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 of a manifold is intuitively a height function of . Recall the definition of cobordism between two manifolds , it is a quintuple with a disjoint union of . One natural question is: can we separate these two parts? Note that is compact, and are two closed(thus compact) subset of . With the Haudorff property of , we know that there is a function such that . 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 a critical point of a function if . We say also that a critical value of . If at a critical point , we have that the Jacobian of at is not zero, then we say that is non-degenerated at . The problem is, for the function defined above on , does it have all its critical points in the interior of ? 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 is a Morse function, if , and all critical points of are in the interior of and are all non-degenerate. Note that if is non-degenerate at a critical point , then in a neighborhood of can be written as for some chart . It is easy to see that the integer 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 is a Morse function, then all these critical points are isolated, with that being compact, we have that there are only finitely many critical points for . And we call the number of critical points the Morse number of . The fundamental result concerning the Morse function is perhaps the following result:

**For any cobordism , there is a Morse function on .**

The proof of this result is a bit technical. First we will show that there exists a function on such that none of its critical points is in the boundary of . 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 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 with such that all its critical points lie in where is a neighborhood of the boundary .**

Note that we can find a finite open cover of such that for any , can not reach both and . We identify these open sets with its correspondence in . And if reaches , then according to the definition of boundaries of a manifold, we have that lies in the half space with that meets with the hyperplane of . So, we define for some positive constant such that . If reaches , then similarly we define for some positive number such that . Otherwise, we define . Then we use a partition of unity for this cover . So, each can have an extension to the global (just extending by zero), and now is a function on . Since for each point, we have that for all , thus , so there is . What is more, if , then if where is the union of the above open sets meeting , and thus for these , there is . So we have that , a contradiction. And thus . If for all , , since , we must have that for all . Yet , thus there is a such that doesn’t meet and . But note that this time , so again we have that , another contradiction. So, we must have some such that . Note what this means? In the coordinate expression, we recognize that . So, we get that . Similarly, we have that . Next we will show that in a neighborhood of , there is no critical points of this . Suppose that , and then . Note for the first sum, if , that is meets , then we have that . Otherwise, we have that in a neighborhood of , so we can assume that for all , we have in a neighborhood of . Thus the first sum is reduced to . Since this reduction is valid in a neighborhood of , and , the first term is just zero. Then we consider the second term. Note that , so we have that , thus showing that the derivative of at boundary points is not zero, and concluding the proof of this result.

Next we need the following result,

**If is a mapping of an open set to , then for almost all linear mapping (‘almost’ means a full measure set in ), 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 , and whose critical points correspond to the degenerate critical points of . The following proof is a genius idea. We consider . And the projection . We admit the fact that is a sub-manifold. Note that determines , so we have a natural injection which is also surjection. We can show that is a diffeomorphism, and thus is the same thing as .Then a point is critical of , it is the same as , 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 such that for some is critical(equivalently, ) is of zero measure. This gives the proof of the above result.

The next result says that we can define a 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 is an open set under this topology. We know easily how to define a topology. Yet for 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 locally into Euclidean spaces. So, we can choose a finite cover of with compact refinement (each is compact in ). Note that the space 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 , we set . We can use these to form a base of topology and thus to generate a topology on . We call this topology the Whitney topology on . Note that in the above definition, we have chosen a particular open covering of . In fact, this topology is independent of this choice, since it is easy to verify that if these derivatives of 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 is an open set, and is compact. If having only non-degenerate critical points in and being , then there exists a positive number such that for any (in other words, ), has only non-degenerate critical points in . **

This is not hard to show. We have to show that can not be zero at the same time.Since is so, and is a compact set, we can find such a .

Noe let’s try to prove the theorem:

**For any cobordism , there is a Morse function .**

Note that we have shown that there exists a function such that it has no critical points in a neighborhood of the boundary. We can choose this neighborhood with being compact and has no critical points on a neighborhood of . Then under the topology, we have that there is a neighborhood of such that all its elements have no critical points on . Now may have degenerate critical points on . Next we want to eliminate these points by modifying a little on a local coordinate level. Suppose that has a finite open cover such that each is just an open set in and each has a compact closure. Now look at the first open set . We know that, as the above result says, for almost all linear functionals , the map has no degenerate critical points. Clearly we can choose those with sufficiently small coefficients to guarantee that will not be too far from . So now we have constructed a on which has no degenerate critical points. But the problem is, how to extend to other ? We can extend by zero, this reminds us that we can construct a function such that in a compact neighborhood of and outside another compact neighborhood of (in other words, there is with the closures of each open set being compact, and ), and we set . Note that on we still have that , thus has no degenerate critical points on . Choosing still smaller, now has no critical points on and has no degenerate critical points on . Now, we continue this process to . We can suppose that . Then we can choose a linear functional and a function just as such that has no degenerate critical points on and has no critical points on . We have to show that has no degenerate critical points on , either. This is not hard to show, since we can choose to be small enough such that those derivatives of are still near to those of . So, in this way we have constructed a which has no degenerate critical points on and has no critical points on . This process can be continued and so the finally we can get a which has no degenerate critical points on and has no critical points on . So, we have proven that this has only non-degenerate critical points and has no critical points in a neighborhood of . Note that in each step above, we have that . What is more, for , we can choose such that , since is a compact set, we can choose such that the sup and inf of on is not . We can do similarly for other steps such that these share this property. And thus we have that and . So, this is just the Morse function that we are seeking. In fact, we see that is still not far from . So, we can say that the set of Morse functions on is a dense subset in . What is more, the set of Morse function is open in , so it is also open in .

In fact, we can say something more. We can modify this such that the new has no two critical points with the same value. Indeed, suppose that such that . Then using the Morse lemma, we see that in a neighborhood of there is only one point(that is the point ) such that at this point(recall that near , there is ), so we can choose two neighborhoods such that is compact and is bounded from below by on this compact set, and of course in there is only this critical point . Now choose a smooth function such that and . We can choose such that on and is smaller than all the possible non-zero differences of the values of the critical points. Then, we set that . It is clear that . What is more, on , , on , . And on , 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 is a Morse function on . Then for any such that contains no critical points(there are plenty of this kind of points, since there are only finitely many critical points), then we see that is an open set of with boundary . Indeed, is a sub-manifold of since is regular here, so the implicit function theorem says that this is so. So, we have that and are both manifolds with boundaries. And the sum of their Morse numbers is equal to that of . So, we can choose successively these such that each manifold in the final decomposition has only Morse number . This is the last result of this post:

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

So, in this post, we have shown that there are in fact many Morse functions, they form a set in the topology.