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.

10 thoughts on “Anabelian puzzle 3: why are there virtual sections?”

1. Richard Kent says:

I like it.

2. Simon says:

I’m not sure I understand the question.

It is true that given a closed normal subgroup H of a profinite group G the canonical projection map has a continuous section — see Proposition 2.2.2 of Ribes and Zalesskii’s book on Profinite groups.

Unfortunately this can’t answer your question as it contradicts your claim that there maybe no section from G_K to pi_1(X). What have I misunderstood?

3. JSE says:

I don’t have Ribes and Zalesskii in front of me but surely there are some conditions: pZ_p is a closed normal subgroup of Z_p, but the quotient Z/pZ doesn’t have a section back to Z_p.

4. I guess the result is about set-theoretic sections; for profinite groups, one needs such things to be continuous for various cohomological constructions, if I remember right.

5. JSE says:

Ah, gotcha. Yes, the sections in the section conjecture are to be homomorphisms of topological groups.

6. Simon says:

Yes. The result I referred to is about set-theoretic sections.

So is the group theoretic question: given a profinite group G and a closed normal subgroup H is there always a finite index open subgroup L of G/H and a continuous group homomorphism from L to G splitting the projection map?

7. JSE says:

That would certainly imply the fact I was mentioning, but surely that’s too strong to be true? I’m envisioning some kind of central extension where the obstruction to a splitting would lie in H^2(G,A) for some abelian profinite group A with an action of another profinite group G, and I guess without any other conditions one can arrange this so as to make the obstruction non-vanishing on any finite-index subgroup of G….?

8. bhargav says:

Yes, one can make such things in H^1 and take cup products (as everyone probably realised). For example, G = A = Z-hat, with the trivial action H^1(G,A) = Hom(G,A) = A. Kunneth gives H^2(G x G,A) = A \otimes A = A. Restricting to finite index subgroups can’t kill this class. Here’s a topological proof: what we have is a generator of H^2(torus,Z) (approriately completed). Restricting to finite index subgroups corresponds to pulling back to finite etale covers of the torus. Such covers are themselves tori, and the pullback map on H^2 is simply multiplication by degree when both groups are identified with A, and thus non-zero as A is torsion free.

9. JSE says:

Ah, right — this is just like saying that you can’t make a section from the abelianization of the Heisenberg group back to the Heisenberg group, and this problem doesn’t go away after passing to a finite index subgroup.

10. Ben Wieland says:

Extensions K -> G -> Q, are classified by H^2(Q; Z(K)), where Z(K) is the center of K; and given merely Q -> Out(K), an element in H^3(Q; Z(K)) is the obstruction to the existence of extensions. In the anabelian case, we expect Z(K) to be trivial, and thus these obstructions to vanish.

In these situations, given Q -> Out(K), there is a unique extension of Q by K, so you might think of it as “trivial,” but it need not be a semidirect product. Splitting is equivalent to lifting Q -> Out(K) to Aut(K). For example, if K is the fundamental group of a surface then Out(K)=pi_1(M_g) and Aut(K)=pi_1(M_g,1) and we certainly do not expect a section from M_g to M_g,1, even on passing to a finite cover (for g>2).

So the existence of Q-bar points from the section conjecture requires special properties of the Galois group (or geometrically realizable actions), and not just special properties of anabelian geometric fundamental groups.