In §5.3 of Kontsevich's *[Formality Conjecture](http://www.ihes.fr/~maxim/TEXTS/Formality%20conjecture.pdf)* he writes:



> 
> This (...) gives a remarkable vector field on the space of bi-vector fields on $\mathbf{R}^d$. The evolution with respect to the time $t$ is described by the following non-linear partial differential equation: $$\frac{{\rm d}\alpha}{{\rm d}t} = \sum\_{i,j,k,l,m,k',l',m'} \frac{\partial^3 \alpha\_{ij}}{\partial x\_k \partial x\_l \partial x\_m} \frac{\partial \alpha\_{kk'}}{\partial x\_{l'}} \frac{\partial \alpha\_{ll'}}{\partial x\_{m'}} \frac{\partial \alpha\_{mm'}}{\partial x\_{k'}} \left( \frac{\partial}{\partial x\_i} \wedge \frac{\partial}{\partial x\_j} \right),$$
>  where $\alpha = \sum\_{i,j} \alpha\_{ij}(x) \frac{\partial}{\partial x\_i} \wedge \frac{\partial}{\partial x\_j}$ is a bi-vector field on $\mathbf{R}^d$.
> 
> 
> 


What does ${\rm d}\alpha/{\rm d}t$ here mean exactly? We presume it gives a system of $d(d-1)/2 = {d \choose 2}$ partial differential equations, looking at the equation above component-wise (modulo anti-symmetry of the wedge), and a solution would be a flow $t \mapsto \alpha(t)$, where each $\alpha(t)$ is a bi-vector field.


Next he writes:



> 
> Also, in dimension $d = 2$ the direct calculation shows that the evolution operator gives a conjugation of bi-vector field $\alpha$ by a vector field whose coefficients are differential polynomials in coefficients of $\alpha$.
> 
> 
> 


In this case we have one equation:


$$\frac{{\rm d}\alpha\_{12}}{{\rm d}t} = \sum\_{k,l,m,k',l',m'} \frac{\partial^3 \alpha\_{12}}{\partial x\_k \partial x\_l \partial x\_m} \frac{\partial \alpha\_{kk'}}{\partial x\_{l'}} \frac{\partial \alpha\_{ll'}}{\partial x\_{m'}} \frac{\partial \alpha\_{mm'}}{\partial x\_{k'}}, $$
where the $\alpha\_{ij}$ in the nonzero terms are all $\pm \alpha\_{12}$. Explicitly, writing $u = \alpha\_{12}$, $x\_1 = x, x\_2= y$, we get $$u\_t = u\_{xxx}(u\_y)^3 - u\_{yyy}(u\_x)^3 - 3u\_{xxy}u\_x(u\_y)^2 + 3u\_{xyy}(u\_x)^2u\_y.$$


How do we proceed to show that the solution is a conjugation by a vector field?




---


**Edit (17/06/15)**: Or is it the right hand side of the equation that is a conjugation?


The Schouten bracket of a bivector $\alpha = \alpha^{12}(x) \partial\_1 \wedge \partial\_2$ and a vector $X = X^1\partial\_1 + X^2\partial\_2$ is a bivector: $$\begin{align\*}[\![ \alpha, X ]\!] &= [\![\alpha, X^1\partial\_1]\!] + [\![\alpha, X^2\partial\_2]\!]\\ &=[\alpha^{12}(x)\partial\_1, X^1\partial\_1] \wedge \partial\_2 - [\partial\_2, X^1\partial\_1] \wedge \alpha^{12}(x)\partial\_1 \\ & \quad + [\alpha^{12}(x)\partial\_1, X^2\partial\_2] \wedge \partial\_2 - [\partial\_2, X^2\partial\_2] \wedge \alpha^{12}(x)\partial\_1\\ &=(\alpha^{12}(x)\partial\_1X^1 - X^1\partial\_1\alpha^{12}(x) - X^2\partial\_2 \alpha^{12}(x) + \alpha^{12}(x)\partial\_2X^2)\partial\_1 \wedge \partial\_2.\end{align\*}$$


Putting $u = \alpha^{12}(x), f = X^1, g = X^2,$ this is: $$uf\_x - fu\_x - gu\_y + ug\_y = u(f\_x + g\_y) - fu\_x - gu\_y.$$


~~But this involves $u$'s without derivatives, so we can't match it directly to the RHS above~~.


The Jacobi identity is of no use ~~either~~: every bivector field on $\mathbf{R}^2$ is Poisson.


(Strikethrough on 02/07/15)



Kontsevich's flow on the space of Poisson structures
By looking at scaling, both $f$ and $g$ have to be cubic in $u$. Moreover, monomials in $f$ contain 3 derivatives in $y$ and 2 derivatives in $x$; similarly for $g$.


There are 12 possible monomials in each $f$ and $g$. E.g. the ones in $g$ are $u\_{xxxyy}u^2, u\_{xxxy}u\_yu, u\_{xxyy}u\_xu, u\_{xxx}u\_y^2,u\_{xxx}u\_{yy}u,u\_{xxy}u\_xu\_y,u\_{xxy}u\_{xy}u,u\_{xyy}u\_x^2,u\_{xyy}u\_{xx}u,u\_{yy}u\_{xx}u\_x,u\_{xy}u\_{xx}u\_y,u\_{xy}^2u\_x.$


The coefficients can be determined using a computer algebra system. For instance, the following works:
$$
f=u\_{yyy}u\_x^2+3u\_{yy}u\_xu\_{xy}+u\_yu\_{xy}^2+u\_yu\_xu\_{xyy}+2u\_yu\_{yy}u\_{xx}+u\_y^2u\_{xxy}, \\
g=-u\_xu\_{xy}^2-u\_x^2u\_{xyy}-2u\_{yy}u\_xu\_{xx}-3u\_yu\_{xy}u\_{xx}-u\_yu\_xu\_{xxy}-u\_y^2u\_{xxx}.
$$


Note that $f\_x + g\_y = 0$, so the terms containing $u$ are absent.


