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.
 A: 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.
