De Rham cohomology of $S^1$

Question

I want to calculate $H^1_{dR}(S^1)$.

I'm stuck at the beginning. I know that

$$H^1_{dR}(S^1) = Z^1/B^1 = \frac{\{\alpha\in\Omega^1\ \vert\ X_0(\alpha(X_1)) - X_1(\alpha(X_0)) = \alpha([X_0,X_1])\}}{\{df\ \vert\ f\in C^\infty(S^1)\}}$$

I know it is supposed to be $\mathbb{R}$ but I don't see how to find it.

Any hints to get me started?

Answer 1

Since $S^1$ is 1-dimensional, it follows that $\{\alpha\in\Omega^1\ \vert\ X_0(\alpha(X_1)) - X_1(\alpha(X_0)) = \alpha([X_0,X_1])\}$ is actually all of $\Omega^1$, which you can easily check by working in a local coordinate.

So it remains to decide which 1-forms are exact. When can the equation $\alpha=df$ be solved. Let's write this in a local coordinate as well: $\alpha(\theta)\,d\theta=f'(\theta)\,d\theta.$ So we need to know when we can find an antiderivative of the smooth periodic function $\alpha(\theta).$

The periodic functions $e^{in\theta}$ have antiderivative $\frac{1}{in}e^{in\theta},$ except when $n=0.$ Then for the $n=0$ case, we can put $f(\theta)=\theta,$ and we do have $f'(\theta)=\alpha(\theta)=e^0=1.$ But this function is not periodic, so not an element of $C^\infty(S^1).$ Finally when we observe that the functions $e^{in\theta}$ are dense in smooth periodic functions, we conclude that a basis for the only non-exact 1-forms is $d\theta.$

Answer 2

Hint: Define $p: \mathbb R \to S^1$ by $p(t) = e^{it}$. This map induces a pullback homomorphism $p^*: \Omega^1(S^1) \to \Omega^1(\mathbb R)$ so that if $\alpha \in \Omega^1(S^1)$, then $p^*(\alpha) = f(t)dt$ for some $2\pi$-periodic function $f:\mathbb R \to \mathbb R$. Hence, we can define a map $\Omega^1 \to \mathbb R$ by $\alpha \mapsto \int_0^{2\pi} f(t)dt$. Prove that this map is a surjective homomorphism with kernel precisely $\{df \mid f\in C^{\infty}(S^1)\}$.

Answer 3

The first cohomology group is the dual of the first homology group. You know the first homology group via Hurewicz since the fundamental group $\pi_1(S^1)$ is known.