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I asked this question over at Math.StackExchange and despite having had a bounty on it I did not receive an answer.

Suppose that $G$ is a finite group and we have a unitary irreducible representation $\rho:G\rightarrow \hom V$. Suppose we fix a basis $\{e_i\}_{i\geq 1}$ of $V$ and with respect to this basis we have that

$$\rho(s)e_j=\sum_i \rho_{ij}(s)e_i.$$

The elements of $F(G)$, $\rho_{ij}$, are called the matrix elements of $\rho$.

We can define a representation conjugate to $\rho$ denoted by $\bar{\rho}$. This is a map $\bar{\rho}:G\rightarrow \hom \bar{V}$ where $\bar{V}$ is the conjugate vector space.

This is defined on a basis $\{\bar{e}_i\}$ of $\bar{V}$ by

$$\bar{\rho}(s)\bar{e}_j=\sum_i\overline{\rho_{ij}(s)}\bar{e}_i.$$

Two group representations $\rho_1:G\rightarrow \hom V_1$ and $\rho_2:G\rightarrow \hom V_2$ are said to be equivalent, $\rho_1\equiv\rho_2$, if there exists an invertible map $T:V_1\rightarrow V_2$ such that

$$T\circ \rho_1(s)=\rho_2(s)\circ T.$$

Such a map is known as an intertwiner and in this case where $\rho_1$ and $\rho_2$ are irreducible and unitary I think that we can find a unitary intertwiner.

If we have $\overline{\rho_{ij}}=\rho_{ij}$ (i.e. $\rho_{ij}:G\rightarrow \mathbb{R}$), then we see that $c:V\rightarrow \bar{V}$, $e_i\mapsto \bar{e_i}$, is an invertible intertwiner and so $\rho\equiv \bar{\rho}$.

My question is

Suppose that $\rho$ is a unitary irreducible representation such that $\rho\equiv \bar{\rho}$. What can we say about intertwiners from $\rho$ to $\bar{\rho}$? Moreover, what can we say about the matrix elements of $\rho_{ij}\in F(G)$ in this case?

Edit: Some of this might be helpful:

If $\rho$ is equivalent to $\overline{\rho}$ then there is an invertible linear $T:V\rightarrow\overline{V}$ such that

$$\bar{\rho}\circ T=T\circ \rho.$$

Note that the conjugation map, $c:V\rightarrow \overline{V}$, $e_i\mapsto \overline{e_i}$, is an intertwiner from $\rho^{-t}=(\rho^{-1})^t$ to $\overline{\rho}$. This is because the matrices $\rho$ are unitary and of course $\rho(s^{-1})=\rho(s)^{-1}=\rho(s)^*$.

We show this. Let $e_j\in V$. We have $c(e_j)=\overline{e_j}$ and $$\overline{\rho}(s)c(e_j)=\sum_i\overline{\rho_{ij}(s)}\overline{e_i}.$$

Note $$\rho^{-t}(s)e_j=\rho^{-1}(s)^te_j=(\rho(s)^*)^te_j=\sum_i\overline{\rho_{ij}(s)}e_i.$$

We get $\overline{\rho}(s)c(e_j)$ when we apply $c$ to get $c\circ \rho^{-t}=\overline{\rho}\circ c$.

Now we have $\bar{\rho}=T\circ\rho\circ T^{-1}$ and $\overline{\rho}=c\circ \rho^{-t}\circ c^{-1}$ so that

$$c^{-1}\circ T\circ\rho=\rho^{-t}\circ c^{-1}\circ T,$$

that is $\rho\equiv \rho^{-t}$.

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The following is more of a long comment than a proper answer.

Let $\chi$ be the character of $\rho$. Look at the character $\overline\chi$ of $\overline\rho$. We have $\overline\chi(s)=\overline{\chi(s)}$ for $s\in G$. As the trace is preserved under matrix conjugation, and as $\overline\chi(s)=Tr(\overline\rho(s))$, we have $\overline\chi(s)=Tr(\overline\rho(s))=Tr(T^{-1}\rho(s)T)=Tr(\rho(s))=\chi(s)=\overline{\chi(s)}$.

Thus $\chi$ has its values in $\mathbb{R}$. By the Frobenius-Schur theorem (Thm 31 in Sect 13.2 of Jean-Pierre, Serre (1977). Linear Representations of Finite Groups. Springer.), this means that $\rho$ preserves a non-degenerate bilinear form on $V$. In fact, depending on whether this form is symmetric or alternating, $\rho$ can or cannot be realisible over $\mathbb{R}$ (cf. Prop. 38 in [loc.cit.]).

To summarise: in the interesting (i.e. non-real) case, your group preserves an alternating non-deg. bilinear form.

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  • $\begingroup$ you might like to investigate what $T$ does to these invariant forms (it should map the one for $\rho$ to the one for $\overline\rho$ --- or perhaps the other way around...) $\endgroup$ – Dima Pasechnik Mar 23 '15 at 11:55
  • $\begingroup$ ...this does the trick thank you... the answer to my question is that in the quaternionic case, the matrix elements need not be real. $\endgroup$ – JP McCarthy Apr 21 '15 at 11:54

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