The n-Category Café

In case some of my statements are misleading, I should point out that the classification of torsors for $\hat{\mathbb{Z}}(1)$ itself is actually not hard. Using Hilbert’s theorem 90, one finds the classifying set $({\mathbb{Q}}^{*}{)}^{\wedge }$, the pro-finite completion of the multiplicative group of $\mathbb{Q}$. In virtually all other cases of fundamental groups of curves, the classifying space of torsors is difficult.

Tom writes:

So go on, what is the abelianization of $\mathrm{Gal}(\overline{\mathbb{Q}}\mid \mathbb{Q})$?

First of all, it’s $\mathrm{Gal}({\mathbb{Q}}^{\mathrm{ab}}\mid \mathbb{Q})$, where ${\mathbb{Q}}^{\mathrm{ab}}$ is the ‘maximal abelian extension of $\mathbb{Q}$’ — the biggest algebraic extension of $\mathbb{Q}$ whose Galois group over $\mathbb{Q}$ is abelian. Concretely, we get ${\mathbb{Q}}^{\mathrm{ab}}$ by taking $\mathbb{Q}$ and throwing in all roots of unity.

Second of all, $\mathrm{Gal}({\mathbb{Q}}^{\mathrm{ab}}\mid \mathbb{Q})$ has a nice concrete description. Everything I know and love is in This Week’s Finds — or will be eventually. So, to see how we get this concrete description, try this quote by Edward Frenkel in week221:

…recall the classical Kronecker-Weber theorem which describes the maximal abelian extension ${\mathbb{Q}}^{\mathrm{ab}}$ of the field $\mathbb{Q}$ of rational numbers (i.e., the maximal extension of $\mathbb{Q}$ whose Galois group is abelian). This theorem states that ${\mathbb{Q}}^{\mathrm{ab}}$ is obtained by adjoining to $\mathbb{Q}$ all roots of unity; in other words, ${\mathbb{Q}}^{\mathrm{ab}}$ is the union of all cyclotomic fields $\mathbb{Q}(1^{1/N})$ obtained by adjoining to $\mathbb{Q}$ a primitive $N$th root of unity

$$1^{1/N}$$

The Galois group $\mathrm{Gal}(\mathbb{Q}(1^{1/N})\mid \mathbb{Q})$ of automorphisms of $\mathbb{Q}(1^{1/N})$ preserving $\mathbb{Q}$ is isomorphic to the group $(\mathbb{Z}/N{)}^{*}$ of units of the ring $\mathbb{Z}/N$. Indeed, each element $m$ in $(\mathbb{Z}/N{)}^{*}$, viewed as an integer relatively prime to $N$, gives rise to an automorphism of $\mathbb{Q}(1^{1/N})$ which sends

$$1^{1/N}$$

to

$$1^{m/N}$$

Therefore we obtain that the Galois group $\mathrm{Gal}({\mathbb{Q}}^{\mathrm{ab}}\mid \mathbb{Q})$, or, equivalently, the maximal abelian quotient of $\mathrm{Gal}(\overline{\mathbb{Q}}\mid \mathbb{Q})$, where $\overline{\mathbb{Q}}$ is an algebraic closure of $\mathbb{Q}$, is isomorphic to the projective limit of the groups $(\mathbb{Z}/N{)}^{*}$ with respect to the system of surjections

$$(\mathbb{Z}/N{)}^{*}\to (\mathbb{Z}/M{)}^{*}$$

for $M$ dividing $N$. This projective limit is nothing but the direct product of the multiplicative groups of the rings of $p$-adic integers, ${\mathbb{Z}}_p^{*}$, where $p$ runs over the set of all primes. Thus, we obtain that

$$\mathrm{Gal}({\mathbb{Q}}^{\mathrm{ab}}\mid \mathbb{Q})=\prod _p{\mathbb{Z}}_p^{*}$$

If you stare at this and don’t let your eyes glaze over, you’ll see that at the end he’s computing the inverse limit of the multiplicative groups $(\mathbb{Z}/N{)}^{*}$ and getting

$$\prod _p{\mathbb{Z}}_p^{*}$$

So, the group you’re asking about has many nice descriptions, which is one reason I like it so much.

Another reason is that a Galois group is like a first homotopy group (${\pi }_1$), so an abelianized Galois group is like an abelianized first homotopy group, namely a first homology group ($H_1$). Homotopy groups are hard, but homology groups are easier. Here we see that very nicely. $\mathrm{Gal}(\overline{\mathbb{Q}}\mid \mathbb{Q})$ is big, scary and mysterious — at least one mathematician has gone insane contemplating it. Its abelianization is big, impressive — yet friendly and eminently comprehensible.

This is the idea behind class field theory: don’t mess with general Galois groups; stay in the safe waters of abelian Galois groups, where you’re secretly doing homology theory instead of homotopy theory.

Minhyong is leaving these safe waters, so what’s he’s doing is deeper and — to me, at least — more frightening.

I’ve never tried too hard to understand the GT theory. But in case some people haven’t seen it before, I will remark that the connection with G (=$\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$) comes exactly from the embedding $${\rho }^{\mathrm{na}}:G\hookrightarrow \mathrm{Out}({\pi }_1^{\mathrm{et}}(\overline{X}))$$ alluded to above, where $X$ is the projective line minus three points. (Actually, I spoke of the automorphisms rather than the outer automorphisms, but it’s still an injection when you pass to the quotient. This allows us to suppress reference to a basepoint.) One definition of the profinite GT group is as a subgroup of $\mathrm{Out}({\pi }_1^{\mathrm{et}}(\overline{X}))$ satisfying certain natural constraints that come from viewing $X$ as the configuration space of four points on ${\mathbb{P}}^1$, and then examining relations to the configuration space of five points. Roughly speaking, you require the automorphisms to act on the ${\pi }_1$ of both configuration spaces, and be compatible with natural maps between them. Sorry I’m not being more precise at the moment. But I will say that these constraints define a subgroup $$\mathrm{GT}\subset \mathrm{Out}({\pi }_1^{\mathrm{et}}(\overline{X})).$$It’s not too hard to see that $\mathrm{Im}({\rho }^{\mathrm{na}})$ is contained in GT. Some people then conjecture that this inclusion is actually an equality.

Thanks for reminding me, Bruce!

Grothendieck–Teichmüller theory is one of those amazing conjectural connections between beautiful fundamental structure that are exactly what I like to talk about on This Week’s Finds…. so I must come back to it someday. I’ve been meaning to, off and on. I just never feel I understand it well enough.

I guess you know, Bruce, that the ‘projective line minus three points’, which Minhyong is talking about, is an algebraic version of the ‘pair of pants’ that string theorists and TQFT dudes are always talking about:

So, somehow Grothendieck–Teichmüller theory is secretly all about a relation between $\mathrm{Gal}(\overline{\mathbb{Q}}\mid \mathbb{Q})$ and string theory! There should be some stunning implications for mathematical physics, but I’ve never succeeded in guessing what they are.

Close, but not quite!

Let me say something about your last point. When you say “affine line” in algebraic geometry, you technically have to say what field (or ring, or scheme, or…) you want your affine line to be over. This is kind of like saying “trivial fibration” in usual geometry—you have to say what the base is. Also the etale fundamental group depends on what the base is. This is just like usual geometry where the fundamental group of the total space of a fibration depends on what the base is. And just like in usual geometry, the fundamental group is an extension of the fundamental group of the base by the fundamental group of the fiber.

So, it’s perfectly fine to talk about the fundamental group of the $A^1-\{0\}$ over $Q$. It’s just the semi-direct product of the fundamental group of $A^1-\{0\}$ over $\overline{Q}$ and $\mathrm{Gal}(\overline{Q}/Q)$. (As Minhyong mentioned above, it’s a split extension because the fibration has a section, i.e. the affine line minus one point still has a rational point.) Which brings us to the action…

The etale fundamental group of a variety over $\overline{Q}$ is just the pro-finite completion of the usual fundamental group of the corresponding complex-analytic space. In the case of the affine line minus one point, we get $\hat{Z}$, not ${\hat{Z}}^{*}$. This is good. $\hat{Z}$ is much more like $Z$ than ${\hat{Z}}^{*}$ is.

The ${\hat{Z}}^{*}$ comes in with the Galois action. ${\hat{Z}}^{*}$ acts on $\hat{Z}$ by multiplication, and there is a perfectly natural map (called the “cyclotomic character”) $\mathrm{Gal}(\overline{Q}/Q)\to {\hat{Z}}^{*}$. Composing these two then gives the action of the absolute Galois group on the etale fundamental group of the affine line minus one point.

It might be worth looking at this on the finite quotient $Z/\mathrm{nZ}$. Then $\mathrm{Gal}(Q({\zeta }_n)/Q)=(Z/\mathrm{nZ}{)}^{*}$, and again the action on $Z/\mathrm{nZ}$ is given by multiplication. The original thing is just a way of collecting all this data as $n$ varies into one convenient package where you don’t have to keep track of which finite quotient you’re talking about.

By the way, there is one other fundamentally abelian example—elliptic curves with “complex multiplication”. Ignoring algebraic structure, these are just tori of the form $C/Z+\tau Z$, where $\tau$ is a imaginary quadratic irrational. Maybe if there’s enough public demand, Minhyong will be inspired to explain the Galois action there…

What you write is correct. If we put $X=A^1\setminus \{0\}$, then it is the ${\pi }_1^{\mathrm{et}}$ of $\overline{X}$ that becomes $\hat{\mathbb{Z}}$. The distinction between the fundamental groups over $\mathbb{Q}$ and $\overline{\mathbb{Q}}$ has to do with the fact that $X$ has etale covers $X_F$ coming from just constant field extensions $F$ of $\mathbb{Q}$. That is, if we define $X_F$ to be the pull-back of $X$ via the map $$\mathrm{Spec}(F)\to \mathrm{Spec}(\mathbb{Q}),$$ then $$X_F\to X$$ is an etale map. Going to the limit over all such $F$, we get $\overline{X}$. And then ‘composing’ with a system of coverings for $\overline{X}$ will finally yield the universal cover of $X$ itself. This is why we get the exact sequence $$0\to {\pi }_1(\overline{X})\to {\pi }_1(X)\to G\to 0$$ I mentioned before. So ${\pi }_1(X)$ actually subsumes the Galois group $G$ as a quotient, and hence, is horribly complicated. ${\pi }_1(\overline{X})$, on the other hand, can be computed using ‘standard geometry.’ To see that the system of $n$-th power maps gives a universal cover, you just need to use the so-called ‘Riemann-Hurwitz formula.’ In that article, the formula is given for a Riemann surface, but it’s identical for an algebraic curve over $\overline{\mathbb{Q}}$. Now if you have a covering $$Y\to \overline{X},$$ it extends to a covering of the compactifications $$Y^{\prime }\to {\mathbb{P}}^1$$ that is ramified only over $0$ and $\mathrm{\infty }$. It is a relatively simple exercise using the formula to see that $Y^{\prime }$ must be of genus zero and that the inverse image of $\{0,\mathrm{\infty }\}$ is a two-point set. From this, one quickly deduces that it is isomorphic to an $n$-th power map.

When I said that abelian class field theory is the study of the Galois action on ${\pi }_1^{\mathrm{et}}(\overline{X})$, I meant provisionally just that this action is the same as the action on the roots of unity. But there are more precise things one can say, for which I will try eventually to muster up the energy!

But for now, I will repeat that non-abelian class field theory in the sense of Langlands attempts to understand Galois actions on $H_i^{\mathrm{et}}(\overline{V},\hat{\mathbb{Z}})$ for general varieties $V$ in situations where the image of the Galois group inside $\mathrm{Aut}(H_i^{\mathrm{et}}(\overline{V},\hat{\mathbb{Z}}))$ is non-abelian. (Note that for $X$, ${\pi }_1(\overline{X})=H_1(\overline{X})$.) Wiles’ great theorem deals with the case where $V$ is an elliptic curve over $\mathbb{Q}$ . However, replacing abelian homology by non-abelian fundamental groups as the object of the action is genuinely an additional layer of non-commutativity that is expected to yield more information about the arithmetic geometry of the entities involved.

Thanks, James and Minhyong, for gently correcting some of my silly mistakes. I’ve always been vaguely puzzled about the interaction between $\hat{\mathbb{Z}}$ and ${\hat{\mathbb{Z}}}^{*}$ in number theory, and I got them tangled up in my last comment. I’m too tired to think clearly about this now, but I see you’ve offered me the royal road to forever quelling this puzzlement: we’ve got both $\hat{\mathbb{Z}}$ and ${\hat{\mathbb{Z}}}^{*}$ sitting nicely in the same situation here, the latter acting on the former.

I’m sorry for not diving directly into the more excitingly nonabelian examples, Minhyong — I’ve just got to get some basics straightened out first. But, I’m not immune to the lure of nonabelian Galois representations. For the last month or so, James Dolan have spent our biweekly afternoon chats discussing elliptic curves and the Modularity Theorem. I’m finally getting a clear picture of what modular curves look like! We’re drawing nice pictures of lots of them, relating them to dessins d’ enfants, and so on. Eventually this should connect with some of the things you’re trying to explain here… but not, alas, tonight.