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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?

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    $\begingroup$ It cannot be true as stated: there may exist a morphism $M\rightarrow \mathbb{P}^n$, with $n=\frac{1}{2}\dim M $. A smooth variety has symplectic singularities ... $\endgroup$ – abx Mar 14 '14 at 14:51
  • $\begingroup$ Yeah this is one of the many reasons I've found the statement strange. I'm guessing they really meant that the symplectic form is non-degenerate and holomorphic. In this case, $\mathbb P^n$ certainly doesn't contradict this. $\endgroup$ – HNuer Mar 14 '14 at 15:23
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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.

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