# Abstract Nonsense

## Complex Conjugate Representation (Characterization of Complex Conjugates)

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

We’d now like to find necessary and sufficient conditions for an irrep to be a self-conjugate and when this is true whether it is real or quaternionic. But, first we need a theorem which can be used to show that all complex conjugates, given the right basis, have a particularly nice form and which will be integral in this aforementioned characterization of quaternionic and real self-conjugate irreps.

Theorem: Let $J$ be a complex conjugate on a pre-Hilbert space $\mathscr{V}$ of dimension $n$. Then, $\mathscr{V}$ admits an orthonormal basis $\{v_1,\cdots,v_n\}$ such that $J(v_k)=v_k,\;\; k=1,\cdots,n$.

Proof: Consider the vector space $\mathscr{V}^\mathbb{R}$ which is $\mathscr{V}$ but considered as a vector space over $\mathbb{R}$ instead of a vector space over $\mathbb{C}$. It’s an elementary fact that $\dim_\mathbb{R} \mathscr{V}^{\mathbb{R}}=2\dim_{\mathbb{C}}\mathscr{V}=2n$. We can define an inner product on $\mathscr{V}^\mathbb{R}$, denoted $\langle\cdot,\cdot\rangle$ by

$\langle\cdot,\cdot\rangle=\text{Re}\langle\cdot,\cdot\rangle$

where $\langle\cdot,\cdot\rangle$ is the given inner product on $\mathscr{V}$. This is indeed a symmetric inner product on $\mathscr{V}^\mathbb{R}$. To see that it’s linear in the first column note that (recall that this is a $\mathbb{R}$-space) since

\begin{aligned}\langle \alpha x+\beta y,z\rangle_\mathbb{R} &=\text{Re}\langle \alpha x+\beta y,z\rangle\\ &=\text{Re}\left(\alpha\langle x,z\rangle+\beta\langle y,z\rangle\right)\\ &=\alpha\text{Re}\langle x,z\rangle+\beta\text{Re}\langle y,z\rangle\\ &=\alpha\langle x,z\rangle_\mathbb{R}+\beta\langle y,z\rangle_\mathbb{R}\end{aligned}

It’s symmetric since

$\langle x,y\rangle_\mathbb{R}=\text{Re}\langle x,y\rangle=\text{Re}\overline{\langle y,x\rangle}=\text{Re}\langle y,x\rangle=\langle y,x\rangle_\mathbb{R}$

Lastly, to see that it’s positive definite we merely note that $\langle x,x\rangle_\mathbb{R}=\text{Re}\langle x,x\rangle=\langle x,x\rangle$ since for any inner product $\langle x,x\rangle\in\mathbb{R}$ and the conclusion follows since $\langle\cdot,\cdot\rangle$ is positive definite.

Now, with respect to this new pre-Hilbert space $\left(\mathscr{V},\langle\cdot,\cdot\rangle_\mathbb{R}\right)$ we claim that the map $J$ is linear, involutory, and self-adjoint. This first part of this statement is clear since

$J(\alpha x+\beta y)=\overline{\alpha}J(x)+\overline{\beta}J(y)=\alpha J(x)+\beta J(y)$

(once again recalling that we are now dealing with an $\mathbb{R}$-space). The second is also clear by our initial assumptions. The last statement follows since for any $x,y\in\mathscr{V}^\mathbb{R}$ we have that

\begin{aligned}\left\langle J(x),y\right\rangle_\mathbb{R} &=\text{Re}\left\langle J(x),y\right\rangle\\ &=\text{Re}\left\langle y,x\right\rangle\\ &=\text{Re}\langle x,y\rangle\\ &=\text{Re}\left\langle J(y),x\right\rangle\\ &=\text{Re}\left\langle x,J(y)\right\rangle\\ &=\left\langle x,J(y)\right\rangle_\mathbb{R}\end{aligned}

With this, define $\mathscr{X}_1=\text{im }\left(\mathbf{1}+J\right)$ and $\mathscr{X}_2=\text{im}\left(\mathbf{1}-J\right)$. We claim that $\mathscr{V}^\mathbb{R}=\mathscr{X}_1\oplus\mathscr{X}_2$. Indeed, to see that $\mathscr{V}^\mathbb{R}=\mathscr{X}_1+\mathscr{X}_2$  we note that for any $x\in\mathscr{V}^\mathbb{R}$ one has that

$x=\frac{1}{4}\left(\mathbf{1}+J\right)^2(x)+\frac{1}{4}\left(\mathbf{1}-J\right)^2(x)$

Indeed, expansion merely gives

$\frac{1}{4}\left(\mathbf{1}+J+J^2+\mathbf{1}-J+J^2\right)=\frac{1}{4} 4\mathbf{1}=\mathbf{1}$

from where the rest is trivial. To see that $\mathscr{X}_1\cap\mathscr{X}_2$ is trivial, assume that $x\in\mathscr{X}_1\cap\mathscr{X}_2$ then $x=(\mathbf{1}+J)(y)=(\mathbf{1}-J)(z)$ for some $y,z\in\mathscr{V}^\mathbb{R}$. But,

$\left\langle (\mathbf{1}+J)(y),(\mathbf{1}-J)(z)\right\rangle_\mathbb{R}=\left\langle y,\left(\mathbf{1}+J\right)\left(\mathbf{1}-J\right)(y)\right\rangle_\mathbb{R}=0$

from where it follows that $x=\bold{0}$ (note we made use of the fact that $(\mathbf{1}+J)(\mathbf{1}-J)=\mathbf{1}-J^2=\bold{0}$).

Next note that if $x\in\mathscr{X}_1$ then $J(x)=x$ since $x=(\mathbf{1}+J)(y)$ for some $y\in\mathscr{V}^\mathbb{R}$ and thus

$J(x)=J(\mathbf{1}+J)(y)=\left(J+J^2\right)(y)=\left(\mathbf{1}+J\right)(y)=x$

similarly one shows that if $x\in\mathscr{X}_2$ then $J(x)=-x$. With this we claim that the map $I:\mathscr{X}_2\to\mathscr{X}_1$ by $x\mapsto ix$ is an isomorphism. Indeed, $I$ is clearly linear. To see that $I$ is bijective we note that $JI(x)=J(ix)=\bar{i}K(x)=-IJ(x)$ and thus $I$ is invertible. It follows then that $I$ is an isomorphism and thus by a counting argument $\dim_\mathbb{R}\mathscr{X}_1=\dim_\mathbb{R}\mathscr{X}_2=n$.  Let $\{x_1,\cdots,x_n\}$ be an orthonormal basis for $\mathscr{X}_1$.  But, recalling the simple identity $-\text{Re}(iz)=\text{Im}(z)$ one sees that

\begin{aligned}\langle x,y\rangle &=\text{Re}\langle x,y\rangle+i\text{Im}\langle x,y\rangle\\ &=\text{Re}\langle x,y\rangle-i\text{Re}\left(i\langle x,y\rangle\right)\\ &=\text{Re}\langle x,y\rangle-i\text{Re}\langle ix,y\rangle\\ &= \langle x,y\rangle_\mathbb{R}-i\langle ix,y\rangle_\mathbb{R}\end{aligned}

So, note then that if $i\ne j$ then by assumption of the way the basis was chosen $\langle x_i,x_j\rangle=0$. That said, this implies that $\text{Re}\langle x_i,x_j\rangle=0$ so that $\langle x_i,x_j\rangle=ir$ for some real number $r$. It follows then that $\text{Re}i\langle x_i,y_j\rangle=-r$ and thus $ir=-r$ from where it follows that $r=0$ and thus $\langle x_i,x_j\rangle=0$. That said, note that since $\langle x_i,x_i\rangle\in\mathbb{R}$ we have that $\text{Re}\left(i\langle x_i,x_i\rangle\right)=0$ and thus

$\langle x_i,x_i\rangle=\langle x_i,x_i\rangle_\mathbb{R}=1$

by assumption. It follows that $\{x_1,\cdots,x_n\}$ is orthonormal in $\left(\mathscr{V},\langle\cdot,\cdot\rangle\right)$ and since $\dim\mathscr{V}=n$ it follows that $\{x_1,\cdots,x_n\}$ is an orthonormal basis for $\left(\mathscr{V},\langle\cdot,\cdot\rangle\right)$.

That said, since $x_1,\cdots,x_n\in\mathscr{X}_1$ by prior discussion we know that $K(x_j)=x_j,\;\; j=1,\cdots,n$ from where the theorem follows. $\blacksquare$

From this we get the following interesting corollary:

Corollary: Let $\mathscr{V}$ be a pre-Hilbert space and $J$ a complex conjugate on $\mathscr{V}$. Then, there exists some orthonormal basis $\{x_1,\cdots,x_n\}$ such that

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

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 -

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