Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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


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?

share|improve this question

5 Answers 5

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.

share|improve this answer
That's what I was also thinking, but maybe it would be safer to assume that both curves have positive genus, to rule out the possibility of cones over plane curves. In fact, just abelian surfaces might work. –  damiano Jul 22 '10 at 11:54
Good point. The example of a a cone shows that this nonsense as written. –  Donu Arapura Jul 22 '10 at 12:00
I think that there must be a counterexample, too. But, as it is clear from your interesting answers, finding it could be rather difficult. –  Francesco Polizzi Jul 23 '10 at 13:09

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.

share|improve this answer
@aginensky Can you please add some references to your answer? Where is it phrased as an unsolved problem? Who made the conjecture? –  JME Jul 26 '10 at 12:44
@JME I will try and get some references. As I said, it has been a while since I was looking at these things. –  aginensky Jul 29 '10 at 0:30

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).

share|improve this answer
Oops, this is also nonsense as written, because $S'-C$ and $S-D$ aren't isomorphic. There is a proof along these lines but I am unable to recall it now. –  Tony Scholl Jul 22 '10 at 13:21
I like the argument in principle, but I do have one concern: why is $S'-C\cong S-D$? It seems that the argument could be applied when $S'$ is a cone, leading to a false conclusion. –  Donu Arapura Jul 22 '10 at 13:29
OK, I guess my comment was redundant. –  Donu Arapura Jul 22 '10 at 13:30
It seems this problem is quite hard: in Nakamura and Umezu, Tokyo J Math (18) 1995, the authors show there is no normal quintic surface in $P^3$ birational to an abelian surface. They conjecture the result in arbitrary degree. –  Tony Scholl Jul 22 '10 at 21:22
@Tony I did not know this reference, thank you for pointing it out. –  Francesco Polizzi Jul 23 '10 at 13:02

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?

share|improve this answer
If you blow up the original surface S, you will always have curves with negative self-intersections. The question is asking for BIRATIONAL! –  CYXU Dec 13 '12 at 19:57
I think there may be a confusion between two meanings of "minimal" here: surfaces do admit minimal resolutions (i.e., resolutions through which all other factor) but those needn't be minimal (i.e., they may contain $-1$-curves). –  algori Dec 13 '12 at 20:59
@Dmitri: the question was not whether there exists a birational morphism to $\mathbb P^3$ whose image has the desired properties, but whether any surface is birational to such a surface in $\mathbb P^3$. The point being that your proof shows that for certain surfaces there is no such morphism, but as CX points out, there still could be a rational map that is not everywhere defined. –  Sándor Kovács Dec 14 '12 at 0:46
ps: a minimal surface by any definition may have a curve with negative self-intersection. For instance a (smooth) K3 surface may contain a $(-2)$-curve, yet it is minimal in any which way you like to define minimal. (In particular, it is the minimal resolution of the surface you get when you blow down that $(-2)$-curve). –  Sándor Kovács Dec 14 '12 at 0:50
Sandor, sure I understand what you say. I will rewrite this "answer" so there are no doubts in this. –  Dmitri Dec 14 '12 at 10:07

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...

share|improve this answer
why should the desired birational map factor through an imbedding into $\mathbb{P}^4$? –  Vivek Shende May 10 '13 at 22:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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