Let me recall some quick definitions. A projective hyperkahler manifold is a simply connected smooth projective variety $M$ such that $H^0(M,\Omega\_M^2)=\mathbb C\sigma$, with $\sigma$ an everywhere non-degenerate holomorphic 2-form. 


A normal projective variety $M$ is said to have $\textit{symplectic singularities}$ if the smooth part $M\_{reg}$ admits a symplectic 2-form $\omega$ such that for any resolution $\pi:\tilde{M}\rightarrow M$, $(\pi|\_{\pi^{-1}(M\_{reg})})^\*\omega$ extends to a holomorphic 2-form on all of $\tilde{M}$.


I've seen it stated that given a hyperkahler manifold $M$ and a dominant rational map $M \dashrightarrow \overline{M}$, with $\overline{M}$ a normal projective variety with symplectic singularities, then $\dim M=\dim \overline{M}$. Moreover, supposedly this follows from the definitions. 


I was wondering why this fact is true? 



Dominant map from hyperkahler manifolds to normal projective varieties with symplectic singularities
So apparently I was just being silly and it does indeed follow from the definitions. Let $f:M\dashrightarrow \overline{M}$ be our dominant rational map between a hyperkahler manifold $M$ and a projective normal variety $\overline{M}$ with symplectic singularities. Let $\pi:\tilde{M}\rightarrow \overline{M}$ be a resolution of singularities being an isomorphism over $\overline{M}\_{reg}$. Then we get a dominant rational map $g=\pi^{-1}\circ f:M\dashrightarrow \tilde{M}$. We know $(\pi|\_{\pi^{-1}(\overline{M}\_{reg})})^\*\omega$ extends to a global holomorphic form $\tilde{\omega}\in H^2(\tilde{M},\Omega^2\_{\tilde{M}})$. Moreover, $g$ is defined on an open set $U$ with complement codimension at least 2, so pulling back $\tilde{\omega}$ gives a nonzero holomorphic form $$g^\*\tilde{\omega}\in H^2(U,\Omega^2\_M|\_U)=H^2(M,\Omega^2\_M)=\mathbb C \sigma.$$ Thus it must be non degenerate. But restricting to the smooth locus $V\subset \tilde{M}$ of $g$, we get that for any $p\in g^{-1}(V)$, $$dg\_p:T\_{M,p}\rightarrow T\_{\tilde{M},g(p)}$$ is surjective. Thus for any $v\in\ker dg\_p,u\in T\_{M,p}$, we have $$g^\*\tilde{\omega}(v,u)=\tilde{\omega}(dg\_p(v),dg\_p(u))=\tilde{\omega}(0,dg\_p(u))=0.$$ By non-degeneracy, we must have $v=0$. Thus $\ker dg\_p=0$, so $$\dim M=\dim T\_{M,p}=\dim T\_{\tilde{M},g(p)}=\dim \tilde{M}=\dim \overline{M},$$ as claimed.


