# If $M$ is diffeomorphic to $N$, then $\mathbf{H}_{DR}^p(M)$ is isomorphic to $\mathbf{H}_{DR}^p(N)$.

I thought I got this, but no.....

Given $\mathbf{H}_{DR}^2(S^3)$ is trivial but $\mathbf{H}_{DR}^2(T^3)$ is not, how can I show $S^3$ and $T^3$ are not diffeomorphic?

I am also wondering about the general case, that if $M$ is diffeomorphic to $N$, then $\mathbf{H}^p_{DR}(M)$ is isomorphic to $\mathbf{H}^p_{DR}(N)$.

Where $\mathbf{H}^p_{DR}(N)$ is the DeRham Cohomology group.

Thank you...

• $\mathbf{H}^p(M)$ is the de Rham cohomology group? – Daniel Fischer Aug 9 '13 at 22:00
• Yes @DanielFischer, I will revise. Thank you. – WishingFish Aug 9 '13 at 22:01

Say that $M,N$ are smooth manifolds and $F:M\to N$ is a diffeomorphism. For every $k$-form $\omega$ on $N$, the pullback $F^*\omega$ is a $k$-form on $M$. I claim the pullback descends to an isomorphism of de Rham cohomology groups.

Let $Z^p(N)$ and $Z^p(M)$ denote the closed $p$-forms on $N$ and $M$, respectively. Moreover, let $B^p(N)$ and $B^p(M)$ denote the exact $p$-forms on $N$ and $M$, respectively.

I claim that $F^*:Z^p(N)\to Z^p(M)$ and $F^*:B^p(N)\to B^p(M)$, i.e. $F^*$ maps closed forms to closed forms and exact forms to exact forms. This follows quickly from the fact that pullback commutes with exterior differentiation: $dF^*=F^*d$. So if $\omega$ is a closed $p$-form on $N$, then $d(F^*\omega)=F^*(d\omega)=F^*(0)=0$, so $F^*\omega$ is a closed $p$-form on $M$. Likewise, if $\omega$ is exact, that is, $\omega=d\nu$ for some $p-1$ form $\nu$ on $N$, then $F^*(\omega)=F^*(d\nu)=d(F^*(\nu))$. So $F^*\omega$ is exact.

Thus, $F^*$ descends to a linear transformation (still written) $F^*:H_{dR}^p(N)\to H_{dR}^p(M)$. It'll be of great use to you to check that $Id^*=(F\circ F^{-1})^*=(F^{-1})^*\circ F^*$. This shows, among other things, that $F^*$ is an invertible linear transformation, i.e. an isomorphism of vector spaces.

• That's so helpful, thank you very much!! – WishingFish Aug 9 '13 at 22:33

If $M$ and $N$ are diffeomorphic, then they have isomorphic de Rham cohomology.

If $\varphi:M\to N$ is a smooth map, then for each $p\geq 0$ there is an induced map $\varphi^*:\Omega^p(N)\to \Omega^p(M)$ where $\Omega^p(\bullet)$ denotes the vector space of smooth $p$-forms for that manifold. This is known as the pullback.

The pullback $\varphi^*$ takes closed forms to closed forms and it takes exact forms to exact forms, which follows from the equation $d(\varphi^*\alpha) = \varphi^*(d\alpha)$, and it therefore gives a well-defined map $\varphi^*:H^p_{DR}(N)\to H^p_{DR}(M)$. If $\varphi$ is a diffeomorphism then $\varphi^*$ is an isomorphism, with inverse $(\varphi^{-1})^*$; this follows from the identities $(\varphi\circ\psi)^* = \psi^*\circ\varphi^*$ and $\operatorname{id}_M^*=\operatorname{id}$.

• Oh, I should have said $\mathbf{H}_{DR}^2(S^3)$ is trivial but $\mathbf{H}_{DR}^2(T^3)$ is not. – WishingFish Aug 9 '13 at 22:22