The n-Category Café

My pleasure. Two more remarks.

a. In #1 above, I forgot the word ‘etale’ at an important point: every *etale* $\bar{Q}$-algebra is a finite product of copies of $\bar{Q}$. It is obviously false otherwise.

b. I see now you were talking about all algebraic integers. I’m not really sure how to picture this. One complication is that there are two kinds of universal cover. There is the usual one, but there is also the universal profinite cover, which is the inverse limit of all the finite-to-one covering spaces. For example, the universal profinite cover of the circle is a $\hat{Z}$-bundle over the circle. (In fact, it corresponds to the usual addition action of the usual fundamental group $Z$ on $\hat{Z}$, rather than on $Z$ itself.)

This second universal cover is really the better analogue of the algebraic closure. The problem with it is that it’s hard to visualize, and the whole point of this endeavor is to use geometric intuition to explain in some sense the algebra.

Now consider the field $C(z)$ of ratios of complex polynomials. The picture is basically the same as for $Q$, but it is one dimension lower. We picture its spectrum as the Riemann sphere sewn from fly-screen cloth. The/An analogue of the ring of algebraic integers is the ring $A$ of elements in $\overline{C(z)}$ which satisfy monic polynomials with coefficients in $C[z]$. Since $A$ is the direct limit of its finitely generated subrings, which are necessarily contained in finite extensions of $C(z)$, we can view Spec $A$ as being the inverse limit of all the branched covers of the Riemann sphere. In other words, we have filled in the punctures on the profinite universal cover. This is a very accurate analogy, but is essentially impossible to picture. At least it is for me.

It’s much easier to visualize the usual universal cover, but I’m not sure if we can fill in the punctures. For instance, let’s suppose we have just two punctures. Then our space is $C^*$. It has a nice universal cover $C\to C^*$ given by the exponential map. But it doesn’t look possible to fill that in to a branched cover of $C$. To prove this, you’d have to make the statement itself precise.

So I give up. I can sort of visualize the analogue of the ring of algebraic integers in the case where I can’t really visualize the universal cover. In the case where I can visualize the universal cover, I can’t make sense of the algebraic integers. It’s a good thing we can always go back to the algebra.

I just caught up with this thread, but I’m afraid I still can’t contribute much. To a certain extent, it seems there would be too much to say with no hope of getting anything right.

But here are a few superficial comments, in any case.

1. Although $Spec(\bar{Q})$ is probably contractible in some sense, I’m not sure the rationale given is correct. Any simply connected variety over $C$, for example, satisfies the condition that etale covers split. That is, the splitting of finite extensions by itself doesn’t say anything more more less than that the space is simply connected. Perhaps the more relevant fact is the $Spec(Q)$ does behave like a $K(\pi,1)$: cohomology of sheaves on $Spec(Q)$ can be computed using group cohomology of its fundamental group. A space like $P^1_C$, on the other hand, is obviously not the universal cover of any sort of $K(\pi,1)$.

2. Perhaps the best way to provisionally visualize certain infinite coverings is to pretend that we have a feeling for one basic case: If we have a space $X$ that we can visualize, and $$Y\rightarrow X$$ is a connected Galois etale covering with Galois group `some profinite version of $Z$,’ then we imagine that we can see a loop on the base permuting the sheets of the covering. Starting from this, it’s not too much bigger a leap of the imagination to contemplate a $Y\rightarrow X$ that is ramified over a few points in $X$, or to covering groups that are topologically finitely-generated.

A situation that number theorists have some feeling for is when $X=Spec(Z)$, $S$ is a finite set of prime ideals in $Z$ containing some fixed prime $p$, and $Y\rightarrow X$ is the covering corresponding to the pro-$p$ fundamental group of $Spec(Z[1/S])$. This last space is the open subscheme of $Spec(Z)$ where all the points in $S$ have been removed (by inverting the primes in $S$, hence the notation). In arithmetic lingo, $Y$ is the ring of integers in the maximal pro-$p$ extension of $Q$ unramified outside $S$. The construction of such a thing is also complicated. We would have to contemplate rings like $$Z[x]/(f(x))$$ with irreducible $f$ having discriminant divisible only by primes in $S$ and having the added property that the splitting field of $f$ has a $p$-group as Galois group, and then take the union of all these (say in $C$). Nevertheless, because the pro-$p$ fundamental group $\Pi$ in question is topologically finitely-generated, and one has some control of its structure, we can try to geometrize our intuition. One common approach is to look for usual topological coverings $$N\rightarrow M$$ whose covering group has pro-$p$ completion isomorphic to $\Pi$ and then examine the extent to which $M$ and $N$ can give insight on the geometry of $X$ and $Y$.

In other words, $Spec(\bar{Z})\rightarrow X=\Spec(Z)$ is indeed hard. I think it’s safe to say that no one in the world has a good feeling for this covering. But we can probe its nature by studying various subcoverings

$Spec(\bar{Z}) \rightarrow Y \rightarrow X.$

The two key properties of such coverings that make them tractable are:

A. bounded ramification; and

B. restrictions on the Galois group, making them quite close to $p$-adic Lie groups for some prime $p$.

The ring of integers in the abelian extension that John mentioned is already quite hard because it’s ramified over every prime. However, if we let $Y=Spec(R)$ where $R$ is the ring of integers in $Q(\mu_{p^{\infty}})$, the field of $p$-power roots of unity, then the covering transformations of $Y\rightarrow X$ is $Z_p^*$ as has been pointed out. This fits into an exact sequence $$0\rightarrow [1+pZ_p] \rightarrow Z_p^*\rightarrow F_p^*\rightarrow 0$$ where the group $F_p^*$ correponds to the subextension $Q(\mu_p)$ generated by the $p$-roots (not allowing any powers). Its ring of integers is $Z[e^{2\pi i/p}]$, so that we have the tower $$Y \stackrel{1+Z_p}{\rightarrow} Spec(Z[e^{2\pi i/p}])\stackrel{F_p^*}{\rightarrow}X.$$ The covering $$Spec(Z[e^{2\pi i/p}])\stackrel{F_p^*}{\rightarrow}X$$ is of course finite, and ramified only over the prime $p$. In fact, there is only one point $P=(1-e^{2\pi i/p})$ lying above $(p)$. (The same is true for any finite layer of the covering $Y\rightarrow X$.) The covering $$Y \stackrel{1+Z_p}{\rightarrow} Spec(Z[e^{2\pi i/p}])$$ is infinite, but the Galois group is isomorphic to $Z_p$ (via the log map). Furthermore, it is ramified only over $P$. So if we restrict the covering to $Spec(Z[e^{2\pi i/p}]))-P$, it is unramified and we pretend we can visualize it. The whole thing has all the sheets coming together over $P$, which are then permuted by a generator of $Z_p$. After a sufficient amount of time thinking about such examples, you can trick yourself into acquiring geometric intuition. On the other hand, if either A or B is not satisfied, even trickery doesn’t help much.