*I asked this question over at [Math.StackExchange](https://math.stackexchange.com/questions/1182799/matrix-elements-of-real-represententations) 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](http://en.wikipedia.org/wiki/Complex_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 $\rho\equiv \rho^{-t}$.



Matrix Elements of Real Representations
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.


