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Consider a Kaehler degeneration $\mathcal X\to \Delta$ of smooth manifolds: Here $\Delta$ is the unit disc, $\pi$ a proper flat map, smooth over $\Delta^∗=\Delta−\{0\}$. The general fibres are $X_t=\pi^{−1}(t),\; $ and the central fibre is $ \; X_0=\pi^{−1}(0) $. Assume that all of the fibres admit Kaehler-Einstein metrics: $Ric(\omega_s)=\lambda\omega_s$.

Is there a method to compute $$h^1\left(Π(X_0)\right) $$

where $Π(X_0)$ is the dual graph of the central fibre $X_0$?

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    $\begingroup$ When $\mathcal X$ is 3-fold. It is known result of around 70. But I don't have any idea about your question in general . Let $X_0=\sum^n_{i=1}V_i$ be the decomposition of $X_0$ into irreducible components and let $C_k, k=1,⋯,m,$ be the double curves in $X_0$.Define $ε_{ki}=−1$, if $C_k=V_j∩V_i \; (j<i)$, $\epsilon _{ki}=+1$ if $C_k=V_j\cap V_i (j>i)$, $\epsilon _{ki}=0$ otherwise. We can define $\alpha _i:\bigoplus _kH^0(C_k)→H^0(V_i)$ then we have $$h^1\left(Π(X_0)\right)=\{(λ_1,⋯,λ_m)\in\mathbb Z_m:\sum ^m_{k=1}λ_k\epsilon_{ki}α_i[C_k]=0,\; i=1,⋯,n\}.$$. $\endgroup$
    – user21574
    Jul 26, 2017 at 17:07
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    $\begingroup$ In special case , In a Type III degeneration of K3 surfaces the dual graph of the central fibre is a triangulation of $\mathbb S^2$. $\endgroup$
    – user21574
    Jul 31, 2017 at 2:37

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