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Let $G_1,G_2$ be $p$-adic groups. Let $\rho_1,\pi_1$ be smooth representations of $G_1$ and $\rho_2,\pi_2$ be smooth representations of $G_2$. Assume that $\pi_2$ is admissible.

I am wondering that if $\text{Hom}_{G_1}(\rho_1,\pi_1)=0$, then $\text{Hom}_{G_1 \times G_2}(\rho_1 \boxtimes \rho_2,\pi_1 \boxtimes \pi_2 )=0$?

If it does not true in general, then is it true when $\pi_1$ is 1-dimensional representation?

Improve this question
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  • $\begingroup$ Maybe I am being very stupid here: under the assumption, wouldn't $\mathrm{Hom}_{G_1}(\rho_1\otimes V_1,\pi_1\otimes V_2)=0$ for any vector spaces $V_1,V_2$ (in particular possibly infinite-dimensional) as the image of $v\otimes w$ is necessarily zero for any $v\in\rho_1$, $w\in V_1$, without even using the smoothness? $\endgroup$
    – Cheng-Chiang Tsai
    Jul 5, 2020 at 1:13
  • $\begingroup$ @Cheng, Thank you for your comment. Why do u think the image of $v \otimes w$ is necessarily zero? $\endgroup$
    – Monty
    Jul 6, 2020 at 5:41
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    $\begingroup$ $\DeclareMathOperator\Hom{Hom}$Assuming, as your title suggests, $\boxtimes$ stands for the external tensor product: Pick two different 1D representations $\rho_1$ and $\pi_1$ of $G$. Then $\Hom_{G_1}(\rho_1, \pi_1)$ is $0$ but $\Hom_{G_1 \times G_1}(\rho_1 \boxtimes \pi_1, \pi_1 \boxtimes \rho_1)$ is non-$0$. $\endgroup$
    – LSpice
    Jul 6, 2020 at 6:17
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    $\begingroup$ @LSpice I believe the map you suggest is only $G_1$-equivariant (diagonally), not $G_1\times G_1$-equivariant: one way goes $(x,y)\mapsto(gx,g'y)\mapsto(g'y,gx)$ and another $(x,y)\mapsto(y,x)\mapsto(gy,g'x)$. Or do you have in mind some other map? $\endgroup$
    – მამუკა ჯიბლაძე
    Jul 6, 2020 at 6:56
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    $\begingroup$ Following my original comment, suppose $\varphi\in\mathrm{Hom}_{G_1}(\rho_1\otimes V_1,\pi_1\otimes V_2)$ is such that $\varphi(v\otimes w)\in\pi_1\otimes V_2$ is non-zero. We can always pick a functional $\phi$ on $V_2$ such that $\phi(\varphi(v\otimes w))\not=0$. But then $\phi\circ\varphi|_{\rho_1\otimes\langle w\rangle}\in\mathrm{Hom}_{G_1}(\rho_1,\pi_1)\not=0$. Or is it? $\endgroup$
    – Cheng-Chiang Tsai
    Jul 7, 2020 at 6:41

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