I have two questions. First, given a finite dimensional complex vector space $V$ and a finite group representation $\rho:G \to GL(V)$, Maschke's theorem tells us that we may decompose $V$ into a direct sum

$$V= \bigoplus_{\lambda}m_{\lambda}V^{\lambda}$$

of irreducible representations $V^{\lambda}$ of multiplicity $m_{\lambda}$. Does this result hold for representations of $G$ on an infinite dimensional vector space?

Of particular interest is the case where the vector space $V$ is graded

$$V=\bigoplus_{n \geq 0} V_n$$

such that there is a representation $\rho_n: G \to GL(V_n)$ for all $n$.

As an example, the symmetric group $S_n$ acts on the polynomial ring $\mathbb{C}[t_1,...,t_n]$ by permuting variables. This gives us a representation of $S_n$ on each graded part of $\mathbb{C}[t_1,...,t_n]$, with the grading given by the degree of the polynomials. Furthermore, on each graded part, the $S_n$ representation is the defining representation.

I'm wondering how to talk about this representation. In other words, in a decomposition of a finite dimensional vector space, the multiplicity of each irreducible representation is a natural number, but in this case we would have infinitely many isomorphic copies of the same representation. Assuming that Maschke's theorem holds in the infinite dimensional case, how do we write or represent the decomposition into irreducibles?

  • $\begingroup$ In the graded case, when all your homogeneous components $V_n$ are finite dimensional (like in your example with $S_n$) and the representation leaves the grading invariant, then you can decompose each $V_n$ into irreducible representations and talk about multiplicities of these in each $V_n$. By the way, if you want to have Maschke's theorem, you should assume that the characteristic of your field does not divide the order of the group $G$. $\endgroup$ – Matthias Klupsch Feb 4 '16 at 6:39
  • $\begingroup$ So in that case would it make sense to talk about a representation on $V$, or just a representation on each $V_n$? And thanks I added that the vector space should be complex since at the moment I'm not concerned with fields of characteristic $p$ $\endgroup$ – leibnewtz Feb 4 '16 at 6:42
  • $\begingroup$ Any homomorphism of $G$ into some general linear group is a representation, regardless of dimension, if that is, what you are asking. $\endgroup$ – Matthias Klupsch Feb 4 '16 at 6:50

Note that one version of Maschke's theorem tells us even more, namely that the group algebra $\mathbb{C} G$ is semisimple. This implies that every $\mathbb{C} G$-module (finite dimensional or not) is a direct sum of simple modules. If $V$ is a $\mathbb{C} G$-module and $M_1,\dots, M_n$ are representatives for the isomorphy classes of the simple $\mathbb{C} G$-modules (recall, that as $G$ is finite, there are only finitely many of those), then $$V \cong \bigoplus_{j = 1}^n M_j^{(I_j)}$$ where the $I_j$ are some indexing sets and $M_j^{(I_j)}$ denotes the 'direct sum of $I_j$ copies of $M_j$'. Note that in the finite dimensional case, all the $I_j$ are finite and we can just write $|I_j| M_j$ instead of $M_j^{(I_j)}$. So it is the cardinality of the $I_j$ which generalizes the concept of multiplicity.

  • $\begingroup$ I think this is exactly the answer I was looking for. So in the case of $\mathbb{C}[t_1,...,t_n]$ as a $\mathbb{C}S_n$ module we have a decomposition $\mathbb{C}[t_1,...,t_n] \cong M_1^{\mathbb{N}} \oplus M_2^{\mathbb{N}}$ where $M_1$ is the representation of degree $n-1$ and $M_2$ is the trivial representation. Am I understanding this correctly? $\endgroup$ – leibnewtz Feb 4 '16 at 7:33
  • $\begingroup$ I believe this decomposition is not correct. If we write $\mathbb{C}[t_1,t_2, \dots, t_n] = \bigoplus_m V_m$, then $V_1$ is indeed the usual permutation representation of $S_n$, so $V_1 = M_1 \oplus M_2$ where $M_1$ is $n-1$ dimensional and $M_2$ is one-dimensional. However, for example for $n = 3$ and $m = 3$, $S_3$ acts by the sign representation on the element $t_1^2 t_2 - t_1^2 t_3 + t_2^2 t_3 - (t_1 t_2^2 - t_1 t_3^2 + t_2 t_3^2)$. $\endgroup$ – Matthias Klupsch Feb 4 '16 at 9:48
  • $\begingroup$ You're right, I had assumed the representation stays the same as the grading goes up. I found this question mathoverflow.net/questions/184509/… which says $\mathbb{C}[t_1,...,t_n]$ has a decomposition in each graded part given by the trivial representation and a free module of rank $n!-1$ over the symmetric polynomials-interesting! $\endgroup$ – leibnewtz Feb 4 '16 at 22:22
  • $\begingroup$ In the link, the statement there is made for the whole $\mathbb{C}[t_1,\dots, t_n]$, not for its graded parts. And 'free module' seems to refer to free over the ring of symmetric functions, so you do not get any statement about the representation theory of $S_n$. However, in the question you linked it is also said that the quotient of the polynomial ring by the ideal spanned by homogeneous symmetric polynomials is isomorphic to the regular representation of $S_n$. $\endgroup$ – Matthias Klupsch Feb 5 '16 at 6:22

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