User philip engel - MathOverflow

Complete curves in $M_g$ and Theta Characteristics

2013-04-30T10:25:26Z

Let $g\geq 3$. Following the reference below, the locus of curves in $M_g$ with an effective even theta characteristic has codimension $1$. (Those are the curves $C$ with an effective line bundle $L$ such that $L\otimes L=K$ and $h^0(L)\equiv 0\mod 2$)

(QUESTION) Are there complete curves in $M_g$ that avoid this locus? I know that this is not the case when $g=3$ because the curves with an effective even theta characteristic are hyper-elliptic and any complete curve in $M_3$ must intersect the hyper-elliptic locus.

The main reason I ask is because I'd like to construct surfaces in $M_{2g}$ as follows: Take a family $F\rightarrow B$ of genus $g$ curves that avoids all theta-null divisors, i.e. all half-canonical bundles have $0$ or $1$ section. Since the parity of $h^0(K^{1/2})$ is constant in families, the ineffective theta characteristics are locally constant and form an etale cover of $B$. There is a map $F\times_BF\rightarrow Jac_BF$ sending $(x,y)\mapsto O(x-y)$ which contracts the diagonal to the zero section. After a suitable base change, one can compose with a map $$Jac_BF\rightarrow Pic_B^{g-1}F$$ sending $L\rightarrow L\otimes K^{1/2}$ for an ineffective $K^{1/2}$. Thus, under the map $$\phi:F\times_BF\rightarrow Pic^{g-1}_BF$$ $\phi(\Delta)$ does not intersect the theta divisor $\Theta\subset Pic^{g-1}_BF$ (excuse the overuse of the word theta!)!

Hence $\phi^{-1}(\Theta)$ parameterizes a surface of distinct pairs of points in $F_b$. After another base change, one can take the double branched cover of $F_b$ branched at these two points and get a surface inside $M_{2g}$. Will this technique ever work to construct surfaces in $M_{2g}$? That, by the way, is the motivation for the question...

http://www.ams.org/journals/tran/1982-271-02/S0002-9947-1982-0654853-6/S0002-9947-1982-0654853-6.pdf

Answer by Philip Engel for Enriques classification of algebraic surfaces

2013-04-30T13:07:05Z

All of the following is from Beauville's wonderful, short, dense book: "Complex Algebraic Surfaces." I really recommend it if you want to learn about the classification or even general techniques in surface theory.

For a minimal surface, $P_{12}=1$ alone is equivalent to $\kappa=0$. It probably is necessary to go through the classification to get this result. The number $12$ is not random: It arises because of the bi-elliptic surfaces, which are quotients of a product of two elliptic curves by a finite group. These finite groups all have elements whose order divides $12$, which is why the number $12$ appears.

The key step is that $P_{12}=0$ implies $S$ is ruled. This follows from two cases:

(1) Suppose $q=0$. Because $P_{12}=0$ we also have $P_2=0$. Castelnuovo's rationality criterion then applies to show $S$ is rational.

(2) Suppose $q\geq 1$. Because $P_{12}=0$ we also have $p_g=0$. One can show relatively easily using the Albanese fibration that the only possibility for $S$ not to be ruled is when $q=1$ and $b_2=2$. Then lots of work shows that if $S$ were not ruled we would have $S=(B\times F)/G$ for curves $B$ and $F$ and a finite group $G$ (and a number of other technical restrictions). By analyzing the canonical bundle on the resulting surface, one can show that $P_{12}$ is never zero. Thus we have a contradiction, so $S$ is ruled. Furthermore, $P_{12}=1$ if and only if both $B$ and $F$ are elliptic curves.

Once this difficult step is out of the way, we know that $\kappa=0$ implies $P_{12}=1$. Conversely, if $P_{12}=1$ then $S$ is non-ruled, and one can show $\chi(O_S)\geq 0$. Since $p_g=0$ or $1$, the list of possibilities for $q$ is finite. After analyzing the five cases (one turns out to be impossible), dealing with the most difficult ones by invoking part (2) from above, one can show that $P_{12}=1$ does in fact imply $\kappa=0$.

Finally, the remaining case is $P_{12}\geq 2$. In this case, it is simple to show that $\kappa=2$ if and only if $K^2>0$. The backward direction is an application of Riemann-Roch, and the forward direction is proven by contrapositive. If $K^2=0$, then the mobile part $M$ of $|nK|$ satisfies $M^2=0$ and hence $\kappa=1$.

Hope this helps. - Phil

Contracting a curve of negative self-intersection on a surface

2013-03-01T21:01:45Z

It is easy to show using birational factorization that the only curves on a surface which can be contracted to get an algebraic, smooth surface are smooth $(-1)$-curves. Furthermore, I know of examples of smooth curves of some other negative self-intersection which can be contracted in the algebraic category to result in a singular surface (where the order of the singularity is equal to the negative self-intersection?)

Question 1: Given a curve of negative self-intersection on a complex surface, what is the construction (in the analytic category) of its contraction?

Question 2: Given the curve, what conditions on it that determine whether the contraction is algebraic or not?

Good, clear, elementary references would be fine, as an alternative to an answer!

Answer by Philip Engel for difference of curve classes

2012-10-25T08:19:36Z

Every divisor class $D$ on a surface is the difference of two smooth, connected curves. Choose a very ample divisor $A$ and an $n$ so that $D+nA$ is also very ample. Then $(D+nA)-nA=D$ so $D$ is the difference of two curves. They may be chosen smoothly by Bertini's Theorem.

ADDED LATER: They may also be chosen to be connected. The Lefschetz hyperplane theorem shows that hyperplane sections of surfaces are connected.

Answer by Philip Engel for Checkmate in $\omega$ moves?

2012-01-05T10:49:04Z

The white queen moves anywhere to the east, then the black rooks force the king east, back rank mate-style, until they've either skewered, pinned, or forked the queen and king. Worst case scenario, black loses 2 rooks, and can still mate. If black ever doesn't check, white will have a perpetual. Note that black can't force mate, as white's strategy can always be "go in a northerly direction to escape check."

P.S. Sorry, I switched the colors.

Answer by Philip Engel for Can any smooth hyperelliptic curve be embedded in a quadric surface?

2011-12-28T18:28:04Z

An explicit realization of degree $2$ and degree $g+1$ maps that separate points can be provided. Suppose the equation of a hyperelliptic curve is $$C:y^2=f(x)$$ with $\deg(f)=2g+2$. "Complete the square" by writing $$f(x)=r(x)^2+q(x)$$ with $\deg(r)=g+1$ and $\deg(q)\leq g$. Then, the maps $$A:(x,y) \mapsto x$$ $$B:(x,y)\mapsto y-r(x)$$ are degree $2$ and degree $g+1$ respectively. The map $B$ is degree $g+1$ because if we assume that $B(x,y)=c$, then we get the equation $c(c+2r(x))=q(x)$, which generically has $g+1$ solutions. Furthermore $(A,B)$ is clearly injective. So, the image $$(A,B):C\mapsto C'\subset\mathbb{P}^1\times\mathbb{P}^1$$ is a degree $(2,g+1)$ curve that $C$ normalizes and the logic of Jack's answer applies.

Is there a way to define Hecke operators "inherently" as certain endomorphisms of the Jacobian?

2010-07-29T04:48:25Z

From the Eichler-Shimura relation, we have a formula for $T_p$ when we reduce $\textrm{End}(\textrm{Jac}(X))$ mod $p$. Explicity, $T_p=\textrm{Frob}_p+p\textrm{Frob}_p^{-1}$. Is there a way to define the Hecke operator as a lift of this operator satisfying certain other properties? Is there a definition of $T_p$ which does not rely on a moduli space interpretation or double coset operators, but "inherently" from the Jacobian? Excuse the vague formulation of this question; I am just learning about this stuff.

Comment by Philip Engel

2013-05-14T22:33:14Z

Haha, my friend and I barely managed to work through the logic with the audience choosing the numbers 2 and 3!

Comment by Philip Engel

2013-05-01T21:14:57Z

Thanks for the reference, this is exactly what I was looking for.

Comment by Philip Engel

2013-05-01T07:58:34Z

One direct way to show $\kappa(S)=-\infty$ implies $S$ is ruled is to use Iitaka's Conjecture: If $S$ is a surface and $S\rightarrow B$ is a fibration with generic fiber $F$ then $\kappa(S)\geq\kappa(B)+\kappa(F)$. Applying this to the Albanese fibration solves case (2) above instantaneously, because it implies that $F$ must be rational. This applies the stronger assumption $P_n=0$ for all $n$ rather than $P_{12}=0$ though. I think there really will be no way to get the specific number $12$ without some classification.

Comment by Philip Engel

2013-03-06T22:25:26Z

Thanks for your answer, it partially resolves the case of my second question, by giving criteria for contractibility in the algebraic category in certain cases. I mainly wanted an explicit construction {\it in the analytic category} of the contraction. (The conditions would of course be weaker if we allow the contraction not to be an algebraic surface). Since the proposition above seems to be the best result about contractibility, I assume it is hard then to determine whether the resulting surface is algebraic...

Comment by Philip Engel

2012-11-01T03:25:07Z

This works whether or not $D$ is torsion, for any $(1,1)$-class in $H^2(X,\mathbb{Z})$.

Comment by Philip Engel

2012-10-26T20:09:43Z

Thanks, Henri. So that settles it for all surfaces!

Comment by Philip Engel

2012-10-25T11:11:36Z

Yeah, dunno what I was thinking... I totally just got it backwards.

Comment by Philip Engel

2012-01-06T01:28:35Z

Good observation; removing the pawn surely goes a long way in proving the existence of a perpetual. A queen sufficiently far away always has at least 5 possible check squares. They can't all be blocked, since there are only 4 rooks.

Comment by Philip Engel

2010-07-29T15:22:13Z

Optimally, $X$ would be any Riemann surface such that the endomorphism ring of its Jacobian is defined over $\mathbb{Q}$. In this case, the definition of $T_p$ couldn't rely an interpretation as a moduli space or quotient of the upper half-plane. This definition would coincide with the one we know for modular curves.