Perverse–Hodge octahedron
Let $X$ be a compact hyperkähler manifold of complex dimension $2n$ with a Lagrangian fibration $f\colon X\to B$. If $\alpha$ is an ample class on $B$ and $\beta=f^{*}\alpha$, then cup product by $\beta$ determines a weight filtration on cohomology. Up to a shift of the indices, this is the perverse filtration $P$ associated with $f$. The construction only uses that $\beta$ is a nonzero isotropic class of type $(1,1)$, so the fibration itself is not essential.
The perverse filtration is compatible with the Hodge decomposition. The entries of the perverse–Hodge octahedron are the integers $h^{i,k,d}=\dim(\operatorname{Gr}^{P}_{d+k}\operatorname{H}^{2d}(X,\mathbb{C}))^{d+i,d-i}$. For odd cohomological degree, the indices $d$, $i$, and $k$ are half-integral, while all indices occurring in the cohomology and filtrations remain integral. Summing over the perverse index recovers the ordinary Hodge numbers: for fixed $i$ and $d$, one has $\sum_k h^{i,k,d}=h^{d+i,d-i}(X)$. Thus the Hodge diamond is a two-dimensional projection of this three-graded table.
The refined $P=F$ symmetry is the identity $h^{i,k,d}=h^{k,i,d}$. It exchanges the Hodge and perverse directions. Geometrically, it comes from the fact that $\beta$ and an anti-symplectic class $\sigma$ are both isotropic for the Beauville–Bogomolov form. The resulting $\mathfrak{so}(6)$-action identifies the entries $h^{i,k,d}$ with the ranks of its eigenspaces, and its Weyl group $S_4$ acts through the rotational symmetries of an octahedron [2409.01800].
The octahedral-support conjecture states that $h^{i,k,d}$ vanishes unless $|i|+|k|+|d-n|\leq n$. Equivalently, the convex hull of the nonzero entries is an octahedron. For $b_2(X)\geq 7$, this support condition is equivalent to the type II case of Nagai’s conjecture on the nilpotency indices of monodromy operators. It is known for every currently known deformation type of compact hyperkähler manifolds.
For an elliptically fibred K3 surface, the middle cohomology has the entry 18 at the centre and the entry 1 at each of the four vertices. The four vertices represent the symplectic form, its complex conjugate, a relatively ample class, and the pullback of an ample class from the base. Summing in the perverse direction gives the usual middle row $(1,20,1)$ of the Hodge diamond.
References
- 2409.01800
- Mauri, M. (2024). Perverse-Hodge octahedron. arXiv:2409.01800v1