How to translate cluster X-coordinates to cluster A-coordinates?

Question

In the paper, (5.28) on page 35 is a formula which translate cluster A-coordinates to cluster X-coordinates: \begin{align} x_i = \prod_{i \to j} a_j. \end{align}

Is there a formula which translate cluster X-coordinates to cluster A-coordinates? Thank you very much.

Edit: as David Speyer pointed out, the correction formula to translate cluster A-coordinates to cluster X-coordinates is \begin{align} x_i = \prod_{i \to j} a_j/\prod_{j \to i} a_j = \prod_{j} a_j^{Q_{ij}}. \end{align}

Answer 1

In general, no.

First of all, I am confused by the formula $\prod_{i \to j} a_j$. The correct formula is $x_i = \prod_{j} a_j^{Q_{ij}}$ (also written in your paper, on the same line) which, if we are going to write it in terms of arrows, is $\prod_{i \to j} a_j / \prod_{i \leftarrow j} a_j$.

If the $Q$ matrix is invertible over the integers, with $Q^{-1} = C$ say, then we can invert this by $a_i = \prod_j x_i^{C_{ij}}$. But if it isn't invertible, you're out of luck.

Answer 2

I don't think so. Look at the cluster algebra discussed in Fock-Goncharov: MODULI SPACES OF LOCAL SYSTEMS AND HIGHER TEICHMULLER THEORY:

Let $S$ be a cusped surface with an ideal triangulation having all vertices in cusps, and let $\widetilde{V}\subset\partial_\infty H^2$ be the lift of the ideal vertices to the ideal boundary of the universal covering $\widetilde{S}=H^2$. Consider (for simplicity) $G=SL(3,{\bf R})$ with its subgroups

• $B$ the group of upper triangular matrices,
• $N\subset B$ the triangular matrices with only $1$'s on the diagonal.

Then the $X$-coordinates parametrise maps $\widetilde{V}\to G/B$ (equivariant for some representation) and the $A$-coordinates parametrise maps $\widetilde{V}\to G/N$ (equivariant for some representation).

Of course, a map $\widetilde{V}\to G/N$ determines a map $\widetilde{V}\to G/B$ but not vice versa. So one would not expect the X-coordinates to determine the A-coordinates.