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I was reading a lemma:

Lemma. Let $G$ be a locally profinite group, and let $H$ be an open subgroup of $G$ of finite index.

  1. If $(\pi, V)$ is a smooth representation of $G$, then $V$ is $G$-semisimple if and only if it is $H$-semisimple.
  2. Let $(\sigma, W)$ be a semisimple smooth representation of $H$. The induced representation $\operatorname{Ind}_H^G \sigma$ is $G$-semisimple

Proof. Suppose that $V$ is $H$-semisimple, and let $U$ be a $G$-subspace of $V$. By hypothesis, there is an $H$-subspace $W$ of $V$ such that $V = U \oplus W$. Let $f \colon V \to U$ be the $H$-projection with kernel $W$. Consider the map $$ f^G \colon v \mapsto (G : H)^{-1} \sum_{g \in G/H} \pi(g) f( \pi(g)^{-1} v ), \quad v \in V. $$ The definition is independent of the choice of coset representatives and it follows that $f^G$ is a $G$-projection $V \to U$. We then have $V = U \oplus \ker f^G$ and $\ker f^G$ is a $G$-subspace of $V$. Thus $V$ is semisimple (cf. 2.2 Proposition).

Conversely, suppose that $V$ is $G$-semisimple. Thus $V$ is a direct sum of irreducible $G$-subspaces (2.2), and it is enough to treat the case where $V$ is irreducible over $G$. As representations of $H$, the space $V$ is finitely generated and so admits an irreducible $H$-quotient $U$. Suppose for the moment that $H$ is a normal subgroup of $G$. By Frobenius Reciprocity (2.4.2), the $H$-map $V \to U$ gives a non-trivial, hence injective, $G$-map $V \to \operatorname{Ind}_H^G U$. As representations of $H$, the induced representation $\operatorname{Ind}_H^G U = c-\operatorname{Ind}_H^G U$ is a direct sum of $G$-conjugates of $U$ (cf. 2.5 Lemma). These are all irreducible over $H$, so $\operatorname{Ind} U$ is $H$-semisimple. Proposition 2.2 then implies that $V \subset \operatorname{Ind} U$ is $H$-semisimple.

In general, we set $H_0 = \bigcap_{g \in G/H} g H g^{-1}$. This is an open normal subgroup of $G$ of finite index. We have just shown that the $G$-space $V$ is $H_0$-semisimple; the first part of the proof shows it is $H$-semisimple.

This completes the proof of 1., and 2. follows readily from the same arguments.

I don't know how $U$ is constructed (after "conversely" in the second paragraph), that is:

Given a locally profinite group $G$, an open subgroup $H$ of finite index, suppose $(\pi, V)$ is an irreducible smooth representation of $G$, we need to construct an $H$-irreducible space $U$ with a nontrivial $H$-homomorphism $V\rightarrow U$.

I had considered $U=V/W$ where $W=$ the $H$-space generated by $\{v-\pi(g_i)v;v\in V,g_i$'s are the right coset representatives of $H$ in $G\}$, but I'm not sure wether $V/W$ is non-trivial or not. Help please.

Jendrik Stelzner
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CYC
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    Take $W$ to be any maximal proper $H$-invariant subspace of $V$. – John M Sep 10 '17 at 14:45
  • @JohnM How to construct such proper $W$? I'm thinking using Zorn's lemma but it doesn't seem to remain proper. – CYC Sep 10 '17 at 14:53
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    See here https://math.stackexchange.com/questions/1260096/submodules-of-a-finitely-generated-module-is-contained-in-a-maximal-ideal – John M Sep 10 '17 at 15:20
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    The idea is to take the collection of proper submodules and find a maximal element. A maximal one will still be proper. – John M Sep 10 '17 at 15:23

1 Answers1

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Consider a totally ordered set $\{U_\alpha\}_{\alpha \in A}$ of proper $H$-invariant subspaces of $V$. Note that this set may be uncountable. The union $$U = \bigcup_{\alpha \in A} U_\alpha$$ is an $H$-invariant subspace of $V$, and can not be all of $V$ because that would contradict the finitely generatedness of $V$.

Therefore the conditions of Zorn's lemma are satisfied, so there is a maximal proper $H$-invariant subspace $W$ of $V$. Then the quotient $V/W$ is an an $H$-irreducible quotient of $V$.

John M
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