# Quadratic algebras and Koszul algebras

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.

• 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. – Benjamin Steinberg Dec 9 '18 at 15:31
• 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. – John Palmieri Dec 9 '18 at 16:21
• @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? – Mare Dec 9 '18 at 17:02
• @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 $V$? – John Palmieri Dec 9 '18 at 18:18
• @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 . – Mare Dec 9 '18 at 19:48