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• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeNov 8th 2010
• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeNov 8th 2010

I made “local coefficients” point to the existing enty local systems.

That entry, though, could do with some polishing and clarification.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeNov 8th 2010
• (edited Nov 8th 2010)

In fact, Mike, could we think about that terminology for a sec?

I believe it should not say “local coefficients” but “constant coefficients” – as in our joint discussion we once archived at nonabelian cohomology.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeNov 8th 2010

I created cohomology with constant coefficients and made homotopy equivalence of toposes point to it.

But let me know if you don’t agree with this state of affairs.

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeNov 8th 2010

I think both are true, although I intended to say what I said; I’ve edited the page to include both. Local systems are sections of a constant stack, but they themselves are locally constant stacks, and so cohomology with coefficients in them is cohomology with local coefficients, not constant coefficients. Moreover, I actually intended that one in the most classical sense of abelian cohomology. Which means I actually don’t know why it’s equivalent to the first one, although I seem to recall that it is.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeNov 8th 2010

Local systems are sections of a constant stack, but they themselves are locally constant stacks, and so cohomology with coefficients in them is cohomology with local coefficients, not constant coefficients.

Yes, but I thought the statement you were after was that you want to say that $X \to Y$ is a homotopy equivalence if for all $\infty$-groupoids $A$ we have that

$Hom(Y, LConst A) \to Hom(X,LConst A)$

is an equivalence. That says it is an equivalence in cohomology with constant coefficients (where each cocycle is a local system, true).

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeNov 8th 2010
• (edited Nov 8th 2010)

Ah, now I looked at the entry and understand what you mean: you are claiming that it is another equivalent condition in terms of cohomology with coefficients in local systems.

1. A geometric morphism which induces an isomorphism on all abelian sheaf cohomology with coefficients in locally constant sheaves (of complexes) of abelian groups (the most classical notion).

Check if you agree.

• CommentRowNumber8.
• CommentAuthorMike Shulman
• CommentTimeNov 8th 2010

Well taking coefficients in complexes of abelian sheaves rather than plain abelian sheaves is certainly also implied by the first two definitions, but it would be a stronger statement to say that it’s sufficient to detect weak equivalences by using cohomology with coefficients in plain abelian sheaves, which is what I thought I remembered. We should look up what’s true.

• CommentRowNumber9.
• CommentAuthorDavidRoberts
• CommentTimeNov 8th 2010

It shouldn’t be just abelian sheaf cohomology, but also $H^1(-,G)$ for sheaves $G$ of nonabelian groups.

• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeNov 8th 2010
• (edited Nov 8th 2010)

I am thinking that one should be able to derive this with some general arguments. The locally constant $(\infty,1)$-sheaf on $X \in \mathbf{H}$ that corresponds to $\tilde \nabla : X \to LConst \mathcal{S}$ but regarded as an object in the little $\mathbf{H}/X$ ought to be the $(\infty,1)$-pullback

$\array{ P_{\tilde \nabla} &\to& LConst \mathcal{Z} \\ \downarrow && \downarrow \\ X &\stackrel{\tilde \nabla}{\to}& LConst \mathcal{S} }$

where the right vertical morphism is the image under $LConst$ of the core of the universal fibration over $Fin \infty Grpd$.

Then the cohomology on $X$ with coefficients in the local system ought to be

$\pi_0 \mathbf{H}_{/X}(X,P_{\tilde \nabla}) \,.$

Now use again that $LConst$ is right adjoint to $\Pi$, work a bit harder than before, and hopefully conclude again that you can throw it all over to the left to get $\Pi(X)$ there and conclude that if $\Pi(X) \to \Pi(Y)$ is an equivalence then so is $\mathbf{H}_{/Y}(Y,P_{\tilde \nabla}) \to \mathbf{H}_{/X}(X,P_{\tilde \nabla})$ and conversely.

But I have to think about the details.

• CommentRowNumber11.
• CommentAuthorDavidRoberts
• CommentTimeNov 8th 2010

I’ve added this to homotopy equivalence of toposes

• CommentRowNumber12.
• CommentAuthorMike Shulman
• CommentTimeNov 8th 2010

Thanks David, I forgot about that. Do you know a reference?

• CommentRowNumber13.
• CommentAuthorUrs
• CommentTimeNov 8th 2010

Well abelian cohomology plus nonabelian in degree 1 is essentially general nonabelian cohomology. That smells like the statement should be true just using general coefficients.

• CommentRowNumber14.
• CommentAuthorMike Shulman
• CommentTimeNov 9th 2010

In what sense is meant “essentially”?

• CommentRowNumber15.
• CommentAuthorUrs
• CommentTimeNov 9th 2010
• (edited Nov 9th 2010)

The fact that I was alluding to is that under Postnikov tower decomposition – which works for $\infty$-stacks formally just as for topological spaces, as you know – every connected $\infty$-stack is obtained from a nonabelian 1-truncated one by successive extension by abelian group objects.

Using this you can decompose any cocycle in general nonabelian cohomology into a 1-cocycle in nonabelian cohomology and abelian higher cocycles on the principal bundle that it classifies.

For low $n = 2$ this is spelled out in a bit of detail at string structure, which is the case that plays probably the main role in the literature.

Now, the “essentially” was mainly there to say “it’s past midnight here and I don’t want to claim before sleeping over it that with using this statement it necessarily follows that your statement above actually holds for all coefficients”.

• CommentRowNumber16.
• CommentAuthorUrs
• CommentTimeNov 9th 2010
• (edited Nov 9th 2010)

Toën in his survey Nonabelian cohomology calls this (very last sentence) the

Whitehead principle Non-abelian cohomology is controlled by non-abelian cohomology in degree one (i.e. torsors theory) and usual abelian cohomology.

• CommentRowNumber17.
• CommentAuthorMike Shulman
• CommentTimeNov 9th 2010

That sounds reasonable, modulo checking of things. Coefficients of complexes don’t seem to be necessary to that argument either.

• CommentRowNumber18.
• CommentAuthorUrs
• CommentTimeNov 9th 2010

Coefficients of complexes don’t seem to be necessary to that argument either.

With putting the “complexes” in paranthesis I did not mean that we need hypercohomology but just wanted to indicate to the potential reader that where the sentence mentions “sheaves of abelian groups” it now suddenly does so in the convention of abelian sheaf cohomology, where this may implicialy mean “shifted up in degree to a complex concentrated in some positive degree” and no longer in the sense of nonabelian cohomology, where it (usually) means really “groupal 0-truncated oo-stack”.

It’s not so important, I just felt there was a potential trap for the reader.

• CommentRowNumber19.
• CommentAuthorDavidRoberts
• CommentTimeNov 9th 2010

@Mike #12 - I read it in Pursuing Stacks, but AG cites Artin-Mazur or Illusie or someone (sorry! can’t remember) as this being the definition of a homotopy equivalence of topoi.

• CommentRowNumber20.
• CommentAuthorMike Shulman
• CommentTimeNov 9th 2010

Moerdijk in “Classifying spaces and classifying topoi” also cites Artin-Mazur.

• CommentRowNumber21.
• CommentAuthorDavidRoberts
• CommentTimeNov 9th 2010

Ah, good.

• CommentRowNumber22.
• CommentAuthorMike Shulman
• CommentTimeNov 9th 2010

And Artin-Mazur is a bit difficult to read, since the specific theorem in question is phrased not in terms of topoi but of pro-homotopy-types, which for them mean pro-objects in the homotopy category, rather than (say) in the homotopy category of pro-spaces. The theorem there is that a map of such pro-homotopy-types induces an isomorphism on all homotopy pro-groups iff it induces an isomorphism on the pro-groups $\pi_0$ and $\pi_1$ and an isomorphism of classical abelian cohomology with coefficients in “twisted coefficient groups.” By the latter they mean an abelian group $A$ together with a map of pro-groups $\pi_1(X) \to Aut(A)$, and cohomology with coefficients in such a thing is the colimit over the defining diagram of a pro-homotopy-type of the ordinary local cohomology of the pieces. (The proof goes by an obstruction-theoretic argument which seems morally about the same as what Urs said about Postnikov decompositions of stacks.) However, they essentially show later that for the pro-homotopy-type associated to a topos, these notions of $\pi_1$ and of twisted coefficient groups agree with the usual definitions in terms of coverings (essentially by definition of their “pro-homotopy-type associated to a topos” in terms of hypercoverings). (Although in fact they do all of this without ever saying “topos”, which is kind of astonishing to me—they always talk in terms of sites.)

This does bring out another interesting question, though: at least for Artin-Mazur’s pro-homotopy-types, even though they consider pro-objects in the homotopy category of CW complexes, a map which induces an isomorphism on all homotopy pro-groups is not necessarily an isomorphism. I expect that probably this is also true of pro-∞-groupoids—that a map of such can induce an isomorphism on all homotopy pro-groups but not be an equivalence? If so, then in the non-locally-∞-connected case, the classical notion of “weak homotopy equivalence of topoi” as in Artin-Mazur would seem not to correspond to an equivalence of shapes qua pro-∞-groupoids, but only to a “weak homotopy equivalence” of shapes (a map inducing isomorphisms on all homotopy pro-groups). This sounds sort of like (non-)hypercompleteness rearing its head again, especially seeing the use of hypercovers…

• CommentRowNumber23.
• CommentAuthorTim_Porter
• CommentTimeNov 9th 2010
• (edited Nov 9th 2010)

There is an interesting paper by Quick which relates to the pro-finite homotopy type in related contexts. The point about Artin-Mazur is that it really relates to pro-finite and étale homotopy types BUT as AG realised is more general (I think). The AM homotopy theory relates to pro-Ho(sSet) not to Ho(pro-sSet), as you note. Quick discusses the different homotopy theories and the pro-finite completions. It may help. (He has other papers on related subjects that look interesting.)

• CommentRowNumber24.
• CommentAuthorMike Shulman
• CommentTimeNov 9th 2010

Thanks Tim, I will try to find some time to look at Quick’s paper. I didn’t see any particular appearance of pro finite things in Artin-Mazur last night, though—why does it relate to that specifically? I guess they state some things for completion relative to “a class of groups” which might be finite groups, but as they say it might also be all groups, which is the case I was thinking about.

Thinking some more about the hypercompleteness issue, I wonder whether the point is rather that the pro-homotopy type defined by Artin-Mazur is actually the shape of the hypercompletion of the topos in question, and to get the shape of the topos itself one should instead use something like bounded hypercovers?

• CommentRowNumber25.
• CommentAuthorTim_Porter
• CommentTimeNov 9th 2010

I am not sure that the pro-finite idea is needed, but Quick has, I seem to remember, some discussion of the various points about A&M in a fairly modern parlance. A&M are looking at étale homotopy theory so the pro-finite stuff comes in because of that. I have a copy of AG’s PS here (one of the originals!!!!) and could try to seek out what he claimed… but it is in a box file, and last time looked to be in a bit of disorder :-(. (The box file is in a pile on the floor of our dining room, next to about 6 other similar piles!) There is an odd fact that your comment makes me think of : Dan Isaaksen has a paper in which he shows that the reindexing lemma in procategory theory (e.g. a morphism between prothings is isomorphic to a pro(morphism between things) ) has some little twists when you look at reindexing a diagram with loops. This means something subtle when looking at prosimplicial sets and simplicial prosets. I can look it out (I saw it recently) if that will help.

All this profinite stuff is getting very popular as it links in with stuff by Morel and A^1-homotopy

• CommentRowNumber26.
• CommentAuthorMike Shulman
• CommentTimeNov 10th 2010

What is it about etale homotopy theory that mandates/suggests profinite things? Is it because the etale topology is locally finite somehow?

• CommentRowNumber27.
• CommentAuthorDavidRoberts
• CommentTimeNov 10th 2010

My guess is that etale covers of schemes are of finite type = have finite fibres.

• CommentRowNumber28.
• CommentAuthorTim_Porter
• CommentTimeNov 10th 2010

Yes. I think what David says is the point. The étale fundamental group of suitable manifolds is the profinite completion of the ordinary one, it therefore ’classifies’ finite covering spaces, and those are the ones that seem to have algebraic/geometric significance.

Notice that Grothendieck’s idea for $Gal(\overline{\mathbb{Q}},\mathbb{Q})$ is to do with the geometri of the Teichmuller groupoids and the profinite completions of the braid groups. (I forget the details, which are in Esquisse).

• CommentRowNumber29.
• CommentAuthorDavidRoberts
• CommentTimeNov 10th 2010
• (edited Nov 10th 2010)

But from a topological/homotopical viewpoint, I also don’t see why we need to restrict to pro_finite_, rather than prodiscrete.

For example, if we take the profinite fundamental group (pro object in FinGrp) of a space, considered as a pro-group, and compare it to the pro fundamental group (calculated using all covering spaces, not just finite covering spaces), are they isomorphic (or whatever counts as equivalence of progroups)?

• CommentRowNumber30.
• CommentAuthorTim_Porter
• CommentTimeNov 10th 2010

It is too early here to give an answer. (not enough coffee yet!) One point is that limits behave well on profinite things, they do not generally.

I think the profinite one is the profinite completion of the other. That causes no problem here but may do if one goes up dimensions. The profinite completion of a prospace is discussed in Quick’s thesis and the papers he derived from that.

• CommentRowNumber31.
• CommentAuthorUrs
• CommentTimeNov 10th 2010
• (edited Nov 10th 2010)

That’s pretty much the issue we just discussed in another thread, as to why a covering space is defined to be a section of the constant stack $LConst Core Fin Grpd$ on the groupoid of finite groupoids. Because otherwise that constant stack does not exist in this universe, but only in the universe next door.

• CommentRowNumber32.
• CommentAuthorMike Shulman
• CommentTimeNov 10th 2010

@Urs, I don’t think that’s the issue, because many people would prefer to define a covering space in an intrinsic way, and only later prove that it is the same as a section of some sufficiently large constant stack. In any case, that argument says nothing about why “finite” rather than “countable” or “of size less than κ”.

It also seems right to me that the profinite fundamental group should be the profinite completion of the full fundamental (pro)group. I don’t understand why it’s only finite covering spaces that have “algebraic/geometric significance,” though, although I’ve seen similar restrictions to finite things in lots of different places.

• CommentRowNumber33.
• CommentAuthorDavidRoberts
• CommentTimeNov 11th 2010

I don’t understand why it’s only finite covering spaces that have “algebraic/geometric significance,”

for algebraic geometry it’s because these are really only the sensible ones to talk about (going back to rational maps between projective varieties). Maybe in differential topology this is also pretty common, but I’m only guessing.

• CommentRowNumber34.
• CommentAuthorMike Shulman
• CommentTimeNov 11th 2010

Can you say anything about why finite ones are really the only sensible ones? My knowledge of algebraic geometry is very superficial in some ways.

• CommentRowNumber35.
• CommentAuthorGuest
• CommentTimeNov 11th 2010
Mike, all maps in algebraic geometry are defined locally by polynomials, and if a fiber of a polynomial map is zero-dimensional, it must be finite. So only finite-to-one covering spaces can be represented by polynomial maps. For instance, if X=C^*, then you can get the finite covering spaces C^* -&gt; C^* by sending z to z^n. But the universal covering space C -&gt; C^* is given by sending z to exp(z), which is not a polynomial map.

Jim Borger
• CommentRowNumber36.
• CommentAuthorMike Shulman
• CommentTimeNov 11th 2010

Thanks Jim, that is pretty obvious. (To get nicely forrmatted math, choose “Markdown+itex” under the input box and then write your math as usual inside dollar signs.) That seems to say that the only covering spaces in algebraic geometry are finite. (Or, at least, the only connected ones—surely we can have infinite disconnected covering spaces?) So why do we need to manually restrict to only considering finite things when we define, say, the fundamental group? Why wouldn’t the general definition of a fundamental progroup just happen to turn out to be profinite in the algebraic case where all covering spaces are finite? Or does the topos definition of a fundamental progroup include information about some “non-algebraic” covering spaces even of an algebraic gadget?

• CommentRowNumber37.
• CommentAuthorTim_Porter
• CommentTimeNov 11th 2010

Surely it depends on the topology used, Mike. The A & M LN100 stuff is explicitly for the étale topology and there it is built in.

• CommentRowNumber38.
• CommentAuthorMike Shulman
• CommentTimeNov 11th 2010

So, if finiteness is built in by the etale topology, then why do A&M need to take the further step of profinitely completing their pro-homotopy-types? Shouldn’t they already be profinite? Sorry if I’m being dense….

• CommentRowNumber39.
• CommentAuthorTim_Porter
• CommentTimeNov 11th 2010

The Grothendieck fundamental group of a complex variety (I forget the details) is the profinite completion of the classical topological one. That is one reason for the profinite completion functor. (I am afraid that I forget their other reasons…. bother … and I thought I had a copy of the book… cannot find it.. double bother!@) I think they also wanted to bring in constructions from ordinary homotopy theory and to profinitely complete them so as to be useful in the étale context. (This is not a good reply… sorry!)

• CommentRowNumber40.
• CommentAuthorMike Shulman
• CommentTimeNov 11th 2010

Well, it’s something – at least you confirmed my expectation in one case, that the topos-theoretic notion for the etale topology gives you something that’s already profinite. If the profinite completion is there only in order to relate the “naturally profinite” etale notions with the non-profinite classical ones, then that makes me happy.

• CommentRowNumber41.
• CommentAuthorTim_Porter
• CommentTimeNov 11th 2010

Don’t be too sure. You are relying on my memory from years back. I never trust that!!!!

• CommentRowNumber42.
• CommentAuthorMike Shulman
• CommentTimeNov 12th 2010

Ah, I think now I recall that “the Grothendieck fundamental group” is defined to be profinite, and that the AM fundamental progroup and pro-homotopy-type are a “version” of it for non-finite things, but not the same; probably “the Grothendieck” one is their profinite completion.

And indeed, unless I’m mistaken, you can have a locally constant sheaf on an algebraic variety with infinite fibers. The universal cover of $S^1$ is locally constant over a Zarski-open cover (removing one point suffices). So it seems like the unmodified topos-theoretic definition would not necessarily give you a profinite group, even for an “algebraic” topos, without some further restriction. Unless I’m confused, which is entirely possible.

• CommentRowNumber43.
• CommentAuthorTim_Porter
• CommentTimeNov 12th 2010

Perhaps it is worth thinking back to Galois theory. The Galois group of a field extension is profinite. (The extension should be a Galois extension if I remember rightly.) Have a look at Borceux and Janeldize. They define Galois descent and then a Galois groupoid which is profinite i think. Later Ch 7, they look at more general effective descent I do not know that stuff as well as I should, but it looks as if there are the two theories Galois-Grothendieck and Joyal-Tierney and A & M look to be working mostly with the first situation. (That would suggest it is back to the topology being used or the type of hypercovers.)

• CommentRowNumber44.
• CommentAuthorMike Shulman
• CommentTimeNov 13th 2010

Yes, clearly I need to go learn more about the meaning of “Galois” in this context and its relation to the original meaning that I learned in undergraduate algebra. I will come back when I have digested some of Borceux-Janelidze.

Returning to the original topic of this thread, I have edited homotopy equivalence of toposes so that I currently believe everything that it claims to be true. Anyone who knows more should feel free to edit further.

• CommentRowNumber45.
• CommentAuthorGuest
• CommentTimeNov 14th 2010

Here is my understanding of the situation. I’ve never carefully thought out the details, though, so caveat emptor.

1. There are two versions of the algebraic fundamental group for (connected) schemes. The usual one is a pro-finite topological group. It was defined by Grothendieck in SGA 1. It classifies finite ’covering spaces’ (i.e. finite etale maps to our scheme). The second one is called the enlarged fundamental group. It is actually not a group, but a pro-group. It classifies G-torsors for any (discrete) group G, finite or not. This was defined by Grothendieck in SGA 3. If the scheme in question is sufficient nonsingular (geometrically unibranch, I think), then the two agree in the sense that the pro-group of discrete quotients of the usual algebraic fundamental group is isomorphic to the enlarged fundamental group. So often it doesn’t matter which fundamental group people use. And since the SGA 1 group came first and since the theory in SGA 3 is hidden in some other sections, that’s what people know and write about. So that’s what people mean by the algebraic fundamental group 99% of the time.

An example where the two differ is the following: Let $X$ be two copies of the projective line with the two points 0 glued together transversely and with the two points 1 glued together transversely. So the set of complex points of $X$ in the ’ball’ topology is homeomorphic to two 2-spheres glued together at two pairs of points. Therefore the topological fundamental group is $\mathbf{Z}$. The universal cover is an infinite chain of 2-spheres glued together. This topological space also has an algebraic structure (an infinite chain of projective lines glued together transversely), and in fact the enlarged fundamental group is $\mathbf{Z}$. But the usual (SGA 1 style) fundamental group is the profinite completion of $\mathbf{Z}$.

1. I think it’s true that the enlarged fundamental group is the ’right’ one. You might then ask why the other one exists. I think the reason is probably that some of the basic definitions in scheme theory weren’t finalized at the time of SGA 1. For instance, the concept of etale map (which is the analogue of a local isomorphism in usual covering space theory and which is therefore completely fundamental here) is defined in SGA 1 only for noetherian schemes and for morphisms of finite type. (This condition means is that the map is assumed to be compact (i.e. base change takes Zariski-compact objects to Zariski-compact objects), and locally the maps $A \arrow B$ of rings it induces make the $B$’s finitely generated algebras over the $A$’s. It wasn’t realized till later that it’s best to drop the compactness and noetherian requirements and change finite generation of the ring maps to finite presentation.) In most practical situations, the SGA 1 definitions are OK. But it’s probably the case that the theory of the enlarged fundamental group only works if you have exactly the right definitions, so there was probably a period of a few years in which the usual fundamental group was the only thing they had.

2. So, to answer your questions explicitly… a) Yes, you can always have infinite-to-one covering spaces in algebraic geometry by taking disconnected covers, but the ’finite type’ condition in the SGA 1 definition of etale does not allow that. The correct definition of etale that came along a few years later does allow them, however. It also allows some connected infinite-to-one covering spaces if the base scheme is sufficiently singular. b) You don’t need to manually restrict to finite covering spaces, but that was how the theory was set up in SGA 1. (This was also before the concepts of site and topos existed, so I hope you can forgive them!) c) When the scheme is sufficiently nonsingular, the enlarged fundamental group is just a pro-finite group. You can still get infinite covering spaces using disconnectedness, but in some sense that’s the only way, because the enlarged fundamental group is pro-finite. d) I’m not sure why Artin and Mazur complete their homotopy types. It could be that they worked everything out before the enlarged fundamental group existed, or maybe they thought non-pro-finite phenomena weren’t worth bothering with. e) Regarding your remark about $S^1$: $S^1$ is not the set of complex points of an algebraic variety. It is the set of real points of the real projective line, but that’s completely irrelevant – the topological gadget corresponding to the real projective line is the Riemann sphere equipped with the involution $z\mapsto\bar{z}$. Many people don’t realize that a real variety (in scheme theory) is a complex variety with extra structure. Some people investigate topological spaces given by the real zeros of families of real polynomials, but this has nothing to do with schemes, Grothendieck, Hartshorne, etc.

I hope that helps!

Jim

• CommentRowNumber46.
• CommentAuthorGuest
• CommentTimeNov 14th 2010
Hmm. The formatting and numbering didn't come out as planned....
• CommentRowNumber47.
• CommentAuthorDavidRoberts
• CommentTimeNov 14th 2010

That’s fine - it’s great to see the answer to my (vague) question above in #29. And forgive me for being stupid; Jim who?

• CommentRowNumber48.
• CommentAuthorMike Shulman
• CommentTimeNov 15th 2010

Thanks Jim!

the two agree in the sense that the pro-group of discrete quotients of the usual algebraic fundamental group is isomorphic to the enlarged fundamental group

Huh. While that’s nice, it’s definitely not the sort of agreement I would have expected. I would have expected rather that the profinite group of finite quotients of the enlarged fundamental group would be isomorphic to the “usual” profinite fundamental group. I’m also confused by this:

the enlarged fundamental group is Z. But the usual (SGA 1 style) fundamental group is the profinite completion of Z.

since the profinite completion of Z is (by definition) the profinite group of finite quotients of Z, but I don’t see how Z is the progroup of discrete quotients of its profinite completion.

• CommentRowNumber49.
• CommentAuthorMike Shulman
• CommentTimeNov 15th 2010

Oh, I guess you were saying that that space X is not sufficiently nonsingular for the type of “agreement” you quoted to hold. Are there any conditions under which the type of agreement I would expect (the “usual” fundamental group is the profinite completion of the enlarged one) holds? It seems like if I have a progroup which classifies arbitrary covering spaces, then its profinite completion ought to classify finite covering spaces more or less by definition… but my intuition for pro-things is still not as good as it could be.

• CommentRowNumber50.
• CommentAuthorMike Shulman
• CommentTimeNov 15th 2010

Many people don’t realize that a real variety (in scheme theory) is a complex variety with extra structure.

Reading Eisenbud+Harris led me to believe that a real variety was a (certain type of) scheme equipped with a map to $Spec \mathbb{R}$. Is that wrong? If so, then does that mean that there is not a definition of “variety over K” for a field K, of which both “real variety” and “complex variety” are special cases?

• CommentRowNumber51.
• CommentAuthorDavidRoberts
• CommentTimeNov 15th 2010

@Mike - I think the idea with the real variety is that the complex scheme with involution is analogous to a ’homotopy quotient’ in the same way an action groupoid is a good replacement for a bad quotient.

• CommentRowNumber52.
• CommentAuthorMike Shulman
• CommentTimeNov 15th 2010

So a scheme over $Spec \mathbb{R}$ is a “bad quotient”?

That’s weird; I think of homotopy quotients as something you do when taking the “ordinary” quotient loses too much data, but a scheme over $Spec \mathbb{R}$ still contains all the information about the complex points, since you can just pull it back to $Spec \mathbb{C}$. Can you (or someone) give an example of why it’s better to think of a real variety as a complex variety with an involution?

• CommentRowNumber53.
• CommentAuthorDavidRoberts
• CommentTimeNov 15th 2010

No, it was just an attempt at an analogy.

Perhaps the machinery of fixed points of G-schemes is better at counting points/other things….

• CommentRowNumber54.
• CommentAuthorGuest
• CommentTimeNov 15th 2010

"Oh, I guess you were saying that that space X is not sufficiently nonsingular for the type of "agreement" you quoted to hold."

Yes. Sorry for the confusion.

"Are there any conditions under which the type of agreement I would expect (the "usual" fundamental group is the profinite completion of the enlarged one) holds? It seems like if I have a progroup which classifies arbitrary covering spaces, then its profinite completion ought to classify finite covering spaces more or less by definition… but my intuition for pro-things is still not as good as it could be."

I believe that what you said is true more or less by definition, but again I’ve never actually gone over the details.

"Reading Eisenbud+Harris led me to believe that a real variety was a (certain type of) scheme equipped with a map to Spec R. Is that wrong?"

No, that’s correct. But a "real variety" is much more than its locus of real points. For instance the real schemes defined by the one-variable equations $x^2+1=0$ and $1=0$ have the same loci of real points (=empty), but the real schemes are different. They are both affine schemes, one given by the spectrum of $R[x]/(x^2+1)$ and the other by $R[x]/(1)$. Neither has any maps to $R$ (i.e., neither has any real points), but they are nonisomorphic rings. This is very different to the situation when you’re working over an algebraically closed field $K$, where (by the Nullstellensatz) a variety is essentially determined by its locus of $K$-points. (’Essentially’ means modulo nilpotent elements.)

To interpret the real schemes above in traditional topological terms, you need to look at their loci of complex points, together with the conjugation involution. In the first case, you get two points which are swapped by conjugation, and in the second case you get no complex points at all. Thus a real scheme is a complex scheme with extra structure, not with less structure. Some people might use the term ’real variety’ to mean the locus of real points of a real scheme, but in scheme theory world, it means a scheme over the real numbers (with certain properties).

Another way of putting things is this: a scheme is essentially just a list of equations (perhaps with coefficients in some given base ring, like $R$). So it’s a syntactic object. The set of $K$-points, for some ring $K$, is the set of solutions for these equations. So it’s a model. By the Nullstellensatz, when $K$ is algebraically closed the model determines the syntax (up to nilpotents), but otherwise it does not.

Jim

• CommentRowNumber55.
• CommentAuthorGuest
• CommentTimeNov 15th 2010
I can't get the math to work. Here is the post again as raw text....

"Oh, I guess you were saying that that space X is not sufficiently nonsingular for the type of "agreement" you quoted to hold."

Yes. Sorry for the confusion.

"Are there any conditions under which the type of agreement I would expect (the "usual" fundamental group is the profinite completion of the enlarged one) holds? It seems like if I have a progroup which classifies arbitrary covering spaces, then its profinite completion ought to classify finite covering spaces more or less by definition... but my intuition for pro-things is still not as good as it could be."

I believe that what you said is true more or less by definition, but again I've never actually gone over the details.

"Reading Eisenbud+Harris led me to believe that a real variety was a (certain type of) scheme equipped with a map to Spec R. Is that wrong?"

No, that's correct. But a "real variety" is much more than its locus of real points. For instance the real schemes defined by the one-variable equations $x^2+1=0$ and $1=0$ have the same loci of real points (=empty), but the real schemes are different. They are both affine schemes, one given by the spectrum of $R[x]/(x^2+1)$ and the other by $R[x]/(1)$. Neither has any maps to $R$ (i.e., neither has any real points), but they are nonisomorphic rings. This is very different to the situation when you're working over an algebraically closed field $K$, where (by the Nullstellensatz) a variety is essentially determined by its locus of $K$-points. ('Essentially' means modulo nilpotent elements.)

To interpret the real schemes above in traditional topological terms, you need to look at their loci of *complex* points, together with the conjugation involution. In the first case, you get two points which are swapped by conjugation, and in the second case you get no complex points at all. Thus a real scheme is a complex scheme with extra structure, not with less structure. Some people might use the term 'real variety' to mean the locus of real points of a real scheme, but in scheme theory world, it means a scheme over the real numbers (with certain properties).

Another way of putting things is this: a scheme is essentially just a list of equations (perhaps with coefficients in some given base ring, like $R$). So it's a *syntactic* object. The set of $K$-points, for some ring $K$, is the set of solutions for these equations. So it's a model. By the Nullstellensatz, when $K$ is algebraically closed the model determines the syntax (up to nilpotents), but otherwise it does not.

Jim
• CommentRowNumber56.
• CommentAuthorMike Shulman
• CommentTimeNov 17th 2010

Hmm. At one level, I knew all of that, but I guess I hadn’t internalized it sufficiently so as to give me the correct intuition about what real varieties “look like” geometrically. More indications that I still have a long way to go before I can think like an algebraic geometer. Thanks for explaining patiently!

• CommentRowNumber57.
• CommentAuthorMike Shulman
• CommentTimeNov 17th 2010

One more question. Back in #45, you said:

the two agree in the sense that the pro-group of discrete quotients of the usual algebraic fundamental group is isomorphic to the enlarged fundamental group.

I am a bit confused: since the “usual” algebraic fundamental group is profinite, can it have any non-finite discrete quotients? And if not, isn’t the pro-group of its (finite) discrete quotients just equal to itself?

• CommentRowNumber58.
• CommentAuthorGuest
• CommentTimeNov 17th 2010
Mike, yes, all the discrete quotients are finite (but I've heard that sometimes you can make some silly non-discrete finite quotients using choice). So yes, I was just saying that the two algebraic fundamental groups agree. The only reason I put it the way I did was because the first one is usually taken to be a topological group, whereas the second apparently must be taken to be a pro-group (i.e. taking the limit can lose information). I was just spelling out how to go from a pro-finite group, viewed as a topological group, to a pro-group.
• CommentRowNumber59.
• CommentAuthorMike Shulman
• CommentTimeNov 17th 2010

Thanks! I suspected that might be it. (To me the “real” definition of a profinite group is a filtered inverse limit of finite groups, and it’s only a nice accident (having to do with the fact that every finite Boolean algebra is atomic) that you can fully-faithfully represent them as Stone-topological groups.)

The page fundamental group of a topos needs some love, and would be a good place to record this discussion. Maybe I’ll have some time to do that one of these days…

• CommentRowNumber60.
• CommentAuthorTim_Porter
• CommentTimeNov 17th 2010

@Mike Do you mean that a profinite group is really a pro (finite group). The limit is not really needed as it is one of the few cases where the progroup can be rebuilt from the limit.

• CommentRowNumber61.
• CommentAuthorMike Shulman
• CommentTimeNov 17th 2010

Yes, the page profinite group describes the way I think of it. (I see from that page that we also have a stub algebraic fundamental group.)

• CommentRowNumber62.
• CommentAuthorTim_Porter
• CommentTimeNov 25th 2010
• (edited Nov 25th 2010)

@Mike (and others) I have just come across the [paper] (http://www.dpmms.cam.ac.uk/~jpp24/weiln.pdf) by J. Pridham. He has quite a lot that may be relevant in his archiv submissions. He seems to get a variant of some of Toen’s stuff. This may be worth following up. What do you think?