TITLE: A character identity
QUESTION [18 upvotes]: This is related to my question, but it concerns a specific point of the proof of Schur's Theorem.
Let $G$ be a finite group and $\chi$ an irreducible character of $G$. Is it true that
$$\forall g\in G,\qquad\sum_{h\in G}\overline{\chi(h)}\chi(gh)=\frac{|G|}{\chi(e)}\,\chi(g)\quad?$$
This does not seem to be related to the orthogonality properties of the table of characters. This obviously true if $\chi$ is a linear character, or if $g=e$. I checked it for $G=\frak S_3$ and for one nonlinear character of $\frak S_4$.

REPLY [3 votes]: It is perhaps interesting to note that the Schur relations themselves do follow from the second orthogonality relations alone, in fact from orthogonality with the identity column. Start from that relation in the form $$\sum_{\chi\in\mathrm{Irr}(G)}\chi(1)\chi(gh)=\delta_{gh,1}|G|,$$ where $g,h\in G.$ Fix a matrix representation affording each $\chi\in\mathrm{Irr}(G)$ and label its matrix entries $\chi_{ij}(g)$ for $g\in G$ and $1\le i,j\le \chi(1).$ (The notation is slightly abusive as $\chi_{ij}$ depends on the choice of representation.) Let $I\subseteq\mathrm{Irr}(G)\times\mathbb N^2$ be the set of triples $(\chi,i,j)$ where $\chi\in\mathrm{Irr}(G)$ and $1\le i,j\le\chi(1),$ noting $|I|=|G|$ (from the above orthogonality relation with $gh=1$). Then $$\sum_{(\chi,i,j)\in I,(\psi,k,l)\in I}\frac{\chi(1)}{|G|}\chi_{ij}(g)\delta_{i,l}\delta_{j,k}\delta_{\chi,\psi}\psi_{kl}(h)=\delta_{gh,1}$$ for $g,h\in G,$ where $\delta_{a,b}$ is 1 if $a=b$ and 0 otherwise. This is just the above orthogonality relation again, as can be seen on collapsing the deltas.  Define $I\times G,$ $I\times I$ and $G\times G$ matrices $A,B$ and $\Delta$ by $$A_{(\chi,i,j),g}=\chi_{ij}(g), B_{(\chi,i,j),(\psi,k,l)}=\delta_{i,l}\delta_{j,k}\delta_{\chi,\psi}\frac{\chi(1)}{|G|}\text{ and }\Delta_{g,h}=\delta_{gh,1}.$$ Note $A,$ $B$ and $\Delta$ are all $|G|\times|G|$ matrices. The above is equivalent to $$A^TBA=\Delta.$$ Since $\Delta^2=1_{|G|\times |G|},$ all the matrices are invertible. Hence $B^{-1}=A\Delta A^T.$ But
$$B^{-1}_{(\chi,i,j),(\psi,k,l)}=\delta_{i,l}\delta_{j,k}\delta_{\chi,\psi}\frac{|G|}{\chi(1)}\text{ and }A\Delta A^T_{(\chi,i,j),(\psi,k,l)}=\sum_{g\in G}\chi_{ij}(g)\psi_{kl}(g^{-1}).$$ Equating these gives the Schur relations: $$\sum_{g\in G}\chi_{ij}(g)\psi_{kl}(g^{-1})=\delta_{\chi,\psi}\delta_{il}\delta_{jk}\frac{|G|}{\chi(1)}.$$ (If we choose the matrices $\psi_{kl}$ to be unitary, as we always may for a finite group, we can replace $\psi_{kl}(g^{-1})$ with $\bar\psi_{lk}(g).$)
As noted by Geoff Robinson, the idempotent formula follows from the Schur relations by multiplying by $\chi_{ji}(h^{-1}),$ summing over $i$ and $j$: $$\sum_{g\in G}\chi(gh)\psi_{kl}(g^{-1})=\delta_{\chi,\psi}\chi_{kl}(h)\frac{|G|}{\chi(1)}$$ and finally taking the trace on both sides.