$H^1_{dR}(S^1 \times S^1) = H^1_{dR}(S^1) \oplus H^1_{dR}(S^1)$ without Künneth?

Question

As in the title: I am trying to derive $H^1_{dR}(S^1 \times S^1) = H^1_{dR}(S^1) \oplus H^1_{dR}(S^1)$.

First let me share my thoughts:

I am trying to derive that

$$ H^1_{dR}(S^1 \times S^1) = \mathbb R \oplus \mathbb R$$

I can do it by applying the Künneth theorem:

$$ H^n_{dR}(S^1 \times S^1) = \bigoplus_{i + j = n} H^i_{dR}(S^1) \otimes H^j_{dR}(S^1)$$

For the right hand side I get

$$ H^0(S^1)\otimes H^1(S^1) \bigoplus H^1(S^1)\otimes H^0(S^1) $$

And since $ H^1(S^1) = H^0(S^1) = \mathbb R$, this is equal to

$$ \mathbb R \otimes \mathbb R \bigoplus \mathbb R \otimes \mathbb R = \mathbb R \oplus \mathbb R$$

But this is an overkill.

How to directly derive it?

Answer 1

Here are two ways of seeing that $H^1_{dR}(S^1\times S^1)$ is two-dimensional without using the Künneth Theorem.

1. For any smooth connected manifold $X$, $H^1_{dR}(X) \cong \operatorname{Hom}(\pi_1(X), \mathbb{R})$ (see Lee's Introduction to Smooth Manifolds (second edition) Theorem $17.17$ and Problem $18$-$2$), so

$$H^1_{\text{dR}}(S^1\times S^1) \cong \operatorname{Hom}(\pi_1(S^1\times S^1), \mathbb{R}) \cong \operatorname{Hom}(\mathbb{Z}^2, \mathbb{R}) \cong \mathbb{R}^2.$$

2. For any smooth manifold $X$, we define the Euler characteristic of $X$ by $\chi(X) := \sum_{k=0}^n(-1)^k\dim H^k_{\text{dR}}(X)$; for a two-dimensional manifold, this is just the usual Euler characteristic for surfaces.

   The two-dimensional manifold $S^1\times S^1$ is the torus which has Euler characteristic zero. As $S^1\times S^1$ is connected, $H^0_{\text{dR}}(S^1\times S^1) \cong \mathbb{R}$, and as $S^1\times S^1$ is a compact orientable two-dimensional manifold, $H^2_{\text{dR}}(S^1\times S^1) \cong \mathbb{R}$. Therefore,

   \begin{align*} \chi(S^1\times S^1) &= \dim H^0_{\text{dR}}(S^1\times S^1) - \dim H^1_{\text{dR}}(S^1\times S^1) + \dim H^2_{\text{dR}}(S^1\times S^1)\\ 0 &= 1 - \dim H^1_{\text{dR}}(S^1\times S^1) + 1, \end{align*}

   so $\dim H^1_{\text{dR}}(S^1\times S^1) = 2$.

Answer 2

You can define two $1$-forms which give you a basis of the cohomology, and show by hand that every other $1$-form is cohomologous to a linear combination of them. This is not hard to do.