Changing coordinates to adapt to a map of constant rank

Transcription

1 Introduction to Submanifolds Most manifolds of interest appear as submanifolds of others e.g. of R n. For instance S 2 is a submanifold of R 3. It can be obtained in two ways: 1 as the image of a map into R 3 or 2 as the level set of a function with domain R 3. We shall examine both methods below. Our considerations are local at this point. Changing coordinates to adapt to a map of constant rank Suppose F : M 1 M 2 is smooth with dimm i = n i. Let U x j and V y k be charts around p M 1 and F p M 2 respectively. Then F is given locally by the functions y k = F k x 1... x n 1 ; 1 k n 2. Definition: The rank of the differentiable function F at p is rankdf p = dimrangedf p. Suppose that the rank of F at p is m. By permuting the coordinates if necessary we may suppose that F a det 1 a b m is non-zero. x b p Now define new coordinates u k in a neighborhood of p by u a = F a x 1... x n 1 1 a m u r = x r m < r n 1 Then where 1 a b m. u F a / x b = x 0 I n1 m This is non-singular at p and hence in a neighborhood of p so the u i are legitimate coordinates because of the inverse function theorem. In terms of these new coordinates the map F is now given by Ψ 1... Ψ n 2 where So in these coordinates Ψ a u 1... u n 1 = u a ; 1 a m. Im m 0 DF p = Ψ j / u n k 2 n 1 where in the lower right block m + 1 j n 2 m + 1 k n 1. 1

2 We now make the further assumption that the rank is constant = m in a neighborhood of p. Since the upper left hand block has rank m in this neighborhood the lower right hand block must vanish identically otherwise the rank would increase. So near p we have DF p = I 0 0 which means that Ψ a = Ψ a u 1... u m so we ve chosen coordinates in which the map F depends only on the first m of them. Corollary: If F : M N is smooth and of constant rank its image can locally be represented as the graph of a smooth function. Proof: In the above notation we have the image of F as the graph of the function u 1... u m t Ψ m+1... Ψ n 2 t. Exercise: Consider the map F : R 2 R 3 given by F φ θ = sin θ cos φ sin θ sin φ cos θ Carry out the procedure above to exhibit the image of F as a graph.. Continuing along we can now fix up the coordinates in the range near F p by defining with z a = y a : 1 a m z r = y r Ψ r y 1... y m : m + 1 r n 2 z y = Im 0 I n2 m And so the map F is given in terms of u 1... u n 1 and z 1... z n 2 by. z i = Φ i u 1... u n 1 : 1 i n 2 with z a = u a : 1 a m z r = Ψ r u Ψ r u = 0 for i > m Summarizing we ve proven the rank theorem: Let F : M 1 M 2 be smooth and of constant rank k in some neighborhood of p. Then there exist charts U u j at p and Z z i at F p such that F is locally represented by 2

3 z a = u a : 1 a m z a = 0 : a > m. That is the image of the map F is represented locally by a coordinate slice. Exercise: Suppose f : V R k R m is smooth and define F : V R m+k by setting F x = x fx the graph of f. Then DF = Ik k f/ x m+k k has constant rank k. Finish the construction above. Submanifolds We are interested in the specific case where the rank is constant and equal to the dimension of the domain of F. Definition: Suppose F : M N is smooth and dimm dimn. If rankf = dimm at all point of M then F is said to be an immersion and F M is an immersed submanifold of N. By the inverse function theorem F is locally 1-1 onto its image. F is an embedding if F is an immersion and F is a homeomorphism onto its image in the induced topology of N. Definition: F : M N is proper if F 1 K is compact in M for any compact K N. If F is a proper embedding the F M is called a closed submanifold of N. An embedding of M onto an open subset of N is called an open submanifold. Remark: Immersions embeddings and proper embeddings are locally indistinguishable. Exercise: Consider the set GLn R = {A n n matrices over R with deta 0}. Then GLn R is an open submanifold of R n2. GLn R is called the general linear group of order n. It s our first example of a Lie group a differentiable manifold which is also a group and for which the group operations are smooth functions. GLn R has two connected components. Exercise: Consider the torus T 2 obtained from R 2 by identifying points according to x y u v x u y v Z 2. Now consider the map f : R T 2 given by ft = [ta tb] where [ ] denotes the equivalence class. If a b Z 2 i.e. if b/a is rational then the image under f is a closed curve. In 3

4 this case φ is an immersion; it would be an embedding if we d obtained it as the image of a circle but we didn t. If the slope of the image is irrational then the map is injective and thus f is injective but it s still not an embedding since the image is dense in T 2. What s wrong here is that φ is not a homeomorphism onto its image in the induced topology. Exercise: For those with some knowledge of differential equations Consider the nonlinear 2-d ODE in polar coordinates: dr = µr r2 dt dθ dt = ω 0 > 0 The equilibrium point at r = 0 is unstable repelling for µ > 0. There is also a periodic orbit at r = µ. Evidently if r < µ dr/dt > 0 so r is strictly increasing along any orbit starting in the interior of the periodic orbit. Along any such orbit as t the orbit spirals out to the closed orbit at r = µ but never gets there. As it spirals out it circles the origin infinitely often. As t the orbit approaches the unstable equilibrium point at 0. The orbit is homeomorphic to R in the induced topology and hence the image is an embedded submanifold of R 2. But it is not a closed submanifold of R 2. In light of the above the following makes sense: Definition: A subset M of the differentiable manifold N is a k-dimensional regular submanifold of N if at each point p M there is a coordinate neighborhood of p in N U u i such that M U is given by u k+1 = = u n = 0. Normally when one uses the word submanifold this is what s meant and this will be assumed from now on. Submanifolds as level sets The most common way of constructing submanifolds is a consequence of the following: Theorem: Suppose F : M N is smooth and of constant rank k on M. If p F M then F 1 p is an m k-dimensional submanifold of M. Proof: Let A = F 1 p M and let q A. By the rank theorem there exist coordinates x i : 1 i m in U M centered at q such that xq = 0 and coordinates y j : 1 j n in N centered at p with yp = 0 such that F is given locally by y a = x a : 1 a k y a = 0 : k + 1 a n. 4

5 Since all the points of A U map to 0 = yp they must be given by x 1 = x 2 = = x k = 0 and this is the submanifold condition. Corollary: Suppose F : M N is smooth dimn = n m = dimm and the rank of F = n maximal at every point of A = F 1 a. Then A is a closed regular submanifold of dimension m n of M. Proof: At any p A F is of maximal rank. Since locally the rank can only jump up this means that there s a neighborhood of A in M in which F has the constant rank n so the theorem applied to this neighborhood gives the result: A is a regular closed submanifold of this open set and hence of M. Exercise: Consider the group SL2 R = {A 2 2 : deta = 1} the special linear group of order 2. Writing a b A = c d and fa = ad bc we find DfA = d c b a which has constant rank 1. So SL2 R is a regular submanifold of GL2 R. As an example of the coordinates mentioned in the theorem put xa b c d = ad bc 1 and retain the coordinates a b c. Then a b c x a b c d = d c b a which is non-singular provided that a 0. In this chart the submanifold is given by x = 0. This together with the obvious other three charts exhibit the submanifold structure. Exercise: The function f : R n R given by fx = n 1 xi 2 has rank 1 maximal on f 1 1 = S n 1 ; so S n 1 is a regular closed submanifold of R n. Exercise: Regarding the matrix A n n as an element of X = R n2 the map f : X X given by fa = A t A has constant rank on the open submanifold GLn R; hence f 1 I = {A GLn R : A t A = I} is a closed regular submanifold of GLn R. This is the Lie group of orthogonal matrices. Verify this for n = 2 or larger. 5