Degeneration of Kaehler-Einstein metric of negative Ricci curvature

Question

Let $π:X→Δ$ be a family of compact complex manifolds such that the fibre $X_t:=π^{−1}(t)$ admits a Kaehler-Einstein metric of negative Ricci curvature for all $t≠0$. Then does the special fiber $X_0:=π^{−1}(0)$ also admit a Kaehler-Einstein metric of negative Ricci curvature?

Answer 1

I am posting my comment above as an answer. It appears that the answer is negative.

From what I can find, if $X_0$ is a compact, complex, Kähler manifold that admits a Kähler-Einstein metric, then $c_1(T X_0)$ is either positive, zero, or negative. In particular, if $c_1(T X_0)$ has negative intersection with some closed, analytic subcurves, yet it has zero intersection with other closed, analytic subcurves, then the first Chern class is neither negative nor zero. Thus, there is no Kähler-Einstein metric.

Now let $d\geq 5$ be an integer, and let $Y\subset \mathbb{P}^1\times \mathbb{P}^n$ be a general hypersurface of bidegree $(1,d)$, i.e., the degree of a general fiber of $\text{pr}_{\mathbb{P}^1}:Y\to \mathbb{P}^1$ is a hypersurface of degree $d$, and the general fiber of the other projection is a hypersurface of degree $1$. For a general such hypersurface, there are $4(d-1)^3$ points $s_i\in \mathbb{P}^1$ such that the corresponding fiber $Y_{s_i}$ of $\text{pr}_{\mathbb{P}^1}$ has a single ordinary double point $y_i$, and no other singularities. Let $f:C\to \mathbb{P}^1$ be a double cover that is ramified over each of the points $s_i$ (and possibly also over finitely many additional points). Denote by $t_i\in C$ the unique point that maps to $s_i$ under $f$.

Consider the base change $X'=C\times_{\mathbb{P}^1} Y$ with its projection $$\pi:X\to C.$$ The total space $X'$ is a complex, projective variety of dimension $3$ that has an ordinary double point over every point $x_i=(s_i,y_i)$. An ordinary double point of a $3$-fold is complex analytically isomorphic to a cone over a smooth quadric surface. Thus, there is a proper, holomorphic morphism $$\nu:X\to X',$$ that is a small resolution of each of these singular points $x_i$. In particular, every fiber $X_{t_i}$ is a compact, complex manifold. Moreover, it is a small resolution of the singular fiber $X'_i$. Thus, it is a complex projective surface (every smooth, compact Moishezon surface is projective). So $X_{t_i}$ is a Kähler manifold of dimension $2$.

However, the exceptional curve in $X_{t_i}$ is a smooth, genus $0$ curve whose intersection number with $c_1(T X_{t_i})$ is zero. Thus, the compact, complex, Kähler manifold $X_{t_i}$ does not admit any Käher-Einstein metric.

Answer 2

As Jason points out, the special fiber does not have a smooth negatively curved KE metrics in general (you can blow up the family along a subscheme supported on the special fiber without changing the assumptions you are working with, so certainly you can kill the property that the central fiber is canonically polarized).

For simplicity, I assume that $\pi$ is projective. Then $X_0$ admits a singular Kähler-Einstein metric (in the sense of Boucksom-Eyssidieux-Guedj-Zeriahi). This is a smooth Kähler-Einstein metric on a Zariski open subset of $X_0$ (the complement of the augmented base locus $\mathbb B_+(K_{X_0})$), with some additional global properties. The reason is that $K_{X_0}$, even if it may not be ample, is certainly big, for instance by semicontinuity of the function $t\mapsto \dim H^0(X_t, K_{X_t}^{\otimes m})$ for any integer $m>0$. This function is actually constant by a deep theorem of Siu, but it is irrelevant here.