Let $k$ be a field, and $G$ a finite group such that the characteristic of $k$ does not divide $|G|$. Then $kG$ is a semisimple $k$-algebra, and the representation theory of $G$ over $k$ is semisimple.

Fix an irreducible representation $V$ of $kG$. Then for any representation $W$, define the isotypic component of $W$ as the sum of all subspaces isomorphic to $V$, denoted $W[V]$. $$ W[V] = \sum_{\substack{U \subseteq W \\ U \cong V}} U$$ One can check that the assignment $W \mapsto W[V]$ is functorial. Define the element $\psi_V \in kG$ by the formula $$ \psi_V := \frac{\dim V}{|G|} \sum_{g \in G} \chi_V(g) g^{-1}$$ where $\chi_V: G \to k$ is the character of $V$, defined as $\chi_V(g) = \operatorname{tr} \rho_V(g)$. Then $\psi_V$ acts in $W$ as the $G$-equivariant projector onto $W[V]$, in the sense that $\rho_W(\psi_V) \in \mathrm{End}_k(W)$ is $G$-equivariant, and projects to the subspace $W[V]$.

Question: Is there a "nice" explanation of the formula defining $\psi_V$?

Here is the direction I would like "nice" to go in. The functor $W \mapsto W^G = W[\text{trivial}]$ taking a representation to its subspace of invariants can equivalently be thought of as the map $W \mapsto \psi_1 W$, where $\psi_1 \in kG$ is the element $$ \psi_1 = \frac{1}{|G|} \sum_{g \in G} g $$ and checking this fact is very easy. We also have that the "$V$-isotypic component functor" $W \mapsto W[V]$ is isomorphic to the functor $$ W \mapsto V \otimes_k \operatorname{Hom}_{kG}(V, W)$$ where we consider the $k$-vector space $\operatorname{Hom}_{kG}(V, W)$ to have trivial $G$-action. We can follow this second functor up with the evaluation map $$ \operatorname{ev}: V \otimes_k \operatorname{Hom}_{kG}(V, W) \to W, \quad v \otimes f \mapsto f(v) $$ to get an actual subspace of $W$ as the image of $\operatorname{ev}$. Moreover, using the various isomorphisms/identities:

  • $\operatorname{Hom}_k(V, W)$ is a $G$-representation via $(g \cdot f)(v) = g f(g^{-1} v)$.
  • $\operatorname{Hom}_k(V, W)^G = \operatorname{Hom}_{kG}(V, W)$
  • $\operatorname{Hom}_k(V, W) \cong V^* \otimes W$ as $G$-representations.

We can get to the fact that the image of $\operatorname{ev}$ must be the image of

$$ V \otimes V^* \otimes W \twoheadrightarrow V \otimes (V^* \otimes W)^G \hookrightarrow V \otimes V^* \otimes W \xrightarrow{\operatorname{ev} \otimes 1} W$$ which on a pure tensor $v \otimes f \otimes w$ is (by using $\psi_1$) $$v \otimes f \otimes w \mapsto \frac{1}{|G|} v \otimes \sum_{g \in G} gf \otimes gw \mapsto \frac{1}{|G|} \sum_{g \in G} f(g^{-1} v) (gw)$$

This looks so close to $\psi_V$ that I feel like this must be on the right track, but I'm quite stuck taking this further. I would very much like the map $W \mapsto V \otimes \operatorname{Hom}_{kG}(V, W)$ to just lead to a formula for $\psi_V$, without having to know a bunch of orthogonality-of-characters stuff. (I am happy to accept the category is semisimple).

  • $\begingroup$ I don't have time to write much right now, but I think this is a good direction. In general, if we use this averaging to turn a linear map $f$ into a homomorphism (as we do here) and focus on the case where the representation is irreducible, we end up with a scalar, and that scalar is precisely $\frac{1}{\dim(V)}\operatorname{Tr}(f)$. $\endgroup$ Oct 4, 2018 at 14:06
  • $\begingroup$ Old post, but your formula for $\psi_V$ doesn't hold in the stated level of generality. It's easy to see that your formula implies the orthogonality relations by inspecting coefficients of $\delta_{ij} \psi_{V_i} = \psi_{V_j} \psi_{V_i}$. However, if you start with $V$ not absolutely irreducible and pass to the algebraic closure, you get $\langle V, V\rangle > 1$, a contradiction. The trouble is you need to use $\psi_V := \frac{\dim(V)}{|G|} \sum_{g \in G} \frac{\chi_V(g^{-1})}{\dim_k \mathrm{End}(V)} g$ where $\mathrm{End}(V)$ is the division ring of $kG$-linear endomorphisms of $V$. $\endgroup$ Jul 23, 2019 at 7:46
  • $\begingroup$ ...and I can't find a reference for this written down anywhere. If you inspect the argument here, you find the algebraic closure assumption is used to get the division rings to be just $k$. (Characteristic zero only seems important when deducing the orthogonality relations--you need to cancel some $\dim V_i$'s.) $\endgroup$ Jul 23, 2019 at 7:50
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    $\begingroup$ @JSwanson Thanks for that. As you pointed out, it's important for the argument that the characters of irreducible representations form an orthonormal basis rather than just an orthogonal basis, and we need the representations of $G$ to completely split over the base field for that. Alternatively we could correct by dividing each character by $\dim_k \operatorname{End}_{kG}(V)$, which should always work even in characteristic $p$ (I think...) $\endgroup$
    – Joppy
    Jul 24, 2019 at 1:57
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    $\begingroup$ Yes, my formula works even in characteristic p in the non-modular case by essentially the argument in the PDF I linked--you only have to divide by $|G|$. Looks like you'd have to use Brauer characters in the modular case or for the orthogonality relations outside of characteristic 0. I'm not sure if this extra term would help your original argument. For what it's worth, the Artin-Wedderburn argument in the PDF I linked does derive the formula very elegantly "without having to know a bunch of orthogonality-of-characters stuff" from semisimplicity. Wish it was in a textbook.... $\endgroup$ Jul 24, 2019 at 5:13


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