5
$\begingroup$

In J.P. Serre's Linear Representations of Finite Groups, two representations $(\rho, V)$ and $(\rho', V')$ of a finite group $G$ are called isomorphic if there exists a linear transformation $T: V \to V'$ such that $T \cdot \rho = \rho' \cdot T$. In the case where $V = V'$, we would call these representations isomorphic if $\rho' = T \cdot \rho \cdot T^{-1} = c_T\circ \rho$, where $c_T$ is conjugation by $T$. Conceptually this makes sense to me because if we identify these two vector spaces using any choice of bases, then this definition essentially says that a change of basis, i.e. how we write the same transformations shouldn't change the representation.

However, if we consider $k$-vector spaces where $GL_n(k)$ has non-trivial outer automorphisms, i.e. automorphisms which are not conjugation by some element, we could call two representations $\rho, \rho':G \to GL_n(k)$ isomorphic if $\rho = \phi\circ \rho'$ for any $GL_n(k)$ automorphism $\phi$. This seems similar and a good candidate because in this case too we can always recover a representation from any isomorphic copies, if we know what the isomorphism is, and because the actions of the group elements maintain their relations under such an isomorphism. An example of this sort of outer automorphism is complex conjugation in $GL_2 (\mathbb{C})$; this is not inner because for any matrix with a non-real determinant, the determinant changes, whereas inner automorphisms must preserve determinants.

I can see that we don't actually define isomorphisms in this fashion because if we did we would lose the uniqueness of the characters up to isomorphism (at the very least). My questions are the following:

  1. If two representations are related in this fashion (i.e. $\rho = \phi \circ \rho'$), what can we say about them?

  2. If one is irreducible, is the other irreducible too? If so, how exactly do outer automorphisms act on the set of irreducible representations of a fixed dimension?

  3. Can we calculate the decomposition of one representation using the decomposition of the other? Is there any nice correspondence?

  4. What does the action of outer automorphisms do at the level of characters? Can this action be described intrinsically on the vector space of class functions?

I think that there are a lot of other questions one could ask in this set-up; any remarks about this situation would be welcome. If there are better answers for similar cases (wherever the question makes sense), for instance locally compact or reductive groups, they would be welcome too.

$\endgroup$
  • $\begingroup$ I've stated Q1 as a catch-all for any useful facts about quasi-equivalent representations which might not explicitly be an answer to this question. How do you prove the first part of Q2 in the affirmative - could you quote any standard source or give some idea if this is easy to see? By the second part of Q2, I mean to ask whether the group action is free/transitive/faithful etc. $\endgroup$ – Anamay Chaturvedi Feb 17 '17 at 22:24
  • $\begingroup$ By Q3, I mean to ask whether there is any way to calculate the character of a quasi-equivalent copy of a sum of representations given their characters, or is the only way to go back to the level of representations and calculate the character of the latter (which wouldn't be using the work done in calculating the first representation's decomposition, in some sense)? A nice correspondence would be any way that does better than the latter method. $\endgroup$ – Anamay Chaturvedi Feb 17 '17 at 22:30
  • $\begingroup$ Sorry, I am going to delete my previous comment, because I completely misread the question. I thought you were talking about representations that became equivalent after applying an automorphism of $G$. Such representations are called quasi-equivalent. But I see now that you are applying the automorphism to ${\rm GL}(n,k)$ rather than to $G$. $\endgroup$ – Derek Holt Feb 18 '17 at 2:09
  • $\begingroup$ I think you need to know something about the automorphism group of ${\rm GL}(n,K)$ for the field in question in order to make progress. For many familiar fields, the automorphism group is induced by automorphism of the field $k$ together with the inverse-transpose (duality) automorphism. For those types of automorphisms there is no problem. Over the complex numbers, duality has the same effect on characters as complex conjugation, so you need only consider automorphisms induced by those of $k$, which simply permute the entries of the character. $\endgroup$ – Derek Holt Feb 18 '17 at 2:26
4
$\begingroup$

One way this can happen is if the ground field an automorphism $\psi:k \to k$, this will induce automorphisms $\psi_n: GL_n(k) \to GL_n(k)$ for any $n$ that you can then twist your representations by.

In particular if $K$ is a field extension of $k$ then we get a whole $Gal(K:k)$ worth of such automorphisms to twist things by. If $K = \mathbb{C}$ and $k = \mathbb{R}$ this is just the identity map and complex conjugation.

It turns out that for any $g \in Gal(K:k)$, we get a $k$-linear auto-equivalence of categories from $Rep(G,K)$ to itself (the inverse is just given by $g^{-1}$). In particular it preserves irreducibility, and allows you to match up the decompositions accordingly.

Moreover this equivalence of categories commutes with the forgetful functor to vector spaces over $K$ (and its $k$-linear auto-equivalence induced by $g$). In particular this means that if you twist a representation by $g \in Gal(K:k)$ and then take characters it is the same as twisting the character values by $g$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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