Show that $S^2$ is not diffeomorphic to $T^2$

Question

I know this question have already been answered many times. However I want to prove this using the De Rham cohomology of them, showing that they are not isomorphic and thus $S^2$ and $T^2$ are not diffeomorphic. I know that

$H_{d R}^r\left(\mathbb{S}^m, \mathbb{R}\right) \approx\left\{\begin{array}{ll}\mathbb{R} & \text { se } \quad r=0, r=m \\ \{0\} & \text {otherwise}\end{array}\right.$

which follows by the Poincarè Lemma. I'm not quite sure how to calculate the De Rham cohomology of the Torus though. Do I use that $T^2 = S^1 \times S^1$ somehow?

Answer 1

You do not need to work this hard to answer your question. You don't need to actually prove that $H^1_{\text{dR}}(T^2) \cong \Bbb R^2$. Since $H^1_{\text{dR}}(S^2) \cong \Bbb 0$, all you have to do is show that $H^1_{\text{dR}}(T^2)$ is nontrivial. Indeed, we can show it is at least two-dimensional.

Write $T^2 = S^1_1\times S^1_2$, with projections $\pi_i$ to the two factors. We know $H^1_{\text{dR}}(S^1)\cong\Bbb R$, with basis $d\theta$ (which is not exact, notwithstanding all appearances to the contrary). Now it is easy to check that $\pi_1^*(d\theta_1)$ and $\pi_2^*(d\theta_2)$ are closed forms on $T^2$; they cannot be exact (as you have closed curves over which their integrals are non-zero). Moreover, by the same logic, they are linearly independent (although we do not even need this here).

Answer 2

The most direct way to compute the De Rham cohomology of the torus is writing the Mayer-Vietoris sequence for two cylinders which glued together form the torus. This requires leveraging a bit the inclusion maps between these spaces. Here you can find a computation made for singular homology, but you can obtain the same result for the De Rham cohomology just by reversing the arrows.

As the torus is a product of CW-complexes, and $\mathbb{R}$ is a field, you can also use the Künneth isomorphism to compute the cohomology, this gives you $$ H^r(S^1 \times S^1) \simeq \bigoplus_{p+q=r} H^p(S^1) \otimes H^q(S^1), $$ thus $H^1(T^2) \simeq \mathbb{R}^2$ and $H^2(T^2) \simeq \mathbb{R}$. This obstructs any homeomorphism between $S^2$ and $S^1 \times S^1$.

Answer 3

I would like to give a more elementary answer:
Your goal is to prove non-diffeomorphy of $2$-sphere and torus using De Rham cohomology. In order to achieve this, it is sufficient to show that the former's first De Rham cohomology group vanishes whereas the latter's does not. This can be seen as follows:
Suppose we have a diffeomorphism $\Phi\colon S^2\to (S^1)^2$, then the restriction $\Phi'\colon S^2\setminus\{P\}\to (S^1)^2\setminus\{\Phi(P)\}$ where we remove a point $P\in S^2$ is still a diffeomorphism. In particular, we have $$\mathrm H^1(S^2\setminus\{P\})\cong\mathrm H^1(\mathbb R^2)\cong\{0\}.$$ Now write $\Phi(P)=(z,w)$ with $z,w\in S^1$ and pick a $v\in S^1$ with $v\neq w$ (tbe such a $v$ exists, as we could e.g. take $-w$). Then note $$S^1\times\{v\}\subset(S^1\times S^1)\setminus\{(z,w)\}=(S^1)^2\setminus\{\Phi(P)\}$$ and the smooth map $$f\colon S^1\to(S^1)^2\setminus\{\Phi(P)\},x\mapsto(x,v)$$ is well-defined and has got the left-inverse $$g\colon(S^1)^2\setminus\{\Phi(P)\}\to S^1,(x,y)\mapsto x.$$ Therefore $g\circ f=\operatorname{id}_{S^1}$ and since we have got a contravariant functor at hand $$(g\circ f)^*=f^*\circ g^*=\operatorname{id}_{\mathrm H^1(S^1)}.$$ You know that $\mathrm H^1(S^1)\cong\mathbb R$ and since $f^*$ is surjective we conclude $$\mathrm H^1((S^1)^2\setminus\{\Phi(P)\})\neq\{0\}$$ in contradiction to the diffeomorphism assumption.