# Schur's lemma and Invariant subspaces of direct sums of irreducible representations

There is a corollary to Schur's lemma which says that : If $V$ is a finite dimensional irreducible complex representation of a group G or Lie algebra and $\phi :V \rightarrow V$ is an intertwining map, then $\phi=\lambda I$ for some $\lambda \in \mathbb{C}$. In the proof that I know, we use the fact that the $V$ is a complex vector space and not just a real one.

1.) Please give a counterexample to the above if V is real.

I am reading Brian Hall's book on representation theory of Lie groups & Lie algebras and wish to settle this question in context of Problem 16, Ch.4.

Suppose that $V$ is an irreducible finite dimensional representation of a group or Lie algebra and consider the associated representation $V \bigoplus V$. Show that every nontrivial irreducible invariant subspace of $V \bigoplus V$ is of the type $\{ (\lambda_1v,\lambda_2v)| v\in V\}$ for some constants $\lambda_1, \lambda_2$ not both zero.

My second question is

2.) Do we need to assume in this problem that $V$ is complex or is it true for real vector space representations as well ? If it is valid for real vector space representations, how does one prove it ?

• By an intertwining map, do you simply mean a homomorphism? – Tobias Kildetoft Oct 15 '13 at 18:08
• @TobiasKildetoft An intertwining map is a linear map which commutes with the group action on V. – user90041 Oct 15 '13 at 19:32
• On some more thought, I could think of the following answer to Q.1 :Consider the action of $S^1$ on $\mathbb{R}^2$ and let the intertwining map be rotation by 90 degrees. Clearly the corollary to Schur's lemma does not hold in this case as we are dealing with a real vector space. In general take any abelian group with an irreducible representation and choose the intertwining map to be the image of any group element suitably chosen so that the map is not a multiple of identity to get a counterexample. I still do not know how to do question number 2 if V is a real vector space. – user90041 Oct 15 '13 at 19:38
• For the second one, note that any isomorphism $V\to V$ will give rise to a submodule of $V\oplus V$, so once you have one which is not multiplication by a scalar, you get an example of what you need. – Tobias Kildetoft Oct 16 '13 at 6:58
• BTW, what source are you using? I am not sure I have seen the term intertwining map used for this before (usually they are just called homomorphisms, at least the places I usually read these things). – Tobias Kildetoft Oct 16 '13 at 6:59