In §5.3 of Kontsevich's Formality Conjecture 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)


1 Answer 1


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.

  • 2
    $\begingroup$ I think your strategy is sound, but when I try it, your $f$ and $g$ do not give the right expression for the RHS from $u_t = \cdots$ in OP's question. I find the following solution: $f = h_y$, $g = -h_x$, where $$h = (u_x)^2 u_{yy} + (u_y)^2 u_{xx} - 2 u_x u_y u_{xy}.$$ $\endgroup$ Jun 22, 2015 at 20:50
  • $\begingroup$ You're totally right, I made a mistake in the calculation (my $h$ had a $+1$ in the last term). $\endgroup$ Jun 23, 2015 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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