$G$ is an p-adic group, and $\pi$ is an irreducible representation of $G$, then do we naturally have $Hom_G(C_c^{\infty}(G),\pi)\cong Hom_{\mathbb{C}}(\pi^{\vee},\mathbb{C})$? I think it is true, but I do not have found the detailed proof.
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$\begingroup$ This is true for any smooth representation $\pi$: you do not need to suppose it is irreducible. Cf. my post below. $\endgroup$– Paul BroussousCommented Jan 8, 2019 at 12:36
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For two smooth representations $\pi_i$, $i=1,2$, of $G$, one has $\mathrm{Hom}_G (\pi_1 ,\pi_2 ) \simeq \mathrm{Hom}_G (\pi_2^\vee ,\pi_1^\vee )$. On the other hand the contragredient of $C_c^\infty (G)$ is $C^\infty (G)$ (the space of smooth functions with arbitrary support), the pairing being given by $\langle f, g \rangle =\int_G fg\, d\mu$, $f\in C_c^\infty (G)$, $g\in C^\infty (G)$ (for some fixed Haar measure $\mu$ on $G$). Moreover the space $C^\infty (G)$ is the induced representation $\mathrm{Ind}_{\{ 1\}}^G {\mathbb C}$. All together, we get $$ \mathrm{Hom}_{G}(C_c^\infty (G), \pi )\simeq \mathrm{Hom}_G(\pi^\vee , \mathrm{Ind}_{\{ 1\}}^G {\mathbb C}) \simeq \mathrm{Hom}_{\{ 1\}} (\pi^\vee ,{\mathbb C}), $$ where the last isomorphism follows from Frobenius reciprocity for induction.
answered Jan 7, 2019 at 20:11
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2$\begingroup$ Concretely, I think that the isomorphism is $T \mapsto (v^\vee \mapsto \lim_{K \downarrow \{1\}} v^\vee(T[K]))$, where $K$ runs over compact, open subgroups, and $T[K]$ is the value of $T$ at the characteristic function of the identity. $\endgroup$– LSpiceCommented Jan 7, 2019 at 22:36
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2$\begingroup$ With the inverse map given by $S \mapsto \bigl([g K] \mapsto (v^\vee \mapsto \int_K g k S(v^\vee)\,\mathrm d\mu(k))\bigr)$ for all $g \in G$ and all compact open subgroups $K$, where we have identified $\pi$ with its double contragredient. $\endgroup$– LSpiceCommented Jan 7, 2019 at 22:40
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