I'm seeking to understand the de Rham cohomology of a Hilbert scheme $K3^{[4]}$ of the K3 surface. By Beauville, this 8-dimensional compact manifold is Kaehler, irreducible, holomorphically symplectic (that is, it possesses a nondegenerate form $\rho$ of type (2,0)), and simply-connected. By the Kodaira Embedding Theorem, this manifold embeds in $CP^{N}$ for N sufficiently large, and hence, in particular, $K3^{[4]}$ admits a Kaehler form $\phi$ that represents an integral de Rham class (it is a projective manifold). As an algebraic-geometric novice, I am not aware of all the properties that extend from varieties to their schemes, but it makes sense that many would. Specifically, since Beauville generalized the Beauville-Bogomolov form $$q_{X,\omega}(\alpha):=n\int_{X}(\omega\bar{\omega})^{n-1}\alpha^{2}+(1-2n)\left(\int_{X}\omega^{n}\bar{\omega}^{n-1}\alpha\right)\left(\int_{X}\omega^{n-1}\bar{\omega}^{n}\alpha\right)$$ from compact, Kaehler, irreducible symplectic varieties to compact, Kaehler, irreducible symplectic manifolds, I'm wondering if a $q_{X,\omega}$-orthogonal decomposition applicable on varieties may also extend to the more general manifold case. The decomposition to which I'm referring is described in Proposition 22 from a paper by Martin Schwald, which may be found here: arXiv:1701.09069v1 [math.AG] 31 Jan 2017. I'll reproduce it here:

Let $(X,\omega)$ be a 2n-dimensional, irreducible symplectic variety with normed Beauville-Bogomolov form $q_{X}=q_{X,\omega}, \omega\in H^{2}(X,\mathbf{C})$, and an ample class $a\in H^{2}(X,\mathbf{R})$. Restricting $q_{X}$ gives a real quadratic form $H^{2}(X,\mathbf{R})\rightarrow \mathbf{R}$ and $H^{2}(X,\mathbf{R})$ splits into the following $q_{X}$-orthogonal subspaces $$V_{+}:=\mathbf{R}\left<\omega+\bar{\omega},i\omega-i\bar{\omega},a\right>, V_{-}:=\{d\in H^{1,1}(X)\cap H^{2}(X,\mathbf{R})|q_{X}(d,a)=0\}$$ such that $q_{X}$ is positive definite on $V_{+}$, negative definite on $V_{-}$ and $dim_{\mathbf{R}}V_{+}=3$.

By taking $\omega:=\rho$ and $a:=[\phi]$, I sense that perhaps this theorem may be adapted to $K3^{[4]}$, but if possible, I'd like to hear what some algebraic geometry experts here think about this idea.

  • $\begingroup$ I am afraid that you are confused by the term "Hilbert scheme". A $K3^{[4]}$ is an irreducible symplectic variety. $\endgroup$ – abx May 27 '18 at 5:14
  • $\begingroup$ @abx So, as I understand it, parts of scheme theory can be reduced to variety theory, but what properties of K3^{[4]} enable that easy transferrence? I apologize for this perhaps elementary question, but my sources fall short of explicitly confirming or explaining the issue. $\endgroup$ – zenith90 May 30 '18 at 18:41
  • $\begingroup$ Variety = scheme of finite type over a field. In particular, a $K3^{[4]}$ is a variety. $\endgroup$ – abx May 31 '18 at 8:24

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