Consider a symmetric algebra $H$ over a field $k$. By definition, this is a $k$-algebra $H$ with a *symmetrizing trace* $\tau$, which is a $k$-linear map $\tau:H\to k$ such that $\tau(hh')=\tau(h'h)$ for all $h,h'\in H$ and the corresponding bilinear form is non-degenerate. I have been using chapter 7 of Characters of finite Coxeter groups and Iwahori-Hecke algebras by Geck and Pfeiffer as a reference for symmetric algebras.

According to Theorem 7.26 of this reference, if $H$ is split, then $H$ is semisimple if and only if all Schur elements are non-zero. The Schur element $c_V$ of a split simple $H$ module $V$ can be defined by $$ \sum_{b \in \mathcal{B}}\chi_V(b)\chi_V(b^\vee)=c_V \dim_k V, $$ where $\mathcal{B}$ is a $k$ basis of $H$, and $\lbrace b^\vee:b \in \mathcal{B}\rbrace$ is the basis dual to $\mathcal{B}$ under $\tau$.

If $H$ is split and $k$ has characteristic 0, are the Schur elements always non-zero? Equivalently, is a split symmetric algebra over a field of characteristic 0 always semisimple?

I assume that the answer is no, because I would have seen a result along these lines if it were true, but I can't find a counterexample.