# When is mean curvature flow a Hamiltonian isotopy?

Assume we have a compact immersed Lagrangian $L$ in a Kaehler manifold $X$. Recall that a normal vector field $v \in \Gamma(L, N)$ is called Hamiltonian iff $\omega(v, \bullet)$ is an exact 1-form. My question is: what are possible conditions on the geometry of $L$ or $X$ that guarantee that mean curvature flow is a Hamiltonian isotopy (for short time)? I think a necessary condition is that Maslov class is zero. I don't think it's sufficient if $H^1(L, \mathbb{R})\neq 0$.

My specific interest is in Lagrangian tori in $\mathbb{C}P^2$, although any facts in the general direction would be appreciated.

P.S.: this question has been asked on math.stackexchange.com, though I did not have much success with it there.

When $M$ is Kahler Einstein (which is the situation where MCF preserves Lagrangian condition), we have

$$\frac{\partial H}{\partial t} = -d d^*H.$$

Taking cohomology class gives

$$\frac{\partial [H]}{\partial t} = [-d d^*H] = 0.$$

So if $L$ is of zero maslov class ($[H_{t=0}] = 0$), then so is $L_t$ all $t$ and thus the necessary condition is indeed sufficient.

• John, kind of late, but it appears that the variation equation you've written does not quite hold for non-CY manifolds. Proposition 1.6.1 in Smoczyk's Habilitation thesis gives formula with a term proportional to Ricci curvature of the ambient manifold. This final conclusion is still correct though
– user74900
May 28 '18 at 12:33

You might find interest in the following work by I. Castro and A. M. Lerma on Lagrangian mean curvature flow and the Clifford torus.

• This works did not address the problem discussed here, if I remember correctly? Aug 11 '17 at 12:51