# Abstract Nonsense

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

Point of post: In this post we describe a certain subset of the set of all $\mathbb{C}$-representations of a finite group $G$ and show that this subset is in bijective correspondence with the set of all $\mathbb{R}$-representations of $G$. Moreover, we shall show that this bijection naturally restricts to a subset of the set of all $\mathbb{C}$-reps of $G$ to the the $\mathbb{R}$-irreps of $G$.

Motivation

In our last post we discussed the notion of $\mathbb{R}$representations for a finite group $G$. Naturally our first desire would be to see if we could, in some way, connect $\mathbb{R}$representations of $G$ to the $\mathbb{C}$-representations  which has held the center of our attention for so long. We begin this process in this post by showing that there is a natural place for which these $\mathbb{R}$-representations occur. Namely, we shall see that every $\mathbb{C}$-representation $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ for which there is a complex conjugate $J$ for which $J\rho(g)J=\rho(g)$ for every $g\in G$ naturally admits a $\mathbb{R}$-representation $\rho_{\Re}$. We shall show then that in fact the reverse is true–namely that for every $\mathbb{R}$-representation $\psi$ there is a natural way to produce a $\mathbb{C}$-representation $\rho$ such that $\rho_{\Re}=\psi$. Moreover, we’ll show that if $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ is a $\mathbb{C}$-representation satisfying the aforementioned conditions and the added condition that there does not exist $\mathscr{W}\leqslant \mathscr{V}$ such that $J\left(\mathscr{W}\right)=\mathscr{W}$ and $\mathscr{W}$ is $\rho$-invariant then we shall see that $\rho_{\Re}$ is an irreducible $\mathbb{R}$-representation.

A Characterization of $\mathbb{R}$– Representations

Our first goal is to show that a certain class of $\mathbb{C}$-representations naturally admits a $\mathbb{R}$representation. Indeed, say that a $\mathbb{C}$-representation $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ satisfies the real condition if there exists a complex conjugate $J:\mathscr{V}\to\mathscr{V}$ such that $J\rho(g)J=\rho(g)$ for every $g\in G$. We call such a $J$ the ‘realizer’ of $\rho$. With this in mind:

Theorem: Let $G$ be a finite group and $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ be a $\mathbb{C}$-representation satisfying the real condition with realizer $J$. Then, the space $\mathscr{V}_\Re\overset{\text{def.}}{=}\left\{v\in \mathscr{V}:J(v)=v\right\}$ is a real vector space with $\dim_{\mathbb{R}}\mathscr{V}_\Re=\dim_\mathbb{C}\mathscr{V}$ and the same inner product as $\mathscr{V}$. Moreover, the map $\rho_\Re:G\to\mathcal{U}\left(\mathscr{V}_\Re\right)$ given by $g\mapsto \rho(g)\mid_{\mathscr{V}_\Re}$ is a $\mathbb{R}$-representation of $G$. Further more, $\rho_\Re$ is irreducible if and only if there does not exist a $\rho$-invariant $\{\bold{0}\}<\mathscr{W}<\mathscr{V}$ for which $J\left(\mathscr{W}\right)=\mathscr{W}$.

Proof: We first show that $\mathscr{V}_\Re$ is a real inner product space of real dimension $n=\dim_\mathbb{C}\mathscr{V}$. The fact that $\mathscr{V}_\Re$ is a real vector space with the same addition as $\mathscr{V}$ and scalar multiplication restricted to $\mathbb{R}$ is clear, thus it remains to prove then that with this definition of operations $\dim_\mathbb{R}\mathscr{V}_\Re=n$. To see this recall from our past characterizations of complex conjugates that $\mathscr{V}$ admits a basis $\left\{x_1,\cdots,x_n\right\}$ such that $J(v_k)=v_k$ for $k=1,\cdots,n$. We claim then that $\{x_1,\cdots,x_n\}$ is similarly a basis for $\mathscr{V}_\Re$. Indeed, the fact that $\{x_1,\cdots,x_n\}$ is linearly independent is evident and thus it suffices to prove that $\text{span}_\mathbb{R}\{x_1,\cdots,x_n\}=\mathscr{V}_\Re$. To see this let $v\in\mathscr{V}_\Re$ be arbitrary. Then, since $\{x_1,\cdots,x_n\}$ is a basis for $\mathscr{V}$ there exists $\zeta_1,\cdots,\zeta_n\in\mathbb{C}$ such that $\displaystyle \sum_{k=1}^{n}\zeta_k x_k=v$. Note though that by assumption

$\displaystyle \sum_{k=1}^{n}\zeta_k v_k=v=J(v)=\sum_{k=1}^{n}\overline{\zeta_k}x_k$

and thus $\zeta_k=\overline{\zeta_k}$ and thus $\zeta_k\in\mathbb{R}$ it clearly follows then that $v\in\text{span}_\mathbb{R}\{x_1,\cdots,x_n\}$. Since $v$ was arbitrary it follows that $\{x_1,\cdots,x_n\}$ is a real basis for $\mathscr{V}_\Re$ and thus $\dim_\mathbb{R}\mathscr{V}_\Re=n$ as desired. Thus, it remains to show that the restriction of the inner product $\langle\cdot,\cdot\rangle$ to $\mathscr{V}_\Re\times\mathscr{V}_\Re$ is really a real inner product. But, all the axioms are clear except for possibly that $\langle\cdot,\cdot\rangle$ maps $\mathscr{V}_\Re\times\mathscr{V}_\Re$ into $\mathbb{R}$ (there’s not ostensible reason this must be true). To see this merely note that for any $u,v\in\mathscr{V}_\Re$ one has that

$\displaystyle \left\langle u,v\right\rangle=\left\langle J(u),J(v)\right\rangle=\left\langle v,u\right\rangle=\overline{\left\langle u,v\right\rangle}$

from where the result follows.

Now, we prove that $\rho_\Re$ as defined in the theorem statement is really a $\mathbb{R}$-representation. The first thing to show is that for every $g\in G$ one has that $\rho(g)\mid_{\mathscr{V}_\Re}$ really is an element of $\mathcal{U}\left(\mathscr{V}_\Re\right)$. The first, but crucial step in this procedure is to show that, while silly, $\rho_g\left(\mathscr{V}_\Re\right)\subseteq\mathscr{V}_\Re$ since this isn’t absolutely clear. But, this is where the real condition comes into play. Namely, since $J^2=\mathbf{1}$ we may reinterpret the real condition as saying $J\rho(g)=\rho(g)J$ for each $g\in G$. Thus, in particular for any $g\in G$ and any $v\in\mathscr{V}_\Re$ one has that $J(\rho_g(v))=\rho_g(J(v))=\rho_g(v)$ so that $\rho_g(v)\in\mathscr{V}_\Re$ from where the fact that $\rho_g\left(\mathscr{V}_\Re\right)\subseteq\mathscr{V}_\Re$ follows. Now, the fact that $\rho_g$ is a $\mathbb{R}$-homomorphism and unitary clearly follow from definition. Thus, to prove that $\rho_\Re$ is a $\mathbb{R}$-representation it merely suffices to prove that it is a homomorphism, but this is clear since $\rho$ is a representation. Thus, it follows that $\rho_\Re$ is a homomorphism $G\to\mathcal{U}\left(\mathscr{V}_\Re\right)$ from where the fact that it is a $\mathbb{R}$-representation of $G$ clearly follows.

Finally we prove that $\rho_\Re$ is irreducible if and only if $\mathscr{V}$ doesn’t admit any non-trivial proper subspaces $\mathscr{W}$ for which $J\left(\mathscr{W}\right)=\mathscr{W}$. Indeed, suppose that $\rho_\Re$ is reducible so that there exists $\{\bold{0}\}<\mathscr{W}<\mathscr{V}_\Re$ such that $\mathscr{W}$ is $\rho_\Re$ invariant. Define then $\mathscr{W}_\Im=\text{span}_{\mathbb{C}}\mathscr{W}$. Clearly by definition one has that $\mathscr{W}_\Im\leqslant\mathscr{V}$. Moreover, we claim that $J\left(\mathscr{W}_\Im\right)=\mathscr{W}_\Im$. Indeed, let $u\in\mathscr{W}_\Im$ then $\displaystyle u=\sum_r \alpha_r w_r$ where $w_r\in\mathscr{W}$ and so

$\displaystyle J(u)=\sum_r \overline{\alpha_r}J(w_r)=\sum_r \overline{\alpha_r}w_r\in\mathscr{W}_\Im$

Conversely, if $\displaystyle \sum_r\alpha_r w_r\in\mathscr{W}_\Im$ where $w_r\in\mathscr{W}$ then $\displaystyle \sum_{r}\overline{\alpha_r}w_r\in\mathscr{W}_\Im$ and evidently

$\displaystyle J\left(\sum_r\overline{\alpha_r}w_r\right)=\sum_r \alpha_r J(w_r)=\sum_r \alpha_r w_r$

and so $\displaystyle \sum_r\alpha_r w_r\in J\left(\mathscr{W}_\Im\right)$ from where the fact that $J\left(\mathscr{W}_\Im\right)=\mathscr{W}_\Im$ follows. We now claim that $\mathscr{W}_\Im$ is $\rho$-invariant. Indeed, let $g\in G$ and $\displaystyle \sum_{r}\alpha_r w_r\in\mathscr{W}_\Im$ be arbitrary (where $w_r\in\mathscr{W})$. Then, one has that

$\displaystyle \left(\rho(g)\right)\left(\sum_r \alpha_r w_r\right) =\sum_r \alpha_r \left(\rho(g)\right)\left(w_r\right)= \sum_r \alpha_r \left(\rho_\Re(g)\right)(w_r)$

but by assumption one has that $\left(\rho_\Re(g)\right)(w_r)\in\mathscr{W}$ so that we may conclude that $\displaystyle \left(\rho(g)\right)\left(\sum_r \alpha_r w_r\right)\in\text{span}_\mathbb{C}\mathscr{W}=\mathscr{W}_\Im$. Since $g\in G$ and $\displaystyle \sum_r \alpha_r w_r\in\mathscr{W}_\Im$ was arbitrary we may conclude that $\mathscr{W}_\Im$ is $\rho$-invariant as claimed. Lastly, noting that since $\{\bold{0}\}<\mathscr{W}<\mathscr{V}_\Re$ we clearly have that $\text{span}_\mathbb{C}\{\bold{0}\}<\mathscr{W}_\Im<\text{span}_\mathbb{C}\mathscr{V}_\Re$ which (recalling that $\mathscr{V}_\Re$ contains a complex basis for $\mathscr{V}$) implies that $\{\bold{0}\}<\mathscr{W}_\Im<\mathscr{V}$ from where it follows that $\mathscr{W}_\Im$ is a $\rho$-invariant, non-trivial proper subspace of $\mathscr{V}$ for which $J\left(\mathscr{W}_\Im\right)=\mathscr{W}_\Im$. It follows then that if $\mathscr{V}$ does not admit such a subspace then $\rho_\Re$ must be irreducible.

Conversely, suppose that $\{\bold{0}\}<\mathscr{W}<\mathscr{V}$ is $\rho$-invariant and $J\left(\mathscr{W}\right)=\mathscr{W}$. Define then $\mathscr{W}_\Re=\left\{w\in\mathscr{W}:J(w)=w\right\}$. Evidently, it’s true then that $\{\bold{0}\}<\mathscr{W}_\Re<\mathscr{V}_\Re$. We claim that $\mathscr{W}_\Re$ is $\rho_\Re$ invariant. Indeed, let $g\in G$ and $g\in G$ and $w\in\mathscr{W}_\Re$ then $\left(\rho_\Re(g)\right)(w)=\left(\rho(g)\right)(w)\in \mathscr{W}$ and moreover $J\left(\rho(g)\right)(w)=\left(\rho(g)\right)\left(J(w)\right)=\left(\rho(g)\right)(w)$ so that in fact $\left(\rho(g)\right)(w)\in\mathscr{W}_\Re$. Since $g\in G$ and $w\in\mathscr{W}_\Re$ were arbitrary we have that $\mathscr{W}_\Re$ is $\rho_\Re$-invariant and thus $\rho_\Re$ is reducible. Thus, if $\rho_\Re$ is irreducible then $\mathscr{V}$ cannot admit a non-trivial proper subspace which is invariant under both $\rho$ and $J$.

Putting this all together the conclusion follows. $\blacksquare$

References:

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

March 29, 2011 -