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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Introducing TypedAutobahn

Moved here.

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

I came across the following statement while reading notes in a course about p-adic representations:

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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