Character of an $\mathbb{R}G$-module constructed from a $\mathbb{CG}$-module

Question

I have been reading Representations and Characters of Groups by Gordon James and Martin Liebeck. I encountered the following construction of an $\mathbb{R}G$-module from a $\mathbb{C}G$-module.

Next, there's a proposition related to this on the next page.

The proof of Proposition 23.6(1) is clear from the first image. I don't understand the proof of Proposition 23.6(2). How does the author break down $V_{\mathbb{R}}$ into a direct sum of $U$ and $W$? Are we using the fact that $V$ is irreducible as a $\mathbb{C}G$-module?

Answer 1

First of all, to make things more comfortable for you I will do all calculations on $\mathbb{C}G$-modules using the following observation: if $V_0$ is an $\mathbb{R}G$-module of real dimension $n$ with character $\chi$, the complexification $\mathbb{C}\otimes_\mathbb{R} V_0=V$ is a $\mathbb{C}G$-module ($G$ acting only on the right factor) of complex dimension $n$ with the same character $\chi$ (a basis of $V$ over $\mathbb{C}$ is $1\otimes v_i$ for $v_i$ an $\mathbb{R}$-basis of $V_0$ and $G$ acts trivially on the left factor).

Thus suppose your $\mathbb{R}G$-module $V_R$ splits as $V_R = U\oplus W$ and let $\phi$ and $\psi$ be the characters of $U$ and $W$ respectively. Then by distributivity $$\mathbb{C}\otimes V_R = (\mathbb{C}\otimes U)\oplus (\mathbb{C}\otimes W)$$ and denoting $\mathbb{C}U$, $\mathbb{C}W$ the corresponding $\mathbb{C}G$-modules, we get $\mathbb{C}V_R = \mathbb{C}U\oplus \mathbb{C}W$.

By the first paragraph, the characters of these $\mathbb{C}G$-modules did not change, so we still have the equality of characters of $\mathbb{C}G$-modules $$\chi+\overline{\chi}=\phi + \psi.$$ But now we are in that comfortable place of algebra of characters of complex representations, so we can feel free to use the orthogonality relations. Since $V$ is an irreducible $\mathbb{C}G$-module, $\chi$ is an irreducible character, and similarly for $\overline{\chi}$.

Now characters of irreducible representations form an orthonormal basis. Assume $\chi\neq \overline{\chi}$ for a contradiction and expand $\phi = a\chi +b\overline{\chi} + S_\phi$ and $\psi = a'\chi + b'\overline{\chi}+S_\psi$, where the $S$s are sums of irreducible characters orthogonal to those two. Thus we have $$\chi+\overline{\chi} = (a+a')\chi+(b+b')\overline{\chi} + S_\phi + S_\psi.$$ Taking inner products with $\chi$ and using $||\chi||^2=1$ we get $1= a+ a'$. Similarly, taking inner products with $\overline{\chi}$ we get $1 = b+b'$. Taking inner products with all characters involved in $S_\phi$ and $S_\psi$ we conclude $S_\phi = S_\psi = 0$. Finally, since $a,a',b,b'$ are non-negative integers, we have without loss of generality $a=1$,$a'=0$, $b=0,b'=1$ and $S_\phi = S_\psi = 0$.

This shows that $\chi = \phi$ and $\overline{\chi} = \psi$; in particular, $\chi=\overline{\chi}$ since $\phi$ is real and we have a contradiction. Thus $\chi=\overline{\chi}$ and we can still use the orthogonality relations in the equation $2\chi = \phi+\psi$ to conclude as before that $\phi = \psi = \chi$.

Answer 2

Indeed the argument used is not quite clear as it stands, because context is missing. Apparently something is already known about uniqueness of decomposition of characters into irreducibles. What is known here is that we have a $\Bbb R G$-module that is reducible, and whose character is $\chi+\overline\chi$; its character must break up as the sum of at least two characters as submodules. And since $\chi$ and $\overline\chi$ are irreducuble characters (though a priori only characters of $\Bbb C G$-modules) the author concludes that the breaking up of the character can only correspond to its decomposition as $\chi+\overline\chi$. And then this implies that $\chi$ and $\overline\chi$ are actually characters of $\Bbb R G$-modules (indeed of the submodules $U,V$).