neverendingbooks – Page 187

Let us
take a hopeless problem, motivate why something like non-commutative
algebraic geometry might help to solve it, and verify whether this
promise is kept.

Suppose we want to know all solutions in invertible
matrices to the braid relation (or Yang-Baxter equation)

$X Y X = Y X Y$

All such solutions (for varying size of matrices)
form an additive Abelian category $\mathbf{rep}~B_3$ , so a big step forward would be to know all its
simple solutions (that is, those whose matrices cannot be brought in
upper triangular block form). A literature check shows that even this
task is far too ambitious. The best result to date is the classification
due to Imre Tuba and
Hans Wenzl of simple solutions of which the matrix size is at most
5.

For fixed matrix size n, finding solutions in $\mathbf{rep}~B_3$ is the same as solving a system of $n^2$ cubic
polynomial relations in $2n^2$
unknowns, which quickly becomes a daunting task. Algebraic geometry
tells us that all solutions, say $\mathbf{rep}_n~B_3$ form an affine closed subvariety of $n^2$ -dimensional affine space. If we assume that $\mathbf{rep}_n~B_3$ is a smooth variety (that is, a manifold) and
if we know one solution explicitly, then we can use the tangent space in
this point to linearize the problem and to get at all solutions in a
neighborhood.

So, here is an idea : assume that $\mathbf{rep}~B_3$ itself would be a non-commutative manifold, then
we might linearize our problem by considering tangent spaces and obtain
new solutions out of already known ones. But, what is a non-commutative
manifold? Well, by the above we at least require that for all integers n
the commutative variety $\mathbf{rep}_n~B_3$ is a commutative manifold.

But, there
is still some redundancy in our problem : if $(X,Y)$ is a
solution, then so is any conjugated pair $(g^{-1}Xg,g^{-1}Yg)$ where $g \in GL_n$ is a basechange matrix. In categorical terms, we are only
interested in isomorphism classes of solutions. Again, if we fix the
size n of matrix-solutions, we consider the affine variety $\mathbf{rep}_n~B_3$ as a variety with a $GL_n$ -action
and we like to classify the orbits of simple solutions. If $\mathbf{rep}_n~B_3$ is a manifold then the theory of Luna slices
provides a method, both to linearize the problem as well as to reduce
its complexity. Instead of the tangent space we consider the normal
space N to the $GL_n$ -orbit
(in a suitable solution). On this affine space, the stabilizer subgroup
$GL(\alpha)$ acts and there is a natural one-to-one
correspondence between $GL_n$ -orbits
in $\mathbf{rep}_n~B_3$ and $GL(\alpha)$ -orbits in the normal space N (at least in a
neighborhood of the solution).

So, here is a refinement of the
idea : we would like to view $\mathbf{rep}~B_3$ as a non-commutative manifold with a group action
given by the notion of isomorphism. Then, in order to get new isoclasses
of solutions from a constructed one we want to reduce the size of our
problem by considering a linearization (the normal space to the orbit)
and on it an easier isomorphism problem.

However, we immediately
encounter a problem : calculating ranks of Jacobians we discover that
already $\mathbf{rep}_2~B_3$ is not a smooth variety so there is not a
chance in the world that $\mathbf{rep}~B_3$ might be a useful non-commutative manifold.
Still, if $(X,Y)$ is a
solution to the braid relation, then the matrix $(XYX)^2$
commutes with both X and Y.

If $(X,Y)$ is a
simple solution, this means that after performing a basechange, $C=(XYX)^2$ becomes a scalar matrix, say $\lambda^6 1_n$ . But then, $(X_1,Y_1) = (\lambda^{-1}X,\lambda^{-1}Y)$ is a solution to

$XYX = YXY , (XYX)^2 = 1$

and all such solutions form a
non-commutative closed subvariety, say $\mathbf{rep}~\Gamma$ of $\mathbf{rep}~B_3$ and if we know all (isomorphism classes of)
simple solutions in $\mathbf{rep}~\Gamma$ we have solved our problem as we just have to
bring in the additional scalar $\lambda \in \mathbb{C}^*$ .

Here we strike gold : $\mathbf{rep}~\Gamma$ is indeed a non-commutative manifold. This can
be seen by identifying $\Gamma$
with one of the most famous discrete infinite groups in mathematics :
the modular group $PSL_2(\mathbb{Z})$ . The modular group acts by Mobius
transformations on the upper half plane and this action can be used to
write $PSL_2(\mathbb{Z})$ as the free group product $\mathbb{Z}_2 \ast \mathbb{Z}_3$ . Finally, using
classical representation theory of finite groups it follows that indeed
all $\mathbf{rep}_n~\Gamma$ are commutative manifolds (possibly having
many connected components)! So, let us try to linearize this problem by
looking at its non-commutative tangent space, if we can figure out what
this might be.

Here is another idea (or rather a dogma) : in the
world of non-commutative manifolds, the role of affine spaces is played
by $\mathbf{rep}~Q$ the representations of finite quivers Q. A quiver
is just on oriented graph and a representation of it assigns to each
vertex a finite dimensional vector space and to each arrow a linear map
between the vertex-vector spaces. The notion of isomorphism in $\mathbf{rep}~Q$ is of course induced by base change actions in all
of these vertex-vector spaces. (to be continued)

why nag? (1)

make your own mopp

3 yrs. later

neverendingbooks Posts

why nag? (1)

make your own mopp

3 yrs. later