Irreducible group representations: Does every endomorphism on the vector space represent an element of the group algebra?

Question

Let $G$ be a finite group. Let $V$ be a finite-dimensional complex vector space, and let $\rho : G \to \text{GL}(V, \mathbb C)$ be an irreducible representation of $G$ on $V$.

If $\mathbb C[G]$ denotes the group algebra of $G$ over $\mathbb C$ and if $\text{Hom}_{\mathbb C}(V, V)$ denotes the endomorphisms of $V$, then there exists a natural $\mathbb C$-algebra homomorphism $\theta : \mathbb C[G] \to \text{Hom}_{\mathbb C}(V, V)$ defined by $$ \theta \left( \sum_{g \in G} a_g \cdot g \right) = \sum_{g \in G} a_g \rho (g).$$

My question is: Is $\theta$ guaranteed to be surjective?

I ask this purely out of curiosity. This feels like something basic that I ought to know.

Some ideas:

• Since $\rho : G \to \text{GL}(V, \mathbb C)$ is an irreducible representation of $G$, it must be the case that $|G| \geq (\text{dim}_{\mathbb C} V)^2 = \text{dim}_{\mathbb C} \left( \text{Hom}_{\mathbb C} (V, V) \right)$. This suggests that my claim is likely to be true, but of course, this is not a proof.
• If $v$ is an arbitrary vector in $V$, then $\{ \rho(g) \cdot v : g \in G \}$ must span $V$ (otherwise $\rho$ wouldn't be an irreducible representation). Therefore, if $v'$ is any other vector in $V$, then there exists an element $x \in \mathbb C[G]$ such that $\theta\left(x\right) \cdot v = v'$. This is good to know, but this isn't enough to answer my question.
• If $g \in G$ is an element of order $k$, then the eigenvalues of $\rho(g)$ are $k$th roots of unity. Defining $x_l := \tfrac 1 k \sum_{n = 0}^{k - 1} (e^{-2\pi i l / k} g)^{n} \in \mathbb C[G]$, where $l \in \{ 0, 1, \cdots , k - 1 \}$, we find that $\theta(x_l)$ is a projection onto the eigenspace of $\rho(g)$ associated with the eigenvalue $e^{2\pi i l / k}$. Using this idea, we can construct quite a lot of projection maps from elements in $\mathbb C[G]$. This may turn out to be a useful strategy for generating elements in $\text{Hom}_{\mathbb C}(V, V)$ from elements of $\mathbb C[G]$, but I haven't been able to use this idea to my advantage.

Answer 1

Here are four perspectives, most mentioned in the comments.

$\bullet$ Tensor-Hom Adjunction. We have $\mathrm{End}(V)\cong V\otimes V^\ast$ as $(G\times G)$-reps. Since $V$ is a $G$-irrep, so is $V^\ast$, which means their tensor product is a $(G\times G)$-irrep. Showing the general implication $U$ $G$-irrep, $V$ $H$-irrep $\implies U\otimes V$ $(G\times H)$-irrep is a straightforward character theory calculation.

$\bullet$ Artin-Wedderburn Theorem. Putting the projections $\mathbb{C}[G]\to\mathrm{End}(U)$ together (over all irreps $U$) yields an isomorphism $\mathbb{C}[G]\cong\bigoplus\mathrm{End}(V)$. This also implies the Burnside irreducibility criterion: $U$ is an irrep iff $\mathbb{C}[G]\to\mathrm{End}(U)$ is onto. See a proof of AW here.

$\bullet$ Jacobson Density Theorem. This says that if $U$ is a simple right $R$-module and $D=\mathrm{End}_R(U)$ (which is a division ring by Schur's lemma) then for any $D$-linearly independent elements $u_1,\cdots,u_n$ and arbitrary elements $v_1,\cdots,v_n$ there is a $r\in R$ for which $v_1=ru_1,\cdots,v_n=ru_n$. We can apply this by treating a $G$-irrep $U\cong\mathbb{C}^n$ as a right $\mathbb{C}[G]$-module via $u\cdot g:=\rho(g)^{-1}u$, then $D$ is comprised of scalar matrices and we can pick $u_1,\cdots,u_n$ to be the standard basis and $v_1,\cdots,v_n$ to be the columns of any matrix.

$\bullet$ Double-Centralizer Theorem. Define $E=\mathrm{End}(U)$ for a $G$-irrep $U$. Let $R$ be the image of $\mathbb{C}[G]$ within $E$. Then $C_E(R)=\mathbb{C}\cdot I$ are the scalar matrices, by Schur's Lemma, so $C_E(C_E(R))=E$, but the DC Thm. says this is $R$ itself. (Note the DC Thm can be used to prove AW or JD).