Here the story of an idea to construct new examples of non-commutative compact manifolds, the computational difficulties one runs into and, when they are solved, the white noise one gets. But, perhaps, someone else can spot a gem among all gibberish…
Qurves (aka quasi-free algebras, aka formally smooth algebras) are the \’affine\’ pieces of non-commutative manifolds. Basic examples of qurves are : semi-simple algebras (e.g. group algebras of finite groups), path algebras of quivers and coordinate rings of affine smooth curves. So, let us start with an affine smooth curve $X$ and spice it up to get a very non-commutative qurve. First, we bring in finite groups. Let $G$ be a finite group acting on $X$, then we can form the skew-group algebra $A = \mathbfk[X] \bigstar G$. These are examples of prime Noetherian qurves (aka hereditary orders). A more pompous way to phrase this is that these are precisely the one-dimensional smooth Deligne-Mumford stacks.
As the 21-st century will turn out to be the time we discovered the importance of non-Noetherian algebras, let us make a jump into the wilderness and consider the amalgamated free algebra product $A = (\mathbf k[X] \bigstar G) \ast{\mathbf k G} \mathbfk H$ where $G \subset H$ is an interesting extension of finite groups. Then, $A$ is again a qurve on which $H$ acts in a way compatible with the $G$-action on $X$ and $A$ is hugely non-commutative… A very basic example : let $\mathbb{Z}/2\mathbb{Z}$ act on the affine line $\mathbfk[x]$ by sending $x \mapsto -x$ and consider a finite simple group $M$. As every simple group has an involution, we have an embedding $\mathbb{Z}/2\mathbb{Z} \subset M$ and can construct the qurve $A=(\mathbfk[x] \bigstar \mathbb{Z}/2\mathbb{Z}) \ast{\mathbfk \mathbb{Z}/2\mathbb{Z}} \mathbfk M$ on which the simple group $M$ acts compatible with the involution on the affine line. To study the corresponding non-commutative manifold, that is the Abelian category $\wis{rep}~A$ of all finite dimensional representations of $A$ we have to compute the one quiver to rule them all for $A$. Because $A$ is a qurve, all its representation varieties $\wis{rep}n~A$ are smooth affine varieties, but they may have several connected components. The direct sum of representations turns the set of all these components into an Abelian semigroup and the vertices of the \’one quiver\’ correspond to the generators of this semigroup whereas the number of arrows between two such generators is given by the dimension of $Ext^1A(Si,Sj)$ where $Si,Sj$ are simple $A$-representations lying in the respective components. All this may seem hard to compute but it can be reduced to the study of another quiver, the Zariski quiver associated to $A$ which is a bipartite quiver with on the left the \’one quiver\’ for $\mathbfk[x] \bigstar \mathbb{Z}/2\mathbb{Z}$ which is just (where the two vertices correspond to the two simples of $\mathbb{Z}/2\mathbb{Z}$) and on the right the \’one quiver\’ for $\mathbf k M$ (which just consists of as many verticers as there are simple representations for $M$) and where the number of arrows from a left- to a right-vertex is the number of $\mathbb{Z}/2\mathbb{Z}$-morphisms between the respective simples. To make matters even more concrete, let us consider the easiest example when $M = A5$ the alternating group on $5$ letters. The corresponding Zariski quiver then turns out to be The Euler-form of this quiver can then be used to calculate the dimensions of the EXt-spaces giving the number of arrows in the \’one quiver\’ for $A$. To find the vertices, that is, the generators of the component semigroup we have to find the minimal integral solutions to the pair of equations saying that the number of simple $\mathbb{Z}/2\mathbb{Z}$ components based on the left-vertices is equal to that one the right-vertices. In this case it is easy to see that there are as many generators as simple $M$ representations. For $A5$ they correspond to the dimension vectors (for the Zariski quiver having the first two components on the left) $\begin{cases} (1,2,0,0,0,0,1) \ (1,2,0,0,0,1,0) \ (3,2,0,0,1,0,0) \
(2,2,0,1,0,0,0) \ (1,0,1,0,0,0,0) \end{cases}$ We now have all info to determine the \’one quiver\’ for $A$ and one would expect a nice result. Instead one obtains a complete graph on all vertices with plenty of arrows. More precisely one obtains as the one quiver for $A5$

with the number of arrows (in each direction) indicated. Not very illuminating, I find. Still, as the one quiver is symmetric it follows that all quotient varieties $\wis{iss}n~A$ have a local Poisson structure. Clearly, the above method can be generalized easily and all examples I did compute so far have this \’nearly complete graph\’ feature. One might hope that if one would start with very special curves and groups, one might obtain something more interesting. Another time I\’ll tell what I got starting from Klein\’s quartic (on which the simple group $PSL2(\mathbb{F}7)$ acts) when the situation was sexed-up to the sporadic simple Mathieu group $M{24}$ (of which $PSL2(\mathbb{F}_7)$ is a maximal subgroup).

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Posted in geometry
Written on Fri, 10 June 2005 at 12:35 pm
Tags: Artin, groups, Klein, Mathieu, non-commutative, quivers, representations, simples
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