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.

  • $\begingroup$ 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. $\endgroup$
    – u184
    Nov 11 '19 at 11:39
  • $\begingroup$ @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. $\endgroup$
    – AmorFati
    Sep 11 '20 at 7:17
  • 1
    $\begingroup$ @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. $\endgroup$ Sep 11 '20 at 9:32
  • $\begingroup$ @RobertBryant Thank you! $\endgroup$
    – AmorFati
    Sep 11 '20 at 21:40

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