Let $(M, J, h)$ be an almost Hermitian manifold, where $J$ is an almost complex structure and $h$ is a Hermitian metric. Let $D$ be the unique $h$-connection compatible with $J$, i.e. $Dh = 0$ and $(0,1)$-part of $D$ coincides with the Cauchy-Riemann operator $\overline{\partial}_J$. Let $\tau$ be the torsion of $D$. If we decompose our connection $D$ into (1,0) and (0,1) parts $D = D' + D''$, then the torsion $\tau$ of $D$ will also be decomposed $\tau = \tau' + \tau''$. It is not hard to see that $\tau' = N$, the Nijenhuis tensor for $J$, which is exactly the obstruction for an integrable complex structure. What about the other part $\tau''$? Is there any geometric meaning?

So far, I find that it will be an obstruction for being almost Kähler, i.e. $d\omega=0$, where $\Im h:=\omega$. I mean, the following holds: $d\omega=0 \Rightarrow \tau''=0$.

My question is about the converse. Is it true that $\tau''=0 \Rightarrow d\omega=0 $?

By the way, if $M$ itself a Hermitian manifold, it is well-known that $\tau''=\tau$, which will be the exact obstruction for being Kähler.