Skip to content →

why nag? (2)

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,

\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}

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

Published in featured


Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.