# Tag Info

## Hot answers tagged algebraic-surfaces

24

The "cyclic cover trick" can refer to more than one thing. One example is as follows. Let $L$ be an invertible sheaf on a smooth projective scheme such that some power $L^{\otimes d}$ has a global section $s$ whose zero scheme $D$ is a smooth Cartier divisor (all of these smoothness conditions are not strictly necessary). Let $\nu:Y\to X$ be the ...

18

I am surprised nobody mentioned the result of Lu and Miyaoka (Math. Res. Letter 2, 663-676 (1995)) which implies indeed that there are only finitely many smooth rational curves on a surface of general type, thus answering the question of the OP.

15

It turns out that condition (T) is, indeed, sufficient for the $27$ lines (distinct and intersecting as expected) to lie on a cubic surface. To see this, consider the lines $a_1,a_2,a_3,a_4,a_5$ and $b_6$, where the labeling is as in note (2) of the question: $a_1$ through $a_5$ are pairwise skew, and $b_6$ intersects all of them. Choose $4$ distinct ...

14

Edit. I edited the argument below to make it work in all characteristics. By SGA $7_{II}$ Exposé XVII, this requires working with a sufficiently general pencil of divisors in $\mathcal{O}_{\mathbb{P}^3}(e)|_S$ for $e\geq 2$ (as before, $e=1$ suffices if $d\geq 3$ and the characteristic is sufficiently large). These kinds of arguments used to be "...

14

Yes, with precisely one exception. If $K^2 \neq 8$, then the del Pezzo surface is the blow-up of the plane at $9-K^2$ points, so it is homeomorphic to the connected sum of $\mathbb{CP}^2$ with $9-K^2$ copies of $\overline{\mathbb{CP}^2}$. If $K^2=8$, then we have either the quadric $\mathbb{P}^1 \times \mathbb{P}^1$, which is clearly homeomorphic to $S^2 \... 14 Let$k$be an algebraically closed field. Let$f:X\to \mathbb{P}^1$be a smooth proper morphism with fibres of dimension one. Note that the fibres of$f$are geometrically connected by Stein factorization and the fact that$\mathbb{P}^1$is simply connected (Riemann-Hurwitz). Theorem. The morphism$f$is isotrivial. Proof. Let$g$be the ... 13 The answer is no in general. For an easy$2$-dimensional case, take for example$X=\mathbb{P}^1\times \mathbb{P}^1$, in which case$d=2$. Then, only the multiples of$d$are possible for$d'$. Indeed, take a morphism$X\to \mathbb{P}^2$, and denote by$D$the divisor of its linear system, which satisfies$D^2=d'$. The Picard group of$X$is generated by$f_1,...

12

A preprint by Stefan Schröer came out today with the answer to this question: arXiv:2004.07025. No such Enriques surface exists. In fact, there is no classical Enriques surface over $\mathbb F_2$ with 25 $\mathbb F_2$-points (and with the extension of $\mathbb Z^{10}$ by $\mathbb Z/2$ split in the Picard group, which you can also deduce).

12

These are indeed good questions, and while there is a very good corpus of answers to them, the analogy is not perfect. 0. The non-archimedean analogy First of all, I would like to go back to the relative situation of a surface $\mathcal X\to B$ fibered over a germ of curve $(B,b)$. Then any local function $f$, resp. local section $s$ of a line bundle $\... 11 The rigidity of quotient singularities in dimension greater or equal than$3$was established by Schlessinger in his paper Rigidity of quotient singularities, Invent. Math. 14 (1971). Roughly speaking, he proved that if$(X, \,x)$is a local scheme with an isolated singularity at$x$and$\dim X \geq 3$, then deforming$X$is equivalent to deform the ... 11 Not exactly. The quadratic transformation commutes with the action of$S_3$, and they both act on$(\mathbb{C}^*)^2$; so the automorphism group is$(\mathbb{C}^{*})^{2}\rtimes (S_3\times S_2)$. You'll find a detailed study of the automorphisms of del Pezzo surfaces in Chapter 8 of Dolgachev's book "Classical Algebraic Geometry: a modern view". 11 My local library has the paper version, here is a scan. 10 I am only posting this as an answer because it annoys me to see a question like this listed as "unanswered", thus "hovering" near the top of the list of unanswered questions. If dhy wants to write up his comment as an answer, then I will delete this answer. The surface given by the OP is as far as possible from being "general type". Just to remind, a ... 10 There are several different notions of "rigidity" (local rigidity, global rigidity, infinitesimal rigidity, étale rigidity and strong rigidity) and it is possible to provide examples for each of them. This topic is discussed in the paper by I. Bauer and F. Catanese On rigid compact complex surfaces and manifolds, you can have a look at it for more details ... 9 Let$S' \subset \mathbb{P}^3$be the birational projection of a smooth surface$S \subset \mathbb{P}^4$. The generic projection theorem of Gruson-Peskine (http://arxiv.org/abs/1010.2399v2) tells you that either$S'$is smooth or has a curve of double points. For instance, if$S$is the Veronese surface in$\mathbb{P}^4$, then its projection in$\mathbb{P}^3$... 9 Here are the arguments to exclude polydisk and the ball (there are no other complex 2-dimensional bounded symmetric domains: In fact, one can do without this and argue that any domain other than the ball would have rank$\ge 2$and, hence, Margulis superrigidity theorem would apply). Kefeng Liu ("Geometric height inequalities", Math. Research Letters, 3 (... 9 This is more or less what Jason has done, but maybe a bit more direct, and it is so elementary that it's hard not to call it an exercise that possibly does not belong on MO. Start with$y^2=x^4+z^6$. Change variables$(x,y,z)=(zx_1,z^2y_1,z)$to get$y_1^2=x_1^4+z^2$. Next change variables$(x_1,y_1,z)=(x_1,x_1^2y_2,x_1^2z_2)$to get$y_2^2+1=z_2^2$. This ... 9 There is another inequality which says the following: Easy addition (Using the same notation): $$\kappa(X)\leqslant \kappa(X_y) + \dim Y$$ Consequently, if$Y$is of general type, i.e.,$\kappa(Y)=\dim Y$, then the subadditivity conjecture is equivalent with equality instead of inequality. Subadditivity is also known in the case$X_y$is of ... 9 I think the Proposition is not true if$D$is singular. Take a smooth curve$C$of genus 2, and$X=JC$; embed$C$in$X$(say, by choosing a point of$C$). Let$\alpha$be a point of order 2 in$X$; take for$Y$the quotient of$X$by the translation$x\mapsto x+\alpha $, and put$D=f(C)$. Then$f^*D=C+C'$, with$C':=C+\alpha $, and$C\cdot C'=C^2=2$. Here$...

9

On any surface $X$ with non-negative Kodaira dimension the $(-1)$-curves (i.e, the smooth rational curves $D$ with $D^2=-1$) are isolated, in other words any two of them do not intersect. From this one can deduce the uniqueness of the minimal model of $X$, see the proof of Proposition 4.6, p. 79 in the book W. Barth, C. Peters, A. Van de Ven: Compact ...

9

The diagonal $\Delta$ is linearly equivalent to $\{p\}\times \mathbb{P}^1 +\mathbb{P}^1\times \{p\}$ for any $p$ in $\mathbb{P}^1$. Therefore $X$ is the zero locus in $S\times S'$ of a section of $L:=\pi^*\mathcal{O}(1) \boxtimes \pi'^*\mathcal{O}(1)$. On the other hand, standard theory of elliptic surfaces gives $\omega _S=\pi ^*\mathcal{O}(-1)$ and $\... 9 If$2$is a cube mod$p$then you can take$(x,y) = (ct^2, t^6-1)$where$c^3 = -4$. This works for every odd$p \equiv -1 \bmod 3$, but since you specified$p \equiv +1 \bmod 3$the first case is$p=31$, with$c \in \{-3,-15,18\}$. The next few such$p$are$43, 127, 223, 283, 307, 439, 499, 643, 691$. By Cubic Reciprocity, the condition on$p$is ... 9 You can observe that the elliptic curve$E_2: y^2=x^3+(t^3+1)^2$is a generic fibre of a rational elliptic surface. Over the algebraically closed field$k$the group of$k(t)$points on the curve has rank equal to 2 (by the Shioda-Tate formula) and from the classification of possible groups of$k(t)$-rational points by Oguiso-Shioda (case 39 because we have ... 9 I am posting this answer because the following linear algebra proposition is too long for a comment. Lemma. Let$A$, respectively$B$, be an invertible$\mathbb{R}$-linear operator on$\mathbb{C}^2$that is$\mathbb{C}$-linear, resp.$\mathbb{C}$-conjugate linear. (1). For every$1$-dimensional$\mathbb{C}$-linear subspace$V\subset \mathbb{C}^2$such that ... 8 Yes, it is true. This is easily seen by looking for example at the Picard group, generated by$C$and$D$, the two fibres of the projections. Since$C^2=D^2=0$and$C\cdot D=1$, the only curves of self intersection$0$are multiple of$C$or$D$, and the irreducible ones are equivalent to$C$or$D$. Composing an automorphism by the exchange of coordinates ... 8 The answer to the last question (and therefore to the others) is yes. An étale cover of a variety$X$is dominated by an étale Galois$G$-cover, for some finite group$G$; and this is given by a homomorphism$u:\pi _1(X)\rightarrow G$. If$X=X_1\times \ldots \times X_p$, this gives homomorphisms$u_i:\pi _1(X_i)\rightarrow G$, and there are étale coverings$\...

8

Borrowing from 4 different comments to make a complete answer (and making it community wiki): We can construct a bijection between the set of del Pezzo surfaces and $\pi_4(S_3)$ using topology. However, the proof that this is a bijection depends on the classification of del Pezzo surfaces and does not give an explanation for why there are exactly two. Both ...

8

A “formula” is a lot to ask for, but there are algorithms based on Morse theory. E.g. §5 of Basu (1999), or §3 of Fortuna-Gianni-Luminati (2004).

8

Here is a simple method for constructing projective examples: Assume there exist maps $f:C \to \mathbb{P}^1$ and $g:T \to \mathbb{P}^1$ of the same degree which are totally ramified at $c$ and $t$. Let $X = C \times T$, $Y = \mathbb{P}^1 \times \mathbb{P}^1$, and consider the map $p:=(f,g): X \to Y$. Let $X'$ be the blowup of $X$ at $(c,t)$ and $Y'$ the ...

8

At least for effective divisors, the answer is strongly related to Zariski decomposition. If $D$ is an effective divisor on a smooth surface $X$, Zariski proved in [Z62] that there exists a unique decomposition $D=P + N$, where $P$ is a nef $\mathbb{Q}$-divisor $N$ is an effective $\mathbb{Q}$-divisor $PC=0$ for every curve $C$ appearing in \$\operatorname{...

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