Let $M$ be a compact smooth $d$-dimensional oriented manifold. The natural pairing of $d$-forms $\omega^{(d)}$ with the fundamental class is given by integration $\int_M \omega^{(d)}$. Let us also assume that all homology classes of $M$ are also represented by smooth submanifolds.

On the other hand, in singular (co-)homology we have a cap product $ \frown: H_k(M)\times H^\ell (M) \to H_{k-\ell}(M)$ defined on simplices as $\sigma\frown \phi =\phi(\sigma\circ [v_0\cdots v_\ell]) \sigma \circ[v_\ell\cdots v_k]$, where $[v_0\cdots v_m]:\Delta^m\to \Delta^k$, and we extend bilinearly.

My question is, assuming as above that all classes are represented by smooth submanifolds, is there a way to write the pairing $\phi(\sigma\circ [v_0\cdots v_\ell])$ as an integration and also make sense of the remaining cycle $\sigma \circ[v_\ell\cdots v_k]$ in the case of de Rham cohomology? In other words, what is the cap product in the case of de Rham cohomology?

A related question Cap product in De Rham cohomology? .

  • $\begingroup$ In de Rham cohomology, if you have some $k$-form $\alpha$ on some $k+l$-dimensional submanifold $N$ which is the Cartesian product $N_1\times N_2$ of two further submanifold of dimension $k$ and $l$, then it is reasonable to guess that the outcome should be the $\big(\int_{N_1}\alpha\big) [N_2]$, where $[N_2]$ denotes the fundamental class of $N_2$. Perhaps the main difficulty is generalizing this to arbitrary $N$... $\endgroup$
    – Danu
    Feb 21, 2019 at 13:49
  • $\begingroup$ Maybe think about chains as currents. $\endgroup$ Feb 21, 2019 at 14:11

1 Answer 1


After a few days of pondering I think I understood the answer: Let $M$ be a $d$-dimensional manifold as above, and let $i: N\hookrightarrow M$ be the embedding of a submanifold of dimension $\ell$. The cap product between $[N]\in H_\ell (M)$ and a $k$-form $[\omega]\in H^k(M)$ is an element in $\phi\in H_{\ell-k}(M)\cong \big(H^{\ell-k}(M)\big)^*$ (universal coefficient theorem). The element is specified by the formula:

$$ \phi(\zeta)=\int_{N} i^*\omega\wedge i^*\zeta, \quad \zeta\in H^{\ell-k}(M) $$

This is follows from the fact the isomorphism $H_m(M)\cong \big(H^m(M)\big)^*$ is precisely given by $\frown:[C]\mapsto [C]\frown\cdot$ for $[C]\in H_m(M)$, and the cap product satisfies $[C]\frown ([\alpha]\smile[\beta])=([C]\frown[\alpha])\frown[\beta]=([C]\frown[\alpha])([\beta])$ for $[\alpha]\smile[\beta]\in H^m(M)$.

Here we use de Rham (co-)homology classes, but we may also work on a (co)-chain level.


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