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.

  • $\begingroup$ What is a split algebra? $\endgroup$ – Roman Fedorov Jun 27 '11 at 9:53
  • $\begingroup$ @Roman: A $K$-algebra $A$ is called split in Geck-Pfeiffer iff $End(S)=K$ for all simple modules of $A$. In other words: Iff $K$ is a splitting field for $A$. $\endgroup$ – Johannes Hahn Jun 27 '11 at 11:15

The answer is no. The reason is that every algebra can be embedded into a symmetric algebra, the so called trivial extension:

If $A$ is a $K$-algebra, then define $D(A):=A\oplus Hom_K(A,K)$. $I:=Hom_K(A,K)$ is a $A$-$A$-bimodule via

$a\cdot \phi \cdot b:=x\mapsto \phi(bxa)$

Hence you can define an $K$-algebra structure on $D(A)$ such that $I$ becomes an ideal with $I^2=0$. The multiplication is explicitly given by:

$(a+\phi)(b+\psi) := ab+a\cdot\psi+\phi\cdot b$

The trace form is given by

$(a+\phi,b+\psi) := \psi(a)+\phi(b)$

Now $I$ is a nilpotent ideal and therefore contained in the Jacobsen radical of $D(A)$. In particular do $A$ and $D(A)$ have the same simple modules and $D(A)$ is split iff $A$ is. But because $J(D(A))\neq 0$ the $K$-algebra $D(A)$ is never semisimple regardless of what field $K$ you started with.

  • $\begingroup$ Thanks for the nice answer! In the case $A = K$, $D(A)$ is just $K[x]/x^2$ and the trace is $(1,1) = (x,x) = 0, (1,x) = (x,1)=1$. I didn't realize this is a symmetric algebra. I will post a follow-up question about the case that the trace form is positive definite. $\endgroup$ – Jonah Blasiak Jun 27 '11 at 14:12
  • $\begingroup$ Here is the follow-up mathoverflow.net/questions/68947/… $\endgroup$ – Jonah Blasiak Jun 27 '11 at 17:40

As a concrete version of Johannes Hahn's answer:

Let $H=k[x]/(x^2)$ and $\tau:H\times H\to k:(a+bx,c+dx) \mapsto ad+bc$. Clearly $\tau$ is symmetric and it is non-degenerate. The only simple module is $H/(x) \cong k$, and it is absolutely irreducible, so $H$ is split. Of course $H$ is not semisimple, since $J(H) = Hx \cong k \neq 0$. This works for any field $k$, including those of characteristic 0 like $k=\mathbb{Q}$.

If one wants the single variable $\tau$, then $\tau:H\to k:a+bx\mapsto b$ has $\ker(\tau) = \{a \in k \}$ which contains no nonzero ideal, and $\tau((a+bx)(c+dx))=\tau(ac+(ad+bc)x) = ad+bc$ is again symmetric.

Basically this takes a non-semisimple group algebra like $\operatorname{GF}(2)[C_2]$, the group ring of a cyclic group of order 2 over a field of size 2, and rewrites it as an analogous algebra over another field.


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