Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The character of a representation of a finite group or a finite-dimensional Lie group determines the representation up to isomorphism.

Is there an algorithmic way of recovering the representation given the character and the conjugacy structure of the group?

share|improve this question
1  
Let me know if you want this expanded to an answer: The GAP package repsn implements some important speedups for recovering the representation from a character of a concrete finite group. The basic idea is due to Dixon (with improvements from Schneider). I believe Dixon's original paper talks about unitary representations and handling them in floating point, but this is rarely done in the computational group theory community. It may not be possible to recover the representation only from the character and conjugacy structure, since it may not be possible to recover the group itself. –  Jack Schmidt Jul 28 '12 at 18:11
    
@JackSchmidt: I am actually not interested in a computer program but in the conceptual ideas behind such an algorithm. So a reference to the original papers would be a useful answer. –  Arnold Neumaier Jul 28 '12 at 18:13
1  
Depending on the detail you'd like, ams.org/mathscinet-getitem?mr=2630014 its predecessor by both authors, the referenced papers by the second author, and the first author's dissertation are all good. I believe the source code includes a reasonable amount of theory, but I read the papers at the same time. –  Jack Schmidt Jul 28 '12 at 18:18
    
@JackSchmidt: Thanks. At the moment, this is more than enough for the finite case. –  Arnold Neumaier Jul 28 '12 at 18:37

1 Answer 1

up vote 2 down vote accepted

Below by "representation" I mean "finite-dimensional complex continuous representation."

This is false for Lie groups in general; for example, the character of a representation of $\mathbb{R}$ does not distinguish the $2$-dimensional trivial representation from the representation $$r \mapsto \left[ \begin{array}{cc} 1 & r \\ 0 & 1 \end{array} \right].$$

(Technically speaking, any countable discrete group is also a Lie group as it is a $0$-dimensional manifold for any reasonable definition of manifold, but now I'm being far too picky.)

The correct statement is for compact groups (not necessarily Lie). It suffices to address the problem for irreducible representations. If $G$ is such a group, $\mu$ is normalized Haar measure, and $\chi_V$ is the character of an irreducible representation $V$, let $L_g : L^2(G) \to L^2(G)$ be the map which translates a function by $g$ (that is take $L_g(f(h)) = f(g^{-1}h)$). Then $$\dim(V) \int_G \overline{ \chi_V(g) } L_g \, d \mu$$

is a projection $L^2(G) \to L^2(G)$ whose image is the $V$-isotypic component of $L^2(G)$ (exercise). Taking any nonzero vector in this image and applying suitably many elements of $G$ to it will give you explicit vectors spanning a subspace of $L^2(G)$ isomorphic to $V$; turn these into a basis, and then you get explicit matrices for the elements of $G$.

Since $L^2(G)$ is a bit large, an alternate method (which only works for Lie groups) is to start with a faithful representation $W$ and take a tensor power $W^{\otimes n} (W^{\ast})^{\otimes m}$ containing $V$ (this is always possible, see this MO question), then apply the projection $$\dim(V) \int_G \overline{\chi_V(g)} g \, d \mu.$$

share|improve this answer

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.