# Invariant subspace $U$ of direct sum $V\oplus V$ of irreducible representation $V$ is isomorphic to $V$

I am studying from Hall's book "Lie Groups, Lie Algebras, and Representations" and I'm stumped on the following question:

Suppose that $$V$$ is an irreducible finite-dimensional representation of a group or Lie algebra over $$\mathbb{C}$$, and consider the associated representation $$V\oplus V$$. Show that every nontrivial invariant subspace $$U$$ of $$V\oplus V$$ is isomorphic to $$V$$ and is of the form $$U=\{(\lambda_1v,\lambda_2v)|v\in V\}$$ for some constants $$\lambda_1$$ and $$\lambda_2$$, not both zero.

From the looks of it, I think I should be using Schur's lemma, which tells me that if $$V$$ is an irreducible complex representation of a group or Lie algebra and $$\phi:V\to V$$ is an intertwining map of $$V$$ with itself, then $$\phi=\lambda I$$ for some $$\lambda\in\mathbb{C}$$. An intertwining map is a linear map which commutes with the action. My understanding is that other authors just call this a homomorphism of representations.

However, I do not see how to connect the end result with Schur's lemma. Perhaps the projection maps will come in handy here? Any help is appreciated.

Here's a more detailed answer. First assume that $$U$$ is non-trivial irreducible invariant subspace of $$V \oplus V$$. For $$i: U \hookrightarrow V \oplus V$$ the inclusion and $$pr_1 , pr_2: V \oplus V \twoheadrightarrow V$$ the projections on the first and second component, repectively, look at the maps $$\phi_j:=pr_j \circ i: U \rightarrow V.$$

By Schur's Lemma, $$\phi_j$$ is with zero or isomorphism of representation. We cannot have both $$\phi_j =0$$. So we know for sure that $$U$$ is isomorphic to $$V$$.

The missing part in Torsten's answer is to show that $$U$$ has the form given in the question $$U=\{(\lambda_1v,\lambda_2v)|v\in V\}$$

• If $$\phi_1 = 0$$, then we know $$U$$ is the second copy of $$V$$ in $$V\oplus V$$ so $$U$$ takes the form required in the question with $$\lambda_1 = 0, \lambda_2 = 1$$.
• Similar argument if $$\phi_2 = 0$$.
• If both $$\phi_1$$ and $$\phi_2$$ are non-zero, we can apply the third statement of Schur's Lemma (as in Hall's Theorem 4.29) (note here we used the fact that we are working over $$\mathbb{C}$$) and conclude that $$\phi_1 = \lambda \phi_2$$ for some $$\lambda\in\mathbb{C}$$. This gives us $$U=\{(v,\lambda v)|v\in V\}$$.

Now what happens in general if $$U$$ is not irreducible, non-zero and invariant? Then it must contain a non-zero, invariant, irreducible subspace (this is an earlier question in Hall, but it's also straightforward to prove). The only such latter subspaces are all isomorphic to $$V$$, so $$\dim U > \dim V$$.

Consider $$U\cap V$$ where $$V$$ here is the first copy in $$V\oplus V$$. By considering dimensions, we must have $$U\cap V$$ non-zero. But $$U\cap V$$ is invariant. Since $$V$$ is irreducible, we must have $$U\cap V$$ equals to the first copy of $$V$$. A same argument gives $$U$$ intersect second copy of $$V$$ equals the second copy of $$V$$. Therefore, $$U$$ contains both copies of $$V$$ and therefore $$V\oplus V$$. $$U$$ must be the whole space.

[Alternative to the last paragraph: By proposition 4.26, we can write $$V\oplus V$$ as $$U\oplus W$$ where $$W$$ is another invariant subspace. If $$\dim U > \dim V$$, then $$\dim W < \dim V$$ and that can only mean $$W = 0$$.]

A hint would be to write down the short exact sequence $$0\rightarrow U \rightarrow V \oplus V \rightarrow W \rightarrow 0.$$

We now have an intertwining map (i.e. homomorphism of representations) $$\phi: V \oplus V \rightarrow W$$.

This map, of course, is a direct sum of two maps $$\phi_1: V \rightarrow W$$ and $$\phi_2: V \rightarrow W$$.

What does Shur's lemma say about $$\phi_1$$ and $$\phi_2$$?

• If $W$ is irreducible, then it says that $\phi_1$ and $\phi_2$ are each either zero or an isomorphism. So I suppose that I'm supposed to choose $W$ in such a way that this helps me? Commented Nov 27, 2020 at 0:29

Step 1: Assume $$U$$ is irreducible. For $$i: U \hookrightarrow V \oplus V$$ the inclusion and $$pr_1 , pr_2: V \oplus V \twoheadrightarrow V$$ the projections on the first and second component, repectively, look at the maps $$pr_j \circ i: U \rightarrow V.$$ Use Schur's Lemma. Notice what happens if both are zero.

Step 2: Now, assume $$U$$ contains two different irreducible subrepresentations $$U_1 \neq U_2$$. Using the result of step 1, show that $$U_1+U_2$$ contains both $$V \oplus 0$$ and $$0 \oplus V$$, hence is all of $$V \oplus V$$.

• I see what you mean, but I don't know that $U$ is irreducible so I'm not sure that I can apply Schur's lemma here. I know that there must be an irreducible invariant subspace contained in $U$, call it $W$. So applying Schur's lemma to your suggested maps with $W$ in place of $U$ shows that $W$ is isomorphic to $V$. Can I conclude that $U$ is isomorphic to $V$? Commented Nov 27, 2020 at 17:21
• You're right, sorry I overlooked that. See my addition for an attempted fix. Commented Nov 28, 2020 at 6:18
• Consider how multiplicities of irreps in $U$ are used to simplify $\dim\hom_G(U,V\oplus V)$ in order to get that $U$ can only be a (proper, nontrivial) subrep if $U\cong V$.
– anon
Commented Nov 28, 2020 at 6:27