I want to prove that the sphere $\mathbb{S}^{n+m}$ is not homeomorphic to the product of N and M, orientable manifolfs with $\textit{dim}\;N=n$ and $\textit{dim}\;M=m$. I know that I have to use the de Rham cohomology of the sphere to prove it, but I don´t know how to do this. Any hint?

  • $\begingroup$ Do you know Poincare duality? $\endgroup$ – Jason DeVito Dec 4 '18 at 19:38
  • $\begingroup$ Yes, but I don't know how to apply it here $\endgroup$ – davidivadful Dec 4 '18 at 19:39
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    $\begingroup$ Perhaps more pointed: do you know the Kunneth theorem? $\endgroup$ – user98602 Dec 4 '18 at 19:40
  • $\begingroup$ I don't know Kunneth theorem or any theorem about products $\endgroup$ – davidivadful Dec 4 '18 at 19:41

We want to find two closed forms $\omega_1$ and $\omega_2$ on $M \times N$ so that $\omega_1 \wedge \omega_2$ is a closed form of top degree with $\int \omega_1 \wedge \omega_2 \neq 0$. Therefore, $[\omega_1] \wedge [\omega_2]$ is nonzero in cohomology, and therefore each $[\omega_i]$ must have been as well.

Write $\omega_M$ for a volume form on $M$ - a nonvanishing top-dimensional form - which necessarily has $\int_M \omega_M \neq 0$. Similarly for $\omega_N$. Our desired forms are $\pi_M^* \omega_M$ and $\pi_N^* \omega_N$.

Now the Fubini theorem on integrals of functions on $\Bbb R^{n + m}$ of the form $f \cdot g$, where $f$ is a compactly supported function on $\Bbb R^n$ (respectively, $\Bbb R^m$) states that $$\int_{\Bbb R^{n+m}} fg = \int_{\Bbb R^n} f \cdot \int_{\Bbb R^m} g.$$

The usual argument to take a theorem about compactly supported integrals in $\Bbb R^n$ and make it into a theorem about integrals of differential forms is to apply a partition of unity to your atlas and them sum up the result. Nothing changes here (you should use two partitions of unity: one for $M$ and one for $N$), and you find that $$\int_{M\times N} \pi_M^*\omega_M \wedge \pi_N^* \omega_N = \int_M \omega_M \cdot \int_N \omega_N \neq 0,$$ as desired.

  • $\begingroup$ But aren't M and N are too general to find a closed form? $\endgroup$ – davidivadful Dec 4 '18 at 19:55
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    $\begingroup$ @davidivadful Nope! It might not be completely explicit, but that doesn't matter. There is a crucial assumption on $M$ and $N$ you have to use. :) $\endgroup$ – user98602 Dec 4 '18 at 19:57
  • $\begingroup$ But a form in $M$ gives me a form in $M\times N$? And closed? $\endgroup$ – davidivadful Dec 4 '18 at 20:07
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    $\begingroup$ @davidivadful You'll have to think about how to do that part: it only uses simple operations on forms that you already know, and their properties. $\endgroup$ – user98602 Dec 4 '18 at 20:09
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    $\begingroup$ You want to consider $\pi_M^* \omega_M \wedge \pi_N^* \omega_N$. By Fubini's theorem, the integral is $\int_M \omega_M \int_N \omega_N$. $\endgroup$ – user98602 Dec 5 '18 at 13:54

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