Let $X$ and $Y$ be generic cubic and quadric hypersurfaces in $\mathbb{CP}^4$. Let $Z$ be their intersection, which is a K3 surface. Then by adjunction formula, $N_{Z/X}=-K_X|_Z$ and $N_{Z/Y}=-K_Y|_Z$ are both ample. On the other hand, Kawamata-Namikawa deformation theorem claims that $X\cup Y$ is smoothable iff $N_{Z/X}\otimes N_{Z/Y} \cong \mathcal{O}_Z$. This should hold in our case because $X\cup Y$ is smoothable to a quintic threefold, but it conflicts with the above ampleness. So it seems I misunderstand something. Could anyone point it out?
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1$\begingroup$ I think that your condition $N_{Z/X}\otimes N_{Z/Y} \cong \mathcal{O}_Z$ is equivalent to the existence of a semistable smoothing, in particular the total space of the deformation must be smooth (at least, this is the requirement in Friedman's paper Global smoothings of varieties with normal crossings). Have you checked if this is the case for the smoothing of $X \cup Y$ to a quintic threefold? $\endgroup$– Francesco PolizziCommented Feb 15, 2017 at 9:47
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$\begingroup$ Francesco is completely correct. The total space of the deformation will not be smooth. A first order deformation as hypersurfaces in $\mathbb{P}^4$ is equivalent to an element in $H^0(X\cup Y, \mathcal{O}_{\mathbb{P}^4}(\underline{X}+\underline{Y})|_{X\cup Y})$. This section vanishes on a Cartier divisor $(X\cup Y)\cap Z$ in $X\cup Y$, where $Z$ is a quintic hypersurface. The total space of the deformation is singular along $(X\cap Y\cap Z)$. This is a curve in the sextic K3 surface of degree $30$. $\endgroup$– Jason StarrCommented Feb 15, 2017 at 11:27
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1 Answer
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Let me explain more concretely what's happening.
Choose a smooth quintic hypersurface $W$ in $\mathbb C \mathbb P^4$ that intersect transversely with $X \cap Y$. Then $W$ and $X+Y$ are linearly equivalent divisors. Choose $\sigma, \sigma_0$ from $H^0(\mathbb C \mathbb P^4, \mathcal O_{\mathbb C \mathbb P^4}(W))$ such that $$div(\sigma) = W, div(\sigma_0) = X+Y.$$ Let $$\mathcal X = \{(x,t) \in {\mathbb C \mathbb P^4} \times \mathbb C | x \in div(\sigma_0 + t \sigma) \}$$ and let $\pi : \mathcal X \rightarrow \mathbb C$ be one of the projection. Then $$\mathcal X_0 = X' \cup Y'$$ and $$\mathcal X_\infty = W',$$ where $\mathcal X_t = \pi^{-1}(t)$, $X'=X\times \{0\}$, $Y'=Y\times \{0\}$ and $W'= W\times \{\infty\}$. So it is smoothing of $X' \cup Y' \simeq X \cup Y $ to $W' \simeq W$.
But the total space $\mathcal X$ is not smooth and it has singularities along the smooth curve $C:=(X'\cap Y' \cap W') \times \{0 \}$. The singularities locally are the product of a smooth curve and a three-dimensional ordinary double point singularity. If we blow up $\mathcal X$ along $C$, then the exceptional locus is a $\mathbb C \mathbb P^1 \times \mathbb C \mathbb P^1 $-bundle over $C$. It is a usual procedure to contract one of the ruling of the bundle smoothly to get $\mathcal X'$ and have another smoothing $ \mathcal X' \rightarrow \mathbb C$.
We can choose the ruling of the contraction such that $$\mathcal X'_0 = X'' \cup Y''$$ where $X'' \simeq X'$ and $Y''$ is the blow-up of $Y'$ along $C$. Now the total space $\mathcal X'$ is smooth. For the central fiber $X'' \cup Y''$, we have the $d$-semistable condition $$N_{Z''/X''} \otimes N_{Z''/Y''} \simeq \mathcal O_{Z''},$$ where $Z'' = X'' \cap Y''$.
