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Home , Local diffeomorphism, Tangent bundle, Tangent space, Tangent vector, Vector space

4 The

4.1 Tangent

n For embedded M R , the tangent TpM at p M can be defined ✓ 2 as the of all vectors v = g˙(0), where g : J M is a smooth with ! g(0)=p; here J R is an open around 0. ✓

n It turns out (not entirely obvious!) that TpM becomes a vector subspace of R . (Warn- ing: In pictures we tend to draw the as an affine subspace, where the has been moved to p.)

Example 4.1. Consider the Sn Rn+1, given as the set of x such that x 2 = 1. ✓ || || A curve g(t) lies in Sn if and only if g(t) = 1. Taking the of the equation || || g(t) g(t)=1 at t = 0, we obtain (after dividing by 2, and using g(0)=p) · p g˙(0)=0. · n+1 3 That is, TpM consists of vectors v R that are orthogonal to p R 0 . It is not 2 2 \{ } hard to see that every such vector v is of the form g˙(0),⇤ hence that n TpS =(Rp)?, the orthogonal to the through p.

Given v, take g(t)=(p +tv)/ p +tv . ⇤ || || 70 4 The

2 n n Exercise 45. What is T0R ? What is T0R ? How about TpR for some other p Rn? 2 To extend this idea to general , note that the vector v = g˙(0) defines a “” C•(M) R: ! v : f d f (g(t)). 7! dt |t=0

Exercise 46. Fix some p Rn Show that the directional derivative in Rn 2 f (p +tv) f (p) v( f )=lim t 0 t ! n n • n (for v TpR = R and f C (R )) has the following properties. For any 2 ⇠ • n 2 v TpM and f ,g C (R ): 2 2 a) v : C• R is well-defined (that is, the limit exists for any f C•). ! 2 b) v : C• R is linear: v(l f + µg)=lv( f )+µv(g). ! c) v : C• R is a derivation: v( fg)=v( f )g + fv(g). ! For a general , we will define TpM as a set of directional . Definition 4.1 (Tangent spaces – first definition). Let M be a manifold, p M. The • 2 tangent space TpM is the set of all linear maps v : C (M) R of the form ! v( f )= d f (g(t)) dt |t=0 for some smooth curve g C•(J,M) with g(0)=p. 2 The elements v T M are called the tangent vectors to M at p. 2 p The following local coordinate description makes it clear that TpM is a of the L(C•(M),R) of linear maps C•(M) R, of ! equal to the dimension of M. Theorem 4.1. Let (U,j) be a coordinate chart around p. A linear v : C•(M) ! R is in TpM if and only if it has the form, m ∂( f j 1) v( f )= ai  ∂ui u=j(p) i=1 for some a =(a1,...,am) Rm. 2 Proof. Given a v of this form, let g˜ : R j(U) be a curve with g˜(t)= ! j(p)+ta for t sufficiently small. Let g = j 1 g˜. Then | | d d 1 f (g(t)) = ( f j )(j(p)+ta) dt t=0 dt t=0 m 1 ∂( f j ) = ai ,  ∂ui u=j(p) i=1 4.1 Tangent spaces 71

by the chain rule. Conversely, given any curve g with g(0)=p, let g˜ = j g be the corresponding curve in j(U) (defined for small t ). Then g(0)=j(p), and | | d d 1 f (g(t)) = ( f j )(g˜(te)) dt t=0 dt t=0 m 1 i ∂( f j ) = Âa i u=g(p), i=1 ∂u |

where a = dg˜ . dt t=0 ut We can use this result as an alternative definition of the tangent space, namely: Definition 4.2 (Tangent spaces – second definition). Let (U,j) be a chart around • p. The tangent space TpM is the set of all linear maps v : C (M) R of the form ! m ∂( f j 1) v( f )= ai (4.1)  ∂ui u=j(p) i=1 for some a =(a1,...,am) Rm. 2 Remark 4.1. From this version of the definition, it is immediate that TpM is an m- dimensional vector space. It is not immediately obvious from this second definition that TpM is independent of the choice of coordinate chart, but this follows from the equivalence with the first definition. Alternatively, one may check directly that the subspace of L(C•(M),R) characterized by (4.1) does not depend on the chart, by studying the effect of a change of coordinates. According to (4.1), any choice of coordinate chart (U,j) around p defines a vector m 1 m space TpM = R , taking v to a =(a ,...,a ). In particular, we see that m ⇠ if U R is an open , and p U, then TpU is the subspace of the space of ✓ • 2 linear maps C (M) R spanned by the partial derivatives at p. That is, TpU has a ! ∂ ∂ ,..., ∂x1 |p ∂xm |p m identifying TpU R . Given ⌘ ∂ v = ai  ∂xi |p the coefficients ai are obtained by applying v to the coordinate functions x1,...,xm : U R, that is, ai = v(xi). ! We now describe yet another approach to tangent spaces which again charac- terizes “directional derivatives” in a coordinate-free way, but without reference to g. Note first that every satisfies the , also called the Leibniz rule: Lemma 4.1. Let v T M be a tangent vector at p M. Then 2 p 2 v( fg)= f (p)v(g)+v( f )g(p) (4.2) for all f ,g C•(M). 2 72 4 The Tangent Bundle

Proof. Letting v be represented by a curve g, this follows from d d d f g(t) g g(t) = f (p) g g(t) + f g(t) g(p). dt t=0 dt t=0 dt t=0 ⇣ ⌘ ⇣ ⌘ ⇣ ⌘ ut Alternatively, in local coordinates it is just the product rule for partial derivatives. It turns out that the product rule completely characterizes tangent vectors:

Exercise 47. Suppose that v : C•(M) R is a linear map satisfying the product ! rule (4.2). Prove the following two facts, which will be used in the proof of Theorem 4.2 below: a) v vanishes on constants. That is, if f C•(M) is the constant map, then 2 v( f )=0. b) Suppose f ,g C•(M) with f (p)=g(p)=0. Then v( fg)=0. 2 • Theorem 4.2. A linear map v : C (M) R defines an element of TpM if and only if ! it satisfies the product rule (4.2). The proof of this result will require the following fact from multivariable calcu- lus: m Lemma 4.2 (Hadamard Lemma). Let U = BR(0) R be an open ball of radius ✓ R > 0 and h C•(U) a smooth . Then there exist smooth functions h 2 i 2 C•(U) with m i h(u)=h(0)+Â u hi(u) i=1 for all u U. Here h (0)= ∂h (0). 2 i ∂ui Proof. Let h be the functions defined for u =(u1,...,um) U by i 2 1 h(u1,...,ui,0,...,0) h(u1,...,ui 1,0,0,...,0) if ui = 0 ui 6 h (u)= i 8 ∂h ( 1,..., i 1, , ,..., ) i = < ∂ui u u 0 0 0 if u 0

Using Taylor’s: formula with remainder, one sees that these functions are smooth†. i m i If all u = 0, then the sum  u hi(u) is a telescoping sum, equal to h(u) h(0). By 6 i=1 continuity, this result extends to all u. Finally, evaluating the derivative

m ∂h ∂ i k ∂hk i = i h(0)+Â u hi(u) = hi(u)+Âu i ∂u ∂u i=1 ! k ∂u

at u = 0, we see that ∂h = h (0). ∂ui u=0 i ut † It is a well-known fact from calculus (proved e.g. by using Taylor’s theorem with remain- der) that if f is a smooth function of a real x, then the function g, defined as g(x)=x 1( f (x) f (0)) for x = 0 and g(0)= f (0), is smooth. 6 0 4.1 Tangent spaces 73

Proof (Theorem 4.2). Let v : C•(M) R be a linear map satisfying the product rule ! (4.2).

Step 1: If f1 = f2 on some open neighborhood U of p, then v( f1)=v( f2).

Equivalently, letting f = f1 f2, we show that v( f )=0 if f = 0 on U. Choose • a ‘’ c C (M) with c(p)=1, with c M U = 0. Then f c = 0. The 2 | \ product rule tells us that

0 = v( f c)=v( f )c(p)+v(c) f (p)=v( f ).

Step 2: Let (U,j) be a chart around p, with U = j(U). Then there is unique linear map v : C•(U) R such that v( f )=v( f ) whenever f agrees with ! f j 1 on some neighborhood of p. e Given f , we cane alwayse find a function fesuche that f agrees withe f j 1 on some neighborhood of p. Given anothere such function g, it follows from Step 1 that v( f )=v(g)e. e Step 3: In a chart (Ue,j) around p, the map v : C•(M) R is of the form (4.1). ! Since the condition (4.1) does not depend on the choice of chart around p, we may assume that p = j(p)=0, and that U is an open ball of some radius R > 0 around 0. Define v as in Step 2. Since v satisfies the product rule on C•(M), the map v satisfies the product rule on C•(U).e Given f C•(M), consider the Taylor e 2 expansion of the coordinate expression f˜ = f j 1 near u = 0: e e e ∂ f˜ f˜(u)= f (0)+ ui + r˜(u) Â ∂ui u=0 i The remainder termr ˜ is a smoothe function that vanishes at u = 0 together with its i first derivatives. By Lemma 4.2, it can be written in the formr ˜(u)=Âi u r˜i(u) where r˜i are smooth functions that vanish at 0. Let us now apply v to the formula for f . Since v vanishes on products of functions vanishing at 0 (by Exercise 4.1), we have that v(r)=0. Since it also vanishes on constants, we obtain e e e ∂ f˜ e e v( f )=v( f )= ai , Â ∂ui u=0 i e where we put ai = v(ui). e

To summarize, we have the following alternative definition of tangent spaces: e Definition 4.3 (Tangent spaces – third definition). The tangent space TpM is the space of linear maps C•(M) R satisfying the product rule, ! v( fg)= f (p)v(g)+v( f )g(p) for all f ,g C•(M). 2 74 4 The Tangent Bundle

At first sight, this characterization may seem a bit less intuitive then the defini- tion as directional derivatives along curves. But it has the advantage of being less redundant – a tangent vector may be represented by many curves. Also, as in the co- ordinate definition it is immediate that TpM is a linear subspace of the vector space L(C•(M),R). One may still want to use local charts, however, to prove that this vector subspace has dimension equal to the dimension of M. The following remark gives yet another characterization of the tangent space. Please read it only if you like it abstract – otherwise skip this!

Remark 4.2 (A fourth definition). There is a fourth definition of TpM, as follows. • For any p M, let Cp (M) denotes the subspace of functions vanishing at p, and let • 2 2 • C (M) consist of finite sums  fi gi where fi,gi C (M). We have a p i 2 p decomposition C•(M)=R C•(M), p where R is regarded as the constant functions. Since any tangent vector v : C•(M) • ! R vanishes on constants, v is effectively a map v : Cp (M) R. By the product • 2 • ! rule, v vanishes on the subspace Cp (M) Cp (M). Thus v descends to a linear map • • 2 ✓ • • 2 C (M)/C (M) R, i.e. an element of the (C (M)/C (M) )⇤. The p p ! p p map • • 2 T M (C (M)/C (M) )⇤ p ! p p just defined is an isomorphism, and can therefore be used as a definition of TpM. This may appear very fancy on first sight, but really just says that a tangent vector is a linear on C•(M) that vanishes on constants and depends only on the first order Taylor expansion of the function at p. Furthermore, this viewpoint lends itself to generalizations which are relevant to algebraic and non-commutative • geometry: The ‘vanishing ideals’ Cp (M) are the maximal ideals in the of • 2 smooth functions, with Cp (M) their second power (in the sense of products of ide- als). Thus, for any maximal ideal I in a A one may regard 2 (I /I )⇤ as a ‘tangent space’.

After this lengthy discussion of tangent spaces, observe that the velocity vec- tors of curves are naturally elements of the tangent space. Indeed, let J R be an ✓ open interval, and g C•(J,M) a smooth curve. Then for any t J, the tangent (or 2 0 2 velocity) vector g˙(t ) T M. 0 2 g(t0) at time t0 is given in terms of its action on functions by d (g˙(t0))( f )= f (g(t)) dt t=t0 We will also use the notation dg (t ) or dg to denote the velocity vector. dt 0 dt |t0 4.2 The Tangent Map 75 4.2 The Tangent Map

4.2.1 Definition of the tangent map, basic properties

For smooth maps F C•(U,V) between open U Rm and V Rn of Eu- 2 ✓ ✓ clidean spaces, and any given p U, we considered the derivative to be the linear 2 map m n d DpF : R R , a F(p +ta). ! 7! dt t=0 The following definition generalizes the derivative to smooth maps between mani- folds. Definition 4.4. Let M,N be manifolds and F C•(M,N). For any p M, we define 2 2 the tangent map to be the linear map T F : T M T N p p ! F(p) given by T F(v) (g)=v(g F) p for v T M and g C•(N). 2 p 2 We leave it as an exercise to check that the right hand side does indeed define a tangent vector:

Exercise 48. Show that for all v T M, the map g v(g F) satisfies the prod- 2 p 7! uct rule at q = F(p), hence defines an element of TqN.

Proposition 4.1. If v T M is represented by a curve g : J M, then (T F)(v) is 2 p ! p represented by the curve F g. Exercise 49. Prove Proposition 4.1.

Remark 4.3 (Pull-backs, push-forwards). For smooth maps F C•(M,N), one can 2 consider various ‘pull-backs’ of objects on N to objects on M, and ‘push-forwards’ of objects on M to objects on N. Pull-backs are generally denoted by F⇤, push-forwards by F . For example, functions on N pull back ⇤ • • g C (N) F⇤g = g F C (M). 2 2 Curves on M push forward:

g : J M F g = F g : J N. ! ⇤ ! Tangent vectors to M also push forward,

v TpM F (v)=(TpF)(v). 2 ⇤ The definition of the tangent map can be phrased in these terms as (F v)(g)=v(F⇤g). ⇤ Note also that if v is represented by the curve g, then F v is represented by the curve ⇤ F g. ⇤ 76 4 The Tangent Bundle

Proposition 4.2 (Chain rule). Let M,N,Q be manifolds. Under composition of maps F C•(M,N) and F C•(N,Q), 2 0 2 T (F0 F)=T F0 T F. p F(p) p Exercise 50. Prove Proposition 4.2. Give two different proofs: one using the action of tangent vectors on functions, and one using Proposition 4.1.

Exercise 51. a) Show that the tangent map of the identity map id : M M M ! at p M is the identity map on the tangent space: 2

Tp idM = idTpM.

b) Show that if F C•(M,N) is a , then T F is a linear iso- 2 p morphism, with inverse

1 1 (TpF) =(TF(p)F ).

c) Suppose that F C•(M,N) is a constant map, that is, F(M)= q for 2 { } some element q N. Show that T F = 0 for all p M. 2 p 2

4.2.2 Coordinate description of the tangent map To get a better understanding of the tangent map, let us first consider the special case where F C•(U,V) is a smooth map between open subsets U Rm and V Rn. 2 ✓m ✓ For p U, the tangent space TpU is canonically identified with R , using the basis 2 ∂ ∂ ,..., TpU ∂x1 p ∂xm p 2 of the tangent space (cf. Remark 4.1). Similarly, T V = n, using the basis given F(p) ⇠ R ∂ by partial derivatives ∂y j F(p). Using this identifications, the tangent map becomes a m | n linear map TpF : R R , i.e. it is given by an n m-. This matrix is exactly ! ⇥ the Jacobian: Proposition 4.3. Let F C•(U,V) be a smooth map between open subsets U Rm n 2 ✓ and V R . For all p M, the tangent map TpF is just the derivative (i.e., Jacobian ✓ 2 matrix) DpF of F at p. Proof. For g C•(V), we calculate 2 ∂ ∂ (TpF) (g)= (g F) ∂xi p ∂xi p ⇣ ⌘ n ∂g ∂F j = Â ∂y j F(p) ∂xi p j=1 n j ∂F ∂ = Â i j (g). = ∂x p ∂y F(p) ⇣ j 1 ⌘ 4.2 The Tangent Map 77

This shows ∂ n ∂F j ∂ (TpF) = . ∂xi p  ∂xi p ∂y j F(p) j=1 Hence, in terms of the given bases of TpU and TF (p)V, the matrix of the linear map ∂F j TpF has entries i . ∂x p • Remark 4.4. For F C (U,V), it is common to write y = F(x), and accordingly j 2 ( ∂y ) write ∂xi i, j for the Jacobian. In these terms, the derivative reads as ∂ ∂y j ∂ TpF i =  i j . ∂x p j ∂x p ∂y F(p) ⇣ ⌘ This formula is often used for explicit calculations.

Exercise 52. Consider R2 with coordinates x,y. Introduce a new , polar coordinates r,q given by

x = r cosq , y = r sinq

on the (r,q) (0,•) ( p,p). Express the tangent vectors 2 ⇥ ∂ ∂ , ∂r ∂q p p as a combination of the tangent vectors ∂ ∂ , ∂x ∂y p p (with coefficients are C• functions of x,y).

For a general smooth map F C•(M,N), we obtain a similar description once 2 we pick coordinate charts. Given p M, choose charts (U,j) around p and (V,y) 2 around F(p), with F(U) V. Let U = j(U), V = y(V), and put ✓ 1 F = y F j : U V. e e ! Since the coordinate map j : U Rm is a diffeomorphism onto U, It gives an e ! e e isomorphism (cf. Exercise 51)

m e Tpj : TpU T (p)U = R . ! j n Similarly, TF(p)y gives an isomorphism of TFe(p)V with R . Note also that since U M is open, we have that T U = T M. We obtain, ✓ p p 1 T F = T y T F (T j) . j(p) F(p) p p e 78 4 The Tangent Bundle which may be depicted in a

D F m j(p) / n RO RO e T j T y p ⇠= ⇠= F(p)

TpM = TpU / TF(p)V = TF(p)N TpF

Now that we have recognized TpF as the derivative expressed in a coordinate- free way, we may liberate some of our earlier definitions from coordinates: Definition 4.5. Let F C•(M,N). 2 The of F at p M, denoted rank (F), is the rank of the linear map T F. • 2 p p F has maximal rank at p if rank (F)=min(dimM,dimN). • p F is a if T F is surjective for all p M, • p 2 F is an if T F is injective for all p M, • p 2 F is a if T F is an isomorphism for all p M. • p 2 p M is a critical point of F is T F does not have maximal rank at p. • 2 p q N is a regular value of F if T F is surjective for all p F 1(q) (in particular, • 2 p 2 if q F(M)). 62 q N is a singular value (sometimes called critical value) if it is not a regular • 2 value.

Exercise 53. As an example of the advantage of an intrinsic (i.e. coordinate free) definition, use this new definitions to show that the compositions of two submersions is again a submersion, and that the composition of two immersions is an immersion.

4.2.3 Tangent spaces of submanifolds Suppose S M is a , and p S. Then the tangent space T S is canon- ✓ 2 p ically identified as a subspace of T M. Indeed, since the inclusion i : S , M is an p ! immersion, the tangent map is an injective linear map, T i : T S T M, p p ! p and we identify TpS with the subspace given as the image of this map‡. As a special case, we see that whenever M is realized as a submanifold of Rn, then its tangent n n spaces TpM may be viewed as subspaces of TpR = R . Proposition 4.4. Let F C•(M,N) be a smooth map, having q N as a regular 2 2 value, and let S = F 1(q). For all p S, 2 TpS = ker(TpF), as subspaces of TpM. ‡ Hopefully, the identifications are not getting too confusing: S gets identified with i(S) M, ✓ hence also p S with its image i(p) in M, and T S gets identified with (T i)(T S) T M. 2 p p p ✓ p 4.2 The Tangent Map 79

Proof. Let m = dimM, n = dimN. Since TpF is surjective, its has dimension m n. By the form for submersions, this is also the dimension of S, hence of T S. It is therefore enough to show that T S ker(T F). Letting i : S M be he p p ✓ p ! inclusion, we have to show that T F T i = T (F i) p p p is the zero map. But F i is a constant map, taking all points of S to the constant value q N. The tangent map to a constant map is just zero (Exercise 51). Hence 2 T (F i)=0. p ut As a special case, we can describe the tangent spaces to level sets: Corollary 4.1. Suppose V Rn is open, and q Rk is a regular value of F • k ✓ 2 1 2 C (M,R ), defining an embedded submanifold M = F (q). For all p M, the tan- n n 2 gent space TpM TpR = R is given as ✓ T M = ker(T F) ker(D F). p p ⌘ p n Example 4.2. Recall that at the beginning of the chapter we have calculated TpS directly from the curves definition of the tangent space. Here is another way of doing so. We Let F : Rn+1 R be the map F(x)=x x =(x0)2 + +(xn)2. Then, for all ! · ··· p F 1(1)=Sn, 2 d d (DpF)(a)= F(p +ta)= (p +ta) (p +ta)=2p a, dt t=0 dt t=0 · · hence n n+1 TpS = a R a p = 0 = span(p)?. { 2 | · } As another typical application, suppose that S M is a submanifold, and f ✓ 2 C•(S) is a smooth function given as the restriction f = h of a smooth function |S h C•(M). Consider the problem of finding the critical points p S of f , that is, 2 2 Crit( f )= p S T f = 0 . { 2 | p } Letting i : S M be the inclusion, we have f = h = h i, hence T f = T h T i. ! |S p p p It follows that Tp f = 0 if and only if Tph vanishes on the range of Tpi, that is on TpS: Crit( f )= p S T S ker(T h) . { 2 | p ✓ p } m If M = R , then Tph is just the Jacobian Dph, whose kernel is sometimes rather easy to compute – in any case this approach tends to be much faster than a calculation in charts.

Exercise 54. Find the critical points of

f : S2 R, f (x,y,z)=xy. ! (a) Write f = h i for the appropriate map h. Then find T h = D h. p p (b) Examine what happens when Dph = 0. 80 4 The Tangent Bundle

(c) Suppose D h = 0. Find kerT h, and find when does T S2 ker(T h). p 6 p p ✓ p (d) Summarize your finding by listing the critical points (you should have found six of them).

Exercise 55. Let S R3 be a . Show that p S is a critical point of the ✓ 2 function f C•(S) given by f (x,y,z)=z, if and only if T S is the xy-. 2 p

Exercise 56. Show that the equations

x2 + y = 0 , x2 + y2 + z3 + w4 + y = 1

define a two dimensional submanifold S of R4, and find the equation of the tangent space at the point (x ,y ,z ,w )=( 1, 1, 1, 1). 0 0 0 0 (a) Reduce the problem to showing that (0,1) is a regular value of a certain function F C•(R4,R2). 2 (b) Compute TpF = DpF, and find the critical points of F. Show that (0,1) is a regular value. (c) Compute D F at ( 1, 1, 1, 1) and use Corollary 4.1 to find the equa- p tion of the tangent space.

Example 4.3. We had discussed various matrix Lie groups G as examples of mani- folds. By definition, these are submanifolds G Mat (n), consisting of invertible ✓ R matrices with the properties

1 A,B G AB G, A G A G. 2 ) 2 2 ) 2 The tangent space to the identity ( unit) for such matrix Lie groups G turns out to be important; it is commonly denoted by lower case Fraktur letters:

g = TIG.

Some concrete examples: 1. The matrix

GL(n,R)= A Mat (n) det(A) = 0 { 2 R | 6 }

of all invertible matrices is an open subset of MatR(n), hence

gl(n,R)=MatR(n) is the entire space of matrices. 4.2 The Tangent Map 81

2. For the group O(n), consisting of matrices with F(A) := A>A = I, we have com- puted TAF(X)=X>A + AX>. For A = I, the kernel of this map is

o(n)= X Mat (n) X> = X . { 2 R | }

3. For the group SL(n,R)= A MatR(n) det(A)=1 , given as the level set 1 { 2 | } F (1) of the function det : Mat (n) R, we calculate R !

d d d 1 1 DAF(X)= F(A+tX)= det(A+tX)= det(I +tA X)=tr(A X), dt t=0 dt t=0 dt t=0 where tr : Mat (n) R is the (sum of diagonal entries). (See exercise R ! below.) Hence sl(n,R)= X Mat (n) tr(X)=0 . { 2 R | }

Exercise 57. Show that for every X Mat (n), 2 R d det(I +tX)=tr(X).

dt t=0

both and trace are invariant under .

Hint: Use that every matrix is similar to an upper triangular matrix, and that

4.2.4 Example: Steiner’s surface revisited

As we discussed in 3.5.4, Steiner’s ‘Roman surface’ is the image of the map 1 RP2 R3, (x : y : z) (yz, xz, xy). ! 7! x2 + y2 + z2

(We changed notation from a,b,g to x,y,z.) Given a point p RP2, is the map an 2 immersion on an open neighborhood of p? To investigate this question, one can ex- press the map in local charts, and compute the resulting Jacobian matrix. While this approach is perfectly fine, the resulting expressions will become rather complicated. A simpler approach is to consider the composition with the local diffeomorphism p : S2 RP2, given as ! S2 R3, (x,y,z) (yz, xz, xy). ! 7!

In turn, this map is the restriction F 2 of the map |S F : R3 R3, (x,y,z) (yz, xz, xy). ! 7! 2 We have T (F 2 )=T F 2 , hence ker(T (F 2 )) = ker(T F) T S . p |S p |TpS p |S p \ p 82 4 The Tangent Bundle

Exercise 58.

(a) Compute TpF = DpF, and find its determinant. Conclude that the kernel is empty except when one of the coordinates is 0. (b) Suppose p lies in the yz-plane. Find the kernel of D F, and ker(T F) T S. p p \ p Repeat with p lying in the xz-plane, and the xy-plane. (c) What are the critical points of the map RP2 R3? (You should have found ! a total of six critical points.)

4.3 The Tangent Bundle

In the next chapter we define vector fields as a family of tangent vectors, one for each point on the manifold X T M, depending smoothly on the point p M. This p 2 p 2 allows us to study dynamics on the manifold, demonstrating the utility of the concept of tangent spaces. Another advantage of this concept is that now we have a vector space associated to each point of the manifold. This allows us to extend many of the natural constructions of to manifolds. For example, in one defines an inner product , on T M for each point of the manifold h· ·ip p p M, depending smoothly on the base point p M. This allows the introduction of 2 2 many familiar geometric concepts like and . The philosophy behind these constructions is that they are defined locally, for each TpM, but “patch together” globally by requiring smooth dependence on the base point p M. There are several ways to make the concept of “smooth dependence 2 on base points” rigorous, and we will see some of them in the next chapter. The most elegant is one is simply to use our existing definition of a smooth function between manifolds by turning the collection of all tangent spaces into a manifold. That manifold is called the tangent bundle.

Proposition 4.5. For any manifold M of dimension m, the tangent bundle

TM = TpM p M (disjoint union of vector spaces) is a manifold of dimension 2m. The map

p : TM M ! taking v T M to the base point p, is a smooth submersion, with fibers the tangent 2 p spaces.

Proof. The idea is simple: Take charts for M, and use the tangent map to get charts for TM. For any open subset U of M, we have

1 TU = TpM = p (U). p U G2 4.3 The Tangent Bundle 83

m (Note TpU = TpM.) Every chart (U,j) for M, with j : U R , gives vector space ! m m Tpj : TpM T (p)R = R ! j for all p U. The collection of all maps T j for p U gives a , 2 p 2 m Tj : TU j(U) R , v (j(p),(Tpj)(v)) ! ⇥ 7! for v T U TU. The images of these are the open subsets 2 p ✓ m m (Tj)(TU)=j(U) R R2 , ⇥ ✓ hence they define charts. We take the collection of all such charts as an for TM:

TU / j(U) Rm Tj ⇥ p (u,v) u 7! ✏ ✏ / ( ) U j j U

We need to check that the transition maps are smooth. If (V,y) is another coordinate chart with U V = /0,the transition map for TU TV = T(U V)=p 1(U V) is \ 6 \ \ \ given by, 1 m m Ty (Tj) : j(U V) R y(U V) R . (4.3) \ ⇥ ! \ ⇥ But T y (T j) 1 = T (y j 1) is just the derivative (Jacobian matrix) for the p p j(p) change of coordinates y j 1; hence (4.3) is given by 1 1 (x,a) (y j )(x), D (y j )(a) 7! x ⇣ ⌘ Since the Jacobian matrix depends smoothly on x, this is a smooth map. This shows that any atlas A = (U ,j ) for M defines an atlas (TU ,Tj ) for TM. Taking { a a } { a a } A to be countable the atlas for TM is countable. The Hausdorff property is easily checked as well. ut Proposition 4.6. For any smooth map F C•(M,N), the map 2 TF : TM TN ! given on T M as the tangent maps T F : T M T N, is a smooth map. p p p ! F(p) Proof. Given p M, choose charts (U,j) around p and (V,y) around F(p), with 2 F(U) V. Then (TU,Tj) and (TV,Ty) are charts for TM and TN, respectively, ✓ with TF(TU) TV. Let F = y F j 1 : j(U) y(V). The map ✓ ! 1 m n TF = Ty TF (Tj) : j(U) R y(V) R e ⇥ ! ⇥ is given by e (x,a) (F)(x), D (F)(a) . 7! x ⇣ ⌘ It is smooth, by smooth dependence ofe the differentiale DxF on the base point. Con- sequently, TF is smooth, ut e