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I have a family of finite dimensional algebras that are directed quasihereditary. I think they might be Koszul algebras and I am wondering what approaches there are to check Koszulness or even quadraticity. I know the quivers of these algebras and can compute Ext^n between simple modules for all n, but I do not have a quiver presentation. I know that there are paths of length 2 and of higher lengths between all vertices of the quiver with nonvanishing Ext^2 so I cannot prove or eliminate quadraticity for trivial reasons. Any thoughts?

I should add that I do not have explicit minimal projective resolutions of the simple modules.

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    $\begingroup$ I think this is honestly too vague to get useful answers. How are you computing Ext^n of all the simples without finding minimal projective resolutions? Are your algebras graded, or are you looking for a grading that makes things Koszul? $\endgroup$
    – Ben Webster ♦
    Commented Jul 20, 2011 at 18:38
  • $\begingroup$ I am looking for a grading. I was hoping to use the grading coming from path length. I compute the Ext^n using classifying spaces. My algebras are monoid algebras. We can show that Ext between simple modules can be computed as the cohomology of certain submonoids with coefficients in some nice module. This in turns out to be the cohomology of a certain finite category with coefficients in the field. This we compute by using Quillen's theorem A to get to a nice simplicial complex. So basically we have no nice resolution. If anything we are implicitly using bar resolutions. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 21, 2011 at 1:34
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    $\begingroup$ Can you share a typical example of your algebras? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jul 21, 2011 at 20:54

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One incredibly useful fact is that it suffices to find linear resolutions of standard modules, not simples. These are usually much easier to find "by hand." Strictly speaking this is stronger than Koszul (the term is "standard Koszul") but in practice it seems rare for a quasi-hereditary algebra to be Koszul and not standard Koszul.

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  • $\begingroup$ Thanks. In my case the quasihereditary structure is directed, so the standard modules are projective indecomposables. I am not a specialist in finite dimensional algebras, but it would seem to me that in this case standard Koszul and Koszul should be the same. Is that correct? $\endgroup$
    – Benjamin Steinberg
    Commented Jul 21, 2011 at 1:38
  • $\begingroup$ I looked at the paper of Dlab et al, where standard Koszul is introduced. They say you need a linear resolution of both left and right standard modules. For directed quasihereditary algebras the standard modules on one side are projective and on the other side simple. So if I understood rightly, the standard Koszul is equivalent to Koszul in the directed case. They also point out in the paper it suffices to show the Ext algebra is generated in degrees zero and one. I might be able to determine this. I know how to explicitly construct all extensions of one simple by another. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 21, 2011 at 2:56

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