# 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, 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.