Direct proof that the wedge product preserves integral cohomology classes?

Let $H^k(M,\mathbb R)$ be the De Rham cohomology of a manifold $M$.

There is an approved map $H^k(M;\mathbb Z) \to H^k(M;\mathbb R)$ from the indispensable cohomology to the cohomology with coefficients in $\mathbb R$, which is isomorphic to the De Rham cohomology. As a previous question currently disclosed, the photos of this map are specifically the courses of differential $k$ - kinds $[\omega]$ that generate integers when incorporated over a $k$ - cycle $\sigma$,

$$ \int_{\sigma} \omega \in \mathbb{Z} \quad\text{ whenever } d\sigma = 0$$

Let us call them "integral forms".

Encouraged by the mug item on cohomology, my question/request is the following:

Give a straight evidence that the wedge item $[\omega\wedge\eta]\in H^{k+l}(M,\mathbb R)$ of 2 indispensable kinds $\omega\in \Omega^k(M)$ and also $\eta\in \Omega^l(M)$ is once more an indispensable kind.

This need to hold true due to the fact that the mug item is mapped to the wedge item, yet the factor of the workout is to confirm this declaration straight, without creating the single cohomology $H^k(M,\mathbb Z)$ or homology first.

Possibly I additionally need to see to it that the problem of being an indispensable kind is something that can be "checked effectively" without single homology ; this could be based on a new inquiry.

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2022-06-06 15:30:34
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Answers: 1

I primarily make use of cohomology as a blackbox so I might be forgeting technological troubles yet below is just how I would certainly attempt to confirm your declaration.

First I would certainly keep in mind that being indispensable is a neighborhood building: the class $[\omega]$ of a shut kind is indispensable iff its constraints $[\omega|_{U_i}]$ to a covering $M = \bigcup U_i$ is indispensable. One means is noticeable: if $\sigma \subset U_i$, after that $\int_\sigma \omega|_{U_i} = \int_\sigma \omega$. For the various other, we can picked a (limited) triangulation $\sigma = \bigcup \sigma_\alpha$ such that each $\sigma_\alpha$ is consisted of in among the $U_i$ to make sure that we can write $\int_\sigma \omega = \sum_\alpha \int_{\sigma_{\alpha}} \omega|_{U_{i(\alpha)}}$.

The factor is that you can examine integrality by utilizing the triangulation you such as.

After that I would certainly write the mug item as $H^i_{dR}(X) \otimes H^j_{dR}(X) \to H^{i+j}_{dR}(X\times X) \to H^{i+j}_{dR}(X)$ the exterior item adhered to by constraint to the angled. And also I would certainly examine that each map sends out indispensable indispensable courses to indispensable courses making use of triangulations.

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2022-06-06 15:56:03
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