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I am puzzled if we can establish differential relations about anti-self-dual 2-forms on the Kähler space similar to ones for self-dual forms?

Let $(\mathcal{M},g,J,\omega = J^{(1)})$ be a Kähler space. We will pick the orientation such that the Kähler form $J^{(1)}$ is self-dual. Then it is known that the covariant derivatives of self-dual forms satisfy a set of simple relations.

Denote by $J^{(i)}$, $i=1,2,3$ a basis on the space of self-dual 2-forms $\Omega^+(\mathcal{M})$. Then the following two facts are known:

  • We can further specify a basis on $\Omega^2(\mathcal{M})$ that satisfies quaternionic algebra relations \begin{equation} J^{(i)\,p}_{m} J^{(j)}_{\quad p n} = -\delta^{ij}\, J^{(i)}_{mn} + \epsilon^{ijk} \,J^{(k)}_{mn}. \tag{1}\label{1} \end{equation} In fact we can always locally pick such a basis for any four-dimensional Riemann manifold.
  • The covariant derivatives of $J^{(i)}$ are \begin{equation} \begin{aligned} &\nabla_X J^{(1)} = 0\\ &\nabla_X J^{(2)} = P(X) J^{(3)}\\ &\nabla_X J^{(3)} = - P(X) J^{(2)} \end{aligned} \tag{2}\label{2} \end{equation} where $X\in T\mathcal{M}$ and $P\in \Omega^1(\mathcal{M})$ is the Ricci potential: $\mathcal{R} = d P$.

Now similarly denote by $I^{(i)}$, $i=1,2,3$ a basis on the space of anti-self-dual 2-forms $\Omega^-(\mathcal{M})$. Again, we can pick the basis to satisfy quaternionic-like relations $$ I^{(i)\,p}_{m} I^{(j)}_{\quad p n} = -\delta^{ij}\, I^{(i)}_{mn} - \epsilon^{ijk} \,I^{(k)}_{mn}.$$ Are there similar nice relations for $\nabla_X I^{(i)}$ ?


P.S. Since I have never seen a proof of \eqref{2} in the literature, I will sketch it here. First of all, notice from \eqref{1} that $J^{(2)}, J^{(3)}$ are anti-J-invariant forms. In fact, they are the only anti-J-invariant 2-forms, as all ASD forms (and, obviously, $J^{(1)}=J$) are J-invariant. We can then define J-conjugation $\phi: \beta(X,Y)\mapsto \beta(JX,JY)$ which is an automorphism of $\Omega^2(\mathcal{M})$ which commutes with covariant derivative $$\phi (\nabla_X \beta) = \nabla_X \phi (\beta).$$ Consequently, the covariant derivative of $J^{(2)}, J^{(3)}$ must be a section of $T^*(\mathcal{M}) \times \Omega^{(-J)}(\mathcal{M})$, where the latter is spanned by $J^{(2)}, J^{(3)}$. Furthermore, notice that $$J^{(i) mn} \nabla_X J^{(j)}_{mn}= - J^{(j) mn} \nabla_X J^{(i)}_{mn}$$ which implies that \begin{align*} &\nabla_X J^{(2)} = k(X) J^{(3)}\\ &\nabla_X J^{(3)} = - k(X) J^{(2)} \end{align*} for some 1-form $k$. It can be found from integrability constraint $$J^{(3)pq}[\nabla_m,\nabla_n] J^{(2)}_{pq} = 4 \mathcal{R}_{mn} = 4 (dk)_{mn}.$$ Using $\mathcal{R}= dP$ we find that $k=P$ up to a closed form.

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    $\begingroup$ Your post referenced (1) and (2) but did not seem to define them, so I guessed at what seemed to be the most sensible definition. I hope that it was correct. $\endgroup$
    – LSpice
    Oct 3, 2022 at 16:50
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    $\begingroup$ Thank you! Your correction is right. I did not know that you have to do \tag{1} for numeration in MathJax $\endgroup$ Oct 3, 2022 at 17:34

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