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why nag? (3)

Published March 29, 2005 by lievenlb

Here is
the construction of this normal space or chart $\mathbf{chart}_{\Gamma}$ . The sub-semigroup of $Z^5$ (all
dimension vectors of Q) consisting of those vectors $\alpha=(a_1,a_2,b_1,b_2,b_3)$ satisfying the numerical condition $a_1+a_2=n=b_1+b_2+b_3$ is generated by six dimension vectors,
namely those of the 6 non-isomorphic one-dimensional solutions in $\mathbf{rep}~\Gamma$

$S_1 = \xymatrix@=.4cm{ & & & & \vtx{1} \\ \vtx{1} \ar[rrrru]^1 \ar[rrrrd] \ar[rrrrddd] & & & & \\ & & & & \vtx{0} \\ \vtx{0} \ar[rrrruuu] \ar[rrrru] \ar[rrrrd] & & & & \\ & & & & \vtx{0}} \qquad S_2 = \xymatrix@=.4cm{ & & & & \vtx{0} \\ \vtx{0} \ar[rrrru] \ar[rrrrd] \ar[rrrrddd] & & & & \\& & & & \vtx{1} \\\vtx{1} \ar[rrrruuu] \ar[rrrru]^1 \ar[rrrrd] & & & & \\ & & & & \vtx{0}}$

$S_3 = \xymatrix@=.4cm{ & & & & \vtx{0} \\ \vtx{1} \ar[rrrru] \ar[rrrrd] \ar[rrrrddd]^1 & & & & \\ & & & & \vtx{0} \\ \vtx{0} \ar[rrrruuu] \ar[rrrru] \ar[rrrrd] & & & & \\ & & & & \vtx{1}} \qquad S_4 = \xymatrix@=.4cm{ & & & & \vtx{1} \\ \vtx{0} \ar[rrrru] \ar[rrrrd] \ar[rrrrddd] & & & & \\ & & & & \vtx{0} \\ \vtx{1} \ar[rrrruuu]^1 \ar[rrrru] \ar[rrrrd] & & & & \\ & & & & \vtx{0}}$

$S_5 = \xymatrix@=.4cm{ & & & & \vtx{0} \\ \vtx{1} \ar[rrrru] \ar[rrrrd]^1 \ar[rrrrddd] & & & & \\ & & & & \vtx{1} \\ \vtx{0} \ar[rrrruuu] \ar[rrrru] \ar[rrrrd] & & & & \\ & & & & \vtx{0}} \qquad S_6 = \xymatrix@=.4cm{ & & & & \vtx{0} \\ \vtx{0} \ar[rrrru] \ar[rrrrd] \ar[rrrrddd] & & & & \\ & & & & \vtx{0} \\ \vtx{1} \ar[rrrruuu] \ar[rrrru] \ar[rrrrd]^1 & & & & \\ & & & & \vtx{1}}$

In
particular, in any component $\mathbf{rep}_{\alpha}~Q$ containing an open subset of
representations corresponding to solutions in $\mathbf{rep}~\Gamma$ we have a particular semi-simple solution

$M = S_1^{\oplus g_1} \oplus S_2^{\oplus g_2} \oplus S_3^{\oplus g_3} \oplus S_4^{\oplus g_4} \oplus S_5^{\oplus g_5} \oplus S_6^{\oplus g_6}$

and in
particular $\alpha = (g_1+g_3+g_5,g_2+g_4+g_6,g_1+g_4,g_2+g_5,g_3+g_6)$ . The normal space
to the $GL(\alpha)$ -orbit of M in $\mathbf{rep}_{\alpha}~Q$ can be identified with the representation
space $\mathbf{rep}_{\beta}~Q$ where $\beta=(g_1,\ldots,g_6)$ and Q is the quiver of the following
form

$\xymatrix{ & \vtx{g_1} \ar@/^/[ld]^{C_{16}} \ar@/^/[rd]^{C_{12}} & \\ \vtx{g_6} \ar@/^/[ru]^{C_{61}} \ar@/^/[d]^{C_{65}} & & \vtx{g_2} \ar@/^/[lu]^{C_{21}} \ar@/^/[d]^{C_{23}} \\ \vtx{g_5} \ar@/^/[u]^{C_{56}} \ar@/^/[rd]^{C_{54}} & & \vtx{g_3} \ar@/^/[u]^{C_{32}} \ar@/^/[ld]^{C_{34}} \\ & \vtx{g_4} \ar@/^/[lu]^{C_{45}} \ar@/^/[ru]^{C_{43}} & }$

and we can
even identify how the small matrices $C_{ij}$ fit
into the $3 \times 2$ block-decomposition of the base-change matrix B

$B = \begin{bmatrix} \begin{array}{ccc|ccc} 1_{a_1} & 0 & 0 & C_{21} & 0 & C_{61} \\ 0 & C_{34} & C_{54} & 0 & 1_{a_4} & 0 \\ \hline C_{12} & C_{32} & 0 & 1_{a_2} & 0 & 0 \\ 0 & 0 & 1_{a_5} & 0 & C_{45} & C_{65} \\ \hline 0 & 1_{a_3} & 0 & C_{23} & C_{43} & 0 \\ C_{16} & 0 & C_{56} & 0 & 0 & 1_{a_6} \\ \end{array} \end{bmatrix}$

Hence, it makes sense
to call Q the non-commutative normal space to the isomorphism problem in
$\mathbf{rep}~\Gamma$ . Moreover, under this correspondence simple
representations of Q (for which both the dimension vectors and
distinguishing characters are known explicitly) correspond to simple
solutions in $\mathbf{rep}~\Gamma$ .

Having completed our promised
approach via non-commutative geometry to the classification problem of
solutions to the braid relation, it is time to collect what we have
learned. Let $\beta=(g_1,\ldots,g_6)$ with $n = \gamma_1 + \ldots + \gamma_6$ , then for every
non-zero scalar $\lambda \in \mathbb{C}^*$ the matrices

$X = \lambda B^{-1} \begin{bmatrix} 1_{g_1+g_4} & 0 & 0 \\ 0 & \rho^2 1_{g_2+g_5} & 0 \\ 0 & 0 & \rho 1_{g_3+g_6} \end{bmatrix} B \begin{bmatrix} 1_{g_1+g_3+g_5} & 0 \\ 0 & -1_{g_2+g_4+g_6} \end{bmatrix}$

$Y = \lambda \begin{bmatrix} 1_{g_1+g_3+g_5} & 0 \\ 0 & -1_{g_2+g_4+g_6} \end{bmatrix} B^{-1} \begin{bmatrix} 1_{g_1+g_4} & 0 & 0 \\ 0 & \rho^2 1_{g_2+g_5} & 0 \\ 0 & 0 & \rho 1_{g_3+g_6} \end{bmatrix} B$

give a solution of size
n to the braid relation. Moreover, such a solution can be simple only if
the following numerical relations are satisfied

$g_i \leq g_{i-1} + g_{i+1}$

where indices are viewed
modulo 6. In fact, if these conditions are satisfied then a sufficiently
general representation of Q does determine a simple solution in $\mathbf{rep}~B_3$ and conversely, any sufficiently general simple n
size solution of the braid relation can be conjugated to one of the
above form. Here, by sufficiently general we mean a Zariski open (hence
dense) subset.

That is, for all integers n we have constructed
nearly all (meaning a dense subset) simple solutions to the braid
relation. As to the classification problem, if we have representants of
simple $\beta$ -dimensional representations of the quiver Q, then the corresponding
solutions $(X,Y)$ of
the braid relation represent different orbits (up to finite overlap
coming from the fact that our linearizations only give an analytic
isomorphism, or in algebraic terms, an etale map). Such representants
can be constructed for low dimensional $\beta$ .
Finally, our approach also indicates why the classification of
braid-relation solutions of size $\leq 5$ is
easier : from size 6 on there are new classes of simple
Q-representations given by going round the whole six-cycle!

why nag? (2)

Published March 28, 2005 by lievenlb

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 lievenlb

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.

Author: lievenlb

why nag? (3)

why nag? (2)

seen this quiver?