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Exchange pattern, see Section 2 in "cluster algebras I: foundations" by Fomin and Zelevinsky or How to understand exchange pattern?

Given an example $\cdots \overset{2}{-} t_1 \overset{1}{-} t_2 \overset{2}{-} t_3 \overset{1}{-} t_4 \overset{2}{-} \cdots$, where \begin{align*} \mathbf{x}(t_1) = (x_1(t_1), x_2(t_1)) = (x,y); \ (M_1(t_1), M_1(t_2)) = (y,1); \\ \mathbf{x}(t_2) = (x_1(t_2), x_2(t_2)) = (\frac{y+1}{x},y); \\ \mathbf{x}(t_3) = (x_1(t_3), x_2(t_3)) = (\frac{y+1}{x},\frac{x+y+1}{xy}); \\ \mathbf{x}(t_4) = (x_1(t_4), x_2(t_4)) = (\frac{x+1}{y},\frac{x+y+1}{xy}). \end{align*}

By E3, we have \begin{align*} (M_2(t_2), M_2(t_3)) = (\frac{y+1}{x},1), \ (M_1(t_3), M_1(t_4)) = (\frac{x+y+1}{xy},1). \end{align*}

But \begin{align*} M_0 = (M_2(t_2) + M_2(t_3))|_{\frac{y+1}{x} = 0} = (\frac{y+1}{x} +1 )|_{\frac{y+1}{x} = 0} = 1; \\ \frac{M_1(t_3)}{M_1(t_4)} = \frac{M_1(t_2)}{M_1(t_1)}|_{y \leftarrow \frac{1}{y}} \Rightarrow \frac{x+y+1}{xy} \neq \frac{1}{y}|_{y \leftarrow \frac{1}{y}} = y. \end{align*} This contradicts E4. I don't know where is wrong. Any help is needed.

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Using Fomin-Zelevinsky's notation, we can start with $(M_1(t_1),M_1(t_2))=(x_2,1)$, $(M_2(t_2),M_2(t_3))=(x_1,1)$. This can be considered initial data, considering $t_2$ as the initial cluster: I don't think the axioms can be used to deduce the values of the second pair from the first. However, note that this choice is consistent with equation 2.6 in Fomin-Zelevinsky (which I am assuming to be E3 in your notation): we have $x_2|M_1(t_1)$ and $x_1|M_1(t_2)$, for example.

We can then use the fourth axiom (equation 2.7 in FZ, i.e. E4 in your notation) to compute the pair $(M_1(t_3),M_1(t_4))$. We have $$M_0=(M_2(t_2)+M_2(t_3))|_{x_1=0}=(x_1+1)|_{x_1=0}=1.$$ So $$\frac{M_1(t_3)}{M_1(t_4)}=\frac{M_1(t_2)}{M_1(t_1)}|_{x_2\leftarrow M_0/x_2}=\frac{1}{x_2}|_{x_2\leftarrow 1/x_2}=x_2.$$ Hence $M_1(t_3)=x_2$ and $M_1(t_4)=1$. Note that this then gives the correct exchange from $t_3$ to $t_4$. We have $$x_1(t_3)x_1(t_4)=M_1(t_3)(\mathbf{x}(t_3))+M_1(t_4)(\mathbf{x}(t_4)),$$ i.e. (using the values for $\mathbf{x}(t_3)$ and $\mathbf{x}(t_4)$ above): $$\frac{y+1}{x}\cdot \frac{x+1}{y}=\frac{x+y+1}{xy}+1.$$ I think the problem arises from writing the $M_i(t)$ in terms of the initial cluster and not in terms of separate variables $x_1,x_2$ as in Fomin-Zelevinsky.

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  • $\begingroup$ @ Robert J. Marsh Thank you very much for your reply. I have one more question. You mean that for any $t \in \mathbb{T}_n$, \begin{align*} M_j(t) = p_j(t)\prod_{i \in I} x^{b_i}_i \end{align*} is only dependent on the coefficient $p_j(t)$. When for each $t \overset{j}{-} t'$ we compute \begin{align*} x_j(t) x_j(t') = M_j(t) (\mathbf{x}(t)) + M_j(t') (\mathbf{x}(t')), \end{align*} we use the variable $x_j(t)$ (resp. $x_j(t')$) in $\mathbf{x}(t)$ (resp. $\mathbf{x}(t')$) instead of $x_j$ (resp. $x_j$) in $M_{j}(t)$ (resp. $M_{j}(t')$). Is it right? Thank you again. $\endgroup$
    – bing
    Commented Sep 20, 2016 at 2:37
  • $\begingroup$ For the first point, I'd prefer to say that $M_j(t)$ depends on the $x_i$ as well as $p_j(t)$. For the question, that's right - the notation means to substitute $x_1(t),\ldots ,x_n(t)$ for $x_1,\ldots ,x_n$ respectively. (I wasn't notified by the system of your reply, so only just saw it). $\endgroup$
    – Robert J. Marsh
    Commented Oct 2, 2016 at 20:13
  • $\begingroup$ @ Robert J. Marsh I got it. Thank you very much. $\endgroup$
    – bing
    Commented Oct 4, 2016 at 4:25

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