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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?

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    $\begingroup$ 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
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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.

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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 $0$. $\square$.

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  • $\begingroup$ 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. $\endgroup$ – DES-SupportsMonicaAndTransfolk 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 $\left( \sum g \right) V = 0$. $\endgroup$ – DES-SupportsMonicaAndTransfolk 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
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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 $\Bbb C[G]$).

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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).

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  • $\begingroup$ 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 $\mathrm{End}(V_i)$? $\endgroup$ – DES-SupportsMonicaAndTransfolk 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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