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Let $Mat_3$ be the set of all 3 by 3 matrices. I have some questions on the cluster algebra structure on the coordinate ring of $Mat_3$.

We use $\Delta_{j_1\ldots j_n}^{i_1\ldots i_n}$ to denote the minor a of a matrix consisting of the $i_1,\ldots, i_n$-th columns and $j_1,\ldots, j_n$-th rows.

A basis of $\mathbb{C}[Mat_3]$ is $x_{ij}$, $i,j=1,2,3$, where $x_{ij}(a) = a_{ij}$, $a \in \mathbb{C}[Mat_3]$.

Is $$\Delta_2^2, \Delta_{23}^{23}, \Delta_{3}^{2},\Delta_{2}^{3},\Delta_{1}^{3},\Delta_{3}^{1},\Delta_{12}^{23},\Delta_{23}^{12},\Delta_{123}^{123}$$ a basis of $\mathbb{C}[Mat_3]?$ How to obtain $x_{11}$ from $$\Delta_2^2, \Delta_{23}^{23}, \Delta_{3}^{2},\Delta_{2}^{3},\Delta_{1}^{3},\Delta_{3}^{1},\Delta_{12}^{23},\Delta_{23}^{12},\Delta_{123}^{123}?$$

The 9-tuple $$(\Delta_2^2, \Delta_{23}^{23}, \Delta_{3}^{2},\Delta_{2}^{3} \mid \Delta_{1}^{3},\Delta_{3}^{1},\Delta_{12}^{23},\Delta_{23}^{12},\Delta_{123}^{123})$$ is an extended cluster ($\Delta_2^2, \Delta_{23}^{23}, \Delta_{3}^{2},\Delta_{2}^{3}$ are cluster variables and the rest are frozen variables). Using the mutation: $$\Delta_{12}^{12} \Delta_{23}^{23} = \Delta_{12}^{23}\Delta_{23}^{12} + \Delta_{2}^{2} \Delta_{123}^{123},$$ we obtain $\Delta_{12}^{12}$ and another extended cluster $$(\Delta_2^2, \Delta_{12}^{12}, \Delta_{3}^{2},\Delta_{2}^{3} \mid \Delta_{1}^{3},\Delta_{3}^{1},\Delta_{12}^{23},\Delta_{23}^{12},\Delta_{123}^{123}).$$ How many clusters do this cluster algebra have? Thank you very much.

Edit: I learn this example from the beginning of a talk by M. Shapiro.

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    $\begingroup$ Could you add some motivation? Is this mutation used in some paper? And what does one get from explicitly computing? $\endgroup$ – ThiKu Jul 25 '16 at 13:02
  • $\begingroup$ For the first question: $x_{11}$ shows up in $\Delta_{123}^{123}$ in a summand of the form $x_{11}\Delta_{23}^{23}$ and this is the only way to get hold on it. There is also no problem in getting the third summand of $\Delta_{123}^{123}$, because it is $x_{31}\Delta_{12}^{23}$ and you know $x_{31}=\Delta^3_1$. But it's not clear to me how to get the second summand $x_{21}\Delta_{13}^{23}$. One half of it is computable from $\Delta_{23}^{12}\Delta_1^3$ but I don't see how to get the other half. $\endgroup$ – ThiKu Jul 25 '16 at 13:11
  • $\begingroup$ @ThiKu, thank you very much. I edited the post. This is an example in the beginning of a talk by M. Shapiro. $\endgroup$ – Jianrong Li Jul 25 '16 at 13:30
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Let $$F = \big\{ \Delta_2^2, \Delta_{23}^{23}, \Delta_{3}^{2}, \Delta_{2}^{3}, \Delta_{1}^{3}, \Delta_{3}^{1}, \Delta_{12}^{23}, \Delta_{23}^{12}, \Delta_{123}^{123} \big\}.$$

The claim in the linked talk is not that $F$ is a linear basis for $\mathbb{C}[Mat_3] = \mathbb{C}[x_{ij}]$, but that each $x_{ij}$ can be expressed as a rational function in the $\Delta^J_I \in F$. The claim is actually a bit more, it is that each minor of the matrix $M = (x_{ij})$ is a subtraction-free rational function in the elements of $F$. This means if each minor in $F$ is positive, then the matrix is totally positive.

Here is how to recover $x_{11}$. Notice we are given $x_{22}, x_{23}, x_{32}, x_{13}$, and $x_{31}$. We can then recover $$x_{33} = \frac{\Delta^{23}_{23} + x_{23}x_{32}}{x_{22}}$$ since we when also given $\Delta^{23}_{23}$. We can then similarly recover $x_{12}$ and $x_{21}$ since we have the minors $\Delta^{23}_{12}$ and $\Delta^{12}_{23}$ along with all but one relevant coordinate in each case. We can then recover $$\Delta_{12}^{12} = \frac{x_{22}\Delta_{123}^{123} + \Delta^{12}_{23}\Delta^{23}_{12}}{\Delta^{23}_{23}}$$ using the Lewis Carroll identity. Finally we get $x_{11}$ using $\Delta_{12}^{12}$ along with $x_{12}, x_{21}$, and $x_{22}$.

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  • $\begingroup$ thank you very much. What is the cluster type of this cluster algebra? How many clusters and cluster variables are there in this cluster algebra? $\endgroup$ – Jianrong Li Jul 25 '16 at 16:14
  • $\begingroup$ It is type D4. I forget the number of clusters but it is in Keller's survey and in a number of other papers. It has the same cluster type as Gr(3,6), incidentally. $\endgroup$ – Jan Grabowski Jul 25 '16 at 17:28
  • $\begingroup$ @Jan Grabowski, thank you very much. $\endgroup$ – Jianrong Li Jul 25 '16 at 18:53

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