Why Kähler 4-manifold with two compatible complex structure are Hyperkähler

Question

Let $(M,g)$ be a 4-dimensional Riemannian manifold which is Kähler with respect to two independent complex structures $J_1, J_2$. Why if these induce the same complex orientations then $(M,g)$ is hyperkähler?

The only reference I found for this fact is the mathoverflow question https://mathoverflow.net/a/436867/508697, but the answer is quite obscure to me. In particular I cannot see neither why having two independent parallel and self-dual $(1,1)$-forms forces the 3-bundle of self-dual forms to be flat, nor why this fact implies the existence of an hyperkähler structure on $M$.

Answer 1

The MO answer uses linear algebraic facts in dimension $4$ and considerations about parallel sections of some bundle.

Lets unpack them, the calculations for the linear algebra part are kind of annoying so I will just give the statements. Let $(M,g)$ be an oriented Riemannian manifold of dimension $4$. For $x\in M$ let $*_x:\Lambda^k(T_xM)\to \Lambda^{4-k}(T_xM)$ denote the Hodge star induced by the orientation. Then:

1. The metric $g$ induces a linear isomorphism of $\Lambda^2(T_xM)$ to the space of anti-symmetric linear maps $T_xM\to T_xM$ via $\omega(v,w)=g(A_\omega v,w)$. Under this isomorphism the scalar product on $\Lambda^2(T_xM)$ becomes $$\langle \omega_1,\omega_2\rangle = \mathrm{Tr}((A_{\omega_1})^T A_{\omega_2})=-\mathrm{Tr}(A_{\omega_1}A_{\omega_2}).$$
2. $\omega\in \Lambda^2(T_xM)$ satisfies $*_x(\omega) = \pm\omega$, if and only if $A_\omega^2=-\|\omega\|^2\Bbb1$.
3. If $\omega_1, \omega_2\in\Lambda^2(T_xM)$ are both self-dual, (meaning $*(\omega_i)=\omega_i$) then they are orthogonal if and only $A_{\omega_1}A_{\omega_2}+A_{\omega_2}A_{\omega_1} =0$, ie if and only if they anti-commute.

So if $J_1, J_2$ are two complex units on $T_xM$ let $\omega_1,\omega_2$ be the associated $2$-forms. If $J_1, J_2$ induce the same orientation you can check $*_x(\omega_i)=\omega_i$ for this orientation. By Gramm-Schmidt there is some linear combination that makes the $2$-forms orthogonal to each other and norm $1$. Since linear combinations preserve self-duality you get anti-symmetric $I_1, I_2$ with $(I_i)^2=-\Bbb1$ and $I_1I_2+I_2I_1=0$. This implies that $I_1, I_2, I_1I_2$ are $3$ complex units satisfying the same relations as $i,j,k$.

Doing this at every point gives you an almost-hyperkähler structure. Why are the units parallel? The MO answer argues that this follows from dimensional reasons (if an $n$ dimensional euclidean vector bundle has an $n-1$ dimensional flat subbundle then the bundle itself is flat). Perhaps it is simpler to argue as follows:

1. If $A\in \Gamma(TM\otimes TM^*)$ is anti-symmetric then it is parallel if and only if the associated form $\omega\in \Omega^2(TM)$ is parallel.
2. Linear combinations of parallel forms are parallel.
3. If $A, B\in \Gamma(TM\otimes TM^*)$ are parallel then $AB$ is parallel.

Then from $J_1, J_2$ being parallel you get that $I_1, I_2, I_1I_2$ are parallel and the structure is actually hyperkähler.