De Rham Cohomology of Sphere

Theorem
Let $S^n$ denote the n-Sphere.

Then the de Rham Cohomology of $S^n$ are:


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

and Higher de Rham Cohomology Vanishes.