# to test equivalence of representations under automorphism

Hi all,

I would like to propose the following problem:

Given two representations $\rho$ and $\tau$ of a group $G$ over complex number, we would like to know if there exists an automorphism $\phi$, such that $\rho\circ\phi$ and $\tau$ are equivalent.

Is there any mathematical results concerning this problem? It seems that to understand the action of automorphisms on the set of irreducible representations is crucial.

Thank you!

Youming

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One way to make this more precise is to ask whether the orbits of Aut(G) on the irreps can be described using only the character table. – Mariano Suárez-Alvarez Aug 20 '10 at 10:58

I assume that $\phi$ is an automorphism of $G.$ Note that if $\phi$ is inner then trivially $\rho$ and $\rho\circ\phi$ are equivalent, thus the answer depends only on the image of $\phi$ in the outer automorphism group $Out(G).$

If $G$ is a finite group (or, more generally, compact group) and representations are finite-dimensional, so that they are determined up to isomorphism by their characters, then this problem admits a complete theoretical solution using the character theory. The automorphism group $Aut(G)$ acts on the set $\{C_i\}$ of the conjugacy classes of $G$, this action factors through the action of $Out(G),$ and

$$\chi_{\rho\circ\phi}(C)=\chi_{\rho}(\phi(C)),\qquad (*)$$

where $\chi_\rho$ is the character of $\rho$ and $C$ is any conjugacy class. Since representations are determined by their characters,

$$\rho\circ\phi\simeq\sigma \iff \chi_{\rho\circ\phi}=\chi_\sigma,$$

which can be tested using $(*).$ Of course, applying this method in practice requires good knowledge of the character table of $G$ and the outer automorphism group $Out(G).$

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Victor's answer shows that it is important to understand the action of $Out(G)$ on the conjugacy classes of $G$. This can be interesting even in the abelian case, where the problem amounts to calculating the orbits of $Aut(G)$ on $G$. Even though the answer was well-known, we found a combinatorial approach to the question quite interesting.

The idea (which can be formulated for any group) is to define a pre-order on elements: $x\leq y$ if an endomorphism maps $y$ to $x$. The equivalence classes associated to this pre-order ($x\sim y$ if $x\leq y$ and $y\leq x$) are unions of outer equivalence classes. In the finite abelian case, these classes are precisely the $Aut(G)$-orbits.

The question of what other classes of groups this method can be applied to remains open.

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