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Is it true that if $f \in D'(\Omega_1), g \in D'(\Omega_2)$ then $$\mathrm{sing\,supp}(f \times g) = \mathrm{sing\,supp}(f) \times \mathrm{supp}(g) \cup \mathrm{supp}(f) \times \mathrm{sing\,supp}(g)?$$

I can see the inclusion $\fbox{$\subset$}$. Denote by $(-)^{C_1}$ complement in $\Omega_1$ and by $(-)^{C_2}$ complement in $\Omega_2$. Let $f \equiv \widetilde f \in C^\infty(\mathrm{sing\,supp}(f)^{C_1}), g \equiv \widetilde g \in C^\infty(\mathrm{sing\,supp}(g)^{C_2})$. Then $f \times g \equiv \widetilde f(x) \widetilde g(y)$ on

$$C^\infty(\mathrm{sing\,supp}(f)^{C_1} \times \mathrm{sing\,supp}(g)^{C_2}) = C^\infty(\Omega_1 \times \Omega_2 \setminus (\Omega_1 \times \mathrm{sing\,supp}(g) \cup\mathrm{sing\,supp}(f) \times \Omega_2)),$$ and $f \times g \equiv 0 \equiv \widetilde f(x) \widetilde g(y)$ on

$$C^\infty(\mathrm{supp}(f)^{C_1} \times \Omega_2 \cup \Omega_1 \times \mathrm{supp}(g)^{C_2}) = C^\infty(\Omega_1 \times \Omega_2 \setminus \mathrm{supp}(f) \times \mathrm{supp}(g)).$$

Hence, \begin{align*} \mathrm{sing\,supp}(f \times g) &\subset (\Omega_1 \times \mathrm{sing\,supp}(g) \cup \mathrm{sing\,supp}(f) \times \Omega_2) \cap \mathrm{supp}(f) \times \mathrm{supp}(g) = \\ & = (\Omega_1 \times \mathrm{sing\,supp}(g) \cap \mathrm{supp}(f) \times \mathrm{supp}(g)) \cup (\mathrm{sing\,supp}(f) \times \Omega_2 \cap \mathrm{supp}(f) \times \mathrm{supp}(g)) = \\ &= \mathrm{sing\,supp}(f) \times \mathrm{supp}(g) \cup \mathrm{supp}(f) \times \mathrm{sing\,supp}(g), \end{align*} since $\mathrm{sing\,supp}(f) \subset \mathrm{supp}(f), \mathrm{sing\,supp}(g) \subset \mathrm{supp}(g)$.

The other direction seems natural to my, but I don't how to formalize the reasonings.

My question is motivated by the fact that $\mathrm{sing\,supp}(\Theta(x) \Theta(y)) = \{0\} \times \mathbb R_+ \cup \mathbb R_+ \times \{0\}$.

I would appreciate any help.

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  • $\begingroup$ Better terminology and notation: "direct product"$->$"tensor product", $f\times g$ $->$ $f\otimes g$. $\endgroup$ Oct 27, 2021 at 15:14

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