Notation: Let $k$ be a field and let $A$ be a finite-dimensional semisimple $k$-algebra. For a finite-dimensional $A$-module $M$ with corresponding $k$-algebra homomorphis $\rho_M \colon A \to \operatorname{End}_k(M)$ its character is the $k$-linear map $\chi_M \colon A \to k$ given by $\chi_M(a) = \operatorname{tr}(\rho_M(a))$. Then $\chi_M|_{[A,A]} = 0$ and $\chi_M$ can be identified with a $k$-linear map $A/[A,A] \to k$, i.e. an element of $(A/[A,A])^*$.
Question: Can we construct on $(A/[A,A])^*$ a $k$-bilinear form $(-,-)$ such that $$ (\chi_M, \chi_N) = \dim_k \operatorname{Hom}_A(M,N) $$ for all finite-dimensional $A$-modues $M, N$?
This question is motivated by the following two special cases:
If $G$ is a finite group with $\operatorname{char}(k) \nmid |G|$ then for $A = k[G]$ we can identify $(A/[A,A])^*$ with the space of class functions on $G$ and define such a billinear form by $$ (\chi, \chi') = \frac{1}{|G|} \sum_{g \in G} \chi(g) \chi'(g^{-1}). $$
If $k$ is algebraically closed then the irreducible characters $\chi_1, \dotsc, \chi_n$ of $A$ form a $k$-basis of $(A/[A,A])^*$. We can then extend $(\chi_i, \chi_j) = \delta_{ij}$ (which needs to hold by Schur’s lemma) to a bilinear form $(-,-)$ on $(A/[A,A])^*$. Because every $A$-module is semisimple it then follows that the desired equality holds for all finite-dimensional $A$-modules $M, N$.
(This example feels somewhat unsatisfying because it shows the existence of such a bilinear form, but gives no better description of it.)
Some thoughts:
It sufficies that the desired equality holds for simple $A$-modules because every $A$-module is semisimple.
If the irreducible characters of $A$ were linearly independent then one could at least show the existence of $(-,-)$ as in the second example. But I only know that this holds for $k$ algebraically closed or $\operatorname{char}(k) = 0$.
The bilinear form $(-,-)$ should probably be symmetric because $\dim_k \operatorname{Hom}_A(M, N)$ is symmetric in $M, N$ (because they are semisimple).
Any help is welcome.
As you say, to get the existence of such a bilinear form, it would suffice to know that the irreducible characters are linearly independent. But actually, you need a bit less than this: you just need to know that the nonzero irreducible characters are linearly independent, and that if $M$ is a simple module with $\chi_M=0$, then $\dim_k \operatorname{End}(M)$ is divisible by the characteristic. This follows easily from Artin-Wedderburn. Specifically, by Artin-Wedderburn, $A$ is isomorphic to a product $\prod_iM_{n_i}(D_i)$ of matrix rings over division algebras $D_i$ over $k$, and the simple $A$-modules are $M_i=D_i^{n_i}$ with the canonical action of the matrix ring, with $\operatorname{End}(M_i)\cong D_i$. So, the character $\chi_{M_i}$ is supported on the $i$th coordinate of the product $\prod_iM_{n_i}(D_i)$, and it follows immediately that the $\chi_{M_i}$ which are nonzero are linearly independent. Moreover, if $\chi_{M_i}=0$, then this means the trace of every element of $M_{n_i}(D_i)$ over $k$ is $0$. Now if $e\in M_{n_i}(D_i)$ is the matrix over $D_i$ with $1$ in the top left corner and $0$ everywhere else, the trace of $e$ over $k$ is just $\dim_k D_i$, so $\dim_k \operatorname{End}(M_i)$ is divisible by the characteristic, as desired.