A character identity 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 $\frak S_4$.
 A: This is correct. It does not follow immediately from the orthogonality relations for group characters, but both they and the above formula follow from the orthogonality relations for individual matrix coefficients with respect to irreducible unitary representations which follow from Schur's Lemma (and were known to Schur). I won't give all details as they can be found in most texts ( and basically generalize to finite dimensional unitary representations of Lie groups), but the basic ideas are: Let $A$ and $B$ be finite dimensional unitary irreducible representations of a finite group $G$ and set $A(g) = [a_{ij}(g)],B(g) = [b_{ij}(g)]$ for each $g \in G.$
For any (suitably sized) matrix $M$, set $T(A,B) = \sum_{g \in G} A(g^{-1})MB(g).$ Then $A(h^{-1})T(A,B)B(h) = T(A,B)$ for all $h \in G,$ and it follows from Schur's Lemma that $T(A,B)$ is the zero matrix if $A,B$ are inequivalent, and that $T(A,B)$ is a scalar matrix if $A,B$ are equivalent.
It follows ( taking different choices for $M$ with exactly one non-zero entry $1$), that we obtain $\sum_{h \in G} a_{ij}(h)\overline{b_{k\ell}(h))} = \frac{|G|}{{\rm dim} A}\delta_{ik}\delta_{j\ell}$ if $A,B$ are equivalent, or always $0$ if $A,B$ are inequivalent.
Now apply these finer relations in the case that $A = B$ affords character $\chi$, and observe that $\chi(h) = \sum_{ i= 1}^{\chi(1)} a_{ii}(h)$ while $\chi(gh) = \sum_{i=1}^{\chi(1)} \sum_{k = 1}^{\chi(1)} a_{ik}(g)a_{ki}(h),$ and you get the formula you want.
Note also that your formula only depends on the character afforded by the representation, so only uses the equivalence type of the representation, so it is fine to choose a unitary representation.
Later remarks: Note that the above method actually gives that $\sum_{h \in G} \chi_{i}(h^{-1})\chi_{j}(gh) = \delta_{ij}\frac{|G|\chi_{i}(g)}{\chi_{i}(1)}$ for irreducible characters $\chi_{i}$ and $\chi_{j},$ which (applying with $g = e$ (the identity of $G$) justifies my assertion that the finer Schur orthogonality relations for matrix coefficients imply the usual orthogonality relations for group characters.
Much later edit: I realised that a better way to answer this question is to use the semisimplicity of the group algebra $\mathbb{C}G$ and its decomposition into a direct sum of full matrix algebras and the decomposition of $1_{G}$ as a sum of mutually orthogonal primitive idempotents of $Z(\mathbb{C}G).$
Wedderburn theory tells us that if $G$ is a finite group with $k$ conjugacy classes, the $\mathbb{C}G$ is a direct sum $\bigoplus_{ \chi \in {\rm Irr}(G)}
M_{\chi(1)}(\mathbb{C})$ as $\chi$ runs through the distinct complex irreducible characters of $G.$
To each irreducible character $\chi$ of $G,$ there corresponds a primitive idempotent $e_{\chi}$ of $Z(\mathbb{C}G)$ which is just the identity element of the matrix algebra summand (labelled as) $M_{\chi(1)}(\mathbb{C}).$
This idempotent $e_{\chi}$ is $\frac{\chi(1)}{|G|}\left( \sum_{g \in G} \chi(g^{-1})g \right).$ This follows from the fact that the regular representation of $G$ affords character $\sum_{\chi} \chi(1)\chi.$ On the other hand, direct computation shows that (for any $g$) the coefficient of $g$ in $e_{\chi}e_{\chi}$ is given by $\frac{\chi(1)^{2}}{|G|^{2}}\left( \sum_{h \in G} \chi(h^{-1})\chi(hg^{-1}) \right).$
This gives ( for each $g \in G$) that 
$\frac{|G|\chi(g^{-1})}{\chi(1) } = \sum_{h \in G} \chi(h^{-1})\chi(hg^{-1}),$ as required.
A: 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.
A: Yes. Let 
$$ F(g) = \sum_{h \in G} \overline{\chi(h)}\chi(gh).$$
I claim that $F$ is a class function. Indeed,
$$F(s^{-1} g s) = \sum_{h \in G} \overline{\chi(h)}\chi(s^{-1}gsh)=\sum_{h \in G} \overline{\chi(h)}\chi(g(shs^{-1})) = \sum_{h' \in G} \overline{\chi(s^{-1}h's )}\chi(gh') = \sum_{h' \in G} \overline{\chi(h' )}\chi(gh') = F(g).$$
Thus, we can write $F$ as a combination of irreducible characters:
$$F(g) = \sum_{i} a_{\chi_i} \chi_i(g).$$
Let $\rho_{\chi_i}$ be the representation with character $\chi_i$. Then
$$a_{\chi_i} = \frac{1}{|G|} \sum_{g \in G} \overline{\chi_i(g)}F(g) = \frac{1}{|G|} \sum_{g,h} \overline{\chi_i(g)} \overline{\chi(h)}\chi(gh)= \frac{1}{|G|} \sum_{h} \overline{\chi(h)} (\mathrm{Tr}(  [\sum_{g} \overline{\chi_i(g)}\rho_{\chi}(g)]\rho_{\chi}(h) ).  $$
Letting
$$A_i = \sum_{g} \overline{\chi_i(g)}\rho_{\chi}(g),$$
we have $$(*)a_{\chi_i}  = \frac{1}{|G|} \sum_{h} \overline{\chi(h)} (\mathrm{Tr}( A_i\rho_{\chi}(h) ).$$ 
I claim $A_i$ commutes with $\rho_{\chi}(h)$ for all $h$-s:
$$A_i \rho_{\chi}(h) =\sum_{g} \overline{\chi_i(g)}\rho_{\chi}(gh) =\sum_{g'} \overline{\chi_i(g'h^{-1})}\rho_{\chi}(g')=\sum_{g'} \overline{\chi_i(h^{-1} g')}\rho_{\chi}(g')=\sum_{g} \overline{\chi_i(g)}\rho_{\chi}(hg)= \rho_{\chi}(h)A_i.$$
Thus, by Schur's lemma, $A_i$ is a multiple of the identity:
$$A_i = c_i I$$
for some constant $c_i$, which can be extracted by taking traces:
$$c_i \cdot \dim(\rho_i) = \mathrm{Tr}(A_i) = \sum_{g \in G}\overline{\chi_i}(g) \chi(g) = 1_{\chi_i = \chi} |G|.$$ 
Thus, when $\chi_i \neq \chi$ we have $A_i=0$ and so $a_{\chi_i}=0$ in that case by $(*)$, and $F(g)$ must be proportional to $\chi(g)$. To find the constant of proportionality, note that if $\chi_i=\chi$, we have from $(*)$ that
$$a_{\chi_i}=\frac{1}{\dim \rho_{\chi_i}}\sum_{h} \overline{\chi_i}(h) \chi(h) = \frac{|G|}{\chi(e)},$$
as needed.
