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Let $C \subset {\bf CP}^2$ be an irreducible algebraic smooth (projectively) planar curve over the complex numbers of degree $d$ (we allow finitely many points to be deleted from $C$ to make it smooth). Then a generic complex line in ${\bf CP}^2$ intersects $C$ in $d$ distinct points $z_1,\ldots,z_d$.

Question 1: for any permutation $\sigma: \{1,\ldots,d\} \to \{1,\ldots,d\}$, is it possible to continuously deform the $d$ points $z_1,\ldots,z_d$ to $z_{\sigma(1)},\ldots,z_{\sigma(d)}$ while keeping all the points collinear, distinct, and on $C$ at all times? In other words, does there exist continuous maps $z_i: [0,1] \to C$ with $z_i(0) = z_i$, $z_i(1) = z_{\sigma(i)}$, and the $z_1(t),\ldots,z_d(t)$ distinct and collinear for all $t\in [0,1]$?

I believe the answer to this question is yes, because one can transpose any two of the $z_i,z_j$ while keeping the other $z_k$ unchanged by bringing them close together near a generic point of $C$ (and making the collinear line close to the line of tangency of this point to $C$, which generically will not be tangent to anywhere else in $C$), performing the transposition, and then returning back to the original position. One can also phrase the question as follows: if we let $S \subset C \times Gr(2,1)$ be the set of incidences $(p,\ell)$ with $p \in C$ and $\ell \in Gr(2,1)$ a line through $p$, then (after passing to an open dense subset of $Gr(2,1)$), $S$ is a covering space over (most of) $Gr(2,1)$ whose fibre has $d$ points, and the question asserts that the fundamental group of the base acts completely transitively on the fibre (in that every permutation of the fibre shows up); equivalently, the question asserts that the $d^{th}$ exterior power of $S$ over $Gr(2,1)$ is irreducible.

However, I'm having more trouble with proving the following generalisation of Question 1. Keep the same setup as before, but now suppose we also have another irreducible curve $C'$ covering $C$ with fibres of cardinality $d'$. Thus, above a generic $z_i$ in $C$ we have $d'$ points $w_{i,1},\ldots,w_{i,d'}$ in $C'$.

Question 2: for any permutation $\sigma: \{1,\ldots,d\} \to \{1,\ldots,d\}$, and any $j_1,\ldots,j_d,k_1,\ldots,k_d \in \{1,\ldots,d'\}$ is it possible to continuously deform the $d$ points $w_{1,j_1},\ldots,w_{d,j_d}$ to $w_{\sigma(1),k_1},\ldots,w_{\sigma(d),k_d}$ while keeping all the points on $C'$, and their projections onto $C$ collinear and distinct?

Like Question 1, one can phrase this question in terms of the transitivity properties of the action of the fundamental group of a suitable open dense subset of $Gr(2,1)$ on some fibre, but this formulation becomes rather complicated to state and I won't give it here.

I believe I can resolve this problem in the affirmative if $C$ has a smooth closure, in which case I can use ad hoc arguments to move around each of the $j_i$ independently, but encountered a problem if $C$ has a singular point in its closure, and $C'$ ramifies above this point, as then the points $z_i$ appear to become entangled with each other near this point in a manner that I was not able to analyse by ad hoc topological arguments.

[My motivation for this problem was to understand arcs in a finite plane ${\bf F}_p^2$, that is to say sets that do not contain any collinear triples, and to classify when these arcs can be constructed as the ${\bf F}_p$-points of a low-degree algebraic curve $C$ (or as the projection of the ${\bf F}_p$-points of a low-degree algebraic curve $C'$). Using a Lefschetz principle argument, one can transfer the problem (in the large $p$ limit) to a problem in the complex plane related to the questions above.]

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up vote 20 down vote accepted

The answer to question 1 is yes, and is known as Harris's Uniform Position Lemma. It was proved in Harris's 1980 paper Galois groups of enumerative problems. You can find a nice exposition in Chapter 9 of Solving Polynomial Equations, by Bronstein, Dickenstein and Emiris.

You should be warned that the analogous statement is not true in characteristic $p$, which you say is your motivation. For example, consider the curve $y=x^p$ in characteristic $p$. Any line $y=mx+b$ meets this curve at $p$ points, whose $x$ coordinates are the roots of $x^p=mx+b$. Note that the roots of this polynomial form an arithmetic progression: If $r$ and $s$ are roots, so is $r+k(s-r)$ for every $k \in \mathbb{Z}/p$. That means that the monodromy, as you move the line, must preserve this structure, so it is a subgroup of $\mathbb{Z}/p^{\ast} \ltimes \mathbb{Z}/p$.


UPDATE I think question 2 is false! Let $C$ be a smooth cubic curve in $\mathbb{P}^2$, and choose a flex on $C$ to be the origin, so $C$ has a group structure. Take $C'$ to be equal to $C$, but with the map $C' \to C$ being the doubling map in the group law. So $d'=4$.

If $L$ is a line in $\mathbb{P}^2$, and $L \cap C = \{ x,y,z \}$, then $x+y+z=0$ in the group law. If $2 x'=x$, $2y' =y$ and $2 z'=z$, then $x'+y'+z'$ is a $2$-torsion point of $C'$. If we move $(L, x',y',z')$ continuously, we can't change which $2$-torsion point $x'+y'+z'$ is. So there are (at least) four separate connected components in the space of $(x',y',z')$ triples.

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Thanks! And now I see you blogged about this lemma at sbseminar.wordpress.com/2008/03/08/… . If Question 1 is already a publishable lemma, I guess Question 2 is going to be nontrivial! As for characteristic p, I'm mostly (though not exclusively) interested in the regime where the degree is bounded and the characteristic goes to infinity, so in this case the Lefschetz principle kicks in and the sort of counterexamples you mention don't appear. But it does suggest that a purely algebraic argument may be difficult... –  Terry Tao Nov 15 '13 at 1:18
    
Well, there is a lot more in that paper; I doubt Harris would have published it with just the lemma. But yes, this doesn't seem trivial to me. –  David Speyer Nov 15 '13 at 1:42
    
Thanks! In retrospect, I knew this counterexample in a different context (one can use an index 2 coset of an elliptic curve over a finite field to make an arc). It's still potentially possible that the result still holds when C has degree bigger than 3 (so that it is no longer an abelian variety), but I guess I would have to be much more skeptical now... –  Terry Tao Nov 15 '13 at 3:16
    
... and now I see where I went wrong in thinking that Question 2 was true in the smooth case; I had thought the fundamental group was generated purely by small loops around punctures (and their conjugates), but of course there are global generators as well, which is what is powering the elliptic curve example... –  Terry Tao Nov 15 '13 at 3:19
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I don't think there's much hope if the genus is greater than $1$. There will always be the Abel-Jacobi map to its Jacobian, an abelian variety. You can pull back the multiplication-by-two cover of the abelian variety to get a connected etale cover of the curve, and the same argument works. –  Will Sawin Nov 15 '13 at 17:04

While both of my original questions have been answered by David, I wanted to record the fact that a slightly weaker version of Question 2 is in fact true (and good enough for my particular application), in which one allows one of the points to move freely:

Question 2': for any permutation $\sigma: \{1,\ldots,d\} \to \{1,\ldots,d\}$, and any $j_1,\ldots,j_{d-1},k_1,\ldots,k_{d-1} \in \{1,\ldots,d'\}$ is it possible to continuously deform the $d-1$ points $w_{1,j_1},\ldots,w_{d-1,j_{d-1}}$ to $w_{\sigma(1),k_1},\ldots,w_{\sigma(d-1),k_{d-1}}$ while keeping all the points on $C'$, and their projections onto $C$ collinear and distinct?

We sketch the proof as follows. Let $V \subset C^d$ be the set of all $k$-tuples of distinct collinear points $(z_1,\ldots,z_d)$ in $C$; by the positive solution to Q1, this is irreducible. The embedding $V \subset C^d$ induces a homomorphism $\pi_1(V) \to \pi_1(C)^d$; by abstract nonsense, Q2' follows from (and is basically equivalent to) the assertion that the resulting image of $\pi_1(V)$ in $\pi_1(C)^d$ projects surjectively onto $\pi_1(C)^{d-1}$.

Now, since any two points determine a line, $C^2$ lifts to $V$, which implies that the projection of $\pi_1(V)$ to $\pi(C)^2$ is surjective. Also, the image of $\pi_1(V)$ in $\pi_1(C)^d$ is symmetric with respect to permutations of the $d$ coordinates. This is enough to imply the surjectivity onto $\pi_1(C)^{d-1}$ by elementary group theory (this is a trick I learned from a paper of Furstenberg and Weiss). Indeed, for any $g \in \pi_1(C)$, the surjectivity onto the final $\pi_1(C)^2$ implies that the image of $\pi_1(V)$ in $\pi_1(C)^d$ contains an element of the form $(g_1,\ldots,g_{d-1},g,1)$ for some $g_1,\ldots,g_{d-1}$; by symmetry it also contains $(g_1,\ldots,g_{d-1},1,g)$,and by dividing it thus contains $(1,\ldots,1,g,g^{-1})$. Thus the projection onto $\pi_1(C)^{d-1}$ contains $(1,\ldots,1,g)$; permuting and composing we obtain the required surjectivity.

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I'm confused, although this seems like excellent progress. I assume that $V$ is the set of ordered $d$-tuples, since you want an embedding $V \hookrightarrow C^d$. But when you use the fact that any two points determine a line, this seems to only lift $C^2$ to the space of unordered $d$-tuples, $V/S_d$. Am I missing something? –  David Speyer Nov 15 '13 at 17:05
    
Ugh. I think the argument still works, because any loop in $C^2$ (starting near the diagonal) can lift to a path in $V$ which loops back to itself in the first two components but not necessarily the last $d-2$, and then the rest of the argument (now working on fundamental groupoids instead of fundamental groups) still gives $(1,\ldots,1,g,g^{-1})$ in the image of $\pi_1(V)$. Not sure how to phrase this argument cleanly though. –  Terry Tao Nov 15 '13 at 18:39
    
I think now that these arguments, combined with the Abel-Jacobi observation of Will, now give a complete description of the image of $\pi_1(V)$ to $\pi_1(C)^d$, namely this consists of those elements whose projection to $H_1(C)^d$ by the Hurewicz homomorphism has entries summing to zero; thus Q2 has a positive answer precisely when there exist curves from each $w_{i,j_i}$ to $w_{\sigma(i),k_i}$ whose lift to the Jacobi torus have total displacement summing to zero. –  Terry Tao Nov 16 '13 at 17:51

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