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!


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.

  • $\begingroup$ 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

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