$\newcommand{\refone}{\textbf{(1)}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\tr}{\operatorname{Tr}} \newcommand{\kk}{\mathbf{k}}$ Let $\kk$ be a field. Let $A$ be a $\kk$-algebra which is finite-dimensional as a $\kk$-vector space. Let $\tau : A \to \kk$ be a $\kk$-linear map. Define a $\kk$-bilinear form $\left<\cdot, \cdot\right> : A \times A \to \kk$ by $\left< a, b\right> = \tau\left(ab\right)$. **Assume that this form $\left<\cdot, \cdot\right>$ is symmetric and nondegenerate.** Let $\left(a_i\right)_{i \in I}$ and $\left(a'_i\right)_{i \in I}$ be two mutually dual bases of $A$ with respect to this form (so that $\left<a_i, a'_j\right>$ equals $1$ whenever $i = j$ and $0$ otherwise).

**Facts:** (See below for the proofs of these claims.)

**(1)** We have $\sum\limits_{i\in I} a_i \tau\left(a'_i b\right) = \sum\limits_{i\in I} a'_i \tau\left(a_i b\right) = b$ for every $b \in A$.

**(2)** We have $\sum\limits_{i\in I} a_i \tau\left(a'_i\right) = \sum\limits_{i\in I} a'_i \tau\left(a_i\right) = 1$.

**(3)** The tensor $\sum\limits_{i\in I} a_i \otimes a'_i \in A \otimes A$ (over $\kk$) does not depend on the choice of bases $\left(a_i\right)_{i \in I}$ and $\left(a'_i\right)_{i \in I}$.

**(3')** For any $b \in A$, we have $\sum\limits_{i\in I} b a_i \otimes a'_i = \sum\limits_{i\in I} a_i \otimes a'_i b$.

**(3'')** For any $b \in A$, we have $\sum\limits_{i\in I} a_i b \otimes a'_i = \sum\limits_{i\in I} a_i \otimes b a'_i$.

**(4)** Consider the dual space $A^*$ of $A$ as an $A$-$A$-bimodule in the usual way (so $\left(af\right)\left(b\right) = f\left(ba\right)$ and $\left(fa\right)\left(b\right) = f\left(ab\right)$ for all $a \in A$, $b \in A$ and $f \in A^*$). Let $J : A \to A^*$ be the map which sends every $b \in A$ to the $\kk$-linear map $\beta : A \to \kk$ given by $\beta\left(c\right) = \left<b,c\right>$. Then, $J$ is an isomorphism of $A$-$A$-bimodules. This makes $A$ into a so-called **symmetric Frobenius algebra**. (Conversely, any symmetric Frobenius algebra has a $\tau$ satisfying the conditions we imposed on $\tau$ above.)

From now on, let $E$ be a left $A$-module which is finite-dimensional as a $\kk$-vector space. For every $x \in A$, let $\tr_E x$ be the trace of the $\kk$-vector space endomorphism of $E$ given by the action of $x$ on $E$. Let $\gamma_E$ be the element $\sum\limits_{i\in I} \tr_E \left(a_i\right) a'_i$ of $A$.

**(5)** This element $\gamma_E$ does not depend on the choice of bases $\left(a_i\right)_{i \in I}$ and $\left(a'_i\right)_{i \in I}$.

**(6)** This element $\gamma_E$ lies in the center of $A$.

**(7)** If $E'$ is a simple left $A$-module such that $\Hom_A\left(E',E\right) = 0$ (this happens, for instance, if $E$ and $E'$ are nonisomorphic simple $A$-modules), then $\gamma_E E' = 0$.

**Question:**

Are the following true?

**(8)** If $E$ is a simple left $A$-module, then the element $\gamma_E$ is quasi-idempotent, meaning that $\gamma_E^2 = \lambda_E \gamma_E$ for some $\lambda_E \in \kk$.

**(9)** If $E$ is a simple left $A$-module, then the action of $\gamma_E$ on $E$ is a scalar multiple of the identity map.

**Remarks:**

One example of a situation as described above is when $A$ is the group algebra $\kk\left[G\right]$ of a finite group $G$, and $\tau$ is the map sending every element $a$ of this group algebra to the coefficient of $1 \in G$ in $a$. One can then take $\left(a_i\right)_{i\in I} = \left(g\right)_{g\in G}$ and $\left(a'_i\right)_{i\in I} = \left(g^{-1}\right)_{g\in G}$. This works no matter whether $A$ is semisimple or not.

The question was born in my attempt to lift the "split semisimple" condition in Proposition 19.2 of George Lusztig's *Hecke algebras with unequal parameters* (arXiv:math/0208154v2). Specifically I am trying to generalize parts (b) and (e) of said proposition; part (a) trivially generalizes and parts (c) and (d) don't generalize this far (at least semisimplicity cannot be disposed of). My main hope with this question is to obtain an alternative route to a piece of representation theory (specifically, character orthogonality formulas) in a greater generality than is done in standard texts, and in a lightweight way (no algebraic closure requirements, no use of Artin-Wedderburn, no characteristic-$0$ assumptions, and maybe even allowing $\kk$ to be any commutative ring).

Notice that the conjectural claims **(8)** and **(9)** are easily seen to be equivalent. I possibly can show **(8)** and **(9)** under the assumption that $A$ is semisimple, using some basic Galois theory (reducing the problem to the algebraic closure of $\kk$ and there applying Artin-Wedderburn); but I haven't fully worked out a proof and to be honest I am not interested in doing so (see my above motivation).

It is **not** generally true that the scalar in **(9)** is nonzero.

**Proofs:**

For the sake of completeness, here are my proofs for the facts above.

**(1)** Every $b \in A$ satisfies $\sum\limits_{i\in I} a_i \left<a'_i, b\right> = b$ (since $\left(a_i\right)_{i \in I}$ and $\left(a'_i\right)_{i \in I}$ are two mutually dual bases of $A$). Since $\left<a'_i, b\right> = \tau\left(a'_i b\right)$, this rewrites as $\sum\limits_{i\in I} a_i \tau\left(a'_i b\right) = b$. The same argument, with $\left(a_i\right)_{i \in I}$ and $\left(a'_i\right)_{i \in I}$ switched, shows that $\sum\limits_{i\in I} a'_i \tau\left(a_i b\right) = b$ for every $b \in A$ (because being dual bases with respect to a symmetric bilinear form is a symmetric relation). This proves **(1)**.

**(2)** This follows from setting $b = 1$ in **(1)**.

**(4)** The definition of $J$ shows that $\left(J\left(b\right)\right)\left(c\right) = \left<b, c\right>$ for all $b \in A$ and $c \in A$.

The map $J$ is injective. (In fact, if $b \in A$ is such that $J\left(b\right) = 0$, then every $c \in A$ satisfies $\left<b, c\right> = \underbrace{\left(J\left(b\right)\right)}_{=0}\left(c\right) = 0$, whence $b = 0$ (by the nondegeneracy of the form $\left<\cdot, \cdot\right>$).) It is also easy to see that the map $J$ is $\kk$-linear (since the form $\left<\cdot, \cdot\right>$ is $\kk$-bilinear).

Every $a \in A$, $b \in A$ and $c \in A$ satisfy

$\left(a J\left(b\right)\right) \left(c\right) = \left(J\left(b\right)\right) \left(ca\right) = \left<b, ca\right>$ (by the definition of $J$)

$= \tau\left(bca\right) = \left<bc, a\right> = \left<a, bc\right> $ (by the symmetry of the form $\left<\cdot,\cdot\right>$)

$= \tau\left(abc\right)$

and

$\left(J\left(ab\right)\right) \left(c\right) = \left< ab, c\right> = \tau\left(abc\right)$.

Hence, every $a \in A$, $b \in A$ and $c \in A$ satisfy $\left(a J\left(b\right)\right) \left(c\right) = \tau\left(abc\right) = \left(J\left(ab\right)\right) \left(c\right)$. In other words, every $a \in A$ and $b \in B$ satisfy $a J\left(b\right) = J\left(ab\right)$. Hence, the map $J$ is left $A$-linear. A similar (but simpler!) argument shows that $J$ is right $A$-linear. Hence, $J$ is a $A$-$A$-bimodule homomorphism. Since $J : A \to A^*$ is $\kk$-linear and injective, we conclude that $J$ is a $\kk$-vector space isomorphism (since $A$ and $A^*$ are finite-dimensional $\kk$-vector spaces of the same dimension). Therefore, $J$ is an $A$-$A$-bimodule isomorphism (since $J$ is a $A$-$A$-bimodule homomorphism). This proves **(4)**.

**(3)** Let $\rho : A \otimes A \to \kk$ be the $\kk$-linear map induced by the $\kk$-bilinear map $\left<\cdot,\cdot\right> : A\times A \to \kk$ (according to the universal property of a tensor product). Thus, $\rho\left(b\otimes c\right) = \left<b, c\right>$ for all $b \in A$ and $c \in A$. Let $J$ be as in claim **(4)**.

Since $A$ is a finite-dimensional $\kk$-vector space, we can identify $\left(A\otimes A\right)^*$ with $A^* \otimes A^*$. Thus, for example, $\left(J \otimes J\right) \left(\sum\limits_{i\in I} a_i \otimes a'_i\right) \in A^* \otimes A^*$ becomes an element of $\left(A\otimes A\right)^*$, that is, a $\kk$-linear map $A \otimes A \to \kk$.

Now, every $b \in A$ and $c \in A$ satisfy

$\underbrace{\left(\left(J \otimes J\right) \left(\sum\limits_{i\in I} a_i \otimes a'_i\right)\right)}_{=\sum\limits_{i\in I} J\left(a_i\right) \otimes J\left(a'_i\right)} \left(b \otimes c\right) = \left(\sum\limits_{i\in I} J\left(a_i\right) \otimes J\left(a'_i\right)\right) \left(b \otimes c\right)$

$= \sum\limits_{i\in I} \underbrace{\left(J\left(a_i\right)\right)\left(b\right)}_{= \left<a_i, b\right>} \otimes \underbrace{\left(J\left(a'_i\right)\right)\left(c\right)}_{= \left<a'_i, c\right> = \tau\left(a'_i c\right)} = \sum\limits_{i\in I} \left<a_i, b\right> \otimes \tau\left(a'_i c\right)$

$= \sum\limits_{i\in I} \left<a_i, b\right> \tau\left(a'_i c\right)$ (since we are identifying $\kk \otimes \kk$ with $\kk$)

$= \left<\underbrace{\sum\limits_{i\in I} a_i \tau\left(a'_i c\right)}_{= c \text{ (by }\refone{ )}}, b\right> = \left<c, b\right> = \left<b, c\right>$ (since the form $\left<\cdot,\cdot\right>$ is symmetric)

$= \rho\left(b \otimes c\right)$.

In other words, the two $\kk$-linear maps $\left(J \otimes J\right) \left(\sum\limits_{i\in I} a_i \otimes a'_i\right)$ and $\rho$ are identical on each pure tensor. Hence, these two $\kk$-linear maps must be equal. Thus, $\left(J \otimes J\right) \left(\sum\limits_{i\in I} a_i \otimes a'_i\right) = \rho$. Since $J \otimes J$ is an isomorphism (because so is $J$), we thus have $\sum\limits_{i\in I} a_i \otimes a'_i = \left(J \otimes J\right)^{-1} \left(\rho\right)$, and thus $\sum\limits_{i\in I} a_i \otimes a'_i \in A \otimes A$ (over $\kk$) does not depend on the choice of bases $\left(a_i\right)_{i \in I}$ and $\left(a'_i\right)_{i \in I}$. This proves **(3)**.

**(3')** Let $J$ be as in claim **(4)**. Let $b \in A$. Every $c \in A$ and $d \in A$ satisfy

$\underbrace{\left(\left(J \otimes J\right) \left(\sum\limits_{i\in I} b a_i \otimes a'_i\right)\right)}_{= \sum\limits_{i\in I} J\left(b a_i\right) \otimes J\left(a'_i\right)} \left(c \otimes d\right) = \left(\sum\limits_{i\in I} J\left(b a_i\right) \otimes J\left(a'_i\right)\right) \left(c \otimes d\right)$

$= \sum\limits_{i \in I} \underbrace{\left(J\left(b a_i\right)\right)\left(c\right)}_{= \left< b a_i, c\right> = \tau\left(b a_i c\right)} \otimes \underbrace{\left(J\left(a'_i\right)\right)\left(d\right)}_{= \left< a'_i, d\right> = \tau\left(a'_i d\right)} = \sum\limits_{i \in I} \tau\left(b a_i c\right) \otimes \tau\left(a'_i d\right)$

$= \sum\limits_{i\in I} \tau\left(b a_i c\right) \tau\left(a'_i d\right)$ (since we are identifying $\kk \otimes \kk$ with $\kk$)

$= \tau\left(b \underbrace{\sum\limits_{i\in I} a_i \tau\left(a'_i d\right)}_{= d \text{ (by }\refone{ )}} c\right) = \tau\left(bdc\right)$

and

$\underbrace{\left(\left(J \otimes J\right) \left(\sum\limits_{i\in I} a_i \otimes a'_i b\right)\right)}_{= \sum\limits_{i\in I} J\left(a_i\right) \otimes J\left(a'_i b\right)} \left(c \otimes d\right) = \left(\sum\limits_{i\in I} J\left(a_i\right) \otimes J\left(a'_i b\right)\right) \left(c \otimes d\right)$

$= \sum\limits_{i \in I} \underbrace{\left(J\left(a_i\right)\right)\left(c\right)}_{= \left< a_i, c\right> = \tau\left(a_i c\right)} \otimes \underbrace{\left(J\left(a'_i b\right)\right)\left(d\right)}_{= \left< a'_i b, d\right> = \tau\left(a'_i b d\right)} = \sum\limits_{i \in I} \tau\left(a_i c\right) \otimes \tau\left(a'_i b d\right)$

$= \sum\limits_{i\in I} \tau\left(a_i c\right) \tau\left(a'_i b d\right)$ (since we are identifying $\kk \otimes \kk$ with $\kk$)

$= \tau\left(\underbrace{\sum\limits_{i\in I} a_i \tau\left(a'_i b d\right)}_{= b d \text{ (by }\refone \text{, applied to } bd \text{ instead of } b \text{)}}\right) = \tau\left(bdc\right)$.

Thus, every $c \in A$ and $d \in A$ satisfy $\left(\left(J \otimes J\right) \left(\sum\limits_{i\in I} b a_i \otimes a'_i\right)\right) \left(c \otimes d\right) = \tau\left(bdc\right) = \left(\left(J \otimes J\right) \left(\sum\limits_{i\in I} a_i \otimes a'_i b\right)\right) \left(c \otimes d\right)$.

In other words, the two $\kk$-linear maps $\left(J \otimes J\right) \left(\sum\limits_{i\in I} b a_i \otimes a'_i\right)$ and $\left(J \otimes J\right) \left(\sum\limits_{i\in I} a_i \otimes a'_i b\right)$ are identical on each pure tensor. Hence, these two $\kk$-linear maps must be equal. Thus, $\left(J \otimes J\right) \left(\sum\limits_{i\in I} b a_i \otimes a'_i\right) = \left(J \otimes J\right) \left(\sum\limits_{i\in I} a_i \otimes a'_i b\right)$. Since $J \otimes J$ is an isomorphism (because so is $J$), we thus have $\sum\limits_{i\in I} b a_i \otimes a'_i = \sum\limits_{i\in I} a_i \otimes a'_i b$. This proves **(3')**.

**(3'')** One can prove **(3'')** in a similar way to **(3')** (with a little twist: one has to use symmetry of $\left< \cdot, \cdot \right>$ at one step). But we can alternatively prove **(3'')** using **(3')** as follows: Fix $b \in A$. Being dual bases with respect to a symmetric bilinear form is a symmetric relation. Thus, $\left(a'_i\right)_{i\in I}$ and $\left(a_i\right)_{i\in I}$ are two mutually dual bases of $A$ with respect to the form $\left<\cdot, \cdot\right>$. Hence, applying **(3')** to $a'_i$ and $a_i$ instead of $a_i$ and $a'_i$, we obtain $\sum\limits_{i\in I} b a'_i \otimes a_i = \sum\limits_{i\in I} a'_i \otimes a_i b$. Applying the $\kk$-linear map $A \otimes A, \ x \otimes y \mapsto y \otimes x$ to both sides of this equality, we obtain $\sum\limits_{i\in I} a_i \otimes b a'_i = \sum\limits_{i\in I} a_i b \otimes a'_i$. This proves **(3'')**.

**(5)** Let $\tr_E^{(1)}$ denote the $\kk$-linear map $A \otimes A \to A$ sending every $b \otimes c$ to $\tr_E\left(b\right) c$. (This is clearly well-defined.) Then, $\gamma_E = \sum\limits_{i\in I} \tr_E \left(a_i\right) a'_i$ is the image of the tensor $\sum\limits_{i\in I} a_i \otimes a'_i \in A \otimes A$ under the map $\tr_E^{(1)}$. Since said tensor does not depend on the choice of bases $\left(a_i\right)_{i \in I}$ and $\left(a'_i\right)_{i \in I}$ (by **(3)**), the same must hold for $\gamma_E$. This proves **(5)**.

**(6)** For every $b \in A$, we have $\gamma_E = \sum\limits_{i\in I} \tr_E \left(a_i\right) a'_i$ and thus

$\left< \gamma_E, b\right> = \left< \sum\limits_{i\in I} \tr_E \left(a_i\right) a'_i, b\right> = \sum\limits_{i\in I} \tr_E \left(a_i\right) \left<a'_i, b\right> = \tr_E \left(\sum\limits_{i\in I} a_i\underbrace{\left<a'_i, b\right>}_{= \tau\left(a'_i b\right)}\right) $

**(10)** $= \tr_E \left(\underbrace{\sum\limits_{i\in I} a_i \tau\left(a'_i b\right)}_{= b \text{ (by }\refone\text{)}}\right) = \tr_E \left(b\right) $.

Now, if $u$ and $v$ are two elements of $A$, then

$\left<\gamma_E u, v\right> = \tau\left(\gamma_E u v\right) = \left< \gamma_E, uv\right> = \tr_E \left(uv\right)$ (by **(10)**)

$ = \tr_E \left(vu\right)$ (since any two endomorphisms $\alpha$ and $\beta$ of a finite-dimensional $\kk$-vector space satisfy $\tr\left(AB\right) = \tr\left(BA\right)$)

$ = \left< \gamma_E, vu\right>$ (again by **(10)**)

$ = \left< vu, \gamma_E\right>$ (since $\left<\cdot,\cdot\right>$ is symmetric)

$ = \tau\left(vu\gamma_E\right) = \left< v, u \gamma_E\right> = \left< u\gamma_E, v\right>$ (since $\left<\cdot,\cdot\right>$ is symmetric).

Since the form $\left<\cdot,\cdot\right>$ is nondegenerate, this shows that $\gamma_E u = u \gamma_E$ for every $u \in A$. In other words, the element $\gamma_E$ of $A$ is central. This proves **(6)**.

**(7)** This proof is somewhat weird and I am not sure if I like it. Let $E'$ be a simple left $A$-module such that $\Hom_A\left(E',E\right) = 0$.

First of all,

**(11)** if $F$ is any simple **right** $A$-module, then the right $A$-module $F$ is isomorphic to a quotient of the right $A$-module $A$.

To prove this, let $F$ be any simple right $A$-module. Then, $F$ is nonzero (because it is simple). Fix any nonzero $x \in F$ (such an $x$ exists, since $F$ is nonzero). Then, the map $A \to F$ sending each $b \in A$ to $xb$ is right $A$-linear; its image is thus an $A$-submodule of $F$. Since the only $A$-submodules of $F$ are $0$ and $F$, this yields that the image of this map is $0$ or $F$. But it cannot be $0$ since it contains $x1 = x \neq 0$ (the image of $1$ under this map). Hence, it must be $F$, so that this map is surjective. We thus have found a surjective $A$-linear map $A \to F$; therefore, $F$ is isomorphic to a quotient of the right $A$-module $A$. This proves **(11)**.

Similarly to **(11)**, we can see that:

**(12)** if $F$ is any simple left $A$-module, then the left $A$-module $F$ is isomorphic to a quotient of the left $A$-module $A$.

We know that $E'$ is a simple left $A$-module, so that **(12)** shows that $E'$ is isomorphic to a quotient of the left $A$-module $A$. Hence, $\dim_{\kk} E' \leq \dim_{\kk} A < \infty$. Hence, the dual $E'^*$ of $E'$ is a simple right $A$-module (because $E'$ is a simple left $A$-module). We can thus apply **(11)** to $F = E'^*$, and conclude that the right $A$-module $E'^*$ is isomorphic to a quotient of the right $A$-module $A$. Dualizing again (we can do this because $\dim_{\kk} E' \leq \dim_{\kk} A < \infty$), we conclude that the left $A$-module $E'$ is isomorphic to a submodule of the left $A$-module $A^*$. Since the left $A$-module $A^*$ is isomorphic to the left $A$-module $A$ (by the map $J$, as **(4)** shows), this yields that the left $A$-module $E'$ is isomorphic to a submodule of the left $A$-module $A$. Since the claim that we want to prove is invariant under isomorphism of $E'$, we can thus WLOG assume that $E'$ **is** a submodule of the left $A$-module $A$. Assume this.

Let $w \in E'$ and $u \in A$. We are going to prove that $\left<\gamma_E w, u\right> = 0$. To do so, let us first check that $wuv = 0$ for every $v \in E$.

In fact, let $v \in E$. Then, $guv \in E$ for every $g \in E'$ (because $E$ is a left $A$-module). Hence, we can define a left $A$-linear map $E' \to E$ by sending each $g \in E'$ to $guv$. This map belongs to $\Hom_A\left(E',E\right) = 0$, and thus is zero. In other words, $guv = 0$ for every $g \in E'$. Applied to $g = w$, this yields $wuv = 0$.

Now, forget that we fixed $v \in E$. We have shown that $wuv = 0$ for every $v \in E$. In other words, $wu \in A$ acts as $0$ on the left $A$-module $E$. In particular, this yields $\tr_E \left(wu\right) = 0$.

Now,

$\left<\gamma_E w, u\right> = \tau\left(\gamma_E w u\right) = \left<\gamma_E, wu\right> = \tr_E\left(wu\right)$ (by **(10)**, applied to $b = wu$)

$= 0$.

Forget that we fixed $u$. We have shown that $\left<\gamma_E w, u\right> = 0$ for every $u \in A$. Hence, $\gamma_E w = 0$ (since the form $\left<\cdot,\cdot\right>$ is nondegenerate). Since this holds for all $w \in E'$, this yields $\gamma_E E' = 0$, and so **(7)** is proven.