# Why is the Brauer Loop Scheme Not a Variety?

I am trying to grapple with the basics of scheme theory. Is the scheme defined by Spec[C[x,y,z]/(xy,yx,zx)] a variety? What do the points look like?

I suspect it represents points satisfying xy = yz = zx = 0, so it should have three irreducible components {x = y = 0}, {y = z = 0} and {z = x = 0}. The motivation for this example comes from statistical mechanics and it has quite a bit more content:

Consider the space of 3x3 matrices (entries in C) with the following deformation of the matrix product: $P \circ Q = \sum_{i \leq j \leq k, cyc} P_{ij} P_{jk}$. Here we summing over j such that i, j, k appear in cyclic order mod 3. It appears in a set of slides on The Combinatorics of the Brauer Loop Scheme.

The paper then proceeds to define a scheme using equations in matrices. In the space of matrices with 0's along the diagonal, we consider the matrices with $M \circ M = 0$. In coordinates, the matrix product therefore looks like: $\left( \begin{array}{ccc} 0 & b_{12} & b_{13} \\\\ b_{21} & 0 & b_{23} \\\\ b_{31} & b_{32} & 0 \end{array} \right) \circ \left( \begin{array}{ccc} 0 & b_{12} & b_{13} \\\\ b_{21} & 0 & b_{23} \\\\ b_{31} & b_{32} & 0 \end{array} \right) = \left( \begin{array}{lll} 0 & 0 & b_{12}b_{23} \\\\ b_{23}b_{31} & 0 & 0 \\\\ 0 & b_{31}b_{12} & 0 \end{array} \right)$ As all the entries on the right side vanish, this defines three equations in six unknowns. (Actually, only the "clockwise" matrix entries seem to be involved.)

• Varieties are usually defined to be irreducible, and your scheme is not irreducible. – Mariano Suárez-Álvarez Jan 14 '10 at 5:18

• The equations "$P$ has zero diagonal, $P\circ P = 0$", from his second ideal, taken from our paper. – Allen Knutson Jan 14 '10 at 15:40