# Reference for this theorem in representation theory?

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 $\chi^n$.

1. Do you know any reference for this theorem?

2. Is it also working in positive characteristic?

3. Is it also working for some infinite groups? (maybe some special classes: reductive, Lie type, etc)

Thank you very much!

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You may get more answers if you edit the title of the question so that it gives some idea what sort of mathematics it is about. –  Michael Lugo Dec 30 '09 at 16:12
@Michael: done. –  Tom Leinster Dec 30 '09 at 16:52
See mathoverflow.net/questions/58633/… for the case of compact Lie groups. –  Qiaochu Yuan Apr 17 '11 at 8:47

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.

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Your argument seems to shows something more; namely that any character of G is a polynomial of the faithful character. Can you give me a reference for the version of Stone-Weierstrass that is used here? I don't remember if the functions necessarily have to be given on a group? I am asking this since I proved by a different methods a similar result for Hopf algebras. –  Sebastian Burciu Dec 30 '09 at 14:13
One of standard versions of Stone-Weierstrass is "Suppose X is a compact Hausdorff space and A is a subalgebra of C(X) which contains a non-zero constant function. Then A is dense in C(X) if and only if it separates points." I think you can find it in Rudin's analysis textbook in this form. However, in the case of finite groups, you don't need that much (since matrix elements of U separate points of G/H, we embed G/H in the (dim(U))^2-dimensional coordinate vector space and do polynomial interpolation), but for the case of SU(n) or something, Stone-Weierstrass is handy. –  Vladimir Dotsenko Dec 30 '09 at 14:31
P.S. A related remark: we all know that polynomial finite-dim irreps of GL(V) are contained in V^{\otimes p}\otimes (V^*)^{\otimes q}, and this is also suggested by the Stone-Weierstrass viewpoint, at least in the complex case, as we need polynomials in z and z-bar, which precisely means polynomials in matrix elements of V and V^*. –  Vladimir Dotsenko Dec 30 '09 at 14:40
I guess what I said in the previous comment is not true. Any character is a polynomial in all matrix elements $U_{ij}$ but not a polynomial in the character $\chi=\sum_{i}U_{ii}$. But I remember seeing somewhere showing that (at least) the regular character of $G/H$ is a polynomial in $\chi$. –  Sebastian Burciu Dec 30 '09 at 16:05

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).

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Thanks for the answer! Do you also know a reference (or proof) for the following related fact: If $\chi$ is a faithful irreducible character of $G$ then the regular character of $G$ is a polynomial in $\chi$? I know this fact is true since there is a generalization of it for Hopf algebras in Corollary 19 of the paper FSU96-08 from math.fsu.edu/~aluffi/eprint.archive.html. I was wondering if someone knows a simpler proof for this fact. –  Sebastian Burciu Dec 30 '09 at 20:48
Let's think of characters as functions on conjugacy classes. Then $\chi(1)=n={\rm dim}(V)$, and $\chi(g)$ for $g\ne 1$ has smaller absolute value than $n$ (since the representation is faithful and eigenvalues of $g$ in $\chi$ are roots of 1). In particular, $\chi(g)\ne n$. Now let $P$ be the interpolation polynomial such that $P(n)=|G|$ and $P(x)=0$ for any other value of $\chi$. Then $P(\chi)$ is the regular character. –  Pavel Etingof Feb 1 '10 at 15:57

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 element of $G$ acts as scalars n the original module (and a version of such a result is also proved in J.Alperin's book "Local Representation Theory").

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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).

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That power series argument that Fulton-Harris do is just a fancy way to write out your Vandermonde argument. Both times, it's the same lemma: If $v_1$, $v_2$, ..., $v_n$ and $w_1$, $w_2$, ..., $w_n$ are elements of some field, and we have $\sum_{i=1}^n v_iw_i^k=0$ for every $k\in\mathbb N$, then $\sum_{i\in\left\{1,2,...,n\right\};\ w_i=x}v_i=0$ for every $x$ in the field, so that particularly, $v_i=0$ for every "unique" $i$ (that is, for every $i$ which satisfies $w_i\neq \left\{w_1,w_2,...,w_{i-1},w_{i+1},w_{i+2},...,w_n\right\}$). –  darij grinberg Dec 30 '09 at 12:54
I meant $w_i\not\in \left\{w_1,w_2,...,w_{i-1},w_{i+1},w_{i+2},...,w_n\right\}$ instead of $w_i\neq \left\{w_1,w_2,...,w_{i-1},w_{i+1},w_{i+2},...,w_n\right\}$. –  darij grinberg Dec 30 '09 at 12:55
In the above situation, after re-indexing $w_C=\chi(C)$ and $v_C=|C|\bar{\phi(C)}$. I guess the situation $\chi(C)=\chi(D)$ is possible but $\sum_{D;\chi(C)=\chi(D)}v_D=0$ is not possible. –  Sebastian Burciu Dec 30 '09 at 13:06