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!


2 Answers 2


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.

  • 1
    $\begingroup$ 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
    Commented Dec 16, 2010 at 22:10
  • $\begingroup$ Are we sure it's not a minus sign between the two integrals? $\endgroup$ Commented Dec 1, 2019 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. $\endgroup$ 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).


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