# Inverse Hodge and inverse Betti problems for Kähler manifolds

A Betti sequence is a map $$\mathbb{Z}_{\geq 0}\to \mathbb{Z}_{\geq 0}$$.

A Betti sequence $$b$$ is realizable if there is a connected closed Kähler manifold $$M$$ such that $$b(k)=b_k(M)$$.

A Hodge diamond is a map $$\mathbb{Z}_{\geq 0}\times \mathbb{Z}_{\geq 0}\to \mathbb{Z}_{\geq 0}$$. To any Hodge diamond $$h$$ we associate a Betti sequence $$b$$ given by $$b(k)=\sum_{i=0}^k h(i, k-i)$$.

A Hodge diamond $$h$$ is realizable if there is a connected closed Kähler manifold $$M$$ such that $$h(i, j)=h^{i, j}(M)$$. Realizable Hodge diamonds have realizable Betti sequences.

A Hodge diamond is naively realizable if $$h(0, 0)=1$$ and there is an integer $$n\geq 0$$ such that

• $$h(i, j)=h(j, i)=h(n-i, n-j)$$ if $$0\leq i, j\leq n$$
• $$h(i, j)\geq h(i-1, j-1)$$ if $$i, j\geq 1$$ and $$i+j\leq n$$
• $$h(i, j)=0$$ if $$i, j\geq 0$$ and $$\mathrm{min}(i, j)\geq n+1$$

Is there a naively realizable, non-realizable Hodge diamond with a realizable Betti sequence?

For example, the naively realizable Hodge diamond with $$n=2$$ and $$h(0, 1)=h(1, 1)=1$$, $$h(0, 2)=0$$ is not realizable but its Betti sequence is not realizable either.

The Hodge diamond $$\begin{array}{ccccc}&&1&&\\&0&&0&\\a&&1&&a\\&0&&0&\\&&1&&\end{array}$$ is naively realisable.
Suppose $$M$$ is a compact Kähler surface with the given Hodge diamond with $$a \geq 2$$. As $$h^{2,0}(M) > 1$$, the Kodaira dimension of $$M$$ is either $$1$$ or $$2$$. Note that $$c_1(M)^2 = 2\chi(M) + 3\sigma(M) = 2(2a + 3) + 3(2a + 1) = 10a + 9.$$ It follows from the Kodaira-Enriques classification that $$c_1(M)^2 \leq 0$$ for a surface of Kodaira dimension $$1$$, so we must have $$\kappa(M) = 2$$. Note however that $$c_1(M)^2 = 10a + 9 > 6a + 9 = 3(2a + 3) = 3c_2(M)$$ which contradicts the Bogomolov-Miyaoka-Yau inequality. Therefore, the above Hodge diamond with $$a \geq 2$$ is not realisable.
Finally, the corresponding Betti sequence is realisable as the blowup of $$\mathbb{CP}^2$$ at $$2a$$ points demonstrates.
The same arguments apply to the Hodge diamond $$\begin{array}{ccccc}&&1&&\\&0&&0&\\a&&b&&a\\&0&&0&\\&&1&&\end{array}$$ with $$a \geq b \geq 1$$ and $$a \geq 2$$.
The case $$a = b = 1$$ provides another example. The same arguments as above can be used to rule out Kodaira dimensions $$-\infty$$, $$1$$, and $$2$$. To rule out $$\kappa(M) = 0$$, note that $$M$$ must be minimal and hence must be finitely covered by a K3 surface or a complex torus. As $$\chi(M) = 5$$ and the Euler characteristic is multiplicative under finite covers, this is impossible.