Let $G$ be a finite group and $\chi$ be an irreducible character of $G$ (characteristic zero algebraically closed base field). If $H$ is the kernel of $\chi$ then the irreducible representations of $G/H$ are exactly all the irreducible constituents of all tensor powers $\chi^n$.
Do you know any reference for this theorem?
Is it also working in positive characteristic?
Is it also working for some infinite groups? (maybe some special classes: reductive, Lie type, etc)
Thank you very much!
I am not quite sure about the reference :( I always thought of this fact as follows.
Matrix elements of tensor powers of a representation U are all possible monomials in matrix elements of U, so the space of all their linear combinations are values of all possible polynomials in the matrix elements of U. Now, by definition of H, values of matrix elements of U separate elements of G/H, so every function on G/H (including all irreducible characters) can be written as a polynomial in the matrix elements of U in the case of finite groups, or can be approximated by polynomials with arbitrary precision in the case of compact infinite groups and the ground field being R or C (Stone-Weierstrass).
Now, to complete the proof, we may use orthogonality of matrix elements: if E_{ij} are matrix elements of an irrep V, and F_{ij} --- matrix elements of an irrep W (all thought of as functions on the group), then for the standard bilinear form on the ring of functions C(G) we have (E_{ij},F_{kl})=0 unless V is isomorphic to W and, in the latter case, i=l, j=k (in which case the value is 1) - here I probably want the order of the group to not be divisible by char(k) in the finite case, or the group to be compact, and the field be real/complex in the infinite case. Since irreducible characters can be approximated by polynomials in matrix elements, such a character cannot be orthogonal to all matrix elements of tensor powers and is, therefor, contained in one of them.
The easiest proof I know of this result for finite groups is due to H. Blichfeldt ( I believe), and I think it is easier than Brauer's proof, which was itself easier than the power series type proof which maybe first appeared in W. Burnside's book. I am not sure about a textbook reference for it, though. It is certainly mentioned in papers by P. Cameron about "sharp" characters. I think the proof, which follows, was also rediscovered by D. Chillag. Let $\chi$ be a faithful character of a finite group $G$, and let "1" denote the trivial character.Let $a_1,a_2,\ldots,a_n$ be the distinct values taken by $\chi$ on non-identity elements. Then $\prod_{i=1}^{n} (\chi - a_{i}1)$ is an integral combination of powers of $\chi$ (you need a little Galois theory and number theory here). On the other hand, the product clearly takes value zero on each non-identity element of $G$, so is an integer multiple of the regular character. In particular, every irreducible character $\mu$ must be a constituent of $\chi^{m}$ for some $m.$ As for positive characteristic $p$, there are various directions in which to generalize it.If you work with Brauer characters ( and take the $a_i$ as the values taken on non-identity p-regular elements), and we take a faithful module in characteristic $p$, then the above argument directly generalizes to show that every irreducible module occurs as a composition factor of some tensor power of that module. This generalization may be due to L.G. Kovacs. There is a different generalization in a paper of Bryant and Kovacs, where one works in the Green ring, rather that the character ring, and proves that there is a free summand of some tensor power, provided no non-identity element of $G$ acts as scalars on the original module (and a version of such a result is also proved in J.Alperin's book "Local Representation Theory").
This is equivalent to the statement that any irreducible representation of a group $G$ is contained in some tensor power of a faithful representation $V$.
A proof, due to Brauer, is given e.g. here. In fact, it is possible to make the size of the tensor power that one has to take explicitly bounded (the number of distinct values taken by the character of $V$ minus 1), using vanderMonde determinants.
I learned the theorem from Curtis and Reiner, Methods of Representation Theory.
In the case of $GL(V)$, it doesn't work because $V$ is a faithful representation, but not every irreducible of $GL(V)$ is contained in a tensor power (you have to look at the duals as well).
I think this (or an equivalent version of this) was an exercise in Harris & Fulton - "Representation Theory : A First course". Specifically on pg $25$, at the end of Chapter $2$, Problem $2.37$. It's only stated for finite groups there - specifically it's stated that given any faithful character of $G$, $\rho$, then any irreducible representation of $G$ is contained in some tensor power of $\rho$. Your version is equivalent, since $\chi$ will be irreducible on the quotient $G/H$.
This problem also has a fairly complete answer/hint on pg $517$. That approach in the answers is very beautiful and elegant, when I was working through the book last year I came up with something much less elegant (I think I used Vandermonde determinants, and I tried to do it the "obvious" way by looking at inner products of the tensor with the irreducible representations, i.e. character theory, so that's another way of doing it, if you want to try that).
Explicit references:
Generalizations:
