MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
What is a split algebra? – Roman Fedorov Jun 27 '11 at 9:53
@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 $A$. – Johannes Hahn Jun 27 '11 at 11:15
up vote 4 down vote accepted

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.

share|cite|improve this answer
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. – Jonah Blasiak Jun 27 '11 at 14:12
Here is the follow-up… – 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.