# Cap product and de Rham cohomology

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? .

• 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 $N$...
– Danu
Feb 21, 2019 at 13:49
• Maybe think about chains as currents. Feb 21, 2019 at 14:11

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 $$[\alpha]\smile[\beta]\in H^m(M)$$.