# Colorful mutation

A follow-up to a February post about quiver mutations. Quivers naturally correspond to skew-symmetric integer matrices: the $(i,j)$ entry is the number of oriented edges from $i$ to $j$, with the understanding that negative numbers mean opposite orientation.

However, in the theory of cluster algebras one also considers more general sign-skew-symmetric matrices, which are integer matrices $B$ such that $\mathrm{sign}\, B = (\mathrm{sign}\, b_{ij})$ is skew-symmetric. How do we draw such a matrix? We can represent positive entries $b_{ij}>0$ by $b_{ij}$ black arrows from $i$ to $j$, and negative entries $b_{ij}<0$ by $-b_{ij}$ red arrows from $i$ to $j$. So, instead of reversing orientation we mark negative numbers in red. The sign-skew-symmetric prevent edges of different colors from going in the same direction, (and edges of same color from going in opposite directions). It also stipulates that each pair of vertices is connected either by edges of both colors, or by neither. Mutation at vertex v is now done as follows.

1. v is removed and each monochromatic oriented path of length two through v is contracted into an edge. That is, the stopover at v is eliminated.
2. Step 1 may create black-red pairs of parallel edges, which are deleted. That is, we cancel the pairs of different colored arrows going in the same direction.
3. The replacement vertex v’ is inserted, connected to the rest of the graph in the same way that v was, except the red edges become black and vice versa. In practice, one simply reuses v for this purpose.

Problem is, such mutation may create monochromatic oriented 2-cycles, and the rules do not allow for their elimination. More precisely, we may lose the sign-skew-symmetric (SSS) property during mutations. A matrix is called totally sign-skew-symmetric (TSSS) if this never happens.

Given an SSS matrix, how can we tell if it’s TSSS? The above quiver isn’t one; mutation at vertex 2 leaves 1 and 3 connected by only one color, which is not allowed.