Talk:De Rham Cohomology of Sphere

Do you mean:

$\map {H^k} {S^n} = \begin {cases} \Z^2 & : k = 0, n = 0 \\ \Z & : k = 0, n > 0 \\ 0 & : 0 < k < n \end {cases}$

?

And what happens when $0 < k \le n$ and $k > n?$ --prime mover (talk) 16:10, 5 November 2021 (UTC)

Ah yes, but missing $\Z : k = n > 0$. Higher cohomology vanishes. --Z423x5c6 (talk) 16:16, 5 November 2021 (UTC)