Surfaces in $\mathbb{P}^3$ with isolated singularities - MathOverflow

Surfaces in $\mathbb{P}^3$ with isolated singularities

2010-07-22T11:16:43Z

It is classically known that every smooth, complex, projective surface $S$ is birational to a surface $S' \subset \mathbb{P}^3$ having only ordinary singularities, i.e. a curve $C$ of double points, containing a finite number of pinch points and a finite number of triple points, which are triple also for $S'$. The proof is obtained by embedding $S$ in $\mathbb{P}^5$ and by taking a projection

$\pi_{L} \colon S \to \mathbb{P}^3$,

where $L \subset \mathbb{P}^5$ is a general line. This is the method originally used by M. Noether in order to prove his famous formula

$\chi(\mathcal{O}_S)=\frac{1}{12}(K_S^2+c_2(S))$,

see Griffiths-Harris "Principles of Algebraic Geometry", p. 600.

My question is now the following:

is it also true that every smooth, complex, projective surface $S$ is birational to a surface $S' \subset \mathbb{P}^3$ having only isolated singularities? And if not, is there any counterexample?

Answer by Donu Arapura for Surfaces in $\mathbb{P}^3$ with isolated singularities

2010-07-22T11:50:26Z

My instinct tells me no. Here's a pseudo-proof. Take $X$ to be a surface with nontrivial fundamental group e.g. a product of two curves $X=C_1\times C_2$ where say $C_2$ has positive genus. Suppose it's birational to surface $Y\subset \mathbb{P}^3$ with isolated singularities at $p_1,\ldots p_N$. Let $U=\mathbb{P}^3-\{p_1,\ldots p_N\}$ . Then $\pi_1(U)=\pi_1(\mathbb{P}^3)$ is trivial. Then some version of Zariski-Lefschetz should give $\pi_1(S\cap U)=\pi_1(U)=1$ (this is the "pseudo" part, since I'm too lazy to track this down). We have a diagram $$U\leftarrow U''\to U'\subset X$$ where the arrows are blow ups of points, and the inclusion is as an open set. Then $$\pi_1(U)\cong \pi_1(U'')\cong \pi_1(U')$$ is trivial. However $\pi_1(U')$ would surject onto $\pi_1(X)$ QED.

Remark: It suffices to use $H_1$ in the place of $\pi_1$. I think this would be easier to justify.

Remark 2: As is clear from remarks below, this argument is insufficient even with $H_1$, but perhaps there is a germ of a correct idea.

Some hours later: I no longer feel that this approach is viable. Nevertheless, I believe for whatever irrational reason that there must be a counterexample. One thing is certainly clear, and that is that this is a damn good problem.

Answer by Tony Scholl for Surfaces in $\mathbb{P}^3$ with isolated singularities

2010-07-22T12:58:01Z

Suppose $S$ is a smooth surface birational to an abelian surface, and $f:S\to S'\subset\mathbb{P}^3$ is a birational morphism. Then $S'$ cannot have isolated singularities.

If it did, one could find a smooth hyperplane section $C\subset S^'$ missing the singular points. Since $V=\mathbb{P}^3- S'$ is smooth and affine, $H^i_c(V)=0$ for $i<3$ and so $H^1(S')=H^1(\mathbb{P}^3)=0$ by the exact sequence for $H_c$. On the other hand, let $U=S'-C\simeq S-D$ where $D=f^{-1}(C)$. Consider the long exact cohomology sequences for the pairs $(S,D)$ and $(S',C)$: $$ 0 \to H^1(S) \to H^1(U) \to H^2_D(S) \to\dots $$ and $$ 0 \to H^1(S') \to H^1(U) \to H^2_C(S') \to \dots $$ As $H^1(S')=0$, these imply that $H^1(S)$ injects into $H^2_C(S')$. But $C$ is irreducible, and contained in the smooth part of $S'$, so $H^2_C(S')$ is 1-dimensional. (Or you can argue with weights).

Remark: as indicated below this argument is false (and any cohomological argument along the same lines runs into the same problem).

Answer by aginensky for Surfaces in $\mathbb{P}^3$ with isolated singularities

2010-07-24T22:21:22Z

To the best of my knowledge this is a long standing open problem. I cannot recall a reference, as this is something I studied in the 1980's, but I recall this being phrased as an unsolved problem from the 19th century Italian school. The conjecture is that no normal surface in P^3 is birational to a smooth surface which has two dimensional image in it's Albanese. One specific case of this that has been studied more extensively are Zariski surfaces:z^n = f(x,y) where f is a polynomial of degree n with only cusps and nodes as singularities. There are lots of information about when such a surface is irregular, but beyond that not much is known. I believe that even if f is a sextic polynomial it is unknow whether or not the resulting surface can have 2 dimensional image in it's Albanese. I have heard Catanese ask about the case where S is an abelian surface.

Answer by Dmitri for Surfaces in $\mathbb{P}^3$ with isolated singularities

2012-12-13T17:18:12Z

This answer is completely rewritten. This is not an actual answer but a thought related to the question. I decided to leave it hear since it is short.

Note first that if there is a regular map from a surface $X$ to $\mathbb P^3$ whose image has only isolated singularities, then $X$ has curves with negative self-intersection. In particular, if $X$ has no such curves then its image in $\mathbb P^3$ is smooth.

Now, suppose we have a surface $X$ with isolated singularities in $\mathbb CP^3$, say of general type and consider the question:

Question. Let $X'$ be the minimal resolution of singularities on $X$. Can we say something about $X$ if $X'$ contains rational $-1$ curves?

Answer by Johan for Surfaces in $\mathbb{P}^3$ with isolated singularities

2013-05-10T19:08:39Z

Let $S' \subset \mathbb{P}^3$ be the birational projection of a smooth surface $S \subset \mathbb{P}^4$. The general projection theorem of Gruson-Peskine (http://arxiv.org/abs/1010.2399v1) tells you that $S'$ is either smooth or has a curve of double points.

For instance if $S$ is the Severi surface in $\mathbb{P}^4$, then its projection on $\mathbb{P}^3$ (the Steiner surface) has a curve of double points.

So the answer to your question is "never true", unless your surface naturally lives in $\mathbb{P}^3$.

Edit Ok after some times, I realize that you are interested in a smooth surface (say $S \subset \mathbb{P}^5$) which maps birationnaly to a surface $S' \subset \mathbb{P}^3$ but for which the map does not come from the ambiant projective space. So my answer is useless...