# What is the de Rham cohomology of $S^k\times S^n$?

I am trying to do this calculus and I have a guest, but I don't if it is right:

Supposing $n >k$ then:

If $r=0$ then

$$H_{dR}^0(S^k\times S^n) = \mathbb{R}$$

$$H_{dR}^r(S^k\times S^n) = 0,$$ if $0 <r<k.$

If $r=k$ then $$H_{dR}^k(S^k\times S^n)= \mathbb{R}$$

If $r >k$ and $r < n$ then

$$H_{dR}^r(S^k\times S^n) = 0$$

If $r=n$

$$H_{dR}^n(S^k\times S^n) = \mathbb{R}$$

Is this right???

Thanks!

• What a complicated way of saying «$H^r(S^k\times S^n)$ is isomorphic to $\mathbb R$ if $r\in\{0, k,n\}$ and zero otherwise»! – Mariano Suárez-Álvarez Jun 1 '16 at 16:00
• In any case, all the groups you wrote are correct. There is one more $r$ for which $H^r$ is nonzero, though. – Mariano Suárez-Álvarez Jun 1 '16 at 16:01
• @MarianoSuárez-Alvarez, $r=k+n?$ – L.F. Cavenaghi Jun 1 '16 at 16:03
• Indeed. ${}{}{}$ – Mariano Suárez-Álvarez Jun 1 '16 at 16:06
• @MarianoSuárez-Alvarez, thanks a lot! – L.F. Cavenaghi Jun 1 '16 at 16:11

## 1 Answer

Intuitively do you have closed forms that are not exact?

Certainly the dimensions must be $0, n, k , n+K$. For example, the volume (or surface area) form in each of those dimensions.

To prove there are no other examples requires Mayer Vietoris.