Consider the vector space $\mathbb C[G]$ of functions $f: G \longrightarrow \mathbb{C}$ where $G$ is a finite group, or equivalently a vector space of all formal linear combinations of elements of $G$ over the complex numbers. How can I prove the map $$\phi : \mathbb{C} [G] \longrightarrow \prod_{\rho} \text{End}(V_{\rho})$$ $$f \mapsto (\rho(f) )_{\rho}$$ where $\rho(f) = \sum_{g \in G} f(g) \rho(g)$ (where $\rho$ runs over all the isomorphism classes of irreducible representations of $G$) is both a ring isomorphism and also an intertwining isomorphism of $G \times G$ where our action, given $T \in \text{End}(V_{\rho})$, is $(g,h)\cdot T := \rho(g) \circ T \circ \rho(h)^{-1}$? We define ring multiplication by $(f_1 \ast f_2)(g) = \sum_{g = x y} f_1(x) f_2(y)$ in $\mathbb{C}[G]$. Does this isomorphism lead to a more natural decomposition of the regular representation $\mathbb{C}[G]$?

I know how to show that each map $\rho : G \longrightarrow \text{GL}(V_{\rho})$ extends to $\phi$ by linearity. I can also see how ring multiplication is respected by $\phi$. I also know that we can show the map is injective because $\dim \mathbb{C}[G] = |G|$, but also $$\dim (\prod_{\rho} \text{End}(V_{\rho}))=\sum_\rho \dim (\text{End}(V_\rho))=\sum_\rho \dim(V^{\ast}_\rho \otimes V_\rho) =$$ $$\sum_\rho \dim(V^{\ast}_\rho) \dim(V_\rho) =\sum_\rho \dim(V_\rho)^2 = |G|$$ by the theorem that the sum of the squares of the degrees of the irreducible representations is equal to the order of the group.

I can't seem to work out how as a linear transformation it intertwines with the action on $G \times G$ described above or how the convolution multiplication defined above is respected by $\phi$ as a ring isomorphism.

  • $\begingroup$ " is a not only a ring isomorphism, but also an intertwining isomorphism of [...]" - do you mean it is or isn't also an intertwining isomorphism of ...? $\endgroup$ Apr 21, 2014 at 5:19
  • $\begingroup$ I claim it is both a ring isomorphism and a vector space isomorphism that also commutes with the action described above on $\text{End}(V_\rho)$ (as a vector space) $\endgroup$
    – Matt R
    Apr 21, 2014 at 5:28
  • $\begingroup$ If the dimension of domain and codomain of a linear map between vector spaces are equal, then the map is a bijection. You were able to see injectivity this way but surjectivity follows just as well. As for how to check that the map is intertwining, you merely have to do the obvious: check that it satisfies the property of being intertwining! The criteria for determining when a map is intertwining is given right in the definition! $\endgroup$
    – anon
    Apr 21, 2014 at 5:40

1 Answer 1


$\def\CC{\mathbb C}$Notice that to prove injectivity of the map it is enough to show that for each $f\in\CC G$ there is a $\rho$ such that $\rho(f)\neq0$ in $\operatorname{End}(V_\rho)$.

There is a decomposition $\CC G=\bigoplus_{i=1}^r I_i$ of the algebra $\CC G$ as a direct sum of minimal left ideals. Indeed, $\CC G$ is a representation of $G$ on th left in the obvious way —the left regular representation— so it decomposes as a direct sum of irreducible subrepresentations, and you can easily check that these subrepresentations are in fact left ideals of $\CC G$.

Let $1$ be the unit element in $\CC G$. In view of the above decomposition, there are uniquely determined elements $e_i\in I_i$ for all $i$ such that $1=\sum_{i=1}^re_i$.

Now let $f\in\CC G\neq0$. Since $f=f\cdot 1=f\cdot\sum_ie_i=\sum_if\cdot e_i$, we see that there is an $i_0$ such that $f\cdot e_i\neq0$. This tells us that $f$ acts as non-zero on some irreducible representation of $G$..

  • $\begingroup$ OP stated they already had injective, and wanted to know about surjective. If they're using the fact that $\sum(\dim V)^2=|G|$ right from the outset then the fact that $\phi$ is bijective is a simple fact about vector spaces and requires no algebra. However this gives an elementary argument for injectivity that doesn't require something as powerful as that equation. $\endgroup$
    – anon
    Apr 21, 2014 at 5:57
  • $\begingroup$ Well, I assumed he is aware that in his case injectivity and surjectivity are equivalent (I want to believe that by the time one gets to representation theory of groups one is sufficiently acquainted with basic linear algebra!) so I chose to understand what he wrote as saying that he knows it is enough to show either one. $\endgroup$ Apr 21, 2014 at 5:59
  • $\begingroup$ OP explicitly says they know how to prove injectivity and is explicitly asking how to prove surjectivity, so I doubt while writing the question OP was aware of an equivalence. $\endgroup$
    – anon
    Apr 21, 2014 at 6:01
  • $\begingroup$ He does not explicitly say that at all; he merely observed that the domain and the codomain have the same dimension (and we can also assume that he is perfectly aware that this does not imply injectivity!) $\endgroup$ Apr 21, 2014 at 6:02
  • $\begingroup$ Hmm, so when OP says «I also know that we can show the map is injective», rather than explicitly telling us they know how to prove injectivity, they are implicitly telling us that they know proving injectivity is sufficient to proving surjectivity. That does make sense, and seems to be a clever deduction on your part. $\endgroup$
    – anon
    Apr 21, 2014 at 6:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .