Let $(X,\omega,J)$ be a 4-dim almost Kahler manifold, equivalently, $(X,\omega)$ is an 4-dim symplectic manifold, $J$ is an almost complex structure on $X$ which is compatible with $\omega$, $g$ is the Riemann metric $\omega(X,Y)=g(JX,Y)$.


Denote $D$ to be the Levi-Civita connection which is compatible with $g$. 


A $(0,2)$-tensor $B$ is defined by $B\_{ij}=g^{kl}g\_{mn}D\_kJ\_i^m D\_lJ\_j^n$.


I want to know how to deduce the equation $B=\frac{1}{4}|DJ|^2 g$ under the conditions given above?


REMARK: It seems that the condition '$X$ is 4-dim' is crucial.



How to deduce this equation for a 4-dim almost Kahler manifold?
Here is one way to do this computation:


Choose a local coframing $\eta = (\eta\_i)$ 
in which $g = {\eta\_1}^2{+}{\eta\_2}^2{+}{\eta\_3}^2{+}{\eta\_4}^2$
and $\omega = \eta\_1\wedge\eta\_2{+}\eta\_3\wedge\eta\_4$. Let $\nabla$
be the Levi-Civita connection of $g$ and orient $M$ so that $\tfrac12\omega^2$
is the positive volume form for $g$. 


The complex structure $J$ defined by $\omega$ is compatible with $g$, 
so $\omega$ is a self-dual $2$-form of constant norm. It follows that
$\nabla\_X\omega$ must be self-dual for any vector field $X$ and it must
satisfy $\omega\wedge(\nabla\_X\omega) = 0$ as well. Since $\kappa=\eta\_1\wedge\eta\_3+\eta\_4\wedge\eta\_2$ 
and $\lambda=\eta\_1\wedge\eta\_4+\eta\_2\wedge\eta\_3$, together
with $\omega$, form a basis for the self-dual $2$-forms on the domain of
the coframing $\eta$, it follows that
$$
\nabla\omega = \alpha\otimes\kappa+\beta\otimes\lambda
$$ 
where $\alpha$ and $\beta$ are $1$-forms on the domain of the coframing.


Since $\omega$ is closed and $\nabla$ is torsion-free, we have
$$
0 = d\omega = \alpha\wedge\kappa+\beta\wedge\lambda.
$$
Thus, if $
\alpha = a\_1\ \eta\_1 + a\_2\ \eta\_2 + a\_3\ \eta\_3 + a\_4\ \eta\_4\ ,
$
then
$$
\beta = a\_2\ \eta\_1 - a\_1\ \eta\_2 + a\_4\ \eta\_3 - a\_3\ \eta\_4\ . 
$$


At a point where all of the $a\_i$ vanish, $\nabla\omega$
vanishes, so that the $(0,2)$-tensor $B$ vanishes there as well,
so there is nothing to prove.


Near a point where not all of the $a\_i$ vanish, one can choose
the coframing $\eta$ in such a way that $a\_2=a\_3=a\_4=0$ while $a\_1 = a >0$.
(NB: This is because the set of $g$-orthonormal coframings $\eta$ for which $\omega$, $\kappa$, and $\lambda$ have the given expressions in terms of $\eta$ is the set of (local) sections of 
an $SU(2)$-bundle, and this $SU(2)\subset SO(4)$ acts (simply) transitively on the 
unit sphere in $\mathbb{R}^4$.)


In this coframing, 
$$
\nabla\omega = a\eta\_1\otimes(\eta\_1\wedge\eta\_3{+}\eta\_4\wedge\eta\_2)
 -a\eta\_2\otimes(\eta\_1\wedge\eta\_4{+}\eta\_2\wedge\eta\_3).
$$
Thus, $D\_3J = D\_4J=0$ while $D\_1J = a K$ and $D\_2J = -a L$, 
where $K$ and $L$ are the almost complex structures 
that are compatible with $g$ and satisfy
$$
g(KX,Y) = \kappa(X,Y)\qquad\text{and}\qquad g(LX,Y) = \lambda(X,Y).
$$


Because the coframe $\eta$
is $g$-orthonormal, the matrix $(g\_{ij})$ is the identity matrix.


Because $K$ and $L$ are skew-symmetric, the tensor $B$, regarded
as a (symmetric) endomorphism, is of the form
$$
B = -a^2 K^2 - a^2 L^2 = 2a^2\ Id,
$$
so, regarded as a $(0,2)$ tensor, $B = 2a^2\ g$.


Meanwhile, $\|\nabla\omega\|^2 = 4a^2$, so $\|\nabla J\|^2 = 8a^2$. 


