# Show that $\Omega^*(M)\otimes \Omega^*(N)$ is isomorphic to $\Omega^*(M\times N)$

Let $M, N$ be compact manifolds and $\Omega^*$ its algebra exterior. How to prove that $\Omega^*(M)\otimes \Omega^*(N)$ is isomorphic to $\Omega^*(M\times N)$?

I thought about the function $f(\omega,\eta) = \pi_M^*(\omega)\wedge \pi_N^*(\eta)$, where $\omega \in \Omega^*(M)$, $\eta \in \Omega^*(N)$ and $\pi_M$, $\pi_N$ the canonical projection.

The problem is to construct the linear function define on the tensor product. How to proceed?

• Can you think in a bilinear map from $\Omega(X)\times\Omega(Y) \to \Omega(X \times Y)$? If yes, then you should use the universal property of the tensor product to finish... think you can proceed from here? EDIT: Tipo Jun 13, 2016 at 22:48

Such an isomorphism is not true. What really happens is that, under the bilinear map from $\Omega^*(M)\times\Omega^*(N)$ into $\Omega^*(M\times N)$ you wrote (which, by universality of the tensor product, lifts uniquely to a linear map from $\Omega^*(M)\otimes\Omega^*(N)$ into $\Omega^*(M\times N)$ as user 123456 suggested), $\Omega^*(M)\otimes\Omega^*(N)$ is isomorphic to a dense subspace of $\Omega^*(M\times N)$ in the (Fréchet) topology of uniform convergence of all derivatives in $M\times N$ (or compact subsets thereof if $M$ or $N$ is not compact; it is not difficult to see in either case that this topology does not depend on the choice of atlas).
The trouble begins already at forms of degree zero: it is not true that every $F\in\mathscr{C}^\infty(M\times N)$ is of the form $$F(x,y)=\sum^n_{j=1}f_j(x)g_j(y)\ ,\quad f_j\in\mathscr{C}^\infty(M)\ ,\,g_j\in\mathscr{C}^\infty(N)$$ for some $n\in\mathbb{N}$. Take for instance $M=N=\mathbb{R}/2\pi\mathbb{Z}$ and $$F(x,y)=\sum^\infty_{m,n=0}a_{mn}\cos(mx)\cos(ny)\ ,$$ where the double sequence $(a_{mn})_{m,n\geq 0}$ decreases to zero faster than any negative power of $m+n$ but never becomes zero as $m,n\to+\infty$, guaranteeing that $F$ is smooth and doubly periodic - e.g. $a_{mn}=e^{-mn}$ (the same happens if $M$ or $N$ is not compact, take e.g. $M=N=\mathbb{R}$ and $F(x,y)=e^{xy}$).