# Are there non-projective, but algebraic, hyperkahler varieties?

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 $k$.

• 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. Jun 12 '18 at 21:30
• Maybe the toric hyperkähler varieties (a.k.a. hypertoric varieties) of Hausel-Sturmfels (2002, Def. 6.1)? Jun 12 '18 at 22:02
• FZ: It seems that the OP is interested in compact hyperkahler manifolds
– wnx
Jun 12 '18 at 22:21
• (Hyper)Kaehler or (hyper)Kähler and étale are right! Jun 13 '18 at 11:11

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 $k=\mathbb{C}$.