Suppose that $(M,g, I, J,K)$ is a Hyper Kahler manifold, then the two forms:
\begin{alignat*}{3} \omega_I(v,w)=&g(Iv,w),\qquad &&\omega_J(v,w)=g(Jv,w),\qquad &&\omega_K(v,w)=g(Kv,w) \end{alignat*} are closed, so there are three symplectic structures on $M$. In his book "Monopoles, minimal surfaces, and algebraic curves" Hitchin writes that we can equivalently describe Hyper Kahler manifolds by three symplectic structures as follows:
Let $\omega_1,\omega_2,\omega_3$ be symplectic forms, then as bundle isomorphism $TM\rightarrow T^*M$, there exist inverses $\omega_1^{-1},\omega_2^{1}, \omega_3^{-1}$, which are global sections of $\Lambda^2(TM)$. Then the endomorphism $K$ is given by the contraction: \begin{align*} K=\omega_1^{-1}\omega_2 \end{align*} and $g$ is defined by: \begin{align*} g(v,w)=\omega_3(Kv, w) \end{align*}
I understand how we get $g$, but I do not see how we deduce that $a)$ $K^2=0$, and $b)$ $\nabla K=0$. Once I figure this out, $I$, and $J$ are trivial. In particular if I view $\omega_i$ as the map: \begin{align*} \omega_i:TM&\longrightarrow T^*M\\ v&\longmapsto \omega_i(v,\cdot) \end{align*} and $\omega_i^{-1}$ as the map: \begin{align*} \omega_i^{-1}:T^*M&\longrightarrow TM\\ \eta&\longmapsto \omega_i^{-1}(\eta,\cdot) \end{align*} so: \begin{align*} K(v)=&\omega_1^{-1}\circ \omega_2(v,\cdot)\\ =&\omega_1^{-1}(\omega_2(v,\cdot),\cdot) \end{align*} If I apply $K$ again: \begin{align*} K^2(v)=&\omega_1^{-1}\circ \omega_2(\omega_1^{-1}(\omega_2(v,\cdot),\cdot),\cdot)\\ =&\omega_1^{-1}(\omega_2(\omega_1^{-1}(\omega_2(v,\cdot),\cdot),\cdot),\cdot) \end{align*} this definitely seems like the wrong way to think about this, so does anyone have any pointers here?
Ok, I just misinterpreted the remark, and my statement as written is very much not true, specifically for the reasons I mention in the comments.
Really what the remark is saying is that given a hyper Kahler structure, then two hyper Kahler structures are equivalent if and only if their associated Kahler forms agree, which is much easier to show, probably because it is actually possible to show :).