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Let $G$ be a $p$-adic group, ie. the group of $F$-points of some connected reductive group over $F$, where $F$ is a $p$-adic field. We consider complex smooth representations of $G$. Any irreducible representation of $G$ has a central character. Assume that $\pi$ and $\pi'$ are two irreducible representations of $G$ with different central characters. Is it true that all the Ext groups $\mathrm{Ext}^i_G(\pi,\pi')$ vanish?

It is trivially true for $i=0$, and for $i=1$ it follows from the fact that any short exact sequence $$0 \to V' \to V \to V'' \to 0$$ splits when $V'$ and $V''$ have central characters which are different.
How about the case $i\geq 2$?

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    $\begingroup$ There will be a spectral sequence expressing $Ext^i_G(\pi, \pi')$ in terms of $H^i(G / Z, Ext^j_Z(\pi, \pi'))$, so you can WLOG suppose $G$ is a torus, and this case you can do by hand. $\endgroup$
    – David Loeffler
    Dec 21, 2022 at 9:55
  • $\begingroup$ @DavidLoeffler Thank you for the comment. I understand the spectral sequence and I was able to do the computations when $G$ is a torus. I have just a question regarding the reduction WLOG to the case of tori. I assume that you use the identification $\mathrm{Ext}_{Z}^j(\pi,\pi') \simeq \mathrm{Ext}_{Z^{\circ}}^j(\pi,\pi')^{Z/Z^{\circ}}$ where $Z^{\circ}$ denotes the unit component of the center, which is also the radical of $G$, and which is a torus. [...] $\endgroup$
    – Suzet
    Dec 22, 2022 at 1:27
  • $\begingroup$ [...] However, I imagine it could happen that even though the central characters of $\pi$ and $\pi'$ are different, their restrictions to $Z^{\circ}$ are the same. In such a case, $\mathrm{Ext}_{Z^{\circ}}^j(\pi,\pi')$ may not vanish. So one would need to justify that the $Z/Z^{\circ}$-invariants are zero here, correct? $\endgroup$
    – Suzet
    Dec 22, 2022 at 1:28
  • $\begingroup$ Ah, nevermind. I think that my computations in the case of tori actually work for any abelian locally profinite group, such as $Z$ itself. More precisely, if $G$ is abelian locally profinite and if $\chi,\chi'$ are two different smooth characters of $G$, then $\mathrm{Ext}^{i}_G(\chi,\chi')=0$ for all $i$. So everything is fine! $\endgroup$
    – Suzet
    Dec 22, 2022 at 4:28

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Because of the Ext$^1$ case that you mentioned, it is true that your category of representations $\mathcal{C}$ splits as a direct sum $\bigoplus_{\pi}\mathcal{C}_{\pi}$, where the sum is taken over all central characters $\pi$. If you have now an element in one of the higher ext groups, that is represented by an exact sequence $$0\to V\to M_n\to \cdots\to M_0\to V'\to 0$$ then you can split this sequence according to the central characters. If $V$ and $V'$ have different central characters $\pi$ and $\pi'$ respectively, this sequence will split as the direct sum of exact sequences which have 0 at one of the ends, and will therefore represent a trivial element in the ext group. As a result, all of your ext groups are indeed trivial.

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  • $\begingroup$ Thank you for the answer! I fear however that the category of smooth representations of $G$ does not split along all possible central characters. For instance, if $V = \mathrm{c-Ind}_{\mathrm{GL}_n(\mathbb Z_p)}^{\mathrm{GL}_n(\mathbb Q_p)} \,\rho$, where $\rho$ is any smooth representation of the open compact subgroup $\mathrm{GL}_n(\mathbb Z_p)$, then $V$ does not seem to split as a sum of subrepresentations having central characters. Indeed, the center does not fix any finite dimensional subspace of $V$, unless I'm mistaken. $\endgroup$
    – Suzet
    Dec 21, 2022 at 13:04
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    $\begingroup$ Sadly it is not the case. Many irreducible smooth representations are Infinite dimensional. $\endgroup$
    – Suzet
    Dec 21, 2022 at 13:16
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    $\begingroup$ Interesting question. I actually have no idea if such a statement holds. I would tend to think it does not, but I can't think of a counter example! $\endgroup$
    – Suzet
    Dec 22, 2022 at 4:30
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    $\begingroup$ @EhudMeir It's false as stated. Take $G = \mathbb{G}_m$, and let $V$ be the 2-dimensional representation on which $O_F^\times$ acts trivially and a uniformizer (= generator of $F^\times / O_F^\times$) acts by $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$. $\endgroup$
    – David Loeffler
    Dec 22, 2022 at 15:39
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    $\begingroup$ (It is true that any representation of $G$ splits into a direct sum of character eigenspaces for the maximal compact subgroup of $Z(G)$. In my examples this maximal compact is acting trivially on everything.) $\endgroup$
    – David Loeffler
    Dec 22, 2022 at 15:52

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