# Abstract Nonsense

## Complex Conjugate Representations

Point of post: In this post we discuss the notion of complex conjugates on pre-Hilbert spaces and how to produce new representations based on them

Motivation

Anyone acquainted with basic concepts of complex numbers is well aware of the power of the conjugation map. It would then seem fruitful to try to abstractify the important concepts of this map to general pre-Hilbert spaces. It then seems natural to figure out if one can take these generalizations and produce from them new representations. It turns out that this is indeed fruitful, but much more can be said than is at first apparent. In fact, it turns out that these new representations can be classified into one of three types which model, roughly, the difference between real, complex, and quaternionic numbers.

Complex Conjugate

Let $\left(\mathscr{V},\langle\cdot,\cdot\rangle\right)$ be a pre-Hilbert space. Then a map $J:\mathscr{V}\to\mathscr{V}$ is called a complex conjugate if

\begin{aligned}&\mathbf{(1)}\quad J\left(\alpha x+\beta y\right)=\bar{\alpha}J(x)+\bar{\beta}J(y)\\ &\mathbf{(2)}\quad J^2=\mathbf{1}\quad\\ &\mathbf{(3)}\quad \left\langle J(x),J(y)\right\rangle=\langle y,x\rangle\end{aligned}\quad\begin{aligned}&\left(\text{antilinearity}\right)\\ &\left(\text{involution}\right)\\ &\left(\text{antiunitarity}\right)\end{aligned}

This is evidently a generalization of the conjugation map on $\mathbb{C}$ since the map $\displaystyle z\mapsto\overline{z}$ is a complex conjugate.

Remark: It should be noted that the set of complex conjugates does not form an algebra under composition since the composition of two antilinear maps is not antilinear (in general).

Our first result is that considering complex conjugates gives us the possibility of building many new representations out of old ones. Namely, if $J$ is a specified complex conjugate on a pre-Hilbert space $\mathscr{V}$, $G$ a finite group, and $\rho:G\to\mathcal{U}\left(G\right)$ a representation we define the conjugation representation of $\rho$ with respect to $J$, denoted $\text{Conj}^J_\rho$, to be the map

$\displaystyle \text{Conj}^J_\rho:G\to\mathcal{U}\left(\mathscr{V}\right):g\mapsto J\rho_g J$.

It’s not immediately apparent that this mapping is well-defined (in the sense that $J\rho_gJ$ is always unitary) or that it is representation, that is what we shall now show:

Theorem: Let $\mathscr{V}$ be a pre-Hilbert space with complex conjugate $J$, $G$ a finite group, and $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ a representation. Then, for every $g\in G$ one has that $J\rho_g J$ is unitary, and that the resulting map $g\mapsto J\rho_g J$ is a homomorphism

Proof: To see that $J\rho_g J$ is unitary we merely note that for any $u,v\in\mathscr{V}$ one has that

$\left\langle J\rho_g J(u),J\rho_g J(v)\right\rangle=\left\langle \rho_g(J(v)),\rho_g(J(u))\right\rangle=\left\langle J(v),J(u)\right\rangle=\left\langle u,v\right\rangle$

To see that the map $g\mapsto J\rho_g J$ is a homomorphism it suffices to note that

$J\rho_{gh}J=J\rho_g\rho_h J=J\rho_g J J \rho_h J=\left(J\rho_g J\right)\left(J\rho_h J\right)$

from where the conclusion follows. $\blacksquare$

Corollary: If $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ is a representation then so is $\text{Conj}^J_\rho:G\to\mathcal{U}\left(G\right)$ is a representation.

Our next theorem shows how the irreducibility of a conjugation representation and the irreducibility of the original representation interact, namely:

Theorem: Let $\mathscr{V}$ be a pre-Hilbert space with complex conjugate $J$, $G$ a finite group, and $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ a representation. Then, $\rho$ is an irrep if and only if $\text{Conj}^J_\rho$ is.

Proof: Suppose first that $\rho$ is an irrep and let $\mathscr{W}\leqslant\mathscr{V}$ be $\text{Conj}^J_\rho$-invariant. Then we see that $J(\mathscr{W})$ is $\rho$-invariant. Indeed, for any $g\in G$ and $J(w)\in J\left(\mathscr{W}\right)$ one has that $\rho_g(J(w))=J\left(J\rho_g J(w)\right)\in J\left(\left(\text{Conj}^J_\rho\right)(g)\right)\left(\mathscr{W}\right)\subseteq J(\mathscr{W})$ so that $J(\mathscr{W})$ is $\rho$ invariant. It follows then that $J\left(\mathscr{W}\right)=\{\bold{0}\}$ or $J\left(\mathscr{W}\right)=\mathscr{V}$. Thus, appealing to a dimension argument (and using that $J$ is invertible) we may conclude that $\mathscr{W}=\{\bold{0}\}$ or $\mathscr{W}=\mathscr{V}$. And, since $\mathscr{W}$ was arbitrary it follows that $\text{Conj}_J(\rho)$ is an irrep as desired.

To prove the converse we merely note that if $\text{Conj}^J_\rho$ were an irrep, then by the previous part of the problem we know that $\text{Conj}^J_{\text{Conj}^J_\rho}$ is irreducible. Note though that for every $g\in G$

$\text{Conj}^J_{\text{Conj}^J_\rho}(g)=JJ\rho_gJJ=\rho_g$

and thus $\text{Conj}^J_{\text{Conj}^J_\rho}=\rho$ from where the conclusion follows. $\blacksquare$

It’s clear that different choices of complex conjugates will produce different conjugate representations. More directly, if $J,K$ are distinct complex conjugates and $\rho$ a representation there’s no reason to believe that $\text{Conj}^J_\rho=\text{Conj}^K_\rho$. That said, our next theorem will show that there equivalence class (under the usual definition of representation equivalence)

Theorem: Let $\mathscr{V}$ be a pre-Hilbert space with complex conjugates $J$ and $K$. Then, for any finite group $G$ and representation $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ one has that $\text{Conj}^J_\rho\simeq\text{Conj}^K_\rho$.

Proof: Merely note that $JK$ is unitary since for any $x,y\in\mathscr{V}$ one has that

$\left\langle JK(x),JK(y)\right\rangle=\left\langle K(y),K(x)\right\rangle=\left\langle x,y\right\rangle$

and that $(JK)^{-1}=K^{-1}J^{-1}=KJ$. Thus for any $g\in G$ we have that

$K\rho_gK=KJJ\rho_gJJK=\left(JK\right)^{-1}J\rho_g J (JK)$

And since $g$ was arbitrary it follows that if we define $W=JK$ that $W$ is a unitary endomorphism on $\mathscr{V}$ and $\text{Conj}^K_\rho(g)=W^{-1}\text{Conj}^J_\rho W$ for ever every $g\in G$ and so trivially $\text{Conj}^K_\rho\simeq\text{Conj}^J_\rho$. $\blacksquare$

It follows then that if $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ is an irrep and $J$ and $K$ are any complex conjugates on $\mathscr{V}$ then $\left[\text{Conj}^J_\rho\right]=\left[\text{Conj}^K_\rho\right]$ where $[\cdot]$ is the equivalency class in $\widehat{G}$.

With this information in mind we separate irreps into two classes. Namely, we say that an irrep $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$  is self-conjugate if for any complex conjugate $J$ on $\mathscr{V}$ one has that $\rho\simeq\text{Conj}^J_\rho$. We call if for every complex conjugate $J$, $\rho$ complex if $\rho\not\simeq \text{Conj}^J_\rho$. Because of our previous theorem we see that both of these universal quantifiers may be changed to existential ones.

References:

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

2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print

February 5, 2011 -

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