Surfaces in $\mathbb{P}^3$ with isolated singularities - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T15:17:05Z http://mathoverflow.net/feeds/question/32938 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32938/surfaces-in-mathbbp3-with-isolated-singularities Surfaces in $\mathbb{P}^3$ with isolated singularities Francesco Polizzi 2010-07-22T11:16:43Z 2013-05-10T23:19:14Z <p>It is classically known that every smooth, complex, projective surface $S$ is birational to a surface $S' \subset \mathbb{P}^3$ having only <em>ordinary singularities</em>, 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</p> <p>$\pi_{L} \colon S \to \mathbb{P}^3$,</p> <p>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</p> <p>$\chi(\mathcal{O}_S)=\frac{1}{12}(K_S^2+c_2(S))$,</p> <p>see Griffiths-Harris "Principles of Algebraic Geometry", p. 600.</p> <p>My question is now the following: </p> <p>is it also true that every smooth, complex, projective surface $S$ is birational to a surface $S' \subset \mathbb{P}^3$ having only <em>isolated</em> singularities? And if not, is there any counterexample?</p> http://mathoverflow.net/questions/32938/surfaces-in-mathbbp3-with-isolated-singularities/32942#32942 Answer by Donu Arapura for Surfaces in $\mathbb{P}^3$ with isolated singularities Donu Arapura 2010-07-22T11:50:26Z 2010-07-22T20:59:18Z <p>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 <code>$U=\mathbb{P}^3-\{p_1,\ldots p_N\}$</code>. 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.</p> <p>Remark: It suffices to use $H_1$ in the place of $\pi_1$. I think this would be easier to justify.</p> <p>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.</p> <p><strong>Some hours later:</strong> 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.</p> http://mathoverflow.net/questions/32938/surfaces-in-mathbbp3-with-isolated-singularities/32949#32949 Answer by Tony Scholl for Surfaces in $\mathbb{P}^3$ with isolated singularities Tony Scholl 2010-07-22T12:58:01Z 2010-07-22T21:22:11Z <p>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.</p> <p>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 $i&lt;3$ 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).</p> <p>Remark: as indicated below this argument is false (and any cohomological argument along the same lines runs into the same problem).</p> http://mathoverflow.net/questions/32938/surfaces-in-mathbbp3-with-isolated-singularities/33238#33238 Answer by aginensky for Surfaces in $\mathbb{P}^3$ with isolated singularities aginensky 2010-07-24T22:21:22Z 2010-07-24T22:21:22Z <p>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. </p> http://mathoverflow.net/questions/32938/surfaces-in-mathbbp3-with-isolated-singularities/116300#116300 Answer by Dmitri for Surfaces in $\mathbb{P}^3$ with isolated singularities Dmitri 2012-12-13T17:18:12Z 2012-12-14T11:20:51Z <p>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.</p> <p>Note first that if there is a <em>regular</em> 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.</p> <p>Now, suppose we have a surface $X$ with isolated singularities in $\mathbb CP^3$, say of general type and consider the question:</p> <p><strong>Question.</strong> Let $X'$ be the minimal resolution of singularities on $X$. Can we say something about $X$ if $X'$ contains rational $-1$ curves?</p> http://mathoverflow.net/questions/32938/surfaces-in-mathbbp3-with-isolated-singularities/130292#130292 Answer by Johan for Surfaces in $\mathbb{P}^3$ with isolated singularities Johan 2013-05-10T19:08:39Z 2013-05-10T23:19:14Z <p>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.</p> <p>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.</p> <p>So the answer to your question is "never true", unless your surface naturally lives in $\mathbb{P}^3$.</p> <p><strong>Edit</strong> 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...</p>