## April 23, 2009

### Afternoon Fishing

#### Posted by David Corfield

Fishing about at the Café for material to extract for $n$Lab, I was reminded of a question I had never got around to asking. I was also led to the third of the talks listed here (note caveat there concerning change of sentiments, and here concerning the second half), where Minhyong Kim writes:

For years, I’ve felt the need to deny the popular conception of mathematics that equates it with the study of numbers. It is only recently that I’m returning to a suspicion that mathematics is perhaps about numbers after all. It is said that in ancient Greece the comparison of large quantities was regarded as a very difficult problem. So it was debated by the best thinkers of the era whether there were more grains of sand on the beach or more leaves on the trees of the forest. Equipped now with systematic notation and fluency in the arithmetic of large integers, it is a straightforward (albeit tedious) matter for even a schoolchild to give an intelligent answer to such a question. At present, our understanding of the complex numbers is about as primitive as the understanding of large integers was in ancient Greece.

Earlier in the essay he writes

…in contrast to the continuum picture of the complex plane, a number theorist is more likely to perceive of each individual number or groups of numbers in a discrete fashion, and in nested hierarchies reflecting various complexities, and even attach a symmetry group to each individual number. It is not that number theorists avoid the plane model, since it is also an important tool in much of number theory. It is just that the plane has a much more grainy and elaborate shape, with many levels of microscopic detail and structure.

This neatly leads to the question I wanted to ask.

James shared with us his image of $Spec (\mathbb{Q})$:

I like to picture $Spec (\mathbb{Q})$ as something like a 2-manifold which has had all its points deleted. The extra complication is that what we think of as the points are actually very small circles. So it’s really a three manifold with all of the loops inside it deleted…Maybe some should be seen as bigger than others, corresponding to the fact that there are prime numbers of different magnitudes.

Now, when we were discussing that the fundamental group of $Spec(\mathbb{Q})$ is $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$, the issue of base points arose, which allowed Minhyong to tell us that

…for $\mathbb{Q}$, a choice $\mathbb{Q} \embedsin \bar{\mathbb{Q}}$ of an algebraic closure will correspond to a map $Spec(\bar{\mathbb{Q}}) \to Spec(\mathbb{Q})$ that can then be considered a base-point.

This was in response to Jim Dolan’s remark

…the fundamental groupoid of $Spec(\mathbb{Q})$ is (at least morally) the groupoid of algebraic closures of the field $\mathbb{Q}$.

So how should I square this image with James’ deleted loops?

Perhaps I don’t have long to wait for some useful insight since:

I’ve been helping James Dolan develop some new ways of thinking about algebraic geometry. We’ll release a preliminary version of a paper on that any day now.

Wait a minute, I thought the spectrum of a ring was already a groupoid for you guys. Could Jim have dropped ‘the fundamental groupoid of ’ from his comment?

Posted at April 23, 2009 11:53 AM UTC

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### Re: Afternoon Fishing

Cool stuff! We almost need a whole blog on three dimensional thinking about numbers - oh wait, Lieven Le Bruyn already has one!

Posted by: Kea on April 23, 2009 2:13 PM | Permalink | Reply to this

### Re: Afternoon Fishing

Thomas Riepe appropriately drew our attention today to Analogies between Knots and Primes, 3-Manifolds and Number Rings by Masanori Morishita.

So $Spec(\mathbb{Z})$ with the infinite prime is analogous to the three-sphere. And James’ picture comes from inverting the primes, or in other words, removing the loops corresponding to them.

That’s the process of inverting primes. Now what about the process of algebraic completion? I suppose a first step is to adjoin $i$ to $\mathbb{Q}$. So $Spec(\mathbb{Q}(i))$?

Hmm, is there a way to think of what happens to the ‘mesh’ as algebraic completion takes place?

BTW, more math web-collaboration. There’s a group book on algebraic stacks being written here.

Posted by: David Corfield on April 23, 2009 3:26 PM | Permalink | Reply to this

### Re: Afternoon Fishing

Jim Dolan wrote:

…the fundamental groupoid of Spec(Q) is (at least morally) the groupoid of algebraic closures of the field Q.

David wrote:

So how should I square this image with James’ deleted loops?

It may help if you first replace the fundamental groupoid with the first homology, which is a watered-down ‘abelian’ way of counting holes in a space.

The first homology of a connected space is the abelianization of its fundamental group. The fundamental group of a space $X$ is the same as the set of points sitting over your favorite point $x \in X$ in the universal cover of $X$.

Combining these facts, we see the first homology of $X$ is the set points over $X$ in a certain ‘less universal’ cover of $X$: the universal abelian cover.

If you’ve never thought about this, I guess it’s very good to work out the universal cover of a figure 8, and also the universal abelian cover.

(Hint: if we lift a loop in the base space $X$ to a path in the universal cover, the path gets back where it started iff the loop in the base space is homotopic to the constant loop. If we lift a loop in the base space $X$ to a path in the universal abelian cover, the path gets back where it started iff the loop in the base space is homologous to the constant loop.)

The first homology group is perhaps less fundamental than the fundamental group, but it’s a lot easier to compute.

Applying this idea to $\mathbb{Q}$, it turns out that while the absolute Galois group of $\mathbb{Q}$ is hard to compute, its abelianization is known. And it’s the free profinite abelian group on countably many generators — one generator for each prime!

So, at one level of approximation, we can think of rational numbers as functions on a space that has one hole for each prime. Which makes sense, since we get $\mathbb{Q}$ by taking $\mathbb{Z}$ and inverting each prime. It’s a lot like how $1/z$ is defined on the complex plane with the origin removed: the pole at $0$ is okay, since the pole is in the hole.

This abelianized stuff is called ‘class field theory’ and it’s quite nontrivial, though delightfully simple compared to the full-fledged story.

I said a lot more about the abelianized Galois group of $\mathbb{Q}$ in week201. I explained the idea of ‘poking out a hole for each prime’ in week218. These old articles may be more interesting if you think of this stuff as a preliminary investigation into the deeper issues Minhyong is concerned with.

Posted by: John Baez on April 23, 2009 10:58 PM | Permalink | Reply to this

### Re: Afternoon Fishing

Well I know the universal cover of the figure of eight is that infinite 4-valent graph. And so the universal abelian cover ought to be the graph which looks like a square lattice.

Good to read those old TWFs and Week 257 again. This comment chimes nicely with those of Minhyong I quoted above:

I hope you’re reeling with horror at thought of this vast complicated wilderness of fields containing $\mathbb{Q}$ but contained in $\mathbb{C}$. First there’s a huge infinite thicket of algebraic number fields… and then, there’s an ever scarier jungle of fields that contain transcendental numbers like $\pi$ and $e$! I won’t even talk about that jungle, it’s so dark and scary. Physicists usually zip straight past this whole wilderness and work with $\mathbb{C}$.

But in fact, if you stop and carefully examine all the algebraic number fields and how they sit inside each other, you’ll find some incredibly beautiful patterns. And these patterns are turning out to be related to Feynman diagrams, topological quantum field theory, and so on…

Makes you wonder whether arithmetic is a universal repository of important structure (not via unnatural encoding, as in Gödellian numbering).

I suppose with $Spec(\mathbb{Q}(i))$, one is supposed to think of the ‘3-manifold’ $Spec(\mathbb{Z}[i])$ as a double cover of $Spec(\mathbb{Z})$, then go hole-punching again for each of its primes.

Which raises a question: What does $Spec(A)$, for $A$ the ring of all algebraic integers, look like?

Posted by: David Corfield on April 24, 2009 12:58 PM | Permalink | Reply to this

### Re: Afternoon Fishing

Thanks for the tip of the hat, David. It’s flattering but also slightly scary that I’ve inspired someone to think about these mental doodles.

Here are some responses and comments to what’s above.

1. David wrote

This was in response to Jim Dolan’s remark “…the fundamental groupoid of Spec(ℚ) is (at least morally) the groupoid of algebraic closures of the field ℚ.”

So how should I square this image with James’ deleted loops?

I just picture Spec $\bar{Q}$ as the universal cover of that 3-manifold with deleted loops. (Actually I don’t, because that’s just impossible for me to imagine, but I would if I could. Actually, I usually just picture $\bar{Q}$ as the letter Q with a little line over it, but when I encounter a new homological phenomenon, it’s nice to be able to test it against the more geometric picture.)

This is completely consistent with Jim Dolan’s remark. From the point of view of homotopy theory, a contractible space is the same as a point. So you could think of objects of the fundamental groupoid of $X$ as being contractible spaces equipped with a map to $X$, rather than just points mapping to $X$.

Why is it reasonable to view the universal cover of Spec $Q$ as being contractible, rather than just simply connected? This is because $\bar{Q}$ is algebraically closed, so any $\bar{Q}$-algebra is just a finite product of copies of $\bar{Q}$, and therefore its etale topology is trivial.

I believe this has a reasonable interpretation in terms of the picture. If we think of deleting points from surfaces (instead of loops from threefolds), then once you delete enough points (and if my further doodling is correct), you will get a bouquet of circles, which has contractible universal cover. Is the analogous fact true for threefolds? If $M$ is a threefold, are there finitely many disjoint loops in $M$ such that the complement of their union has contractible universal cover? Perhaps this is obviously true or false to someone with experience with these things. (It appears to be true for the 3-sphere.)

2. John wrote:

Applying this idea to ℚ, it turns out that while the absolute Galois group of ℚ is hard to compute, its abelianization is known. And it’s the free profinite abelian group on countably many generators — one generator for each prime!

The second sentence here is not actually true. The abelianization is canonically isomorphic to $\hat{Z}^*$, the group of multiplicatively invertible profinite integers. This is just another way of expressing the Kronecker-Weber theorem, which says that an extension of $Q$ has abelian Galois group if and only if it is contained in an extension of $Q$ gotten by adjoining a root of unity. And $\hat{Z}^*$ has heaps of torsion (which free profinite abelian groups don’t). You can see this as follows. By the Chinese Remainder theorem, $\hat{Z}^*$ is isomorphic to the product of the groups $Z_p^*$, where $Z_p$ denotes the ring of $p$-adic integers. But $Z_p$ contains the $(p-1)$-st roots of unity, which gives plenty of torsion.

I don’t see off hand if there’s a way to interpret this in terms of threefolds with deleted loops. It may be that this pushes the analogy beyond its breaking point.

3. David wrote:

I suppose with Spec(ℚ(i)), one is supposed to think of the ‘3-manifold’ Spec(ℤ[i]) as a double cover of Spec(ℤ), then go hole-punching again for each of its primes.

That would be exactly my answer, though I would speak of loops rather than holes.

4. David wrote:

Which raises a question: What does Spec(A), for A the ring of all algebraic integers, look like?

Well, the picture for $Z[i]$ works just as well. Spec $A$ would be a degree $n$ cover of the 3-manifold corresponding to Spec $Z$, branched over the loops corresponding to the ramified primes. Here $n$ is the degree of the fraction field of $A$ over $Q$.

5. David wrote:

Makes you wonder whether arithmetic is a universal repository of important structure…

In some sense I believe this to be true, if your structure is rich enough to include addition and multiplication. This is simply because anything mathematical is expressed with finitely many symbols. For example, one can prove theorems about complex projective manifolds as follows: 1. every complex manifold $X$ can be defined by finitely many polynomial equations (Chow’s theorem); 2. since there are finitely many polynomials, the coefficients all lie in a finitely generated subring $R$ of $C$, so $X$ can be defined over $R$; 3. then $X$ can be reduced modulo different primes $p$, and counting / Frobenius arguments can be applied to prove something; 4. since we could use lots of different primes $p$, we can actually conclude something about $X$ from all these different reductions.

Of course, complex projective manifolds are already pretty algebraic. It is much harder to defend this position if we’re talking about more analytic subjects. But it is important to keep in mind, for instance, that only countably many real numbers or real functions can ever be accessed, simply because there are only countably many mathematical papers. Whether this finiteness leads to worthwhile insights (for instance into L-functions) is a completely different matter. But I would say that the steady advance of arithmetic algebraic geometry over the twentieth century shows that no subject that uses addition and multiplication is safe.

But then I would say that.

Posted by: James on April 25, 2009 7:07 AM | Permalink | Reply to this

### Re: Afternoon Fishing

One of the Jameses wrote:
From the point of view of homotopy theory, a contractible space is the same as a point. So you could think of objects of the fundamental groupoid of X as being contractible spaces equipped with a map to X, rather than just points mapping to X.

So the fundamental groupoid id well defined only up to homotopy? Or is there a consensus on the two most obvious defintions in terms of the simplices or the cubes??

Posted by: jim stasheff on April 25, 2009 2:06 PM | Permalink | Reply to this

### fundamental infinity-groupoids

So you could think of objects of the fundamental groupoid of $X$ as being contractible spaces equipped with a map to $X$, rather than just points mapping to $X$.

Incidentally, I was just having a discussion with Tim Porter related to this point, in the context of the role or not played by fundamental groupoids in the context of $n$Lab: local systems and $n$Lab: simplicial local systems.

Tim pointed out that fundamental ($\infty$-)groupoids, for instance in their realization as singular complexes, may be a bit dodgy in that they refer to points.

One remark that this reminds me of is the following:

when realizing the fundamental ($\infty$-)groupoid not just internal to sets, but but as a topological or smooth $(\infty-)$groupoid, then to some extent this seems to be resolved:

the general setup in which fundamental groupoids make sense is this:

let $S$ be some monoidal site of test spaces which is equipped with one cosimplicial object

$\Delta_S : \Delta \to S$.

Then for any test object $U \in S$ its fundamental $\infty$-groupoid would be the (Kan-fibrant replacement of) the simplicial presheaf

$\Pi_\infty(U) : V \mapsto S(V \otimes \Delta_S^\bullet, U) \,.$

This Kan-extends to a definition of fundamental $\infty$-groupoid of generalized spaces in terms of sheaves on $S$ and further of $\infty$-sheaves on $S$:

$\Pi_\infty : \infty-Sh(S) \to \infty-Sh(S) \,.$

Anyway, the point I would like to draw attention to is that in these $S$-parameterized fundamental $\infty$-groupoids $\Pi_\infty(X)$ of generalized spaces $X$, the objects etc. are in a way not single points, but all possible images of test spaces $V$ in the space $X$. Morphisms, similarly, are all possible $V$-families of paths in $X$. Etc.

I am not sure if this resolves or even addresses Tim’s or Jame’s concern, but I thought i might mention it anyway.

Posted by: Urs Schreiber on April 25, 2009 3:46 PM | Permalink | Reply to this

### Re: Afternoon Fishing

Thanks for these great answers, James.

Regarding 4, I meant all the algebraic integers in $\bar{\mathbb{Q}}$. Is that denoted $\bar{\mathbb{Z}}$?

Can there be a branched cover of infinite degree of the 3-manifold?

Posted by: David Corfield on April 25, 2009 5:01 PM | Permalink | Reply to this

### Re: Afternoon Fishing

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.

Posted by: James on April 26, 2009 3:19 AM | Permalink | Reply to this

### Re: Afternoon Fishing

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.

Posted by: Minhyong Kim on April 27, 2009 10:47 AM | Permalink | Reply to this

### Re: Afternoon Fishing

Dear Minhyong,

In your #1, if by ‘etale cover’ you mean ‘finite etale map’, then I agree with everything you wrote. But I think what I said is OK, because I just wrote ‘etale’ (at least in the corrected version) without the finite condition. Of course, these are the same over fields, which is probably why you thought I meant the other one.

In general, if $X$ is a connected scheme with the property that the only etale $X$-schemes are the totally split ones, then the small etale topos of $X$ is equivalent to the topos of sets, and so it really has to be viewed as contractible from the point of view of the etale topology. But this happens if and only if $X$ is the spectrum of a separably closed field. In particular, it rules out the affine line over the complex numbers and so is probably not a very good definition of contractible. But it’s a fine sufficient condition.

Posted by: James on April 27, 2009 1:32 PM | Permalink | Reply to this

### Re: Afternoon Fishing

Oh I see what you mean. I agree that this kind of condition should be sufficient for contractibility. A minor correction is that it’s not just separably closed fields. For example, it could be

$Spec(C[t]/(t^2)).$

I guess the condition is that the reduced subscheme of $X$ should be the spec of a separably closed field.

Posted by: Minhyong Kim on April 27, 2009 11:56 PM | Permalink | Reply to this

### Re: Afternoon Fishing

Oh, you’re right. Thanks!

Posted by: James on April 28, 2009 4:21 AM | Permalink | Reply to this
Read the post The Earth - For Physicists
Weblog: The n-Category Café
Excerpt: A history of the Earth in 3000 words or less, for physicists.
Tracked: April 28, 2009 1:21 AM

### Re: Afternoon Fishing

I just wonder 3-manifolds coming from number rings could be more like “virtual 3-manifolds” to avoid a conflict of the analogy with the Poincare conjecture?

Posted by: Thomas on April 28, 2009 9:23 AM | Permalink | Reply to this

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