For

finite dimensional hereditary algebras, one can describe its

noncommutative topology (as developed in part 2)

explicitly, using results of Markus

Reineke in The monoid

of families of quiver representations. Consider a concrete example,

say

$A = \begin{bmatrix} \mathbb{C} & V \\ 0 & \mathbb{C}

\end{bmatrix}$ where $V$ is an n-dimensional complex vectorspace, or

equivalently, A is the path algebra of the two point, n arrow quiver

$\xymatrix{\vtx{} \ar@/^/[r] \ar[r] \ar@/_/[r] & \vtx{}} $

Then, A has just 2 simple representations S and T (the vertex reps) of

dimension vectors s=(1,0) and t=(0,1). If w is a word in S and T we can

consider the set $\mathbf{r}_w$ of all A-representations having a

Jordan-Holder series with factors the terms in w (read from left to

right) so $\mathbf{r}_w \subset \mathbf{rep}_{(a,b)}~A$ when there are a

S-terms and b T-terms in w. Clearly all these subsets can be given the

structure of a monoid induced by concatenation of words, that is

$\mathbf{r}_w \star \mathbf{r}_{w’} = \mathbf{r}_{ww’}$ which is

Reineke’s *composition monoid*. In this case it is generated by

$\mathbf{r}_s$ and $\mathbf{r}_t$ and in the composition monoid the

following relations hold among these two generators

$\mathbf{r}_t^{\star n+1} \star \mathbf{r}_s = \mathbf{r}_t^{\star n}

\star \mathbf{r}_s \star \mathbf{r}_t \quad \text{and} \quad

\mathbf{r}_t \star \mathbf{r}_s^{\star n+1} = \mathbf{r}_s \star

\mathbf{r}_t \star \mathbf{r}_s^{\star n}$ With these notations we can

now see that the left basic open set in the noncommutative topology

(associated to a noncommutative word w in S and T) is of the form

$\mathcal{O}^l_w = \bigcup_{w’} \mathbf{r}_{w’}$ where the union is

taken over all words w’ in S and T such that in the composition monoid

the relation holds $\mathbf{r}_{w’} = \mathbf{r}_w \star \mathbf{r}_{u}$

for another word u. Hence, each op these basic opens hits a large number

of $~\mathbf{rep}_{\alpha}$, in fact far too many for our purposes….

So, what do we want? We want to define a noncommutative notion of

birationality and clearly we want that if two algebras A and B are

birational that this is the same as saying that some open subsets of

their resp. $\mathbf{rep}$’s are homeomorphic. But, what do we

understand by *noncommutative birationality*? Clearly, if A and B are

prime Noethrian, this is clear. Both have a ring of fractions and we

demand them to be isomorphic (as in the commutative case). For this

special subclass the above noncommutative topology based on the Zariski

topology on the simples may be fine.

However, most qurves don’t have

a canonical ‘ring of fractions’. Usually they will have infinitely

many simple Artinian algebras which should be thought of as being

_a_ ring of fractions. For example, in the finite dimensional

example A above, if follows from Aidan Schofield‘s work Representations of rings over skew fields that

there is one such for every (a,b) with gcd(a,b)=1 and (a,b) satisfying

$a^2+b^2-n a b < 1$ (an indivisible Shur root for A).

And

what is the _noncommutative birationality result_ we are aiming

for in each of these cases? Well, the inspiration for this comes from

another result by Aidan (although it is not stated as such in the

paper…) Birational

classification of moduli spaces of representations of quivers. In

this paper Aidan proves that if you take one of these indivisible Schur

roots (a,b) above, and if you look at $\alpha_n = n(a,b)$ that then the

moduli space of semi-stable quiver representations for this multiplied

dimension vector is birational to the quotient variety of

$1-(a^2+b^2-nab)$-tuples of $ n \times n $-matrices under simultaneous

conjugation.

So, *morally speaking* this should be stated as the

fact that A is (along the ray determined by (a,b)) noncommutative

birational to the free algebra in $1-(a^2+b^2-nab)$ variables. And we

want a noncommutative topology on $\mathbf{rep}~A$ to encode all these

facts… As mentioned before, this can be done by replacing simples with

bricks (or if you want Schur representations) but that will have to wait

until next week.

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