# 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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# Two irreducible representations, a normal subgroup, and a character

This is a variation of a statement I came across while reading the article of Bernstein Zelevinsky regarding representations of $\mathrm{GL}_n(F)$ where $F$ is a non-archimedean local field.

Let $G$ be a group and let $(\pi,V)$ and $(\rho,W)$ be irreducible representations of $G$ over $\mathbb{C}$ . Let $H \triangleleft G$ be a normal subgroup of $G$ such that $G / H$ is commutative (i.e. $H$ contains the commutator subgroup). Suppose that $\mathrm{Hom}_{H}(V,W)$ is finite dimensional and is non-trivial. Then there exists a character $\chi : G \rightarrow \mathbb{C}^\times$ such that $\rho \approx \chi \cdot \pi$ , and such that $\chi$ is trivial on $H$ .

Let me explain the proof. We define a representation $(\tau, \mathrm{Hom}_{H}(V,W))$ of the quotient group $G / H$ by $\tau(g H) T = \rho(g) T \pi(g^{-1})$ . This is well defined, since if $h \in H$ then $\rho(gh) T \pi({(gh)}^{-1}) = \rho(g) \rho(h) T \pi(h^{-1}) \pi(g^{-1})$ , but $T$ is a homomorphism and therefore $\rho(h) T \pi(h^{-1}) = T$ so $\rho(gh) T \pi({(gh)}^{-1}) = \rho(g) T \pi(g^{-1})$ . It is easy to verify that $\tau$ is indeed a representation.

We now consider the set $\tau(G / H)$ . This set is a set of linear automorhpisms of $\mathrm{Hom}_{H}(V,W)$ . Since $G / H$ is commutative, the set $\tau(G / H)$ is commutative. Since $\mathrm{Hom}_{H}(V,W)$ is finite dimensional, we can think of $\tau(G / H)$ elements as matrices. It is a well known fact that a commuting set of matrices over an algebraically closed field, has a (non-zero) common eigenvector. Let $T \in \mathrm{Hom}_{H}(V,W)$ be such a common eigenvector. Then for all $g \in G$ we have $\tau(g H) T = \chi(g) T$ where $\chi(g) \in \mathbb{C}$ . Since $\tau$ is a representation, it follows that $\chi$ is a character. By definition of $\tau$ we have $\tau(g H) T = \rho(g) T \pi(g^{-1}) = \chi(g) T$ which implies $\rho(g) T = T \chi(g) \pi(g)$ . Therefore $T$ is a $G$ -homomorphism between $\chi \cdot \pi$ and $\rho$ . Also since $\tau(hH) T = \tau(1_G \cdot H) T = T$ we have that $\chi$ is trivial on $H$ .

Since $\pi$ is irreducible, so is $\chi \cdot \pi$ . Therefore $T$ is a non-zero homomorphism between the two irreducible representations $\chi \cdot \pi$ and $\rho$ , and therefore it is an isomorphism.

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Moved here.

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# A product of a compact subset and a closed subset is closed

Let $G$ be a topological group and let $A \subseteq G$ be a compact subset and $B \subseteq G$ be a closed subset. Then $A \cdot B \subseteq G$ is closed.

Although the claim doesn’t seem to be too difficult, I didn’t find the proof to be immediate. I tried googling a bit and found the following math.stackexchange question, but the accepted answer there is wrong. Furthermore, the solution assumes that the space is regular, which was fine in my case since I was dealing with a metrizable space, but the statement is true also without this assumption. I’ve found this solution which I want to explain.

We want to show that $G \setminus \left({A \cdot B}\right)$ is open. Let $x \in G \setminus \left({A \cdot B}\right)$ . For every $a \in A$ we have that $a^{-1} \cdot x \notin B$ . Consider the map $m:G \times G \rightarrow G$ defined by $m(g,h) = g^{-1} \cdot h$ . Then $m$ is continuous. Since $B$ is closed, $G \setminus B$ is open, and therefore $m^{-1}(G \setminus B)$ is open. Since for every $a \in A$ , we have $m(a,x) = a^{-1} \cdot x \notin B$ , we have $(a, x) \in m^{-1}(G \setminus B)$ and since the latter is open, there exists open subsets of $G$ with $a \in U_a, x \in V_a$ such that $m(U_a \times V_a) \subseteq G \setminus B$ .

Consider the following open cover of $A$ : $A \subseteq \bigcup_{a \in A} U_a$ . Since $A$ is compact, this cover admits a finite sub-cover $A \subseteq \bigcup_{i \in 1}^N U_{a_i}$ . Define $V = \bigcap_{i = 1}^N V_{a_i}$ . Then $V$ is open with $x \in V$ . We claim that $V \subseteq G \setminus \left({A \cdot B}\right)$ : otherwise there exists $a \in A, b \in B$ with $a \cdot b \in V$ . Since $a \in A$ , we have that $a \in U_{a_i}$ for some $i$ . Then $ab \in V \subseteq V_{a_i}$ and $m(a,ab) = a^{-1} \cdot (ab) = b \in B$ , which is a contradiction since $U_{a_i} \times V_{a_i} \subseteq m^{-1}(G \setminus B)$ .

Note that this proof doesn’t assume that $G$ is Hausdorff.

Also note that the claim isn’t true if the assumption that $A$ is compact is replaced with the assumption that $A$ is closed: we can take in this case $A = \{ (t, \frac{1}{t} ) \mid t > 0 \}$ , $B = \{ (t, -\frac{1}{t} ) \mid t > 0 \}$ . The set $A + B$ clearly contains the sequence $(\frac{1}{n}, 0)$ but $(0,0) \notin A + B$ .

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