h-cobordism-part 2

This series of posts are mainly based on the ‘lectures on the h-cobordism theorem’ by John Milnor.

This second post is concerned with Morse function. A Morse function f of a manifold M is intuitively a height function of M. Recall the definition of cobordism between two manifolds M,M', it is a quintuple (N,m,m;F,F') with N=m\bigcup m' a disjoint union of m,m'. One natural question is: can we separate these two parts? Note that N is compact, and m,m' are two closed(thus compact) subset of N. With the Haudorff property of N, we know that there is a function f:\rightarrow [0,1] such that f^{-1}(\{0\})=m,f^{-1}(\{1\})=m'. We can modify this function such that it is a smooth function and still satisfies the above property. There are some related concepts. We say x\in M a critical point of a function h: M\rightarrow\mathbb{R} if dh_x=0. We say also that h(x) a critical value of h. If at a critical point x, we have that det(\frac{\partial^2 h}{\partial x_j\partial x_i}) the Jacobian of h at x is not zero, then we say that h is non-degenerated at x\in M. The problem is, for the function defined above on N, does it have all its critical points in the interior of N? Are these critical points all non-degenerate? Here we introduce the concept of Morse function, just to include all these conditions into one: a smooth function f:N\rightarrow\mathbb{R} is a Morse function, if f^{-1}(\{0\})=m,f^{-1}(\{1\})=m', and all critical points of f are in the interior of N and are all non-degenerate. Note that if f is non-degenerate at a critical point x\in N, then in a neighborhood U of x f can be written as f(x_1,...,x_n)=-x_1^2-...-x_k^2+x^2_{k+1}+...+x^2_n for some chart \phi:U\subset N\rightarrow \mathbb{R}^n. It is easy to see that the integer k is independent of the choice of the charts. This result is called the Morse lemma. From this lemma, we see easily that the non-degenerate critical points are isolated. And if f is a Morse function, then all these critical points are isolated, with that N being compact, we have that there are only finitely many critical points for f. And we call the number of critical points the Morse number of f. The fundamental result concerning the Morse function is perhaps the following result:

For any cobordism (N,m,m';F,F'), there is a Morse function on N.

The proof of this result is a bit technical. First we will show that there exists a function on N such that none of its critical points is in the boundary of N. Then we will show that after some modification, we can obtain a function that all its critical points are non-degenerate. For this, we can first work in the Euclidean case, and then using a finite cover of N to modify the function step by step until we get a Morse function. For the first step, we show that

There exists a smooth function f:N\rightarrow [0,1] with f^{-1}(\{0\})=m,f^{-1}(\{1\})=m' such that all its critical points lie in N-V where V is a neighborhood of the boundary m\bigcap m'.

Note that we can find a finite open cover \{U_i\}_{1\leq i\leq k} of M such that for any i, U_i can not reach both m and m'. We identify these open sets with its correspondence in \mathbb{R}^n. And if U_i reaches m, then according to the definition of boundaries of a manifold, we have that U_i lies in the half space \mathbb{R}^n_{\geq 0}=\{x\in\mathbb{R}|x_n\geq 0\} with that U_i meets with the hyperplane x_n=0 of \mathbb{R}^n. So, we define f_i: U_i\rightarrow [0,1], x\mapsto c_ix_n for some positive constant c_i such that f_i(U_i)\subset[0,1/2]. If U_j reaches m', then similarly we define f_j:U_j\rightarrow [0,1],x\mapsto 1-c_jx_n for some positive number c_j such that f_j(U_j)\subset [1/2,1]. Otherwise, we define f_l:U_l\rightarrow [0,1], x\mapsto 1/2. Then we use a partition of unity \{\phi_i\}for this cover \{U_i\}. So, each f_i\phi_i can have an extension to the global N(just extending by zero), and now f=\sum_if_i\phi_i is a function on N. Since for each point, we have that 0\leq f_i(x)\leq 1 for all i, thus 0\leq \sum_if_i(x)\phi_i(x)\leq \sum_i\phi_i(x)=1, so there is f:N\rightarrow [0,1]. What is more, if f(x)=0, then if x\in N-\bigcup_{k\in I} where \bigcup_{k\in I} is the union of the above open sets meeting m, and thus for these k\in I, there is \phi_{k}(x)=0. So we have that f(x)=\sum_{i\not\in I}f_i(x)\phi_i(x)\geq 1/2\sum_{i\not\in I}\phi_{i}(x)=1/2\sum_i\phi_i(x)=1/2, a contradiction. And thus x\in \bigcup_{k\in I}V_k. If for all k\in I, f_k(x)>0, since f(x)=0, we must have that \phi_k(x)=0 for all k\in I. Yet \sum_i\phi_i(x)=1, thus there is a k\not\in I such that U_k doesn’t meet m and \phi_k(x)>0. But note that this time f_k(x)\geq1/2, so again we have that f(x)>0, another contradiction. So, we must have some i\in I such that f_i(x)=0. Note what this means? In the coordinate expression, we recognize that x\in m. So, we get that f^{-1}(\{0\})=m. Similarly, we have that f^{-1}(\{1\})=m'. Next we will show that in a neighborhood of m\bigcup m', there is no critical points of this f. Suppose that x\in m', and then \frac{\partial f}{\partial x_n}(x)=\sum_{i}\frac{\partial f_i\phi_i}{\partial x_n}(x)=\sum_i f_i\frac{\partial \phi_i}{\partial x_n}(x)+\sum_{i}\phi_i\frac{\partial f_i}{\partial x_n}(x). Note for the first sum, if x\in U_i, that is U_i meets m', then we have that f_i(x)=1. Otherwise, we have that \phi_i(x')=0 in a neighborhood of x, so we can assume that for all i, we have f_i(x')=1 in a neighborhood of x. Thus the first sum is reduced to \frac{\partial (\sum_i\phi_i)}{\partial x_n}(x). Since this reduction is valid in a neighborhood of x, and \sum_i\phi_i(y)=1(y\in N), the first term is just zero. Then we consider the second term. Note that \frac{\partial f_i}{\partial x_n}(x)=1, so we have that \frac{\partial f}{\partial x_n}(x)>0, thus showing that the derivative of f at boundary points is not zero, and concluding the proof of this result.

Next we need the following result,

If f is a C^2 mapping of an open set U\in\mathbb{R}^n to \mathbb{R}, then for almost all linear mapping L:\mathbb{R}^n\rightarrow\mathbb{R}(‘almost’ means a full measure set in Hom_{\mathbb{R}}(\mathbb{R}^n,\mathbb{R})=\mathbb{R}^n), f+L has only non-degenerate critical points.

This result is due to Morse. Every time there appears the term ‘almost’ in differential topology, we should think of Sard’s theorem. So, perhaps we should construct a mapping whose image domain is U_n=Hom_{\mathbb{R}}(\mathbb{R}^n,\mathbb{R}), and whose critical points correspond to the degenerate critical points of f. The following proof is a genius idea. We consider M=\{(x,L)\in U\times U_n|d(f+L)(x)=0\}. And the projection p:M\rightarrow U_n,(x,L)\mapsto L. We admit the fact that M is a sub-manifold. Note that dL determines L, so we have a natural injection i:U\rightarrow M which is also surjection. We can show that i is a diffeomorphism, and thus p:M\rightarrow U_n is the same thing as q=p\circ i:U\rightarrow U_n,x\mapsto -df(x).Then a point x\in U is critical of q, it is the same as d^2f(x), as a square matrix, is not of full rank, and thus has zero determinant. So according to Sard’s theorem, which says that the set of critical values forms a zero-measure set in the image domain. In other words, the set of L\in U_n such that L=q(x)=-df(x) for some x\in U is critical(equivalently, det(d^2f(x))=0) is of zero measure. This gives the proof of the above result.

The next result says that we can define a C^2 topology, and in this topology for each Morse function, all the elements of some of its neighborhood are Morse function again. That is to say, the set of Morse functions on N is an open set under this topology. We know easily how to define a C^0 topology. Yet for C^1,C^2 things like that, it is not so easy, since we are no longer in an Euclidean space. The solution is to pull the functions on N locally into Euclidean spaces. So, we can choose a finite cover \{U_i\}_{1\leq i\leq k} of N with compact refinement K_i(each K_i is compact in U_i). Note that the space C^2(N,\mathbb{R}) is at least a vector space. So, to make it into a topological vector space, we just need to define the neighborhoods of the origin. For any positive number \delta>0, we set N(\delta)=\{g\in C^2(N,\mathbb{R})| \forall i=1,2,...,k,\forall x\in K_i, |g_{U_i}(x)|<\delta,|\frac{\partial g_{U_i}}{\partial x_j}(x)|<\delta, |\frac{\partial^2 g_{U_i}}{\partial x_j\partial x_l}(x)|<\delta, \forall j,k=1,...,n\}. We can use these N(\delta) to form a base of topology and thus to generate a topology on N. We call this topology the Whitney topology on N. Note that in the above definition, we have chosen a particular open covering of N. In fact, this topology is independent of this choice, since it is easy to verify that if these derivatives of f is bounded on a compact set in one coordinate, then they are also bounded in another coordinate(which sends compacts to compacts). Now we are ready to state our next result:

Suppose U\subset\mathbb{R}^n is an open set, and K\subset U is compact. If f:U\rightarrow \mathbb{R} having only non-degenerate critical points in K and being C^2, then there exists a positive number \delta>0 such that for any g\in N(f,\delta)(in other words, g-f\in N(\delta)), g has only non-degenerate critical points in K

This is not hard to show. We have to show that \frac{\partial g}{\partial x_j}, \frac{\partial^2 g}{\partial x_j\partial x_l} can not be zero at the same time.Since f is so, and K is a compact set, we can find such a \delta.

Noe let’s try to prove the theorem:

For any cobordism (N,m,m';F,F'), there is a Morse function f: N\rightarrow[0,1].

Note that we have shown that there exists a function f:N\rightarrow[0,1] such that it has no critical points in a neighborhood of the boundary. We can choose this neighborhood U\supset m\bigcup m' with \overline{U} being compact and f has no critical points on a neighborhood U' of \overline{U}. Then under the C^2 topology, we have that there is a neighborhood V of f such that all its elements have no critical points on U'. Now f may have degenerate critical points on N-U. Next we want to eliminate these points by modifying f a little on a local coordinate level. Suppose that N-U has a finite open cover \{U_i\}_{1\leq i\leq k} such that each U_i is just an open set in \mathbb{R}^n and each U_i has a compact closure. Now look at the first open set U_1. We know that, as the above result says, for almost all linear functionals L:\mathbb{R}^n\rightarrow\mathbb{R}, the map f+L:U_1\rightarrow\mathbb{R} has no degenerate critical points. Clearly we can choose those L with sufficiently small coefficients to guarantee that f+L will not be too far from f. So now we have constructed a f+L on U_1 which has no degenerate critical points. But the problem is, how to extend f+L to other U_i? We can extend by zero, this reminds us that we can construct a function h_1:N\rightarrow\mathbb{R} such that h_1=1 in a compact neighborhood of U_1 and h_1=0 outside another compact neighborhood of U_1(in other words, there is U_1\subset\overline{U_1}\subset V_1\subset\overline{V_1}\subset W_1 with the closures of each open set being compact, and h_1|_{V_1}=1,h_1|_{N-W_1}=0), and we set f_1=f+h_1L. Note that on U_1 we still have that f_1=f+L_1, thus has no degenerate critical points on U_1. Choosing L still smaller, now f_1 has no critical points on U' and has no degenerate critical points on U_1. Now, we continue this process to U_2. We can suppose that U_1\bigcap U_2\neq \emptyset. Then we can choose a linear functional L and a function h_2:U_2\rightarrow\mathbb{R} just as h_1 such that f_1+h_2L has no degenerate critical points on U_2 and has no critical points on U'. We have to show that f_1+h_2L has no degenerate critical points on U_1, either. This is not hard to show, since we can choose L to be small enough such that those derivatives of f_1+h_2L are still near to those of f_1. So, in this way we have constructed a f_2=f_1+h_2L which has no degenerate critical points on U_1\bigcup U_2 and has no critical points on U'. This process can be continued and so the finally we can get a f_k which has no degenerate critical points on \bigcup_i U_i\supset N-U and has no critical points on U'. So, we have proven that this f_k has only non-degenerate critical points and has no critical points in a neighborhood of m\bigcup m'. Note that in each step above, we have that f_i|_{m\bigcup m'}=f|_{m\bigcup m'}. What is more, for f, we can choose U,U' such that 0<f(x)<1(\forall x\in N-U), since N-U is a compact set, we can choose L such that the sup and inf of f_1 on N-U is not 1,0. We can do similarly for other steps such that these f_i share this property. And thus we have that f_k^{-1}(\{0\})=m and f_k^{-1}(\{1\})=m'. So, this f_k is just the Morse function that we are seeking. In fact, we see that f_k is still not far from f. So, we can say that the set of Morse functions on (N,m,m',F,F') is a dense subset in C'=\{f\in C^2(N,[0,1])|f^{-1}(\{0\})=m, f^{-1}(\{1\})=m'\}. What is more, the set of Morse function is open in C^2(N,\mathbb{R}), so it is also open in C'.

In fact, we can say something more. We can modify this f such that the new f has no two critical points with the same value. Indeed, suppose that x\neq x'\in N such that f(x)=f(x'). Then using the Morse lemma, we see that in a neighborhood of x there is only one point(that is the point x) such that df=0 at this point(recall that near x, there is f(y)=-y_1^2-...-y^2_k+y^2_{k+1}+...+y^2_n+...), so we can choose two neighborhoods x\in V\subset U such that K=\overline{U}-V is compact and |df(y)|=\sum_{i=1,...,n}|\frac{\partial f}{\partial y_i}| is bounded from below by \delta>0 on this compact set, and of course in U there is only this critical point x. Now choose a smooth function h:N\rightarrow[0,1] such that h_V=1 and h_{N-U}=0. We can choose \lambda>0 such that \lambda |dh|<\delta/2 on N and \lambda is smaller than all the possible non-zero differences of the values of the critical points. Then, we set that g=f+\lambda h. It is clear that g(x)\neq g(x'). What is more, on V, dg=df, on N-\overline{U}, dg=df. And on |dg|=|df-\lambda dh|>|df|-\lambda|dh|>1/2|df|>0, so all in all, we don’t create any new critical points and at the same time we isolate two critical points.

Now suppose that f is a Morse function on (N,m,m';F,F'). Then for any c\in(0,1) such that f^{-1}(c) contains no critical points(there are plenty of this kind of points, since there are only finitely many critical points), then we see that f^{-1}([0,c)) is an open set of N with boundary m\bigcup f^{-1}(c). Indeed, f^{-1}(c) is a sub-manifold of N since f is regular here, so the implicit function theorem says that this is so. So, we have that f^{-1}([0,c]) and f^{-1}([c,1]) are both manifolds with boundaries. And the sum of their Morse numbers is equal to that of f. So, we can choose successively these c such that each manifold in the final decomposition has only Morse number 1. This is the last result of this post:

Any cobordism can be decomposed into cobordisms with Morse number 1.

So, in this post, we have shown that there are in fact many Morse functions, they form a G_{\delta} set in the C^2 topology.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s