[proofplan] Choose orthonormal bases of the two representation spaces and write each character as the sum of its diagonal matrix coefficients. The product of the two characters is then a finite sum of products of matrix coefficients, so the integral may be computed term by term. Schur orthogonality for irreducible unitary representations annihilates every term in the non-isomorphic case, and in the isomorphic case we replace one representation by a unitarily equivalent copy and sum the diagonal contributions. [/proofplan]

[step:Write the characters as sums of diagonal matrix coefficients] Let $m=\dim_{\mathbb C}V$ and $n=\dim_{\mathbb C}W$. Choose an [orthonormal basis](/page/Orthonormal%20Basis) $(v_i)_{i=1}^{m}$ of $V$ and an orthonormal basis $(w_a)_{a=1}^{n}$ of $W$, with respect to the invariant Hermitian inner products on $V$ and $W$. Since $\rho(g)$ and $\sigma(g)$ are unitary linear maps, their traces in these orthonormal bases are \begin{align*} \chi_\rho(g)=\sum_{i=1}^{m}(\rho(g)v_i,v_i)_V \end{align*} and \begin{align*} \chi_\sigma(g)=\sum_{a=1}^{n}(\sigma(g)w_a,w_a)_W. \end{align*} Here the inner products are linear in the first argument, so complex conjugation gives \begin{align*} \overline{\chi_\sigma(g)}=\sum_{a=1}^{n}\overline{(\sigma(g)w_a,w_a)_W}. \end{align*} [/step]

[step:Expand the character inner product into matrix coefficient integrals]Because both sums are finite and each matrix coefficient is continuous on the compact group $G$, the products are integrable with respect to the normalized Haar measure $dg$. Hence finite additivity and linearity of the integral give \begin{align*} \int_G \chi_\rho(g)\overline{\chi_\sigma(g)}\,dg=\sum_{i=1}^{m}\sum_{a=1}^{n}\int_G(\rho(g)v_i,v_i)_V\overline{(\sigma(g)w_a,w_a)_W}\,dg. \end{align*}[/step]

[guided]The character is designed to package the diagonal matrix coefficients of a representation. After choosing orthonormal bases, we have \begin{align*} \chi_\rho(g)=\sum_{i=1}^{m}(\rho(g)v_i,v_i)_V \end{align*} and \begin{align*} \overline{\chi_\sigma(g)}=\sum_{a=1}^{n}\overline{(\sigma(g)w_a,w_a)_W}. \end{align*} Multiplying these two finite sums gives \begin{align*} \chi_\rho(g)\overline{\chi_\sigma(g)}=\sum_{i=1}^{m}\sum_{a=1}^{n}(\rho(g)v_i,v_i)_V\overline{(\sigma(g)w_a,w_a)_W}. \end{align*} Each summand is a product of continuous complex-valued functions on the compact topological group $G$, hence is continuous and therefore integrable with respect to the Haar probability measure $dg$. Since the sums are finite, no convergence theorem is needed; ordinary linearity of the integral gives \begin{align*} \int_G \chi_\rho(g)\overline{\chi_\sigma(g)}\,dg=\sum_{i=1}^{m}\sum_{a=1}^{n}\int_G(\rho(g)v_i,v_i)_V\overline{(\sigma(g)w_a,w_a)_W}\,dg. \end{align*} This reduction is the point at which characters become accessible: the desired character integral is now a finite collection of matrix coefficient integrals, exactly the setting of Schur orthogonality.[/guided]

[step:Apply Schur orthogonality in the non-isomorphic case] Assume first that $\rho\not\cong\sigma$. The hypotheses of [citetheorem:9715] apply: $G$ is compact with normalized Haar measure, and $(\rho,V)$ and $(\sigma,W)$ are irreducible finite-dimensional unitary representations. Therefore, for every $i\in\{1,\dots,m\}$ and $a\in\{1,\dots,n\}$, \begin{align*} \int_G(\rho(g)v_i,v_i)_V\overline{(\sigma(g)w_a,w_a)_W}\,dg=0. \end{align*} Substituting this into the finite expansion gives \begin{align*} \int_G \chi_\rho(g)\overline{\chi_\sigma(g)}\,dg=0. \end{align*} [/step]

[step:Reduce the isomorphic case to identical unitary representations] Assume now that $\rho\cong\sigma$. Let $A:V\to W$ be a representation isomorphism, so \begin{align*} A\rho(g)=\sigma(g)A \end{align*} for all $g\in G$. For each $g\in G$, this gives \begin{align*} \sigma(g)=A\rho(g)A^{-1}. \end{align*} Trace is invariant under conjugation by an invertible [linear map](/page/Linear%20Map), hence \begin{align*} \chi_\sigma(g)=\operatorname{tr}(\sigma(g))=\operatorname{tr}(A\rho(g)A^{-1})=\operatorname{tr}(\rho(g))=\chi_\rho(g). \end{align*} Thus the desired integral is unchanged if we replace $(\sigma,W)$ by the unitarily equivalent copy $(\rho,V)$. We may therefore compute in the case $W=V$ and $\sigma=\rho$, using the same orthonormal basis $(v_i)_{i=1}^{m}$ for both characters. [/step]

[step:Sum the diagonal Schur orthogonality contributions] In the reduced case $W=V$ and $\sigma=\rho$, [citetheorem:9715] gives, for all $i,j\in\{1,\dots,m\}$, \begin{align*} \int_G(\rho(g)v_i,v_i)_V\overline{(\rho(g)v_j,v_j)_V}\,dg=\frac{1}{m}(v_i,v_j)_V\overline{(v_i,v_j)_V}. \end{align*} Since the basis is orthonormal, $(v_i,v_j)_V=0$ when $i\ne j$ and $(v_i,v_i)_V=1$. Therefore \begin{align*} \int_G(\rho(g)v_i,v_i)_V\overline{(\rho(g)v_j,v_j)_V}\,dg=0 \end{align*} when $i\ne j$, while \begin{align*} \int_G|(\rho(g)v_i,v_i)_V|^2\,dg=\frac{1}{m} \end{align*} when $i=j$. Hence \begin{align*} \int_G|\chi_\rho(g)|^2\,dg=\sum_{i=1}^{m}\sum_{j=1}^{m}\int_G(\rho(g)v_i,v_i)_V\overline{(\rho(g)v_j,v_j)_V}\,dg=\sum_{i=1}^{m}\frac{1}{m}=1. \end{align*} Combining this with the non-isomorphic case proves \begin{align*} \int_G \chi_\rho(g)\overline{\chi_\sigma(g)}\,dg=\delta_{\rho,\sigma}. \end{align*} [/step]