3

Suppose that $G$ is a finite group and $V$ is a (complex) irreducible representation of $G$, that is, for any $g\in G$ we have $gV=V$ (here we use notation $gv$ for the action of element $g$ on vector $v$ instead of, for example, $\rho(g)v$, where $\rho\colon G\to\operatorname{End} V$).

There is a natural embedding of the group algebra $\Phi\colon\mathbb{C}[G]\hookrightarrow\operatorname{End} V$ which is defined by the formula $$ \Phi\left(\sum_{g\in G}c_g\cdot g\right)(v)=\sum_{g\in G}c_g\cdot gv. $$

I found somewhere (can't remember now) that $V$ is irreducible iff $\Phi$ is surjective, i. e. every linear operator on $V$ can be realised as an element of $\mathbb{C}[G]$, but I don't know how to prove that.

The "if" part is easy. Indeed, if $V$ is reducible, then there is a $G$-invariant subspace $W\subsetneq V$ and we can take projection $P\in\operatorname{End} M$ onto $U$, where $U\in W$ such that $U\oplus W=V$ (and define $P|_{U}=\operatorname{id}_{U}$ and $P|_{W}=0$). It's clear that for any $a\in\mathbb{C}[G]$ we have $\operatorname{Im}\Phi(a)\subset W$, so there is no $a\in\mathbb{C}[G]$ such that $\Phi(a)=P$.

Thus, it remains to prove the "only if" part. Any ideas? I am a beginner in the representation theory, so any help would be appreciated.

richrow
  • 4,092
  • 2
  • 10
  • 32
  • 4
    This follows from the Jacobson density theorem: https://en.wikipedia.org/wiki/Jacobson_density_theorem – Qiaochu Yuan Nov 29 '20 at 08:54
  • So, it's impossible to prove this fact without higly developed theory? I thought that this fact is quite natural and there is a proof using some basic facts (definitions, Schur's lemma, etc.), but I might be wrong. By the way, this proposition can be used as another definition of irreducible representation, right? – richrow Nov 29 '20 at 09:18
  • 2
    The Jacobson density theorem isn't so bad. You can also use some standard facts about semisimple algebras together with Maschke's theorem. – Qiaochu Yuan Nov 29 '20 at 10:13
  • OK, got it. Thanks. – richrow Nov 29 '20 at 10:14

2 Answers2

3

Proving it with the Jacobson Density Theorem is not so bad, in the sense that Wikipedia has a fully self-contained proof. Or you can prove the Artin-Wedderburn Theorem without it and use that.

First note $\mathbb{C}[G]\to\bigoplus\mathrm{End}(V)$ (over all irreps $V$) must be one-to-one: any element of the kernel acts as $0$ on any irrep, hence on any rep, hence sends $1\in\mathbb{C}G$ to $0$, hence is $0$. To see that it is in isomorphism it then suffices to check dimensions match. For this, write $\mathbb{C}G\cong\bigoplus nV$ (where $nV:=V\oplus\cdots\oplus V$) for some multiplicities $n$. Then $\hom_G(\mathbb{C}G,V)\cong V$ ($1\mapsto v$ arbitrarily) so the multiplicities are $n=\dim V$ (recall $\dim\hom_G$ acts as a sort of "inner product" for the multiplicities of irreps in reps). This implies the dimension of $\mathbb{C}[G]$ is $\sum n^2$, same as $\bigoplus\mathrm{End}(V)$.

The map $\mathbb{C}[G]\to\mathrm{End}(U)$ for an irrep $U$ corresponds to a projection map on $\bigoplus\mathrm{End}(V)$. On an arbitrary rep decomposing as $W=\bigoplus mV$ the map $\mathbb{C}[G]\to\mathrm{End}(W)$ lands in the block diagonal subalgebra $\bigoplus m\mathrm{End}(V)$ within (in fact, the diagonal copy of $\mathrm{End}(V)$ in $m\mathrm{End}(V)$) so it is onto iff $W$ is an irrep.

anon
  • 151,657
  • You went a bit fast in $\hom_G(\mathbb{C}G,V)\cong \hom_G(nV,V)\cong n \ \hom_G(V,V)$ and $\hom_G(V,V)=\Bbb{C}$ because it contains the scalars $a\in \Bbb{C}$ and for any other element (since $V$ is irreducible) then $\ker(f-a)$ is either $0$ or $V$, ie. $f$ is a scalar. – reuns Dec 02 '20 at 23:48
  • Thank you for your response. I just wonder if there is a proof wich uses only initial irreducible representation $V$. For me, as a a begiiner, it's both interesting and a bit surprising that proof of this fact about some particular irreducible representation $V$ requires working with all irreducible representations at once. I mean, it's quite natural to expect (at lesat for me) that solution to this problem should be "inside" $V$. – richrow Dec 03 '20 at 15:29
  • 1
    @richrow The AW theorem implies that since you're working with the group algebra $\Bbb C[G]$, you secretly are working with all reps at once. And if $V$ is an arbitrary irrep, then having a proof for a particular $V$ isn't so different from working with all irreps at once. And in any case, the Jacobson density theorem does actually avoid this in its proof. – anon Dec 03 '20 at 23:36
2

From Schur's orthogonality relations we have $$\frac{d}{|G|}\cdot \sum_{g\in G} a_{ji}(g^{-1})\, g = E_{ij}$$ with $1$ at $(i,j)$ and $0$ in rest, and so, any matrix in $M_{d}(\mathbb{C})$ is a combination of the $g$'s.

$\bf{Added:}$ For every $A\in M_{d}(\mathbb{C})$ we have $$ \frac{d}{|G|}\cdot \sum_{g\in G} Tr (g^{-1}A) g = A$$

orangeskid
  • 53,909