Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The following result was mentionned earlier in this thread, I searched a bit in the related threads and couldn't find a proof. I would really like to see a proof of it:

Let $G$ be a finite group and $\rho : G \rightarrow GL(\mathbb{C}, n)$ a faithful representation of $G$. Then every irreducible representation of $G$ is contained in some tensor power of $\rho$.

share|improve this question
2  
I just remembered about mathoverflow.net/questions/10126/… –  darij grinberg Mar 14 '10 at 20:57
add comment

5 Answers

See problem 3.26 in Etingof's "Introduction to representation theory". If you have troubles with understanding the hint, feel free to ask me. (The first sentence uses the fact that if a vector space over an infinite field is the union of finitely many subspaces, then one of these subspaces is the whole vector space. The surjectivity of the map $SV\to F\left(G,\mathbb C\right)$ is because a polynomial can take any arbitrary finite set of values at some given distinct points. In order to conclude from this, note that this map $SV\to F\left(G,\mathbb C\right)$ is a homomorphism of representations of $G$.)

This proof works over any algebraically closed field of characteristic $0$. This can't quite be said about the proof in Fulton-Harris, if I remember it right.

share|improve this answer
    
Another proof is in Curtis-Reiner: Representation theory of finite groups and associative algebras, Thm 32.9 in the first edition. To my mind the proof is very cute, it considers the generating series $\sum_k a_k t^k$, where $a_k$ is the number of times the irreducible representation occurs in the $k$'th tensor power of $\rho$. Using the orthogonality formulas for characters they sum this up as a rational function which, because $\rho$ is faithful, has a simple pole at $t=1/n$ and hence many of the $a_k$ are non-zero. –  Torsten Ekedahl Mar 14 '10 at 18:59
    
I THINK you mean the same proof as in Fulton-Harris. Alas, I don't see how to show that the pole is indeed a pole (i. e., not cancelled by another term) without the use of complex numbers. –  darij grinberg Mar 14 '10 at 19:10
    
There are two additional short proofs in Proc. AMS just after Curtis-Reiner first appeared, avoiding Burnside's use of complex numbers. (These back issues are to appear on the AMS journal page after scanning but haven't yet. I can access them through the UMass library or JSTOR, but these are restricted.) The two papers appear in the collected papers of the two authors: R. Steinberg, Complete sets of representations of algebras, Proc. Amer. Math. Soc. 13 (1962), 746-747 R. Brauer, A note on theorems of Burnside and Blichfeldt, Proc. Amer. Math. Soc. 15 (1964), 31-34 –  Jim Humphreys Mar 14 '10 at 20:01
    
Thanks a lot, I'll look into these. –  darij grinberg Mar 14 '10 at 20:06
    
How does Brauer conclude this: "Since X is faithful, one of the A_j consists only of the unit element"? –  darij grinberg Mar 14 '10 at 20:59
show 11 more comments

Another proof (not really that different from Geoff's, but appealing to a somewhat different intuition): Let $W$ be the representation corresponding to $\rho$, let $\underline{1}$ be the trivial representation, and let $V$ be the representation which we want to appear in some representation of $W^{\otimes N}$. I will show instead that $V$ appears in some representation of $(W \oplus \underline{1})^{\otimes N}$; this is equivalent because $(W \oplus \underline{1})^{\otimes N} = \bigoplus_{k=0}^N \binom{N}{k} W^{\otimes k}$.

Let $\chi$ be the character of $V$ and let $\psi$ be the character of $W$. Then $$\dim \mathrm{Hom}_G(V, (W \oplus \underline{1})^{\otimes N}) = \frac{1}{|G|} \sum_{g \in G} \overline{\chi(g)} (\psi(g)+1)^N. \quad (*)$$ We want to show that this Hom space is nontrivial for large $N$.

We have $|\psi(g)| \leq \dim W$ for all $g \in G$ and, since $W$ is faithful, $\psi(g)$ is $\dim W$ if and only if $g=e$. So $|\psi(g)+1| \leq \dim W + 1$, with equality precisely for $g=e$. So the right hand side of $(*)$ is a finite sum of exponentials, and the term $(\dim V) (\dim W + 1)^N$ has a larger base than any of the others. So the right hand side is positive for large $N$, and we see that the irrep $V$ appears in $(\underline{1} \oplus W)^{\otimes N}$ for sufficiently large $N$.

I explain how to modify this for compact Lie groups in this answer.

share|improve this answer
add comment

As I have said elsewhere on Mathoverflow, in dealing with related questions, I think the simplest and the best proof of this result is due to Blichfeldt. It is simpler than both the power series type argument which appears in Burnside's book and the Vandermonde determinant argument of Brauer. Since we are dealing with characteristic zero representations of finite groups, we need only deal with characters. Let $\chi$ be the character afforded by $\rho$, and let $\alpha_{1},\alpha_{2},\ldots,\alpha_{m}$ be all the distinct values taken by $\chi$ on non-identity elements of $G$. Note that the class function $\chi \prod_{i=1}^{m}(\chi - \alpha_{i}1)$ vanishes on all non-identity elements of $G$, but does not vanish at $1_{G}$. Hence this class function is a non-zero multiple of the regular character (in fact a rational algebraic integer multiple). On the other hand, it may be written in the form $\sum_{j=1}^{m+1} a_{j}\chi^{j}$ for certain rational integers $a_{i}$ (note that $\{\alpha_{1},\ldots \alpha_{m} \}$ is a set of algebraic integers closed under algebraic conjugation). Since any irreducible character $\mu$ of $G$ has non-zero inner product with the regular character, we see that $\langle \chi^{j},\mu \rangle \neq 0$ for some $j$ with $ 1 \leq j \leq m+1$. (It is not necessary to use the factor $\chi$ if any $\alpha_{i}$ is already zero and, in any case, the factor $\chi$ is only used to make sure that we use strictly positive tensor powers of $\rho$, avoiding the issue of the trivial module). Note that Blichfeldt's argument yields that $\prod_{i=1}^{m} (\chi(1) - \alpha_{i})$ is an integer multiple of $|G|$.

share|improve this answer
    
This is an elegant proof, but it doesn't seem to generalize readily to compact groups. –  Qiaochu Yuan Apr 26 '11 at 20:55
    
That's probably correct. It does, however, generalize to representations over fields of finite characteristic in various ways, as I have explained elsewhere. –  Geoff Robinson Apr 26 '11 at 21:10
    
Mmm, presumably you could replace "non-identity elements" with "non-identity elements of order <m" and then let m increase, but it's true that this sort of wrecks the elegant nature of the proof. –  Ben Webster Apr 27 '11 at 0:40
add comment

By Satz 90, the fraction field of the symmetric algebra of a faithful representation of a finite group contains all irreducible representation. It remains to get rid of the denominators, just multiplying by the product of their conjugates.

share|improve this answer
2  
How does this follow from Hilbert 90? –  darij grinberg Apr 19 '13 at 14:33
add comment

Any semilinear Galois representation is trivial (the skew product of a field $K$ with a finite group $G$ of its automorphisms is the endomorphism algebra of $K$ as vector space over the fixed subfield), so extension to $K$ of coefficients of any irreducible representation of $G$ is isomorphic to a direct sum of copies of $K$.

share|improve this answer
add comment

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.