Application of de Rham Theorem on T^2

Question

As an application of the de Rham theorem on $S^n$, it can be determined that all closed $k$-forms are exact because $H^k_{dR}(S^n) = 0$ for $0<k<n$ (referenced from https://scholar.harvard.edu/files/pspark/files/derham.pdf). My question then is, since $H^k_{dR}(T^2) = \mathbb{R}$ for $k=0,2$ and $H^1_{dR}(T^2) = \mathbb{R}^2$, can the opposite be concluded about the torus (there exists closed forms that are not exact)? Or am I missing something?

Answer 1

Recall that for a smooth manifold $M$,

$$H^k_{\text{dR}}(M) = \frac{\{\text{closed}\ k\ \text{- forms}\}}{\{\text{exact}\ k\ \text{- forms}\}}.$$

So if $\alpha$ is a closed $k$-form on $M$, it defines a cohomology class $[\alpha] \in H^k_{\text{dR}}(M)$. Moreover, $[\alpha] = 0$ if and only if $\alpha$ is exact, and so $[\alpha] \neq 0$ if and only if $\alpha$ is not exact. Therefore, $H^k_{\text{dR}}(M) \neq 0$ if and only if $M$ admits a closed $k$-form which is not exact.

So for $T^2$, we see that it admits a closed $k$-form which is not exact for $k = 0, 1, 2$.