# Fibration when central fibre is a Calabi-Yau variety with canonical singularities

Let $f\colon X\to Y$ be a surjective proper holomorphic fibre space such that $X$ and $Y$ are projective varieties and central fibre $X_0$ is Calabi-Yau variety with canonical singularities, then can we say that all the fibres $X_t$ are also Calabi-Yau varieties ? I know that if we replace "Calabi-Yau" with "pseudoeffective" then this this statement is correct.

• What is your definition of Calabi-Yau? If the dimension of the fiber is at least $3$, then the "standard" definition of Calabi-Yau includes vanishing of both $h^{0,1}$ and $h^{0,2}$. Since your central fiber is projective (hence Kaehler), this implies vanishing of $H^1(X_0,\mathcal{O}_X)$ and $H^2(X_0,\mathcal{O}_X)$. Thus, all invertible sheaves on $X_0$ extend uniquely to nearby fibers. In particular, since $\omega_{X_0}$ equals $\mathcal{O}_{X_0}$, this also holds on nearby fibers. – Jason Starr Jul 28 '16 at 18:12
• Even if you don't assume that $h^{0,1}=h^{0,2}$ (in char 0) it still follows that nearby fibers have trivial canonical bundle: Since $X_0$ is smooth, $f:X\to Y$ is smooth on a neighborhood of $0\in Y$ (presumably $Y$ is smooth and hence by standard reductions, we may assume that $Y$ is a curve). By Hodge theory $h^0(K_{X_y})$ is deformation invariant and so $h^0(K_{X_y})>0$. Since $+/-K_{X_0}$ is nef, so is $+/-K_{X_y}$. But the $K_{X_y}$ is numerically equivalent to 0 and effective so that $K_{X_y}$ is linearly equivalent to 0. Does this (or the above) answer your question? – Hacon Aug 1 '16 at 2:14
• Central fibre $X_0$ may be singular Calabi-Yau, with canonical singularities – user21574 Aug 1 '16 at 11:35

Assume for simplicity that $Y$ is a smooth curve. Since $X_0$ has canonical, then by Theorem 1.4 of http://arxiv.org/pdf/math/9809091.pdf, we may assume that $X$ is canonical. Nearby fibers are then also canonical (see eg. Theorem 4.5.1 http://arxiv.org/pdf/alg-geom/9601026.pdf). The proof of Theorem 1.4 actually shows that $O_X(K_X+X_0)\to O_{X_0}(K_{X_0})$ is surjective and so if $K_{X_0}$ is Cartier, then so is $K_X$ (on a neighborhood of $X_0$). By J. Algebr. Geom. 16, No. 1, 1-18 (2007) we even have that $P_m(X_y)$ is deformation invariant for $m\geq 1$ so in fact $h^0(K_{X_y})>0$ for $y\in Y$. Since $K_{X_0}\equiv 0$, then $K_{X_y}\equiv 0$ and so $K_{X_y}\sim 0$. (One could also conclude using Theorem 1.6 of http://arxiv.org/pdf/math/9809091.pdf).