Abstract Nonsense

A Bijection Between A Subset of the Complex Reps of a Finite Group and the Real Reps (Pt. II)

Point of post: This post is a continuation of this one.

The Complexification of a Real Inner Product Space

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Before we get to the main part of this post we must discuss the complexification of a real inner product space. Namely, we define the complexification of the real inner product space $\left(\mathscr{R},\langle\cdot,\cdot\rangle\right)$, denoted $\mathscr{R}^{\mathbb{C}}$,  to be the set $\mathscr{R}\times\mathscr{R}$ with addition defined coordinatewise, scalar multiplication defined $(a+bi)(u,v)=(au-bv,av+bu)$, and inner product defined by $\left\langle (u,v),(u',v')\right\rangle^\mathbb{C}=\langle u,u'\rangle+\langle v,v'\rangle+i\left(\langle v,u'\rangle-\langle u,v'\rangle\right)$. It is easy to verify that $\mathscr{R}^\mathbb{C}$ is indeed a pre-Hilbert space and moreover that $\dim_\mathbb{C}\mathscr{R}^\mathbb{C}=\dim_\mathbb{R}\mathscr{V}$. What we now claim is that the map $\mathbf{1}\oplus -\mathbf{1}$ is a complex conjugate. Indeed:

Theorem: Let $\left(\mathscr{R},\langle\cdot,\cdot\rangle\right)$ be a real inner product space. Then the map $\mathbf{1}\oplus -\mathbf{1}$ is a complex conjugate on $\mathscr{R}^\mathbb{C}$.

Proof: It’s clear that $\left(\mathbf{1}\oplus\mathbf{-1}\right)^2=\left(\mathbf{1}^2\right)\oplus\left(\left(-\mathbf{1}\right)^2\right)=\mathbf{1}\oplus\mathbf{1}=\mathbf{1}_{\mathscr{R}\times\mathscr{R}}$. Next we next claim that $\mathbf{1}\oplus -\mathbf{1}$ is antilinear. Indeed, for any $a+bi\in\mathbb{C}$ and $(u,v)\in\mathscr{V}^\mathbb{C}$ we have that

$\displaystyle \left(\mathbf{1}\oplus-\mathbf{1}\right)\left((a+bi)\left(u,v\right)\right)=\left(\mathbf{1}\oplus-\mathbf{1}\right)\left(au-bv,av+bu\right)=\left(au-bv,-av-bu\right)$

and

$\overline{\left(a+bi\right)}\left(\mathbf{1}\oplus-\mathbf{1}\right)(u,v)=\left(a-bi\right)\left(u,-v\right)=\left(au-bv,-av-bu\right)$

and thus $\left(\mathbf{1}\oplus-\mathbf{1}\right)\left(z(u,v)\right)=\overline{z}\left(\mathbf{1}\oplus-\mathbf{1}\right)(u,v)$ and since $\mathbf{1}\oplus-\mathbf{1}$ is trivially linear antilinearity follows. Lastly, we claim with the proposed inner product that $\mathbf{1}\oplus-\mathbf{1}$ is antiunitary. Indeed, let $(u,v),(u',v')\in\mathscr{V}^\mathbb{C}$ be arbitrary then

\displaystyle \begin{aligned}\left\langle \left(\mathbf{1}\oplus-\mathbf{1}\right)(u,v),\left(\mathbf{1}\oplus-\mathbf{1}\right)(u',v')\right\rangle^\mathbb{C} &= \left\langle (u,-v),(u',-v')\right\rangle^\mathbb{C}\\ &= \langle u,u'\rangle+\langle -v,-v'\rangle+i\left(\langle-v,u'\rangle-\langle u,-v'\rangle\right)\\ &=\langle u,u'\rangle+\langle v,v'\rangle+i\left(\langle u,v'\rangle-\langle v,u'\rangle\right)\\ &= \langle u',u\rangle+\langle v',v\rangle+i\left(\langle v',u\rangle-\langle u',v\rangle\right)\\ &= \left\langle (u',v'),(u,v)\right\rangle^\mathbb{C}\end{aligned}

and since $(u,v)$ and $(u',v')$ were arbitrary antiunitarity follows. Thus, combining these three attributes let us conclude. $\blacksquare$

Inverse Map

In this post we define an inverse (in a sense soon made to be rigorous) to our map $\rho\mapsto \rho_\Re$ discussed previously. Namely:

Theorem: Let $G$ be a finite group and $\rho:G\to\mathcal{U}\left(\mathscr{R}\right)$ be a $\mathbb{R}$-representation on the real inner product space $\left(\mathscr{R},\langle\cdot,\cdot\rangle\right)$. Then the mapping $\rho_\Im:G\to\mathcal{U}\left(\mathscr{R}^\mathbb{C}\right)$ by $\rho_\Im(g)=\rho(g)\oplus\rho(g)$ is a $\mathbb{C}$-representation of $G$ which satisfies the real condition with realizer $\mathbf{1}\oplus-\mathbf{1}$. Moreover, $\mathscr{R}^\mathbb{C}$ has no non-trivial  subspaces $\mathscr{S}$ such that $\left(\mathbf{1}\oplus-\mathbf{1}\right)\left(\mathscr{S}\right)=\mathscr{S}$ which are $\rho_\Im$-invariant  if and only if $\rho$ is an irreducible $\mathbb{R}$-representation.

Proof: We first show that $g\mapsto \rho(g)\oplus \rho(g)$ really is a homomorphism $G\to\mathcal{U}\left(\mathscr{R}^\mathbb{C}\right)$. We first verify that $\rho(g)\oplus \rho(g)$ is actually a unitary linear transformation on $\mathscr{R}^\mathbb{C}$ for each $g\in G$. Indeed, since it’s clear that $\rho(g)\oplus\rho(g)$ is linear it suffices to show that for each $g\in G$ $\rho_g\oplus\rho_g$ is unitary. So, let  $(u,v),(u',v')\in\mathscr{R}^\mathbb{C}$, then one has that

\begin{aligned}\left\langle (\rho_g\oplus\rho_g)(u,v),\left(\rho_g\oplus\rho_g\right)(u',v')\right\rangle^\mathbb{C} &=\left\langle \left(\rho_g(u),\rho_g(v)\right),\left(\rho_g(u'),\rho_g(v')\right)\right\rangle^\mathbb{C}\\ &=\left\langle\rho_g(u),\rho_g(u')\right\rangle+\left\langle\rho_g(v),\rho_g(v')\right\rangle+i\left(\left\langle\rho_g(v),\rho_g(u')\right\rangle-\left\langle\rho_g(u),\rho_g(v')\right\rangle\right)\\ &= \left\langle u,u'\right\rangle+\left\langle v,v'\right\rangle+i\left(\left\langle v,u'\right\rangle-\left\langle u,v'\right\rangle\right)\\ &=\left\langle (u,v),(u',v')\right\rangle^\mathbb{C}\end{aligned}

from where unitarity of $\rho(g)\oplus\rho(g)$ follows, and since $g\in G$ was arbitrary it follows that $\rho_\Im$ really is a mapping $G\to\mathcal{U}\left(\mathscr{R}^\mathbb{C}\right)$.

We now show that $\rho_\Im$ satisfies the real condition with realizer $\mathbf{1}\oplus-\mathbf{1}$. But, this is evident since

$\left(\mathbf{1}\oplus-\mathbf{1}\right)(\rho_g\oplus\rho_g)=(\mathbf{1}\rho_g)\oplus(-\mathbf{1}\rho_g)=\left(\rho_g\mathbf{1}\right)\oplus\left(\rho_g(-\mathbf{1})\right)=\left(\rho_g\oplus\rho_g\right)\left(\mathbf{1}\oplus-\mathbf{1}\right)$

Therefore, it remains to verify our last claim. Suppose first that $\rho$ wasn’t reducible so that there existed some $\{\bold{0}\}<\mathscr{S}<\mathscr{R}$ such that $\mathscr{S}$ is $\rho$-invariant. Note then that $\mathscr{S}\times\mathscr{S}$ is a subspace of $\mathscr{R}^\mathbb{C}$ and clearly it is $\rho_\Im$ and $\mathbf{1}\oplus-\mathbf{1}$ invariant. Conversely, suppose that $\{\bold{0}\}<\mathscr{W}<\mathscr{R}^\mathbb{C}$ then by assumption for every $(u,v)\in\mathscr{W}$ we have that $(u,-v)\in\mathscr{W}$ and so $(u,\bold{0})\in\mathscr{W}$ and for every $a\in\mathbb{R}$ we have that $(a+0i)(u,\bold{0})\in\mathscr{W}$ and similarly for every $(u,0),(u',0)\in\mathscr{W}$ one has that $(u+u',\bold{0})\in\mathscr{W}$. Thus, we may conclude that $\mathscr{S}=\left\{w:(w,\bold{0})\in\mathscr{W}\right\}$ is a subspace of $\mathscr{R}$ which is clearly $\rho$-invariant by assumption since for every $g\in G$ and $w\in\mathscr{S}$ we have that $(w,\bold{0})\in\mathscr{W}$ and thus by assumption $(\rho_g\oplus\rho_g)(w,\bold{0})=(\rho_g(w),\bold{0})\in\mathscr{W}$ so that $\rho_g(w)\in\mathscr{S}$. The conclusion follows. $\blacksquare$

References:

1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.