If $(\pi,V)$ is irreducible, then $\pi(G)$ spans $\operatorname{End}(V)$

Question

Let $(\pi,V)$ be a finite dimensional complex representation of a finite group $G$. If $\pi$ is irreducible, then the linear span of $\pi(g) : g \in G$ is equal to $\operatorname{End}_{\mathbb C}(V)$. One way to see this is by considering the linear map of the group algebra $\mathbb C[G]$ into the direct sum of the $\operatorname{End}_{\mathbb C}(V)$ for $V$ irreducible. It follows from the Peter-Weyl theorem that this map is an isomorphism. But this takes a bit of work.

Is there a simple proof that $\mathbb C[G] \rightarrow \operatorname{End}_{\mathbb C}(V)$ is surjective when $(\pi,V)$ is irreducible?

Is not this called Burnside's theorem (or lemma)? It is in Lang's Algebra, no Peter-Weyl theorem or Lie groups is needed. — P Vanchinathan, Apr 06 '19 at 04:39
I think this is sometimes called the “density theorem” for irreducible representations. — Joppy, Apr 06 '19 at 12:27
@PVanchinathan Indeed, it is a direct consequence of the Burnside's theorem on matrix algebras: the only irreducible subalgebra of the algebra of linear transformations on a finite-dimensional vector space over an algebraically closed field is the entire algebra. An elementary proof based on rank 1 matrices, which goes back to Halperin and Rosenthal, is in The simplest proof of Burnside's theorem on matrix algebras. See also Shapiro's notes. — Conifold, Sep 09 '23 at 12:08

score 4 · Answer 1 · answered Apr 06 '19 at 14:06

One way to see this should be the following: given two groups (you don't need them to be finite) $G,H$ with representation $(\rho ,V), (\sigma,W)$ one can define the representation $\tau=\rho \times \sigma$ (I think this is a terrible notation but I do not know the canonical one sorry).

The representation is defined as this $$\tau: G \times H \to End(V \otimes W)$$ $$(g,h) \to \rho(g) \otimes \sigma(h) .$$

If $\rho ,\sigma$ are irreducible so is $\tau$: this is not trivial but let's try this for a moment (with finite groups you can check this using characters but it's true without hte finiteness hypothesis).

So , now consider $H=G$ and $\sigma=\rho^*$ the dual representation: what you get is that the representation $\rho \times \rho^*$ of $G \times G$ on $V \otimes V^*$ is irreducible.

You have a canonical isomorphism $V \otimes V^* \cong End(V)$ and this turns out to be an isomorphism of representations (with $G \times G$ acting on $End(V)$ as $((g,h)f)(v)=g(f(h^{-1}v))$ for $v \in V$ and $f \in End(V)$.

So ,now this latter representation on $End(V)$ is irreducible: if you take the $G \times G$ span of the identity endomorphism, you will get the entire $End(V)$. Let's look what happens with $f=id$. One can $((g,h)f)(v)=(gh^{-1})(v)$ for every $v \in V$ so that taking the $G \times G$ span of the identity is just taking the span of the elements $\rho(g) \in End(V)$.

I like this proof because it is very slick, but I don't like it because it hardly generalizes to representations of algebras (for which the result still holds). At the moment I don't know how to prove the irreducibility result without using the desired result though — Maxime Ramzi, Apr 06 '19 at 15:41

score 2 · Answer 2 · edited Jun 12 '20 at 10:38

Here is a proof that works for representations of arbitrary algebras.

More precisely,

Let $A\subset \operatorname{End}(V)$ be a subalgebra, for any finite dimensional $k$-vector space, $k$ any algebraically closed field. Assume $V$ has no nontrivial $A$-stable subvector space. Then $A=\operatorname{End}(V)$.

You can then apply this to $\pi\mathbb{C}[G]$.

Here's a proof using the Jacobson density theorem. Let $f:V\to V$ be $A$-linear, that is, it commutes with every element of $A$. Then for $\lambda \in k$, $\ker (f-\lambda id)$ is $A$-stable. However, since $V$ is finite dimensional and $k$ is algebraically closed, there is $\lambda\in k$ such that it's not $0$. Therefore it must be $V$ for this $\lambda$: $f=\lambda id$. Therefore if $D=End(_A V)$, $D=k id$, and $D$-linearly independent means linearly independent. The Jacobson density theorem then allows us to conclude by taking $X$ in the above link to be any basis of $V$.

You can of course find more elementary proofs. A possible approach to a more elementary proof is to prove that by irreducibility, if $A$ contains one rank $1$ operator, it contains them all (by taking a look at what rank $1$ operators look like), thus if it contains one rank $1$ operator, then we're done. Then you can prove that if $a\in A\setminus \{0\}$ is not a homotethy, and $a$ has rank $r>1$, then there is $b\in A\setminus \{0\}$ with $\mathrm{rk}(b) < r$, and thus you are done. This second step uses eigenvalues, as above when we used the Jacobson density theorem.

Note that for the theorem to hold, you really need $k$ to be algebraically closed. Otherwise, let $P$ be an irreducible polynomial, and let $A= k[X]/(P)$, with the regular left action of $A$ on itself. This can be realized as group algebra (take $M$ to be the Compagnon matrix of $P$ over $k$, and $G= \langle M\rangle$, though $G$ may not be finite. If $P\mid X^n-1$ for some $n$, you can choose it to be finite). Then this is clearly irreducible (and finite dimensional over $k$), as a submodule would be an ideal of $A$, which is a field; but the image of $A$ is not the whole algebra $\operatorname{End}_k(k[X]/(P))$ : for starters, it's commutative, and it doesn't have a big enough dimension

If $(\pi,V)$ is irreducible, then $\pi(G)$ spans $\operatorname{End}(V)$

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