5
$\begingroup$

Let $A=KQ/I$ be a finite dimensional quiver algebra of finite global dimension. Is it true that the dimension of $A/[A,A]$ is equal to the number of simples of $A$? Here $[A,A]$ is the vector space generated by all elements of the form $ab-ba$.

Note that it is known that in general for such $A$ that the dimension of $A/([A,A]+rad(A))$ is equal to the number of simple $A$-modules. Thus the question should be equivalent to asking whether we have $rad(A) \subseteq [A,A]$ in case $A$ has finite global dimension.

$\endgroup$
  • $\begingroup$ I am maybe missing something, but what if your quiver has one vertex, and your algebra $A$ is the algebra of polynomials? What does your suggested result want to say then? $\endgroup$ – Vladimir Dotsenko Mar 19 '20 at 8:31
  • 2
    $\begingroup$ @VladimirDotsenko A quiver algebra is finite dimensional for me. I add that assumption, thanks. $\endgroup$ – Mare Mar 19 '20 at 8:31
6
$\begingroup$

Yes. See the result of Section 2.5 of a wonderful paper of Bernhard Keller :

https://webusers.imj-prg.fr/~bernhard.keller/publ/ilc.pdf

(and the references therein).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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