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I'm reading M. Audin and M, Damian's Morse Homology book right now and i have some issue regarding its section about Poincare Duality (Section 4.3, page 83). In this version of "Poincare Duality", it state that $HM_k(V,\Bbb{Z}/2) \cong HM_{n-k}(V,\Bbb{Z}/2)$, where $HM_k(V,\Bbb{Z}/2)$ is the $k$-th Morse homology of a manifold without boundary $V$ with coefficient $\Bbb{Z}/2$.

I'm not able to show this since i can't find the appropriate chain map, $$ \require{AMScd} \begin{CD} \cdots @>>> C_{k}(f) @>\partial_{k}>> C_{k-1}(f) @>>> \cdots \\ @. @V?VV @VV?V \\ \cdots @<<< C_{n-k}(-f) @<\partial_{n-k+1}<< C_{n-k+1}(-f) @<<< \cdots \end{CD} $$

but i have shown another version of this theorem that is $MH_k(V,\Bbb{Z}/2) \cong MH^{n-k}(V,\Bbb{Z}/2)$, where $HM^{k}$ is the $k$th Morse cohomology as defined in here. After done some reading, like Schwarz's Morse Theore book and some parts of this nice paper, i can't found any similar result as in Audin. Also, if the result in Audin is true, then we have $HM_k(V) \cong HM^k(V)$ which is a bit strange to me.

Would anyone can clarify this to me ? Am I wrong ? How to show Audin's version if it is true ? Thank you.

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If $f$ is a Morse function on the closed manifold $V$, then so is $-f$. The index $k$ critical points of $f$ are index $(n-k)$ critical points of $-f$, and the boundary map $$\partial_k: C_k(V,f) \to C_{k-1}(V,f)$$ dualizes to give the boundary map $$\partial_{n-k}: C_{n-k+1}(V,-f) \to C_{n-k}(V,-f).$$ Check this using the definitions.

Now note that if $W_1 \xrightarrow{g_1} W_2 \xrightarrow{g_2} W_3$ is a complex of finite-dimensional vector spaces (so the composite is zero), then the homology $\text{ker}(g_2)/\text{im}(g_1)$ is isomorphic to the homology $\text{ker}(g_1^*)/\text{im}(g_2^*)$ of $$W_3^* \xrightarrow{g_2^*} W_2^* \xrightarrow{g_1^*} W_1.$$

This gives $H_k(V,f;k) = H_{n-k}(V,-f;k),$ with coefficients in any field $k$. Because Morse homology is independent of Morse function we have the desired result.

This is a straightforward calculation and I encourage you to try it. It is crucial here that we are working with vector spaces and not free abelain groups, or otherwise the above result would be false. That is why we have $H^k(V;\Bbb Z/2) = H_k(V;\Bbb Z/2)$ --- this does not hold over the integers.

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  • $\begingroup$ Thank you so much. I really appreciate it. $\endgroup$ – Si Kucing Jan 26 at 6:55

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