Let $A$ be a quadratic algebra and $B$ the Ext-algebra of $A$. In case $A$ is a Koszul algebra, we should have that the global dimension of $A$ plus one is equal to the Loewy length of $B$ (is there a reference for this?).

Namely we should have $gldim(A)= \sup \{ i \geq 0 | Ext_A^i(A_0,A_0) \neq 0 \} = LL(B)-1$, where LL stands for Loewy length and $A_0$ is the degree zero part of the graded algebra $A$. Im not sure in general about the first equality here (it should at least hold for $A$ finite dimensional), but the second equality should be correct since $B$ is generated in degree 0 and 1.

Thus $gldim(A)+1=LL(B)$.

Question 1: Is $gldim(A)= \sup \{ i \geq 0 | Ext_A^i(A_0,A_0) \neq 0 \} $ true in general or under some restrictions? Is there a reference?

Question 2: Is a quadratic algebra Koszul iff $gldim(A)+1=LL(B)$ holds?

Maybe on needs to assume further restrictions for question 1, but I think in some form it will be true. One should be able to apply this in two nice examples:

a) $A=K[x_1,...,x_n]$ the polynomial ring in $n$ variables. Here $B$ is the Grassmann algebra in $n$ variables which has Loewy length $n+1$ and this shows that $A$ has global dimension $n$.

b)$A=kQ$ the quiver algebra of an arbitrary quiver with finitely many points and at least one arrow (that may be infinite dimensional). Then $B$ is the algebra with the same quiver and radical square zero and thus Loewy length 2. Thus the formula would give here that $A$ has global dimension one and I think the proof of this is actually quite complicated without those tools.

  • $\begingroup$ If you are assuming $A$ is finite dimensional and the grading is by path length of the quiver then the first equality is always true in question 1. I am not sure what conditions you need for the radical of the Ext algebra to be generated in degree one. $\endgroup$ – Benjamin Steinberg Dec 9 '18 at 15:31
  • $\begingroup$ Suppose $A$ is an exterior algebra on two generators, so $B$ is polynomial on two generators. The equality in question 2 will hold. If both generators are in degree 1, $A$ is Koszul, but if they are in different degrees, it is not. $\endgroup$ – John Palmieri Dec 9 '18 at 16:21
  • $\begingroup$ @JohnPalmieri Im not very experienced with this, but when one generator has degree larger than 1, then the relations seem to be non-quadratic or? $\endgroup$ – Mare Dec 9 '18 at 17:02
  • $\begingroup$ @Mare: you may be right. Does quadratic mean that the relations are in degree 2 or that the relations are in $V \otimes V$, where the algebra is a quotient of the tensor algebra on $V$? $\endgroup$ – John Palmieri Dec 9 '18 at 18:18
  • $\begingroup$ @JohnPalmieri I use the definition of quadratic as in definition 1.2.2. of ams.org/journals/jams/1996-9-02/S0894-0347-96-00192-0 . $\endgroup$ – Mare Dec 9 '18 at 19:48

A useful reference for answering your questions at least partially is Theorem 1.7 (especially part (5) of it) in the notes http://inmabb.criba.edu.ar/revuma/pdf/v48n2/v48n2a05.pdf .


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