An application of topological equivalence to Morse theory

Abstract

In a previous paper, under the assumption that the Riemannian metric is special, the author proved some results about the moduli spaces and CW structures arising from Morse theory. By virtue of topological equivalence, this paper extends those results by dropping the assumption on the metric. In particular, we give a strong solution to the following classical question: Does a Morse function on a compact Riemannian manifold give rise to a CW decomposition that is homeomorphic to the manifold?

Appendix

In this appendix, we shall prove Lemma 8.2.

Suppose M is an n dimensional manifold. Suppose L is a connected and closed k dimensional submanifold of M. Let U be a closed tubular neighborhood of L such that U is diffeomorphic to a closed disk bundle over L via the exponential map. Let $i: L \hookrightarrow U$ be the inclusion and $\pi : U \rightarrow L$ be the smooth projection. Clearly, i and $\pi$ are proper. Thus $\pi ^{*}: H^{k}_{C} (L) \rightarrow H^{k}_{C}(U)$ and $i^{*}: H^{k}_{C} (U) \rightarrow H^{k}_{C}(L)$ are isomorphisms and they are a pair of inverses, where $H^{*}_{C}$ is the cohomology with compact support. Furthermore, $H^{k}_{C} (L) \cong \mathbb {Z}$, its generator is an orientation of L.

Define $H^{n}_{C}(U, U - L) = \lim _{\overrightarrow{K \subseteq L}} H^{n}(U, U-K)$, where K is compact. We can prove the inclusion $H^{n}(U, U-\{x\}) \rightarrow H^{n}_{C}(U, U-L)$ is an isomorphism for any $x \in L$.

Suppose $\alpha \in H^{k}_{C}(L)$ and the Thom class $\beta \in H^{n-k}(U, U-L)$ represent the orientation and the normal orientation of L respectively. Suppose the orientation and the normal orientation define the orientation of M.

Lemma A.1

The following cup product homomorphism is an isomorphism.

Furthermore, for all $x \in L$, via the isomorphism $H^{n}(U, U-\{x\}) \rightarrow H^{n}_{C}(U, U-L)$, we get $\pi ^{*} \alpha \cup \beta \in H^{n}_{C}(U, U-L)$ represents the orientation of M in $H^{n}(U, U-\{x\})$.

Proof

For any $x \in L$, we have a commutative diagram

Here the vertical maps are induced by inclusions and are isomorphisms. The horizontal ones are given by cup product pairings. By excision and the basic property of Thom class, we can localize the argument near x. However, the disk bundle near x has a product structure. Now apply Künneth Formula to the upper horizontal map, which completes the proof. $\square$

Proof of Lemma 8.2

It suffices to prove the special case of that $L_{i}$ is connected.

Let $U_{2}$ be a closed tubular neighborhood of $L_{2}$ with the smooth projection $\pi _{2}: U_{2} \rightarrow L_{2}$. Let $\alpha _{2} \in H^{k}_{C}(L_{2})$ be the orientation of $L_{2}$, by the above lemma, we have $\pi _{2}^{*} \alpha _{2} \cup \beta _{2}|_{U_{2}} = \gamma _{2} \in H^{n}_{C}(U_{2}, U_{2} - L_{2})$ represents the orientation of $M_{2}$ on $L_{2}$. Here $\beta _{2}|_{U_{2}}$ is the image of $\beta _{2}$ under the inclusion $H^{n-k}(M_{2}, M_{2} - L_{2}) \rightarrow H^{n-k}(U_{2}, U_{2} - L_{2})$. It is the restriction of $\beta _{2}$ to $U_{2}$.

Let $U_{1}' = h^{-1}(U_{2})$. Choose a closed tubular neighborhood $U_{1}$ of $L_{1}$ such that $U_{1} \subseteq \mathrm {Int} U_{1}'$ and $\pi _{1}: U_{1} \rightarrow L_{1}$ is a smooth projection. By the above lemma again, we have the following isomorphism

and

$$\begin{aligned} \pi _{1}^{*} \alpha _{1} \cup \beta _{1}|_{U_{1}} = \gamma _{1} \end{aligned}$$

(A.1)

represents the orientation of $M_{1}$ on $L_{1}$, where $\beta _{1}|_{U_{1}}$ is the restriction of $\beta _{1}$ to $U_{1}$.

Consider the following commutative diagram:

where, $i_{1}$, $i_{2}$, j and $\iota$ are inclusions. Since $h^{*} \pi _{2}^{*} \alpha _{2} \cup h^{*} \beta _{2}|_{U_{2}} = h^{*} \gamma _{2}$, we have $\iota ^{*} h^{*} \pi _{2}^{*} \alpha _{2} \cup \iota ^{*} h^{*} \beta _{2}|_{U_{2}} = \iota ^{*} h^{*} \gamma _{2}$. Since h preserves the orientation of $M_{1}$ and the Thom class, we have $\iota ^{*} h^{*} \gamma _{2} = \gamma _{1}$ and $\iota ^{*} h^{*} \beta _{2}|_{U_{2}} = \beta _{1}|_{U_{1}}$. Thus

$$\begin{aligned} \iota ^{*} h^{*} \pi _{2}^{*} \alpha _{2} \cup \beta _{1}|_{U_{1}} = \gamma _{1}. \end{aligned}$$

(A.2)

Since the cup product pairing above is an isomorphism, by (A.1) and (A.2), we infer $\iota ^{*} h^{*} \pi _{2}^{*} \alpha _{2} = \pi _{1}^{*} \alpha _{1}$. So we have

$$\begin{aligned} \alpha _{1}= i_{1}^{*} \pi _{1}^{*} \alpha _{1} = i_{1}^{*} \iota ^{*} h^{*} \pi _{2}^{*} \alpha _{2} = h^{*} i_{2}^{*} \pi _{2}^{*} \alpha _{2} = h^{*} \alpha _{2}. \end{aligned}$$

This completes the proof. $\square$