# Two irreducible representations, and an invariant subspace of their direct sum

The following statement is a variant of a statement from Bernstein Zelevinsky.

Let $G$ be a group and let $\left( \pi, V \right)$ , $\left( \tau, W \right)$ be two irreducible representations of $G$ .
A representation of $G$ denoted by $\pi \oplus \tau$ is defined on the space $V \oplus W$ by $(\pi \oplus \tau)(g) (v, w) = (\pi(g)v, \tau(g)w)$ .
In this post I want to show the proof of the following statement

Suppose $U \subseteq V \oplus W$ is a $\pi \oplus \tau$ invariant subspace (i.e - a subspace of $V \oplus W$ which is closed under $\pi \oplus \tau$ ). Suppose that $U \nsubseteq V, W$ and $V, W \nsubseteq U$ . Then $\left(U, \left( \pi \oplus \tau \right) \restriction_U \right) \cong \left( V, \pi \right)$ and $\left(U, \left( \pi \oplus \tau \right) \restriction_U \right) \cong \left( W, \tau \right)$ . Furthermore, the projections are isomorphisms. In particular we get that $V$ and $W$ are isomorphic.

When we say $V, W \nsubseteq U$ and $U \nsubseteq V, W$ we recognize $V, W$ with their embeddings $V \oplus \{0\}, \{0\} \oplus W$ .

We now move to the proof: first note that for every $0 \ne v \in V$ , we have $\left( v ,0 \right) \notin U$ : Otherwise, since $U$ is $\left(\pi \oplus \tau \right)$ -invariant, $\left(\pi \oplus \tau \right)(g)(v, 0) \in U$ for every $g \in G$ and therefore $(\pi(g) v, 0) \in U$ for every $g \in G$ . But $V$ is irreducible and therefore $\{ \pi(g) v \mid g \in G \} = V$ . This implies $V \subseteq U$ which is a contradiction.
Similarly we have $(0, w) \notin U$ for every $w \in W$ .
We now note that $U \ne \{ (0, 0) \}$ : otherwise $U \subseteq V, W$ .
We show now that the projection map $\mathrm{pr}_V : U \rightarrow V$ is an isomorphism: it is clear that $\mathrm{pr}_V$ is a homomorphism. $\mathrm{pr}_V$ is one-to-one: if $\mathrm{pr}_V(v, w) = 0$ for $(v, w) \in U$ , then $v = 0$ and since $(0, w) \notin U$ for $w \ne 0$ , we have $w = 0$ .
$\mathrm{pr}_V$ is onto: let $(0,0) \ne (v, w) \in U$ (such a vector exists since $U \ne \{0\}$ ). Then $v \ne 0$ and therefore $\{ \pi(g) v \mid g \in G \} = V$ . Finally, $\{ \pi(g) v \mid g \in G \} = \mathrm{pr}_V \left( \{ \left( \pi \oplus \tau \right)\left( g \right)(v,w) \mid g \in G \} \right)$ and therefore $\mathrm{pr}_V$ is onto.

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