I am trying to find an alternate proof for Schur orthogonality relations along the following lines.

Let $G$ be a finite group, with irreducible representations $V_1$, $V_2$, $\cdots$, $V_d$.

Let $V$ denote the adjoint representation ($V = \langle \{e_g\}_{g\in G} \rangle$ with $h\cdot e_g = e_{hgh^{-1}}$). Also, let $W$ denote the representation given by $\bigoplus_{i=1}^{d} \text{End} (V_i)$.

The Schur orthogonality relations tell us these both have the same character, hence are isomorphic. But this requires one to assume the fact that $\{\chi_{V_i} \}_{i=1..d}$ forms a basis for the class functions on $G$.

I would like to show $V \cong W$ by some other means, maybe an explicit isomorphism. This would prove Schur orthogonality without knowing the fact about the basis for class functions. How might this be done?

  • 5
    $\begingroup$ This is a better question for math.stackexchange... but as always with questions about proofs in the beginnings of a theory, you need to tell us what facts you already know and are not afraid to use. Representation theory is a network of interconnected theorems, and every book takes a slightly different route through it, visiting them in a different order. $\endgroup$ – darij grinberg Nov 18 '14 at 5:18
  • $\begingroup$ I am okay with using the fact that the irreducible characters form an orthonormal set, but I want to avoid using that they are a basis for the class functions (as this will provide an alternate proof that they are a basis). However, preferably no character theory at all. $\endgroup$ – Sameer Kailasa Nov 18 '14 at 6:04
  • $\begingroup$ @Sameer: I'd take darij's comment seriously. Also, your terminology "adjoint representation" seems less standard than "conjugating representation" (or "conjugation representation"). $\endgroup$ – Jim Humphreys Nov 18 '14 at 15:48

This follows immediately from the Artin-Wedderburn theorem. The isomorphism is given by the action of the group algebra on each of the simple representations. No character theory is needed.


You are asking about the adjoint (conjugation) action, but a better framing is that $$kG \cong \bigoplus_{i=1}^d \mathrm{End}(V_i)$$ as $G \times G$ reps. Here $G \times G$ acts by left and right multiplication (with $g^{-1}$ on the right, to make a left-action.) Your statement is then restricting to the diagonal $G$ inside $G \times G$. Write $\rho_i : G \to GL(V_i)$ for the representation.

There is an obvious map $\phi: kG \to \bigoplus \mathrm{End}(V_i)$, given by linearly extending the map $g \mapsto (\rho_1(g), \rho_2(g), \dots, \rho_n(g))$, and it is obviously a map of $G \times G$ representations.

Lemma 1 $\mathrm{End}(V_i)$ is irreducible as a $G \times G$ rep.

This doesn't seem to have a really clean proof, as it is the place where we use that $k$ is algebraically closed and that $V_i$ is irreducible. See here for the standard character theory argument.

Lemma 2 If $i \neq j$, then $\mathrm{End}(V_i)$ and $\mathrm{End}(V_j)$ are not isomorphic as $G \times G$ reps.

Proof If they were isomorphic as $G \times G$ reps, they'd be isomorphic as $G \times \{ 1 \}$ reps. But, as $G \times \{ 1 \}$-reps, we have $\mathrm{End}(V_i) \cong V_i^{\oplus \dim V_i}$, so this would violate uniqueness of decomposition into irreps. $\square$.

We now show that $\phi$ is an isomorphism.

Surjectivity Since $\bigoplus_{i=1}^d \mathrm{End}(V_i)$ is a direct sum of nonisomorphic irreps of $G \times G$, the image of $\phi$ must be of the form $\bigoplus_k \mathrm{End}(V_{i_k})$ for some subset $\{ i_1, i_2, \ldots, i_r \}$ of $\{ 1, \ldots, d \}$. So it is enough to show that the projection of $\phi(kG)$ onto $\mathrm{End}(V_i)$ is nonzero for each $i$. The identity of $G$ maps to the identity of $\mathrm{End}(V_i)$, $\square$.

Injectivity Suppose that we had some $\sum_g a_g g$ in $kG$ such that $\sum a_g \rho_i(g) = 0$ for all $i$. Any representation of $G$ is a direct sum of irreps so, by linearity, we would have $\sum_g a_g \rho(g) = 0$ for any representation $\rho: G \to GL(W)$ of $G$. In particular, this would be true for the regular representation. But, if $\sum_g a_g g$ acts by $0$ on the regular representation, then all $a_g$ are $0$. $\square$.

  • $\begingroup$ If $\rho : G \to\text{GL}(v)$ is irreducible, then Schur's lemma tells us $\sum_{g\in G} \rho(g)$ is 0. So wouldn't $\phi(\sum_{g\in G} g) = (0, 0, \cdots, 0)$, contradicting injectivity? $\endgroup$ – Sameer Kailasa Nov 18 '14 at 21:27
  • $\begingroup$ @SameerKailasa, That is not the content of Schur's lemma. Schur's lemma says that $\sum g$ acts as a scalar. That scalar need not (and usually is not) $0$. $\endgroup$ – David Hill Nov 18 '14 at 22:05
  • $\begingroup$ @DavidHill Well, it usually is $0$. More precisely, for $V$ irreducible, we have $\sum \rho_V(g)=0$ if and only if $V$ is not trivial. (Schur's lemma tells us that $\sum \rho_V(g)$ is a scalar, say $a$. Taking traces gives $(\dim V) a = \sum \chi_V(g)$ and the latter is $0$ by character orthogonality if $V$ is not the trivial rep.) But $\sum g$ does not act by $0$ on the trivial rep, so injectivity is saved. $\endgroup$ – David E Speyer Nov 19 '14 at 0:23
  • $\begingroup$ Better proof: $G$ clearly acts trivially on $\left( \sum g \right) V$. So, if $V$ is irreducible and non-trivial, then $\left( \sum g \right) V = 0$. $\endgroup$ – David E Speyer Nov 19 '14 at 0:27
  • $\begingroup$ @DavidSpeyer I think DavidHill's point was that a "generic representation" probably has some invariants. $\endgroup$ – Ben Webster Nov 19 '14 at 13:06

There is a natural map $\Bbb C[G]\to\bigoplus{\rm End}_{\Bbb C}(V)$ induced from the maps $G\to{\rm GL}(V)$. To see that it's an isomorphism, it suffices to check injectivity and equality of dimensions. Injectivity is easy: if any element $x\in{\Bbb C}[G]$ were mapped to $0$ then it would act as $0$ on any irrep, hence any rep, hence act trivially on $\Bbb C[G]$ itself, so $x\cdot e_G=0$ tells us $x=0$. Now for dimensions...

We have $\Bbb C[G]\cong\bigoplus V^{\oplus m(V)}$ for some unknown multiplicities $m(V)$. Convince yourself that

$$\hom_G(\Bbb C[G],W)\cong W$$

are isomorphic via the canonical map $\phi\mapsto\phi(1)$ for any rep $W$. Furthermore $\hom$ is distributive in both of its arguments (e.g. $\hom_G(A,B\oplus C)\cong\hom_G(A,B)\oplus\hom_G(A,C)$; what do you think the natural isomorphism will be?). Therefore we can compute (via Schur's on $\hom_G(V,W)$):

$$W \cong \hom_G(\Bbb C[G],W) \cong\bigoplus_V \hom_G(V,W)^{\oplus m(V)} \cong \Bbb C^{\oplus m(W)} $$

as vector spaces, where the last $\cong$ sign assumes that $W$ is an irreducible representation (the rest holds for every $W$). Therefore $m(W)=\dim W$ for all irreps $W$, and now we compute

$$\begin{array}{lc} \Bbb C[G] & \cong\hom_G(\Bbb C[G],\Bbb C[G]) \cong\bigoplus_{V,W}\hom_G(V,W)^{\oplus m(V)\cdot m(W)} \\[4pt] & \cong\bigoplus_V \Bbb C^{\oplus m(V)^2} \cong \bigoplus{\rm End}_{\Bbb C}(V) \end{array}$$

as vector spaces, yielding equality of dimensions. Thus $\Bbb C[G]\cong\bigoplus{\rm End}_{\Bbb C}(V)$ as algebras.

Fix an irrep $V$ and let $e_V\in{\Bbb C}[G]$ correspond to ${\rm Id}_V\in{\rm End}_{\Bbb C}(V)$ and $0$ in all of the other coordinates. Write $e_V=\sum c_gg$, "rotate" its coefficients as $e_Vh^{-1}=\sum c_g gh^{-1}$, then to both sides apply ${\rm tr}_{\Bbb C[G]}$ to get $m(V)\chi_V(h^{-1})=|G|c_h$, hence $e_V=\frac{\dim V}{|G|}\sum\chi_V(g^{-1})g$.

Now the orthogonality relations simply state $e_Ve_W=\delta_{VW}e_V$ (equate coefficients in $\Bbb C[G]$).


Maybe it's worth teasing out the details of Ben's answer since character theory is lurking beneath Lemma 1 in David's answer. Here, the statement to prove is that $$kG\cong\bigoplus_i\mathrm{End}(V_i)$$ as rings. We'll adopt the notation ${}_{kG}kG$ for the left regular module.

Step 1: There is a ring isomorphism $kG\cong\mathrm{End}_G({}_{kG}kG)^{\mathrm{op}}$ (for a ring $R$, the opposite ring $R^{\mathrm{op}}$ is the same as $R$ as an abelian group, but with multiplication $r*s=sr$).

Proof: The isomorphism is induced by the map $g\mapsto \mu_g$, where $\mu_g(x)=xg$.

Step 2 (Maschke's Theorem) If $\mathrm{char}k$ does not divide $|G|$, then ${}_{kG}kG$ is semisimple.

Proof: Let $X\subset{}_{kG}kG$ be a submodule. If $\mathrm{char}k=0$, then you have $${}_{kG}kG=X\oplus X^{\perp}$$ where $X^{\perp}$ is the orthogonal compliment with respect to the $G$-invariant inner product $\langle g,h\rangle=\delta_{gh}$.

For more general $k$, let $\pi:{}_{kG}kG\longrightarrow X$ be the projection onto $X$ along some vector space compliment. This may not be a $G$-map, so define $$\varpi(x)=|G|^{-1}\sum_{g\in G}g^{-1}\pi(gx).$$ It is straightforward to check that $\varpi$ is another projection onto $X$ which IS a $G$-map. Therefore, ${}_{kG}kG=X\oplus\ker\varpi$ as required.

Step 4: Using Maschke's Theorem, write ${}_{kG}kG=\bigoplus_i V_i^{n_i}$.

Step 5 (Schur's lemma) If $V$ and $W$ are irreducible, then any nonzero $G$-map $\phi:V\to W$ is an isomorphism. In particular, if $k$ is algebraically closed $\mathrm{End}_G(V)\cong k$.

Step 6: Observe that if $V$ is irreducible, then $\mathrm{End}_G(V^{\oplus n})\cong M_n(k)$. Consequently (using Schur's lemma again) \begin{align*} {}_{kG}kG&\cong\mathrm{End}_G\left(\bigoplus_iV_i^{n_i}\right)^{\mathrm{op}}\\ &\cong\left(\bigoplus_i\mathrm{End}_G\left(V_i^{n_i}\right)\right)^{\mathrm{op}}\\ &\cong\bigoplus_iM_{n_i}(k)^{\mathrm{op}} \end{align*}

Two final observations: (1) The map $A\mapsto A^t$ defines an isomorphism $M_n(k)\cong M_n(k)^{\mathrm{op}}$.

(2) It follows from this construction that $\dim V_i=n_i$, so $M_{n_i}(k)\cong\mathrm{End}(V_i)$.

No character theory.

Edit: In light of David Speyer's comment, let me explain observation (2) since it is not trivial. First, observe that for each $i$, $B_i=V_i^{n_i}$ is a minimal 2-sided ideal in $kG$. Indeed, it is a left ideal by definition, and for $i\neq j$, $V_i\cap V_j=0$ so $V_iV_j=V_jV_i=0$. This forces $B_iB_j=B_jB_i=0$, so $B_i$ is a 2-sided ideal.

To see that the $B_i$ are minimal note that if $J$ is any 2-sided ideal, then $J$ contains $V_i$ for some $i$ by the semisimplicty of $kG$. Letting $V_i'$ denote another copy of $V_i$ in $kG$, the composition $$\eta:{}_{kG}kG\twoheadrightarrow V_i\cong V_i'\hookrightarrow _{kG}kG$$ is a $G$-module map (here we take a $G$-module isomorphism $V_i\cong V_i'$). Then, by step 1$V_i'=\eta(V_i)=V_iy\subset J$ for some $y\in kG$. This shows that $B_i\subset J$. Using semisimplicity, you now can show that $J$ is a direct sum of a bunch of $B_i$.

Okay, but now we are done. Since, under the isomorphism above $B_i$ maps isomorphically onto $$\mathrm{End}_{B_i}(B_i)^{\mathrm{op}}=\mathrm{End}_G(V_i^{n_i})^{\mathrm{op}}\cong M_{n_i}(k).$$ (in the first equality we use the fact that $B_i$ commutes with $B_j$ for $j\neq i$). In other words, the minimal left ideals in $B_i$ map isomorphically onto the minimal left ideals in $M_{n_i}(k)$ giving observation (2).

  • $\begingroup$ I'm probably being dumb. At the end of step 6, you know that $kG \cong \bigoplus M_{n_i}(k)$. How do you equate $M_{n_i}(k)$ with $\mathrm{End}(V_i)$? $\endgroup$ – David E Speyer Nov 20 '14 at 18:26
  • $\begingroup$ @DavidSpeyer No, this is a fair question. I've added a proof above. $\endgroup$ – David Hill Nov 20 '14 at 20:09
  • $\begingroup$ The first (characteristic-$0$) proof of Maschke's theorem given in your response does not hold in the generality claimed (not a big deal, since proving Maschke was hardly the point of the question, and you give a better proof a few lines further below). $\endgroup$ – darij grinberg Nov 20 '14 at 21:06
  • $\begingroup$ @darijgrinberg I don't see anything wrong with what I proved. Can you explain? $\endgroup$ – David Hill Nov 20 '14 at 21:08
  • $\begingroup$ $X$ and $X^\perp$ might intersect nontrivially. $\endgroup$ – darij grinberg Nov 20 '14 at 21:09

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