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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.

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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.

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