# Show that $\mathbb{S}^{n+m}$ is not homeomorphic to a product of orientable manifolds

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?

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

• But aren't M and N are too general to find a closed form? – davidivadful Dec 4 '18 at 19:55
• @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. :) – user98602 Dec 4 '18 at 19:57
• But a form in $M$ gives me a form in $M\times N$? And closed? – davidivadful Dec 4 '18 at 20:07
• @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. – user98602 Dec 4 '18 at 20:09
• You want to consider $\pi_M^* \omega_M \wedge \pi_N^* \omega_N$. By Fubini's theorem, the integral is $\int_M \omega_M \int_N \omega_N$. – user98602 Dec 5 '18 at 13:54