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.

  • $\begingroup$ 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

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


(and the references therein).


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