5
$\begingroup$

Let $k$ be an algebraically closed field of characteristic zero. I am not sure what the right definition of a hyperkahler variety over $k$ is, but I think the following might be close enough.

Definition. A smooth proper connected scheme $X$ over $k$ is a hyperkahler variety (over $k$) if $\mathrm{h}^{2,0}(X) = 1$, the etale fundamental group $\pi_1^{et}(X)$ of $X$ is trivial, $\omega_X$ is trivial, and (added later) the generator of $\mathrm{H}^{2,0}(X)$ is everywhere non-degenerate.

With this definition, it is not clear whether every hyperkahler variety over $k$ is projective. If $X$ is a two-dimensional hyperkahler variety over $k$, then $X$ is projective. But what about $\dim X > 2$?

Is every hyperkahler variety over $k$ projective?

PS. Please feel free to correct my definition of a hyperkahler variety over $k$.

$\endgroup$
4
  • 7
    $\begingroup$ That is not the correct definition: a product of a hyperkaehler variety and a Calabi-Yau threefold will satisfy that condition by Kuenneth. You should add the hypothesis that the generator of $H^{2,0}(X)=H^0(X,\bigwedge^2 \Omega_{X/k})$ is everywhere nondegenerate. $\endgroup$ Jun 12 '18 at 21:30
  • $\begingroup$ Maybe the toric hyperkähler varieties (a.k.a. hypertoric varieties) of Hausel-Sturmfels (2002, Def. 6.1)? $\endgroup$ Jun 12 '18 at 22:02
  • 1
    $\begingroup$ FZ: It seems that the OP is interested in compact hyperkahler manifolds $\endgroup$
    – wnx
    Jun 12 '18 at 22:21
  • $\begingroup$ (Hyper)Kaehler or (hyper)Kähler and étale are right! $\endgroup$ Jun 13 '18 at 11:11
2
$\begingroup$

With the correct definition of hyperkähler (which as Jason said requires $H^0(X,\Omega^2_X)$ to be generated by a holomorphic symplectic form), there are examples constructed by Yoshioka in section 4.4 here. The field $k=\mathbb{C}$.

These are non-Kähler compact complex simply connected holomorphic symplectic manifolds which are bimeromorphic to a projective hyperkähler manifold via a Mukai flop. They are therefore Moishezon, but certainly not projective. They are also discussed for example in the book by Gross-Huybrechts-Joyce, example 21.7.

$\endgroup$
2
  • 2
    $\begingroup$ This does not anser the question, which asks for hyperkähler schemes. $\endgroup$
    – abx
    Jun 13 '18 at 10:28
  • 1
    $\begingroup$ I see, I was just looking at the title which asked for "algebraic" and I thought the OP wanted an algebraic space. Well, I don't know the answer to the actual question then. $\endgroup$
    – YangMills
    Jun 13 '18 at 16:34

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.