# Abstract Nonsense

## Complex Conjugate Representation (A Characterization of Self-Conjugate Maps)

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

We now proceed, as promised, to give a necessary and sufficient conditions for an irrep of a space to be self-conjugate real and self-conjugate quaternonic. Namely:

Theorem: Let $\mathscr{V}$ be a vector space, $G$ a finite group, and $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ an irrep. Then, $\rho$ is self-conjugate if and only if there exists a $J$ which is antilienar and antiunitary which commutes with $\rho$.  If there exists such a $J$ it is unique up to a constant of modulus one. Moreover, if there exists such a $J$ then $J^2=\pm\mathbf{1}$ with $J^2=\mathbf{1}$ iff $\rho$ is real and $J^2=-\mathbf{1}$ iff $\rho$ is quaternonic.

Proof: Suppose first that $\rho$ is self-conjugate. Then, for any complex conjugate map $J$ on $\mathscr{V}$ there exists some unitary map $U$ such that $WJ\rho_g JW^{-1}=\rho_g$ for every $g\in G$. Note though that $WJ$ is antilinear (indeed, $W(J(\alpha x+\beta y))=W(\overline{\alpha}x+\overline{\beta}y)=\overline{\alpha}W(J(x))+\overline{\beta}W(J(y))$) and antiunitary (indeed, $\left\langle WJ(x),WJ(y)\right\rangle=\left\langle W(y),W(x)\right\rangle=\langle y,x\rangle$) and since $J=J^{-1}$ it follows from this equation that $WJ\rho_g=\rho_g WJ$ and thus the conclusion follows.

Suppose then that there existed an antilinear, antiunitary map $J$ which commutes with $\rho_g$ for every $g\in G$. Then, picking some complex conjugate $K$ on $\mathscr{V}$ we see that $W=KJ$ is unitary (we’ve discussed this before) and by assumption $J\rho_g=\rho_g J$ and thus $KJ\rho_g=K\rho J=K\rho_g KKJ$ and thus $W\rho_gW^{-1}=K\rho_g K$ from where it follows that $\text{Conj}^K_\rho\simeq\rho$ and thus $\rho$ is self-conjugate.

Now, to see that if such a $J$ exists then $J^2=\pm\mathbf{1}$, we merely note that $J^2$ is unitary and $J^2\rho_g=\rho_g J^2$ for every $g\in G$. It follows by the second form of Schur’s lemma that $J^2=\alpha\mathbf{1}$ for some $\alpha$. But, since $J^2$ is unitary we must have that $|\alpha|=1$. Note though that since $\alpha\mathbf{1}=J^2$ we may apply $J$ to both sides to get that $J(\alpha\mathbf{1})=\overline{\alpha}J=J^3=\alpha \mathbf{1}J=\alpha J$ and thus $\alpha=\overline{\alpha}$, from where it follows that $\alpha=\pm 1$.

Now, to see that such a $J$ is unique up to a constant of modulus one, applying the same idea of using Schur’s lemma we can show that if $J_1$ and $J_2$ are any such antiunitary,antilinear commuting maps that  $J_1J_2$ is an intertwinor for $\rho$ and thus by the second form of Schur’s lemma $J_1J_2=\alpha\mathbf{1}$ for some $\alpha$. Note though that since $J_1J_2$ is unitary we must have that $|\alpha|=1$. Note then that by the last part of the proof $J_1J_2=\alpha\mathbf{1}$ implies that $J_1=\pm\alpha J_2$ from where the conclusion follows.

Now, suppose that $J^2=\mathbf{1}$. Then, by our previous theorem we have that there exists an orthonormal basis $\{x_1,\cdots,x_n\}$ for $\mathscr{V}$ for which $J(x_k)=x_k\;\; k\in[n]$. Note though that if $g\in G$ and $\displaystyle \rho_g(x_j)=\sum_{i=1}^{n}\alpha_{i,j}x_i$ then

$\displaystyle \left\langle \rho_g(x_j),x_k\right\rangle=\sum_{i=1}^{n}\alpha_{i,j}\langle x_i,x_k\rangle=\alpha_{k,j}$

That said,

\begin{aligned}\left\langle \rho_g (x_j),x_k\right\rangle &=\left\langle J(x_k),J\left(\rho_g (x_j)\right)\right\rangle\\ &=\left\langle J(x_k),\rho_g\left(J(x_j)\right)\right\rangle\\ &=\left\langle x_k,\rho_g(x_j)\right\rangle\\ &= \overline{\left\langle \rho_g(x_j),x_k\right\rangle}\end{aligned}

Since $j,k\in[n]$ and $g\in G$ was arbitrary it follows that the matrix with respect to this basis is real for each $\rho_g$ and thus $\rho$ is real. Conversely, suppose that $J$ is real. Then, there exists some basis $\{x_1,\cdots,x_n\}$ for which the matrix of $\rho_g$ is real for each $g\in G$. Define $J$ by

$\displaystyle J\left(\sum_{j=1}^{n}\alpha_j x_j\right)=\sum_{j=1}^{n}\overline{\alpha_j}x_j$

it’s easily verified then that $J$ is an antilinear, antinunitary map and $J^2=\mathbf{1}$. But, it easily follows from the fact that $\rho_g$ has a real matrix with respect to the basis $\{x_1,\cdots,x_n\}$ that $J$ commutes with $\rho_g$ for each $g\in G$ and thus the converse follows.

It follows from this that if $\rho$ is quaternonic then any such $J$ is such that $J^2=\pm\mathbf{1}$, but the above excludes the case $J^2=\mathbf{1}$ and thus $J^2=-\mathbf{1}$. Similarly, if $J^2=-\mathbf{1}$ then the above shows that $J$ is not real and thus must be quaternonic. The conclusion follows. $\blacksquare$

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 6, 2011 -

1. […] though our characterization of self-conjugate irreps one can see that is similar to so that as desired. The conclusion […]

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3. […] every . Thus, by one of our previous characterizations of real irreps we may conclude that is real–contradictory to assumption. The conclusion […]

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4. […] and is an irrep which must be real since it must commute with (of course we are appealing to our previous characterization of self-conjugate irreps). Now, suppose that . We claim that if then is a -invariant subspace. […]

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