Is algebra: ac=ca, bd = db , ad - da = cb - bc ("Manin matrix algebra") - a Koszul algebra?

Question

Question: Consider quadratic algebra with four generators $a,b,c,d,d$ and three relations $ac=ca,bd = db, ad-da = cb - bc$ . Is it a Koszul algebra ? (i.e. Koszul complex is resolution of ground field $k$). (Checking algebra is Koszul, is something algorithmic or more kind of creative task ? )

Context: The relations above first appeared in Yuri Manin's papers on "Noncommuative geometry and quantum groups" (around 1986-7). And define "Manin matrices", in the simplest 2x2 case it looks like that, consider 2x2 matrix: $$\begin{pmatrix} a & b \\\ c & d \end{pmatrix}$$ The relations above ensure that despite non-commutativity - such matrices behave like commutative ones - one can define the determinant: $ad-cb$ , it will be multiplicative, Cramer's inversion formula works, Cayley-Hamilton, MacMahon, Newton, etc ... "all" linear algebra statements - holds true in such a relaxed commutativity setup.

Generalizations, applications: There are many applications to representation theory of loop algebras - explicit description of central elements, local Langlands GL-oper, Capelli's identities, etc. See e.g. A. Molev's lectures: https://www.maths.usyd.edu.au/u/alexm/talks/seoul-manin.pdf

The question above have various generalizations - we can consider not 2x2 Manin matrices, but $n\times m$ with generators $M_{ij}$, q-analogues, super-analogues, Manin matrices of types B, C, D, generalization to other qudratic algebras, modifications for algebras with involution and so on. And ask the same question - whether obtained algebras are Koszul. At least when underlying algebra is Koszul.

In particular one the questions - mentioned in A.Molev's lectures (see page 261/78) - for the super-Manin matrices we do not know the Hilbert series and basis generating the algebra. Koszulity might help on that.

Answer 1

If I understand correctly Theorem 4.1.1 of https://www.math.univ-paris13.fr/~vallette/Operads.pdf then the answer is "yes".

I'll take the order $a<b<c<d$ and use lex ordering like they do. Then the leading terms of the relations you list are $ba, dc, da$. In particular, there are no critical monomials, so the theorem automatically applies.

I hesitate here because this seems a bit too easy, but I guess the intuition here is that there are "few relations which don't interact too much" so you get a well-behaved algebra.

EDIT: I did a little more exploration of this example. The observation above also says the 3 relations are a Gröbner basis, so we can compute the Hilbert series by just working with the Koszul algebra $A$ whose relations are the monomials $ba, dc, da$. I can model this as words in $a,b,c,d$ with the following rule: if you see $b$, the very next letter is not $a$, and if you see $d$, the very next letter is not $a$ or $c$.

In particular, words of length $n$ are the same as walks of length $n$ in the directed graph starting at vertex 1 with the following adjacency matrix $\begin{bmatrix} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}$ (the convention here is that the entry in row $i$ and column $j$ is the number of arrows from vertex $i$ to vertex $j$). To elaborate: the walk is reading the letters as input and vertex 2 means we just saw $b$ and vertex 3 means we just saw $d$. Anyway, the transfer matrix method then tells us that the Hilbert series is $\displaystyle \frac{1}{1-4t+3t^2}$.

That also implies that the resolution of the residue field has finite length and looks like:

$0 \to A(-2)^3 \to A(-1)^4 \to A$

(the notation $(-i)$ means grading shift by $i$).