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?

  • 1
    $\begingroup$ 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, 2014 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, 2014 at 15:23

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.