# Ricci flow preserves almost Kahler condition?

I have been unable to find a reference to the following (perhaps too naive) question.

Suppose we have an almost Kahler manifold $$(M^{2n},\omega,J,g)$$ i.e. the almost complex structure $$J$$ is non-integrable but $$\omega$$ is closed. If we run the Ricci flow on the metric i.e. $$\frac{\partial g_t}{\partial t}=-2Ric(g_t)$$ while leaving $$J$$ unchanged, so that we also get a flow for $$\omega_t:=g_t(J\cdot,\cdot)$$, does this imply that $$g_t$$ stays compatible wih $$J$$? The answer is known to be yes if $$J$$ is integrable and this corresponds to the Kahler Ricci flow. Moreover, $$\omega_t$$ stays closed.

Assuming the compatibility with the almost complex structure holds, does it also follow that $$d\omega_t=0$$ for all $$t$$?

I fail to see the difference to proving the above statement whether $$J$$ is integrable or not.

The answer is, in general 'no': Under the Ricci-flow, a metric need not remain compatible with $$J$$ if $$J$$ is not integrable, even if the associated $$2$$-form $$\omega$$ is assumed closed.

I don't see how to see this directly without doing some calculation, but the basic idea is this: Consider an almost-complex $$4$$-manifold $$(M^4,J)$$. The set of $$J$$-compatible metrics on $$M$$ is an open cone in an affine space, the sections of a positive cone in a bundle $$H\subset S^2(T^*M)$$ of real rank $$4$$ over $$M$$. This bundle has a linear embedding $$\Phi:H\to\Lambda^{1,1}(T^*M)$$ into the $$\mathbb{R}$$-valued 2-forms of $$J$$-type $$(1,1)$$ defined by letting $$\Phi(g)(u,v) = g(Ju,v)$$ for any $$u,v\in T_xM$$. The almost-Kähler condition is that $$\mathrm{d}\bigl(\Phi(g)\bigr) = 0$$, so it is a linear condition on the sections of this bundle.

In order for the Ricci-flow starting at a metric $$g$$ to yield a family of metrics compatible with $$J$$, it would at least have to be true that $$\mathrm{Ric}(g)$$ be a section of $$H$$. Unfortunately, in the case that $$J$$ is not integrable, the condition $$\mathrm{d}\bigl(\Phi(g)\bigr) = 0$$ is not sufficient to guarantee that $$\mathrm{Ric}(g)$$ be a section of $$H$$.

The reason is as follows: There is a canonical splitting $$S^2(T^*M) = H \oplus C$$, where $$C$$ is the bundle of quadratic forms that are the real part of a $$J$$-bilinear quadratic form on $$M$$. This $$C$$ is a bundle of real rank $$6$$ over $$M$$. Let $$\pi_H$$ (respectively, $$\pi_C$$) denote the canonical projection from $$S^2(T^*M)$$ to $$H$$ (respectively, $$C$$).

By calculation one can show that, when $$\mathrm{d}\bigl(\Phi(g)\bigr) = 0$$, the tensor $$\pi_C\bigl(\mathrm{Ric}(g)\bigr)$$, which is a section of $$C$$, can be written in the form $$\pi_C\bigl(\mathrm{Ric}(g)\bigr) = L\bigl(\nabla^g(N_J)\bigr)$$ where $$N_J$$ is the Nijnhuis tensor of $$J$$ (a section of $$T\otimes_{\mathbb{C}}\Lambda^{0,2}(T^*)$$), $$\nabla^g$$ is a canonical $$g$$- and $$J$$-compatible connection (with torsion, in general), and $$L$$ is a (surjective) canonical linear operator $$L:T^*\otimes_\mathbb{C}T\otimes_{\mathbb{C}}\Lambda^{0,2}(T^*)\to C$$. In particular, it is easy to show that $$\pi_C\bigl(\mathrm{Ric}(g)\bigr)$$ does not, in general vanish when $$\mathrm{d}\bigl(\Phi(g)\bigr) = 0$$. (Of course, when $$N_J=0$$, i.e., $$J$$ is integrable, the above formula shows that, indeed, $$\pi_C\bigl(\mathrm{Ric}(g)\bigr)=0$$.)

There remains the interesting question as to whether, when $$g$$ satisfies both $$\mathrm{d}\bigl(\Phi(g)\bigr) = 0$$ and $$\pi_C\bigl(\mathrm{Ric}(g)\bigr)=0$$, the Ricci-flow will preserve both of these conditions. I have not done the calculations necessary to check this, but it shouldn't be all that hard to do.

It is not hard to produce examples of non-integrable $$J$$ for which there exist compatible metrics $$g$$ that satisfy both $$\mathrm{d}\bigl(\Phi(g)\bigr) = 0$$ and $$\pi_C\bigl(\mathrm{Ric}(g)\bigr)=0$$ and for which the Ricci-flow does preserve both conditions.

• Many thanks for the great answer. I will try and work out if it is the case that the flow preserves the $d(\Phi(g))=0$ and $\pi_C(Ric(g))=0$ conditions.
– u184
Nov 11 '19 at 11:39
• @Robert This is a very insightful answer. Is there a more straightforward argument when one knows that $J$ is integrable? In the Kähler--Ricci flow literature, I have never seen a justification of the Ricci flow preserving the Kähler condition. Sep 11 '20 at 7:17
• @AmorFati: The reason is that it's a straightforward argument: Specifying a metric $g$ compatible with $J$ is equivalent to specifying a positive $(1,1)$-form $\omega = \Phi(g)$, and, it turns out, when $J$ is integrable, that $$\Phi(\mathrm{Ric}(\Phi^{-1}(\omega))) = R(\omega),$$ where $R$ is a nonlinear second order operator from positive $(1,1)$-forms to exact $(1,1)$-forms. The Kähler–Ricci flow then translates to the equation $$\frac{\partial\omega}{\partial t} = -2R(\omega),$$ which is just a parabolic evolution equation for a closed, positive $(1,1)$-form. Sep 11 '20 at 9:32
• @RobertBryant Thank you! Sep 11 '20 at 21:40