on March 29, 2005 by lieven in geometry, Comments (0)

why nag? (3)

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non-commutative geometry

Here is the construction of this normal space or chart $\wis{chart}_{\Gamma}$ . 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 $\wis{rep}~\Gamma$

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

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

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

In particular, in any component $\wis{rep}_{\alpha}~Q$ containing an open subset of representations corresponding to solutions in $\wis{rep}~\Gamma$ 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 $\wis{rep}_{\alpha}~Q$ can be identified with the representation space $\wis{rep}_{\beta}~Q$ where $\beta=(g_1,\hdots,g_6)$ 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 $\wis{rep}~\Gamma$ . 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 $\wis{rep}~\Gamma$ .

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 $\beta=(g_1,\hdots,g_6)$ with $n = \gamma_1 + \hdots + \gamma_6$ , then for every non-zero scalar $\lambda \in \C^*$ 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 $\wis{rep}~B_3$ 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!

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