Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

2 Answers 2

up vote 13 down vote accepted

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 $Out(G).$

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.