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• CommentRowNumber1.
• CommentAuthorTodd_Trimble
• CommentTimeAug 7th 2015

I added a few words to address an oversight noted by Sridar Ramesh at topological ring, and corrected also a second oversight in the formulation of topological algebra (a standard mistake which would imply that the quaternions are a $\mathbb{C}$-algebra).

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeAug 7th 2015

Thanks! I suppose that was my fault. Thanks for fixing it.

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeAug 7th 2015

Is this notion of “topological field” obtained by internalizing one of the constructive definitions of “field” in the category $Top$?

• CommentRowNumber4.
• CommentAuthorDavidRoberts
• CommentTimeAug 8th 2015

Do you mean in the internal logic (such as it is)? If not, it looks as though it could be a Heyting field.

• CommentRowNumber5.
• CommentAuthorTodd_Trimble
• CommentTimeAug 8th 2015

With regard to the notion of Heyting field (Definition 3 in field), if my calculations are correct, there is some redundancy there: we don’t need to assume addition and multiplication are strongly extensional – that should come for free if the condition “$x - y$ is invertible” defines a (tight) apartness relation. The key fact is that we are working in a local ring, as a consequence of the co-transitivity condition, and then strong extensionality becomes just an easy piece of local algebra.

• CommentRowNumber6.
• CommentAuthorMike Shulman
• CommentTimeAug 8th 2015

@Todd, that sounds right to me. @David yes, I mean in the internal logic, or rather in some internal logic (there might be multiple choices for what that means in $Top$).

• CommentRowNumber7.
• CommentAuthorTodd_Trimble
• CommentTimeAug 9th 2015

I went ahead and recorded the content of #5 at local ring, also observing there that the notion of local homomorphism between local rings is also sensibly construed as strongly extensional homomorphism. Also made the corresponding adjustment to Definition 3 in field.

With regard to internal fields in a topological context, there are probably quite a few ways to go about it. Here might be a cheap way (and please forgive the verbosity that follows): first express the theory of Heyting fields directly as a Heyting hyperdoctrine. As the base category one could take the walking commutative ring $K$ in the doctrine of finitely complete categories. (In other words, take $\mathbf{K}$ to be the opposite of the category of finitely presented commutative rings.) This level of structure would permit one to define the group of units as a suitable equalizer $U \to K \times K$, where here $K$ denotes the generic object. This gives a subobject $U \hookrightarrow K$ by projection (onto the first coordinate, say). By swapping factors of $K \times K$, we easily get a reciprocation function $U \to U$.

Switch doctrines over to full first-order logic, i.e., use the finite limit theory $\mathbf{K}$ to freely generate a Heyting hyperdoctrine $CRing: \mathbf{K}^{op} \to Heyt$. Mod $CRing$ out by axioms forcing $K$ to be a local ring together with the tight apartness condition $\neg U = \{0\} \hookrightarrow K$. This gives a Heyting hyperdoctrine $HF: \mathbf{K}^{op} \to Heyt$ which should be the theory of Heyting fields in hyperdoctrine form.

Now here comes a trick: starting with $Top$, there is a hyperdoctrine $Reg: Top^{op} \to Heyt$ that takes a topological space to the Heyting algebra of its regular subobjects. Then define a topological field to be an interpretation given by a commutative ring, i.e., a finite-limit preserving $F: \mathbf{K} \to Top$, inducing a 2-cell (a suitable natural transformation)

$\array{ \mathbf{K}^{op} & \stackrel{F^{op}}{\to} & Top^{op} \\ & \mathllap{CRing} \searrow \; \Rightarrow & \downarrow \mathrlap{Reg} \\ & & Heyt, }$

subject to the property that it descends to a hyperdoctrine map $\psi: HF \to Reg \circ F$. The reason we take regular subobjects is to make reciprocation $U \to U$ continuous in the subspace topology inherited from the embedding $U \hookrightarrow K$.

If there are no problems with this, it would seem to give a cheap internal notion of “topological field” portable to much wider contexts, say for example for any topological category (over $Set$) in place of $Top$, again using regular subobjects.

• CommentRowNumber8.
• CommentAuthorspitters
• CommentTimeAug 9th 2015

This is nice.

Apartness relations is a bit low on references. Troelstra van Dalen has material on this. Frank Waaldijk has developed quite a lot of topology based solely on apartness relations. It would be nice to have a look at this through categorical glasses. The relation with Hausdorffness has been explored in abstract Stone duality and Escardo’s synthetic topology. Negri and von Plato have a number of proof theoretic papers on the duality with equality. Again it would be interesting to see how this fits with the categorical view point.

• CommentRowNumber9.
• CommentAuthorTodd_Trimble
• CommentTimeAug 10th 2015

I think maybe I was brushing some things under the rug in the hyperdoctrine comments in #7, having to do with the identification between the group of units $U$ appearing as a regular subobject of $K$ with the group of units as a hyperbola subobject of $K \times K$ (I don’t clearly see how what I said would capture such an identification, so let me put the details of that comment on temporary hold).

Something like that should work though. Classically, I think the following works to internalize fields in $Top$ to give topological fields: work in a doctrine of finitely complete, finitely cocomplete categories. Take a commutative ring object $K$. Then

• Define $i: U \hookrightarrow K \times K$ to be the equalizer $x y = 1$ in $K \times K$.

• Add the condition that the monic $j = \pi_1 i: U \to K$ is the equalizer of its cokernel pair.

The point of this condition is that in $Top$, this forces $j$ to be a subspace embedding, so that multiplicative inversion on $U$ is forced to be continuous with respect to the subspace topology inherited from $K$ – the issue raised in the now-erased comment of Sridhar from an earlier revision of topological ring).

Finally,

• The disjoint embeddings $0: 1 \to K$ and $j: U \to K$ induce an arrow $k: 1 + U \to K$. Add the condition that the kernel of the cokernel pair of $k: 1 + U \to K$ is the identity $1_K$.

I don’t think this particular definition of field is found at field (it is not the same as a discrete field; discrete fields internalize in $Top$ to fields with the discrete topology).

This definition is not good for constructive purposes, but if we are regarding $Set$ as a constructivist might (say a complete cocomplete Heyting pretopos), suitable variants are internalizable within the “doctrine” of categories that are topological over $Set$, as I was trying to suggest in #7. The point being that we can form $\neg U \hookrightarrow K$ in $Set$, and then take an initial lift to get the subspace $\neg U \hookrightarrow K$. Then, if you want for example a notion of Heyting field which works correctly in $Top$, replace the third bullet point with the local ring conditions and the tightness condition $\neg U = \{0\}$.

• CommentRowNumber10.
• CommentAuthorMike Shulman
• CommentTimeAug 11th 2015

Haha, very interesting! This turns out to be closely related to the problem of analytic Markov’s principle in real-cohesion, so I am more interested in it than I was previously.

I would expect that we could also use the doctrine of “quasi-coherent categories”, i.e. the things that are to quasitoposes the way coherent categories are to toposes. Does that seem right?

Do we have an example of a topological ring which would be a topological field except that $U\to K$ fails to be a subspace embedding? Or even any topological ring at all where $U\to K$ is not an embedding?

• CommentRowNumber11.
• CommentAuthorTodd_Trimble
• CommentTimeAug 11th 2015
• (edited Aug 11th 2015)

Bas and Mike: thanks for your reactions! I don’t know of an example of such a topological ring, but it’s been on a back burner. We know for example that the ideles are not given the subspace topology coming from the inclusion into the adeles, so perhaps by close study of that one could cook up a similar example but with a field. (Or, one could just ask MO, but first I’d like to give it the old college try.)

I’m not quite sure what a quasi-coherent category is, but my takeaway from Mike’s comment is that $Top$ ought to be an example. Trying to think this through (I woke up not long ago and need more coffee), I find myself a little puzzled by unions in the subobject (not regular subobject!) poset of a topological space, and maybe that’s a further clue – that plus the fact that in a quasitopos subobjects and regular subobjects do not coincide.

• CommentRowNumber12.
• CommentAuthorMike Shulman
• CommentTimeAug 11th 2015

I’m not quite sure what a quasi-coherent category is either, but I feel like such a thing ought to exist. (-:

• CommentRowNumber13.
• CommentAuthorMike Shulman
• CommentTimeAug 11th 2015

Here’s one attempt at a definition of quasi-coherent category: a category with finite limits and pullback-stable (epi, strong-mono) factorizations and finite unions of strong monos.

• CommentRowNumber14.
• CommentAuthorMike Shulman
• CommentTimeAug 11th 2015

Thanks for the example of the ideles, too; maybe I need to finally learn what the heck those are. (-:O

• CommentRowNumber15.
• CommentAuthorTodd_Trimble
• CommentTimeAug 19th 2015
• (edited Aug 19th 2015)

Re #10: this took me an embarrassingly long time, but I finally found an example of a ring topology on a field which is not a field topology. Put a ring topology on the rationals $\mathbb{Q}$ by taking the set of ideals $n\mathbb{Z}$ ($0 \neq n \in \mathbb{Z}$) to be a filter base for the filter of neighborhoods of $0$. That this gives a ring topology follows from three easily verified facts: for any ideal $I$ there is an ideal $J$ such that $J + J \subseteq I$ and $-J \subseteq I$ ($J = I$ works!), (2) if $I$ is any ideal and $q$ is any rational, there is an ideal $J$ such that $q J \subseteq I$, and (3) if $I$ is any ideal, there is an ideal $J$ such that $J J \subseteq I$ (again $J = I$ works). It’s also pretty easy to see that this topology is Hausdorff.

But inversion on the nonzero elements is not continuous. Indeed, for a neighborhood of $1$ not containing $0$, let’s say $1 + 2\mathbb{Z}$, there is no neighborhood $1 + m\mathbb{Z}$ small enough that will fit inside the set of reciprocals $\{\frac1{1 + 2n}: n \in \mathbb{Z}\}$.

Well, that was easy. :-)

• CommentRowNumber16.
• CommentAuthorTodd_Trimble
• CommentTimeAug 20th 2015
• (edited Aug 20th 2015)

I’d like to meditate further on #13, and get more feeling for how relevant it is. I don’t know yet.

Meanwhile, I seem to have bumped into an exactness property of $Top$ that I’d not noticed before: $Top^{op}$ is a regular category! I mean, isn’t it? $Top^{op}$ is finitely complete and cocomplete, and crucially, in $Top$, the pushout of a regular monomorphism along any map is a regular monomorphism: this is proven in the nLab here. Is this fact familiar from the literature? It seems to fit nicely with the situation with locales, as $Loc^{op} \simeq Frm$ is also a regular category.

• CommentRowNumber17.
• CommentAuthorTodd_Trimble
• CommentTimeAug 20th 2015

I have now split off topological field from topological ring, with a view to further discussion.

• CommentRowNumber18.
• CommentAuthorMike Shulman
• CommentTimeAug 20th 2015

Re #16: cute! A quasitopos is also “coregular” (A2.6.3(i) in the Elephant) by an argument involving its strong-subobject classifier; I guess the argument for $Top$ involving the Sierpinski space is somewhat similar.

I need to think more about #15.

• CommentRowNumber19.
• CommentAuthorTobyBartels
• CommentTimeAug 21st 2015
• (edited Aug 21st 2015)

Regarding internal Heyting fields:

I'm inclined to think that the really constructive way to do this is just to talk about internal local rings, which are much easier to make sense of (since you don't have to bring in negation). Once you have the category $LocRing(C)$ of internal local rings, you can then form a quotient category $HeytFld(C)$ by, I guess, formally inverting the surjective local homomorphisms.

If you can represent $HeytFld(C)$ as $C$ with extra structure (as you can when $C$ is $Set$), then that's all well and good, but I wouldn't insist upon it.

• CommentRowNumber20.
• CommentAuthorDavidRoberts
• CommentTimeAug 21st 2015

surjective local homeomorphisms.

is that “(surjective, local) homeomorphisms” or “surjective (local homeomorphisms)”? It’s a cool idea, though!

• CommentRowNumber21.
• CommentAuthorZhen Lin
• CommentTimeAug 21st 2015

Actually, I think it’s “surjective local homomorphism”.

• CommentRowNumber22.
• CommentAuthorDavidRoberts
• CommentTimeAug 21st 2015

@Zhen Lin

I thought that could have been the case. If it is, I suppose we just take plain epimorphisms? Nothing fancy about what sort?

• CommentRowNumber23.
• CommentAuthorTobyBartels
• CommentTimeAug 21st 2015
• (edited Aug 21st 2015)

Yes, I meant ‘homomorphims’. The word ‘local’ made me want to put in that extra ‘e’, I guess. The original comment is now fixed. (It's interesting that ‘local’ weakens ‘homeomorphism’ but strengthens ‘homomorphism’.)

I wasn't sure what ‘surjective’ should mean, and I usually think that it ought to depend: you equip a category with a coverage, and maybe it's canonical in some sense and maybe it isn't, but in any case it defines what ‘surjective’ means (and more generally ‘jointly surjective’).

If ‘surjective’ means regular-epic, then it's notable that a regular epi is invertible iff it's monic, so formally inverting the regular-epic local homomorphisms amounts to saying that local homomorphisms are morally monic. (Compare that field homomorphisms really are always monic.)

On the other hand, since monos in algebraic categories are always regular, maybe I should interpret ‘surjective’ to just mean epic, and the formal invertibility amounts to saying that local homomorphisms are morally regular-monic. But if monos aren't all regular in $C$, then this falls apart.

Another possibility, which I just thought of: If I really want to say that $C$ is equipped with an arbitrary coverage, being about sieves rather than just single morphisms and defining ‘jointly surjective’ more fundamentally than ‘surjective’ itself, then perhaps the rule ought to be to force covering families of homomorphisms to be coproduct diagrams when they are jointly local in some sense. (But this is probably nonsense, since fields in $Set$ don't even have coproducts.)

• CommentRowNumber24.
• CommentAuthorTodd_Trimble
• CommentTimeAug 21st 2015

monos in algebraic categories are always regular

I didn’t know that! Is there a good reference for that (or is it obvious)?

• CommentRowNumber25.
• CommentAuthorTodd_Trimble
• CommentTimeAug 21st 2015

Wait a minute, I don’t think that’s true. If the mono $\mathbb{Z} \to \mathbb{Q}$ were regular, then since it is epic, it would have to be an isomorphism (since epic equalizers are always isos). Maybe you meant something else?

• CommentRowNumber26.
• CommentAuthorTobyBartels
• CommentTimeAug 21st 2015
• (edited Aug 21st 2015)

Sorry, I was mixed up; that's not actually true. (The classic counterexample is $\mathbb{Z} \hookrightarrow \mathbb{Q}$ in rings. ETA: which is what you thought of too!)

This suggests that maybe we should talk about regular epis rather than all epis.

• CommentRowNumber27.
• CommentAuthorDavidRoberts
• CommentTimeAug 21st 2015

I was thinking perhaps extremal epimorphism might be the way to go…

• CommentRowNumber28.
• CommentAuthorTobyBartels
• CommentTimeAug 21st 2015
• (edited Aug 21st 2015)

That is a pretty canonical choice. Do they form a coverage?

• CommentRowNumber29.
• CommentAuthorTodd_Trimble
• CommentTimeAug 21st 2015

Re #18 and thinking more about #15: maybe I was being a little terse in #15. Let me state something more formally:

Let $R$ be a ring. Suppose $\mathcal{T}$ is a topology on the set $R$ that makes the additive group of $R$ a topological group, and let $\mathcal{B}$ be a base of open neighborhoods of $0$ in $\mathcal{T}$. Then $\mathcal{T}$ makes $R$ a topological ring (i.e., multiplication is continuous) if the following two conditions are satisfied:

• For all $U \in \mathcal{B}$ there exists $V \in \mathcal{B}$ such that $V V \subseteq U$;

• For all $U \in \mathcal{B}$ and $r \in R$ there exists $V$ such that $r V \subseteq U$ and $V r \subseteq U$.

Proof: To prove multiplication $R \times R \to R$ is continuous at any point $(a, b) \in R \times R$, write

$x y - a b = (x - a)(y - b) + (x - a)b + a(y - b)$

Then for any $U \in \mathcal{B}$ we can force $x y - a b \in U$ by finding $U'$ such that $U' + U' + U' \subseteq U$, and then finding $V \in \mathcal{B}$ such that $V V \subseteq U'$, $V b \subseteq U'$, and $a V \subseteq U'$ using the two conditions above, and then taking $x - a \in V$ and $y - b \in V$.

• CommentRowNumber30.
• CommentAuthorDavidRoberts
• CommentTimeAug 21st 2015

@Toby

Not sure, but surely one would check this, under the assumption pullbacks of extremal epis exist, by pulling back along monos and extremal epis separately.

• CommentRowNumber31.
• CommentAuthorMike Shulman
• CommentTimeAug 22nd 2015

@Todd: Okay, I think I mostly understand #15 now; thanks for the further explanation! So here the “hyperbola subobject” $U=\{ (x,y)\in \mathbb{Q}^2 \mid x y = 1 \}$ has the topology generated by $(x+n\mathbb{Z})\times(y+m\mathbb{Z})$. Put differently, if we regard $U$ as the space of invertible rationals, then $(n,m)$ generates a neighborhood of $x\in U$ containing those $x'\in U$ such that $x-x' \in n\mathbb{Z}$ and $\frac1x - \frac1{x'} \in m\mathbb{Z}$. Which is noticeably finer than the subspace topology from the weird $\mathbb{Q}$, where the neighborhoods would just be $x + n \mathbb{Z}$ with no condition on the inverses.

Is there a name for “a topological ring such that $U \overset{pr_1}{\hookrightarrow} K$ has the subspace topology” (or I guess equivalently, such that inversion is continuous on the subspace of invertible elements)?

• CommentRowNumber32.
• CommentAuthorTobyBartels
• CommentTimeAug 22nd 2015

@David #30:

In the general case of a coverage, it's not even necessary to require that all pullbacks of covers exist. On the other hand, it's hard to prove anything without some such assumption!

But even supposing that all pullbacks of extremal epis exist, I'm not getting anything. It's not that we want to pull back separately along extremal epis and monos; we pull back an extremal epi, suppose that a mono lives inside the pullback, then need to push that mono forward somehow. But I don't see how.

That may not matter. Just because I think that one should start with a coverage, that doesn't mean that your idea has to fit into my framework.

• CommentRowNumber33.
• CommentAuthorDavidRoberts
• CommentTimeAug 22nd 2015

I was thinking if starting with the easier case where we can at least construct a square as required for a coverage. I guess I was implicitly assuming that image factorisations exist, which are necessarily as (extremal epi)-mono, so one could mimic what happens in a regular category.

I agree that working with a general (singleton?) coverage is the most general situation.

One thing that might be interesting is to see what epimorphisms in a category C underlie extremal epis (say) in the category Rings(C) of rings in C. Using the quasicategory of concrete sheaves might be an good test case.

• CommentRowNumber34.
• CommentAuthorThomas Holder
• CommentTimeAug 22nd 2015

concerning #16: M. Barr, M. C. Pedicchio, $Top^{op}$ is a quasi-variety , Cah. Top. Géom. Diff. Cat. XXXVI no.1 (1995) pp.3-10. (numdam)

• CommentRowNumber35.
• CommentAuthorTodd_Trimble
• CommentTimeAug 22nd 2015

Re #34: thanks, Thomas! That’s an impressive dig.

Re #31: good question, Mike, I’ve not heard of a name for that. Now that you bring it up, perhaps such an extra condition would be appropriate when discussing topological local rings? Already we’re treating the group of units in a local ring as property-like structure when we deem that local homomorphisms are the “right” morphisms (as in the article local ring where we are treating local rings as certain rings in the category of sets equipped with apartness structure), and this just seems like another step in that direction.

I am told this condition holds automatically if the topological ring is locally compact and the group of units is open.

• CommentRowNumber36.
• CommentAuthorMike Shulman
• CommentTimeAug 22nd 2015

perhaps such an extra condition would be appropriate when discussing topological local rings?

Indeed, good point.

I am told this condition holds automatically if the topological ring is locally compact and the group of units is open.

Did whoever told you that also supply or point to a proof?

• CommentRowNumber37.
• CommentAuthorTodd_Trimble
• CommentTimeAug 22nd 2015

I think I’ve seen the statement in more than one source, and have the suspicion it’s true, but I don’t think I saw a proof given or a pointer. (One of the sources has it as an exercise – after I wrote the author, who hadn’t included local compactness in a prior version of that exercise, and this was his fix in response to what I wrote him.) Before I let on more about this, let me see if I have success solving the new-and-improved exercise. :-)

• CommentRowNumber38.
• CommentAuthorTobyBartels
• CommentTimeAug 22nd 2015

@Todd #37: It's certainly true if the topology is Hausdorff, which perhaps was being assumed all along: open subspaces of locally compact spaces are locally compact, and locally compact Hausdorff semitopological groups are topological. The latter result is supposed to be rather difficult, so I don't feel as if I'm spoiling anything for you by revealing it (although it's always possible that the proof is easier for the group of units in a topological ring).

So you don't have to look it up: a group with a topology is semitopological if multiplication is separately continuous in each variable, regardless of the inverse map. Here we have the even stronger property that multiplication is jointly continuous in both variables, which makes the group paratopological. (They also have quasitopological groups, in which multiplication is separately continuous and the inverse map is continuous. Not to mention left-topological and right-topological groups, in which the multiplication need be continuous in one variable only. Isn't it nice to know that these variations have already been named?)

• CommentRowNumber39.
• CommentAuthorTobyBartels
• CommentTimeAug 23rd 2015

While I'm waiting on my laundry, here's a diagram of possible compatibilities between a group structure on a set and a topological structure on that set. There remain two with no names known by me:

$\array { topological\:group \\ \uparrow & \nwarrow \\ quasitopological\:group & & paratopological\:group \\ \uparrow & \nwarrow & \uparrow \\ \uparrow & & semitopological\:group \\ \uparrow & & \uparrow & \nwarrow \\ group\:with\:continuous\:inverse & & left\,topological\:group & & right\,topological\:group \\ & \nwarrow & \uparrow & \nearrow \\ & & group\:with\:topology }$

Then when the topology is locally compact Hausdorff, this simplifies so:

$\array { & & topological\:group \\ & \nearrow & \uparrow & \nwarrow \\ group\:with\:continuous\:inverse & & left\,topological\:group & & right\,topological\:group \\ & \nwarrow & \uparrow & \nearrow \\ & & group\:with\:topology }$

And here's John Baez's picture of all of this:

• CommentRowNumber40.
• CommentAuthorTobyBartels
• CommentTimeAug 23rd 2015
• (edited Aug 23rd 2015)

I guess that this means that my answer to the end of #31, if none has been invented yet, is ‘topological ring with continuous inverse’.

• CommentRowNumber41.
• CommentAuthorTodd_Trimble
• CommentTimeAug 23rd 2015

Thanks, Toby. I think the exercise poser (it’s Pete Clark, page 88 of Algebraic Number Theory II, which you can find here) must have had the same idea as you, since he mentions Ellis just prior, and yes, I think he does assume Hausdorffness throughout, although I think he could be just a teeny bit more explicit on this point. As always I hope for a simpler proof, but maybe I won’t bother this time.

• CommentRowNumber42.
• CommentAuthorTobyBartels
• CommentTimeAug 23rd 2015

Yes, that seems to be what he intended. More direct link.

• CommentRowNumber43.
• CommentAuthorDavidRoberts
• CommentTimeAug 23rd 2015

In particular Proposition 133 b) includes Hausdorffness.

• CommentRowNumber44.
• CommentAuthorTobyBartels
• CommentTimeAug 23rd 2015

H'm, that suggests that Clark does not mean to include Hausdorffness without explicitly saying so. Todd, if we come up with another counterexample, then you can get Exercise 6.9.c named after you too!

• CommentRowNumber45.
• CommentAuthorDavidRoberts
• CommentTimeAug 23rd 2015
• (edited Aug 24th 2015)

In 6.9a) Clark says

Suppose $R$ is a locally compact topological ring in which $R^\times$ is an open subgroup. Show that $R^\times$ is a topological group

does he really mean ’open subgroup’ there?

• CommentRowNumber46.
• CommentAuthorDavidRoberts
• CommentTimeAug 23rd 2015
• (edited Aug 23rd 2015)

Warner’s Topological Fields has Theorem 9.4 saying locally compact semitopological groups are topological groups, with no Hausdorffness and with proof (not easy!). I think Theorem 11.17 is what we are looking for in answer to Clark’s exercise.

Section 14 of his chapter ’topological rings and modules’ is all about continuity of inversion.

• CommentRowNumber47.
• CommentAuthorTobyBartels
• CommentTimeAug 23rd 2015

OK, if Ellis's Theorem doesn't require Hausdorffness, then what Clark has now is correct. (I'm not allowed to look at Theorem 11.17 of Warner right now.)

• CommentRowNumber48.
• CommentAuthorDavidRoberts
• CommentTimeAug 23rd 2015

Sorry, I had it typed out, but lost my comment. 11.17 is pretty much exactly Clark’s exercise.

• CommentRowNumber49.
• CommentAuthorMike Shulman
• CommentTimeAug 23rd 2015

Wow! This is incredible.

Here is a summary of the inversion part of Warner’s Theorem 9.4 in the Hausdorff case: first he goes through some rigamarole to prove that $m:G\times G\to G$ extends to a continuous map $G \times G_\infty \to G_\infty$, where $G_\infty$ is the one-point compactification of $G$. Therefore, the adjoint transpose $G\to (G_\infty)^{G_\infty}$ is also continuous, since $G_\infty$ is locally compact Hausdorff and hence exponentiable. Now for any open $V\subseteq G$, we have $G_\infty \setminus V$ compact in $G_\infty$ and $G_\infty \setminus \{e\}$ open in $G_\infty$, inducing an open subset $C(G_\infty \setminus V, G_\infty \setminus \{e\})$ of the compact-open topology on $(G_\infty)^{G_\infty}$. The inverse image of this open subset is therefore open in $G$. But this inverse image is the set of all $x\in G$ such that multiplication by $x$ maps the complement of $V$ to the complement of $e$, or contrapositively such that if $x y = e$ then $y\in V$, or equivalently such that $x^{-1}\in V$. Thus, the inverse image of $V$ by inversion is open, so inversion is continuous.

The appearance of categorical things like exponential objects makes me wonder (& hope) whether this could be reformulated more abstractly. E.g. suppose we have a monoid $G$ in some topological concrete category $T$ whose underlying set is a group, and which is exponentiable in $T$. Is it (perhaps under more assumptions) a group in $T$?

• CommentRowNumber50.
• CommentAuthorTobyBartels
• CommentTimeAug 23rd 2015

Mike just wrote in part:

since $G_\infty$ is locally compact Hausdorff and hence exponentiable

Oo, but we were wanting this to be true even when $G$ (and hence $G_\infty$) fails to be Hausdorff.

• CommentRowNumber51.
• CommentAuthorTodd_Trimble
• CommentTimeAug 24th 2015

Re #44: I didn’t ask him to name 6.9 b) after me! I found the example in Warner’s book, as I say over here. (I doubt the example is due to Warner, so I don’t know who should ultimately be credited, but I’ll be sure to write Pete again and let him know where I got it. Obviously if anyone here in this discussion finds a counterexample to another claim of Pete’s, credit should go to that person!)

P.S. That math stackexchange post I linked to seems to grab more of Warner’s text than what I was able to see using the link at #46; try it here. In particular, I can see Theorem 11.17 through this link but not David’s. Google books is weird.

• CommentRowNumber52.
• CommentAuthorTobyBartels
• CommentTimeAug 24th 2015

I never thought that you asked him to do that!

• CommentRowNumber53.
• CommentAuthorTodd_Trimble
• CommentTimeAug 24th 2015

One other thing: it was footnote 8, p. 21, that made me think Pete might have been assuming Hausdorffness throughout – I hadn’t noticed David’s observation in #43 to make me second-guess that.

I’ve written Pete on some of the matters brought up here. One other matter is what local compactness is supposed to mean if Hausdorffness is not assumed. Munkres says it means every point has a compact neighborhood, which is weaker than the nLab’s preferred definition unless of course Hausdorffness is assumed.

• CommentRowNumber54.
• CommentAuthorDavidRoberts
• CommentTimeAug 24th 2015

(hmm, now I can’t see the page I posted a link to, nor the one that Todd gives!)

• CommentRowNumber55.
• CommentAuthorMike Shulman
• CommentTimeAug 24th 2015

@Toby #50 I did say “in the Hausdorff case”. I’m personally more interested in having an abstract argument that might work in other contexts than in reaching the maximum possible point-set-topological generality.

• CommentRowNumber56.
• CommentAuthorTobyBartels
• CommentTimeAug 24th 2015

Mike, so you didn't mean to suggest that the proof only works when $G$ is Hausdorff? OK, good, because it seemed to me to work regardless, but I hadn't made sure of every detail. I understand your priorities.

• CommentRowNumber57.
• CommentAuthorTodd_Trimble
• CommentTimeAug 24th 2015

Pete wrote me back to say he was assuming Hausdorffness in exercise 6.9 and in the preceding paragraph. His convention is now firmly stated on page 21: “compact” is to mean “compact Hausdorff”, and “locally compact” is to mean “locally compact Hausdorff”. (In an earlier version of his document, there was some talk about ’quasi-compact’ which algebraic geometers sometimes use when Hausdorffness is not assumed.) However, for general topological spaces or topological groups/rings, etc., he does not take Hausdorffness as a standing assumption. Pete also told me that Ellis also assumed Hausdorffness in his articles (so it seems Warner strengthens Ellis’s results in his book).

Also, my contribution to exercise 6.9 b) is now more accurately described in a footnote. :-)

• CommentRowNumber58.
• CommentAuthorTodd_Trimble
• CommentTimeAug 24th 2015
• (edited Aug 24th 2015)

$G_\infty$ is locally compact Hausdorff and hence exponentiable

It may be well to point out that local compactness (even without the Hausdorff condition) is actually stronger than exponentiability in $Top$, as recounted in the paper by Escardó and Heckmann, page 9 and page 10. So much of the nice summary by Mike would likely go through even if we remove the Hausdorff condition, except possibly for the fact that the exponential topology might be different from the compact-open topology – I’d need to look at this more closely.

• CommentRowNumber59.
• CommentAuthorMike Shulman
• CommentTimeAug 25th 2015

@Toby that’s right. But I haven’t checked the details either (in particular, not the rigamarole I mentioned); I was trying to understand the proof at a high level only.

• CommentRowNumber60.
• CommentAuthorMike Shulman
• CommentTimeAug 28th 2015

If $K$ is a topological field, is $GL_n(K)$ necessarily a topological group? For $n=1$ this is by definition, but it doesn’t seem obvious to me that it follows automatically for $n\gt 1$.

• CommentRowNumber61.
• CommentAuthorSridharRamesh
• CommentTimeAug 28th 2015

Should it not follow automatically by “Cramer’s Rule” for matrix inverses?

• CommentRowNumber62.
• CommentAuthorZhen Lin
• CommentTimeAug 28th 2015

For any topological commutative ring $R$, $Mat_n (R)$ (defined as $R^{n \times n}$ with matrix multiplication) is a topological monoid. If the subspace of units in $R$ is open and is a topological group, then the subspace of invertible elements of $Mat_n (R)$ is also open and a topological group, because the explicit formula for the inverse matrix is polynomial in the entries of the matrix and the inverse of the determinant (which is polynomial in the entries).

• CommentRowNumber63.
• CommentAuthorMike Shulman
• CommentTimeAug 28th 2015

Oh good. I should have thought of that.

• CommentRowNumber64.
• CommentAuthorDavidRoberts
• CommentTimeSep 2nd 2015

Oliver Lorscheid, Cecília Salgado, Schemes as functors on topological rings

Abstract: Let $X$ be a scheme. In this text, we extend the known definitions of a topology on the set $X(R)$ of $R$-rational points from topological fields, local rings and ad`ele rings to any ring $R$ with a topology. This definition is functorial in both $X$ and $R$, and it does not rely on any restriction on X like separability or finiteness conditions. We characterize properties of $R$, such as being a topological Hausdorff ring, a local ring or having $R^\times$ as an open subset for which inversion is continuous, in terms of functorial properties of the topology of $X(R)$. Particular instances of this general approach yield a new characterization of adelic topologies, and a definition of topologies for higher local fields.

• CommentRowNumber65.
• CommentAuthorUrs
• CommentTimeMay 30th 2017

I have made explicit at topological ring that continuity of the additive inverse assignment is implied.

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