Edit: Dear bounty hunters using chatGPT, please do not attempt to answer this question. You may not have enough reputation to see it, but several unfruitful AI-generated answers, all following the same (obviously) wrong pattern, have been posted, then deleted.
A Riemannian metric $g$ on a manifold $M$ induces a pointwise inner product on $\Lambda^2 (TM)$, given on decomposable elements by $$ \langle X\wedge Y, Z\wedge T \rangle = g(X,Z) g(Y,T) - g(X,T) g(Y,Z), $$ and then extended by linearity. Due to its symmetries, the Riemann curvature tensor $R$ induces the so-called curvature operator $\mathcal{R}$, defined by $$ \langle \mathcal{R}(X\wedge Y), Z\wedge T \rangle = R(X,Y,Z,T), $$ which is a symmetric operator. Here I used the convention that the sectional curvature of an orthonormal couple $\{u,v\}$ is $R(u,v,u,v)$. It is a very well known fact that a Riemannian manifold has constant sectional curvature $\kappa$ if and only if the curvature tensor reads $$ R(X,Y,Z,T) = \kappa\big(g(X,Z) g(Y,T) - g(X,T) g(Y,Z) \big), $$ that is, if and only if the associated curvature operator is $\mathcal{R} = \kappa \,\mathrm{Id}_{\Lambda^2(TM)}$.
In the realm of Kähler geometry, it is also known that a Kähler manifold has constant holomorphic sectional curvature if and only if the curvature tensor is given by \begin{align} R(X,Y,Z,T) &= \frac{\kappa}{4} \big(g(X,Z)g(Y,T)-g(X,T)g(Y,Z) \\ & \quad + g(JX,Z)g(JY,T) -g(JX,T)g(JY,Z) + 2g(JX,Y)g(JZ,T) \big). \end{align} Through the scope of the curvature operator $\mathcal{R}$, it appears that the holomorphic sectional curvature is constant equal to $\kappa$ if and only if $$ \mathcal{R} = \frac{\kappa}{4}\left(\mathrm{Id}_{\Lambda^2(TM)} + \bar{J} + A \right) $$ where $\bar{J}(X\wedge Y) = (JX)\wedge (JY)$ and $A$ satisfies $$ \langle A(X\wedge Y), Z\wedge T\rangle = 2g(JX,Y)g(JZ,T) = g(JX,Y)g(JZ,T) + g(X,JY)g(Z,JT). $$
My question is the following: Is there any nice algebraic expression for $A(X\wedge Y)$ solely in terms of $X$, $Y$, and $J$?
Due to the presence of mixed terms between $X$ and $Y$ as well as the complex structure $J$ in the $(Z,T)$ entries of $\langle A(X\wedge Y), Z\wedge T\rangle$, it is somewhat non trivial to find such an algebraic expression. Note that $\bar{J}^2 = \mathrm{Id}_{\Lambda^2(TM)}$, which I interpret as (part of) the $J$-symmetric part of the curvature. My intuition is that $A$ shall be the $J$-anti-symmetric part of the curvature, and such an algebraic expression should then involve $(JX)\wedge Y$ and $X\wedge (JY)$, but my computations are somewhat leading nowhere.
