5
$\begingroup$

Is the exterior wedge product of differential forms

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

injective?

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?

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Olivier Bégassat Jan 20 '18 at 12:19
  • $\begingroup$ What do you think? Have you tried anything? Could you provide some background? $\endgroup$ – Amitai Yuval Jan 20 '18 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 '18 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 '18 at 12:45
  • $\begingroup$ Silly question, but why do you worry about the injectivity of this map? $\endgroup$ – Pedro Tamaroff Jan 20 '18 at 15:56
0
$\begingroup$

I think I answered the question myself.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.