## Anabelian puzzle 5: isogenies between Jacobians and metabelian fundamental groups

Another question that came up at the Newton Insitute:  can two different curves X,Y over F_q have the same geometrically metabelian pro-l fundamental group?

I would think not, and here’s why.  First of all, the actions of Frob_q on H_1(X,Z_ell) and on H_1(Y,Z_ell) agree.  This already implies that Jac(X) and Jac(Y) are isogenous.  Can this actually happen in large genus?  Yes:  a recent arXiv preprint by Ben Smith gives lots of explicit examples of pairs of hyperelliptic curves with isogenous Jacobians.  From Smith’s paper I learned about the recent construction by J. F. Mestre of pairs of hyperelliptic curves in every genus with isogenous Jacobians.

In other words, the geometrically abelian fundamental group need not distinguish X from Y.

But the fact that the geometrically metabelian pro-l fundamental groups agree implies the following much stronger fact.  Let X_n be the maximal abelian cover of X/F_qbar whose Galois group has exponent l^n, and define Y_n similarly.  Then X_n and Y_n have isogenous Jacobians for all n.  I would think this would be impossible if X and Y were not isomorphic; but I don’t have the slightest idea for a proof.

Baby version of this question:  do there exist non-isomorphic curves X and Y of large genus (say, for the moment, over C) whose Jacobians are isogenous, and such that each Prym of X is isogenous to a Prym of Y?

## Anabelian puzzle 4: What is the probability that a set of n points has no 3 collinear?

OK, this isn’t really an anabelian puzzle, but it was presented to me at the anabelian conference by Alexei Skorobogatov.

Let X_n be the moduli space of n-tuples of points in A^2 such that no three are collinear.  The comment section of this blog computed the number of components of X_n(R) back in January.  Skorobogatov asked what I could say about the cohomology of X_n(C).  Well, not a lot!  But if I were going to make a good guess, I’d start by trying to estimate the number of points on X_n over a finite field F_q.

So here’s a question:  can you estimate the number of degree-n 0-dimensional subschemes S of A^2/F_q which have no three points collinear?  It seems very likely to me that the answer is of the form

$P(1/q) q^{2n} + o(q^{2n})$

for some power series P.

One way to start, based on the strategy in Poonen’s Bertini paper:  given a line L, work out the probability P_L that S doesn’t have three points on L.  Now your first instinct might be to take the product of P_L over all lines in A^2; this will be some version of a special value of the zeta function of the dual P^2.  But it’s not totally clear to me that “having three points on L_1” and “having three points on L_2” are independent.

## Anabelian puzzle 3: why are there virtual sections?

Suppose X is a scheme over a field K, and write Xbar for the basechange of X to Kbar, so that as usual we have an exact sequence.

$1 \rightarrow \pi_1(\bar{X}) \rightarrow \pi_1(X) \rightarrow G_K \rightarrow 1$.

Now there may be no section from G_K back to $\pi_1(X)$.  But certainly X has a rational point over some finite extension L/K, which means that there is definitely a section from the finite-index subgroup G_L to $\pi_1(X)$.  This is so easy that I can’t help wondering:  is there a way to see the existence of such a “virtual section” from group theory alone?  My intuition is to say no.  But I just thought I’d mention it, while we’re puzzling anabelianly.

## Anabelian puzzle 2: birational sections in the least non-abelian case

(Note:  this is being posted from an airport shortly before boarding, so this is less edited than usual.  I have some more concrete remarks on the puzzle below but will save them for later.)

Let X/K be a variety over a number field; then we have an exact sequence of etale fundamental groups

$\pi_1(X/\bar{K}) \rightarrow \pi_1(X/K) \rightarrow G_K$

and every point of X(K) induces a section from $G_K$ back to $\pi_1(X/K)$.  Grothendieck’s section conjecture asserts that, at least for certain classes of varieties X including projective smooth curves of genus greater than 1, the group-theoretic sections above are in fact in bijection with X(K).

A point P of X(K) is also a point of U(K), where U is any open subscheme of X containing P.  This gives you a section $G_K \rightarrow \pi_1(U/K)$ for any such U; the limit of all these fundamental groups is the Galois group of the maximal extension of the function field K(X) unramified at P.

In fact, you can associate to $P \in X(K)$ a section from G_K to the whole absolute Galois group of K(X) (or, better, a “bouquet” of related sections.)  The choice of P determines a decomposition group in G_{K(X)}, isomorphic to a semidirect product of its inertia group by a copy of G_K; now a section of this semidirect product is a section from G_K to G_{K(X)}.

So a weaker version of the section conjecture, the birational section conjecture, asserts that all of these sections from G_K to G_{K(X)} come from X(K).  Koenigsmann proved a few years ago that the birational section conjecture holds for K = Q_p.  A recent paper of Florian Pop proves something much stronger; that the birational section conjecture holds over finite extensions of Q_p even when G_{K(X)} is replaced with a really puny quotient.  Namely:  let K’ be the maximal elementary 2-abelian extension of Kbar(X), and let K” be the maximal elementary 2-abelian extension of K”.  Then you have a group Gamma described by an exact sequence

$1 \rightarrow G(K''/\bar{K}(X)) \rightarrow \Gamma \rightarrow G_K$

and what Pop proves is that this group Gamma, which you might call the “geometrically metabelian mod-2 fundamental group of X,” “remembers” enough about the curve X that already the sections from G_K to Gamma are all given by points of X.  Pop calls this a “minimalist birational section conjecture.”

One then wonders:  just how minimal can one get?  Abelianizing Gal(Kbar(X)) is too brutal; the resulting “geometrically abelian” fundamental group has lots of “extra” sections coming from points of Jac(X).  (Note:  just noticed this paper of Esnault and Wittenberg about exactly this, haven’t read it yet.)

Now here’s the puzzle — suppose we let K’ be the maximal elementary 2-abelian extension of Kbar(X) (i.e. the compositum of all quadratic extensions) and K” be the maximal elementary abelian 2-extension of K’ such that Gal(K”/Kbar(X)) has nilpotence class 2.  Then again you have

$1 \rightarrow G(K''/\bar{K}(X)) \rightarrow \Gamma' \rightarrow G_K$

where $\Gamma'$ could be called the “geometrically 2-nilpotent mod 2 fundamental group.”

So what are the sections from G_K to $\Gamma'$?  Is this fundamental group so minimalist that there are tons of extra sections, or is it the maximally minimalist context where a birational section conjecture could hold?

Update:  A bit jetlagged, but let me at least add one concrete question to this post.  I claim that a birational section as above would give you a function

$f: U(\mathbf{Q}) \rightarrow \mathbf{Q}^*/(\mathbf{Q}^*)^2$

for some open subscheme U of P^1, with the property that the Hilbert symbol

$(f(a)/(a-b), f(b)/(b-a))$

always vanishes.  Can you think of any such functions besides f(a) = a + constant, or f(a) = constant?

## Anabelian puzzle 1: lifting Galois representations from symplectic groups to mapping class groups

This is an experimental post:  I’m going to put up something I’ve thought about only vaguely and partially, with the idea that it might be an interesting thing to discuss with people at the upcoming workshop on anabelian geometry at the Newton Institute.  Please forgive (or, better, correct) any and all mistakes.

Here’s an old puzzle of Oort, which I’ve mentioned here previously:

Does there exist, for every g > 3, an abelian variety over Qbar not isogenous to the Jacobian of any smooth genus-g curve?

Here’s one way you could imagine trying to construct an example.  Given a g-dimensional abelian variety A over a number field K, the action of the absolute Galois group G_K on the l-adic Tate module of A provides a Galois representation $\rho: G_K \rightarrow Sp_{2g}(Z_\ell)$.

Now suppose A is in fact the Jacobian of a smooth genus g curve X/K.  Then $\rho$ lifts to a representation $\tilde{\rho}: G_K \rightarrow G$, where G is the automorphism group of the pro-l geometric fundamental group of X.  (In fact, this is the case even if A is only isogenous to a Jacobian, as long as the degree of the isogeny is prime to l.)

So you can ask: are there abelian varieties A such that $\rho$ doesn’t lift from the symplectic group to G?  More:  such that the restriction $\rho | G_L$ doesn’t lift to G, for any finite extension L/K?  This seems quite difficult; it means you have to find an obstruction which isn’t torsion.

My further thoughts on this are even more disorganized and I’ll keep them to myself.  Oh, except I should say:  what makes this puzzle “anabelian” is that it has something to do with the notion that a section of the map

$\pi_1^{et}(M_g/K) \rightarrow G_K$

ought to come from a point of M_g(K).

Yes, there is an anabelian puzzle 2, which I’ll try to post in the next couple of days.