Let $(M,g)$ be a $2n$-dimensional almost Hermitian manifold ($n\geq 2$) with a almost complex structure $\cal J$ (*not necessary integrable*). i.e.,
 $${\cal J}^2=-I,\quad\qquad g({\cal J} X,{\cal J} Y)=g(X,Y).$$
Suppose that $\{X\_i,{\cal J}X\_i\}$ be any local orthonormal ${\cal J}$-frame and the following relations hold for $i\neq j$ 
$$K(X\_i,X\_j)-K(X\_j,{\cal J}X\_j)=\frac{1}{2n-2}\{\rho(X\_i,X\_i)-\rho({\cal J}X\_j,{\cal J}X\_j)\}\tag{1}$$
$$K(X\_i,X\_j)-K(X\_i,{\cal J}X\_j)=\frac{1}{2n-2}\{\rho(X\_j,X\_j)-\rho({\cal J}X\_j,{\cal J}X\_j)\}\tag{2}$$
where $\rho$ and $K$ are Ricci curvature tensor and sectional curvature respectively.
Then 



> 
> Can be deduce that $(M,\cal J,g)$ is of constant curvature?
> 
> 
> 


This question comes from the study of conformally flat almost Hermitian manifolds.


Your suggestions will be appreciated.



Almost Hermitian manifolds of constant curvature
The answer to the question as asked is 'No', at least when $2n=4$. In that case, the conditions (1) and (2) only involve the Weyl curvature of the underlying metric $g$, so, in particular, when $g$ is conformally flat, then the conditions are satisfied, and there are many conformally flat metrics in dimension $4$ that do not have constant curvature. 


To construct an example of $(g,\mathcal{J})$ that does not have constant curvature just choose any conformally flat metric $g$ in dimension $4$ that does not have constant sectional curvature and any $g$-compatible almost complex structure $\mathcal{J}$.


Since the Ricci curvature is not constrained, there are *probably* examples that are not conformally flat, but they do, at least, have to be self-dual, i.e., $W\_-$ must vanish (which is implied by (1) and (2)). Meanwhile, the conditions (1) and (2) imply that $W\_+$ must take values in a certain rank $2$ subbundle of the self-dual Weyl curvatures that is defined using $\mathcal{J}$: If $\omega$ is the the (self-dual) $2$-form defined so that $\omega(v,w) = g(\mathcal{J}\_xv,w)$ for all $v,w\in T\_xM$, then the self-dual $2$-forms can be split as $\Lambda^2\_+(T^\*M) = \mathbb{R}\omega\oplus E$ where $E$ is a bundle of rank $2$. Then (1) and (2) are equivalent to the requirement that the Weyl curvature be self-dual and take values in the rank $2$ subbundle $\omega\circ E\subset S^2\bigl(\Lambda^2\_+(T^\*M)\bigr)$.


