Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 21 / 4

Fix a finite-rank free abelian group $H_{\mathbb{Z}}$. A pure integral Hodge structure of weight $w$ is a decompositionIts Hodge numbers are $h^{p,q}={\dim }_{\mathbb{C}}{H}^{p,q}$, and its Hodge filtration is $F^p={⨁}_{r\ge p}{H}^{r,w-r}$. Equivalently, the filtration has these dimensions and satisfiesThe decomposition is recovered as $H^{p,q}=F^p\cap \overline{F^q}$.

A polarized Hodge structure additionally has a nondegenerate integral bilinear form $Q$ satisfying $Q(u,v)=(-1{)}^{w}Q(v,u)$, extended complex-bilinearly, withWe choose the geometric sign convention in which the positivity condition isThe expression is real by the symmetry and conjugation conditions. In particular it is $iQ(v,\bar{v})>0$ on a weight-one $(1,0)$ summand. Other conventions absorb the constant factor $(-1{)}^{w(w-1)/2}$ into $Q$; a consistent choice matters in weight two.

Fix $H_{\mathbb{Z}}$, $Q$ and the Hodge numbers. The compact dual of a period domain is the projective flag varietyThe orthogonality conditions are algebraic. The Griffiths domain $D$ is the subset whose filtrations also satisfy real opposition and the displayed positivity. It is open in the usual complex topology of $\check{D}$; the inequalities are not additional complex algebraic equations. Thus $\check{D}$ retains the complex flag conditions, whereas $D$ retains genuine polarized Hodge structures.

In weight one, the Hodge numbers are $h^{1,0}=h^{0,1}=r$, and the Hodge filtration is determined by the $r$-plane $H=F^1$. The bilinear orthogonality condition is $\psi (H,H)=0$. An isotropic $r$-plane in a symplectic $2r$-space is a Lagrangian subspace, soThe ordinary Grassmannian has dimension $r^2$. Restriction of the alternating form to an $r$-plane imposes $r(r-1)/2$ independent conditions. Equivalently, near a fixed Lagrangian plane, transverse Lagrangians are graphs of symmetric matrices, with $r(r+1)/2$ independent entries. Therefore

Write ${\mathbb{C}}^{2r}={\mathbb{C}}^r\oplus {\mathbb{C}}^r$ in the block order of the given matrix. On $D$, positivity is $i\psi (v,\bar{v})>0$ for every nonzero $v\in H$. The plane $H$ cannot contain $(x,0)\ne 0$, because that vector has $\psi (v,\bar{v})=0$. Projection onto the second coordinate factor is therefore injective on $H$ and, by equal dimensions, an isomorphism. There is a unique matrix $Z$ such thatFor $v=(Zu,u)$ and $v^{\prime }=(Zw,w)$, direct evaluation givesThus isotropy is exactly $Z=Z^{\mathsf{T}}$. For such $Z$,Hence positivity is precisely positive definiteness of $\mathrm{Im}Z$. Conversely, symmetry and this inequality define a positive Lagrangian plane; it is opposed to its conjugate because $(Z-\bar{Z})u=0$ forces $u=0$. We have explicitly identifiedthe Siegel upper half-space. The choice of the graph $(Zu,u)$, rather than $(u,Zu)$, is what matches the sign of the given symplectic form.

Let an integral element of the symplectic group have block matrixActing on graph vectors givesThe image remains a positive Lagrangian plane, so its second projection is invertible. Thus $CZ+D_0$ is invertible and the induced action isThe symplectic identities give symmetry of this matrix, and preservation of $i\psi (v,\bar{v})$ givesThis is a Hermitian congruence by an invertible matrix, so it proves preservation of the positive domain as well as describing the action. These are coordinates on marked Hodge structures; passing to the integral-group quotient forgets the symplectic marking.

Let $\Lambda =H^2(X,\mathbb{Z})$ with the cup-product pairing $q$ of signature $(3,19)$. For a K3 surface, the Hodge numbers are $(h^{2,0},h^{1,1},h^{0,2})=(1,20,1)$. A nonzero holomorphic two-form $\omega$ spans $F^2$, and its wedge products giveThe remaining filtration is forced by orthogonality:Thus the usual K3 period domain and its quadric compactification areThe quadric is smooth and has complex dimension $21-1=20$. Writing $\omega =x+iy$, its equations say that $x,y$ are orthogonal positive vectors of equal square. A period therefore determines an oriented positive two-plane in ${\Lambda }_{\mathbb{R}}$.

There is a polarization qualification: the full cup-product form is not a polarization in the strict sense of part (a). Its real $(1,1)$ space has signature $(1,19)$, so it cannot satisfy the negative-definiteness requirement there. Fixing an ample integral class $h$ of square $2d>0$ and requiring it to stay of type $(1,1)$ gives the primitive cohomology latticeIts genuine polarized Griffiths domain isIts compact dual is the corresponding smooth quadric of complex dimension nineteen, and $D_h$ has two components, distinguished by the orientation of the positive two-plane. For ample-polarized geometric K3 moduli one also removes the loci $(\omega ,\delta )=0$ with $\delta \in {\Lambda }_h$, ${\delta }^2=-2$, since an ample class cannot be orthogonal to an effective rational curve. Those exclusions concern the ample moduli locus; they are not part of the definition of the polarized Hodge domain.

If one wants a strict polarization on the full $H^2$ of a fixed projective surface, reverse the sign on its rational Kähler line and retain $q$ on the orthogonal complement. For example the integral formhas the required positivity and signature $(2,20)$. It is a different form from the cup product. The twenty-dimensional full period quadric and the nineteen-dimensional primitive polarized domain should therefore be distinguished. The positivity and signature calculations above explain why these two domains describe different data.