3
$\begingroup$

Let $f:X \to Y$ be a proper surjective morphism of projective surfaces such that there exists a curve $C \subset X$ for which $f|_{X\backslash C}$ is an isomorphism and $f(C)$ is a set of points. Suppose $X$ is a closed subscheme of $\mathbb{P}^n$ for some integer $n$ and $C$ contracts to a rational singularity i.e., $f(C)$ are rational singularities on $Y$. Does there exist a closed immersion of $Y$ into $\mathbb{P}^n$ for the same integer $n$?

$\endgroup$

3 Answers 3

5
$\begingroup$

Let $X\subset \mathbb P^3$ be an arbitrary smooth projective surface of degree $d>2$ and assume that $X$ contains a line $\ell\simeq \mathbb P^1$ If $d=4$, assume in addition that $X$ is general among such surfaces. This implies that its Picard number is $2$. A simple adjunction computation shows that $\ell^2=2-d<0$ ($l^2$ computed on $X$). This implies that $\ell\subset X$ is contractible, so we have a morphism $f:X\to Y$ as requested.

Claim. $Y$ cannot be embedded into $\mathbb P^3$. If $d>4$, then $Y$ cannot be embedded into any smooth $3$-fold.

Proof:

  • If $d=3$ (case abx): Since $X$ is rational, so is $Y$, so if it could be embedded to $\mathbb P^3$, its degree would have to be $1,2$ or $3$. Its easy to see using adjunction and that fact that $f$ is the blow up a smooth point that none of these is possible.
  • If $d=4$ (case Olivier): The Kodaira dimension of $X$ is $0$, so if $Y$ were embeddable, it would have to have degree $4$. In this case adjunction does not help, because $0=K_X=f^*K_Y$. However, if $H'\subset Y$ is the hyperplane section of $Y\subset \mathbb P^3$, then $f^*H'\cdot\ell =0$ and since the Picard number of $X$ is $2$, we can compute that in this case $f^*H'=2mH+m\ell$ for some $m\in \mathbb N_+$. But then $H'^2=18m^2\neq 4$, so this leads to a contradiction and hence $Y$ cannot be embedded into $\mathbb P^3$.
  • If $d>4$ (case new):This is actually the easiest. If $Y$ embeds into any smooth $3$-fold, then it is Gorenstein and $K_Y$ is a Cartier divisor. Then we have $K_X=f^*K_Y+ a\ell$ for some $a\in \mathbb Z$. Using adjunction for $\ell$ and our earlier computation tells us that $$2-d=\ell ^2=\frac {-2}{a+1}$$ which means that $$2=(d-2)\cdot(a+1),$$ that is, $d-2$ divides $2$ so $d>4$ is out of the question. This completes the proof.
$\endgroup$
1
  • $\begingroup$ @Kovacs: Thank you for the very detailed and exhaustive answer. $\endgroup$
    – user43198
    Jan 4, 2014 at 14:30
5
$\begingroup$

No. Take a smooth cubic surface $S\subset \mathbb{P}^3$, and a line $E\subset S$. Then $E$ can be contracted to give a smooth Del Pezzo surface of degree 4 (a complete intersection of two quadrics in $\mathbb{P}^4$), which cannot be embedded in $\mathbb{P}^3$.

$\endgroup$
0
3
$\begingroup$

Let $X$ be a projective complex K3 surface whose Néron-Severi-group is generated by two classes $h$ and $l$ such that $h^2=4$, $hl=1$ and $l^2=-2$ and $h$ is ample. One sees that such a surface exists using the surjectivity of the period map for K3 surfaces. General theorems about linear systems on K3 surfaces imply that $h$ is very ample and embeds $X$ in $\mathbb{P}^3$.

Since $l^2=-2$, $l$ or $-l$ is effective. Since $hl=1$, it is $l$ that is effective, and it has to be the class of a line in $\mathbb{P}^3$. This line is a $-2$-curve and may be contracted to a node. Let $f:X\to Y$ be this contraction map. Since $Y$ is a singular K3 surface, if it were possible to embed it in $\mathbb{P}^3$, the degree of the embedding would necessarily be $4$, but this is impossible as no line bundle on $Y$ has this degree.

$\endgroup$
2
  • $\begingroup$ @Benoist: Could you say why $Y$ has no line bundle of degree $4$? $\endgroup$
    – user43198
    Jan 3, 2014 at 19:03
  • $\begingroup$ @user43198: Its pull-back to $X$ would be a degree $4$ line bundle on $X$ that is orthogonal to $l$ (because the line that is a section of $l$ is contracted by $f$). But line bundles on $X$ that are orthogonal to $l$ are proportional to $2h+l$ and $(2h+l)^2=18$. $\endgroup$ Jan 3, 2014 at 21:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.