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

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 $\phi = \psi = \chi$.
