Lecture Note on Elementary Differential Geometry

Home , Holonomic basis, Torsion of a curve

Ling-Wei Luo* Institute of Physics, Academia Sinica

July 20, 2019

Abstract This is a note based on a course of elementary differential geometry as I gave the lectures in the NCTU-Yau Journal Club: Interplay of Physics and Geometry at Department of Electrophysics in National Chiao Tung University (NCTU) in Spring semester 2017. The contents of remarks, supplements and examples are highlighted in the red, green and blue frame boxes respectively. The supplements can be omitted at first reading. The basic knowledge of the differential forms can be found in the lecture notes given by Dr. Sheng-Hong Lai (NCTU) and Prof. Jen-Chi Lee (NCTU) on the website. The website address of Interplay of Physics and Geometry is http: //web.it.nctu.edu.tw/~string/journalclub.htm or http://web.it.nctu. edu.tw/~string/ipg/.

Contents

1 Curve on E2 ...... 1

2 Curve in E3 ...... 6

3 Surface theory in E3 ...... 9

4 Cartan’s moving frame and exterior differentiation methods ...... 31

1 Curve on E2

We define n-dimensional Euclidean space En as a n-dimensional real space Rn equipped a dot product defined n-dimensional vector space.

Tangent vector In 2-dimensional Euclidean space, an( E2 plane,) we parametrize a curve p(t) = x(t), y(t) by one parameter t with respect to a reference point o with a fixed Cartesian coordinate frame. The( velocity) vector at point p is given by p˙ (t) = x˙(t), y˙(t) with the norm Figure 1: A curve. √ √ |p˙ (t)| = p˙ · p˙ = x˙ 2 +y ˙2 , (1)

*Electronic address: [email protected]

1 where x˙ := dx/dt. The arc length s in the interval [a, b] can be calculated by ∫ ∫ ∫ ∫ √ b √ b s = ds = (dx)2 + (dy)2 = x˙ 2 +y ˙2 dt = |p˙ (t)| dt . (2) a a The arc length can be a function of parameter t given by ∫ t s(t) = |p˙ (t′)| dt′ . (3) a From the fundamental theorem of calculus, we have

ds ≠ 0 =⇒ s˙(t) = |p˙ (t)| > 0 . (4) dt

According to the inverse( function) theorem, we have t = t(s). One can parametrize the( curve by arc) length s as p(s) = x(s), y(s) . The corresponding velocity vector should be p′(s) = x′(s), y′(s) , where we have x′ := dx/ds. We can rewrite the derivatives of x and y with respect to s as   dx dx dt dt x′ = = =x ˙ , ds dt ds ds (5)  dt y′ =y ˙ . ds Thus, the norm of the velocity vector parametrized by s can be calculated as √ √ dt dt ds dt |p′(s)| = x′2 + y′2 = x˙ 2 +y ˙2 = |p˙ | = = 1 , (6) ds ds dt ds which implies that the velocity vector p′(s) is a unit vector. We can define a unit tangent vector as a velocity vector parametrized by s

′ T ≡ e1 := p (s) . (7)

′ ′ Normal vector Due to e1 · e1 = p · p = 1, we have ′ · · ′ ⇒ ′ · ⇒ ′ ⊥ e1 e1 + e1 e1 = 0 = e1 e1 = 0 = e1 e1, (8) ′ it indicates that e1 is a normal vector. The principle normal vector is defined by

′ ≡ e1 N e2 := | ′ | (9) e1 | ′ | as a unit normal vector at p(s). The curvature of a curve p(s) is given by κ(s) = e1(s) > 0, which ′ ′′ can be realized as a norm of the acceleration vector a := e1 = p . Therefore, we have a relation

′ e1 = κ(s)e2 . (10)

Remark. If a vector V is an unit vector, |V | = 1, the corresponding derivative vector would be perpendicular to itself, i.e.

V ′ ⊥ V. (11)

2 Osculating plane The plane is spanned by the vectors e1 and e2 is called osculating plane.

Newton’s second law In classical physics, we have a momentum vector p = mT = mp′ with mass m. The force F is defined by Newton’s second law dp dT F = = m = ma = mp′′ (12) ds ds with respect to parameter s.

Frame A set of vector e1, e2 equipped with a point p calls frame. In such of case, a frame at p is denoted by (p; e1, e2).

Frenet-Serret formula in 2D From the orthonormality condition ei · ej = δij (i, j = 1, 2), we have

′ · · ′ ei ej + ei ej = 0 (13a) ⇒ ′ · · ′ · ′ = e1 e2 + e1 e2 = κ + e1 e2 = 0 (13b) ⇒ · ′ − ′ − = e1 e2 = κ (e2 has component κ along e1 direction) (13c) ⇒ ′ − = e2 = κe1 . (13d)

As a result, we have the following relations      ′ ′ ( ) p = +e1 p 1 0 ′  ′    e1 e = +κe2 =⇒ e  =  0 κ (14)  1 1 e ′ − ′ − 2 e2 = κe1 e2 κ 0 called Frenet-Serret formula.

Example (Circle in E2). A circle with radius r can be parametrized by p(t) = (r cos t, r sin t) with 0 ≤ t ≤ 2π.

Figure 2: A circle. The tangent vector is

p˙ (t) = (−r sin t, r cos t) (15)

with norm √ |p˙ | = r2 sin2 t, r2 cos2 t) := r . (16)

3 The arc length s(t) is ∫ ∫ t t s(t) = |p(t′)|dt′ = r dt′ = rt . (17) 0 0 Therefore, the circumference is ∫ ∫ 2π 2π L = |p(t′)|dt′ = r dt′ = 2πr . (18) 0 0 By t = s/r, the circle p(s) and its tangent vector are ( ) s s p(s) = r cos , r sin (19a) r r

and ( ) s s p′(s) = − sin , cos = e = T (19b) r r 1

respectively. From (19b), we have ( ) 1 s 1 s e′ (s) = − cos , − sin . (20) 1 r r r r

The curvature κ can be obtained by √ 1 s 1 s 1 κ = |e′ | = cos2 + sin2 = , (21) 1 r2 r r2 r r which is the inverse of the constant radius r. The normal vector can be calculated by ( ) ( ) ′ e1 − 1 s − 1 s − s s e2 = | ′ | = r cos sin = cos , sin . (22) e1 r r r r r r

Gauss map Gauss map G is a mapping which globally send all the points p of curve to a unit circle S1 (a Gauss circle) centered at c and send the corresponding normal vector e2 to a radius vector from c pointing to S1, which is shown as Fig. 3. Therefore, e2 can be represented as a point on S1. Let’s consider two normal vectors e2(s) and ′ e2(s ) with respect to two infinitesimal points ′ ′ p(s) and p(s ), where s = s + ∆s is infinitesi- Figure 3: The Gauss map G. ′ mal close to s. We can expand e2(s ) at s: ′ e2(s ) = e2(s + ∆s) ≈ ′ e2(s) + e2(s)∆s

= e2(s) + (−κ(s)e1(s))∆s

= e2(s) + (−κ(s)∆s)e1(s) , (23)

4 which is the parametrization of a point under the Gauss map. Thus, we know the distance between ′ two infinitesimal point e2(s) and e2(s ) on Gauss circle given by ′ |e2(s ) − e2(s)| = |∆e2| = κ(s)∆s . (24) And we also have ( ) ∆p ≡ p(s′) − p(s) ≈ p(s) + p′(s)∆s − p(s) = p′(s)∆s =⇒ |p(s′) − p(s)| = |∆p| = ∆s . (25) Therefore, in the local region, the ratio of the length between two points on the Gauss circle and curve, i.e., |∆e2|/|∆p| can be calculate by |e (s′) − e (s)| |∆e | κ(s)∆s 2 2 = 2 = = κ(s) , (26) |p(s′) − p(s)| |∆p| ∆s which measure the curvature of a curve, κ(s), at the neighborhood of a local point p. According to the example of circle, we assume a vector q = p+re2 = p+(1/κ)e2. The derivative of q is 1 1 q′ = p′ + e′ = e + (−κe ) = 0 , (27) κ 2 1 κ 1 which means that q is fixed, i.e., q is the center of the osculating circle with radius 1/κ. By considering the Gauss map of a circle. The radius vector should be e2 and the center c of Gauss circle corresponds to the point q of the osculating circle which is the circle itself. Thus, the Gauss circle can be imaged by rescaling the radius of osculating circle to unity. ( ) Example (Curvature of ellipse). An ellipse is described by p(t) = x(t), y(t) with the parametrization of the coordinates x(t) = a cos t and y(t) = b sin t (a > b > 0), i.e., x2 y2 + = cos2 t + sin2 t = 1 . (28) a2 b2

Figure 4: An ellipse. The tangent vector is ( ) p˙ (t) = − a sin t, b cos t (29)

By changing the parameter to s, we have to calculate ds/dt first: √ ds dt 1 1 = |p˙ | = a2 sin2 t + b2 cos2 t =⇒ = √ = . (30) dt ds a2 sin2 t + b2 cos2 t s˙ Therefore, the tangent vector parametrized by s is obtained by

′ dp dt e1 = p = ( dt ds ) ( ) −a sin t b cos t −a sin t b cos t = √ , √ = , , (31) a2 sin2 t + b2 cos2 t a2 sin2 t + b2 cos2 t s˙ s˙

5 Subsequently, we have ( ) −b cos t −a sin t e = , . (32) 2 s˙ s˙

However de de dt e′ = 1 = 1 = κe . (33) 1 ds dt ds 2 As a result, the curvature is ab κ(t) = . (34) (a2 sin2 t + b2 cos2 t)3/2

If we consider the particular case of a = b, an ellipse reduce to a circle with curvature κ = 1/a.

2 Curve in E3 ( ) In E3, a curve is parametrized as p(t) = x(t), y(t), z(t) and we have to look for an orthonormal ′ frame at p denoted by (p; e1, e2, e3) The vector e1 = p is uniquely defined by the same way. Due ′ ⊥ ′ ′ to e1 e1, vector e1 should be proportional to e2 or e3. Now we can fix e1 = κe2 as the previous section.

Binormal vector Now we define a unit vector orthogonal to T and N called binormal vector

B := T ∧ N

≡ e1 ∧ e2 := e3 , (35) where ∧ is the exterior product or wedge product.

Remark. In 3-dimensional space, the exterior product ∧ is the same to the usual cross product × of two vectors.

By orthonormality condition ei · ej = δij (i, j = 1, 2, 3), we have ′ · · ′ ei ej + ei ej = 0 , (36) which implies: ′ ⊥ ′ (i) If i = j, we have ei ei, e2 should be the combination of e1 and e3. (ii) If i ≠ j, we have { ′ · · ′ · · ′ · ′ 0 = e1 e2 + e1 e2 = (κe2) e2 + e1 e2 = κ + e1 e2 (i = 1, j = 2) , (37a) ′ · · ′ · · ′ · ′ 0 = e1 e3 + e1 e3 = (κe2) e3 + e1 e3 = 0 + e1 e3 (i = 1, j = 3) . (37b) Therefore, with the result (37a), we have to assume that ′ − e2 = κ(s)e1 + τ(s)e3 . (38)

By comparing to (13d), it contains an additional term related to e3. For i = 2, j = 3, we obtain ′ · · ′ − · · ′ · ′ 0 = e2 e3 + e2 e3 = ( κe1 + τe3) e3 + e2 e3 = τ + e2 e3 . (39)

6 ′ Due to (i) and (37b), e3 should be perpendicular to e1 and e3. As a result, we obtain the unique solution that ′ − e3 = τe2 , (40) where τ(s) is called torsion of a curve p(s). The geometric meaning of torsion is that it make the point of the curve leave for the osculating plane spanned by e1 and e2.

Remark. Apparently, the torsion of a curve is always related to the binormal vector B ≡ e3.

Frenet-Serret formula in 3D As a result, we have Frenet-Serret formula:      ′ ′   p = +e1 p 1 0 0  ′     e1 e = +κe e′   0 κ 0    1 2 ⇒  1      ′ = ′ = e2 . (41) e = −κe1 +τe3 e   −κ 0 τ   2 2 e ′ − ′ − 3 e3 = τe2 e3 0 τ 0

∧ ′ − − Remark. If one defines B := N T , then one should assume e2 = κ(s)e1 τ(s)e3 and obtain ′ e3 = +τe2.

Parametrization of a curve in a neighborhood of s0 One can do the Taylor expansion of p(s) at s0.

• First order:

dp p(s) ≈ p(s0) + (s − s0) = p(s0) + e1(s0)(s − s0) . (42) ds s=s0

• Second order: 1 p(s) ≈ p(s ) + p′(s )(s − s ) + p′′(s )(s − s )2 0 0 0 2! 0 0 1 = p(s ) + e (s )(s − s ) + κ(s )e (s )(s − s )2 . (43) 0 1 0 0 2 0 2 0 0 • Third order: 1 1 p(s) ≈ p(s ) + p′(s )(s − s ) + p′′(s )(s − s )2 + p′′′(s )(s − s )3 0 0 0 2! 0 0 3! | {z 0} 0 ′′′ ′′ ′ ′ ′ ′ − p =(p ) =κ e2+κe2=κ e2+κ( κe1+τe3) 1 = p(s ) + e (s )(s − s ) + κ(s )e (s )(s − s )2 0 1 0 0 2 0 2 0 0 1( ) + − κ2(s )e (s ) + κ′(s )e (s ) + κ(s )τ(s )e (s ) (s − s )3 . (44) 6 0 1 0 0 2 0 | 0 {z0 3 0} 0 leading term We only consider the leading term in the third order expansion, then we have 1 1( ) p(s) ≈ p(s ) + e (s )(s − s ) + κ(s)e (s )(s − s )2 + κ(s )τ(s )e (s ) (s − s )3 . 0 1 0 0 2 2 0 0 6 0 0 3 0 0 (45)

7 ( ) Example (Helix in E3). A helix is parametrized as p = x(t), y(y), z(t) with  x(t) = a cos t , y(t) = a sin t , (46)  z(t) = bt .

The tangent vector and the corresponding norm are ( ) ( ) p˙ = x,˙ y,˙ z˙ = − a sin t, a cos t, b (47) and √ √ |p˙ | = a2 sin2 t + a2 cos2 t + b2 = a2 + b2 =s ˙ . (48)

The relation of s and t can be obtained by ∫ ∫ t ds t √ s ′ 2 2 ′ ⇒ s(t) = ′ dt = a + b dt := ct = t = . (49) 0 dt 0 c

Figure 5: A helix. Subsequently, we have tangent vector ( ) a s a s b e = p′ = − sin , cos , (50) 1 c c c c c and ( ) a s a s e′ = p′′ = − cos , − sin , 0 . (51) 1 c2 c c2 c

So the curvature is a a κ = |e′ | = = . (52) 1 c2 a2 + b2 and the normal vector can be obtained by ( ) s s e′ = κe =⇒ e = − cos , − sin , 0 . (53) 1 2 2 c c

8 Finally, we have binormal vector ( ) b s b s a e = e ∧ e = sin , − cos , . (54) 3 1 2 c c c c c

Due to ( ) b s b s e′ = cos , sin , 0 , (55) 3 c2 c c2 c

we can calculate the torsion of p(s) form (40):

b b τ = = . (56) c2 a2 + b2

3 Surface theory in E3

We consider a 2-dimensional( surface M in E)3, we parametrize the surface by two variables u and v written as p(u, v) = x(u, v), y(u, v), z(u, v) .

Remark. If the point p(u, v) moves along u direction, i.e., parametrized by u only, we call the trajectory u-curve. The infinitesimal vector along u is

∂p(u, v) ∆p| = p(u + ∆u, v) − p(u, v) ≈ p(u, v) + ∆u − p(u, v) = p ∆u . (57) u ∂u u Similarly, we have v-curve along v direction and | ≈ ∆p v pv∆v . (58)

Therefore, we have ≈ ∆p pu∆u + pv∆v . (59)

Tangent vector The differential of p is dp = (dx, dy, dz) (60) with   ∂x ∂x dx = du + dv = x du + x dv ,  ∂u ∂v u v  ∂y ∂y dy = du + dv = yudu + yvdv , (61)  ∂u ∂v   ∂z ∂z dz = du + dv = zudu + zvdv , ∂u ∂v Figure 6: A surface. where xu := ∂x/∂u. Therefore, we can write dp as ∂p ∂p dp = du + dv := p du + p dv , (62) ∂u ∂v u v

9 where { p := (xu, yu, zu) u (63) pv := (xv, yv, zv) are called velocity vectors along u and v respectively.

Remark. The vector p in E3 in the Cartesian coordinate system can be written as

a p = x i + y j + z k := x δa (a = 1, 2, 3) , (64)

3 where {δa} is a fixed reference frame of E . So that we have differential

a a dp = (dx )δa + x (dδa) . (65)

Because δa is fixed, i.e. dδa = 0, it leads to the differential of p

a dp = (dx )δa = (dx, dy, dz) . (66)

The general situation for non-fixed frame in space Mn will be discussed in the Sec. 4 of moving frame.

M Tangent space We call a space spanned by pu and pv at point p a tangent space denoted by Tp .

Supplement (Tangent bundle). In tangent space with dimension 2, a vector V has a generalized coordinate transformation GL(2; R), which is uei = uei(u) and gives the transformation for vector

i ‹j e V = V pi = V pj . (67a)

The transformation of the basis and components are given by   ∂ui  pe (ue) = p (u)(Pushforward) , (67b) j ∂uej i  ∂ui V i(u) = V‹j(ue) (Pullback) , (67c) ∂uej where ∂ui (J)i := (68) j ∂uej is an element of the Jacobian matrix J of the general linear transformation GL(2; R). The map pushforward (pullback) means that the covariant (contravariant) quantities expressed in new (old) coordinate system under the generalized coordinate transformation from old (new) coordinate system. We can collect all pairs of the points p on M and their corresponding tangent space TpM.A tangent bundle TM is defined by the collection of TpM, i.e., ∪ TM = TpM . (69) p∈M

A tangent bundle TM is a vector bundle denoted by (E, M, π), which is a special fibre bundle with

10 • base space B: M;

• standard (typical) fibre F over p (an object defined at p): TpM;

• total space E: a collection of all TpM; • bundle projection π (an element u of bundle is projected by the fibre to the corresponding point p): π(u) = p for u ∈ TM; • structure group G: GL(2; R); i • transition function t j: Jacobian matrix J of GL(2; R), −1 and we call Ep = π (p) the fibre of E over point p.

First fundamental (quadratic) form we define

I := dp · dp (70a) = · dudu + 2 · dudv + · dvdv |pu{zpu} |pu{zp}v |pv{zp}v E F G = E dudu + 2F dudv + G dudv (70b) ( )( ) ( ) EF du = du dv . (70c) FG dv called the first fundamental form or metric tensor of surface M, which is a symmetric quadratic form rather than an exterior 2-form.

Remark. In the case of F = 0, the first fundamental form is

I = E dudu + G dvdv (71a) ( )( ) ( ) E 0 du = du dv . (71b) 0 G dv

In such case, we call (u, v) an isothermal coordinates if E = G. Therefore, the component of metric is

· | |2 gij = E δij with E = pi pi = pi > 0 , (72)

and we say that gij is conformally equivalent to δij, which preserved the angle between any two vectors. Because δij gives the flat space, we say that gij is conformally flat.

We consider a curve on the surface, that means( u and v)should be parametrized by one variable t, i.e., u = u(t) and v = v(t). The curve p(t) = p u(t), v(t) . The tangent vector is obtained by

∂p du ∂p dv p˙ = + = p u˙ + p v˙ , (73) ∂u dt ∂v dt u v and the corresponding norm is √ |p˙ | = Eu˙ 2 + 2F u˙v˙ + Gv˙ 2 . (74)

11 We would like to calculate the arc length of a curve by ∫ ∫ s = ds = |p˙ | dt (75) ∫ √ = | Edu2 + 2F{z dudv + Gdv}2 (76) √ ds2 ∫ √ ∫ ′2 ′ ′ ′2 | ′| = | Eu + 2F{z u v + Gv } ds = p ds . (77) 1 Therefore, we have

|p′| = 1 . (78)

We would alwalys write the first fundamental form with u = u1 and v = u2 as

2 i j I ≡ ds ≡ g = gijdu du (i, j = 1, 2) , (79) where ( ) EF g = p · p −→ (80) ij i j FG is the metric tensor represented as a 2 × 2 matrix on the surface M. The inverse of gij is defined by

ki k g gij = δj . (81)

Remark. The first fundamental form describes the distance of two points on the surface M, which gives the intrinsic structure of M.

Supplement (Induced metric). We can regard p as a set of functions defined on the surface M, the differential of p is actually an infinitesimal tangent vector laid on M ( ) ∂p ∂p ∂ ∂ dp = du + dv = du + dv p = (dui∂ )p , (82) ∂u ∂v ∂u ∂u i

i which can be identified as differential operator du ∂i act on a set of functions p. In abbreviated notation, we have

i i ⊗ −→ dp = du ∂i = du ∂i := ϑ (pi ∂i) , (83)

where we use ∂i to abbreviate the basis vector pi = ∂ip, i.e., the vector ∂i should be regarded as a differential operator act on some functions. Here we call ϑ = dp the canonical 1-form or soldering k form, which is a vector-valued 1-form (1-form carries a vector). For any vector V = V ∂k on M, we apply ϑ on V and obtain

k i k i i ϑ(V ) = V du (∂k)∂i = V δk∂i = V ∂i = V . (84)

It is apparent that ϑ is an identity map for a vector.

12 Now we will define the general inner product for two basis vectors ∂i and ∂j instead of dot product as

g(∂i, ∂j) := gij . (85)

i j Therefore, for any two vectors V = V ∂i and W = W ∂j on M, we have

i j i j i j j i g(V , W ) = g(V ∂i,W ∂j) = V W g(∂i, ∂j) = V W gij = VjW = V Wi . (86) In general we also have

g¯(∂a, ∂b) :=g ¯ab , (87) where ∂ ∂ := (a, b = 1, 2, 3) . (88) a ∂xa a We call {x } the Gauss normal coordinates or synchronous coordinates if g¯i3 = 0 and g¯33 = 1, i.e., ∂3 is an unit normal vector of M, which is proportional to n. Furthermore, we can define a metric tensor

a b a b 2 g¯ =g ¯abdx dx = δabdx dx = ds (89)

3 as a line inteval of E , and it is clear that g¯i3 is one of the component of g¯. If we assume that 1 2 3 M ∂ x = x = x(u, v), x = y = y(u, v) and x = z = z(u, v) on . We have basis vectors ∂ui ∂ spanned by ∂xa as   ∂ ∂x ∂ ∂y ∂ ∂z ∂ ∂xa ∂  = + + = a , a ∂u ∂u ∂x ∂u ∂y ∂u ∂z ∂u ∂x ∂ ∂x ∂ a ∂ =⇒ = := h i , (90)  ∂ ∂x ∂ ∂y ∂ ∂z ∂ ∂xa ∂ ∂ui ∂ui ∂xa ∂xa  = + + = , ∂v ∂v ∂x ∂v ∂y ∂v ∂z ∂v ∂xa

a 3 where h i is a projection operator of the vector in E and i, j = 1, 2. The component of metric tensor g of M can be given by ( ) ( ) ( ) ∂ ∂ ∂ ∂ ∂ ∂ g = g , ≡ g¯ ha , hb = ha hb g¯ , = ha hb g¯ , (91) ij ∂ui ∂uj i ∂xa j ∂xb i j ∂xa ∂xb i j ab

3 which is the projection of g¯ab of E onto M. We can define an projection operation P of differential dxa in E3 onto M which is called the pullback (a map for contravariant quantities) of 1-form dxa: ∂xa P(dxa) = dui = ha dui , (92) ∂ui i

a i M 2 M i.e., P(dx ) can be spanned by du on . As a consequence, a line inteval ds M = P(¯g) on is obtained by ( ) ( ) a b a b i j a b i j i j P(¯g) = P(¯gabdx dx ) =g ¯ab h ih jdu du = g¯abh ih j du du := gijdu du = g = I . (93)

2 M Therefore, the first fundamental form I = g = ds M of can be regarded as a projection of metric tensor g¯ with

∂xa ∂xb g :=g ¯ ha hb =g ¯ . (94) ij ab i j ab ∂ui ∂uj We called that g is a induced metric obtained by the pullback of g¯.

13 Supplement (Interior product). We define an anti-derivation on exterior differential p-forms ω for a vector X called interior product with respect to X. It sends an exterior p-form to an exterior (p − 1)-form. We consider an 1-forms ω, the interior product of ω with respect to a vector X is

or ιX ω = X⌋ω := ω(X) (95a) i j i j j = X ωjdu (∂i) = X ωjδi = X ωj . (95b)

Therefore, we have

j j j ι∂i (du ) = du (∂i) = δi . (96)

We note that:

• For 0-form f (a scalar), the interior product is vanished ιX f = 0 because of no (−1)-form. 2 • The second action of ιX = 0. It can be shown that by considering an exterior 3-form ω = i j k (1/3!)ωijkdu ∧ du ∧ du , we have vanished second interior product by X ( ) 1 ι2 ω = ι ι ω dui ∧ duj ∧ duk X X X 3! ijk ( ( 1 = ι Xkω dui(∂ )duj ∧ duk X 3! ijk l )) 1 i j k 2 i j k + (−1) du ∧ du (∂l)du + (−1) du ∧ du du (∂l) ( ( )) 1 = ι Xl ω duj ∧ duk − ω dui ∧ duk + ω dui ∧ duj X 3! ljk |{z}ilk |{z}ijl −ω ( ) lik +ωlij 1 = ι Xlω duj ∧ duk X 2 ljk ( ( )) 1 = XmXl ω duj(∂ )duk + (−1)1dujduk(∂ ) 2 ljk m m ( ) symmetricz }| in{l,m 1 = XmXl ω duk − ω duj = XmXl ω duk = 0 . (97) 2 lmk |{zljm} |{zlmk} −ωlmj anti-symmetric in l,m

However, ιY ιX ≠ 0, e.g. the interior product of an exterior 3-form ω by X and Y should m l k be an 1-form Y X ωlmkdu ≠ 0.

Supplement (Isomorphism between tangent and cotangent space). The 1-form dui is defined on ∂ ≡ the cotangent space which is dual to the basis ∂ui ∂i on the tangent space. We can define a linear map ψ : TM −→ T ∗M. For vectors X, Y ∈ TM and α ∈ T ∗M the 1-form corresponding to vector X, then we define

⟨α, Y ⟩ := α(Y ) = ιY α = g(X, Y ) , (98) where the ⟨•, •⟩ with two slots is a kind of inner product defined between the tangent and cotangent space and

α := ψ(X) . (99)

14 Therefore, we can also write the corresponding 1-form α as

α(•) = g(X, •) , (100) which can be recognized by 1 ( ) α := g dui(X)duj + duiduj(X) 2 ij 1 ( ) = g Xk dui(∂ )duj + duiduj(∂ ) 2 ij k k 1 ( ) = g Xk δi duj + duiδj 2 ij k k i = Xidu . (101)

j By choosing vector X = ∂i, we have an 1-form ψ(∂i) = ψijdu , which leads to ⟨ ⟩ ⟨ k ⟩ k gij = g(∂i, ∂j) = ψ(∂i), ∂j = ψik du , ∂j = ψikδj = ψij . (102) Therefore, we have

j ψ(∂i) = gijdu := dui (103)

i ∗ i called the reciprocal basis of du in T M. It is apparent that gij transforms du to its reciprocal basis duj. In addition, we have a linear inverse map ψ−1 : T ∗M −→ TM such that

X = ψ−1(α) := φ(α) . (104)

i i −1 i ij i For α = du , the inverse map of du can be written as ψ (du ) = φ ∂j. If we take α = du and Y = ∂j in (98), then ( ) ( ) i i ik ik ik δj = du (∂j) = g φ ∂k, ∂j = φ g ∂k, ∂j = φ gkj , (105) therefore, φik = gik and

−1 i ij i ψ (du ) = g ∂j := ∂ (106)

∗ is the reciprocal basis of ∂i. We assume that β = ψ(Y ) and define the inner product in T M which is also denoted by g ( ) g(α, β) := g ψ−1(α), ψ−1(β) . (107)

The definition leads to the following relation by choosing α = dui and β = duj ( ) ( ) ( ) i j −1 i −1 j k j ik jl ij g(du , du ) = g ψ (du ), ψ (du ) = g ∂ , ∂ = g g g ∂k, ∂l = g . (108)

i −1 Now we can clearly express a vector X = X ∂i as an inverse map ψ of an 1-form α with the help of (106): X = Xi∂ = Xig ∂j = X ∂j = X ψ−1(duj) = ψ−1(X duj ) = ψ−1(α) . i ij | {zj } j | j{z } (109) a vector! an 1-form!

ij As a result, we conclude that the metric tensor g (not component gij or g ) turns a vector (1-form) ij into a 1-form (vector). The component of metric tensor g (gij) transforms the a vector ∂i (1-form i i du ) to its corresponding reciprocal basis ∂ (dui).

15 Normal vector of the surface We would like to look for an orthonormal frame (p; e1, e2, e3) of M. Under the Gram-Schmit procedure, we can define

pu e1 := | | (110) pu and − · pv (pv e1)e1 e2 := | − · | . (111) pv (pv e1)e1 Therefore, we have

e3 = e1 ∧ e2 := n , (112) which is an unit normal vector of M. The unit normal vector n = n(u, v) can also be obtained by ∧ pu pv n(u, v) = | ∧ | . (113) pu pv The corresponding differential dn is ∂n ∂n dn = du + dv = n du + n dv . (114) ∂u ∂v u v However, we have: ⊥ ⇒ · (i) n pu = n pu = 0, which have the equations of the partial derivative with respect to u and v are { · · ⇒ · − · ∂u : nu pu + n puu = 0 = n puu = nu pu := L, (115a) · · ⇒ · − · ∂v : nv pu + n puv = 0 = n puv = nv pu := M. (115b) ⊥ ⇒ · (ii) n pv = n pv = 0, we obtain { · · ⇒ · − · ∂u : nu pv + n pvu = 0 = n pvu = nu pv := M, (116a) · · ⇒ · − · ∂v : nv pv + n pvv = 0 = n pvv = nv pv := N. (116b)

Second fundamental (quadratic) form According to (62) and (114), we can define a quadratic form

II := −dp · dn (117a) − = (pudu + pvdv)(nudu + nvdv) − · · · · = (pu nu dudu + pu nv dudv + pv nu dudv + pv nv dvdv) = · dudu + · dudv + · dudv + · dvdv n| {zpuu} n| {zpuv} |n {zpvu} |n {zpvv} L M M N = L dudu + 2M dudv + N dudv (117b) ( )( ) ( ) LM du = du dv (117c) MN dv called the second fundamental form of M. We define the second fundamental form as a tensor given by

i j II = bijdu du (118)

16 with the component of matrix form as ( ) LM b = n · p = −p · n −→ . (119) ij ij i j MN

Remark. The second fundamental form describes the shape of M and how the surface M embed- 3 ded in E . It is an extrinsic property of M and we call the component bij the extrinsic curvature.

Now we would like to discuss decomposition formulas of the derivative vector of frame (p; pu, pv, n). We follow the principle:

• Any vector in the space can be spanned by the basis pu, pv and n.

Gauss formulas We take the partial derivative of pu and pv with respect to u and v:   ∂ u v n puu := pu = (Γu) u pu + (Γu) u pv + (Γu) u n  ∂u  u v  = (Γu) u p + (Γu) u p + (p · n)n , (120a)  u v | u{zu }   L  ∂  u v n puv := pu = (Γu) v pu + (Γu) v pv + (Γu) v n  ∂v  u v  = (Γu) v p + (Γu) v p + (p · n)n , (120b)  u v | u{zv } M   ∂ u v n pvu := pv = (Γv) u pu + (Γv) u pv + (Γv) u n  ∂u  u v ·  = (Γv) u pu + (Γv) u pv + (pvu n)n , (120c)  | {z }  M   ∂  p := p = (Γ )u p + (Γ )v p + (Γ )n n  vv v v v u v v v v v  ∂v  = (Γ )u + (Γ )v + ( · ) .  v v pu v v pv |pv{zv n} n (120d) N We call these set of equations the Gauss formulas. We identify the coefficients

c c (Γa) b ≡ Γ ab , (121)

u 1 e.g., (Γn) v = Γ n2, then Gauss formulas (120) can be written as

k n k pij = Γ ijpk + Γ ijn = Γ ijpk + bijn (i, j, k = 1, 2) , (122) where the coefficients are obtained by

Γ = Γl g = Γl · = · kij ij lk | {zijp}l pk pij pk (123) − pij bij n and n · bij = Γ ij = n pij . (124)

We note that Γkij and bij are symmetric in i, j.

17 Remark. The vectors dp and dpi in terms of the differential form are given by

i i dp = (du ∂i)p = du pi (125)

and

j j k k k n dpi = d(∂ip) = du (∂j∂ip) = du (Γ ijpk + bijn) := Γ ipk + bin = Γ ipk + Γ in (126)

k k j j n j n respectively, where Γ i := Γ ijdu is called connection form and bi := bijdu = Γ ijdu = Γ i.

Weingarten formulas Due to n · n = 1, we have { ∂u : nu · n = 0 =⇒ nu ⊥ n , (127a)

∂v : nv · n = 0 =⇒ nv ⊥ n . (127b)

Therefore, nu and nv do not contain the component of n. We can assume that { u v nu = A pu + B pv = (Γn)u pu + (Γn)u pv , (128a) u v nv = C pu + D pv = (Γn)v pu + (Γn)v pv , (128b) · · which is called the Weingarten formulas. We calculate the inner product of nu pu and nu pv: { nu · p = −L = EA + FB FM − GL FL − EM u =⇒ A = and B = . (129) · − − 2 − 2 nu pv = M = FA + GB EG F EG F

Similarly, we have

FN − GM FM − EN C = and D = . (130) EG − F 2 EG − F 2

The Weingarten formulas (128) can be written by

k nj = Γ njpk , (131) where the coefficients can be calculated by ( ) − · · l l ⇒ k ki − ki − k bij = pi nj = pi Γ njpl = gilΓ nj = Γinj = Γ nj = g Γinj = g bij := b j . (132)

As a result, we obtain

− k nj = b jpk . (133)

Remark. The Weingarten formula written in the differential form is given by

− k j − k k dn = ( b jdu )pk = b pk := Γ npk (134)

18 Acceleration (curvature) vector We have an acceleration (curvature) vector p′′(s) parametrized by s, which can be decomposed by tangential and normal parts

′′ ′′ ′′ p = pt + pn := κg + κn , (135) where the tangential part κg = κgt and normal part κn = κnn are called geodesic curvature and normal curvature respectively.

Remark. We can identify p′′ := a the acceleration vector, therefore, (135) can be read as a = ′′ ′′ aT + aN with the tangent acceleration vector aT := pt and normal acceleration vector aN := pn.

According to (11), we have p′′ · p′ = 0. We would like to discuss the geodesic curvature, we take ′′ ′ the inner product of pt with n and p respectively: { p′′ · n := 0 , t ( ) =⇒ p′′ ∝ t := n ∧ p′ , (136) ′′ · ′ ′′ ′′ · ′ ′′ · ′ t pt p := pt + pn p = p p = 0 , ′′ ∧ ′ then we can have pt = κgt = κg(n p ).

Normal curvature of a curve We would like to discuss normal curvature first, and define κn = κnn, so

κ = κ · = ( ′′ − κ ) · = ′′ · − κ · . n n n p g n p n | g{z n} (137) 0 However we also have

p′ · n = 0 =⇒ p′′ · n + p′ · n′ = 0 . (138)

Therefore, κn can be calculated by

′ ′ κn = −p · n − ′ ′ · ′ ′ = (puu + pvv ) (nuu + nvv ) = L u′u′ + 2M u′v′ + N v′v′ . (139)

However p′ = dp/ds and n′ = dn/ds, which leads to

dp dn −dp · dn II κ = − · = = . (140) n ds ds ds2 I

Remark. We have the norm of tangent vector I 1 = |p′| = Eu′u′ + 2F u′v′ + Gv′v′ = . (141) ds2 According to (83)

dp dui p′ = = ∂ = ui ′∂ , (142) ds ds i i

19 we have

′ ′ k l ′ ′ I(p , p ) = gkldu du (p , p ) i ′ j ′ k l = gklu u du (∂i) du (∂j) i ′ j ′ k l = gklu u δi δj i ′ j ′ = giju u (143a) = Eu′u′ + 2F u′v′ + Gv′v′ = 1 . (143b)

Similarly,

′ ′ i ′ j ′ II(p , p ) = biju u (144a) ′ ′ ′ ′ ′ ′ = L u u + 2M u v + N v v = κn . (144b)

We finally obtain

II(p′, p′) κ = = II(p′, p′) . (145) n I(p′, p′)

We assume that κn has value of λ, which gives the relation

II = λI . (146)

One can divide (146) by ds2 and then obtain II I ( ) = λ =⇒ L u′u′ + 2M u′v′ + N v′v′ = λ Eu′u′ + 2F u′v′ + Gv′v′ , (147) ds2 ds2 where λ can be recognized as the Lagrangian multiplier with constraint Eu′u′ + 2F u′v′ + Gv′v′ = 1. 2 By looking for the extrema λ of κn = II/ds , we take the partial derivative of (147) with respect to ui ′, which leads to the equation of matrix form ( )( ) ( )( ) ( )( ) LM u′ EF u′ L − λE M − λF u′ = λ =⇒ = 0 , (148) MN v′ FG v′ M − λF N − λG v′ which means

j ′ (bij − λgij)u = 0 . (149)

We have to look for the non-trivial solutions, i.e.,

det(b − λg ) = 0 , ( ij ) ij ( ) =⇒ EG − F 2 λ2 − EN + GL − 2FM λ + LN − M 2 = 0 , ( ) =⇒ gλ2 − EN + GL − 2FM λ + b = 0 , (150) where we define { 2 g := det(gij) = EG − F , (151a) 2 b := det(bij) = LN − M . (151b)

As a result, we have the sum and product of two solutions λ1 and λ2 EN + GL − 2FM b λ + λ = and λ λ = . (152) 1 2 g 1 2 g

20 Gauss curvature We define the Gauss curvature (or called total curvature) as product of two curvatures: b K := λ λ = . (153) 1 2 g

Mean curvature The mean curvature is defined by the mean value of sum of two curvatures: 1 EN + GL − 2FM H := (λ + λ ) = . (154) 2 1 2 2 g

Remark. The value λα (α = 1, 2) is called the principal curvature of κn. By substituting λα ′ j ′ j into the equation (149), the corresponding solution of vector p(α) = u(α)pj or dp(α) = du(α)pj is called the principal direction.

Example (Cylindrical surface in E3). A surface parallel with z-axis can be described by ( ) ( ) ( ) ( ) p = x, y, z = x(u), y(u), v =⇒ dp = dx, dy, dz = x′du, y′du, dv . (155)

A cylindrical surface need to have a constraint with ( ) ( ) dx 2 dy 2 + = x′ 2 + y′ 2 = 1 . (156) du du

Figure 7: A cylinder. A cylinder is parametrized by ( ) ( ) ⇒ ′ ′ p(u, v) = x(u), y(u), v = dp = pudu + pvdv = x du, y du, dv , (157)

where { ( ) p = x′du, y′du, 0 , ( ) u ( ) ⇒ ∧ ′ − ′ = pu pv = y , x , 0 . (158) pv = 0, 0, 1 .

and the normal vector is p ∧ p 1 ( ) ( ) n = u v = y′, −x′, 0 = y′, −x′, 0 (159) | ∧ | ′2 ′2 pu pv y + x

21 and ( ) ′′ ′′ dn = nudu + nvdv = y du, −x du, 0 . (160)

with { ( ) ′′ ′′ nu = y , −x , 0 , ( ) (161) nv = 0, 0, 0 .

Then we have first fundamental form ( ) I = dp · dp = x′2 + y′2 du2 + dv2 = du2 + dv2 , (162)

where

E = G = 1,F = 0 . (163)

The second fundamental form can be obtained by

II = −dp · dn = −( · dudu + · dudv + · dudv + · dvdv) pu nu |pu{zn}v p| v{znu} |pv{zn}v ( ) 0 0 0 = − x′y′′ − y′x′′ du2 ( ) = y′x′′ − x′y′′ du2 , (164)

where

L = y′x′′ − x′y′′,M = N = 0 . (165)

So we have

g = 1 , b = 0 ,EN + GL − 2FM = y′x′′ − x′y′′ . (166)

As a result, we obtain

Gauss curvature: K = λ1λ2 = 0, (167) 1( ) 1( ) mean curvature: H = λ + λ = y′x′′ − x′y′′ . (168) 2 1 2 2 We can solve the above equations to obtain

′ ′′ ′ ′′ λ1 = 0, and λ2 = y x − x y . (169)

Geodesic equations The tangent vector parametrized by s is

′ i ′ p (s) = piu , (170)

22 it leads to the acceleration vector is given by p′′ = p′ ui ′ + p ui ′′ ( i i ) = Γk p + b n ui ′uj ′ + p ui ′′ ( ij k ij ) i k ′′ k i ′ j ′ i ′ j ′ = u + Γ iju u pk + biju u n (171) ′′ ′′ = pt + pn (172)

= κg + κn , (173) ′ j k j ′ where we have used pi = pij(du /ds) = (Γ ijpk + bijn)u . Now we call the curve p(s) geodesic if ′′ ′′ p = pn = κn, i.e., the tangential part is vanished ( ) ′′ k ′′ k i ′ j ′ κg = pt = u + Γ iju u pk = 0 , (174) which means that

p only has the normal curvature κn.

Because pks are linear independent, we obtain

k ′′ k i ′ j ′ u + Γ iju u = 0 , (175) which is called geodesic equations.

Supplement (Connection and geodesic). We have differential of p′ given by

dp′ = dp ui ′ + p dui ( i i ) = Γk p + b n ui ′ + p dui ′ ( i k i ) i k ′ k i ′ i ′ = du + Γ iu pk + biu n , (176) E3 i.e., the symbol d is a total or absolute differential with respect to frame (p; p1, p2, n) on . The total or absolute differentiation means that we have to differentiate not only the component ui ′ but ′ ′ also the basis pi of a vector p . We assume that p can be written as

′ i ′ i p = u ∂i := V ∂i = V . (177)

Then (176) can be written as ( ) k k i i dV = dV + Γ iV ∂k + biV n . (178)

and the geodesic equation becomes as ( ) k k i dV + Γ iV ∂k = 0 . (179)

i i We define the connection D = du ⊗ Di on M, which act on the function V and basis ∂i are { i i j i DV = dV = du ⊗ ∂jV , (180a) k j k D∂i = Γ i ⊗ ∂k = du ⊗ Γ ij∂k , (180b)

respectively. We note that D act on a function as a differential d on a function. The connection act on a vector is given by ( ) ( ) ( ) i ⊗ i i ⊗ i k ⊗ k k i ⊗ DV = DV ∂i + V D∂i = dV ∂i + V Γ i ∂k = |dV +{z Γ iV} ∂k . (181) (DV )k

23 The resulting geodesic equation (179) can be read as

DV = 0 . (182)

Therefore, (178) can also written as

⊤ ⊥ i dV = dV + dV = DV + biV n , (183) where ⊤ is the orthogonal projection onto the space spanned by {∂k} and ⊥ means the normal component. If there is no normal space M⊥ of M, the differential

dV = DV (184) would be the total or absolute differential of a vector V on surface M. We can multiply 1/duj to the DV : 1 1 ( ) DV = dV k + Γk V i ∂ duj duj i k 1 ( ) = dV k + Γk dulV i ∂ duj il k ( l ) k k du i = ∂jV + Γ il V ∂k |{z}duj δl ( j) = ∂ V k + Γk V i ∂ ( j ) ij k k := ∇jV ∂k , (185) where we define the component ( ) k ∇ k ∇ k k k i DV j := j |{z}V = ∂j V = ∂jV + Γ ijV (186) component of V

k called the covariant derivative of vector V in ∂j direction. We note that the covariant derivatives k ∇j act on the component of vector V only. It can also be recognized as a vector-valued 1-form DV act on a vector ∂j (( ) ) ( ) ( ) DV (∂ ) = DV kdxi ⊗ ∂ (∂ ) = DV k dxi(∂ ) ⊗∂ = DV k∂ . (187) j i k j i | {z j} k j k i δj

Now we will return to the discussion of the tangential part of acceleration (curvature) vector ( ) ⊤ j ( ) dV DV 1 du 1 (185) a := p′′ = := = DV = DV = V j ∇ V k ∂ := ak∂ , (188) T t ds ds ds ds duj j k T k

′′ which is the orthogonal projection of the acceleration vector p onto the space spanned by {∂k}. As a result, we have tangential acceleration with component

k j∇ k aT := V jV , (189) where

j j ∇ ∇ ∇ j ∇ V j = V ∂j = V ∂j = V . (190)

24 k So we can also write aT as

k ∇ k aT = V V , (191)

which is called the covariant derivative of V k along the direction of V . Then, we call

k ∇ k aT = V V = 0 or DV = 0 (192)

the parallel transport of tangent vector V , which is equivalent to the geodesic equation.

Remark. We would like to remind you the notation of the covariant dereivative in mathematics and physics. Consider a vector V , the covariant dereivative (connection) (D or Dj) of a full vector i V = V ∂i is ( ) ( ) ( ) DV = duj ⊗ D V i∂ = duj ⊗ ∂ V i +V kΓi ∂ := V i duj ⊗ ∂ = DV iduj ⊗ ∂ . j i |j{z} kj i ;j i j i i V,j (193)

i i i k We note that here DjV = ∂jV = V,j and Dj∂i = ∂kΓ ij as shown in (180). In physics, we i always consider a vector represented by its component V , the covariant dereivative (∇j) of a vector V i should be ( ) ∇ i i i k i ≡ i jV = DV j = V,j + V Γ kj V;j . (194)

Christoffel symbols According to (123), we can calculate the coefficients Γkij. Now we would like to derive the coefficients in terms of gij, the components of first fundamental form I. We take the k partial derivative of gijwith respect to u and then interchange the indices of the equations. Therefore, we obtain   ∂  g = p · p + p · p = Γ + Γ , (195a) ∂uk ij ik j i jk jik ijk  ∂ g = Γ + Γ ,  i jk kji jki (195b) ∂u   ∂  g = Γ + Γ . (195c) ∂uj ki ikj kij

Remark. Equations of (195) gives the metric compatibility, which can be written as the covarinat deriavative of gij

l l ∇kgij = ∂kgij − Γ ikglj − Γ jkgli = 0 . (196)

We can define the non-metricity

Qkij := −∇kgij (197)

and (196) would be equivalent to the vanished non-metricity Qkij = 0.

25 ̸ According to (344), which will be shown later that pi = ∂ip in general, we have general case that ( ) ( ) ( ) ( ) ̸ − − k − k − 0 = pij pji = Djpi + bijn Dipj + bjin = Γ ij Γ ji pk + bij bji n . (198)

E3 However, in (120), we have pij = ∂jpi = ∂j∂ip due to the globally fixed frame in , which will be explained in the Sec. 4 by the reduction condition (345). Consequently, if we have { k k − ⇒ Γ ij = Γ ji , (199a) pij pji = 0 . = bij = bji , (199b)

k which give the symmetric condition (torsion-free) for Γ ij and bij.

By computing (195c)+(195b)−(195a), we obtain coefficients ( ) 1 ∂ ∂ ∂ Γ = g + g − g (200) kij 2 ∂uj ki ∂ui jk ∂uk ij and ( ) 1 ∂ ∂ ∂ Γk = gklΓ = gkl g + g − g . (201) ij lij 2 ∂uj li ∂ui jl ∂ul ij

k Remark. According to (180b), the Γ ij is also called the connection coefficients, or simply the connection. Due to metric compatibility (196) and torsion-free (304a), the connection (200) or (201) is a function of the metric tensor gij, we also call this kind of connection the Levi-Civita connection or Riemannian connection. The Levi-Civita connection can also be denoted as { Γ (g) ≡ [ij, k] , (202a) kij { } k ≡ k Γ ij(g) ij , (202b)

which are called the Christoffel symbol of first and second kind respectively, in order to distinguish the general connections.

Supplement (Torsion tensor). For general metric compatible connection, we always do not have k k the symmetric property, i.e., Γ ij ≠ Γ ji. The connection would contain the symmetric and anti- symmetric parts, which is shown as 1( ) 1( ) Γk = Γk + Γk + Γk − Γk ij 2 | ij {z ji} 2 | ij {z ji} symmetric in i,j anti-symmetric in i,j k k = Γ (ij) + Γ [ij] . (203)

We define the torsion tensor

k k k k T ji := Γ ij − Γ ji = 2 Γ [ij] (204)

26 { } k ̸ k as the anti-symmetric part of the connection. We have to note that Γ (ij) = ij . The general connection can be decomposed as { } k k k Γ ij = ij + K ij , (205) and it leads to the relation

k k k T ji = K ij − K ji , (206)

k where K ij is called the contorsion tensor. By permutatiing the indices of (206), it can be show k k that the contorsion K ij can be in terms of torsion tensor T ij as

zsymmetric}| in i,j{ 1( ) 1( ) Kk := − T k + T i − T j = − T k + T i + T j (207) ij 2 ij jk ki 2 |{z}ij jk ik anti-symmetric in i,j or

zsymmetric}| in i,j{ 1( ) Kk := − T k −T k − T k (in pseudo-Riemannian geometry) . (208) ij 2 |{z}ij i j j i anti-symmetric in i,j

So the torsion 2-form can be written as

1 (203) (205) T k = T k duj ∧ dui = Γk duj ∧ dui = Kk duj ∧ dui , (209) 2 ji ij ij

k k j where we define K i := K ijdu the contorsion 1-form. Therefore, we have the form equation

k k i T = K i ∧ du . (210)

Consequently, the symmetric and anti-symmetric parts of the connection are  { }  k k k Γ (ij) = ij + K (ij) ,  1 1  Γk = Kk = − T k = + T k , (211a) [ij] [ij] 2 ij 2 ji { } k k respectively, which shows that the symmetric part Γ (ij) contains Levi-Civita connection ij and k torsion T ij.

Remark. We also note that if we identify

c c (Γa) b ≡ Γ ba (212)

k k i and the connection form is defined by Γ j = Γ ijdu , the torsion tensor will be denoted by

k k k k T ij := Γ ij − Γ ji = 2 Γ [ij] . (213)

27 Christoffel symbols in the orthogonal coordinates If we consider the case in the orthogonal co- 12 21 ordinates, we have g12 = g21 = g = g = 0 and 1 1 1 1 g11 = = , g22 = = . (214) g11 E g22 G The component of Christoffel symbols becomes ( ) k 1 Γ ij = ∂jgki + ∂igjk − ∂kgij (no sum) , (215) 2gkk and we have the following properties: • For j = k, we have ( ) ( ) k 1 ¨¨ ¨¨ 1 1 Γ ik = ¨∂kgki + ∂igkk − ¨∂kgik = ∂igkk = ∂i ln gkk (no sum) . (216) 2gkk 2gkk 2 • For i = j ≠ k, we have ( ) k 1 ¨¨ ¨¨ 1 Γ ii = ¨∂igki + ¨∂igik − ∂kgii = − ∂kgii (no sum) . (217) 2gkk 2gkk • In the general case of dimension > 2, if i ≠ j ≠ k, we have k Γ ij = 0 (in orthogonal coordinates) . (218) In dimension = 2, it is impossible that i, j, k are all distinct, so we have the same consequence k Γ ij = 0. Therefore, for dimension = 2, the christoffel symbols in the orthogonal coordinates are given by E G Γ1 = u , Γ2 = v , 11 2E 22 2G E G Γ1 = Γ1 = v , Γ2 = Γ2 = u , 12 21 2E 21 12 2G −G −E Γ1 = u , Γ2 = v . (219) 22 2E 11 2G Example (Polar coordinates). Consider the first fundamental form in orthogonal coordinates

ds2 = E du2 + 2F dudv + G dv2 = dr2 + r2dθ2 , (220)

where we have u = r, v = θ and  { E = 1 , Er = Eθ = Gθ = 0 , F = 0 , =⇒ (221)  G = 2r . G = r2 . r

The Christoffel symbols are given by 2r 1 −2r Γ2 = Γ2 = = , Γ1 = = −r . (222) 21 12 2r2 r 22 2 The geodesic equations is given by (175). Therefore, each component of the geodesic equations is obtained by   ′′ 2 ′ ′ 2 ′ ′ ′′ 2 ′ ′ θ + Γ 12r θ + Γ 21θ r = θ + r θ = 0 , (223a)  r ′′ 1 ′ ′ ′′ ′ ′ r + Γ 22θ θ = r − rθ θ = 0 . (223b)

28 Gauss-Codazzi equation Now we have Gauss and Weingarten formulas { l pik = Γ ikpl + bikn , (224a) − l nj = b jpl , (224b) which correspond to the derivative vectors of tangent and normal vectors respectively. By taking the j partial derivative of pik with respect to u , we have ∂ p = ∂ Γl p + Γl p + ∂ b n + b n j ik j ik l ik( lj j ik )ik j = ∂ Γl p + Γl Γm p + b n + ∂ b n + b (−bl )p (j ik l ik lj m l)j ( j ik ik ) j l l m l − l l = ∂jΓ ik + Γ ikΓ mj bikb j pl + Γ ikblj + ∂jbik n . (225) By interchanging the indices j and k, we obtain ( ) ( ) l m l − l l ∂kpij = ∂kΓ ij + Γ ijΓ mk bijb k pl + Γ ijblk + ∂kbij n . (226) As a consequence, (225)−(226)= 0, which is − − ∂jpik ∂kpij = ∂j∂kpi ∂k∂jpi = 0 . (227) We define the Riemann(-Christoffel) curvature tensor or simply the curvature tensor as

l l l l m l m R ijk := ∂jΓ ik − ∂kΓ ij + Γ mjΓ ik − Γ mkΓ ij . (228) and m Rlijk := glmR ijk (229) is defined. Therefore, according to (227), we obtain a set of equations called Gauss-Codazzi equation, which are given by 0 = ∂ p − ∂ p (j ik k ij ) ( ) = Rl − b bl + b bl + Γl b − Γl b + ∂ b − ∂ b . | ijk ik{z j ij }k pl | ik lj ij lk{z j ik k ij} n (230) 0 0 The first one is called Gauss equation

l l l R ijk = bikb j − bijb k , (231) and the second one is Codazzi equation

l l ∂jbik − ∂kbij = Γ ijblk − Γ ikblj . (232) We note that there are some symmetries of curvature tensor: • Anti-symmetric in the indices j and k l l R ijk = −R ikj . (233) • Anti-symmetric in l and i (only for Levi-Civita connection)

Rlijk = −Riljk . (234) • Symmetric in the pairs of li and jk (only for Levi-Civita connection)

Rlijk = +Rjkli . (235) We also define the traced curvature tensor called Ricci tensor, which is given by

l Rik = R ilk (236) and the scalar curvature or Ricci scalar ik R = g Rik . (237)

29 Theorem Egregium of Gauss The indices i, j, k, l = 1, 2, by the symmetries of Riemann curvature tensor, the following components are vanished:

R11jk = R22jk = 0 ,Rli11 = Rli22 = 0 , (238) From the Gauss equation (231), the residual component can be given by

R1212 = b22b11 − b12b12 = NL − M 2

= det(bij) = b . (239) Therefore, we can rewrite the Gauss curvature (153) as b R K = = 1212 , (240) g g

3 which is a function of gij only, i.e., a 2-dimensional surface in E is totally determined by it’s intrinsic structure. This is the famous intrinsic geometry of Gauss and we call this the theorem Egregium of Gauss. From the Codazzi equation (232), we only need to consider the case of i = 1, 2 and j = 1 as well as k = 2. This gives  ∂b ∂b  12 − 11 = Γl b − Γl b , (241a) ∂u1 ∂u2 11 l2 12 l1 (l = 1, 2) . ∂b ∂b  22 − 21 = Γl b − Γl b , (241b) ∂u1 ∂u2 21 l2 22 l1

Remark. In general n-dimensional space Mn, we can always choose a orthonormal frame, so that g = 1. Then we call K = Rijij for i ≠ j the sectional curvature of the 2-dimensional surface in Mn, where i, j labeled two components on the surface.

Third fundamental (quadratic) form In analogy we have a Gauss map for a surface, the Gauss sphere S2. A normal vector n will be sent to be a radius vector of S2. Therefore, n on S2 play the same role as p on the surface M. As a result, we can calculate the first fundamental for of S2 by

I|S2 = dn · dn . (242) Now we define the third fundamental form of M to be the first fundamental form of S2

III|M := I|S2 = dn · dn . (243) From the first fundamental form, we can calculate the area element on the surface M and S2, which are denoted by ∆A|M and ∆A|S2 respectively. The results can be obtained by { ≈ ⇒ | | ∧ | | ∧ | ∆p pu∆u + pv∆v = ∆A M = pu∆u pv∆v = pu pv ∆u∆v , (244a)

∆n ≈ nu∆u + nv∆v =⇒ ∆A|S2 = |nu∆u ∧ nv∆v| = |nu ∧ nv|∆u∆v . (244b)

However, the vectors nu and nv are given by the Weingarten formulas (128) with (129) and (130). ∧ We can express nu nv in terms of pu and pv by ∧ ∧ − ∧ nu nv = AD(pu pv) BC(pu pv) FM − GL FM − EN FL − EM FN − GM = (p ∧ p ) − (p ∧ p ) EG − F 2 EG − F 2 u v EG − F 2 EG − F 2 u v LN − M 2 b = (p ∧ p ) = (p ∧ p ) = K(p ∧ p ) . (245) EG − F 2 u v g u v u v

30 Consequently, we can measure the absolute value of Gauss curvature by the ratio

∆A| 2 | ∧ |∆u∆v |K|| ∧ |∆u∆v S nu nv pu pv | | | = | ∧ | = | ∧ | = K . (246) ∆A M pu pv ∆u∆v pu pv ∆u∆v 4 Cartan’s moving frame and exterior differentiation methods

We would like to introduce a very useful lemma of Cartan first.

i Lemma (Cartan’s lemma). Consider a set of linearly independent frame {ei} (or coframe {ϑ }) with i i = 1, . . . , p (p < n) in n-dimensional space M and {Ei} is another set of frame. If e ∧ Ei = 1 p j e ∧ E1 + ··· + e ∧ Ep = 0, then Ei = cije and cij = cji.

Proof. We set the linearly independent frame in M by extending to n-tuple from ei given by

, ,..., , ,..., |e1 e2{z e}p |ep+1 {z en} (247) p n−p with index α labeled components p + 1, p + 2, . . . , n. We assume that Ei is expanded by frame in M as

j α Ei = cije + ciαe . (248)

i According to e ∧ Ei = 0, we have

i i j i α 0 = e ∧ Ei = cije ∧ e + ciαe ∧ e 1( ) = c − c ei ∧ ej + c ei ∧ eα . (249) 2 |ij {z ji} |{z}iα 0 0

Therefore, we obtain the coefficients of Ei { cij = cji , (250a)

ciα = 0 , (250b)

j which means that Ei = cije is constructed by {ej} only.

i Orthonormal frame We have dp = pidu with the basis pi, under the Gram-Schmit procedure, we can obtain an orthonormal frame (p; e1ˆ, e2ˆ, e3ˆ) given by (110), (111) and (112), where we use the hatted indices to label the component of orthonormal frame now. We can expand pi by eˆi shown as ( ) ( )( ) 1ˆ 2ˆ ˆj ˆ ˆ ˆ ˆ p1 a 1 a 1 e1ˆ pi = a ieˆj (i, j = 1, 2) or = 1ˆ 2ˆ , (251) p2 a 2 a 2 e2ˆ

ˆj where we call the expansion factor a i the vielbein (vierbein or tetrad for 4-dimension), which can be regarded as the GL(2, R) transformation of the frame on M. Therefore, we have to obtain the { } differential of frame (p; e1ˆ, e2ˆ, e3ˆ). First we can rewrite dp spanned by frame eaˆ as dp = p du1 + p du2 (1 2 ) ( ) 1ˆ 2ˆ 1 1ˆ 2ˆ 2 = a 1e1ˆ + a 1e2ˆ du + a 2e1ˆ + a 2e2ˆ du ( ) ( ) 1ˆ 1 1ˆ 2 2ˆ 1 2ˆ 2 1ˆ 2ˆ = a 1du + a 2du e1ˆ + a 1du + a 2du e2ˆ := ϑ e1ˆ + ϑ e2ˆ (252)

31 with

ˆi ˆi j ϑ := a jdu (253)

î Then we have to introduce the connection ω ˆj for for eî. As a result, deî can be shown as   1ˆ 2ˆ 3ˆ de1ˆ = ω 1ê1ˆ + ω ˆ1e2ˆ + ω 1ê3ˆ , 1ˆ 2ˆ 3ˆ =⇒ d = ωˆb (â, ˆb = 1ˆ, 2ˆ, 3)ˆ . de2ˆ = ω 2ê1ˆ + ω 2ê2ˆ + ω 2ê3ˆ , eaˆ aêˆb (254)  1ˆ 2ˆ 3ˆ de3ˆ = ω 3ê1ˆ + ω 3ê2ˆ + ω 3ê3ˆ .

ˆb ˆb cˆ In particular, we call the connection form ω aˆ = ω aˆcˆϑ the linear connection form and the coefficient ˆb ω aˆcˆ the Ricci rotation coefficients in the orthonormal (non-coordinate) frame. However, we have ˆb condition for ω aˆ due to the orthogonality · · eî eˆj = δîˆj and eaˆ e3ˆ = δaˆ3ˆ . (255) · We can differentiate the orthogonality condition eî eˆj = δîˆj, thus we have · · · d(eî eˆj) = deî eˆj + eî deˆj kˆ · · kˆ = ω îekˆ eˆj + eî ω ˆjekˆ kˆ kˆ = ω îδkˆˆj + ω ˆjδîkˆ

ˆj ˆi = ω ˆi + ω ˆj = 0 . (256)

Similarly, we have

· 3ˆ î d(eî e3ˆ) = ω î + ω 3ˆ = 0 (257)

· ˆb from eˆi e3ˆ = δˆi3ˆ. Therefore, all the components of ω aˆ are anti-symmetric in the orthonormal frame, we have the consequence:

• The metric compatibility gives the anti-symmetric property for linear connection form in the orthonormal frame, i.e.,

ˆ ˆ ∇ g = ∇ δ = e (δ ) −ωb − ωaˆ = 0 =⇒ ωb = −ωaˆ . (258) cˆ aˆˆb cˆ aˆˆb |cˆ {zaˆˆb} aˆcˆ ˆbcˆ aˆcˆ ˆbcˆ 0

Remark. In pseudo-Riemannian space, we have

· cˆ cˆ d(eaˆ eˆb) = ω aˆηcˆˆb + ω ˆbηaˆcˆ = ωˆbaˆ + ωaˆˆb = 0 , (259)

and the metric compatibility in pseudo-orthonormal frame should be read as

ˆ ˆ ∇ g = ∇ η = e (η ) −ωd η − ωd η = 0 =⇒ ω = −ω . (260) cˆ aˆˆb cˆ aˆˆb |cˆ {zaˆˆb} aˆcˆ dˆˆb ˆbcˆ aˆdˆ ˆbaˆcˆ aˆˆbcˆ 0

32 Finally the equation (254) is reduced to   2ˆ 3ˆ de1ˆ = ω 1ˆe2ˆ + ω 1ˆn , (261a) de = ω1ˆ e + ω3ˆ n , (261b)  2ˆ 2ˆ 1ˆ 2ˆ  1ˆ 2ˆ dn = ω 3ˆe1ˆ + ω ˆ3e2ˆ . (261c) Now we can write down the first, second and third fundamental form in orthonormal frame, which are given by  ( ) ( )  1ˆ 2 2ˆ 2  I = dp · dp = ϑ + ϑ , (262a) − · − 1ˆ 1ˆ − 2ˆ 2ˆ 3ˆ 1ˆ 3ˆ 2ˆ  II = dp dn = ϑ ω 3ˆ ϑ ω 3ˆ = ω 1ˆϑ + ω 2ˆϑ , (262b)  ( ) ( ) ( ) ( ) · 1ˆ 2 2ˆ 2 3ˆ 2 3ˆ 2 III = dn dn = ω 3ˆ + ω 3ˆ = ω 1ˆ + ω 2ˆ . (262c)

3ˆ 3ˆ ˆi i It can be shown that ω 1ˆ and ω 2ˆ are linear combinations of ϑ or du given by (290) due to Cartan’s first structure equation (be introduced later) and Cartan’s lemma { ( ) ( )( ) 3ˆ 1ˆ 2ˆ 3ˆ 1ˆ ω ˆ = bˆˆϑ + bˆˆϑ , ω b b ϑ 1 11 12 =⇒ 1ˆ = 1ˆ1ˆ 1ˆ2ˆ . (263) 3ˆ 1ˆ 2ˆ 3ˆ b b 2ˆ ω 2ˆ = b2ˆ1ˆϑ + b2ˆ2ˆϑ . ω 2ˆ 2ˆ1ˆ 2ˆ2ˆ ϑ

Consequently, the second fundamental form becomes ( ) ( ) 1ˆ 2 2ˆ 1ˆ 1ˆ 2ˆ 2ˆ 2 ˆi ˆj II = b1ˆ1ˆ ϑ + b1ˆ2ˆϑ ϑ + b2ˆ1ˆϑ ϑ + b2ˆ2ˆ ϑ = bˆiˆjϑ ϑ . (264)

We consider the following matrix representation for tensors

−→ î −→ −→ eî e , ϑ ϑ , bîˆj B . (265)

Under the special orthogonal transformation SO(2, R) for frame and coframe, we have a new orthonormal frame { e′ = P e with P T = P −1 , (266) ϑ′ = P Tϑ where P ∈ SO(2, R) and T means the transpose operation for matrix. As a result, we can obtain the diagonal matrix BD from B through P by

T ′ T T ′ II = ϑ Bϑ = (ϑ ) |P {zBP} ϑ , (267)

BD i.e., ( ) T κ1 0 B −→ BD = P BP = . (268) 0 κ2

Therefore, the Gauss curvature and mean curvature can be obtained easily by   T − K = det(BD) = det(P ) det(B) det(P ) = det(B) = b1ˆ1ˆb2ˆ2ˆ b1ˆ2ˆb2ˆ1ˆ , (269a)  1 1 1( ) H = tr B = tr B = b + b , (269b) 2 D 2 2 1ˆ1ˆ 2ˆ2ˆ respectively, where the trace of the matrix B is invariant under the SO(2, R) transformation.

33 Covariant exterior differentiation We define some notation for differential operators for function, vector and 1-form. We use d, d and d∇ for differentiation, exterior differentiation and covariant exterior differentiation respectively. • For a function (0-form) f, the differential df which can also be regarded as the exterior differentiation of 0-form f:

d∇f = df = df . (270)

• For a vector eˆi, we have an absolute differential of vector deˆi which is described by Gauss formulas in differential form formalism (126):

d∇eî = deî = Deî + bîn , (271) is a vector-valued 1-form. If there is no normal space M⊥ of M, i.e., there are no n vector and M bîˆj, the differential is actually equal to the orthogonal projection of vector eî on

d∇eî = deî = Deî . (272)

• For an 1-form ϑˆi, we only do the exterior differentiation on ϑˆi:

ˆi ˆi d∇ϑ = dϑ . (273)

Remark. The covariant exterior differentiation d∇ is a combined operator, which do the exterior differentiation and covariant derivative on an 1-form and vector respectively. For a function f, we also note that the second differentiation is

∂2f ∂2f ∂2f ∂2f d2f(x, y) = dxdx + dxdy + dydx + dydy ≠ 0 , (274) ∂x∂x ∂y∂x ∂x∂y ∂y∂y which should not be confused with the second exterior differentiation ( ) ∂2f ∂2f ∂2f ∂2f d2f(x, y) = ddf = dx ∧ dy + dy ∧ dx = − dx ∧ dy = 0 . (275) ∂y∂x ∂x∂y ∂y∂x ∂x∂y

2 In addition, d∇ would not be vanished in general. Therefore, the second fundamental form is

− · 2 · i j · i j II = dp dn = +d p n = (pijdu du ) n = bijdu du (276)

due to dp · n = 0 which has been shown in the last term in (120). We note that d2p should be realized as a second covariant derivatives of p in (280). For coframe dui, the corresponding exterior differentiation is vanished, which is shown as

ddui = d2ui = 0 . (277)

∂ i We call ∂ui a holonomic frame and du a holonomic coframe which is an exact form according to ˆi ˆi j the Poincaré lemma. For ϑ = a jdu , its exterior differentiation is

î î j î 2 j dϑ = da j ∧ du + a jd u ≠ 0 , (278)

which is called an anholonomic coframe or a Pfaffian form dual to the anholonomic frame eˆi.

34 2 Supplement. We note that the exterior 2-form d∇eˆi will be introduced as the second structure equation through the covariant exterior differentiation and related to the curvature 2-form of (293) and structure constants of (315) later

2 1 ˆl ˆj kˆ d e = d∇(de ) = R ϑ ∧ ϑ ⊗ e (279a) ∇ î î 2 îˆjkˆ ˆl kˆ kˆ kˆ ˆj = d∇(ϑ D e ) = d∇ϑ ⊗ D e −ϑ ∧ ϑ ⊗ D D e kˆ î | {z } |{zkˆ }î ˆj kˆ î (315) ˆ Γl e îkˆ ˆl ( ) 1 1 − kˆ ˆj ∧ mˆ ⊗ ˆl ˆj ∧ kˆ ⊗ − = c ˆjmˆ ϑ ϑ Γ îkêˆl + ϑ ϑ DˆjDkˆ DkˆDˆj eî 2 ( ( 2 ) ) ˆ↔ ˆ 1 ˆ k =mˆ ϑˆj ∧ ϑk ⊗ − cmˆ Γl e + D D − D D e 2 ˆjkˆ | {zîmˆ }ˆl ˆj kˆ kˆ ˆj î D e ( mˆ î ) ˆ 1 = ϑˆj ∧ ϑk ⊗ − cmˆ D + D D − D D e . (279b) 2 ˆjkˆ mˆ ˆj kˆ kˆ ˆj î

However, the second covariant derivatives of a vector eˆi should be

ˆ ˆ ˆ d2e = D2e = D(ϑk ⊗ D e ) = ϑˆj ⊗ D ϑk ⊗ D e +ϑˆj ⊗ ϑk ⊗ D D e î î kˆ î | {zˆj } |{zkˆ }î ˆj kˆ î ˆ ˆl Γk ϑmˆ Γ ˆêˆ ( ˆjmˆ ) ik l ˆ↔ ˆ ˆ k =mˆ ϑˆj ⊗ ϑk ⊗ Γmˆ Γl e +D D e ˆjkˆ | {zîmˆ }ˆl ˆj kˆ î D e ( mˆ î ) ˆj ⊗ kˆ ⊗ mˆ = ϑ ϑ Γ ˆjkˆDmˆ + DˆjDkˆ eî . (280)

2 ̸ 2 Therefor, we conclude that d∇eî = d eî because deî is a vector-valued 1-form. We note that that the wedge product is obtained by anti-symmetrizing the tensor product

1 1 A ∧ B = (A ⊗ B)A = (A ⊗ B − B ⊗ A) = (A B − B A )ϑî ∧ ϑˆj , (281) 2! 2 î ˆj ˆj î

mˆ where A indicates the anti-symmetrization. The anti-symmetrization of DˆjDkêî and Γ ˆjkˆ will be given later, which are shown in (318) and (327b) respectively. As a result, the anti-symmetrization 2 of d eî can be shown as ( ) ˆ 1 ˆ (d2e )A = (D2e )A = D ∧ De = ϑˆj ∧ ϑk ⊗ − T mˆ D + Rl e (cf.(279a) or (293)) . î î î 2 ˆjkˆ mˆ îˆjkˆ ˆl (282)

Canonical 1-form In general case, {eaˆ} does not necessarily be chosen as orthonormal, i.e., the · ̸ { } metric tensor is eaˆ eˆb = gaˆˆb = δaˆˆb. If eaˆ is an orthonormal frame, we have anti-symmetric property of (256) and (257). We would discuss from the differential of frame (p; e1ˆ, e2ˆ, e3ˆ) and write the equations by the covariant exterior differentiation as { aˆ d∇p = ϑ ⊗ eaˆ := ϑ , (283a) ˆb ⊗ d∇eaˆ = ω aˆ eˆb , (283b)

35 where we have defined ϑ := d∇p = dp the canonical 1-form, which is a vector-valued 1-form. We will show that the canonical 1-form ϑ is an identity map of vector in the frame eaˆ. Consider a vector ˆb V = V eˆb, the canonical 1-form act on V gives

aˆ ⊗ ˆb ˆb aˆ ˆb aˆ ˆb ϑ(V ) = ϑ eaˆ(V eˆb) = V ϑ (eˆb)eaˆ = V δˆb eaˆ = V eˆb = V. (284)

Remark. If we consider a point p move on the surface M in E3, the differential would be a vector

spanned by e1ˆ and e2ˆ only, which would be written as

aˆ ˆi 1ˆ 2ˆ d∇p = ϑ eaˆ = ϑ eˆi = ϑ e1ˆ + ϑ e2ˆ (285)

with ϑ3ˆ = 0, it would be reduced to the equation given by (252).

Cartan’s first structure equation Now we do the covariant exterior differentiation on (283). The covariant exterior differentiation of (283a) is

2 aˆ d∇ϑ = d∇p = d∇(ϑ ⊗ eaˆ) aˆ aˆ ¯ = dϑ ⊗ eaˆ + (−1) ϑ ∧ Deaˆ aˆ aˆ ˆb = dϑ e − ϑ ∧ ω eˆ ( aˆ )aˆ b ˆ = dϑaˆ + ωaˆ ∧ ϑb e ( ) ˆb aˆ aˆ = d∇ϑ eaˆ aˆ := T eaˆ = T ̸= 0 , (286) ¯ where D is a connection with respect to eaˆ. Here we have defined T the vector-valued torsion 2-form and the corresponding component the torsion 2-form as ( ) T aˆ aˆ aˆ aˆ aˆ ∧ ˆb := d∇ϑ := D |{z}ϑ := dϑ + ω ˆb ϑ , (287) component of ϑ where D can be identified as operation

D = d + ω∧ (288) act on the differential form which is the component of the corresponding vector-valued form. The equation (287) we obtained is called Cartan’s first structure equation.

M E3 1ˆ 2ˆ 3ˆ Remark. Since p moves on the surface in , we have d∇p = ϑ e1ˆ +ϑ e2ˆ and ϑ = 0. Follow Cartan’s first structure equation (287), we have

3ˆ − 3ˆ ∧ 1ˆ − 3ˆ ∧ 2ˆ − 3ˆ ∧ ˆi 0 = d∇ϑ = ω 1ˆ ϑ ω 2ˆ ϑ = ω ˆi ϑ . (289)

According to Cartan’s lemma, the connection form is obtained as

3ˆ ˆj ω ˆi = bˆiˆjϑ , (290)

which gives the equations (263).

36 (a) Torsion in space. (b) Torsion on surface.

Figure 8: Torsion is related to the translation.

We can consider an infinitesimal contour integral for d∇p infinitesimally around a point as a boundary ∂D of a small region D. By applying Stokes’ theorem to the contour integral of d∇p over ∂D gives I ∫ ∫ 2 d∇p = d∇p = T (291) ∂D D D or equivalent to I ∫ ∫ ϑaˆ = Dϑaˆ = T aˆ . (292) ∂D D D

The integral result implies that the translation of a point or the displacement d∇p is associated with the torsion. If there is no displacement, i.e. d∇p = 0, the space would not be twisted.

Cartan’s second structure equation Similarly, we do the covariant exterior differentiation on (283b) and obtain

2 ˆb ⊗ d∇eaˆ = d∇(ω aˆ eˆb) ˆb ⊗ − ˆb ∧ ¯ = dω aˆ eˆb + ( 1) ω aˆ Deˆb ˆb ˆb cˆ = dω eˆ − ω ∧ ω ˆe ( aˆ b aˆ b) cˆ ˆb ˆb cˆ = dω + ω ∧ ω eˆ ( aˆ ) cˆ aˆ b 2 ˆb = d∇eaˆ eˆb Rˆb R ̸ := aˆeˆb = aˆ = 0 . (293)

Therefore we have the vector-valued curvature 2-form Raˆ with the corresponding component curvature 2-form given by ( ) Rˆb 2 ˆb ˆb ˆb ˆb ∧ cˆ aˆ := d∇eaˆ := D |{z}ω aˆ := dω aˆ + ω cˆ ω aˆ , (294)

component of d∇eaˆ which is called Cartan’s second structure equation.

37 Figure 9: Curvature is related to the rotation.

The similar infinitesimal contour integral for d∇eaˆ gives I ∫ ∫ 2 d∇eaˆ = d∇eaˆ = Raˆ (295) ∂D D D or equivalent to I ∫ ∫ ˆb ˆb ˆb ω aˆ = Dω aˆ = R aˆ , (296) ∂D D D which means that the rotation of a vector is associated with the curvature. If the vector does not change the direction after moving around a contour, i.e. d∇eaˆ = 0, the space would be flat.

First Bianchi identity The exterior differentiation of two structure equations can get more infor- mation of torsion and curvature. The structure equations are { T aˆ aˆ aˆ ∧ ˆb = dϑ + ω ˆb ϑ , (297a) Raˆ aˆ aˆ ∧ cˆ ˆb = dω ˆb + ω cˆ ω ˆb . (297b) We take the exterior differentiation of the first structure equation shown by

aˆ 2 aˆ aˆ ˆb aˆ ˆb dT = d ϑ + dω ˆ ∧ ϑ − ω ˆ ∧ dϑ ( b ) b ( ) Raˆ − aˆ ∧ cˆ ∧ ˆb − aˆ ∧ T ˆb − ˆb ∧ cˆ = ˆb ω cˆ ω ˆb ϑ ω ˆb ω cˆ ϑ ˆ ˆ ˆ ˆ Raˆ ∧ b − aˆ ∧ cˆ ∧ b − aˆ ∧ T b aˆ ∧ b ∧ cˆ = ˆb ϑ ω cˆ ω ˆb ϑ ω ˆb + ω ˆb ω cˆ ϑ , (298) then we obtain the first Bianchi identity

T aˆ T aˆ aˆ ∧ T ˆb Raˆ ∧ ˆb D = d + ω ˆb = ˆb ϑ . (299) If we have torsion-free condition T aˆ = 0, the first Bianchi identity becomes Raˆ ∧ ˆb 0 = ˆb ϑ 1 = Raˆ ϑcˆ ∧ ϑdˆ ∧ ϑˆb 2 ˆbcˆdˆ 1( ) aˆ aˆ aˆ − aˆ − aˆ − aˆ cˆ ∧ dˆ ∧ ˆb = R ˆbcˆdˆ + R cˆdˆˆb + R dˆˆbcˆ R ˆbdˆcˆ R cˆˆbdˆ R dˆcˆˆb ϑ ϑ ϑ (2 ) aˆ aˆ aˆ cˆ ∧ dˆ ∧ ˆb = R ˆbcˆdˆ + R cˆdˆˆb + R dˆˆbcˆ ϑ ϑ ϑ , (300)

38 resulting in

aˆ aˆ aˆ R ˆbcˆdˆ + R cˆdˆˆb + R dˆˆbcˆ = 0 . (301)

Second Bianchi identity Similarly, the exterior differentiation of the second structure equation is

aˆ 2 aˆ aˆ cˆ aˆ cˆ dR ˆ = d ω ˆ + dω ∧ ω ˆ − ω ∧ dω ˆ b ( b cˆ )b cˆ b ( ) Raˆ − aˆ ∧ dˆ ∧ cˆ − aˆ ∧ Rcˆ − cˆ ∧ dˆ = cˆ ω dˆ ω cˆ ω ˆb ω cˆ ˆb ω dˆ ω ˆb ˆ ˆ = Raˆ ∧ ωcˆ −ωaˆ ∧ωd ∧ ωcˆ − ωaˆ ∧ Rcˆ + ωaˆ ∧ ωcˆ ∧ ωd , (302) | cˆ{z }ˆb dˆ cˆ ˆb cˆ ˆb cˆ dˆ ˆb cˆ ∧Raˆ Raˆ +ω ˆb cˆ due to 2-form cˆ which leads to the second Bianchi identity

Raˆ Raˆ − cˆ ∧ Raˆ aˆ ∧ Rcˆ D ˆb = d ˆb ω ˆb cˆ + ω cˆ ˆb = 0 . (303)

Remark. It is essential to consider the geometric structure from Cartan’s viewpoint. The first structure equations in Riemannian geometry is restricted to be torsion-free condition. Then the structure equations are reduced to { aˆ − aˆ ∧ ˆb ˆb ∧ aˆ dϑ = ω ˆb ϑ = +ϑ ω ˆb , (304a) Raˆ aˆ aˆ ∧ cˆ ˆb = dω ˆb + ω cˆ ω ˆb . (304b)

aˆ − ˆb − Because of metric compatibility, we have ω ˆb = ω aˆ (or ωaˆˆb = ωˆbaˆ in pseudo-Riemannian geometry), which gives

aˆ aˆ cˆ − ˆb − ˆb cˆ ⇒ aˆ − ˆb ω ˆb = ω ˆbcˆϑ = ω aˆ = ω aˆcˆϑ = ω ˆbcˆ = ω aˆcˆ . (305)

Due to (304a), we obtain ( ) ˆ ˆ 1 ˆ dϑaˆ = ϑb ∧ ωaˆ = ωaˆ ϑb ∧ ϑcˆ = ωaˆ − ωaˆ ϑb ∧ ϑcˆ . (306) ˆb ˆbcˆ 2 ˆbcˆ cˆˆb However, dϑaˆ is a 2-form, it can be written as

1 ˆ dϑaˆ = aaˆ ϑb ∧ ϑcˆ , (307) 2 ˆbcˆ which leads to

aˆ aˆ − aˆ a ˆbcˆ = ω ˆbcˆ ω cˆˆb . (308)

By permutating the indices aˆ, ˆb and cˆ, we have the equation

aˆ ˆb − cˆ aˆ − ˆb aˆ a ˆbcˆ + a cˆaˆ a aˆˆb = ω ˆbcˆ ω aˆcˆ = 2ω ˆbcˆ . (309)

The resulting connection coefficients are ( ) 1 ˆ ωaˆ = aaˆ + ab − acˆ . (310) ˆbcˆ 2 ˆbcˆ cˆaˆ aˆˆb

39 It can be shown that

aˆ − aˆ a ˆbcˆ = c ˆbcˆ , (311)

aˆ where c ˆbcˆ is called the structure constants or commutation coefficients, which is defined by the commutation relation of the anholonomic frame [ ] a ∂ b ∂ [eaˆ, eˆb] = aaˆ a , aˆb b ∂x( ∂x ) ( ) ∂ ∂ ∂ ∂ a b − b a = aaˆ a aˆb b aˆb b aaˆ a ∂x ( )∂x ∂x ( )∂x ∂ ∂ ∂ ∂ a b − b a = aaˆ a aˆb b aˆb b aaˆ a ∂x ( )∂x ∂x ( )∂x ∂ ∂ a b cˆ − b a cˆ = aaˆ a aˆb a becˆ aˆb b aaˆ a aecˆ [ ∂x ( ) ∂x ( ) ] ∂ ∂ a b cˆ − b a cˆ = aaˆ a aˆb a b aˆb b aaˆ a a ecˆ [ ( ∂x) ( ) ∂x] b cˆ − a cˆ cˆ = eaˆ aˆb a b eˆb aaˆ a a ecˆ := c aˆˆbecˆ . (312)

Here the anholonomic frame eaˆ is identified as the so-called Pfaffian derivative. As a result, we obtain the the structure constants ( ) ( ) cˆ b cˆ − a cˆ c aˆˆb := eaˆ aˆb a b eˆb aaˆ a a (313)

We note that it is apparent that the commutator

[∂a, ∂b] = 0 (314) because two partial derivatives can be interchanged. As a result, we conclude that there is no structure constants in holonomic frame. Therefore, the commutation coefficients can also be called anholonomity which characterizes the property of the anholonomic frame. On the other hand, ( ) dϑcˆ = d acˆ dxb ( b) = d acˆ dxb ( b ) ∂ cˆ a ∧ b = a a b dx dx (∂x ) 1 ∂ ∂ ( ) ( ) cˆ − cˆ a aˆ ∧ b ˆb = a a b b a a aaˆ ϑ aˆb ϑ 2(∂x ∂x ) 1 ∂ ∂ b a cˆ − a b cˆ aˆ ∧ ˆb = aˆb aaˆ a a b aaˆ aˆb b a a ϑ ϑ 2( ∂x ∂x) 1 b cˆ − a cˆ aˆ ∧ ˆb = aˆb eaˆ(a b) aaˆ eˆb(a a) ϑ ϑ 2 ( ) 1 ˆ = − acˆ e (a b) − acˆ e (a a) ϑaˆ ∧ ϑb 2 b aˆ ˆb a ˆb aˆ 1 ˆ = − ccˆ ϑaˆ ∧ ϑb , (315) 2 aˆˆb

40 where we have used a be (acˆ ) = −acˆ e (a b) due to a bacˆ = δcˆ. The above result proves (311) ˆb aˆ b b aˆ ˆb ˆb b ˆb and finally we obtain the linear connection coefficients ( ) 1 ˆ ωaˆ = − caˆ + cb − ccˆ (316) ˆbcˆ 2 ˆbcˆ cˆaˆ aˆˆb or

1( ) ωaˆ = − caˆ − c aˆ − c aˆ (in pseudo-Riemannian geometry) . (317) ˆbcˆ 2 ˆbcˆ ˆb cˆ cˆ ˆb

Supplement (Covariant derivative in anholonomic frame). We can do the calculation in both holonomic and anholonomic frame. However, according to (251), we have ( ) j k DˆjDkêî = aˆj Dj akˆ Dkeî j k j k = aˆj akˆ DjDkeî + aˆj (∂jakˆ )(Dkeî) ˆ = a ja kD D (a ip ) + al (e a k)(D e ) ˆj kˆ j( k î i k ˆj kˆ ) ˆl î j k i i ˆl k = aˆ aˆ Dj aˆ Dkp + (∂kaˆ )p + a k(eâˆ )(Dêˆ) j k ( i i i i j k l i ) j k i i i i = aˆj akˆ aî DjDkpi + (∂jaî )Dkpi + (∂kaî )Djpi + (∂j∂kaî )pi ˆl k + a k(eˆjakˆ )(Dˆleî) . (318) So we can find ( ) ( ) ( ) − i j k − ˆl k − ˆl k DˆjDkˆ DkˆDˆj eî = aî aˆj akˆ DjDk DkDj pi + a k(eˆjakˆ ) a k(ekâˆj ) (Dˆleî) . (319)

ˆl Here we move the last term of (319) to the left-handed side and use the structure constants c ˆjkˆ defined by (313). Then, we also use (190) and (312) to obtain the equation ( ) ˆ a ia ja k(D D − D D )p = D D − D D − cl D e î ˆj kˆ j k k j i ( ˆj kˆ kˆ ˆj ˆjkˆ ˆl ) î − − = DˆjDkˆ DkˆDˆj Dcˆl e eî ( ˆjkˆ ˆl ) = DˆDˆ − DˆDˆ − D eˆ . (320) j k k j [eˆj ,ekˆ] i It gives the general formula of curvature tensor  ( )  l − −  Holonomic frame: R ijkpl = DjDk DkDj |{z}0 pi , (321a)

D[∂ ,∂ ]=0 which is vanished due to (314).  ( j k )  ˆl Anholonomic frame: R ˆeˆ = DˆDˆ − DˆDˆ − D eˆ . (321b) ˆiˆjk l j k k j [eˆj ,ekˆ] i

ˆj kˆ î Therefore, one can consider three vectors X = X eˆj, Y = Y ekˆ and Z = Z eî, then it can be shown that the frame independent formula of curvature tensor is ( ) ˆj kˆ î ˆl − − X Y Z R îˆjkêˆl = DX DY DY DX D[X,Y ] Z := R(X,Y )Z. (322) After the calculation, (321) can be written in terms of the connections and structure constants   l l l l m l m  Holonomic: R ijk = ∂jΓ ik − ∂kΓ ij + Γ mjΓ ik − Γ mkΓ ij , (323a)

 ˆl ˆl − ˆl ˆl mˆ − ˆl mˆ − ˆl mˆ Anholonomic: R îˆjkˆ = eˆjω îkˆ ekˆω îˆj + ω mˆ ˆjω îkˆ ω mˆ kˆω îˆj ω îmˆ c ˆjkˆ . (323b)

41 ˆl Finally (320) gives the transformation formula for curvature tensor by substituting pl = a leˆl to the left-handed side of (320)

ˆl ˆl i j k l R ˆiˆjkˆ = a laˆi aˆj akˆ R ijk . (324)

Similarly, it can( be shown that) by computing the commutator of the covariant derivatives on − function f, i.e., DˆjDkˆ DkˆDˆj f, we obtain the following equations { ( ) Holonomic frame: T i D f = T i ∂ f = D D − D D f , (325a) jk i jk i ( j k k j ) î î − − Anholonomic frame: T ˆjkˆDîf = T ˆjkêîf = DˆjDkˆ DkˆDˆj [eˆj, ekˆ] f , (325b)

ˆj kˆ and the frame independent formula of torsion tensor is given by vectors X = X eˆj and Y = Y ekˆ with eˆi := Dˆif of

ˆj kˆ î − − X Y T ˆjkêî = DX Y DY X [X,Y ] := T (X,Y ) . (326)

In terms of the connections and structure constants, (325) can be written as   i i i  Holonomic: T jk = Γ kj − Γ jk , (327a)

 î î − î − î Anholonomic: T ˆjkˆ = ω kˆˆj ω ˆjkˆ c ˆjkˆ . (327b)

Here we note that a vector X act on a function f and a avector Y are respectively given by

ˆj ˆj kˆ ˆj kˆ X(f) = X eˆj(f) and X(Y ) = X eˆj(Y )ekˆ + X Y eˆjekˆ . (328)

In addition, the transformation formula for torsion tensor should be

ˆi ˆi j k i T ˆjkˆ = a iaˆj akˆ T jk . (329)

n a Non-fixed frame and gauge transformation For general space M , the vector V = V Ea (a = a ∂ 1, . . . , n) under local coordinate X can be spanned by a non-fixed holonomic frame Ea := ∂Xa with ̸ R dEa = 0. The vector Ea can be spanned by another set of anholonomic frame eˆb given by a GL(n, ) transformation

ˆb ˆb ∈ R Ea = A aeˆb with A a GL(n, ) , (330) and we also have coframe

a a ˆb dX = Aˆb ϑ , (331) where we have defined the inverse

a −1 ˆb Aˆb := (A ) a . (332)

Therefore,

a a dV = dV Ea + V dEa (333)

42 b ˆb and we have to introduce the connection form Γ a and ω aˆ for the frame Ea and eaˆ respectively, which gives { b dEa = Γ aEb , (334a) ˆb deaˆ = ω aˆeˆb . (334b) The differential of V can also be expressed as ( ) a a b a b a ¯ a dV = dV Ea + V Γ aEb = dV + V Γ b Ea := (DV ) Ea . (335)

Remark. We note that there is no normal space Mn⊥ of Mn, therefore, we obtain

dV = (dV )⊤ = D¯V , (336) ¯ where the connection D is defined with respect to basis Ea. As a result, the differetial operator d also represents the covariant derivative D¯ on general space Mn.

On the other hand, the differetial ( ) ˆb ˆb ˆb ˆb cˆ ˆb cˆ ˆb dEa = dA aeˆb + A adeˆb = dA aeˆb + A aω ˆbecˆ = dA a + A aω cˆ eˆb . (337) From (330) and (334a), it implies the relation between two connection forms ( ) ( ) c c ˆb b cˆ ˆb ⇒ c c ˆb ˆb cˆ Γ aEc = Γ aA ceˆb = dA a + A aω cˆ eˆb = Γ a = Aˆb dA a + ω cˆA a , (338) which is a frame transformation or GL(n, R) gauge transformation of the connection form.

Remark. We note that the relation (338) comes from the frame transformation or GL(n, R) gauge transformation, rather than metric compatibility. This relation is sometimes called vielbein postulate. However, the equation is still valid even if the frame eaˆ is not orthonormal or the nonmetricity ˆb Qabc = −∇agbc is not vanished. In such case, the connection ω aˆ contains the symmetric or trace ̸ aˆ ̸ part, i.e., ω(ˆaˆb) = 0 or ω aˆ = 0. So it is improper to call the relation postulate. People always implicitly define a total connection D(Γ, ω) of tensor with respect to both the holonomic and an- a aˆ ˆb ⊗ a holonomic basis of Ea, dX , eaˆ and ϑ . By giving a tensor A := A aeˆb dX , the connection D act on A is ( ) ˆb a DA = D A eˆ ⊗ dX ( )a b ( ) ( ) ˆb aˆ ˆb a ˆb a = dA eˆ ⊗ dX + A Deˆ ⊗ dX + A eˆ ⊗ DdX ( aˆ) b a( b ) a b ( ) ˆb a ˆb cˆ a ˆb a c = dA eˆ ⊗ dX + A ω ˆe ⊗ dX + A eˆ ⊗ − Γ dX ( a b a )b cˆ a b c ˆb cˆ ˆb − ˆb c ⊗ a = dA a + A aω cˆ A cΓ a eˆb dX = 0 (339)

due to (338), which is independent of the metric compatibility. As a result, the component gives the vielbein postulate

ˆb ˆb cˆ ˆb ˆb c (DA) a = dA a + A aω cˆ − A cΓ a = 0 (340)

ˆb ˆb cˆ ˆb ˆb c ∇dA a = ∂dA a + A aω cdˆ − A cΓ ad = 0 . (341)

43 Figure 10: Two points on the hypersurface Mn−1 in Mn.

Now we will discuss the connection on the hypersurface Mn−1 of M n. Consider a point p on Mn−1 is identified by a vector V in M n. Simirlarly, a point q on Mn−1 is represented by V ′ in M n. Here we only focus on the connection on Mn−1, we have restricted our case that dV = q−p = ′ − Mn−1 ∂ V V lays on the hypersurface only. So dV can be expanded not only by frame Ea = ∂Xa M n ∂ − Mn−1 (a = 1, . . . , n) on but also by frame ∂i = ∂ui (i = 1, . . . , n 1) on .

Remark. We can consider the case of n = 3 and two infinitesimal closed points q and p with the spherical coordinate Xa = (r, θ, ϕ) in M 3 and polar coordinate ui = (ρ, φ) in M2. Therefore Ea and ∂i are non-fixed frames.

The differential dV a can be given by ∂V a ∂V a dV a = du + dv = ∂ V adui , (342) ∂u ∂v i then all the n-dimensional vectors can be expanded by (n−1)-dimensional ones, the resulting equation of (335) would be rewritten as dV = (D¯V )aE ( a ) a i b a c = ∂iV du + V Γ bcdX Ea ( c ) a i b a ∂X i = ∂iV du + V Γ bc du Ea |∂u{zi} c ( ) h i a b a c i a i = ∂iV + V Γ bch i du Ea := Vi du Ea . (343) Here we define ( ) a a b a c Vi := Vi Ea = ∂iV + V Γ bch i Ea . (344)

Remark. The result of (66) in E3 can be reduced from (335) by   M n −→ E3    V a −→ pa = xa , −→  Ea δa , (345)  a  Γ b −→ 0 ,  (D¯V )a −→ dxa .

44 Due to (334a) and (338), we have

ˆb ˆb cˆ ˆb cˆ ˆb 0 = dδa = dA aeˆb + A aω ˆbecˆ = (dA a + A aω cˆ)eˆb . (346)

ˆb As a consequence, the connection form ω cˆ is obtained by

ˆb a ˆb ˆb a ω cˆ = −Acˆ dA a = +A a dAcˆ . (347)

In addition, within (345), (343) becomes as ( ) ( ) a b a c i a i i dp = ∂ix + x |{z}Γ bc h i du δa = ∂ix du δa := ∂ipdu . (348) 0

M E3 Therefore, (344) reduces to the derivative vector pi := ∂ip on the hypersurface of is

a pi = ∂ip = ∂ix δa = (∂ix, ∂iy, ∂iz) . (349)

Figure 11: Two vectors on the hypersurface Mn−1 in Mn.

Induced connection However, if we move the reference point o on the hypersurface Mn−1, the vector dV should be regarded as the difference between V ′ and V on Mn−1 and is equivalent to ∂ DV with respect to the basis ∂ui as shown in Fig. 11. Then, we have ( ) ∂ dV = DV = D V k ∂uk ( ) ∂ = ∂ V k + V lΓk dui . (350) i li ∂uk ∂ By using chain rule to expand Ea in terms of ∂uk k ∂ ∂u ∂ k ∂ Ea = = = ha (351) ∂Xa |∂X{za} ∂uk ∂uk k ha and substituting the relation into (343), we have ( ) ( ) ∂ dV = ∂ V a + V bΓa hc dui h k i bc i a ∂uk

45 ( ) ( a ) b j ∂X k l ∂X a c k i ∂ = ∂i V ha + V Γ bch iha du |∂u{zj} |∂u{zl} ∂uk a h j hb ( ( l )) ∂ = (∂ V j) ha h k +V l (∂ ha )h k + hb Γa hc h k dui i | {zj a} i l a l bc i a ∂uk k δj ( ( )) ∂ = (∂ V k) + V l (∂ ha )h k + hb Γa hc h k dui . (352) i i l a l bc i a ∂uk

k n−1 Now we compare (350) and (352), the induced connection Γ li on hypersurface M can be obtained a n k from the connection Γ bc on M through the projection ha

k a k b a c k Γ li = (∂ih l)ha + h lΓ bch iha . (353)

We note that the discussion above can be generalized to the case for arbitrary frame (including fixed and non-fixed frame).

Curvature and torsion in subspace Now we would like to consider an n-dimensional space M embedded in a m-dimensional space M . We consider a so-called Darboux frame of M. We label the components by indices a,ˆ ˆb, cˆ = 1ˆ,..., mˆ on M the indices î, ˆj, kˆ = 1ˆ,..., nˆ on M, and the indices p, q, r =n ˆ + 1ˆ,..., mˆ on the normal space M⊥ of M in the orthonormal frame. We define M the geometric objects on specified by barred symbols. The frame ¯eaˆ is extended by the vector eî and epˆ ( ) ( ) î pˆ −→ eî ¯eaˆ = δaˆ eî + δaˆ epˆ e¯ := ¯eaˆ = . (354) epˆ

Similarly, we have ( ) ( ) ¯aˆ aˆ î aˆ pˆ −→ ¯ ¯aˆ î pˆ ϑ = δî ϑ + δpˆ ϑ ϑ := ϑ = ϑ ϑ . (355)

{ } {¯aˆ} ¯î î ¯pˆ pˆ Therefore, the components of ¯eaˆ and ϑ are ¯eî = eî, ¯epˆ = epˆ, ϑ = ϑ and ϑ = ϑ . The metric is defined by the inner product of two vectors, we have the following relations

⊥ g¯(¯eaˆ, ¯eˆb) = δaˆˆb , g(eî, eˆj) = δîˆj , g (epˆ, eqˆ) = δpˆqˆ , g¯(¯eî, ¯epˆ) = 0 , (356) where g¯, g and g⊥ are metrics of the manifolds M , M and M⊥ respectively. Due to the metric ˆb − aˆ compatible condition, we have ω aˆ = ω ˆb and

ˆb ˆb î ˆj ˆb î qˆ ˆb pˆ ˆj ˆb pˆ qˆ ω aˆ = δˆj δaˆω î + δqˆδaˆω î + δˆj δaˆω pˆ + δqˆδaˆω pˆ , (357) which can also be recognized by the matrix as ( ) ( ) ˆj qˆ ˆb ω î ω î ω := ω aˆ = ˆj qˆ . (358) ω pˆ ω pˆ

Now we will discuss the dynamics on subspace M only. The differential of frame on M are given by { ¯aˆ pˆ d∇p = ϑ ¯eaˆ with ϑ = 0 , (359a) ˆb d∇¯eaˆ =ω ¯ aˆ¯eˆb . (359b)

46 or   ˆ      i ˆ  d∇p = ϑ eî , ϑi 0 ( ) (360a) d∇p   ˆj qˆ   ˆj qˆ eî d∇e = ω e + ω e , −→ d∇eˆ = ω ω  . (360b)  î î ˆj î qˆ j î î e  ˆj qˆ qˆ  ˆj qˆ d∇epˆ ω pˆ ω pˆ d∇epˆ = ω pêˆj + ω pêqˆ . (360c) In addition pˆ ⇒ pˆ pˆ pˆ ∧ î pˆ ∧ qˆ T pˆ ϑ = 0 = 0 = Dϑ = d |{z}ϑ +ω î ϑ + ω qˆ |{z}ϑ = , (361) 0 0 Thus, we have

pˆ ∧ ˆi ω ˆi ϑ = 0 . (362) Applying the Cartan’s lemma, we obtain the connection form

pˆ − î pˆ ˆj ω î = ω pˆ = h îˆjϑ , (363) or

ˆj − − ˆj ˆj ωpˆî = hpˆîˆjϑ and ωîpˆ = ωpˆî = hpˆîˆjϑ = hîpˆˆjϑ (in pseudo-Riemannian geometry) . (364) Now we would like to calculate the differential of structure equations. The covarint exterior differentiation of first structure equation is obtained by

2 ¯aˆ î pˆ d∇p = d∇(ϑ ¯eaˆ) = d∇(ϑ eî + ϑ epˆ) î î pˆ pˆ = dϑ eˆ − ϑ ∧ d∇eˆ + dϑ e − ϑ ∧ d∇e i ( i pˆ ) pˆ ( ) î − î ∧ ˆj pˆ pˆ − pˆ ∧ î qˆ = dϑ eî ϑ ω îeˆj + ω îepˆ + dϑ epˆ ϑ ω peî + ω pêqˆ i ˆj î pˆ î pˆ î pˆ qˆ pˆ = dϑ ei + ω ˆ ∧ ϑ eˆ + ω ˆ ∧ ϑ epˆ + dϑ epˆ + ω pˆ ∧ ϑ eˆ + ω pˆ ∧ ϑ eqˆ ( i j i ) ( i ) = dϑî + ωî ∧ ϑî +ωî ∧ ϑpˆ e + dϑpˆ + ωpˆ ∧ ϑqˆ +ωpˆ ∧ ϑî e | {z ˆj } pˆ î | {z qˆ } î pˆ T î T pˆ T aˆ T î T pˆ = ¯eaˆ = eî + epˆ . (365) By using (361), we obtain ( ) 2 T î T î pˆ ∧ qˆ T î d∇p = eî = + ω qˆ |{z}ϑ eî = eî , (366) 0 which leads to the equation

T ˆi = T ˆi . (367)

Remark. For case of M is embedded in M , we have consquence of

T ˆi = T ˆi . (368)

It means that there is no extrinsic torsion contribution in the equation of torsion in embedding structure of geometry (cf. Gauss equation (372a)).

47 The covarint exterior differentiation of second structure equation is 2 Raˆ Rˆj Rpˆ d∇¯eaˆ = aˆ¯eaˆ = aêˆj + aêpˆ , (369) 2 2 which can be calculated separately by d∇eî and d∇epˆ. They are shown by

2 ˆj ˆj pˆ pˆ d∇eˆ = dω êˆ − ω i ∧ d∇eˆ + dω êpˆ − ω ˆ ∧ d∇epˆ i i j ( j i ) i ( ) ˆj − ˆj ∧ kˆ pˆ pˆ − pˆ ∧ ˆj qˆ = dω îeˆj ω î ω ˆjekˆ + ω ˆjepˆ + dω îepˆ ω î ω pêˆj + ω pêqˆ ˆj kˆ ˆj pˆ ˆj pˆ ˆj pˆ qˆ pˆ = dω êˆ + ω ˆ ∧ ω êˆ + ω ˆ ∧ ω êpˆ + dω êpˆ + ω pˆ ∧ ω êˆ + ω pˆ ∧ ω êqˆ ( i j j i k j ) i ( i i j ) i ˆ = dωˆj + ωˆj ∧ ωk +ωˆj ∧ ωpˆ e + dωpˆ + ωpˆ ∧ ωˆj + ωpˆ ∧ ωqˆ e | î {z kˆ }î pˆ î ˆj î ˆj î qˆ î pˆ Rˆj î Rˆj Rpˆ Raˆ = îeˆj + îepˆ = î¯eaˆ (370) and

2 î î qˆ qˆ d∇e = dω eˆ − ω ∧ d∇eˆ + dω e − ω ∧ d∇e pˆ pˆ i pˆ ( i pˆ qˆ) pˆ qˆ ( ) î − î ∧ ˆj qˆ qˆ − qˆ ∧ î rˆ = dω pêî ω pˆ ω îeˆj + ω îeqˆ + dω pêqˆ ω pˆ ω qêî + ω qêrˆ î ˆj î qˆ î qˆ î qˆ rˆ qˆ = dω pêˆ + ω ˆ ∧ ω pêˆ + ω ˆ ∧ ω pêqˆ + dω pêqˆ + ω qˆ ∧ ω pêˆ + ω qˆ ∧ ω pêrˆ ( i i j i ) ( i ) = dωî + ωî ∧ ωˆj + ωî ∧ ωqˆ e + dωqˆ + ωqˆ ∧ ωrˆ +ωqˆ ∧ ωî e pˆ ˆj pˆ qˆ pˆ î | pˆ {z rˆ }pˆ î pˆ qˆ qˆ R pˆ Rî Rqˆ Raˆ = pêî + pêqˆ = pˆ¯eaˆ , (371) respectively. According to the results above, we have the following equations:   Rˆj Rˆj ˆj ∧ pˆ  Gauss equation: î = î + ω pˆ ω î , (372a) Rpˆ pˆ pˆ ∧ ˆj pˆ ∧ qˆ Codazzi equation: î = dω î + ω ˆj ω î + ω qˆ ω î , (372b)  Rqˆ Rqˆ qˆ ∧ î Ricci equation: pˆ = pˆ + ω î ω pˆ . (372c)

Subspace of Em We consider that a space M is embedded in the flat space Em. We can chose the m cartesian coordinate for E , every component of the orthonormal frame {¯eaˆ} is related to the fixed cartesian frame by   ∂ ¯e = a jδ = a j , and ϑ¯i = aî dxj (i = 1, 2, . . . , n) , î î j î ∂xj j (373)  ∂ ¯e = a qδ = a q , and ϑ¯pˆ = apˆ dxq (p = n + 1, . . . , m) , pˆ pˆ q pˆ ∂xq q j î ∈ R M ¯pˆ where aî , a j SO(m, ). Therefore, the differential of the frame on subspace with ϑ = 0 is given by { î ⊗ i ⊗ d∇p = ϑ eî = dx δi , (374a) ˆb ⊗ b ⊗ d∇¯eaˆ = ω aˆ ¯eˆb = Γ a δb , (374b) and we can show that the torsion and curvature are vanished by using (374) in terms of cartesian frame, which gives the following equations for frame with torsion-free and curvature-free on Em   2 T î  d∇p = eî = 0 , (375a) d2 e = Raˆ ¯e = 0 , (375b)  ∇ î î aˆ  2 aˆ d∇epˆ = R pˆ¯eaˆ = 0 . (375c)

48 The equations (367) and (372) turn out to be   T î î î ∧ ˆj  Torsion-free: = dϑ + ω ˆj ϑ = 0 , (376a)   Rˆj − ˆj ∧ pˆ pˆ ∧ pˆ Gauss equation: î = ω pˆ ω î = ω ˆj ω î , (376b)  pˆ pˆ ∧ ˆj p ∧ qˆ Codazzi equation: 0 = dω î + ω ˆj ω î + ω qˆ ω î , (376c)  Rqˆ − qˆ ∧ î qˆ ∧ pˆ Ricci equation: pˆ = ω î ω pˆ = ω î ω î , (376d) because all barred torsion and curvature 2-forms should be vanished in Em. According to (376a), we have torsion-free, it leads us to have Ricci rotation coefficients written as (316). From the consequence of (363), the Gauss equation (376b) is 1 Rˆj = Rˆj ϑkˆ ∧ ϑˆl î 2 îkˆˆl ( ) ˆ ˆ 1 ˆ ˆ = (hpˆ ϑk) ∧ (hpˆ ϑl) = hpˆ hpˆ − hpˆ hpˆ ϑk ∧ ϑl , (377) ˆjkˆ îˆl 2 ˆjkˆ îˆl ˆjˆl îkˆ i.e.,

ˆj pˆ pˆ − pˆ pˆ R îkˆˆl = h ˆjkˆh îˆl h ˆjˆlh îkˆ , (378) or

ˆj − ˆj pˆ ˆj pˆ ˆj pˆ − ˆj pˆ R îkˆˆl = h pˆkˆh îˆl + h pˆˆlh îkˆ = hpˆ kˆh îˆl hpˆ ˆlh îkˆ (in pseudo-Riemannian geometry) . (379)

The Codazzi equation (376c) becomes

pˆ ˆj pˆ kˆ ∧ ˆj pˆ ∧ qˆ ˆj 0 = d(h îˆjϑ ) + (h ˆjkˆϑ ) ω î + ω qˆ (h îˆjϑ ) pˆ ∧ ˆj pˆ ˆj pˆ kˆ ∧ ˆj qˆ pˆ ∧ ˆj = dh îˆj ϑ + h îˆjdϑ + h ˆjkˆϑ ω î + h îˆjω qˆ ϑ . (380)

The Ricci equation can be read as 1 Rqˆ = Rqˆ ϑkˆ ∧ ϑˆl pˆ 2 pˆkˆˆl ( ) ˆ ˆ 1 ˆ ˆ = (hqˆ ϑk) ∧ (hpˆ ϑl) = hqˆ hpˆ − hqˆ hpˆ ϑk ∧ ϑl , (381) îkˆ îˆl 2 îkˆ îˆl îˆl îkˆ i.e.,

qˆ qˆ pˆ − qˆ pˆ R pˆkˆˆl = h îkˆh îˆl h îˆlh îkˆ , (382) or

qˆ − qˆ î qˆ î qˆ î − qˆ î R pˆkˆˆl = h îkˆh pˆˆl + h îˆlh pˆkˆ = h îkˆhpˆ ˆl h îˆlhpˆ kˆ (in pseudo-Riemannian geometry) . (383)

E3 M M M2 E3 ∂ Example (Hypersurface of ). If we consider and to be and and ∂x3 is assume to be aligned to the normal vector n of M, we have   ∂ ¯e = a jδ = a j , and ϑ¯i = aî dxj (i, j = 1, 2 and x1 = x, x2 = y) , î î j î ∂xj j (384a)  ∂ ¯e = a 3δ = a 3 , and ϑ¯3ˆ = a3ˆ dx3 (xp = x3 = z) . 3ˆ 3ˆ 3 3ˆ ∂x3 3

49 Due to the fixed condition p = n + 1 = m = 3, it is impossible to have p ≠ q, which leads to the results for hypersuface with ϑ¯3ˆ = 0 of

qˆ R pˆ = 0 (for hypersuface) , (385)

and

pˆ ω qˆ = 0 (for hypersuface) . (386)

3ˆ M We can identify h îˆj to be bîˆj which is the extrinsic curvature of . The corresponding component equations of (378) and (380) are   ˆj − R îkˆˆl = bˆjkˆbîˆl bˆjˆlbîkˆ , (387a)  ∧ ˆj ˆj ˆj kˆ ∧ ˆl 0 = dbîˆj ϑ + bîˆjdϑ + bˆjkˆω îˆlϑ ϑ . (387b)

If we use the holonomic frame with coordinate {ui} on M, we have

∂xj ∂xj ϑî = aî dxj = aî duk = dui =⇒ eî := aî = δî (388) j j ∂uk k j ∂uk k

î i î ˆj j such that dϑ = ddu = 0 (or c ˆjkˆ = 0) and ω î = Γ i, therefore (387a) and (387b) becomes

j R ikl = bjkbil − bjlbik (389)

and ( ) ( ) k j j k l l l ∂kbij du ∧ du + bjkΓ il du ∧ du = 0 =⇒ ∂kbij − ∂jbik + blkΓ ij − bljΓ ik = 0 , (390)

which have been given by (231) and (232) respectively.

Acknowledgments

Ling-Wei Luo (LWL) would like to thank Prof. Jen-Chi Lee (NCTU) who encurages LWL and gives an opportunity for LWL to give a lecture on elementary differential geometry for both undergraduate and dergraduate students in NCTU. LWL also would like to thans Mr. Shang-Yu Chien who is a PhD student in National Tsing Hua University to check and discuss the contents of this lecture note. The work was supported by Ministry of Science and Technology (MoST-107-2119-M-007-013-MY3) and Academia Sinica Career Development Award Program (AS-CDA-105-M06).

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