# Abstract Nonsense

## Complex Conjugate Representations (An Example)

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

It turns out that this classification is still too broad in the sense that there is a natural way in which self-conjugate maps can be further classified. The self-conjugate class breaks naturally down into whether for a given $\rho$ there exists a basis on $\rho$‘s representation space for which $\rho_g$‘s matrix representation has all real entries for each $g\in G$. To illustrate this concept further we consider the following example

Example: Consider the quaternion group $Q$ with presentation

$\left\langle -1,i,j,k:(-1)^2=1,\;i^2=j^2=k^2=ijk=-1\right\rangle$

which has explicit Cayley Table

$\displaystyle \begin{array}{c|cccccccc}1 & 1 & -1 & i & -i & j & -j & k & -k\\ \hline1 & 1 & -1 & i & -i & j & -j & k & -k\\ -1 & -1 & 1 & -i & i & -j & j & -k & k\\ i & i & -i & -1 & 1 & k & -k & j & -j\\ -i & -i & i & 1 & -1 & -k & k & j & -j\\ j & j & -j & -k & k & -1 & 1 & -i & i\\ -j & -j & j & k & -k & 1 & -1 & -i & i\\ k & k & -k & j & -j & -i & i & -1 & 1\\ -k & -k & k & -j & j & i & -i & 1 & -1\end{array}$

and define the representation $\rho:Q\to\mathbb{C}^2$ by

$\displaystyle \rho(\pm 1)=\pm I_2\quad \rho(\pm i)=\pm\begin{pmatrix}i & 0\\ 0 & -i\end{pmatrix}\quad\rho(\pm k)=\pm\begin{pmatrix}0 &-1\\ 1 & 0\end{pmatrix}\quad \rho(\pm j)=\pm\begin{pmatrix}0 & i\\ i &0\end{pmatrix}$

This is easily seen to be an irrep. Moreover, a quick computation shows that if $J$ is the usual complex conjugate on $\mathbb{C}^2$ (the one which conjugates each entry of a tuple, i.e. $(z,z')\mapsto (\bar{z},\bar{z'})$) then

$\text{Conj}^J_\rho(g)=\rho_k\rho(g)\rho_k^{-1}$

and thus $\rho$ is self-conjugate. That said, suppose that $B$ is a basis for $\mathbb{C}^2$ such that $[\rho(i)]_B$ has real entries. Note then that

$\left\{I_2,\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix},\begin{pmatrix}0 & -i\\ i & 0\end{pmatrix},\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\right\}$

forms a basis for $\text{GL}(2,\mathbb{C})$ and thus there must exists constants $\alpha_1,\cdots,\alpha_4\in\mathbb{C}$ for which

$[\rho(i)]_B=\alpha_1 I+\alpha_2\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}+\alpha_3\begin{pmatrix}0 & -i\\ i & 0\end{pmatrix}+\alpha_4\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}$

But, we know that the trace is conjugation (and thus coordinate invariant) so that taking the trace of both sides, using the linearity of the trace function and the fact that the non-identity matrix elements of this matrix basis are traceless we may conclude that $2\alpha=0$ and thus $\alpha=0$. But, note that

$\alpha_2\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}+\alpha_3\begin{pmatrix}0 & -i\\ i & 0\end{pmatrix}+\alpha_4\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}=\begin{pmatrix}\alpha_4 & \alpha_2-i\alpha_3\\ \alpha_2+i\alpha_3 & -\alpha_4\end{pmatrix}$

from where it easily follows from the assumption that this matrix is real that $\alpha_2,\alpha_4$ are real and $\alpha_3$ is pure imaginary. Note though that $\rho(i)^2=-\mathbf{1}$ from where it follows that $[\rho(i)^2]_B=[\rho(i)]_B^2=-I_2$. But, by assumption

$-I_2=[\rho(i)]_B^2=\begin{pmatrix}\alpha_4 & \alpha_2-i\alpha_3\\ \alpha_2+i\alpha_3 & -\alpha_4\end{pmatrix}^2=(\alpha_2^2+\alpha_3^2+\alpha_4^2)I_2$

and thus $\alpha_2^2+\alpha_3^2+\alpha_4^2=-1$. But, noticing that $\text{tr}(\rho(k))=\text{tr}(\rho(j))=0$ and $\rho(k)^2=\rho(j)^2=-\mathbf{1}$ one may conclude that for $x=i,j,k$ one has that there exists $\beta_1^{(x)},\beta_2^{(x)},\beta_3^{(x)}\in\mathbb{R}$ such that $\left({\beta_2^{(x)}}\right) ^2-\left({\beta_1^{(x)}}\right)^2-\left({\beta_3^{(x)}}\right)^2=1$ and

$[\rho(x)]_B=\beta_1^{(x)}\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}+i\beta_2^{(x)}\begin{pmatrix}0 & -i\\ i & 0\end{pmatrix}+\beta_3^{(x)}\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}$

though a quick check shows that this is impossible. Indeed, noticing that in particular

$\text{tr}\left([\rho(i)]_B[\rho(j)]_{B}\right)=0=\beta_1^{(i)}\beta_1^{(j)}+\beta_3^{(i)}\beta_3^{(j)}-\beta_2^{(i)}\beta_2^{(j)}$

But this is quickly shown to be inconsistent.  It thus follows that no such basis exists.

Thus, it follows that there is a distinction that should be made on the set of self-conjugate irreps. In particular, we call a self-conjugate irrep$\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$  real if there exists a basis $B$ for $\mathscr{V}$ for which each entry of $[\rho(g)]_B$ is real for every $g\in G$. If $\rho$ is self-conjugate but not real it’s called quaternionic. The previous example shows the existence of quaternionic irreps. These shall prove to be interesting in the theory to come.

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

Advertisements

February 5, 2011 -

## 3 Comments »

1. […] Point of post: This post is a continuation of this one. […]

Pingback by Representation Theory: Complex Conjugate Representation (Characterization of Complex Conjugates) « Abstract Nonsense | February 6, 2011 | Reply

2. […] Point of post: In this post we use the techniques we’ve devoloped to construct the character table for the quaternions. […]

Pingback by Representation Theory: Character Table for the Quaternions « Abstract Nonsense | March 23, 2011 | Reply

3. […] from elementary methods whether or not an irrep was real, complex, or quaternionic. Indeed, in our one example of quaternionic irreps the agrument that the irrep in question was, in fact, quaternionic was […]

Pingback by Representation Theory: A Characterization of Real, Complex, and Quaternionic Irreps « Abstract Nonsense | March 23, 2011 | Reply