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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?

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    @JackSchmidt: Thanks. At the moment, this is more than enough for the finite case.2012-07-28

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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 $\dim(V) \int_G \overline{\chi_V(g)} g \, d \mu.$