non-commutative – Page 8 – neverendingbooks

why nag? (2)

Published March 28, 2005 by lieven

Now, can
we assign such an non-commutative tangent space, that is a $\mathbf{rep}~Q$ for some quiver Q, to $\mathbf{rep}~\Gamma$ ? As $\Gamma = \mathbb{Z}_2 \ast \mathbb{Z}_3$ we may
restrict any solution $V=(X,Y)$
in $\mathbf{rep}~\Gamma$ to the finite subgroups $\mathbb{Z}_2$ and $\mathbb{Z}_3$ . Now, representations of finite cyclic groups are
decomposed into eigen-spaces. For example

$V \downarrow_{\mathbb{Z}_2} = V_+ \oplus V_-$

where $V_{\pm} = \{ v \in V~|~g.v = \pm v \}$ with g the
generator of $\mathbb{Z}_2$ . Similarly,

$V \downarrow_{\mathbb{Z}_3} = V_1 \oplus V_{\rho} \oplus V_{\rho^2}$

where $\rho$ is a
primitive 3-rd root of unity. That is, to any solution $V \in \mathbf{rep}~\Gamma$ we have found 5 vector spaces $V_+,V_-,V_1,V_{\rho}$ and $V_{\rho^2}$ so we would like them to correspond to the vertices
of our conjectured quiver Q.

What are the arrows of Q, or
equivalently, is there a natural linear map between the vertex-vector
spaces? Clearly, as

$V_+ \oplus V_- = V = V_1 \oplus V_{\rho} \oplus V_{\rho^2}$

any choice of two bases of V (one
compatible with the left-side decomposition, the other with the
right-side decomposition) are related by a basechange matrix B which we
can decompose into six blocks (corresponding to the two decompositions
in 2 resp. 3 subspaces

$B = \begin{bmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \\ B_{31} & B_{32} \end{bmatrix}$

which gives us 6 linear maps between the
vertex-vector spaces. Hence, to $V \in \mathbf{rep}~\Gamma$ does correspond in a natural way a
representation of dimension vector $\alpha=(a_1,a_2,b_1,b_2,b_3)$ (where $dim(V_+)=a_1,\ldots,dim(V_{\rho^2})=b_3$ ) of the quiver Q which
is of the form

$\xymatrix{ & & & & \vtx{b_1} \\ \vtx{a_1} \ar[rrrru]^(.3){B_{11}} \ar[rrrrd]^(.3){B_{21}} \ar[rrrrddd]_(.2){B_{31}} & & & & \\ & & & & \vtx{b_2} \\ \vtx{a_2} \ar[rrrruuu]_(.7){B_{12}} \ar[rrrru]_(.7){B_{22}} \ar[rrrrd]_(.7){B_{23}} & & & & \\ & & & & \vtx{b_3}}$

Clearly, not every representation of $\mathbf{rep}~Q$ is obtained in this way. For starters, the
eigen-space decompositions force the numerical restriction

$a_1+a_2 = dim(V) = b_1+b_2+b_3$

on the
dimension vector and the square matrix constructed from the arrow-linear
maps must be invertible. However, if both these conditions are
satisfied, we can reconstruct the (isomorphism class) of the solution in
$\mathbf{rep}~\Gamma$ from this quiver representation by taking

$X = B^{-1} \begin{bmatrix} 1_{b_1} & 0 & 0 \\ 0 & \rho^2 1_{b_2} & 0 \\ 0 & 0 & \rho 1_{b_3} \end{bmatrix} B \begin{bmatrix} 1_{a_1} & 0 \\ 0 & -1_{a_2} \end{bmatrix}$

$Y = \begin{bmatrix} 1_{a_1} & 0 \\ 0 & -1_{a_2} \end{bmatrix} B^{-1} \begin{bmatrix} 1_{b_1} & 0 & 0 \\ 0 & \rho^2 1_{b_2} & 0 \\ 0 & 0 & \rho 1_{b_3} \end{bmatrix} B$

Hence, it makes sense to
view $\mathbf{rep}~Q$ as a linearization of, or as a tangent space to,
$\mathbf{rep}~\Gamma$ . However, though we reduced the study of
solutions of the polynomial system of equations to linear algebra, we
have not reduced the isomorphism problem in size. In fact, if we start
of with a matrix-solution $V=(X,Y)$
of size n we end up with a quiver-representation of total dimension 2n.
So, can we construct some sort of non-commutative normal space to the
isomorphism classes? That is, is there another quiver Q whose
representations can be interpreted as normal-spaces to orbits in certain
points?

seen this quiver?

Published March 25, 2005 by lieven

The above quiver on 10 vertices is not symmetric, but has the
interesting property that every vertex has three incoming and three
outgoing arrows. If you have ever seen this quiver in another context,
please drop me a line. My own interest for it is that it is the ‘one
quiver’ for a non-commutative compactification of $GL_2(\mathbb{Z}) $. If
you like to know what I mean by this, you might consult the
Granada-notes which I hope to post over the weekend.

On a
different matter, if you want to know what all this hype on derived
categories and the classification project is about but got lost in the
pile of preprints, you might have a look at the Bourbaki talk by Raphael
Rouquier Categories
derivees et geometrie birationnelle posted today on the arXiv.

B for bricks

Published January 28, 2005 by lieven

Last time we
argued that a noncommutative variety might be an _aggregate_
which locally is of the form $\mathbf{rep}~A$ for some affine (possibly
non-commutative) $C$-algebra $A$. However, we didn't specify what we
meant by 'locally' as we didn't define a topology on
$\mathbf{rep}~A$, let alone on an arbitrary aggregate. Today we will start
the construction of a truly _non-commutative topology_ on
$\mathbf{rep}~A$.
Here is the basic idea : we start with a thick
subset of finite dimensional representations on which we have a natural
(ordinary) topology and then we extend this to a non-commutativce
topology on the whole of $\mathbf{rep}~A$ using extensions. The impatient
can have a look at my old note A noncommutative
topology on rep A but note that we will modify the construction here
in two essential ways.
In that note we took $\mathbf{simp}~A$, the
set of all fnite dimensional simple representations, as thick subset
equipped with the induced Zariski topology on the prime spectrum
$\mathbf{spec}~A$. However, this topology doesn't behave well with
respect to the gluings we have in mind so we will extend $\mathbf{simp}~A$
substantially.

Tag: non-commutative

why nag? (2)

seen this quiver?

B for bricks