Alternate proof of Schur orthogonality relations 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?
 A: 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 $\Bbb C[G]$).
A: 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.
A: 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 $0$. $\square$.
A: 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).
