6
$\begingroup$

First of all I am far from being an expert in representation theory, so it is possible (likely) that the following question is trivial (in fact a trivial reference question):

Let $\Gamma$ be a, let's say finite (abelian) group, and $R$ be an effective representation of $\Gamma$. Is it true, that every irreducible $\Gamma$-representation is a subrepresentation of the $k$-th symmetric power representation $S^kR$, for some $k=0,1,\dots$ ?
(Note that it is enough to show that any regular representations of $\Gamma$ is a subrepresentation of $S^kR$.)

I strongly suspect that this is a well know fact in representation theory, but I couldn't find any reference so far and would therefore appreciate any kind of literature reference.

$\endgroup$
10
$\begingroup$

Edit: The answer is yes for all finite groups $G$ (even, as far as this makes sense, independently of the ground field). A reference is Theorem 1 on page 45 of Alperin: "Local representation theory". While the claim of this theorem only refers to the tensor product $V\otimes \ldots \otimes V$ for some faithful $kG$-module $V$, the proof actually constructs a free submodule in $$ k[V] \cong \bigoplus_{i=0}^{\infty} S^k V $$ which gives you exactly what you need.


For a finite abelian group $G$ the answer is yes. Assume $V=\langle v_1\rangle\oplus \ldots \oplus \langle v_n \rangle$ is the decomposition of a faithful $\mathbb C G$-module $V$ into irreducibles. Denote the character associated to $v_i$ by $\chi_i$. Since $V$ is faithful the $\chi_i$ must generate the character group $\textrm{Hom}(G,\mathbb C^\times)$. So we can write any irreducible character $\psi$ of $G$ as $$ \psi = \chi_1^{k_1}\cdots \chi_n^{k_n} $$ and then the vector $v_1^{k_1}\cdots v_n^{k_n} \in S^{k_1+\ldots+k_n} V$ spans a one-dimensional submodule of this symmetric power with character $\psi$.

$\endgroup$
  • $\begingroup$ Actually, symmetric powers of the natural permutation module of S_n do contain every irreducible. This is one of the few cases of which I'm sure! $\endgroup$ – John Wiltshire-Gordon Aug 18 '12 at 17:09
  • $\begingroup$ @John: Yes, you are right. I misread the paper (they just show that the characters of the symmetric powers do not span $\mathbb C \textrm{Irr}_{\mathbb C}(S_n)$). $\endgroup$ – Florian Eisele Aug 18 '12 at 17:13
  • $\begingroup$ A remark for those who don't own a copy of Alperin's book: The relevant pages can be found at books.google.be/… $\endgroup$ – Florian Eisele Aug 18 '12 at 18:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.