Is the exterior wedge product of differential forms

$$ \begin{align*} \Omega(X) \otimes_{\mathbb{R}} \Omega(Y) &\longrightarrow \Omega(X\times Y) \\ \alpha \otimes \beta &\longmapsto \pi_X^*\alpha \wedge \pi_Y^*\beta \end{align*} $$


UPDATE 1: I think I got it locally:

Suppose $$\sum_{i=1}^n \alpha^i(x)\wedge \beta^i(y) = 0$$ for all $(x,y)\in X\times Y$. If $X=\mathbb{R}^N$, $Y=\mathbb{R}^M$ we write $$ \alpha^i(x) = \sum_I \alpha^i_I(x) \mathrm{d}x^I,\quad \beta^i(y) = \sum_J \beta^i_J(y) \mathrm{d}y^J $$ for $i=1,\ldots,n$. The equation is equivalent to $$ \alpha_I(x) \cdot \beta_J(y) = 0 $$ for all $I$, $J$ and $x,y$, where we collected $\alpha^i_I$'s and $\beta^i_J$'s in vectors $\alpha_I(x), \beta_J(x) \in \mathbb{R}^n$ and used the dot product. It follows that there is an orthonormal basis $v_1, \ldots, v_{k}, w_1, \ldots, w_l\in \mathbb{R}^n$, $k + l =n$ and smooth coefficients $a_I^u(x)$, $b_J^v(y)$ such that $$ \begin{aligned} \alpha_I(x)&= a_I^1(x) v_1 + \ldots + a_I^k(x) v_k \\ \beta_J(y) &= b_J^1(y) w_1+\ldots + b_J^l(y) w_l \end{aligned} $$ for all $I, J$ and $x,y$. Now we have $$ \sum_{i=1}^n \alpha^i \otimes \beta^i = \sum_{I,J,u,p} \bigl(\sum_{i=1}^n v^i_u w^i_p\bigr) (a^u_I (x) \mathrm{d}x^I)\otimes (b^v_J(y) \mathrm{d}x^J) = 0 $$ because $v_u \perp w_p$.

UPDATE 2: For $X$, $Y$ compact we pick coordinate coverings $(U_1,x_1),\ldots,(U_A,x_A)$ of $X$ and $(V_1,y_1),\ldots,(V_B,y_B)$ of $Y$. We pick subordinate partitions of unity $(\lambda_a)$ and $(\mu_b)$ respectively. We do it in such a way that the collections $(\lambda_a\mathrm{d}x^I_a)_{a,I} \subset \Omega(X)$ and $(\mu_b \mathrm{d}y_b^J)_{b,J} \subset \Omega(Y)$ are $\mathbb{R}$-linearly independent. It follows that the collection $\bigl((\lambda_a\mathrm{d}x^I)\otimes(\mu_b \mathrm{d}y^J)\bigr)_{a,I,b,J}$ is linearly independent in $\Omega(X)\otimes_{\mathbb{R}}\Omega(Y)$. We write $\alpha^i$ and $\beta^i$ as linear combinations (with functions as coefficients) of $(\lambda_a\mathrm{d}x^I)$ and $(\mu_b \mathrm{d}y^J)$ respectively and apply exactly the same argument as above replacing $(\mathrm{d}x^I)_I$ by $(\lambda_a\mathrm{d}x^I)_{a,I}$ and $(\mathrm{d}y^J)_J$ by $(\lambda_b\mathrm{d}y^J)_{b,J}$. We see that the exterior wedge product for compact $X$, $Y$ is injective.

UPDATE 3 (reaction to a comment): It is never an isomorphism since e.g. $e^{xy}$ is not equal to a finite sum of products $f(x)g(y)$.

UPDATE 4: I think I got it for non compact $X$, $Y$ as well: Let $U_i$, resp. $V_j$ be exhaustions of $X$, resp. $Y$ by relatively compact open sets. Each of these have a finite atlas, and hence, modifying the proof above, the exterior wedge product $\Omega(U_i)\otimes \Omega(V_j) \rightarrow \Omega(U_i\times V_j)$ is injective. For fixed $i$ the exterior wedge product induces an injection $$\varprojlim_j \Omega(U_i)\otimes \Omega(V_j) \simeq \Omega(U_i) \otimes \varprojlim_j(\Omega(V_j)) \longrightarrow \varprojlim_j \Omega(U_i \times V_j)$$ where we used that the inverse limit commutes with tensor product and preserves exactness. Taking the inverse limit over $i$ we get similarly an injection $$ \varprojlim_i(\Omega(U_i))\otimes \varprojlim_j(\Omega(V_j)) \longrightarrow \varprojlim_i \varprojlim_j \Omega(U_i\times V_j) $$ It is easy to check that the restrictions from $X$ to $U_i$, resp. $Y$ to $V_j$, resp. $X\times Y$ to $U_i\times V_j$ induce embeddings of $\Omega(X)$, resp. $\Omega(Y)$, resp. $\Omega(X\times Y)$ into $\varprojlim_i(\Omega(U_i))$, resp. $\varprojlim_j(\Omega(V_j))$, resp. $\varprojlim_i \varprojlim_j \Omega(U_i\times V_j)$ and that the induced injection restricts to the exterior wedge product $\Omega(X)\otimes \Omega(Y) \rightarrow \Omega(X\times Y)$. Consequently it is injective.

QUESTION LEFT: Is the above proof correct? Shall I close the question?

  • 3
    $\begingroup$ I think it is, one should do local calculations to prove it. I think it may even be an isomorphism in the compact case, and likely in the non compact case aswell but might require a more subtle argument. $\endgroup$ Jan 20, 2018 at 12:19
  • $\begingroup$ What do you think? Have you tried anything? Could you provide some background? $\endgroup$ Jan 20, 2018 at 12:23
  • $\begingroup$ Suppose $X=\mathbb{R}^n, Y=\mathbb{R}^m$. If we write $\alpha^i(x) = \sum_I \alpha^i_I(x) \mathrm{d}x^I$ and $\beta^i(y) = \sum_J \beta^i_J(y) \mathrm{d}y^J$ for $i=1,\ldots,n$, then from $\sum_{i=1}^n \alpha^i(x)\wedge \beta^j(y) = 0$ for all $(x,y)\in X\times Y$ follows $\sum_{i=1}^n \alpha^i_I(x)\beta^j_J(y) = 0$ for all multiindices $I$, $J$ and points $(x,y)\in X\times Y$. In case $n=1$ it follows clearly that either $\alpha=0$ or $\beta=0$. I do not what to do for $n>1$. $\endgroup$
    – Pavel
    Jan 20, 2018 at 12:34
  • $\begingroup$ There is a typo above: Let me rather denote the dimensions by capitals $N, M$ so that it does not colide with $n$ as a limit value of $i$. Also there should be $\beta^i$ and not $\beta^j$ in the sums. $\endgroup$
    – Pavel
    Jan 20, 2018 at 12:45
  • $\begingroup$ Silly question, but why do you worry about the injectivity of this map? $\endgroup$
    – Pedro
    Jan 20, 2018 at 15:56

1 Answer 1


I think I answered the question myself.


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