1
$\begingroup$

This question comes from Example 4.2 of the Gekhtman-Shapiro-Vainshtein book Cluster Algebras and Poisson Geometry which I have attached. My goal is to understand how to compute $\{x_1’,x_2’\}=-\lambda x_2^2/x_1-\mu x_2x_3/x_1-\lambda x_2x_3/x_1$ knowing already that $\{x_1,x_2\}=\lambda x_1x_2,\{x_2,x_3\}=\mu x_2x_3,$ and $\{x_1,x_3\}=\nu x_1x_3$ (by log-canonicality).

My thought was to use the Leibniz identity $\{f_1f_2,f_3\}=f_1\{f_2,f_3\}+\{f_1,f_3\}f_2$ by taking $f_1=1/x_1, f_2=x_2+x_3,$ and $f_3=x_2$ which gives $\{x_1’,x_2’\}=\{x_2+x_3,x_2\}/x_1+\{1/x_1,x_2\}(x_2+x_3).$ That first bracket term will give me the $-\mu x_2x_3$ but for the second bracket term I do not know how to compute $\{1/x_1,x_2\}$ from knowing $\{x_1,x_2\}.$

Any guidance is appreciated as I am very new to Lie brackets and Poisson brackets.

enter image description here

$\endgroup$
1
$\begingroup$

In local coordinates $(x^1, \ldots, x^d)$ you can always calculate the Poisson bracket of arbitrary functions in terms of Poisson brackets of the coordinates $\{x^i, x^j\}$ and first derivatives of the functions (by the fact that we have a skew-symmetric bi-derivation, combined with Taylor's theorem, see e.g. [Lectures on Poisson Geometry, Proposition 14]): $$\{f,g\} = \sum_{i,j=1}^d \{x^i, x^j\} \frac{\partial f}{\partial x^i} \frac{\partial g}{\partial x^j} = \sum_{1 \leq i < j \leq d} \{x^i, x^j\} \bigg(\frac{\partial f}{\partial x^i} \frac{\partial g}{\partial x^j} - \frac{\partial f}{\partial x^j} \frac{\partial g}{\partial x^i}\bigg)$$

In particular in coordinates $(x_1,x_2,x_3)$ we have $\{1/x_1, x_2\} = \{x_1,x_2\}\frac{\partial}{\partial x_1}\big(1/x_1\big)\frac{\partial x_2}{\partial x_2} = -\{x_1,x_2\}/x_1^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.