# Pairing used in Lefschetz duality

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.

• 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. Commented Dec 16, 2010 at 22:10
• Are we sure it's not a minus sign between the two integrals? Commented Dec 1, 2019 at 5:19
• @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. Commented May 21, 2020 at 15:27

I cannot find this pairing in terms of differential forms in any literature, so let me answer it myself.

Let $$M$$ be a compact with boundary $$\partial M$$. Let $$M^{\circ}=M\setminus\partial M$$ be the interior. Then there is an isomorphism between relative (de Rham) cohomology and the cohomology with compact support

$$H^k(M,\partial M)\cong H_c^k(M^{\circ}). \label{1}\tag{1}$$

On the other hand, there is a natural pairing

$$H^k_c(M^{\circ})\times H^{n-k}(M)\to \mathbb R, \ (\phi,\psi)\mapsto \int_{M}\phi\wedge\psi.\label{2}\tag{2}$$

So if we can find an explicit isomorphism for \eqref{1}, then the pairing \eqref{2} can be transformed into a pairing between $$H^k(M,\partial M)$$ and $$H^{n-k}(M)$$.

First of all, there is a different version of relative cohomology, obtained as cohomology of complex $$\Omega^*(M,\partial M)$$ of differential forms on $$M$$ that vanishes on $$\partial M$$ (cf. Originally from [Godbillion] defined for a pair of a smooth manifold and a smooth closed submanifold, but should apply to a manifold with boundary as well). We denote this cohomology theory as $$H^*(M,\partial M)_G$$. It turns out that the natural inclusion $$H^k(M,\partial M)_G\to H^k(M,\partial M)$$ is an isomorphism and the inverse map is given by $$(\alpha,\theta)\mapsto \alpha-d(\pi^*\theta\wedge\eta),$$ where $$\eta$$ is a bump function supported on $$T$$ and has constant value $$1$$ in a smaller neighborhood $$T'$$. (See this post)

Second, since $$\partial M$$ is homotopy equivalent to its tubular neighborhood, it turns out that the natural inclusion $$H^k_c(M^{\circ})\to H^k(M,\partial M)_G$$ is an isomorphism (cf. [Godbillion, Cha. XII, Theorem 3.1]) and the inverse map is $$\omega'\mapsto \omega'-d(u\wedge\eta)$$, where $$du=\omega'$$ on $$T'$$. Note one can choose $$u$$ to vanish on $$\partial M$$.

Now let $$(\alpha,\theta)\in H^k(X,Y)$$, and $$\beta\in H^{n-k}(M)$$. Write $$\lambda=\pi^*\theta-u$$ where $$du=\omega-d\pi^*\theta$$ on $$T'$$. Combine the isomorphisms $$H^k_c(M^{\circ})\cong H^k(M,\partial M)_G\cong H^k(M,\partial M)$$, then the pairing \eqref{2} becomes

$$\int_{M}\big(\alpha-d(\lambda\wedge\eta)\big)\wedge\beta=\int_M\alpha\wedge \beta-\int_Td(\lambda\wedge\eta)\wedge\beta.$$

Apply the Stokes' theorem to the second term and use the fact that $$\eta$$ is supported on $$T$$, we have

$$\int_Td(\lambda\wedge\eta)\wedge\beta=\int_Td(\lambda\wedge\eta\wedge\beta)=\int_{\partial M}\lambda\wedge\beta.$$

Now since $$u$$ vanishes on $$\partial M$$, we obtain the pairing formula

$$((\alpha,\theta),\beta)\mapsto \int_{M}\alpha\wedge\beta-\int_{\partial M}\theta\wedge\beta_{|\partial M}.$$

Note there is a minus sign on the second term. Here the orientation of $$\partial M$$ is the induced orientation from $$M$$ (normal vector points outward).