Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
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. –  jvkersch Dec 16 '10 at 22:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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