16
$\begingroup$

I am considering the following two isomorphisms:

First, if $X$ is a reasonably nice topological space, then $X$ has a normal covering space which is maximal with respect to the property of having an abelian group of deck transformations. The group of deck transformations of this covering space is isomorphic to the abelianizatin of the fundamental group of $X$, which can be identified with the singular homology group $H_1(X,\mathbb{Z})$.

Second, if $K$ is a number field, class field theory gives an isomorphism between the Galois group of the maximal abelian unramified extension of $K$ (the Hilbert class field) and the ideal class group of $K$. The ideal class group can be identified with the sheaf cohomology group $H^1(\mathrm{Spec}(\mathcal{O}_K),\mathbf{G}_m)$.

Given the apparent similarity between these two theorems, is there some more general theorem which implies both of these results as special cases?

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ mathoverflow.net/questions/546/… $\endgroup$
    – Alex B.
    Commented Nov 26, 2010 at 7:21
  • 2
    $\begingroup$ Dear Julian: the above link doesn't address abelian unramified case. There are really two theorems, and one bridge. Compact Riemann surfaces and global fn fields are subsumed by "geometric class field theory" of Lang-Rosenlicht (using Jacobians to classify finite abelian coverings); see Serre's book. Global fn fields and number fields are subsumed by the usual CFT (together with Kummer sequence to relate cohomology of $\mathbf{G}_m$ to that of $\mu_n$, and the link between Jacobian and Pic). So global fn fields are the bridge. Weil wrote to his sister about it. Talk to K.P. or M.Z. $\endgroup$
    – BCnrd
    Commented Nov 26, 2010 at 8:54
  • $\begingroup$ Dear Brian, maybe I am reading too much into Minhyong's answer, but I think he does mention the connection between the fundamental group of a variety and the Galois group of the maximal unramified extension of the function field. He also mentions that "Weil was fully aware that homology and class groups are somehow the same". The rigorous connection to the number field case, given by étale fundamental groups, is explained in Szamuely's book that Davidac897 links to, and particularly in the 2 extra chapters, linked to by AS. Or am I missing something and that's not what this question is about? $\endgroup$
    – Alex B.
    Commented Nov 26, 2010 at 9:44
  • $\begingroup$ Dear Alex: Minhyong's essay is largely focused on the saga of base points, which is not so relevant in the abelian case, and although he mentions Weil's awareness of the abelian analogy, he doesn't get into the perspective of Jacobians that is needed to make it precise. It seems to me that the story of Jacobians and Picard functors (which is somewhat different from that of etale fundamental groups) is at the heart of unifying the two situations Julian asks about. That's the sense in which Minhyong's essay seems to be about something else (though it is still important information!). $\endgroup$
    – BCnrd
    Commented Nov 26, 2010 at 14:42

1 Answer 1

Reset to default
12
$\begingroup$

As BCnrd says, the theorem you want is geometric class field theory. One version says that the abelianization of the fundamental group of a curve over an algebraically closed field is the fundamental group of its Jacobian. One can use this to derive class field theory for curves over finite fields. Over the complex numbers, this is a topological statement, since the Jacobian of a Riemann surface can be constructed by topological methods, such as $J(C)=H_1(C;\mathbb R/\mathbb Z)$. Also, consider Weil's construction of the Jacobian by using Riemann-Roch to recognize a high symmetric power of a curve as a $\mathbb C\mathbb P^n$ bundle over the Jacobian. The projective space bundle is probably not a topological invariant, but the symmetric power is and it already has the right fundamental group. That has an extensive topological generalization, the Dold-Thom theorem, that the homology of a reasonable space is the homotopy groups of its infinite symmetric power.

The key unifying ingredient is, as BCnrd says, the Jacobian, even though it is missing from both of your statements.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .