# Exercise 3.39 of Fulton & Harris

I would like some help with exercise 3.39 from Fulton & Harris' 'Representation Theory A First Course':

Let $V_0$ be a real vector space on which $G$ acts irreducibly, $V=V_0 \otimes \mathbb{C}$ is a complex representation of $G$. Show that if $V$ is not irreducible, then it has two irreducible factors, and they are conjugate complex representations of $G$.

First of all I would like to remark that conjugate complex representations were not covered before in the book, but after a bit of research I believe the statement in the exercise is equivalent to: $V=W \oplus W^*$ for some (complex) irreducible sub-representation $W \subset V$. If I am mistaken please do correct me!

This question was asked here: Question about Real Representations, but the post got bogged down with notations and since it was so old I decided to ask it again here. It was suggested in the comments of that post that we consider a irreducible (proper) sub-representation $W \subset V$, let $V=W \oplus U$, and show that $U=W^*$, but I don't really see how we can go on...

The only way I have learnt to compute the amount of irreducible factors inside a representation $V$ is by computing $(\chi_V,\chi_V)=\frac{1}{|G|} \sum_{g \in G} |\chi_V(g)|^2$, but since we don't have a corresponding formula (not one that I am aware of) in the real case, I'm not sure how we can prove there are 2 irreducible factors in $V$ here. I am also failing to see how the second statement about the relation between the two factors can follow from the first part.

Any help is appreciated!

## 2 Answers

There is actually a much more elementary way to see this. We write elements of $V_0\otimes \mathbb{C}$ as $x+iy$ for $x,y\in V_0$.

Claim: If W is a nonzero submodule of V, then $W\oplus W^* = V$.

Proof: It is trivial that $W+ W^*$ is a submodule of $V$. Now suppose $x+iy\in W$ is nonzero, where $x,y\in V_0$. Note that $x-iy\in W^*$, so $(x+iy)+(x-iy) = 2x\in W+ W^*$. Then if $x\neq 0$, since $V_0$ is irreducible, this implies that $V_0\subseteq W+W^*$ (using the natural embedding of $V_0$ into $V$). Then $iV_0\subseteq W+ W^*$ and therefore $V = W+ W^*$. If instead $x = 0$, then since we assumed $x+iy\neq 0$, we have $y\neq 0$. So then irreducibility gives $iV_0\subseteq W$ and we proceed similarly to above. We are done proving this claim.

Now if we assume that $W$ is a proper nonzero irreducible submodule of $V$, then since we can easily show $W\cap W^*$ is a submodule of $W$, we have either $W\cap W^* = W$ or $W\cap W^* = 0$. The first case would contradict the claim proven above, since we assumed $W$ was a proper submodule of $V$. So $W\cap W^* = 0$ and $V = W\oplus W^*$.

In the context of Fulton and Harris' book, here is how I would explain what is going on:

Suppose $V$ is not irreducible as a $\mathbf{C}[G]$-module and let $0 \neq W \neq V$ be a proper non-zero sub-module. By restricting to $\mathbf{R}[G]$ we see that $W$ is a proper non-zero $\mathbf{R}[G]$-submodule of $V_0^{\oplus 2}$ (here we are using the relation between the real numbers and the complex numbers).

If $W_0$ is an irreducible $\mathbf{R}[G]$-submodule of $W$ then by Schur's lemma, $W_0 \cong V_0$ as $\mathbf{R}[G]$-modules. By comparing dimensions we must have $W=W_0$. We have proved that every proper non-zero $\mathbf{C}[G]$-submodule of $V$ is isomorphic, upon restriction to $\mathbf{R}[G]$, to $V_0$. In particular the dimension of $W$ is half the dimension of $V$.

Now as in the discussion on page 40, a positive definite $G$-invariant inner product on $V_0$ induces a non-degenerate symmetric $\mathbf{C}$-bilinear form $(\cdot,\cdot)$ on $V$. Using what we proved in the previous paragraph there are only two possibilies: $W \cap W^\perp=0$ or $W=W^\perp$ (in the second case, $W$ is "Lagrangian").

In the first case $W \cong W^*$ and $W^\perp \cong (W^\perp)^*$, while in the second case the pairing induces an isomorphism $W^* \cong W'$ for any $G$-stable complement $W'$ to $W$ in $V$. It remains only to observe that in the first case the characters of $W$ and $W^\perp$ are real, implying that they are both equal to half the character of $V_0$ and hence $W^\perp \cong W \cong W^*$ as desired.

• I follow until the last sentence: why $\chi_W$ is real implies $\chi_W=\chi_{V_0}$? I understand $W$ considered as a real representation is $V_0$, but does this play a role in the above argument? Commented Aug 31, 2016 at 11:46
• @Tsang For your first question: given a complex representation, its character when you regard it as a real representation is twice the real part of its complex character (exercise with matrices). So actually $2 \chi_W=\chi_{V_0}=2 \chi_{W^\perp}$ (I dropped the factor of two in my answer, and am about to edit it in) implying $W \cong W^\perp$. Commented Aug 31, 2016 at 17:27
• The role that the isomorphism $W \cong V_0$ (as real representations) plays is as follows: it is used as in my previous comment to establish the duality statement, and it is used via considering dimensions to obtain $W=W^\perp$ in the second case considered in paragraph four above, when only $W \subseteq W^\perp$ is evident a priori. Commented Aug 31, 2016 at 17:30
• As a small aside: it might be a useful mnemonic to consider the endomorphism ring of $V_0$ as an $\mathbf{R}[G]$-module. By Schur's lemma it is a division algebra; by the classification of division algebras it is either $\mathbf{R}$, $\mathbf{C}$, or the quaternions $\mathbf{H}$. The first case corresponds to $V$ irreducible, the second to $V=W \oplus W^*$ with $W$ and $W^*$ not isomorphic, and the last to $V=W \oplus W$ with $W \cong W^*$. Commented Aug 31, 2016 at 17:36
• I understand your answer now, thank you! Commented Sep 1, 2016 at 1:11