x |y = 0,x < 0 }.
The parameters θ and φ are called longitude and latitude, the latter is zero at the North pole, ak ∈ S, and π/2 at the equator, S ∩ {x |z = 0 }.
An alternative parametrization of the sphere is given by the Mercator projection k. Later we will see that k is conformal or angle preserving and for this reason is found useful by cartographers.
cosθ u a 3 k : R × (−π,π) 3 7→ h(u,θ) = sinθ ∈ R θ coshu sinhu
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Figure 3.1: two parametrizations of each of the saddle and monkeysaddle
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Figure 3.2: sphere and torus surfaces
3.4 saddle and monkey saddle
The saddle surface D pervades all of mathematics, see figure(3.1). The explicit and implicit equations are z = x2 − y2 = r2 cos(2θ) and x2 − y2 − z = 0. A parameterization is r cosθ θ h : (−π,π) × [0,∞) 3 7→ h(θ,r) = r sinθ ∈ D ⊂ R3 r r2 cos(2θ) A weird cousin is the monkey saddle,see figure(3.1), M with explicit and implicit equations, z = x3 −3xy2 = r3 cos(3θ) and x3 − 3xy2 − z = 0 and parameterization r cosθ θ h : (−π,π) × [0,∞) 3 7→ h(θ,r) = r sinθ ∈ M ⊂ R3 r r3 cos(3θ)
3.5 surface of revolution
Let Γ be a curve in xz-space R2 ⊂ R3. Rotating Γ about the z−axis often generates a surface of revolution Σ ⊂ p R3. If z = g(x) and f (x,z) = d are explicit and implicit equations for the curve Γ ⊂ R2 then z = g( x2 + y2) p and f ( x2 + y2,z) = d are explicit and implicit equations of the surface Σ ⊂ R3. If X(t) c(t) = t ∈ [a,b] Z(t) is a parameterization of Γ then x(t,θ) X(t)cosθ h(t,θ) = y(t,θ) = X(t)sinθ (t,θ) ∈ [a,b] × (−π,π) z(t,θ) Z(t) is a parameterization of Σ.
In the following subsections we treat the sphere, torus, cone and one and two sheeted hyperbolas as surfaces of rotation. Other examples are the cylinder, tractroid etc. see exercises ??. In §§3.6 we will meet ruled surfaces, i.e. which are composed of straight lines. It is easy to see that the cylinder and cone are ruled, it is not so obvious that the single sheeted hyperboloid is a (twice) ruled surface as is also the saddle.
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3.5.1 the sphere revisited
2 The sphere S √(a) is a surface of revolution generated by Γ a semi-circular curve with explicit and implicit equations z = a2 − x2 and x2 + z2 = a2. And with parameterization
asint a 1 c(t) = , 0 ≤ t ≤ π or c(t) = , t ∈ R acost cosht sinht see figure(3.2).
3.5.2 the torus as surface of revolution x b Let 0 < a < b. Let C be the circle in R2 (xz-space) with center at = and radius a. Thus z 0 x 2 2 2 C = (x − b) + z = a . z
T = T 2(a,b) known as the torus is obtained as as the surface of revolution of the curve C about the z-axis, see figure(3.2). The explicit and implicit equations of C are q z = a2 − (x − b)2 and (x − b)2 + z2 = a2 and a parametric equation of C is
x(φ) b + sin(φ) c : (−π,π) 3 φ 7→ c(φ) = = ∈ C ⊂ R2 z(φ) cos(φ)
The explicit and implicit equations for T are therefore
q p p z = g(x,y) = a2 − ( x2 + y2 − b)2 , f (x,y,z) = ( x2 + y2 − b)2 + z2 = a2 and the corresponding parameterization of T is
(b + asinφ)cosθ φ h : Q 3 7→ h(φ,θ) = (b + asinφ)sinθ ∈ R3 θ acosφ
Then T = Im(h), the image of h in R3. The parametric equations of T are
x(φ,θ) = (b + asinφ)cosθ y(φ,θ) = (b + asinφ)sinθ , −π < φ < π, −π < θ < π. z(φ,θ) = acosφ
In fact Im(h) is only part of T, namely T − (S ∪ S0). where S and S0 are each a circle on T.
The parameters φ and θ are called (toroidal) latitude and longitude.
3.5.3 hyperboloids and cone as surfaces of revolution two sheeted hyperboloid Consider the curve with implicit, explicit and parametric equations
p x(t) sinh(t) x2 − z2 = −1 , z = ± x2 + 1 , c(t) = = z(t) ±cosh(t)
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Figure 3.3: single and double sheeted hyperboloids, cone, hyperbolic system
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an hyperbola with two components. Rotation about the z axis forms the two sheeted hyperboloid surface [see figure(3.3)] consisting of two hyperbolic bowls with implicit, explicit and parametric equations x(t,θ) sinh(t)cos(θ) 2 2 2 p x + y − z = −1 , z = ± x2 + y2 + 1 , h(t,θ) = y(t,θ) = sinh(t)sin(θ) z(t,θ) ±cosh(t) asymptotic cone Replace −1 by 0 we obtain two lines meeting at 0 being asymptotes to the preceding (and following) hyperbola x(t) t x2 − z2 = 0 , z = ±x , c(t) = = z(t) ±t which rotate to form the cone surface [see figure(3.3)] x(t,θ) t cos(θ) 2 2 2 p x + y − z = 0 , z = ± x2 + y2 , h(t,θ) = y(t,θ) = t sin(θ) z(t,θ) t single sheeted hyperboloid Replace −1 this time by +1 we again obtain an hyperbola p x(t) cosh(t) x2 − z2 = 1 , z = ± x2 − 1 , c(t) = = z(t) sinh(t) which rotates to form the single sheeted hyperboloid [see figure(3.3)] x(t,θ) cosh(t)cos(θ) 2 2 2 p x + y − z = 1 , z = ± x2 + y2 − 1 , h(t,θ) = y(t,θ) = cosh(t)sin(θ) z(t,θ) sinh(t)
See the hyperbolic system, figure(??), of single and double sheeted hyperbolic surfaces with asymptotic cone. This is a contour diagram of R3 which is viewed as a disjoint union of level surfaces of the function f (x,y,z) = x2 + y2 − z2. disjoint x 3 [ R = y f (x,y,z) = d
d∈R z
3.6 ruled surfaces
A varying line passing through a point which moves along a curve in R3 forms a ruled surface. A ruled surface is one which admits a C 2 parameterization of the form t h : [a,b] × [c,d] 3 7→ c(t) + sv(t) ∈ Σ ⊂ R3 (3.1) s where c : [a,b] → Γ ⊂ R3 is a C 2 parameterization of the curve Γ and v : [a,b] → R3 is a C 2 mapping assigning a non zero vector at each point c(t) ∈ Γ; v is said ro be a vector field defined over Γ.
3.6.1 cylinder as ruled surface The cylinder C = x x2 + y2 = a2 can be viewed as a surface of revolution, here we present it as a ruled surface: one figure3.4 shows the cylinder in both guises. The direction vector v(t) ≡ k is constant along the curve Γ which is a a circle. acos(t) 0 t h : [−π,π] × R 3 7→ c(t) + sv(t) = asin(t) + s 0 ∈ C ⊂ R3 (3.2) s 0 1
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Figure 3.4: cylinder ruled and rotated, saddle doubly ruled
3.6.2 cone as ruled surface The cone K = x x2 + y2 − z2 = 0 has been viewed in §§3.5.3 as a surface of revolution, here we present it as a ruled surface, a single figure(3.3) shows the cone in both guises. 0 cost t h : [−π,π] × R 3 7→ c(t) + sv(t) = 0 + s sint ∈ K ⊂ R3 (3.3) s 0 1 The curve c is degenerate being a constant point.
3.6.3 single sheeted hyperboloid as ruled surface The single sheeted hyperboloid H = x x2 + y2 − z2 = 1 has been viewed in §§3.5.3 as a surface of revo- lution, see figure(3.3). Perhaps you will be surprised to find that it is also a ruled surface, see fgure(??). cost −sint t h : [−π,π] × R 3 7→ c(t) + sv(t) = sint + s cost ∈ H ⊂ R3 (3.4) s 0 1 The curve Γ is the so called circle of stricture lying in the xy-plane. Join two wire circles by several strings to outline a cylinder. Twist one circle by π/2 to obtain a single sheeted hyperboloid: the strings form the generating lines.
3.6.4 saddle surface is doubly ruled Consider the cube [−1,1] × [−1,1] × [−1,1] ⊂ R3 Let Σ be the ruled surface c(t) + sv(t) t 0 0 1 t t h : [−1,1]×[−1,1] 3 7→ or = 0 +s 1 or s +t 1 = s ∈ Σ ⊂ R3 s c(s) + tv(s) 0 t 0 s st (3.5) The curve Γ can be taken to be the x-axis in which case the generating lines meet the faces x = ±1 in the lines z = ±y. Alternatively Γ can be taken to be the y-axis in which case the (alternative) generating lines meet the faces y = ±1 in the lines z = ±x. This surface has explicit equation z = xy and is the saddle in disguise which can be seen by putting x = X −Y,y = X +Y the explicit equation then becoming z = X2 −Y 2. Figure(3.4) shows a doubly ruled saddle, compare with figure(3.1)
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Figure 3.5: the single sheeted hyperbola as ruled surface
3.6.5 the right helicoid or screw surface is ruled The Helicoid surface known as the screw S = {x |z = tanθ = arctan(y/x) } is a ruled surface, see figure(??) . 0 cost t h : R × [0,∞) 3 7→ c(t) + sv(t) = 0 + s sint ∈ S ⊂ R3 (3.6) s t 0 The curve Γ is the z-axis from which emanates horizontal generating lines which spin with ascent of Γ.
3.6.6 the tangent developable helicoid ruled surface The tangent lines along a curve Γ trace out a ruled surface Σ called the tangential developable of Γ. In particular when Γ is the helix we obtain
acost −asint t h : R × R) 3 7→ c(t) + sv(t) = asint + s acost ∈ Σ ⊂ R3 (3.7) s bt b
A computer drawing reveals the details, see ??.
3.7 charts, affine linear approximate 3.7.1 chart, surface element In §3.1 we studied examples of surfaces in the large; we did not worry to much about the precise definition of surface. Now we confine ourselves to parameterizing a small patch on a surface. definition A C 1 injective mapping x(s,t) s f : R2 ⊃ A 3 r = 7→ y(s,t) ∈ R3 (3.8) t z(s,t) from an open set in R2 into R3 is called a chart or surface element. f can be viewed as parameterizing a small surface patch Σ = f(A) ⊂ R3
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Figure 3.6: right and tangent developable helicoid ruled surfaces
3.7.2 affine linear approximate Let q ∈ A be a fixed point. Write
a x(α,β) α q = , p = f(q) = f(α,β) = b = y(α,β) β c z(α,β) for the corresponding point in Σ. Amongst all affine linear mappings there is one F which most closely approximates f for r close to q. F is the first order Taylor expansion of f at the point q.
F : R2 3 r 7→ F(r) = f(q) + Df(q)(r − q) ∈ R3 (3.9)
Here Df(q) = Df(α,β) is the differential 3 × 2 matrix of f computed at the point q ∈ A. In full detail
∂x ∂x ∂x ∂x a + (s − α) + (t − β) ∂s ∂t ∂s ∂t ∂y ∂y s − α ∂y ∂y F(s,t) = f(α,β) + = b + (s − α) + (t − β) (3.10) ∂s ∂t t − β ∂s ∂t ∂z ∂z ∂z ∂z c + (s − α) + (t − β) ∂s ∂t ∂s ∂t where all partial derivatives have been taken at q. If we write F(r) − p = Df(q)(r − q) we see that F is essentially the linear (matrix) mapping Df(q) preceded by a change of origin in R2 and followed by a change of origin in R3. F is said to be affine linear.
3.7.3 the tangent plane The image set f(A) of the surface element mapping f is of course surface the patch Σ ⊂ R3. The image set F(R2) of the linear approximate mapping F is the (affine) linear surface or affine plane which most closely approximates the surface Σ near p = f(q). Tp = Im(F) is called the tangent plane at the point p to the surface Σ. The parametric equation of Tp is
x(s,t) = F(s,t)
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s − α ⇔ x(s,t) = f (α,β) + D f (α,β) t − β ∂f1 ∂f1 1 x(s,t) f (α,β) ∂s ∂t ∂f2 ∂f2 s − α ⇔ y(s,t) = f2(α,β) + t − β z(s,t) f3(α,β) ∂s ∂t ∂f3 ∂f3 ∂s ∂t ⇔ ∂x ∂x x(s,t) = a + (s − α) + (t − β) ∂s ∂t ∂y ∂y y(s,t) = b + (s − α) + (t − β) ∂s ∂t ∂z ∂z z(s,t) = c + (s − α) + (t − β) ∂s ∂t
Since q,q + i and q + j lie in R2 the three points
F(q) = p , F(q + i) = p + Df(i) = p + D~f1 and F(q + j) = p + Df(j) = p + D~f2 (3.11) lie in Tp. In full detail
a a ∂x/∂s a ∂x/∂t b , b + ∂y/∂s , b + ∂y/∂t ∈ Tp (3.12) c c z/ s c z/ t ∂ ∂ (α,β) ∂ ∂ (α,β)
3.7.4 normal vector
Using the three points in Tp we will obtain an expression for the unit normal vector to the tangent plane Tp and also the implicit equation of Tp. Recall from linear algebra that (b − a) × (c − a) is normal to the plane through the three points a,b,c. A normal vector to the tangent plane is therefore
i j k ∂x ∂y ∂z ~n = D f1 × D f2 = det ∂r ∂r ∂r ∂x ∂y ∂z ∂s ∂s ∂s (α,β)
Thus
Tp 3 = x ∈ R < x − p,~n >= 0 3 = x ∈ R < x − p,D f1(α,β) × D f2(α,β) >= 0 x − a y − b z − c x ∂x ∂y ∂z
= y ∂s ∂s ∂s = 0 z ∂x ∂y ∂z ∂t ∂t ∂t (α,β)
Here we see the implicit equation for the tangent plane Tp.
Divide by the norm to obtain the unit normal vector n to Tp and thus also to the surface patch Σ at the point p = f(q). ~n Df × Df n = = 1 2 (α,β) (3.13) ||~n|| ||Df1 × Df2||
c jbquig-UCD September 23, 2009 3.7. CHARTS, AFFINE LINEAR APPROXIMATE 77
In full detail
∂y/∂s ∂y/∂t ∂z/∂s ∂z/∂t ∂x/∂s ∂x/∂t i + j + k ∂z/∂s ∂z/∂t ∂x/∂s ∂x/∂t ∂y/∂s ∂y/∂t n = (α,β) (3.14) s 2 2 2 ∂y/∂s ∂y/∂t ∂z/∂s ∂z/∂t ∂x/∂s ∂x/∂t + + ∂z/∂s ∂z/∂t ∂x/∂s ∂x/∂t ∂y/∂s ∂y/∂t eg. parabolic bowl In xz − space z = x2 is the explicit equation of a parabola. Rotate this curve about the z − axis to obtain P a parabolic bowl surface (of revolution) with explicit equation z = x2 + y2. Use this explicit equation to generate the following parametrization of P x(s,t) s s f : R2 3 7→ y(s,t) = t ∈ P ⊂ R3 (3.15) t z(s,t) s2 +t2
α Let q = ∈ R2 so that f(q) = f(α,β) = αi + βj + (α2 + β2)k ∈ P The differential matrix β
∂x ∂x ∂s ∂t 1 0 ∂y ∂y Df(α,β) = (α,β) = 0 1 ∂s ∂t ∂z ∂z 2α 2β ∂s ∂t The affine linear approximate mapping F to f near q is s α F(s,t) = f(q) + Df(q)( − ) t β α 1 0 s − α = β + 0 1 t − β α2 + β2 2α 2β α + 1 · (s − α) = β + 1 · (t − β) α2 + β2 + 2α(s − α) + 2β(t − β) s = t 2αs − α2 + 2βt − β2
P The parametric equation of the tangent plane T = Tp = Tp is x(s,t) = s s y(s,t) = t ∈ R2 (3.16) t z(s,t) = 2αs + 2βt − α2 − β2 A normal vector at p ∈ P at is i j k −2α
~n = Df1 × Df2 = 1 0 2α = −2β
0 1 2β 1 and the unit vector there is −2α 1 n = −2β p 2 2 1 + 4α + 4β 1
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The implicit equation of the tangent plane T is
< x − p , Df1(q) × Df2(q) > = 0
⇔
x − α y − β z − (α2 + β2))
1 0 2α = 0
0 1 2β
⇔ −2α(x − α) − 2β(y − β) + 1 · (z − (α2 + β2)) = 0
⇔ 2αx + 2βy − z = α2 + β2 eg. saddle surface z = xy is the explicit equation of a saddle surface D with the following parametrization. x(s,t) s s f : R2 3 7→ y(s,t) = t ∈ D ⊂ R3 (3.17) t z(s,t) st α Let q = ∈ R2 so that f(q) = f(α,β) = αi + βj + (αβ)k ∈ D The differential matrix β ∂x ∂x ∂s ∂t 1 0 ∂y ∂y Df(α,β) = (α,β) = 0 1 ∂s ∂t ∂z ∂z β α ∂s ∂t The affine linear approximate mapping F to f near q is s α F(s,t) = f(q) + Df(q)( − ) t β α 1 0 s − α = β + 0 1 t − β αβ β α α + 1 · (s − α) = β + 1 · (t − β) αβ + β(s − α) + α(t − β) s = t βs + αt − αβ D The parametric equation of the tangent plane T = Tp is x(s,t) = s s y(s,t) = t ∈ R2 (3.18) t z(s,t) = βs + αt − αβ A normal vector at p ∈ D at is i j k −β
~n = Df1 × Df2 = 1 0 β = −α
0 1 α 1
c jbquig-UCD September 23, 2009 3.8. GAUSS AND WEINGARTEN MAPS 79
and the unit vector there is −β 1 n = −α p 2 2 1 + α + β 1 The implicit equation of the tangent plane T is
< x − p , Df1(q) × Df2(q) > = 0
⇔
x − α y − β z − αβ)
1 0 β = 0
0 1 α
⇔ −β(x − α) − α(y − β) + 1 · (z − αβ) = 0
⇔ βx + αy − z = αβ
3.8 Gauss and Weingarten maps
The surface n o 2 3 S = S (1) = x ∈ R ||x|| = 1 (3.19) is known as the unit sphere in R3 or for short simply as the sphere. At each point x ∈ Σ the unit normal n(x) lies on this sphere. Thus we obtain the Gauss mapping
n : R3 3 Σ 3 x 7→ n(x) ∈ S ⊂ R3 (3.20)
The linear approximate mapping at p ∈ Σ to the Gauss mapping is denoted L and is known as the Weingarten mapping. Σ S L : Tp → Tn(p) maps the tangent plane of Σ at p to the tangent plane of S at n(p). But the normal vector to S at n is n itself and so these tangent planes are one and the same (after translation of origins). Thus the Weingarten mapping or linearized Gauss normal map can be regarded as mapping the tangent plane of Σ to itself
Σ Σ L : Tp → Tp (3.21) remark The details of ”curvature” of Σ will be extracted from the mapping L. After choice of a basis for the tangent plane L, which is linear, will be represented by a 2×2 matrix, denoted L. We will be interested in the eigenvalues, eigenvectors, trace,and especially in the determinant K(p) of L. In fact K(p) is the Gaussian curvature at p ∈ Σ. Since L is the linearization of the Gauss map n : Σ → S it follows that K(p) is the ratio of, area swept out on S by the normal vector n(p), to area swept out by x as this point moves through Σ close to p. remark Computing the matrix L directly is not trivial. Instead one defines and readily computes three Σ bilinear forms I, II, III on Tp : from these L is readily computed. The bilinear forms are of interest in them- selves and indeed I forms the starting point for study of intrinsic geometry of manifolds.
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3.9 self adjointness of L
The next result is of great importance in the study of curvature of surfaces. The Weingarten or linearized normal mapping L : Tp → Tp is self adjoint with respect to the restriction to Tp of the standard inner product <,> on R3. In other words theorem 6 3 < Lv,w > = < v,Lw >, for all v,w ∈ Tp ⊂ R (3.22) proof Write {e1,e2} for the vectors {D f1,D f2} which form a basis for the vector space (plane) Tp. Let v = ae1 + be2 and w = ce1 + de2 be vectors in Tp, then
< Lv,w >=< v,Lw >
⇔ < L(ae1 + be2),ce1 + de2 >=< ae1 + be2,L(ce1 + de2) >,
⇔ < aLe1 + bLe2,ce1 + de2 >,=< ae1 + be2,cLe1 + dLe2 >, by linearity of L ⇔ ac < Le1,e1 > − < e1,Le1 > + ad (< Le1,e2 > − < e1,Le2 >) +
bc(< Le2,e1 > − < e2,Le1 >) + bd (< Le2,e2 > − < e2,Le2 >) = 0 by bi-linearity of <,> ⇔ ac 0 + ad (< Le1,e2 > − < e1,Le2 >) + bc(< e1,Le2 > − < Le1,e2 >) + bd (0) = 0 by symmetry of <,>
⇔ (ad − bc)(< Le1,e2 > − < e1,Le2 >) = 0 which is true if we can prove 3.22 for e1 and e2, but
< Le1,e2 > = < e1,Le2 >
⇔ < L(Df1),Df2 > = < Df1,L(Df2) >
⇔ < Dn(Df1),Df2 > = < Df1,Dn(Df2) >
⇔ < D(n ◦ f )1),Df2 > = < Df1,D(n ◦ f)2 > , by the chain rule of advanced calculus ∂(n ◦ f ) ∂f ∂f ∂(n ◦ f) ⇔ < , > = < , > ∂s ∂t ∂s ∂t ∂2f ∂2f ⇔ < n ◦ f , > = < ,n ◦ f > the crucial step, see 3.26 ∂t∂s ∂s∂t
∂2 f ∂2 f which is true by symmetry of <,> and since = . ∂s∂t ∂t ∂s Above we used ∂f < ,n ◦ f > = 0, since n ⊥ T 3 Df (3.23) ∂s p 1 ∂ ∂f ⇒ < ,n ◦ f > = 0 (3.24) ∂t ∂s ∂2f ∂f ∂n ◦ f ⇒ < ,n ◦ f > + < , > = 0 (3.25) ∂s∂t ∂s ∂t ∂2f ∂f ∂n ◦ f ⇒ − < ,n ◦ f > = < , > (3.26) ∂s∂t ∂s ∂t
c jbquig-UCD September 23, 2009 3.10. EIGENVALUES AND EIGENVECTORS OF A SELF ADJOINT MAPPING 81
3.10 eigenvalues and eigenvectors of a self adjoint mapping
Let (V,B) be an inner product of dimension 2. Let A : (V,B) → (V,B) be a self adjoint (i.e. w.r.t. B) linear mapping. Let (e1,e2) be an orthonormal basis for (V,B), (it is always possible to find such, say by applying the Gramme-Schmidt orthonormalization process to an arbitrary basis being a trivial calculation in a b the present dimension 2). Let A = be the matrix of A with respect to this basis. c d
A(e1) = ae1 + ce2 and A(e2) = be1 + de2 Then
B(e1,Ae2) = B(Ae1,e2)
⇔ B(e1,be1 + de2) = B(ae1 + ce2,e2)
⇔ bB(e1,e1) + dB(e1,e2) = aB(e1,e2) + cB(e2,e2) ⇔ b · 1 + d · 0 = a · 0 + c · 1 ⇔ b = c a c A is a symmetric matrix A = . The characteristic polynomial of A is c d
a + d 2 a − d 2 p(λ) = det(λId −A) = λ2 −tr Aλ+detA = λ2 −(a+d)λ+(ad −c2) = λ2 − − −c2 2 2 In particular p(λ) has real roots s a + d a − d 2 λ = α,β = + ± + c2 2 2 The eigenvalues of A are real. Next suppose α 6= β are eigenvalues with corresponding eigenvectors Av = αv and Aw = βw. Then αB(v,w) = B(αv,w) = B(Av,w) = B(v,Aw) = B(v,βw) = βB(v,w). Thus (α − β)B(v,w) = 0 and since α − β 6= 0 then B(v,w) = 0. It is now an easy exercise to prove that there is a basis for (V,B) of orthogonal eigenvectors of A. summary Let A : (V,B) → (V,B) be a self adjoint linear mapping on an inner-product space of dimension 2. (i) The matrix of A with respect to an orthonormal basis is symmetric. (ii) The eigenvalues of A are real. (iii) Eigenvectors of A corresponding to distinct eigenvalues are orthogonal w.r.t. the inner product B. (iv) We can find an orthonormal basis for (V,B) consisting of eigenvectors of A. In particular all this applies to the Weingarten mapping L.
3.11 invariants of the Weingarten mapping
3 Let p ∈ Σ ⊂ R and L : Tp → Tp be the Weingarten mapping. We seek invariants of L, i.e. scalars and vector properties of L which do not depend on any basis dependent matrix used to represent and (to compute with) L. We have • α and β eigenvalues called the principle curvatures of Σ at p. • the corresponding eigenvectors v ⊥ w where Av = αv and Aw = βw: these are called principle directions on Σ at p. • H(p) = tr L = (α + β)/2 called the mean curvature of Σ at p. • K(p) = det L = αβ called the Gaussian curvature of Σ at p.
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3.12 geometric meaning of Weingarten mapping
This section has been included to answer the question raised in class, ”What is the meaning of the Weingarten mapping L?”. Before answering the question we remark that the Gauss normal mapping n : Σ → S carries a point on Σ to a point on S the spherical surface: L(p) is the linear approximation to n near p ∈ Σ so that detLp is the ratio at of area swept out by n(x) on S to area swept out by x on Σ as x moves on Σ close to p. This quantity detLp is the famous Gauss curvature K(p).
Again before answering the main question we need to remind ourselves of the concept ”directional deriva- tive” from advanced calculus. Let f : R3 → R is a function of three variables and p ∈ R3 and v = ai+bj+ck ∂ f ∂ f ∂ f be a vector. We have the concept of the differential D f (p) = (p), (p), (p) at the point p being ∂x ∂y ∂z the linear mapping which most closely approximates f near p. We also have the concept of the directional derivative Dv f (p), which is the rate of change of f in the direction v at the point p. In other words, if c is a f ◦ c curve with c(0) = p and c0(0) = v then D f (p) = (0) v dt We have the formula d( f ◦ c) D f (p)(v) = D f (p) = (0) v dt The formula can be elaborated to many variants as follows
D f (p)(v) a ∂ f ∂ f ∂ f = (p), (p), (p) b x y z ∂ ∂ ∂ c ∂ f ∂ f ∂ f = a (p) + b (p) + c (p) ∂x ∂y ∂z = Dv f (p) f ◦ c = (0) dt = < D f t (p),v > = grad f (p),v ∂ f /∂x(p) a = < ∂ f /∂y(p) , b > ∂ f /∂z(p) c
Now we are ready to address the question— TO BE CONTD a.s.p. jbq 22/04/05
c jbquig-UCD September 23, 2009 3.13. PROBLEM SET 83
3.13 problem set
1. Here are some basic facts about the Gramme-Schmidt-Ortho-Normalization-Process(GSONP). We need it only in (V,B) a two dimensional IPS. Here it is for (V,B) a three dimensional IPS.
Given an arbitrary basis { f1, f2, f3} for the inner product space V,B.
(i) Replace { f1, f2, f3} by {e1, f2, f3} where f e = 1 1 1/2 B( f1, f1)
(ii) Replace {e1, f2, f3} by {e1,e2, f3} where
f2 − B( f2,e1)e1 e2 = B(numerator,numerator)1/2
(iii) Replace {e1,e2, f3} by {e1,e2,e3} where
f3 − B( f3,e1)e1 − B( f3,e2)e2 e3 = B(numerator,numerator)1/2
(iv) Check that at each stage we have a basis for V, i.e. that each of
{ f1, f2, f3}, {e1, f2, f3}, {e1,e2, f3}, {e1,e2,e3},
is a basis.
(v) Check that {e1,e2,e3} is an ORTHONORMAL basis for V w.r.t the inner product B. (vi) Carry out the GSONP in R3,<> starting with 1 2 3 { f1, f2, f3} = −2 −2 −1 1 0 1
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c jbquig-UCD September 23, 2009 Chapter 4 bilinear forms
The Blurb on the topic TODO.
4.1 definition of bilinear forms I, II and III
Let f : R2 ⊃ A → Σ ⊂ R3 be a surface element and f(q) = p as in §§(3.7.1)(3.7.2). We will define three bilinear forms, denoted I, II and III on the tangent plane Tp. All three definitions are in terms of (a) the 3 standard inner product <,> restricted from R to Tp and (b) the Weingarten mapping L : Tp → Tp. definition three bilinear forms
• I: Tp × Tp 3 (v,w) 7→ I(v,w) =< v,w >∈ R (4.1)
• II : Tp × Tp 3 (v,w) 7→ II(v,w) =< v,Lw >∈ R (4.2)
• III : Tp × Tp 3 (v,w) 7→ III(v,w) =< Lv,Lw >∈ R (4.3)
Since <,> is an bilinear, symmetric and positive definite the bilinear form I is also bilinear, symmetric and positive definite I is an inner product called the first fundamental form on the tangent space Tp.
II is bilinear (since <,> is bilinear and L linear). II is symmetric using self adjointness of L, 3.22. In general II is not positive definite; consider for example the saddle surface. II is a symmetric bilinear form called the second fundamental form on Tp.
III is bilinear and symmetric (since <,> has these properties) III is positive definite (since <,> is p.d.) but only if L is nondegenerate. In fact III(v,v) ≥ 0 always but III(v,v) =< Lv,Lv >= 0 if Lv = 0. III is a positive (not always definite) symmetric bilinear form called the third fundamental form on Tp.
4.2 matrix representation of the three fundamental forms
The linear mapping L is not easily expressed in matrix form. However, we readily compute matrix repre- sentatives g,h and e for the bilinear forms I,II and III. L is then accessible being the matrix composite g−1 · h.
85 86 CHAPTER 4. BILINEAR FORMS
4.2.1 the matrix of a form We remind the reader of some basic linear algebra concerning bilinear forms. Let B : V × V → R be a symmetric bilinear form and let {e1,e2} be a basis for the vector space V assumed to be of dimension 2. Define Bi, j, 1 ≤ i, j ≤ 2 by
B B B(e ,e ) B(e ,e ) B = 1,1 1,2 = 1 1 1 2 (4.4) B2,1 B2,2 B(e2,e1) B(e2,e2) The matrix B = Bi, j 1 ≤ i, j ≤ 2 represents the abstract bilinear form B w.r.t. the basis {e1,e2}. a c Consider vectors v = ae + be and w = ce + de ∈ V. Write v = and w = . The column 1 2 1 2 b d vectors v and w represent the abstract vectors v and w with respect to the basis.
Computation with B involves the observation
B(v,w) = wt Bv (4.5) proof
B(v,w)
= B(ae1 + be2,ce1 + de2)
= abB(e1,e1) + acB(e1,e2) + bcB(e2,e1) + bdB(e2,e2) = abB1,1 + acB1,2 + bcB2,1 + bdB2,2 a = (c,d)B b = wt Bv
4.2.2 g,h,e matrices representing I,II,III theorem 7 Let f : U ⊂ R2 → R3 parameterize the surface element Σ ⊂ R3, q ∈ U and p = f(q) ∈ Σ. Let n : Σ → S denote the normal mapping of Gauss, L = Dn : Tp → Tp the Weingarten mapping. With respect to the basis {Df1,Df2} of Tp the matrices g(q),h(q),e(q) respectively of the three fundamental forms I,II,III on Tp are
< ∂f/∂s,∂f/∂s > < ∂f/∂s,∂f/∂t > g(q) = (q) = Df(q)t · Df(q) < ∂f/∂t,∂f/∂s > < ∂f/∂t,∂f/∂t >
∂f ∂n ∂f ∂n ∂2f ∂2f < , > < , > < ,n > < ,n > h(q) = ∂s ∂s ∂s ∂t (q) = − ∂2s ∂s∂t (q) ∂f ∂n ∂f ∂n ∂2f ∂2f < , > < , > < ,n > < ,n > ∂t ∂s ∂t ∂t ∂t∂s ∂2t < ∂n/∂s,∂n/∂s > < ∂n/∂s,∂n/∂t > e(q) = (q) < ∂n/∂t,∂n/∂s > < ∂n/∂t,∂n/∂t > proof Take {i,j} as the basis for the tangent space to q in R2. This tangent space is R2 itself translated to q. Then {Df(i),Df(j)} = {Df1,Df2} is our choosen basis for Tp. Thus
I(Df , Df ) I(Df , Df ) g(q) = 1 1 1 2 (q) I(Df2 , Df1) I(Df2 , Df2)
c jbquig-UCD September 23, 2009 4.2. MATRIX REPRESENTATION OF THE THREE FUNDAMENTAL FORMS 87
= < Df , Df > < Df , Df > 1 1 1 2 (q) < Df2 , Df1 > < Df2 , Df2 >
= < ∂f/∂s, ∂f/∂s > < ∂f/∂s, ∂f/∂t > (q) < ∂f/∂t , ∂f/∂s > < ∂f/∂t , ∂f/∂t >
= ∂x/∂s ∂x/∂t ∂x/∂s ∂y/∂s ∂z/∂s (q) ∂y/∂s ∂y/∂t (q) ∂x/∂t ∂y/∂t ∂z/∂t ∂z/∂s ∂z/∂t
= Df(q)t · Df(q) and II(Df(i), Df(i)) II(Df(i), Df(j)) h(q) = (q) II(Df(j), Df(i)) II(Df(j), Df(j))
= < Df(i), L(Df(i)) > < Df(i), L(Df(j)) > (q) < Df(j), L(Df(i)) > < Df(j), L(Df(j)) >
= < Df(i), D(n ◦ f)(i) > < Df(i), D(n ◦ f)(j)) > (q) < Df(j), D(n ◦ f)(i) > < Df(j), D(n ◦ f)(j)) >
= < ∂f/∂s, ∂n/∂s > < ∂f/∂s, ∂n/∂t > (q) < ∂f/∂t , ∂n/∂s > < ∂f/∂t , ∂n/∂t >
= < ∂2f/∂2s, n > < ∂2f/∂s∂t , n > (q) < ∂2f/∂t∂s, n > < ∂2f/∂2t , n > and III(Df(i), Df(i)) III(Df(i), Df(j)) e(q) = (q) III(Df(j), Df(i)) III(Df(j), Df(j))
= < L(Df(i)), L(Df(i)) > < L(Df(i)), L(Df(j)) > (q) < L(Df(j)), L(Df(i)) > < L(Df(j)), L(Df(j)) >
= < D(n ◦ f)(i), D(n ◦ f)(i) > < D(n ◦ f)(i), D(n ◦ f)(j)) > (q) < D(n ◦ f)(j), D(n ◦ f)(i) > < D(n ◦ f)(j), D(n ◦ f)(j)) >
= < ∂n/∂s, ∂n/∂s > < ∂n/∂s, ∂n/∂t > (q) < ∂n/∂t , ∂n/∂s > < ∂n/∂t , ∂n/∂t >
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4.2.3 the matrix of the Weingarten mapping
Let L denote the 2 × 2 matrix representing the Weingarten mapping L : Tp → Tp. with respect to the basis {D f1,D f2}. Recall II < v,w > = I < Lv,w >, for all v,w ∈ Tp In terms of representative matrices and vectors this becomes t t t t w hv = w g(Lv) ⇔ w hv = w (gL)v, for all v,w ∈ Tp Thus h = gL and we have finally an expression for the matrix of L L = g−1h
4.2.4 Examples, computing g,h,e,L eg. computations for the sphere Let r > 0 and write 2 2 2 2 2 S = S (r) = x x + y + z = r Recall the spherical polar parameterization of the spherical surface S. r sinφcosθ φ x f : 3 (0,π) × (−π,π) 7→ z = r sinφsinθ ∈ S ⊂ R3 (4.6) θ y r cosφ The differential matrix Df is ∂x/∂φ ∂x/∂θ r cosφcosθ −r sinφsinθ Df(φ,θ) = ∂y/∂φ ∂y/∂θ = r cosφsinθ r sinφcosθ (4.7) ∂z/∂φ ∂z/∂θ −r sinφ 0 We can now compute the matrix g of the first fundamental I form. (Note that the normal n at x ∈ S to the sphere S is x/||x||, this is used below.) 2 < Df1,Df1 > < Df1,Df2 > r 0 g(φ,θ) = = 2 2 < Df2,Df1 > < Df2,Df2 > 0 r sin φ and the matrix h of the second fundamental form II h(θ,φ)
= ∂2f ∂2f < ,n > < ,n > ∂2φ ∂φ∂θ − ∂2f ∂2f < ,n > < ,n > ∂θ∂φ ∂2θ
= −r sinφcosθ sinφcosθ −r cosφsinθ sinφcosθ < −r sinφsinθ , sinφsinθ > < r cosφcosθ , sinφsinθ > −r cosφ cosφ 0 cosφ − −r cosφsinθ sinφcosθ −r sinφcosθ sinφcosθ < r cosφcosθ , sinφsinθ > < −r sinφsinθ , sinφsinθ > 0 cosφ 0 cosφ
= −r 0 − 0 −r sin2 φ r 0 0 r sin2 φ
c jbquig-UCD September 23, 2009 4.2. MATRIX REPRESENTATION OF THE THREE FUNDAMENTAL FORMS 89
Since n = f/r it is easy to check that e = g/r2 i.e. the matrix of the third fundamnetal form is
1 r2 0 1 0 e(φ,θ) = = r2 0 r2 sin2 φ 0 sin2 φ The matrix of the Weingarten mapping is
1/r2 0 r 0 1 1 0 L = g−1h = = 0 1/r2 sin2 φ 0 r sin2 φ r 0 1 eg. computations for the saddle Write D for the saddle surface D = {x |z = xy } which can be parameterized
x r r f : s 3 R2 7→ y = s ∈ D ⊂ R3 (4.8) , z rs
The differential matrix Df is ∂x/∂r ∂x/∂s 1 0 Df(r,s) = ∂y/∂r ∂y/∂s = 0 1 (4.9) ∂z/∂r ∂z/∂s s r
We can now compute the matrix g of the first fundamental form I. Computing n = Df1 × Df2/||Df1 × Df2|| The Gauss mapping is
x −s r 1 n : R2 3 7→ y = √ −r ∈ D ⊂ R3 s 2 2 z 1 + r + s 1
2 < Df1,Df1 > < Df1,Df2 > 1 + s rs g(r,s) = = 2 < Df2,Df1 > < Df2,Df2 > rs 1 + r and the matrix h of the second fundamental form II is
h(r,s)
= ∂2f ∂2f < ,n > < ,n > − ∂2r ∂r∂s ∂2f ∂2f < ,n > < ,n > ∂s∂r ∂2s = 0 −s 0 −s 1 1 < 0 , √ −r > < 0 , √ −r > 2 2 2 2 0 1 + r + s 1 1 1 + r + s 1 − 0 −s 0 −s 1 1 < 0 , √ −r > < 0 , √ −r > 2 2 2 2 1 1 + r + s 1 0 1 + r + s 1
= −1 0 1 √ 1 + r2 + s2 1 0
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The matrix of the Weingarten mapping is
L = g−1h
= − 1 + s2 rs 1 −1 0 1 2 √ rs 1 + r 1 + r2 + s2 1 0
= 1 1 + r2 −rs −1 0 1 2 √ 1 + r2 + s2 −rs 1 + s 1 + r2 + s2 1 0
= 1 −rs 1 + r2 (1 + r2 + s2)3/2 1 + s2 −rs
To compute the matrix e of the third fundamental form III we will need
−s 0 rs ∂n 1 = −r(1 + r2 + s2)−3/2 −r + (1 + r2 + s2)−1/2 −1 = −1 − s2 r 2 2 3/2 ∂ 1 0 (1 + r + s ) −r and , by a similar computation
−1 − r2 ∂n 1 = rs s 2 2 3/2 ∂ (1 + r + s ) −s
e(r,s) < ∂n/∂r,∂n/∂r > < ∂n/∂r,∂n/∂s > = (q) < ∂n/∂s,∂n/∂r > < ∂n/∂s,∂n/∂s > 1 (1 + r2)(1 + r2 + s2) −rs(1 + r2 + s2) = (1 + r2 + s2)3 −rs(1 + r2 + s2)(1 + s2)(1 + r2 + s2) 1 1 + r2 −rs = (1 + r2 + s2)2 −rs 1 + s2 eg. computations for the right helicoid surface Write H for the (right) helicoid surface which rises to height b/2π in one revolution. b y H = x z = θ = arctan 2π x
H is a ruled surface with generating lines running off the helicoid curve Γ ⊂ R3.
acost 3 c : [0,2π] 3 t 7→ asint ∈ Γ ⊂ R bt
The surface H will be the same for all a so from now on we take a = 1. Now we give a parameterization of H scost 3 f : [0,∞) × [0,2π) 3 s,t 7→ ssint ∈ H ⊂ R (4.10) bt
c jbquig-UCD September 23, 2009 4.2. MATRIX REPRESENTATION OF THE THREE FUNDAMENTAL FORMS 91
The differential matrix Df is ∂x/∂s ∂x/∂t cost −ssint Df(s,t) = ∂y/∂s ∂y/∂t = sint scost (4.11) ∂z/∂s ∂z/∂t 0 b
We next compute a normal vector on H
~n
= Df1 × Df2 i j k = det ∂x/∂s ∂y/∂s ∂z/∂s ∂x/∂t ∂y/∂t ∂z/∂t i j k = det cost sint 0 −ssint scost b bsint = −bcost s
The Gauss mapping is bsint 2 s ~n(s,t) 1 3 n : R 3 7→ = √ −bcost ∈ H ⊂ R (4.12) t ||~n(s,t)|| 2 2 b + s s
We next compute the matrix g of the first fundamental form I. < Df1,Df1 > < Df1,Df2 > 1 0 g(s,t) = = 2 2 < Df2,Df1 > < Df2,Df2 > 0 s + b
In passing we remark that the parameterization mapping f : R2 → H ⊂ R3 is not conformal (not angle pre- serving).
The matrix h of the second fundamental form II is
h(r,s)
= ∂2f ∂2f < ,n > < ,n > − ∂2s ∂s∂t ∂2f ∂2f < ,n > < ,n > ∂t∂s ∂2t = 0 bsint −sint bsint 1 1 < 0 , √ −bcost > < cost , √ −bcost > 2 2 2 2 0 b + s s 0 b + s s − −sint bsint −scost bsint 1 1 < cost , √ −bcost > < −ssint , √ −bcost > 2 2 2 2 0 b + s s 0 b + s s
= 1 0 b √ b2 + s2 b 0
September 23, 2009 c jbquig-UCD 92 CHAPTER 4. BILINEAR FORMS
= b 0 1 √ b2 + s2 1 0
The matrix of the Weingarten mapping is
L = g−1h
= − 1 0 1 b 0 1 √ 0 s2 + b2 b2 + s2 1 0
= 1 s2 + b2 0 b 0 1 √ s2 + b2 0 1 b2 + s2 1 0
= b 0 b2 + s2 (b2 + s2)3/2 1 0
To compute the matrix e of the third fundamental form III we will need
0 bsint −ssint ∂n 1 s b = √ 0 − −bcost = +scost s 2 2 2 2 3/2 2 2 3/2 ∂ b + s 1 (b + s ) s (b + s ) b and , by a similar computation −bcost ∂n 1 = √ −bsint t 2 2 ∂ b + s 0
e(s,t) < ∂n/∂s,∂n/∂s > < ∂n/∂s,∂n/∂t > = (q) < ∂n/∂t,∂n/∂s > < ∂n/∂t,∂n/∂t > 1 1 0 = (b2 + s2)2 0 (b2)(b2 + s2)
Note that Gaussian curvature on H is
−b2 K(s,t) = detL(s,t) = (b2 + s2)2 and is negative everywhere, each point on H is hyperbolic, see below 4.4. The eigenvalues α,β of L are the roots of the characteristic polynomial p(λ) = λ2 − b2/(b2 + s2)2,
±b α,β = . b2 + s2
Neither Df1 nor Df2 is a principal direction on H. In particular the direction Df1 of a generating line is not a principal direction. The reader should compute the principal directions and include them with Df1 and Df2 in a drawing of H.
c jbquig-UCD September 23, 2009 4.3. EXPLICIT PRESENTATION AND CURVATURE 93
4.3 explicit presentation and curvature
It is instructive to compute matrices g,h,e,L representing the three fundamental forms and the Weingarten mapping starting from an explicit presentation of a surface Σ
x g : R2 ⊃ U 3 7→ g(x,y) ∈ Σ ⊂ R3 y
We use a standard method and create the parameterization of Σ
x(r,s) r r f : R2 ⊃ U 3 7→ f (r,s) = y(r,s) = s ∈ Σ ⊂ R3 s z(r,s) g(r,s)
We next compute the differential matrix
∂x/∂r ∂x/∂s 1 0 D f = ∂y/∂r ∂y/∂s = 0 1 ∂z/∂r ∂z/∂s ∂g/∂r ∂g/∂s
Using this we can compute a (non unit as yet) normal vector i j k −gx ~n = D f1 × D f2 = ∂ f /∂r × ∂ f /∂s = 1 0 gx = −gy 0 1 gy 1
We divide by the norm to obtain the unit normal or gauss map to the surface
−g ~n 1 x n = = −gy ||~n|| q 2 2 gx + gy + 1 1
Next we compute the matrix g which represents the first fundamental form I. Notation:- No confusion, g is the function in the explicit representation: g is the matrix representing I.
g
= < D f1,D f1 > < D f1,D f2 > < D f2,D f1 > < D f2,D f2 >
= < ∂ f /∂r,∂ f /∂r > < ∂ f /∂r,∂ f /∂s > < ∂ f /∂s,∂ f /∂r > < ∂ f /∂s,∂ f /∂s >
= 1 1 1 0 < 0 , 0 > < 0 , 1 > gr gr gr gs 0 1 0 0 < 1 , 0 > < 1 , 1 > gs gr gs gs
= 2 1 + gr grgs 2 gsgr 1 + gs
September 23, 2009 c jbquig-UCD 94 CHAPTER 4. BILINEAR FORMS
Next we compute the matrix h representing the second fundamental form II. h
= < D f ,n ◦ f > < D f ,n ◦ f > − 1,1 1,2 < D f2,1,n ◦ f > < D f2,2,n ◦ f >
= < ∂ f 2/∂r2,n ◦ f > < ∂ f 2/∂r∂s,n ◦ f > − < ∂ f 2/∂s∂r,n ◦ f > < ∂ f 2/∂s2,n ◦ f >
= −1 · p 2 2 1 + gr + gs 0 −gr 0 −gr < 0 , −gs > < 0 , −gs > gr,r 1 gr,s 1 0 −gr 0 −gr < 0 , −gs > < 0 , −gs > gs,r 1 gs,s 1
= −1 gr,r gr,s p 2 2 g g 1 + gr + gs s,r s,s Next we find a matrix L representing the Weingarten mapping L. L
= g−1h
= − 1 + g2 g g 1 −1 g g r r s · r,r r,s g g 1 + g2 p 2 2 g g s r s 1 + gr + gs s,r s,s = 2 1 1 + gs −grgs 2 2 2 · (1 + gr )(1 + gs ) − (grgs)(gsgr) −gsgr 1 + gr −1 gr,r gr,s p 2 2 g g 1 + gr + gs s,r s,s = 2 1 1 + gs −grgs 2 2 2 · 1 + gr + gs −gsgr 1 + gr −1 gr,r gr,s p 2 2 g g 1 + gr + gs s,r s,s = −1 1 + g2 −g g g g s r s · r,r r,s 2 2 3/2 2 (1 + gr + gs ) −gsgr 1 + gr gs,r gs,s
c jbquig-UCD September 23, 2009 4.4. SURFACE AS A GRAPH OVER THE TANGENT PLANE AT A POINT 95
It would be onerous to compute all the surface invariants in this general context. They can readily be com- puted for any particular explicitely presented surface. However one invariant K(p), the Gauss curvature, is readily computed in general.
K(r,s)
= det L
= −1 2 1 + g2 −g g g g det s r s det r,r r,s 2 2 3/2 2 (1 + gr + gs ) −gsgr 1 + gr gs,r gs,s
= 1 2 2 gr,r gr,s 2 2 3 (1 + gr + gs )det (1 + gr + gs ) gs,r gs,s
= 2 gr,rgs,s − gr,s 2 2 2 (1 + gr + gs ) 4.4 surface as a graph over the tangent plane at a point
Let Σ be a surface. Let p be a point in Σ. Let Tp denote the tangent plane at the point p. After a translation 3 we may assume p = 0 and afer a rotation of R we may assume that T0 is the xy-plane. Now consider the explicit equation z = g(x,y) of Σ. We are viewing the surface as a graph over its tangent plane at a point.
We will compute matrices g,h,L representing fundamental forms I and II and the Weingarten mapping L at −gx 0 ∈ Σ. Since T0 is the xy-plane k = n ≡ −gy so that gx(0,0) = gy(0,0) = 0. Thus 1
1 + g2 g g 1 0 g(0,0) = x x y = g g 1 + g2 0 1 y x y (0,0) and −1 g g g g h(0,0) = x,x x,y = − x,x x,y q 2 2 gy,x gy,y gy,x gy,y 1 + gx + gy (0,0) (0,0) and g g L = g−1(0,0)h(0,0) = h(0,0) = − x,x x,y gy,x gy,y (0,0) At this point we notice that
g g h(0,0) = L(0,0) = − x,x x,y gy,x gy,y (0,0)
But this matrix will soon appear in a third guise, as the Hessian matrix D2g(0,0) of the function of two variables g(x,y). Consider the Taylor expansion to the second order near 0 of the function z = g(x,y) of two variables. Before we begin note 0 ∈ Σ implies g(0,0) = 0 and recall that gx(0,0) = gy(0,0) = 0. Thus, near 0 ∈ R2,
g(x,y)
September 23, 2009 c jbquig-UCD 96 CHAPTER 4. BILINEAR FORMS
Figure 4.1: (i) elliptic (ii) hyperbolic (iii) parabolic (iv) umbilic
x − 0 x − 0 x − 0 ≈ g(0,0) + Dg + D2g( , ) y − 0 y − 0 y − 0 by Taylor’s theorem to degree 2 Dg and D2g taken at 0 x 1 gxx gxy x = g(0,0) + (gx,gy) + (x,y) y 2 gyx gyy y all p.d.s taken at 0 x 1 g g x = 0 + (0,0) + (x,y) xx xy y 2 gyx gyy (0,0) y 1 g g x = (x,y) xx xy 2 gyx gyy (0,0) y
Next let v ⊥ w ∈ T0 be eigen vectors of the Weingarten mapping L : T0 → T0, say Lv = αv and Lw = βw and take these two eigenvectors as basis for T0 the xy-plane. Then at 0 the matrices g,h and L assume the forms 1 0 α 0 g = , h = L = − 0 1 0 β As a graph over the tangent plane, with eigenvectors of L as basis, the explicit equation of Σ is, to an approximation of the second order, αx2 + βy2 g(x,y) = . 2 We emphasize that this analysis has been carried out at an arbitrary point p on an arbitrary surface Σ.
We say that the point p ∈ Σ is • elliptic iff αβ > 0 iff K(p) > 0 • hyperbolic iff αβ < 0 iff K(p) < 0
c jbquig-UCD September 23, 2009 4.4. SURFACE AS A GRAPH OVER THE TANGENT PLANE AT A POINT 97
• parabolic iff one only of α,β is zero iff rank(L) = 1
• umbilic iff α = β iff L = αId The reason for these terms is to be found in the contour diagrams of the mapping g expressing Σ as a graph 2 over Tp. Recall that we view this map as g : R → R and that the domain plane is disected into preimage sets forming the so called contour diagram made up of of level curves of g.
dis joint R2 = [ g−1(d) d∈Im(g)⊂R
The implicit equations of a level curve is g(x,y) = αx2 +βy2 = d ∈ R and the contour diagram is made up of elliptic curves , hyperbolae, straight lines or circles according as the point p is elliptic, hyperbolic, parabolic or umbilic: see figure 4.1. Such a level curve is known as a Dupin indicatrix of the point p ∈ Σ. A very down to earth view is that the indicatrix curve is the intersection of the surface Σ with the tangent plane Tp displaced bya small amount in the normal direction. In three of the above cases Σ lies completely on one side of the tangent plane, only in the hyperbolic case (K(p) < 0) does the tangent plane cut through the surface.
September 23, 2009 c jbquig-UCD 98 CHAPTER 4. BILINEAR FORMS
4.5 problem set
Attempt a representative selection only of the following 13 questions. The problems all ask for the same computations, but concerning 13 different surfaces. All the following questions can be processed using a surface parametrization. In certain cases computations will be less onerous if use of an explicit surface presentation is possible.
ALL these questions are ANSWERED in http://mathsci.ucd.ie/courses/math40060/maple/answers/chap04 Some have a second solution using the explicit presentation in http://mathsci.ucd.ie/courses/math40060/maple/answers/chap04/explicit method p 1. The cone C ⊂ R3 has implicit equation x2 +y2 −z2 = 0, explicit equation z = ± x2 + y2 and parametriza- tion. scos(r) r f : R × [0,2π] 3 C ⊂ R3 7→ ∈ ssin(r) s s Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that the cone is flat. N.B This surface can be studied using either (or both) of the parametric and explicit equations.
p 2. The cylinder C ⊂ R3 has implicit equation x2 + y2 = 0, no explicit equation z = ± x2 + y2 and parametrization. acos(t) 3 f : R × [0,2π] 3 C ⊂ R 7→ r,s ∈ asin(t) s Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that the cylinder is flat.
3. The double sheeted hyperbola Σ ⊂ R3 has implicit equation x2 + y2 − z2 = −1, explicit equation z = p ± 1 + x2 + y2 and parametrization.
sinh(s)cos(r) r f : R × [0,2π] 3 Σ ⊂ R3 7→ ∈ sinh(s)sin(r) s cosh(s)
Compute g,h,e and L the mean and Gaussian curvatures. Prove that the d-s-h Σ has everywhere posi- tive curvature. N.B This surface can be studied using either (or both) of the parametric and explicit equations.
4. The monkey saddle surface Σ ⊂ R3 has implicit equation x3 − 3xy2 − z = 0, explicit equation z = x3 − 3xy2 and parametrization.
r r f : R × R 3 Σ ⊂ R3 7→ ∈ s s r3 − 3rs2 or scos(r) r f : [0,2π] × [0,∞) 3 Σ ⊂ R3 7→ ∈ ssin(r) s s3 cos(3r)
c jbquig-UCD September 23, 2009 4.5. PROBLEM SET 99
Compute g,h,e and L the mean and Gaussian curvatures. Prove that the monkey saddle has everywhere negative curvature. N.B This surface can be studied using either (or both) of the parametric and explicit equations.
5. The sphere S ⊂ R3 has several useful parametrization, one is the Mercator projection used in geograpgy and which is conformal i.e. angle preserving sech (s)cos(r) r f : R × [0,2π] 3 S ⊂ R3 7→ ∈ sech (s)sin(r) s tanh(s) Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions using this parametrization. Show that the sphere has constant Gauss curvature and only one principle curva- ture (with multiplicity 2). Prove that f is conformal, i.e. that g and <> yield the same measure of angle. This is so iff g is a scalar multiple of Id, prove that.
6. The paraboloid or parabolic bowl B ⊂ R3 has implicit equation x2 + y2 − z = 0, explicit equation z = x2 + y2 and parametrization. r r f : R × R 3 B ⊂ R3 7→ ∈ s s r2 + s Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that the Gauss curvature is positive everywhere. N.B This surface can be studied using either (or both) of the parametric and explicit equations.
7. The pseudosphere Σ ⊂ R3 (also called the trumpet) is a rotated tractoid curve and has parametrization. sech (s)cos(r) r f : R × [0,2π] 3 Σ ⊂ R3 7→ ∈ sech (s)sin(r) s s − tanh(s) Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that the pseudosphere has constant gauss curvature -1. The principle curvatures α and β obey β = −1/α.
8. The right helicoid surfave H ⊂ R3 explicit equation xtanz = y = and parametrization. scos(r) r f : R × [0,2π] 3 C ⊂ R3 7→ ∈ ssin(r) s r Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that H is NOT flat.
9. The saddle surface Σ ⊂ R3 has implicit equation x2 − y2 − z = 0, explicit equation z = x2 − y2 and parametrization. r r f : R × R 3 Σ ⊂ R3 7→ ∈ s s r2 − s2
September 23, 2009 c jbquig-UCD 100 CHAPTER 4. BILINEAR FORMS
or scos(r) r f : [0,2π] × [0,∞) 3 Σ ⊂ R3 7→ ∈ ssin(r) s s2 cos(2r) Compute g,h,e and L the mean and Gaussian curvatures. Prove that the saddle has everywhere nega- tive curvature. N.B This surface can be studied using either (or both) of the parametric and explicit equations.
10. The sphere S is the arcytypal surface and has constant positive Gaussian curvature. S ⊂ R3 has implicit p equation x2 + y2 + z2 = a2, explicit equation z = ± a2 − x2 − y2 and parametrization.
asin(s)cos(r) r f : R × [0,2π] 3 S ⊂ R3 7→ ∈ asin(s)sin(r) s acos(s)
Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that the the Gaussian curvature is constant at 1/a2. The principle curvatures are equal. Any direction is principal.
11. The single sheeted hyperbola Σ ⊂ R3 has implicit equation x2 + y2 − z2 = 1 and parametrization.
cosh(s)cos(r) r f : R × [0,2π] 3 Σ ⊂ R3 7→ ∈ cosh(s)sin(r) s sinh(s)
Compute g,h,e and L the mean and Gaussian curvatures. Prove that the s-s-h Σ has everywhere nega- tive curvature.
12. The tangent developable helicoid surface D ⊂ R3 has parametrization.
cos(r) − ssin(r) r f : R × [0,2π] 3 D ⊂ R3 7→ ∈ sin(r) + scos(r) s r + s
Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that D IS a flat surface (the right helicoid is not flat).
p 13. The Torus T = T 2(a,b) ⊂ R3 (0¡a¡b) is the surface with implicit equation ( x2 + y2 − b)2 + z2 = a2 and parametrization.
(b − asin(s))cos(r) r f : [0,2π] × [0,2π] 3 T ⊂ R3 7→ ∈ (b − asin(s)sin(r) s acos(s)
Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that the Gaussian curvature is positive and negative on two regions of T separated by two circles on both of which the Gaussian curvature is zero.
c jbquig-UCD September 23, 2009 Part I
Answers to problem sets of part I
101
Chapter 5
Answers to questions in Chapter1
2 2 If n = ai + bj ∈ R is a unit vector (||n|| = 1) then L = x ∈ R < x,n >= 0 is a line through the origin and n is the normal vector to this line. 5.1 question and answer 5.1.3 answer
(i) If A ∈ SO(3,R) then det(A) = 1. Let C be a reflection: we must prove that CtC = Id ⇔ C ∈ O(3,R) and that det(C) > 0 is false. (ii) If A ∈ O(3,R) then det(A) = ±1. But for some unit vector (iii) Each reflection is orthogonal but not special. t (iv) Every orthogonal matrix is either a rotation or is C = Id − 2nn the product of a rotation and a reflection. Then (v) Give a 3× orthogonal matrix A which is not special and not a reflection. Prove your assertion. Express Ct A as the product of a rotation and a reflection. = ( Id − 2nnt )t = Idt − 2(nt )t (n)t , since (CD)t = DtCt = Idt − 2nnt , since (Ct )t = V 5.1.1 answer = Recall We have so far proven that A ∈ SO(3,R) ⇔ Aat = Id and det(A) > 0 t Thus C = C, if C is a reflection
1 Next = det(Id) CtC = det(AAt ) = CC = det(A)det(At ), since det(CD) = det(C)det(D) = (Id − 2nnt )(Id − 2nnt ) = (det(A))2since det(At ) = det(A) = Id − 4nnt + 4(nnt )(nnt ) Thus det(A) = ±1. But det(A) > 0. Conclusion = Id − 4nnt + 4n(nt n)nt ) t t det(A) = 1 = Id − 4nn + 4n(< n,n >)n ) = Id − 4nnt + 4n(1)nt ) 5.1.2 answer = Id − 4nnt + 4nnt ) Recall = Id A ∈ O(3,R) ⇔ Aat = Id We have proven that Ct = C−1 so that C ∈ O(3,R). The preeceding argument shows passing remark det(A) = ±1 We have proven more, that
[A C = Ct = C−1
103 104 CHAPTER 5. ANSWERS TO QUESTIONS IN CHAPTER1
Next we must prove that C ∈/ SO(3,R), i,e, that det > 0 is (v) Compute the matrix of rotation Sn,2π/3. Compute it false. But (see lecture notes) by four different methods (when we finish Chapter 1). C = ECE−1
Thus For the first 3 parts we will use the formulae det(C) t = det(E)det(C)det(E−1) An = nn t = det(E)det(C)(det(E))−1 Bn = Id − nn t = det(C) Cn = Id − 2nn 1 0 0 Dn = [n×] = det 0 1 0 Sn,θ) = An + cos(θ)Bn + sin(θ)Dn 0 0 −1 = −1 5.2.1 answer Alternatively using eigenvalues α,β,γ of C With n − i it is trivial to calculate (C) = = ( )( )(− ) = − det αβγ 1 1 1 1 1 0 0 Ai = 0 0 0 5.1.4 answer 0 0 0 A ∈ ( ,R) (A) = Let O 3 . On one hand assume det 1 then 0 0 0 A ∈ SO(3,R) and we are done. On the other hand assume Bi = 0 1 0 A = − C C2 = Id det 1. Let be a (any) reflection. Then . 0 0 1 Thus A = A · Id = A(CC) = (AC)C. Thus A = BC where C is a reflection and B = AC ∈ SO(3,R) is a rotation. −1 0 0 Ci = 0 1 0 5.1.5 answer 0 0 1 0 0 0 Take A = −Id. It is easy to check that A ∈ O(3,R) that D = 0 0 −1 det(A) = −1 so A is not special. Also A is not a reflection i 0 1 0 since Av = −v for all v ∈ R3 and so +1 is not an eigenvalue. 1 0 0 Ei = 0 cos(θ) −sin(θ) 0 sin(θ) cos(θ) A −1 0 0 5.2.2 answer = 0 −1 0 With n − j it is trivial to calculate 0 0 −1 −1 0 0 1 0 0 0 0 0 = 0 −1 0 0 1 0 Aj = 0 1 0 0 0 1 0 0 −1 0 0 0
= Sk,πCk 1 0 0 Here A (orthogonal but neither special nor a reflection) is Bj = 0 0 0 expressed as a product of a rotation and a reflection. 0 0 1 1 0 0 5.2 question and answer Cj = 0 −1 0 0 0 1 Let n be the unit vector in the direction i + j + k. Let 0 0 1 θ ∈ R. Dj = 0 0 0 (i) Compute Ai,Bi,Ci,Di,Si,θ. −1 0 0 (ii) Compute Aj,Bj,Cj,Dj,Sj,θ. cos(θ) 0 sin(θ) (iii) Compute Ak,Bk,Ck,Dk,Sk,θ. Ej = 0 1 0 −sin(θ) 0 cos(θ) (iv) That was to warm up. Now compute An,Bn,Cn,Dn.
c jbquig-UCD September 23, 2009 5.2. QUESTION AND ANSWER 105
5.2.3 answer Writing a With n − k it is trivial to calculate n = b 0 0 0 c √ Ak = 0 0 0 where of course a = b = c = 1/ 3 0 0 1 Dn 1 0 0 0 −c b B = 0 1 0 k = c 0 −a 0 0 0 −b a 0 1 0 0 0 −1 1 Ck = 0 1 0 1 = √ 1 0 −1 0 0 −1 3 −1 1 0 0 −1 0 Dk = 1 0 0 5.2.5 answer 0 0 0 We wish to compute S by four different methods. In cos(θ) −sin(θ) 0 n,θ this example θ = 2π/3 so that cos(θ) = −1/2 and E = sin(θ) cos(θ) 0 √ √ k sin(θ) = 3/2. In this example n = (i + j + k)/ 3. 0 0 1 First method
5.2.4 answer Sn,θ) = An + cos(θ)Bn + sin(θ)Dn The case where A,B and D have been computed above 1 √ ~n = i + j + k = 1 S(i+j+k)/ 3,2π/3 1 is more demanding, we will give full details. First we need = the unit vector √ A(i+j+k)/ 3 1 ~n 1 + n = = √ 1 ||~n|| cos(2π/3)B √ 3 1 (i+j+k)/ 3 + ( / )D √ An sin 2π 3 (i+j+k)/ 3 = nnt 1 = 1 1 = √ 1 √ (1,1,1) 1 1 1 3 3 1 1 1 1 1 3 1 1 1 1 1 = 1 (1,1,1) + 3 2 −1 −1 1 −1 1 −1 2 −1 1 1 1 2 3 1 −1 −1 2 = 1 1 1 3 1 1 1 + √ 0 −1 1 3 1 √ 1 0 −1 Bn 2 3 −1 1 0 = Id − An 2 −1 −1 1 = = −1 2 −1 1 1 1 3 1 −1 −1 2 1 1 1 3 1 1 1 Cn − 2 −1 −1 = Id − 2An 1 −1 2 −1 1 −2 −2 6 1 −1 −1 2 = −2 1 −2 3 −2 −2 1 +
September 23, 2009 c jbquig-UCD 106 CHAPTER 5. ANSWERS TO QUESTIONS IN CHAPTER1
0 −1 1 1 method 4 We use the method of Euler angles 1 0 −1 2 −1 1 0 S(i+ j+k,2π/3) 0 0 1 = Sk,α ◦ Sj,β ◦ Sk,γ = 1 0 0 Determination of the three Euler angles is postponed, see 0 1 0 below, they tutn out to be method 2 The second method uses the formula π/2 t α = β and γ = 0 Sn,θ = (l,mn)Sk,θ(l,m,n) = Here E = (l,m,n) is a positively oriented orthonormal we put them to use. triple (with n) in position 3) froming a special orthogonal
(basis changing) matrix E. In our example first take S(i+ j+k,2π/3) ~n = i + j + k Second take l =~i − j which is ⊥ to~l. Third put ~m =~n ×~l = i + j − 2k. We have = 1 1 1 cos(α) −sin(α) 0 ~ (l,~m,~n) = ( 1 , −1 , 1 sin(α) cos(α) 0 · 1 0 −2 0 0 1 a positively oriented orthogonal triple. Dividing each cos(α) 0 sin(α) vector by its norm we have an p.o.o.n triple with n as third 0 1 1 · entry. −sin(α) 0 cos(α) √ √ √ cos(γ) −sin(γ) 0 1/√2 1/√6 1/√3 sin(γ) cos(γ) 0 E = (l,m,n) = −1/ 2 1/√6 1/√3 0 −2/ 6 1/ 3 0 0 1 Also = √ √−1/2 − 3/2 0 cos(π/2) −sin(π/2) 0 S = k,2π/3 3/2 −1/2 0 sin(π/2) cos(π/2) 0 · 0 0 1 0 0 1 We leave the reader to multiply matrices and obtain cos(π/2) 0 sin(π/2) √ t 0 1 0 · S(i+j+k)/ 3,2π/3 = ESk,2π/2E −sin(π/2) 0 cos(π/2) and check that the result is cos(0) −sin(0) 0 0 0 1 sin(0) cos(0) 0 √ S(i+j+k)/ 3,2π/3 = 1 0 0 0 0 1 0 1 0 = third method 0 −1 0 Sn,θ 1 0 0 · = exp(θ[n×]) 0 0 1 0 0 1 = exp(θ)Dn) 0 1 0 · = exp(EθD Et ) k −1 0 1 t = E exp(θDk)E 1 0 0 0 −θ 0 0 1 0 t = E exp θ 0 0 E 0 0 1 0 0 0
exp(θDk) is well known = cos(θ) −sin(θ) 0 0 −1 0 0 0 1 t · = E sin(θ) cos(θ) 0 E ) 1 0 0 0 1 0 0 0 1 0 0 1 −1 0 0 0 0 1 = = 1 0 0 0 −1 0 0 1 0 0 0 1 matrix slog multiplication omited −1 0 0
c jbquig-UCD September 23, 2009 5.3. QUESTION AND ANSWER 107
3 Errors to be fixed and details to be supplied soon, jbq. (ii) Find n ∈ R , ||n|| = 1 so that Bn is isometric typing fatigue!!! projection (this is the unique orthogonal projection in which all 3 principle directions (EW,NS,UD) are scaled equally). Compute the matrix B = Bn. 5.3 question and answer (iii) (See expanded version of (iii) in next question, added sept 2009) Prove that, for a 3 × 3 reflection matrix The eight vertices of the standard cube At = A = A−1 C = [−1,1] × [−1,1] × [−1,1] of volume 8 are ±1 ±i ± j ± k = ±1 . Compute B(±i ± j ± k). ±1 Express all 8 as linear combinations of l,m where 5.3.1 answer {l,m,n} is a PONB. Actually draw the isometric projection of the standard cube in the 2-D image Let A be a reflection: we must prove that plane Pn. A = At = A−1
For some unit vector
A = Id − 2nnt 5.4.1 answer
Then Bi t A = Id − iit = ( Id − 2nnt )t 1 0 0 1 = Idt − 2(nt )t (n)t , since (CD)t = DtCt = 0 1 0 − 0 (1,0,0) 0 0 1 0 = Idt − 2nnt , since (Ct )t = V = A 1 0 0 1 0 0 = 0 1 0 − 0 0 0 We have so far proven that 0 0 1 0 0 0 0 0 0 t A = A = 0 1 0 0 0 1 Next Simililarly At A 1 0 0 = AA B = 0 0 0 t t j = (Id − 2nn )(Id − 2nn ) 0 0 1 = Id − 4nnt + 4(nnt )(nnt ) and t t t = Id − 4nn + 4n(n n)n ) 1 0 0 t t = Id − 4nn + 4n(< n,n >)n ) Bk = 0 1 0 = Id − 4nnt + 4n(1)nt ) 0 0 0 = Id − 4nnt + 4nnt ) = Id 5.4.2 answer
t −1 We have proven that A = A . The trick here is to find suitable n so that Bn is isometric Putting both above results together we have proven projection. Such n must be symmetric in all 3 principle directions (this is the crucial point) and so must be of the A = At = A−1 form ai + bj + ck with a = b = c. There are only two possible suitable vectors√ of length 1: n = ±(i + j + k)/ 3. Both will lead to the same matrix 5.4 question and answer Bn, we take the ’+’ option.
(i) Find the matrices of orthogonal projection 1 1 Bk,Bi,Bj; in mechanical drawing these are called n = √ 1 plan, frontal and side elevation. 3 1
September 23, 2009 c jbquig-UCD 108 CHAPTER 5. ANSWERS TO QUESTIONS IN CHAPTER1
Then 5.5 question and answer
|| || = , ∈ 3 P = { | < , >= } ⊂ 3 Bn Let n 1 n R . Let x x n 0 R . Let B : R3 → R3 denote orthogonal projection into the = Id − nnt plane P. Choose h,v ∈ P (many choices are possible) so 1 0 0 1 1 1 that {h,v,n} is a P.O.N.Basis for R3. Then {h,v} is an = 0 1 0 − √ 1 √ (1,1,1) 3 3 O.N. Basis for the plane P. Any vector in P can be written 0 0 1 1 as a linear combination Xh +Yv. Denote by B the 2 × 3 1 0 0 1 1 1 matrix which represents B : R3− > P w.r.t. bases {i,j,k} 1 = 0 1 0 − 1 1 1 and {h,v} for R3 and R2 respectively. B is the most useful 3 0 0 1 1 1 1 matrix representing O.P into the plane P. t 1 0 0 1 1 1 h 1 (i) Prove that B = t . = 0 1 0 − 1 1 1 v 3 0 0 1 1 1 1 (ii) Argue that h = v × n with v = B(k)/||k|| is the 2 −1 −1 most apt choice of the basis of the picture plane P. 1 = −1 2 −1 (iii) Compute h,v and the 2 × 3 matrix B for isometric 3 −1 −1 2 projection. (iv) Now easily do the last part of the preceeding 5.4.3 answer question.
The eight vertices of the standard cube are the columns of the following 3 × 8 matrix. 5.5.1 answer −1 1 1 −1 −1 1 1 −1 Ax = nnt x = n < n,x > = αn −1 −1 1 1 −1 −1 1 1 −1 −1 −1 −1 1 1 1 1 a scalar multiple of n where the scalar α =< n,x >.
The eight images under isometric projection of the eight 5.5.2 answer vertices of the standard cube are the columns of the product matrix We use, from the preeceding part, Ax = αn where α =< n,x >. Then Bx=x − Ax = x − αx −1 1 1 −1 −1 1 1 −1 < Ax,Bx > B −1 −1 1 1 −1 −1 1 1 = < αn,x − αn > −1 −1 −1 −1 1 1 1 1 = < αn,x > − < αn,αn > = = α < n,x > −α2 < n,n > 2 −1 −1 2 1 = αα − α (1) −1 2 −1 · 3 = 0 −1 −1 2 −1 1 1 −1 −1 1 1 −1 Next −1 −1 1 1 −1 −1 1 1 < Ax,Dx > − − − − 1 1 1 1 1 1 1 1 = < αn,n × x > = = α < n,n × x > 2 −1 −1 = α(0) 1 −1 2 −1 · 3 = 0 −1 −1 2 Next 0 4 2 −2 2 2 0 −4 0 −2 2 4 −2 −4 0 2 < Bx,Dx > 0 −2 −4 −2 4 2 0 2 = < x − αn,n × x > = < x,n × x > − α < n,n × x > It is difficult to visualize the isometric image of the cube = 0 − α(0) from this data. Given a p.o.o.n.√ basis l,m,n whose last vector is n = (i + j + k)/ 3 all projected vectors lie in the = 0 plane < x,n >= 0 and so can be expressed as a linear We have proved as required that combination of l and n. to be finished!! typing fatigue!! < Bx,Dx > = < Dx,Ax > = < Ax,Bx >
c jbquig-UCD September 23, 2009 5.6. QUESTION AND ANSWER 109
5.5.3 answer 5.6.1 answer
It suffices to prove that < Bx,Bx > = < Dx,Dx > but 0 0 0 I = 0 0 −1 < Bx,Bx > 0 1 0 = < x − αn,x − αn > 0 0 1 = < x,x > − 2α < n,x > +α2 < n,n > J = 0 0 0 = < x,x > −2αα + α2(1) −1 0 0 = < x,x > − α2 0 −1 0 K = 1 0 0 = < x,x > − < n,x >2 0 0 0 = < n × x,n × x > by theorem(??) 5.6.2 answer = < Dx,Dx > To prove that (R3,×) is a Lie Algebra one must check closure Given v,w ∈ R3 then v × w ∈ R3 by the very 5.5.4 answer definition of the cross product. bilinearity The cross product can be expressed as a determinant and determinants are tri-linear in their det(Bx,Dx,n) rows. = det(x − αn,n × x,n) anti-commutativity w × v = −v × w since a determinant = det(x,n × x,n) − αdet(n,n × x,n) is negated if two rows are swapped. by an elementary col op (E.C.O.) cyclic associativity This is the only one of these four checks which is not completely trivial. We have = det(x,n × x,n) − α · 0 recourse to the bac-cab rule. cols equal makes determinant zero a × (b × c) = b < a,c > − c < a,b > = det(x,n × x,n) = det(n,x,n × x) which we apply three times by an E.C.O. u × (v × w) = v < u,w > − w < u,v > = < n × x,n × x > w × (u × v) = u < w,v > − v < w,u > > 0 unless n × x = 0 v × (w × u) = w < v,u > − u < v,w > Then adding 3 L.H.Sides and equating to the sum of 3 R.H.Sides we get 5.6 question and answer u × (v × w) + w × (u × v) + v × (w × u) = 0 Let n ∈ R3 with ||n|| = 1. Write A = An, B = Bn, D = Dn. Prove the following facts which 5.6.3 answer were used during lectures to justify the formula To prove that (so(3,R),[,]) is a Lie Algebra one must check Sn,θ = A + cos(θ)B + sin(θ)D closure Given A and B ∈ so(3,R) we must check that Given any x ∈ R3 [A,B] = AB − BA ∈ so(3,R). In other word, given that At = −A and Bt = −B we must prove that (i) Ax is axial, i.e. a scalar multiple of n. [A,B]t = −[A,B]. But (ii) < Bx,Dx > = < Ax,Bx > = < Ax,Dx > = 0, i.e. [A,B]t = (AB − BA)t the three vectors Ax,Bx and Dx are mutually = (AB)t − (BA)t orthogonal. = Bt At − At Bt (iii) ||Bx|| = ||Dx||, i.e. the two vectors Bx and Dx = BA − AB have the same length. = −(AB − BA) (iv) The triple Bx,Dx,n is positively oriented, = −[A,B] det(Bx,Dx,n) > 0. (NB this part has been corrected: in a previous erroneous version, Ax was bilinearity This is trivial. in the place of n.) anti-commutativity [B,A] = (BA − AB) = −(AB − BA) = −[a,B]
September 23, 2009 c jbquig-UCD 110 CHAPTER 5. ANSWERS TO QUESTIONS IN CHAPTER1
cyclic associativity and [A,[B,C]] + [C,[A,B]] + [B,[C,A]] [F (u),F (v)]
= = [A,(BC −CB)] + [C,(AB − BA)] + [B,(CA − AC)] F (u)F (v)] − F (v)F (u)
= = A(BC −CB) − (BC −CB)A 0 −c b 0 − f e c 0 −a f 0 −d + −b a 0 −e d 0 C(AB − BA) − (AB − BA)C − + 0 − f e 0 −c b B(CA − AC) − (CA − AC)B f 0 −d c 0 −a −e d 0 −b a 0
= = ABC − ACB − BCA +CBA −c f − be bd cd + ae −c f − ad ce a f b f −be − ad CAB −CBA − ABC + BAC − + −c f − be ae a f BCA − BAC − CAB + ACB bd −c f − ad b f cd ce −be − ad = = 0 0 −ae + bd cd − a f ae − bd 0 ce − b f 5.6.4 answer a f − cd b f − ce 0 The mapping in question is By observation F : R3 → so(3,R) F (u × v) = [F (u),F (v)] a 0 −c b b 7→ c 0 −a 5.6.6 answer c −b a 0 We now have a bijection of Lie algebras It is trivial note that F is a bijective mapping between two 3-dimensional vector spaces and is linear, i.e. F : (R3,×) → (so(3,R),[,]) F (αu + βv) = αF (u) + βF (v) which preserves linearity by (iv) and preserves Lie products by (v). That is what is meant by an isomorphism 3 for all u,v ∈ R and α,β ∈ R. of Lie Algebras. In common language, both Lie algebras are the same. 5.6.5 answer Let 5.7 question and answer a d u = b and v = e (i) Find skew symmetric matrices I,J,K which c f represent unit angular velocity about the i,j,k axes. Then (ii) Prove that (R3,×) is a Lie Algebra. F (u × v) (iii) Prove that (so(3,R),[ , ]) is a Lie Algebra. You b f − ce must check = F ( cd − a f ) (i) closure of so(3,R) under the operation [ , ] ae − bd (ii) bilinearity of the operation [ , ] 0 −ae + bd cd − a f (iii) anticommutativity of the operation [ , ] = ae − bd 0 −b f + ce −cd + a f b f − ce 0 (iv) cyclic (Lie) associativity of the operation [ , ]
c jbquig-UCD September 23, 2009 5.8. QUESTION AND ANSWER 111
(iv) Prove that the mapping 5.8 question and answer F : R3 3 ai + bj + ck 7→ aI + bJ + cK ∈ so(3,R) Let A,B : R3 → R3 be reflections in the plane is a vector space isomorphism. < x,n >= 0 and < x,m >= 0 respectively. Prove that BA (v) Prove that F (u × v) = [F (u),F (v)] for all is a rotation. Find the angle and axis of rotation. (Hint, try u,v ∈ R3. reflections in lines in R2 to start with.) (vi) Conclude that (R3,×) and (so(3,R),[ , ]) are isomorphic as Lie algebras. 5.8.1 ?? 5.7.1 answer A ∈ O(3,R) and B ∈ O(3,R) ⇒ BA ∈ O(3,R). AA detA = −1 = detB ⇒ det(BA) = (−1)(−1) = 1. = (nnt )(nnt ) Thus BA ∈ SO(3,R) and by Eulers theorem is a rotation. = n(nt n)nt t The vector m × n is perpendicular to n and so = n(< n,n >)n = n(1)nt A(m × n) = m × n = nnt Similarly = A B(m × n) = m × n Thus and then BA(m × n) = A(m × n) = m × n Thus m × n is an axial vector (not necessarily of unit BB length) for the rotation BA. = (Id − A)(Id − A) It remains to find the angle of rotation under the action of = Id − 2A + AA BA. Note that n is perpendicular to the axis of rotation = Id − A, since AA = A m × nand so lies in the plane P of rotation associated to BA. The sought for angle of rotation θ is the angle by = B which n turns in the plane P. In fact and cosθ < BAn,n > = ||BAn|| · ||n|| CC < BAn,n > = (Id − 2A)(Id − 2A) = 1 · 1 = Id − 4A + 4AA note n is a unit vector = Id, since AA = A and BA ∈ SO(3,R) ⇒ BA preserves length = < BAn,n > For the next proof we will appeal to the bac cab rule for = < B(An),n > the cross product = < B(−n),n > a × (b × c) = b < a,c > − c < a,b > = − < (I − 2mmt )n),n > = − < − t , > n 2mm n n Let x ∈ R3 be an arbitrary vector. Then = − < n − 2m < n,m >,n > = − < n,n > + 2 < n,m >2 (DD)(x) = 2cos2 φ − 1 = D(Dx) where φ is the angle between m and n = n × (n × x) = cos(2φ) = n < n,x > − x < n,n > = nt x > n − (1)x = −x + nt nx BA is rotation about axis m × n by twice the angle = −(Id − nt n)x between m and n. For example old fashioned dressing tables have 2 hinged = −Bx mirrors. Double reflection results in a rotation by twice the angle between the mirror planes. We have proven that DD = −B
September 23, 2009 c jbquig-UCD 112 CHAPTER 5. ANSWERS TO QUESTIONS IN CHAPTER1
5.8.2 answer also
Ct AB = (Id − 2A)t = A(Id − A) = Idt − 2At = A − A2 = Id − 2A = A − A = C = 0 −1 t Similarly BA = 0. To prove that C = C we need only prove that C C = Id.
CtC Let x ∈ R3 be an arbitrary vector. Then = (Id − 2A)t (Id − 2A) (AD)(x) = (Idt − 2At )(Id − 2A) = A(D(x)) = (Id − 2A)(Id − 2A) = A(n × x) = Id − 4A + 4AA = nnt (n × x) = Id − 4A + 4A = n < n,n × x > = Id = < n,n × x > n That Dt = −D is clear from the matrix form of D = 0n = 0 0 −c b D = c 0 −a This proves that AD = 0. Also −b a 0
(DA)(x) = D(nnt x)) 5.8.4 answer = D(n < n,x >) We must prove that St S = Id = < n,x > D(n) = < n,x > n × n (A + cos(θ)B + sin(θ)D)t (A + cos(θ)B + sin(θ)D) = < n,x > 0 = (At + cos(θ)Bt + sin(θ)Dt )(A + cos(θ)B + sin(θ)D) = 0 = (A + cos(θ)B − sin(θ)D)(A + cos(θ)B + sin(θ)D) = (A + cos(θ)B)2 − (sin(θ)D)2 This proves that DA = 0. since BD = DB and AD = DA Next = A2 + 2cos(θ)AB + cos2(θ)B2 − sin2(θ)D2 = A + 2cos(θ)0 + cos2(θ)B − sin2(θ)(−B) BD = A + (cos2(θ) + sin2(θ))B = (Id − A)D = A + (1)B = D − AD = (A + B) = D − 0 = Id = D
Similarly DB = D. 5.9 question and answer
5.8.3 answer Let n ∈ R3 with n = 1 and θ ∈ R. For short write A,B,C,D,S for An,Bn,Cn,Dn,Sn,θ respectively. Prove t A (i) AA = A , BB = B , CC = I , DD = −B = (nnt )t (ii) AB = BA = AD = DA = 0 , BD = DB = D = (nt )t (n)t (iii) At = A , Bt = B , Ct = C−1 = C , Dt = −D = nnt = A (iv) S−1 = St
c jbquig-UCD September 23, 2009 5.10. QUESTION AND ANSWER 113
5.9.1 answer 5.9.2 answer
Before the actual solution we give a helpful way of Here A is 5 × 5 thus looking at A (which is of course the very important Jordan Form). We write exptA A = aI + N = exp(taI +tN) Here I is the 5 × 5 identity matrix given by = exp(taI)exp(tN) 1 i = j since I commutes with N Ii = j 0 i 6= j = etaI exp(tN) trivially, from defn of exp(taI) and N is called nilpotent (the reason for the name will soon be clear) and is given by. exp(taI) = exp(ta)I ∞ k ta t 1 i = j − 1 = e I ∑ Ni = k! j 0 i 6= j − 1 k=0 4 tk = etaI In fact ∑ k! 0 1 0 0 0 k=0 k 0 0 1 0 0 since N = 0, k ≥ 5 N = 0 0 0 1 0 = 0 0 0 0 1 t t2 t3 t4 etaI I + N + N2 + N3 + N4 0 0 0 0 0 1! 2! 3! 4! All entries in N are zero except on that super diagonal = 2 3 4 entries are 1. It is easy to compute the powers of N in Nk 1 t t /2 t /6 t /24 all entries are zero except for 1’s on the k-th super 0 1 t t2/2 t3/6 at diagonal. For example e 0 0 1 t t2/2 0 0 0 1 t 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 N3 = 0 0 0 0 0 0 0 0 0 0 5.9.3 answer 0 0 0 0 0 There is nothing new here, just the above in a general We have the formula context. 1 i = j − k (Nk)i = j 0 i 6= j − 1 5.9.4 answer A consequence of this is that Nk = 0 for k > 4 (recall we There is nothing new here, just the above in a general are discussing 5× matrices here); thus the name nilpotent. context. Now we are ready for computations. A4 5.10 question and answer = (aI + N)4 Let a ∈ R. Let A be the basic k × k Jordan block matrix we can use the binomial th given by a i = j since I and N commute i A j = 1 i = j − 1 = 0 otherwise 4 4 4 4 a4I + a3N + a2N2 + aN3 + N4 1 2 3 4 For example when k = 5 = a4 4a3 6a2 4a 1 a 1 0 0 0 4 3 2 0 a 4a 6a 4a 0 a 1 0 0 4 3 2 0 0 a 4a 6a A = 0 0 a 1 0 0 0 0 a4 4a3 0 0 0 a 1 0 0 0 0 a4 0 0 0 0 a
September 23, 2009 c jbquig-UCD 114 CHAPTER 5. ANSWERS TO QUESTIONS IN CHAPTER1
(i) For k = 5 prove that These 3 simultaneous equations are readily solved, we obtain a4 4a3 6a2 4a 1 4 3 2 x = z = −y 0 a 4a 6a 4a 4 4 3 2 A = 0 0 a 4a 6a We can take n = i − j + k and normalizing 0 0 0 a4 4a3 0 0 0 0 a4 1 1 1 n = √ −1 , cos(θ) = − (ii) For k = 5 prove that 2 3 1 1 t t2/2 t3/6 t4/24 0 1 t t2/2 t3/6 To sort out the ambiguity in θ we must find sin(θ). We at exptA = e 0 0 1 t t2/2 study entry K1 of the matrix K:( use the notation 3 0 0 0 1 t n = ai + bj + ck) 0 0 0 0 1 0 (iii) Compute An then exp(tA) for general n ∈ N and 1 k = 5. = K3 2 3 4 n t 1 t 1 1 (iv) Compute A ,A ,A then A then exp(tA) for = (nn )3 + cos(θ)(Id − nn )3 + sin(θ)[n×]3 general k = ac + cos(θ)(−ac) + sin(θ)(c) 1 1 1 1 = + (− )(− ) + sin(θ) √ 3 2 3 3 5.10.1 answer 1 sin(θ) + √ 6 3 It is trivial to check that the columns K1,K2,K3 of K are each of length 1 and mutually orthogonal. Thus √ 3 K ∈ O(3,R). Trivially also det(K) = 1 so K ∈ SL(3,R). Thus sin(θ) = − 2 and since we have cos(θ) = −1/2 we Thus K ∈ O(3,R)TSL(3,R) = SO(3,R) and by have unambiguously, θ = −2π/3. Finally theorem(??) is a rotation. 1 1 2π n = √ −1 and θ = − 3 5.10.2 answer 3 1 To find theta we use formula(??) 1 + 2cos(θ) 5.10.3 answer = tr (K) Take~n = (i − j + k). Take any convenient~l which is 1 2 3 = K1 + K2 + K3 orthogonal to~n; say~l = (i + j). Take = 0 + 0 + 0 −1 This cos(θ) = −1/2 and ~m =~n ×~l = 1 2π θ = ± 2 3 This ambiguity in θ will be sorted out, see below. Normalizing these three vectors we obtain a p.o.o.n. triple from which we form the required special orthogonal The convention way to find n = xi + yj + zk is to solve the matrix eigenvector equation, i.e. solve 3 simultaneous linear eqns √ √ √ in 3 variables. −1/√3 1/√2 1/√3 E = (l,m,n) = 1/ 3 1/ 2 −1/ 3 An = 1n ⇔ (1 · Id − A)n = 0 √ √ 2/ 3 0 1/ 3 For other matrices, solving the eigenvector eqn can involve a lot of tedious computation; for example in Since E ∈ SO(3,R) the inverse is easily found E−1 = Et . question(??). See an alternative less onerous and clever Also φ = θ = −2/pi/3 with √ technique, used in the solution of question(??). cos(φ) = −1/2 and sin(φ) = − 3/2. Thus (1 · Id − A)n = 0 K 1 1 0 x 0 0 −1 0 = 0 1 1 y = 0 = 0 0 −1 −1 0 1 z 0 1 0 0 x + y = 0 = E ◦ § φ ◦ E−1 = y + z = 0 k,2π/3 −x + z = 0
c jbquig-UCD September 23, 2009 5.11. QUESTION AND ANSWER 115
= = √ √ √ −1/ 3 1/ 2 1/ 3 t √ √ √ Sj,π/2Sk,−π/2K 1/√3 1/ 2 −1/√3 2/ 3 0 1/ 3 = √ √−1/2 − 3/2 0 0 0 1 − 3/2 −1/2 0 0 1 0 0 0 1 −1 0 0 √ √ √ −1/√6 1/√6 2/ 6 0 1 0 1/√2 1/√2√ 0 −1 0 0 1/ 3 −1/ 3 1/ 3 0 0 1 0 −1 0 5.10.4 answer 0 0 −1 1 0 0 K = = exp(φ[n×]) √ √ 1 0 0 0 1/ 3 1/ 3 2π √ √ 0 1 0 = exp( −1/ 3 0 1/ 3 ) 3 √ √ 0 0 1 −1/ 3 −1/ 3 0 We deduce that γ = 0. Finally 0 1 1 2π = exp √ −1 0 1 K 3 3 −1 −1 0 = 5.10.5 answer Sk,π/2Sj,−π/2Sk,0 K carries k to the point
K(k) = 0 −1 0 = K~3 1 0 0 0 0 0 1 = −1 0 0 0 −1 0 1 0 sin(α)cos(β) 1 0 0 = sin(α)sin(β) cos(α) 1 0 0 0 1 0 sin(π/2)cos(−π/2) 0 0 1 = sin(π/2)cos(−π/2) cos(π/2) 5.11 question and answer which we have expressed using spherical polar coordinates. From this we immediatly obtain two Euler Let angles π π −12/325 −316/325 3/13 α = , β = − S = / / / 2 2 309 325 12 325 4 13 A little work is needed to obtain the third Euler angle γ. −4/13 3/13 12/13 (i) Prove that K is a rotation. K = Sk,αSj,βSk,γ (ii) By (i) K is rotation about the axis n, (||n|| = 1) by ⇔ S = (S S )−1K γ k,α j,β angle θ, i.e. K = Sn,θ. Find n and θ. −1 ⇔ Sγ = (Sk,αSj,β) K (iii) Express K in the form −1 −1 ⇔ Sγ = Sj,β Sk,αK cosφ −sinφ 0 −1 −1 K = E ◦Sφ ◦E = E sinφ cosφ 0 E ⇔ Sγ = Sj,−βSk,−αK 0 0 1 Thus Find φ and give E and E−1 in full detail. Sγ (iv) Express K in the form exp(J). Give the matrix J in full detail.
September 23, 2009 c jbquig-UCD 116 CHAPTER 5. ANSWERS TO QUESTIONS IN CHAPTER1
−12 + 12 + 300 (v) Find (Euler) angles α,β,γ such that = 325 300 K = 325 = Sk,α ◦ S ◦ Sk,γ j,β 12 = cosγ −sinγ 0 13 = sinγ cosγ 0 · Thus cos(θ) = (12/13 − 1)/2 = (−1/26). 0 0 1 −1 sinβ 0 cosβ cos(θ) = 26 0 1 0 · At this stage θ is not uniquely determined. We will (see cosβ 0 −sinβ below) compute sin(θ) and then θ will be unambiguously cosα −sinα 0 determined. sinα cosα 0 0 0 1 Write n = ai + bj + ck. The usual way to find a,b,c is to [Footnote solve the eigenvalue equation Kn = 1 · n. In the present In this question an angle 0 ≤ θ < 2π should be specified case this reduces to solving three strenous simultaneous by giving both cosine and sin as fractions. For example equations for the three unknowns a,b,c. We present a clever method which involves much less work and as a θ = arccos(3/5) = arcsin(−4/5 bonus also determines sin(θ) ⇔ cos(θ) = 3/5 and sin(θ) = −4/5 Write K = A + cos(θ)B + sin(θ)D Then Note that K − Kt K = (A − At ) + cos(θ)(B − Bt ) + sin(θ)(D − Dt ) − / − / / 12 325 316 325 3 13 = (0) + cos(θ)(0) + sin(θ)(2D) = 309/325 12/325 4/13 −4/13 3/13 12/13 since A,B are symmetric and D is skew-symmetric −12 −316 75 1 = 309 12 100 = 2sin(θ)[n×] 325 −100 75 300 Applying this observation we obtain 0 −c b 5.11.1 answer sin(θ) c 0 −a −b a 0 Recall K1,K2,K3 denote the columns of K. It is onerous but straightforward to check that = (K − Kt )/2 < K2,K3 >=< K3,K1 >=< K1,K2 >= 0 and that −12 −316 75 −12 309 −100 < K ,K >=< K ,K >=< K ,K >= 1 1 1 1 2 2 3 3 1. Thus = 309 12 100 − −316 12 75 650 650 K ∈ O(3,R). It is also onerous and straightforward to −100 75 300 75 100 300 prove that det(K) = 1 > 0, thus K ∈ SL(3,R). Thus 0 −625 175 K ∈ SO(3,R) ∩ SL(3,R) = SO(3,R) and K is proven to be 1 = 625 0 25 a rotation. Computations are ommited and left to the 650 reader. Here is one example −175 −25 0 0 −25 7 1 < K1,K1 >= 1 = 25 0 1 26 ⇔ 122 + 3092 + 1002 = 3252 −7 −1 0 ⇔ 144 + 95481 + 10000 = 105625 Thus a −1 1 which is true sin(θ) b = 7 26 c 25 We have calculated cos2(θ) = 1/676 Thus 5.11.2 answer 2 2 2 2 sin (θ) = 1 − cos (θ)√ = (675/676) = 3 · (15) /26 . Let us take sin(θ) = +15 3/26. Then 1 + 2cos(θ) a −1 √ −1 = tr (k) 26 1 3 n = b = √ 7 = 7 1 2 3 15 3 26 45 = K1 + K2 + K3 c 25 25
c jbquig-UCD September 23, 2009 5.11. QUESTION AND ANSWER 117
We sum up, 5.11.4 answer √ −1 √ 3 −1 15 3 K n = 7 ,θ = arccos = arcsin 45 26 26 25 = exp(φ[n×]) √ √ 0 −25 7 footnote Above we could have taken sin(θ) = −15 3/26 3φ = exp( 25 0 1 ) and replaced n by −n. This is explained by the fact that 45 −7 −1 0 S−n,−θ = Sn,θ a point the reader may have already √ −1 15 3 observed. where φ = arccos 26 = arcsin 26
5.11.3 answer 5.11.5 answer Take~n = −1i + 7j + 25k. Take any convenient~l which is K carries k to the point orthogonal to~n; say~l = 7i + j. Take K(k) 7 = K~3 ~m =~n ×~l = 176 75 −49 1 = 100 325 Normalizing these three vectors we obtain a p.o.o.n. triple 300 from which we form the required special orthogonal 3 matrix 1 = 4 13 E 12 = (l,m,n) sin(β)cos(α) 7 7 −1 = sin(β)sin(α) √ √ √ cos(β) 5 2 3 3714 15 3 1 176 7 √ √ √ = which we have expressed using spherical polar 5 2 3 3714 15 3 −49 25 coordinates. From this we immediatly obtain information 0 √ √ about two Euler angles 0 ≤ α ≤ π and 0 ≤ β ≤ 2π 3 3714 15 3 s −1 t 2 Since E ∈ SO(3,R) the inverse is easily√ found E = E . 12 12 5 −1 15 3 cos(β) = , sin(β) = + 1 − = Also φ = θ = arccos 26 = arcsin 26 Thus 13 13 13 K and −12 −316 75 1 4 3 4 = 309 12 100 tan(α) = , cos(α) = , sin(α) = 325 3 5 5 −100 75 300 A little work is needed to obtain the third Euler angle γ. = E ◦ S φ ◦ E−1 k,φ Since K = Sk,αSj,βSk,γ = we have 7 7 −1 √ √ √ 5 2 3 3714 15 3 Sγ 1 176 7 √ √ √ = S−1S−1 K j,β k,α 5 2 3 3714 15 3 −49 25 t t 0 √ √ = Sj,βSk,αK 3 3714 15 3 √ Thus −1 15 3 0 √26 26 cos(γ) 15 3 −1 0 sin(γ) 26 26 0 0 0 1 7 7 −1 t = √ √ √ 5 2 3 3714 15 3 Sk,γi 1 176 7 √ √ √ 5 2 3 3714 15 3 −49 25 = 0 √ √ 3 3714 15 3 t t Sj,βSk,αKi
September 23, 2009 c jbquig-UCD 118 CHAPTER 5. ANSWERS TO QUESTIONS IN CHAPTER1
1200 975 = −500 t t ~ Sj,βSk,αK1 = 12 0 −5 = 1 0 13 0 cos(β) 0 −sin(β) (5) · (13)2 5 0 12 0 1 0 sin(β) 0 cos(β) 48 cos(α) sin(α) 0 39 −20 −sin(α) cos(α) 0 0 0 1 = −12 1 676 309 1 507 325 (5) · (13)2 −100 0
= = 12/13 0 −5/13 4/5 0 1 0 1 3/5 (5) · (13)2 5/13 0 12/13 0 3/5 4/5 0 −4/5 3/5 0 0 0 1 Thus cos(γ) = 4/5 and sin(γ) = 3/5. Finally we have the −12 Euler decomposition of K ∈ SO(3,R). 1 309 325 −100 K = 12 0 −5 1 = 0 13 0 (325) · (5) · (13) −12 −316 75 5 0 12 1 309 12 100 325 3 4 0 −100 75 300 −4 3 0 0 0 5 = −12 Sk,α ◦ Sj,β ◦ Sk,γ 309 −100 = 3 −4 12 5 4 −3 = 0 0 0 5 5 13 13 5 5 12 0 −5 4 3 3 4 1 0 0 1 0 0 0 13 0 5 5 −5 12 5 5 (325) · (5) · (13) 0 5 0 12 0 0 1 13 13 0 0 1
c jbquig-UCD September 23, 2009 Chapter 6
Answers to questions in chapter2
6.1 question and answer 6.1.1 logarithmic spiral The standard spiral has equation in polar co-ords r = θ i.e. For each regular parameterized curve curve. Make a rough parametrization x = t cos(t),y = t sin(t). The arc-length sketch. Prove that s(t) is as given. Reparameterize by arc function for this spiral is just too messy. There are several cost length. (The simple spiral c(t) = t should be spirals with nice arc-length functions. Here we study the sint taken as a starting point when drawing spirals. logarithmic spiral Γ, in polar coords Unfortunately arc length parameterization of the simple r = exp(θ) ⇔ log(r) = θ. We work with the cartesian spiral in very very messy: and so was not included in this parametrization. list.) c : R → Γ ⊂ R2 x(t) exp(t)cos(t) (i) Logarithmic spiral, t 7→ c(t) = = cost y(t) exp(t)(sin(t) c(t) = exp(t) , −∞ < t < ∞, sint √ The velocity vector is s(t) = t exp(t). c0(t) (ii) Archimedean spiral♠, 0 cost −sin(t) x (t) c(t) = t + , 0 < t < ∞, = 0 sint cos(t) y (t) s(t) = t2/2. (exp(t)cos(t))0 = ( (t) (t))0 t exp sin (iii) Catenary♠, c(t) = = cosht exp(t)cos(t) − exp(t)sin(t) = t exp(t)sin(t) + exp(t)cos(t) , 0 ≤ t < ∞, s(t) = sinh(t) (expt + exp−t)/2 thus (iv) Cycloid♠, 0 0 t −sint < c (t),c (t) > c(t) = + , 0 ≤ t < ∞, 1 −cost s(t) = 4 − 4cos(t/2) = (v) Cardioid, z(t) = 2exp(it) − exp(2it), 0 ≤ t < ∞, (exp(t)cos(t) − exp(t)sin(t))2 s(t) = 8 − 8cos(t/2) + (vi) Tractoid (exp(t)sin(t) + exp(t)cos(t))2 0 1 1 c(t) = + , 0 ≤ t ≤ ∞, t cosht sinht s(t) = ( t) = log cosh exp(2t) (cos(t) − sin(t))2 + (sin(t) + cos(t))2
The answers to (i) to (vi) (with curve plots) have been = prepared by the computer program ’maple’ and can be (cos2(t) + sin2(t) − 2sin(t)cos(t)) found off the course home page at exp(2t) + ???????????????????????????????????????????????? (cos2(t) + sin2(t) + 2sin(t)cos(t))
119 120 CHAPTER 6. ANSWERS TO QUESTIONS IN CHAPTER2
Figure 6.1: (i) conical helix curce (ii) cone and screw surfaces intersect
β : [0,L] → [0,2π] s 7→ t = log 1 + √s = 2 exp(2t) (1 − 2sin(t)cos(t)) + (1 + 2sin(t)cos(t)) and so the arc-length parametrization function is
= 2exp(2t) γ(s) = ( (s)) Thus the speed scalar is c β √ √s √s ||c0(t)|| = 2exp(t) exp log 1 + cos log 1 + = 2 2 exp log 1 + √s sin log 1 + √s This logarithmic spiral is an infinite curve and so has no 2 2 start point. From here on We work with the finite curve, 1 + √s cos log 1 + √s the first turn of Γ, i.e. we take 0 ≤ t ≤ 2π. The arc length = 2 2 1 + √s sin log 1 + √s function is 2 2 s(t) = α(t) footnote Both the original and arc length parametrization Z t become more compact in their complex form = ||c0(p)||dp 0 Z t p = (2)exp(t)dp z(t) = exp(t)exp(it) = exp((1 + i)t) 0 √ t = 2 exp(p)|p=0 p = (2)(exp(t) − 1) The total arc length is √ L = α(2π) = 2(exp(2π) − 1) γC(s) −1 = z(β(s)) The√ inverse function α is found by solving s s = 2(exp(t) − 1) for t in terms of s. = exp (1 + i)log 1 + √ 2 −1 s ! t = β(s) = α (s) = log 1 + √ s 1+i 2 = exp log 1 + √ 2 We summarize regarding α and β = α−1 s 1+i α : [0,2π] → [0,L√] = 1 + √ t 7→ s = 2(exp(t) − 1) 2
c jbquig-UCD September 23, 2009 6.1. QUESTION AND ANSWER 121
6.1.2 archimedean spiral The inverse function β = α−1, found by solving s = α(t) = t2/2 for t in terms of s, is A thread is unwound, being held taut, from a spool. The √ end traces out a curve Γ called the Archimedean spiral. t = β(s) = 2s We wish to construct a parametrization c of Γ. Let S(t) and so the arc-length parametrization function is denote a.c.w. rotation of the plane R2 by angle t. Let C be the a.c.w. circle with center at the origin and radius 1 . γ(s) Note that i ∈ C and that the unit tangent vector there is j. = c(β(s)) If thread is unwound starting from the point i ∈ C then √ √ √ cos( 2s) + 2ssin( 2s) after a length t has been unwound the thread is leaving the = √ √ √ sin( 2s) − 2scos( 2s) spool at point S(t)i in the reverse tangent direction −S(t)j. Thus the end of the thread is at the point footnote Both the original and arc length parametrization of the Archimedean spiral become more compact in their c(t) complex form 1 0 = S(t) −tS(t) z(t) = exp(it) · 1 −t exp(it) · (i) = (1 − it)exp(it) 0 1 1 = S(t) γC(s) −t = z(β(s)) √ √ cos(t) −sin(t) 1 = = (1 − i 2s)exp(i 2s) sin(t) cos(t) −t cos(t) +t sin(t) = 6.1.3 catenary sin(t) −t cos(t) The catenary Γ ⊂ R2 is the curve followed by a hanging We have found a regular parametrization of the rope fixed at both ends e.g. a clothes-line, a telegraph line. Archimedean spiral. It has been proven that the catenary is the graph curve of c : R → Γ ⊂ R2 the hyperbolic cosine cos(t) +t sin(t) exp(x) + exp(−x) t 7→ c(t) = y = cosh(x) = . sin(t) −t cos(t) 2 The velocity vector is We use the parametrization 0 2 0 x (t) t cos(t) c : R → Γ ⊂ R c (t) = 0 = x(t) t y (t) t sin(t) t 7→ c(t) = = y(t) cosh(t) and the scalar speed is Velocity and acceleration vectors are ||c0(t)|| x0(t) 1 q c0(t) = = = t2 cos2(t) +t2 sin2(t) y0(t) sinh(t) = t and x00(t) 0 c00(t) = = At this point it is convenient to to restrict attention to the y00(t) cosh2(t) first turn of Γ, i.e. restrict t to the range [0,2π]. and speed is The total arc length is q ||c0(t)|| = 1 + sinh( t) = cosh(t) Z 2π Z 2π 2 2π 0 t 2 2 L = ||c (t)||dt = t dt = = 2π2 (we used the formula cosh −sinh ≡ 1). The catenary as 0 0 2 0 presented here is an infinite curve and so has no end The arc-length function is points. There we pick a point from which to measure arc length s(t) = α(t), we choose c(0) = j as start point. s(t) s(t) = α(t) Z t = α(t) 0 = ||c (p)||dp Z t 0 = ||c0(p)||dp Z t 0 = pdp Z t 0 = cosh(p)dp t p2 0 = t = sinh(p)|p=0 2 p=0 = sinh(t) − sinh(0) t2 = = sinh(t) 2
September 23, 2009 c jbquig-UCD 122 CHAPTER 6. ANSWERS TO QUESTIONS IN CHAPTER2
The inverse function is and speed is
t = β(s) = arcsinh(()s) ||c0(t)|| and so the arc-length parametrization function is q = (1 − cos(t))2 + sin2(t) γ(s) q = c(β(s)) = 1 − 2cos(t) + cos2(t) + sin2(t) arcsinh(s) p = = 2 − 2cos(t) cosh(arcsinh(s)) q = 4sin2(t/2) arcsinh(s) ! = q = 2sin(t/2) 1 + sinh2)(arcsinh(s)) ! arcsinh(()s) = p (we used the double angle formula 1 + sinh2(arcsinh(s)) cos(2A) = 1 − 2sin2(a)). arcsinh(s) = √ 1 + s2 Total arc length is
6.1.4 cycloid L Z 2π A wheel rolls on a straight rail. The cycloid curve Γ is the = ||c0(t)||dt path traced out by that point of the wheel which was 0 Z 2π initially in contact with the rail. = 2sin(t/2)dt 0 A circle C of radius 1 moves to the right on the x-axis by 2π = −4cos(p/2)|0 rolling, it must roll c.w.. The center is initially at the point = −4cos(π) + 4cos(0) j ∈ C and its position at time t is = 8 t j +ti = 1 The arc length function is Relative to the center, the point of contact is initially at −j and at time t is at S(−t)(−j). At time t the absolute position of the point of contact is s(t) = α(t) c(t) Z t t 0 = ||c0(p)||dp = + S(−t) 1 −1 0 Z t p t cos(−t) −sin(−t) 0 = 2sin dp = + 0 2 1 sin(−t) cos(−t) −1 p t = −4cos t cos(t) sin(t) 0 2 p=0 = + 1 −sin(t) cos(t) −1 t = 4 − 4cos t − sin(t) 2 = 1 − cos(t) The inverse function is β = α− found by silving the We have found a regular parametrization of the cycloid equation s = α(t) for t in terms of s curve Γ. The range of t can be taken to be the whole real line. But there are irregular points at {2nπ |n ∈ Z }, so we s restrict attention to one loop, t ∈ (0,2π)). Note we omit t = β(s) = 2arccos 1 − even t = 0 and t = 2π, these values give rise to irregular 4 points and so the arc-length parametrization function is c : (0,2π) → Γ ⊂ R2 x0(t) t − sin(t) t 7→ c(t) = 0 = y (t) 1 − cos(t) γ(s) The velocity vectors is = c(β(s)) 2arccos 1 − s − sin 2arccos 1 − s x0(t) 1 − cos(t) = 4 4 c0(t) = = 1 − cos 2arccos 1 − s y0(t) sin(t) 4
c jbquig-UCD September 23, 2009 6.1. QUESTION AND ANSWER 123
6.1.5 cardioid Finally arclength parameterization. Start by drawing two circles x2 + y2 = 1 and (x − 1)2 + y2 = 1. Both have radius 1 and γ(s) have a point of contact at x = 1,y = 0. Regard the circle = c(β(s)) centered at 0 as fixed. Roll the other circle around it. The x(β(s)) = initial point of contact traces out a heart shaped curve y(β(s)) called the cardoid. The path of the (continuing) point of √ √ cos(4arcsin( s/4)) − 2cos(8arcsin( s/4)) contact is exp(it) on the fixed circle. The path of the = √ √ sin(4arcsin( s/4)) − 2sin(8arcsin( s/4)) center of the rolling circle is exp(it). The clever bit (roll a coin round a coin and see this) is that relative to the center of the rolling circle the initial point of contact moves on 6.1.6 tractoid the curve −exp(2it). The cardoid is therefore the curve
z(t) = 2exp(it) − exp(2it) A weight with a string attached is placed on a table. The weight is dragged along by the string whose end is moved ⇔ along a straight line L. The weight follows a path x(t) 2cost − cos2t c(t) = = (converging to the line L) called a tractrix curve. The y(t) 2sint − sin2t tractrix curve is famous because when rotated it sweeps velocity out a surface called the pseudosphere of constant negative curvature. It can be proven that the following is a regular x˙(t) parametrization of the the tractrix Γ ⊂ R2. c˙(t) = = 2sint − 2sin2t−2cost + 2cos2t y˙(t) c : R → Γ ⊂ R2 speed x0(t) t − tanh(t) t 7→ = ||c˙(t)|| y0(t) sech (t) q 2 2 = x˙ (t) + y˙ (t) We deviate to point out some useful formulae involving q hyperbolic functions = (2sint − 2sin2t)2 + (−2cost + 2cos2t)2 q sinh = 4(sin2 t + cos2 t) − 8(cos(2t)cost + sin(2t)sint) + 4(sin2(2t) + cos2(2t))tanh = cosh p = 4 − 8(cos(2t −t) + 4 1 p sech = = 8 − 8cos(t) cosh q tanh0 = sech 2 = 16sin2(t/2) sech 0 = −sech tanh = 4sin(t/2) sech 2 + tanh2 = 1 arclength The velocity vector is s(t) Z t x0(t) 1 − sech 2(t) = ||c˙(r)||dr c0(t) = = 0 y0(t) −sech (t)tanh(t) Z t = 4sin(r/2)dr 0 and speed is t = −8cos(r/2)|r=0 0 = 8(1 − cos(t/2)) ||c (t)|| 2 q = 16sin (t/4) = (1 − sech 2(t))2 + sech 2(t)tanh2(t) total arclength q = 1 − 2sech 2 + sech 4(t) + sech 2(t)tanh2(t) L = s(2π) = 16sin2(π/2) = 16 q = 1 − 2sech 2 + sech 2(t)(sech 2(t) + tanh2(t)) Next find t = β(s), t in terms of s q = 1 − 2sech 2 + sech 2(t)(1) 2 s = 16sin (t/4) q √ 2 ⇔ sin(t/4) = s/4 = 1 − sech (t) √ q ⇔ t = 4arcsin s/4 = tanh2(t) √ β(s) = 4arcsin s/4 = tanh(t)
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t − sin(t) The tractrix Γ is an infinite curve and so the total arc = length is infinite. Measuring arclength s(t) = α(t) from 1 − cos(t) (the irregular point) c(0) = j we have or z(t) = t + i − iexp(−it) s(t) Z t = ||c˙(r)||dr (i) Compute t(t),n(t) the Serret-Frenet frame at the 0 point c(t) ∈ Γ. Z t = tanh(r)dr (ii) Compute κ(t) the curvature at c(t). 0 Z r=t sinh(r)dr = (iii) Parametrize the evolute curve E(Γ). r=0 cosh(r) (iv) Parametrize the involute curve I(Γ). Z r=t d(cosh(r)) = r=0 cosh(r) (v) Prove that both I(Γ) and E(Γ) are cycloids and make a single rough (sketch) showing all 3 cycloids = log(cosh(r))|r=t r=0 Γ,I(Γ) and E(Γ). = log(cosh(t)) − log(cosh(0)) = log(cosh(t)) − log(1) = log(cosh(t)) − 0 model answer To start we carry out some computations = log(cosh(t)) which will be needed as we go along. First
−1 Next find t as a function of s, i.e. find β = α , by solving x0 1 − cos(t) c0 = = s = α(t) for t in terms of s. y0 sin(t)
t = β(s) = arccosh(exp(s)) Next x00 sin(t) Finally arclength parameterization. c00 = = y00 cos(t) γ(s) Next = c(β(s)) 0 x(β(s)) ||c (t)|| = q y(β(s)) = x02 + y02 arccosh(exp(s)) − tanh(arccosh(exp(s))) = q sech (arccosh(exp(s))) = (1 + cos(t))2 + sin2(t)) p sinh(arccosh(exp(s))) = 2 + 2cos(t) arccosh(exp(s)) − cosh(arccosh(exp(s))) q = 1 = 4sin2(t/2) cosh(arccosh(exp(s))) = 2sin(t/2) p cosh2 −1(arccosh(exp(s))) arccosh(exp(s)) − Next = exp(s) 1 0 00 exp(s) det(c ,c ) pexp(2s) − 1 1 − cost sint arccosh(exp(s)) − = det = exp(s) sint cost exp(−s) = cos(t) − 1 2 arccosh(exp(s)) − p1 − exp(−2s) = −2sin (t/2) = exp(−s) Next, total arc length
6.2 question and answer L Z 2π The cycloid Γ is the curve traced out by a point on the rim = ||c0(t)||dt of a rolling wheel, such as the valve on a bicycle wheel 0 Z 2π and has a parameterization. = 2sin(t/2)dt 0 c(t) t 2π = −4cos( ) t 0 2 0 = + S(−t) 1 −1 = 8
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t − sin(t) sin(t) Next the arc length function = + 2 1 − cos(t) 1 + cos(t) s(t) t + sin(t) Z t = = ||c0(p)||dp 3 + cos(t) 0 Z t = 2sin(p/2)dp 6.2.5 comparison Γ,E(Γ),I(Γ) 0 p t We begin by displaying clearly the three parametrizations = −4cos( ) 2 0 t − sin(t) = 4 − 4cos(t/2) c(t) = 1 − cos(t)
t + sin(t) 6.2.1 t and n e(t) = cos(t) − 1 0 1 x 1 1 − cos(t) t + sin(t) t = p 0 = i(t) = x02 + y02 y 2sin(t/2) sin(t) 3 + cos(t) and It seems likely that both e and i are variants on c i.e. that both E(Γ) and I(Γ) are cycloids: but the details are tricky. 1 −y0 1 −sin(t) n = p 0 = x02 + y02 x 2sin(t/2) 1 − cos(t) Note that c(t + π) 6.2.2 κ t + π − sin(t + π) = 1 − cos(t + π) detc0,c00 −2sin2(t/2) −1 κ(t) = = = π t + sin(t) < c0,c0 >3/2 8sin3 t/2 4sin(t/2) = + 2 cos(t) − 1 π 6.2.3 evolute parametrization, e(t) = e(t) + 2
e(t) i.e. π n(t) e(t) = c(t + π) − = c(t) + 2 κ(t) Conclusion, the evolute E( ) is the original cycloid t − sin(t) 1 −sin(t) Γ = + −4sin(t/2) translated by the vector πi + 2j; the origin of time (the 1 − cos(t) 2sin(t/2) 1 − cos(t) parameter t) here was shifted by π. t − sin(t) sin(t) = + 2 1 − cos(t) cos(t) − 1 Argue similarly for the involute curve I(Γ), check that t + sin(t) = −π cos(t) − 1 i(t) = c(t + π) + 2 6.2.4 involute parametrization i(t) Finally drawing of all three curves can be found at http://mathsci.ucd.ie/courses/math40060/maple The cycloid has irregular points at c(0) and c(2π). To be safe we measure arc length from the regular (half way) point c(π). We suitably adjust s(t) = 4 − 4cos(t/2) to be 6.3 question and answer s(t) = −4cos(t/2). The catenary is the curve formed by a clothesline or i(t) telephone wire, one parameterization is = c(t) − s(t)tt(t) exp(t) + exp(−t) t − sin(t) 4cos(t/2) 1 − cos(t) c(t) = cosh(t) = = + 2 1 − cos(t) 2sin(t/2) sin(t) The tractrix is the curve follows by a resisting dog on a t − sin(t) 4cos(t/2) 2sin2(t/2) = + lease as its owner walks a straight line, one 1 − cos(t) 2sin(t/2) 2sin(t/2)cos(t/2) parametrization is t − sin(t) 2sin(t/2) = + 2cos(t/2) t − tanh(t) 1 − cos(t) 2cos(t/2) d(t) = sech (t) t − sin(t) 2sin(t/2)cos(t/2) = + 2 1 − cos(t) 2cos2(t/2) (i) Show both curves in a single sketch.
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(ii) Compute (t(t),n(t)) and κ(t) at a point c(t) on the 6.3.3 tractrix catenary. We perform some preliminary tractrix computations. (iii) Compute (t(t),n(t)) and κ(t) at a point c(t) on the ( Recall the useful elementary hyperbolic formulae tractrix. cosh2 −sinh2 = 1, tanh2 +sech 2 = 1, tanh0 = sech 2 sech 0 = −sech tanh ) (iv) Prove that the involute of the catenary is the First tractrix. x(t) t − tanh(t) c(t) = (v) Prove that the evolute of the tractrix is the catenary. y(t) sech (t) Next c0(t) model answer x0(t) = y0(t) 1 − sech 2(t) 6.3.1 drawing of curves = −sech (t)tanh(t) Drawing of all both curves can be found at tanh2(t) http://mathsci.ucd.ie/courses/math40060/maple = −sech (t)tanh(t) Next 6.3.2 the catenary c00(t) We perform some preliminary catenary computations. x00(t) = First y00(t) x(t) t c(t) = 2tanh(t)sech (t)tanh(t) y(t) cosh(t) = sech (t)tanh2(t) − sech 3(t) Next 2tanh2(t)sech (t) x0(t) 1 = c0(t) = t 3 t y0(t) sinh(t) sech ( ) − 2sech ( ) Next Next ||c0(t)|| x00(t) 0 c00(t) = = q y00(t) cosh(t) = tanh(t) tanh2(t) + sech 2(t) √ = tanh(t) 1 Next = tanh(t) q q ||c0(t)|| = 1 + sinh2(t) = cosh2(t) = cosh(t) Next we will compute det(c0,c00) Next tanh2 2tanhsech 2 = 1 0 det 3 det(c0,c00) = det = cosh(t) −sech tan sech − 2sech sinh(t) cosh(t) = tanh2 sech − 2tanh2 sech 3 + 2tanh2 sech 3 Next the arc-length function on the catenary starting from = tanh2(t)sech (t) c(0) = 0i + j. We are ready to compute t(t),n(t) and κ(t) as required Z t Z t t(t) s(t) = ||c0(p)||dp = cosh(p)dp = sinh(t) 0 0 1 x0 = p 0 x02 + y02 y We are ready to compute t(t),n(t) and κ(t) as required 1 tanh2 = 1 x0 1 1 sech (t) p 4 2 2 −sech tanh t(t) = = = tanh +sech tanh p 02 02 y0 (t) (t) x + y cosh(t) sinh tanh 1 tanh2 = q −sech tanh tanh tanh2 +sech 2) 1 −y0 −tanh(t) n(t) = = p 02 02 x0 sech (t) 1 tanh2 x + y = tanh −sech tanh det(c0,c00) cosh(t) κ(t) = = = sech 2(t) tanh(t) 0 0 3/2 3 = < c ,c > cosh (t) −sech (t)
c jbquig-UCD September 23, 2009 6.4. QUESTION AND ANSWER 127
n(t) and this is the parametrization of the catenary as 1 −y0 requested. = p 0 x02 + y02 x sech (t) = 6.4 question and answer tanh(t) When a circle of radius 1 rolls round inside a circle of radius 3 the initial point of contact traces out a deltoid κ(t) curve ∆. det(c0,c00) = (i) Prove that the following is a parametrization of the < c0,c0 >3/2 deltoid. tanh2(t)sech (t) = 2cos(t) + cos(t) 3 c(t) = tanh (t) 2sin(t) − sin(t) 1 = or sinh(t) z(t) = 2exp(it) + exp(−it) = cosech(t) (ii) Sketch the deltoid curve inside the circle of radius 3. 6.3.4 involute of catenary (iii) Compute (t(t),n(t)) and κ(t) at a point c(t) on ∆. Parametrization of the involute of the catenary is (iv) Prove that the evolute E(∆) is also a Deltoid 3 i(t) times the size of the original. = c(t) − s(t)t(t) (v) Prove that the involute I(∆) is also a Deltoid one third the size of the original t sech (t) = − sinh(t) cosh(t) tanh(t) t − tanh(t) = model answer cosh(t) − sinh(t)tanh(t) t − tanh(t) = 6.4.1 deltoid parametrization (cosh2(t) − sinh2(t))/cosh(t) When a circle of radius 1 rolls round inside a fixed circle t − tanh(t) = of radius 3 the initial point of contact traces out a deltoid 1/cosh(t) curve ∆. The center of the rolling circle moves a.c.w. on a t − tanh(t) = circle of radius 2 so the parametrization of the path of the sech (t) rolling center is and this is the parametrization of the tractrix as required. cos(t) t 7→ 2 sin(t) 6.3.5 evolute of the tractrix The rolling circle rotates c.w. with triple angular speed, i.e. with angular speed −3. BUT relative to the radius Parametrization of the evoluteof the tractrix is joining the center of the fixed circle to the center of rolling e(t) circle it rotates with angular speed -2. The parametrization n(t) of the deltoid is thus = c(t) + κ(t) c(t) t − tanh(t) 1 sech (t) cos(t) cos(−2t) = + = 2 + sech (t) 1/sinh(t) tanh(t) sin(t) sin−2t t − tanh(t) tanh(t) 2cos(t) + cos(2t) = + = sech (t) sinh(t)tanh(t) sin(t) − sin(2t) t = The complex version is of course sech (t) + sinh(t)tanh(t) t z(t) = 2exp(it) + exp(−2it) = (1 + sinh2(t))/cosh(t) t 6.4.2 drawing = 2 cosh (t)/cosh(t) Drawing of ∆ the deltoid curve inside the circle can be t found at = cosh(t) http://mathsci.ucd.ie/courses/math40060/maple
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6.4.3 t, n and κ 8(sin2(2t) + cos2(2t)) To start we carry out some computations which will be + needed as we go along. First 4(cos(2t)cos(t) − sin(2t)sin(t))
x0 −2sin(t) − 2sin(2t) c0 = = y0 2cos(t) − 2cos(2t) = 4 − 8 + 4cos(2t +t) Next x00 −2cos(t) − 4cos(2t) c00 = = = y00 −2sin(t) + 4sin(2t) −4 + 4cos(3t) Next = ||c0(t)||2 3t 02 02 −8sin2 = x + y 2 = (−2sin(t) − 2sin(2t))2 + (2cos(t) − 2cos(2t))2() Now we are ready to compute the Serret-Frenet frame 2 2 2 2 = 4(sin (t) + cos (t)) + 4(sin (2t) + cos (2t)) (t(t),n(t)) and the curvature κ(t) at the point c(t) on the −8(cos(2t)cos(t) − sin(2t)sin(t)) deltoid. = + − ( ( t +t)) 4 4 8 cos 2 t(t) = 8(1 − cos(3t)) c0(t) 2 = = 8(1 − ( 1 − 2sin (3t/2)) ) ||c0(t)|| 2 = 16sin (3t/2) 1 −2sin(t) − 2sin(2t) = 3t 4sin(3t/2) 2cos(t) − 2cos(2t) = 4sin2( 2 1 −sin(3t/2)cos(t/2) = sin(3t/2)sin(t/2) Thus sin(3t/2) 0 3t −cos(t/2) ||c (t)|| = 4sin = 2 sin(t/2) Next
det(c0,c00) n(t) = S(π/2)t(t) 0 −1 −cos(t/2) = = −2sin(t) − 2sin(2t) −2cos(t) − 4cos(2t) 1 0 sin(t/2) det 2cos(t) − 2cos(2t) −2sin(t) + 4sin(2t) sin(t/2) = − cos(t/2) = or n(t) sin(t) + sin(2t) cos(t) + 2cos(2t) 4det 1 −cos(t) + cos(2t) −cos(t) + cos(2t) sin(t) − 2sin(2t) = 2sin(3t/2) −sin(t) − sin(2t) = 4(sin(t) + sin(2t))(sin(t) − 2sin(2t)) κ(t) 0 00 − det(c (t),c (t)) = 0 3 4(cos(t) + 2cos(2t))(−cos(t) + cos(2t)) ||c (t)|| −8sin2(3t/2) = (4sin(3t/2))3 = −1 4(sin2(t) − sin(t)sin(2t) − 2sin2(2t)) = 8sin(3t/2) − 4(−cos2(t) − cos(t)cos(2t) + 2cos2(2t)) 6.4.4 evolute of the deltoid ∆
= e(t) 4(sin2(t) + cos2(t)) − =
c jbquig-UCD September 23, 2009 6.4. QUESTION AND ANSWER 129
n(t) Z t c(t) + = 4sin(3p/2)dp κ(t) π/3 t 8 3p = = − cos( ) 3 2 2cos(t) + cos(2t) π/3 8 3t π 2sin(t) − sin(2t) = − cos − cos + 3 2 2 8 3t 1 1 −cos(t) + cos(2t) = − cos −1/8sin(3t/2) 2sin(3t/2) −sin(t) − sin(2t) 3 2 Now we are ready to parametrize the involute ∆. = 2cos(t) + cos(2t) −cos(t) + cos(2t) i(t) − 4 2sin(t) − sin(2t) −sin(t) − sin(2t) = = 2cos(t) + cos(2t) 4cos(t) − 4cos(2t) c(t) − s(t)t(t) + 2sin(t) − sin(2t) +4sin(t) + 4sin(2t) = = 2cos(t) + cos(2t) 6cos(t) − 3cos(2t) 2sin(t) − sin(2t) 6sin(t) + 3sin(2t) − 8 3t −cos(t/2) It is trivial to check that − cos 3 2 sin(t/2) e(t) = 3S(π)c(t − π) = From this we see that the evolute E(∆) is another deltoid: 2cos(t) + cos(2t) In fact to obtain the parametrization e(t) of E(∆) from the 2sin(t) − sin(2t) parametrization c(t) of ∆, first change the time origin by + π, second rotate by Π, third scale by a factor of 3. 4 −2cos(3t/2)cos(t/2) 3 2cos(3t/2)sin(t/2) 6.4.5 involute of the deltoid curve = We do some preliminary computations. We calculate the 2cos(t) + cos(2t) total arc length and the arc length function. The deltoid is 2sin(t) − sin(2t) regular except for 3 cusps at c(0),c(2π/3) and c(4π/3). + We compute L the total arc length between two cusps. 4 −cos(2t) − cos(t)] 3 sin(2t) − sin(t) L Z 2π/3 = ||c0(t)||dt = 0 1 6cos(t) + 3cos(2t) Z 2π/3 = 4sin(3t/2)dt 3 6sin(t) − 3sin(2t) 0 + 2π/3 2 1 −4cos(t) − 4cos(2t)] = 4 − cos(3t/2) 3 0 3 −4sin(t) + 4sin(t) 8 = ((−cosπ) − (−cos0)) 3 = 8 1 2cos(t) − cos(2t) = (−(−1) − (−1)) 3 3 2sin(t) + sin(2t) 16 = It is trivial to check that 3 1 When computing s(t) we avoid irregular points, we start at i(t) = S(π)c(t − π) 3 c(π/3), i.e. half way between two cusps. From this we see that the involute I(∆) is another deltoid: s(t) In fact to obtain the parameterization i(t) of I(∆) from c(t) Z t the parametrization of ∆, first change the time origin by π, = ||c0(p)||dp π/3 second rotate by Π, third scale by a factor of 1/3.
September 23, 2009 c jbquig-UCD 130 CHAPTER 6. ANSWERS TO QUESTIONS IN CHAPTER2
6.5 question and answer, helix 6.5.3
Let a,b,α > 0. Let Γ be the cylindrical helix curve which 0 −aαsinαt −asinαt turns 1 radian, anticlockwise while ascending height b, on c (t) 1 1 t(t) = = √ aαcosαt = √ acosαt 2 2 2 ||c0t|| 2 2 2 2 the cylinder C,x + y = a , in time period 1/α of a α a + b bα a + b b second, starting from position ai ∈ C when t = 0. A parametrization of Γ is It is easiest next to compute
x(t) acos(αt) b(t) c(t) = y(t) = asin(αt) z(t) bαt = c0 × c00 (i) Sketch the curve. ||c0 × c00|| (ii) Compute c0,c00,c000,s0(t). √ = (iii) Prove that s(t) = tα a2 + b2. Find the arc length i j k paramaterization γ(s) of the curve c. det −aαsinαt aαcosαt bα (iv) Compute t(t), n(t), b(t), κ(t),τ(t). −aα2 cosαt −aα2 sinαt 0 ||top|| (v) Prove that κ and τ are constants and express both in terms of a and b. = (vi) Find the eigenvector v(t) (with associated i j k eigenvalue 0) of the Serret-Frenet matrix (i.e.the det −asinαt acosαt b instantaneous axis of rotation of the S-F-frame). −acosαt −asinαt 0 Express v(t) in terms of the moving S-F-frame. ||top|| Express v(t) in terms of the fixed standard basis (i,j,k); hence prove that v(t) is a constant vector. = absinαt −abcosαt 6.5.1 answer a2 See figure 2.3. ||top|| = 6.5.2 answer bsinαt −bcosαt a 2 acosαt −aαsinαt −aα||topcos||αt 0 00 c(t) = asinαt , c (t) = aαcosαt , c (t) = −aα2 sinαt bαt bα = 0 bsint 1 q √ −bcost 0 2 2 2 2 p 2 2 2 2 s (t) = (−aαsinαt) + (aαcosαt) + b α = α a + b a + b a Thus Finally n(t) is easily computed as Z t Z t 0 p 2 2 p 2 2 s(t) = s (t)dt = α a + b dt = αt a + b n(t) 0 0
Also s = t(s) = √ α a2 + b2 b(t) × t(t) and arc-length parameterization is = γ : R → Γ ⊂ R3 i j k s 1 acos √ det bsinαt −bcosαt a α a2+b2 a2 + b2 s 7→ γ(s) = asin √ s −asinαt acosαt b α a2+b2 b √ s α a2+b2 =
c jbquig-UCD September 23, 2009 6.6. QUESTION AND ANSWER, CONICAL HELIX 131
−cosαt 0 κ(t) 0 −sinαt −κ(t) 0 τ(t) 0 0 −τ(t) 0
Finally = 0 a 0 1 t(t),n(t),b(t) √ −a 0 b 2 2 a + b 0 −b 0 = −√asinαt √bsinαt 6.5.5 a2+b2 −cosαt a2+b2 √acosαt , −sinαt , −√bcosαt Express the Darboux vector v(t) (i.e.the eigenvector of a2+b2 a2+b2 √ b 0 √ a F(t) with associated eigenvalue 0) as a linear 2 2 2 2 a +b a +b combination of t(t),n(t) and b(t).
τ(t) b 6.5.4 1 v(t) = 0 = 0 a2 + b2 Compute the curvature and torsion κ(t) and τ(t). Write κ(t) a down the Seret-Frenet matrix F(t). Recall from above that This column refers to the moving S-F frame so to be abα3 sinαt concise see (*) below. c0 × c00 = −abα3 cosαt v(t) a2α3 we can compute = √ ||c0 × c00|| aα3 a2 + b2 a τ(t) κ(t) = = = ||c0||3 α3(a2 + b2)3/2 (a2 + b2) (t(t),n(t),b(t)) 0 κ(t) We compute = τ(t) τ(t)t(t) + κ(t)b(t)(∗)
= = 0 00 000 < c × c ,c > −asinαt < c0 × c00,c0 × c00 > b 1 √ acosαt a2 + b2 2 2 a + b b = + bsinαt 3 3 a 1 abα sinαt aα sinαt √ −bcosαt 3 3 a2 + b2 2 2 < −abα cosαt , −aα cosαt > a + b a a2α3 0 α6a2(a2 + b2) = o 1 = √ 0 2 2 a + b 1 a2b 2 2 2 = a (a + b ) 1 √ k 2 2 = a + b
b 6.6 question and answer, conical a2 + b2 helix The Serret-Frenet matrix is Let Γ be the conical helix curve with parametrization F(t) x(t) t cos(t) c(t) = y(t) = t sin(t) = z(t) t
September 23, 2009 c jbquig-UCD 132 CHAPTER 6. ANSWERS TO QUESTIONS IN CHAPTER2
(i) Find the equation of a conical surface K and a We compute s0(t) screw surface S such that Γ = K TS. 0 (ii) Sketch Γ = K TS ⊂ R3. s (t) 0 0 1/2 (iii) Compute c0,c00,c000,s0(t). = < c ,c > 1/2 (iv) Compute t(t), n(t), b(t), κ(t),τ(t). = < v(t) +tw(t) + k,v(t) +tw(t) + k > p = 1 +t2 + 1 + 0 + 0 + 0 6.6.1 answer using orthonormality 6 times p See figures 6.1 and 6.1. = 2 +t2
6.6.2 answer 6.6.3 answer First write, for convenience, To compute t,n,b,κ,τ we first need the following v(t) = costi + sintj and w(t) = −sinti + costj quantities < c0,c0 >,< c0,c00 >,c0 × c00 and < c0 × c00,c0 × c00 >. Note that (v,w,k) is a positively oriented orthonormal triple: all inner products and all cross products amongst these three vectors are known (trivial). < c0,c0 > = < v(t) +tw(t) + k,v(t) +tw(t) + k > c0 2 x0 = 2 +t = y0 z0 < c0,c00 > (t cost)0 = < v(t) +tw(t) + k,−2w(t) −tv(t) > = (t sint)0 t0 = −3t cost −t sint = sint +t cost 1 c0 × c00 = v(t) +tw(t) + k = (v(t) +tw(t) + k) × (−2w(t) −tv(t)) = −2v × w −tv × v − 2tw × w −t2w × v − 2k × w −tkv c00 = −2k −t0 − 2t0 +t2k2v −tw x00 2 = y00 = 2v −tw + (−2 +t )k z00 (cost −t sint)0 = (sint +t cost)0 < c0 × c00,c0 × c00 > 10 = < 2v −tw + (−2 +t2)k,2v −tw + (−2 +t2)k > −2sint −t cost = 4 +t2 + (−2 +t2)2 = 2cost −t sint 0 = 8 − 3t2 +t4 = −2w(t) −tv(t) Now we are ready to compute t,n,b,κ,τ. c000 x000 = y000 t z000 c0 = (−2sint −t cost)0 ||c0|| = (2cost −t sint)0 v +tw + k = √ 0 2 +t2 (−3cost +t sint) 2cost +t sint 1 = (−3sint −t cost) = 2sint −t cost 8 − 3t2 +t4 0 2t2 − 2 = −3v(t) −tw(t) see expressions for v,w
c jbquig-UCD September 23, 2009 6.7. QUESTION AND ANSWER, VIVIANI’S CURVE 133
The unit binormal vector is (i) Prove that the following is a parametrization of Γ.
b x(t) cos(2t) + 1 c0 × c00 = c = y(t) = sin(2t) ||c0 × c00|| z(t) 2sin(t) 2v −tw + (−2 +t2)k = 3 8 − 3t2 +t4 (ii) Roughly sketch Γ ⊂ S ∩C ⊂ R showing the curve 2cost +t sint and both surfaces. 1 = 2sint −t cost (iii) Compute c0, c00, c000 and < c0(t),c0(t) >. 8 − 3t2 +t4 2t2 − 2 (iv) Compute see expressions for v,w c0 × c00, < c0 × c00,c0 × c00 >, and < c0 × c00,c000 >. The unit normal is (v) Compute t(t), n(t), b(t), κ(t),τ(t). n = b × t 6.7.1 (d0 × d00) × d0 = We must prove that ||d0 × d00|| · ||d0|| 2 2 2 2 2 (2v −tw + (t2 − 2)k) × (v +tw + k) x (t) + y (t) + z (t) = 4 and (x(t) − 1) + y (t) = √ √ 2 4 2 8 − 3t +t · 2 +t but (2t − (−t))v × w(−t − (t2 − 2)t)w × k(t2 − 2 − 2)k × v = √ √ 2 2 2 8 − 3t2 +t4 · 2 +t2 x (t) + y (t) + z (t) since v × v = w × w = k × k = 0 (cos(2t) + 1)2 + sin2(2t) + 4sin2(t) and v × w = −w × v etc. = (cos2(2t) + sin2(2t)) + 2cos(2t) + 1 + 4sin2(t) (3t)k + (t −t2)v + (t2 − 4)w 2 = √ √ = 2 + 2cos(2t) + 4sin (t) 2 4 2 8 − 3t +t · 2 +t = 2 + 2(2cos2(t) − 1) + 4sin2(t) 2 2 (t −t )cost − (t − 4)sint 2 2 1 = 4(cos (t) + sin (t)) = √ √ (t −t2)sint + (t2 − 4)cost 2 4 2 8 − 3t +t · 2 +t 3t = 4 and
κ 2 2 2 ||c0 × c00|| (x(t) − 1) (+y (t) + z (t) = 2 2 ||c0||3 cos (2t) + sin (2t) < c0 × c00,c0 × c00 >1/2 = 1 = ||c0||3 √ 8 − 3t2 +t4 6.7.2 answer = 2 3/2 ||2 +t || See figures(2.4)(2.4).
τ 6.7.3 answer < c0 × c00,c000 > = < c0 × c00,c0 × c00 > −2sin2t −4cos2t < 2v −tw + (−2 +t2)k,−3v(t) −tw(t) > 0 00 = c (t) = 2cos2t , c (t) = −4sin2t 8 − 3t2 +t4 2cost −2sint −6 +t2 = and 8 − 3t2 +t4 8sin2t 000 c (t) = −8cos2t 6.7 question and answer, −2cost Viviani’s curve < c0(t),c0(t) > 3 Let S ⊂ R be the spherical surface of radius 2, = 4sin2 2t + 4cos2t + 4cos2 t x2 + y2 + z2 = 4. Let C ⊂ R3 be the cylindrical surface = 4 + 4cos2 t (x − 1)2 + y2 = 1 and Γ = S ∩C be the curve of 2 intersection, known as Viviani’s curve. = 4(1 + cos t)
September 23, 2009 c jbquig-UCD 134 CHAPTER 6. ANSWERS TO QUESTIONS IN CHAPTER2
6.7.4 answer +(8)(−2cos(t))
0 00 c (t) × c (t) = 64(2cos3(t) + cos(t))(1 − cos2(t)) = + (− 3(t))(− 2(t) + ) 64 cos 2cos 1 i j k −16cos(t) −2sin(2t) 2cos(2t) 2cos(t)
−4cos(2t) −4sin(2t) −2sin(t) = = 5 3 −128cos (t) + 64cos (t) + 64cos(t)1 i j k +128cos5(t) − 64cos3(t) 4 −sin(2t) cos(2t) cos(t)
−2cos(2t) −2sin(2t) −sin(t) −16cos(t)
= = −cos(2t)sin(t) + 2sin(2t)cos(t) 48cos(t) 4 −2cos(2t)cos(t) − sin(2t)sin(t) 2 6.7.5 answer = 2 2 −(2cos (t) − 1)sin(t) + 4sin(t)cos (t) t(t) 4 −2(2cos2(t) − 1)cos(t) − 2sin2(t)cos(t) = c0(t)/ < c0(t),c0(t) >1/2 2 −sin2t 1 = = p cos2t 1 + cos2(t) (2cos2(t) + 1)sin(t) cost 2 2 4 −2(2cos (t) − 1)cos(t) − 2(1 − cos (t))cos(t) It is easiest to next compute 2 b(t) = c0(t) × c00(t) = 4(2cos2(t) + 1)sin(t) < c0(t) × c00(t),c0(t) × c00(t) >1/2 3 −8cos (t) 4(2cos2(t) + 1)sin(t) 8 1 3 = p −8cos (t) 4 5 + 3cos3(t) Next 8 2 0 00 0 00 (2cos (t) + 1)sin(t) < c × c ,c × c > 1 3 = p −2cos (t) 2 2 2 5 + 3cos3(t) = 4(2cos2(t) + 1)sin(t) + − 8cos3(t) + − 8 2 Next we compute n as b × t = 16 (4cos4(t) + 4cos2(t) + 1)(1 − cos2(t)) + 4cos6(t) + 4 n(t) 6 4 2 = 16 0 · cos (t) + 0 · cos (t) + 3cos (t) + 5 = b(t) × t(t) = 16(3cos2(t) + 5) 1 = p p 5 + 3cos2(t) 1 + cos2(t) Next i j k 0 00 000 < c × c ,c > det (2cos2(t) + 1)sin(t) −2cos3(t) 2 −sin2t cos2t cost = Thus 4(2cos2(t) + 1)sin(t)(8sin(2t)) q q 5 + 3cos2(t) 1 + cos2(t) n(t) +(−8cos3(t))(−8cos(2t)) +(8)(−2cos(t)) i j k = det (2cos2(t) + 1)sin(t) −2cos3(t) 2 −sin2t cos2t cost = −2cos4(t) − 2cos(2t) 4(2cos2(t) + 1)sin(t)(16sin(t)cos(t)) = −(2cos2(t) + 1)sin(t)cos(t) − 2sin2(t) 3 2 +(−8cos (t))(−16cos (t) + 8) (2cos2(t) + 1)sin(t)cos(2t) − 2cos3(t)sin(2t)
c jbquig-UCD September 23, 2009 6.7. QUESTION AND ANSWER, VIVIANI’S CURVE 135
−2cos4(t) − 2(1 − 2cos2(t) < c0(t) × c00(t),c0(t) × c00(t) >1/2 3 = = −2cos t sin(t) − 5sin(t)cos(t) < c0(t),c0(t) >3/2 (2cos2(t) + 1)(2cos2(t) − 1)sin(t) − 4cos4(t)sin(t) p 4 5 + 3cos2(t) 2 2 = −2(1 − cos (t)) 8(1 + cos2(t))3/2 2 = −sin(t)cos(t)(5 + 2cos (t)) p 5 + 3cos2(t) −sin(t) = 2(1 + cos2(t))3/2 Thus −2(1 − cos2(t))2 Finally the torsion (scalar) is p 2 p 2 5 + 3cos (t) 1 + cos (t) − (t) (t)( + 2(t)) τ(t) sin cos 5 2cos n(t) = p p (6.1) 0 00 000 5 + 3cos2(t) 1 + cos2(t) < c (t) × c (t),c (t) > = 0 00 0 00 −sin(t) < c (t) × c (t),c (t) × c (t) p p 5 + 3cos2(t) 1 + cos2(t) 48cos(t) = 16(5 + 3cos2(t)) Next we compute the curvature (scalar) 3cos(t) = κ(t) (5 + 3cos2(t))
September 23, 2009 c jbquig-UCD 136 CHAPTER 6. ANSWERS TO QUESTIONS IN CHAPTER2
c jbquig-UCD September 23, 2009 Chapter 7
Answers to questions in chapter3
There are no questions or solutions in chapter(3). See the questions at the end of chapter(4).
137 138 CHAPTER 7. ANSWERS TO QUESTIONS IN CHAPTER3
c jbquig-UCD September 23, 2009 Chapter 8
Answers to questions in chapter(4)
These answers were all generated automatically by ’maple’.
8.1 question and answer p The cone C ⊂ R3 has implicit equation x2 +y2 −z2 = 0, explicit equation z = ± x2 + y2 and parametrization.
scos(r) r f : R × [0,2π] 3 C ⊂ R3 7→ ∈ ssin(r) s s
Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that the cone is flat. N.B This surface can be studied using either (or both) of the parametric and explicit equations.
8.1.1 answer See http://mathsci/courses/math40060/maple/answers/chap04/gc cone.html and http://mathsci/courses/math40060/maple/answers/chap04/explicit method/explct gc cone.html
8.2 question and answer p The cylinder C ⊂ R3 has implicit equation x2 +y2 = 0, no explicit equation z = ± x2 + y2 and parametriza- tion. acos(t) 3 f : R × [0,2π] 3 C ⊂ R 7→ r,s ∈ asin(t) s Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that the cylinder is flat.
8.2.1 answer See http://mathsci/courses/math40060/maple/answers/chap04/gc cylinder.html
139 140 CHAPTER 8. ANSWERS TO QUESTIONS IN CHAPTER(??)
8.3 question and answer p The double sheeted hyperbola Σ ⊂ R3 has implicit equation x2 +y2 −z2 = −1, explicit equation z = ± 1 + x2 + y2 and parametrization. sinh(s)cos(r) r f : R × [0,2π] 3 Σ ⊂ R3 7→ ∈ sinh(s)sin(r) s cosh(s) Compute g,h,e and L the mean and Gaussian curvatures. Prove that the d-s-h Σ has everywhere positive curvature. N.B This surface can be studied using either (or both) of the parametric and explicit equations.
8.3.1 answer See http://mathsci/courses/math40060/maple/answers/chap04/gc ds hyperboloid.html
8.4 question and answer
The monkey saddle surface Σ ⊂ R3 has implicit equation x3 − 3xy2 − z = 0, explicit equation z = x3 − 3xy2 and parametrization. r r f : R × R 3 Σ ⊂ R3 7→ ∈ s s r3 − 3rs2 or scos(r) r f : [0,2π] × [0,∞) 3 Σ ⊂ R3 7→ ∈ ssin(r) s s3 cos(3r) Compute g,h,e and L the mean and Gaussian curvatures. Prove that the monkey saddle has everywhere negative curvature. N.B This surface can be studied using either (or both) of the parametric and explicit equations.
8.4.1 answer See http://mathsci/courses/math40060/maple/answers/chap04/gc monkey saddle.html and http://mathsci/courses/math40060/maple/answers/chap04/explicit method/explct gc mnky sddl.html
8.5 question and answer
The sphere S ⊂ R3 has several useful parametrization, one is the Mercator projection used in geograpgy and which is conformal i.e. angle preserving sech (s)cos(r) r f : R × [0,2π] 3 S ⊂ R3 7→ ∈ sech (s)sin(r) s tanh(s) Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions using this parametrization. Show that the sphere has constant Gauss curvature and only one principle curvature (with multiplicity 2). Prove that f is conformal, i.e. that g and <> yield the same measure of angle. This is so iff g is a scalar multiple of Id, prove that.
c jbquig-UCD September 23, 2009 8.6. QUESTION AND ANSWER 141
8.5.1 answer See http://mathsci/courses/math40060/maple/answers/chap04/gc mrctr sphere.html
8.6 question and answer
The paraboloid or parabolic bowl B ⊂ R3 has implicit equation x2 + y2 − z = 0, explicit equation z = x2 + y2 and parametrization. r r f : R × R 3 B ⊂ R3 7→ ∈ s s r2 + s Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that the Gauss curvature is positive everywhere. N.B This surface can be studied using either (or both) of the parametric and explicit equations.
8.6.1 answer See http://mathsci/courses/math40060/maple/answers/chap04/gc paraboloid.html and http://mathsci/courses/math40060/maple/answers/chap04/explicit method/explct gc paraboloid.html
8.7 question and answer
The pseudosphere Σ ⊂ R3 (also called the trumpet) is a rotated tractoid curve and has parametrization.
sech (s)cos(r) r f : R × [0,2π] 3 Σ ⊂ R3 7→ ∈ sech (s)sin(r) s s − tanh(s)
Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that the pseudosphere has constant gauss curvature -1. The principle curvatures α and β obey β = −1/α.
8.7.1 answer See http://mathsci/courses/math40060/maple/answers/chap04/gc pseudosphere.html
8.8 question and answer
The right helicoid surfave H ⊂ R3 explicit equation xtanz = y = and parametrization.
scos(r) r f : R × [0,2π] 3 C ⊂ R3 7→ ∈ ssin(r) s r
Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that H is NOT flat.
September 23, 2009 c jbquig-UCD 142 CHAPTER 8. ANSWERS TO QUESTIONS IN CHAPTER(??)
8.8.1 answer See http://mathsci/courses/math40060/maple/answers/chap04/gc rt helicoid.html and http://mathsci/courses/math40060/maple/answers/chap04/explicit method/explct gc right helicoid.html
8.9 question and answer
The saddle surface Σ ⊂ R3 has implicit equation x2 −y2 −z = 0, explicit equation z = x2 −y2 and parametriza- tion. r r f : R × R 3 Σ ⊂ R3 7→ ∈ s s r2 − s2 or scos(r) r f : [0,2π] × [0,∞) 3 Σ ⊂ R3 7→ ∈ ssin(r) s s2 cos(2r) Compute g,h,e and L the mean and Gaussian curvatures. Prove that the saddle has everywhere negative curvature. N.B This surface can be studied using either (or both) of the parametric and explicit equations.
8.9.1 answer See http://mathsci/courses/math40060/maple/answers/chap04/gc saddle.html and http://mathsci/courses/math40060/maple/answers/chap04/explicit method/explct gc sddl.html
8.10 question and answer
The sphere S is the arcytypal surface and has constant positive Gaussian curvature. S ⊂ R3 has implicit p equation x2 + y2 + z2 = a2, explicit equation z = ± a2 − x2 − y2 and parametrization.
asin(s)cos(r) r f : R × [0,2π] 3 S ⊂ R3 7→ ∈ asin(s)sin(r) s acos(s)
Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that the the Gaussian curvature is constant at 1/a2. The principle curvatures are equal. Any direction is principal.
8.10.1 answer See http://mathsci/courses/math40060/maple/answers/chap04/gc sphere.html and http://mathsci/courses/math40060/maple/answers/chap04/explicit method/explct gc sphere.html
c jbquig-UCD September 23, 2009 8.11. QUESTION AND ANSWER 143
8.11 question and answer
The single sheeted hyperbola Σ ⊂ R3 has implicit equation x2 + y2 − z2 = 1 and parametrization.
cosh(s)cos(r) r f : R × [0,2π] 3 Σ ⊂ R3 7→ ∈ cosh(s)sin(r) s sinh(s)
Compute g,h,e and L the mean and Gaussian curvatures. Prove that the s-s-h Σ has everywhere negative curvature.
8.11.1 answer See http://mathsci/courses/math40060/maple/answers/chap04/gc ss hyperboloid.html
8.12 question and answer
The tangent developable helicoid surface D ⊂ R3 has parametrization.
cos(r) − ssin(r) r f : R × [0,2π] 3 D ⊂ R3 7→ ∈ sin(r) + scos(r) s r + s
Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that D IS a flat surface (the right helicoid is not flat).
8.12.1 answer See http://mathsci/courses/math40060/maple/answers/chap04/gc tngt helicoid.html
8.13 question and answer p The Torus T = T 2(a,b) ⊂ R3 (0¡a¡b) is the surface with implicit equation ( x2 + y2 − b)2 + z2 = a2 and parametrization.
(b − asin(s))cos(r) r f : [0,2π] × [0,2π] 3 T ⊂ R3 7→ ∈ (b − asin(s)sin(r) s acos(s)
Compute g,h,e and L the mean and Gaussian curvatures and principle curvatures and directions. Prove that the Gaussian curvature is positive and negative on two regions of T separated by two circles on both of which the Gaussian curvature is zero.
8.13.1 answer See http://mathsci/courses/math40060/maple/answers/chap04/gc torus.html
September 23, 2009 c jbquig-UCD