MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

Let the bilinear trace form of a finite-dimensional associative algebra be defined as:

$(u,v) \mapsto Tr(L_u L_v)$

For $L_u$ the linear map given by multiplication on the left by $u$. In the literature, there seem to be good characterizations of algebras where this form is non-degenerate (semi-simple, special Frobenius), but what about the other extreme?

Is there a characterization for an associative algebra $A$ whose bilinear trace form, considered as a linear map $V \to V^*$ is rank 1?

In particular, when $A$ is unital and commutative, there seems to be only one possibility for the trace form, if its rank 1:

$(u,v)\mapsto \frac{1}{dimV} Tr(L_u)Tr(L_v)$

This seems like quite the coincidence. Could anyone shed some light on this?

share|cite|improve this question

Note that $Tr(L_uL_v)=Tr(L_{uv})$. Assume characteristic $0$. The kernel of trace has codimension $1$, except in the case of the algebra $0$. This kernel contains the ideal $I$ consisting of all those $u$ such that for all $v$ we have $Tr(L_{uv})=0$, so if that ideal has codimension $1$ then it's equal to $ker(Tr)$. Then $I$, considered as a (nonunital) algebra in its own right, is such that the trace of multiplication by any element is zero. Conversely, given any nonunital algebra with that property, adjoin a $1$ to make it unital and you've got an example of what you're asking about. In such an example, $Tr(L_{uv})$ is proportional to $Tr(L_u)Tr(L_v)$ even without commutativity, but note that it's not equal; the trace of the identity is the vector space dimension.

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.