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Jan 8, 2019 at 1:42 comment added Cooler Panda @LSpice Very helpful!
Jan 8, 2019 at 1:41 vote accept Cooler Panda
Jan 7, 2019 at 22:40 comment added LSpice 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.
Jan 7, 2019 at 22:36 comment added LSpice 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.
Jan 7, 2019 at 22:32 history edited LSpice CC BY-SA 4.0
Missing subscript $G$ and ${\rm…}$ -> $\mathrm{…}$
Jan 7, 2019 at 20:11 history answered Paul Broussous CC BY-SA 4.0