Now, can

we assign such an non-commutative tangent space, that is a for some quiver Q, to ? As we may

restrict any solution

in to the finite subgroups and . Now, representations of finite cyclic groups are

decomposed into eigen-spaces. For example

where with g the

generator of . Similarly,

where is a

primitive 3-rd root of unity. That is, to any solution we have found 5 vector spaces and 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

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

which gives us 6 linear maps between the

vertex-vector spaces. Hence, to does correspond in a natural way a

representation of dimension vector (where ) of the quiver Q which

is of the form

Clearly, not every representation of is obtained in this way. For starters, the

eigen-space decompositions force the numerical restriction

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

from this quiver representation by taking

Hence, it makes sense to

view as a linearization of, or as a tangent space to,

. 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

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?

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