13
$\begingroup$

I am thinking about the precise formulation of the Lefschetz duality for the relative cohomology. If I understand this Wikipedia article correctly, there is an isomorphism between $H^k(M, \partial M)$ and $H_{n-k}(M)$ and hence (I suppose) a non-degenerate pairing $H^k(M, \partial M) \times H^{n-k}(M) \rightarrow \mathbb{R}$. However, I have trouble visualizing this pairing. Let $[(\alpha, \theta)] \in H^k(M, \partial M)$ and $[\beta] \in H^{n - k}(M)$, is it then true that $$ \left< [(\alpha, \theta)], [\beta] \right> = \int_M \alpha \wedge \beta + \int_{\partial M}\theta \wedge \beta_{|\partial M} $$ or am I missing something? If unrelated to Lefschetz duality, does this pairing ever appear in topology?

I can understand how to define a pairing on the homology by counting intersections, but I really don't see how this works for cohomology. Also, a reference on Lefschetz cohomology or just analysis/topology on manifolds with boundary would be greatly appreciated!

$\endgroup$
4
$\begingroup$

Yes, your formula is right. For the intuitive understanding just compute it for 1- and 2- dimensional half-spaces.

See Bott & Tu, Differential forms in Algebraic topology, $\S 5$, Poincaré duality.

I give only sketch of proof for your question.

  1. First of all you need pairing between $H_c^k(M, \partial M)$ and $H^{n-k}(M)$.

  2. Just consider $M= [0,+\infty)$, find $H_c^k(M), H_c^k(M,\partial M), H^k(M), H^k(M,\partial M) $ and check that you have non-generating pairing.

  3. By induction, expand previous statement to $\mathbb R_{+}^n = \{(x_1,x_2\dots,x_n)|x_1\geqslant 0\}$ (read Bott & Tu $\S 4$ and do the same things).

  4. Prove that there is Mayer-Vietoris sequence for $H_c^k(M,\partial M)$ similar to Mayer-Vietoris sequence for $H_c^k(M)$.

  5. Prove duality the same way as in $\S 5$ (check the commutativity of diagram and apply 5-lemma).

That's all, I performed these actions without any troubles.

$\endgroup$
3
  • $\begingroup$ Thanks Nikita! In the months since I asked this question, Bott and Tu has become my bible and I can see how to apply this outline now. $\endgroup$ – jvkersch Dec 16 '10 at 22:10
  • $\begingroup$ Are we sure it's not a minus sign between the two integrals? $\endgroup$ – Chris Gerig Dec 1 '19 at 5:19
  • $\begingroup$ @ChrisGerig this seems to depend on the sign convention used in the differential of the cone. With the convention from Wikipedia, the plus between the integrals is right. $\endgroup$ – Michael Bächtold May 21 '20 at 15:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.