![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
sniffnoySo today I was sitting in my office (because the internet is very spotty at home right now) when I see some old guy standing around outside. At first I just ignore him. Eventually he asks if there's anyone here familiar with French. I say no, the others don't respond. Or any language other than English, he says. What sort of question could possibly require just *some* non-English language, regardless of which?, I think. Again, nobody will say they do (though of course everyone else in my office speaks *some* east Asian language). It's a math question, he insists, which just makes it weirder.
...so he moves on to the next office whose door is open, and there he finds Ari, who will admit to having some familiarity with French. A mistake. And so the guy explains the problem: He's interested in taking a number, writing it down in some language (say English), counting the number of letters used, and iterating this. What happens? Except, of course, he can't say it like that; instead he illustrates it by way of some long example, going through the entire process (I think starting from 121, maybe?). His explanation of what he wants to know about is long, too. Finally I come out, interrupt him, saying, yes, we're familiar with these sorts of things, you don't need to go on at length about it. And Ari and I tell him that, while a language could technically be anything (leaving the question unanswerable in full generality), in any reasonable language, the number of letters to spell out a number will be much smaller (for large numbers) than the number itself; hence no matter what you start with, you will indeed end up in a cycle. However, the specifics of what cycles are possible are impossible to say without knowing the specific language in question. (Of course there's the problem that large numbers don't have a unique name, though I didn't get the chance to point this out, though sometimes this is OK; e.g. it's not too hard to conclude that if we handwave this away, the fixpoint 4 is the unique cycle in English).
But apparently the guy wasn't satisfied and just kept re-explaining the question. Also, he said, he wanted to know why "All cats like mice, and all mice like cheese, therefore all cats like cheese" isn't a valid syllogism. At that point Emily and Nic happened to be walking by so Ari said "Oh, I think Nic will answer your question" and I went back in my office and closed the door. From the sound of it (he was pretty loud), this went on for several more minutes. I have no idea how they got him to leave.
(Here's another odd encounter recently involving the writing of numbers. Late last night, returning home fromthe internet my office, I got hungry, so I stopped at Panchero's. Two other people came in shortly afterward and insisted I come eat with them, so I did. When they found out I did math one of them started claiming that Asians are much better at math than Americans because Asian languages name numbers in a more regular manner. I tried to point out why this was ridiculous but I'm not sure it did much good.)
Let's talk about topological fields!
I mean, everyone knows about valued fields, right? That is, fields equipped with an absolute value. Well, I mean, topological fields are kind of a generalization of them, right? And of ordered fields as well, but I don't know so much about those in topological terms. Anyway, some time ago I was wondering if some of the purely topological facts I either knew or had read somewhere about valued fields still held for general topological fields.
Having now actually read some of this stuff and briefly looked up more of it, I can now say that no, topological fields are not very nice in general. (Apparently much of the specialness of valued fields comes from them being "locally retrobounded"[8]?) Let me give some examples![9]
(Note: The following being a brief summary, I may not distinguish between things I actually know and things I just read somewhere, or the intermediate category of "things I skimmed", or things I intend to read, or... you get the idea. Sources on valued fields: Coppel's "Number Theory", Wiȩsław's "Topological Fields". Sources on more general topological fields: Wiȩsław, also a bit of Warner's "Topological Fields". Sources on other stuff: The internet.)
1. Completion. Every abelian (or more generally, every balanced[0]) group can be completed to a complete abelian (more generally, balanced) topological group. (This can be proved just via abstract nonsense.) Every topological ring can be completed to a complete topological ring. (This unfortunately I don't think can be). Every topological field can be completed to... a complete topological ring. Yup, the completion doesn't have to be a field. Nor even a domain. If it's a field, it's automatically a topological field, for what that's worth. Valued fields complete to fields, we all know, but even what Wiȩsław calls "normed fields"[4] don't necessarily. (I know this because I tried to prove they do, and quickly found that they don't -- take two absolute values on Q and take their maximum; this is a norm, and the completion will not be a domain. (OK, if one of them is the usual absolute value I've only checked that it's not a field, but I'd bet it's not a domain either. In fact I think I recall reading that it'll be the product of the respective completions? Need to check about that.))
2. On the subject of completion: If you have an algebraically closed valued field, and you complete it, the result is again algebraically closed[5]. For a general topological field... will it even be a field? Hell if I know. Maybe if it's a field it'll be algebraically closed? I have no idea. Honestly, this seems like it should be true, but the potential lack of fieldness scares me.
3. Independence: If two nontrivial absolute on a field induce distinct topologies, they induce independent topologies, in the sense that if we let Fi be F with the i'th topology (i=1,2), then the diagonal is dense in F1×F2. (I.e., given any x1,x2∈F, you can find a sequence in F that converges to x1 wrt the first absolute value and x2 wrt the second. Hence, "independence". Obviously sequences aren't enough in general but hey, metric.)[6] So in particular they're necessarily incomparable! But more general topologies on a field (staying compatible with the field structure, of course) form a complete lattice! (I guess we have to include the discrete and indiscrete topologies for that to work.) So they certainly need not be incomparable, and even if incomparable they can still be dependent.
4. Finite dimensional topological vector spaces: It's a well-known fact that given a complete valued field K, the only Hausdorff topology on K^n making it a topological vector space over K is the product topology. In the one-dimensional case, you don't even need completeness. Well, the latter statement is definitely false about more general topological fields, because if you have a strictly coarser nontrivial field topology, it'll make K a topological vector space over the original topology. I haven't yet found a counterexample which is complete, but I'd bet there is one. (Terminology note: A field satisfying the one-dimensional version of this is known as "strictly minimal" (since as noted above it must be minimal) or "straight".) Note: This also works for ordered fields, or fields with an "absolute value" taking values in the nonnegative elements of some ordered field that need not be R. Also I think I read it works for locally retrobounded fields? Maybe such fields as I described *are* locally retrobounded? That would surprise me, from what little I know of this notion.
5. Schmidt's theorem: Schmidt proved that if a field is complete with respect to two independent absolute values, it must be algebraically closed. More precisely, a field F is complete wrt two independent absolute values iff it is algebraically closed and |F|^ℵ0=|F|. (This has the nice corollary that any algebraic automorphism of a complete but not algebraically closed field is automatically a continuous automorphism.) For two independent field topologies generally...? No idea. (There's a version also for a different special type of topological fields -- also among the "locally retrobounded" -- but I'm not talking about that here, because I don't much care. Also it's not exactly the same I don't think?)
6. Probably lots more but this is what comes to mind right now.
Incidentally, one of the most important facts about topological fields is that any locally compact field comes from an absolute value, and then once you know that you can use various techniques to go ahead and classify them. The latter I've seen proved (in Coppel), the former I hadn't. Wiȩsław also gives a proof of this which I haven't read yet but seems like it should actually be understandable, and very unlike the more "natural" proof that Wikipedia hints at. Well, assuming you can prove it that way; I assume you can. What WP suggests is, the field is locally compact, so take a Haar measure; then multiplication by a nonzero element just scales the Haar measure, so define |x| to be the scaling factor (and |0|=0). Then |x|=0 iff x=0, and the map is obviously multiplicative... however it isn't necessarily an absolute value, just a power of one. (Of course, if we already know the classification of local fields, we know the only case where it isn't is C, where it's the square of one.) Well, somehow you show that fact, and that it induces the original topology. But those two steps sound scary[7].
Conclusion: Now when I see a purely topological/algebraic fact about valued fields, I know that "Does this generalize to topological fields?" is not necessarily a good question.
-Harry
[0]By this, I mean that the right and left uniformities are the same. Apparently there are unbalanced groups that cannot be completed to a topological group (though I guess we have to ask, completed in which of the 3 uniformities[3]? I no longer remember or much care).
[3]I suppose there's a fourth if it's locally compact, but I get the idea nobody cares about that uniformity, and, more relevantly, in that case it's already complete!
[4]By a "norm" on a field F, Wiȩsław means |•|:F→[0,∞) satisfying:
[5]Worth noting: Wiȩsław gives a single proof for this that works for any valued algebraically closed field, whereas (though I hadn't seen an actual proof before) from what descriptions I've heard it seemed like everyone else does it by doing it one way for the non-Archimedean case, and presumably handling the Archimedean case by just noting that the only complete Archimedean fields are R and C.
[6]This is known as the "weak approximation theorem" or the "Artin-Whaples approximation theorem". Well, really, that's for any finite number of them, not just two, but I didn't want to bother stating that.
[7]OK, it's a general fact that when considering absolute values on a field, the triangle inequality can actually be replaced by the weaker requirement that |x+y|≤2max(|x|,|y|). (See this earlier post for a proof.) So if you have |x+y|≤C*max(|x|,|y|), raising |•| to a power just does the same to C, hence some power of |•| is an absolute value. So I assume the way to do it is to show that this inequality holds for some C, but again, no idea how you'd do that. It'd be funny if it just worked by abstractly showing that there is some C, because if we already know the classification of local fields, we know the highest C that will occur is 4!)
[8]A.k.a. "of type V".
[9]OK, so some of these are just examples of "I haven't seen it stated to be true so if it is I doubt anyone's proved it", rather than examples of "It's false", but...
...so he moves on to the next office whose door is open, and there he finds Ari, who will admit to having some familiarity with French. A mistake. And so the guy explains the problem: He's interested in taking a number, writing it down in some language (say English), counting the number of letters used, and iterating this. What happens? Except, of course, he can't say it like that; instead he illustrates it by way of some long example, going through the entire process (I think starting from 121, maybe?). His explanation of what he wants to know about is long, too. Finally I come out, interrupt him, saying, yes, we're familiar with these sorts of things, you don't need to go on at length about it. And Ari and I tell him that, while a language could technically be anything (leaving the question unanswerable in full generality), in any reasonable language, the number of letters to spell out a number will be much smaller (for large numbers) than the number itself; hence no matter what you start with, you will indeed end up in a cycle. However, the specifics of what cycles are possible are impossible to say without knowing the specific language in question. (Of course there's the problem that large numbers don't have a unique name, though I didn't get the chance to point this out, though sometimes this is OK; e.g. it's not too hard to conclude that if we handwave this away, the fixpoint 4 is the unique cycle in English).
But apparently the guy wasn't satisfied and just kept re-explaining the question. Also, he said, he wanted to know why "All cats like mice, and all mice like cheese, therefore all cats like cheese" isn't a valid syllogism. At that point Emily and Nic happened to be walking by so Ari said "Oh, I think Nic will answer your question" and I went back in my office and closed the door. From the sound of it (he was pretty loud), this went on for several more minutes. I have no idea how they got him to leave.
(Here's another odd encounter recently involving the writing of numbers. Late last night, returning home from
Let's talk about topological fields!
I mean, everyone knows about valued fields, right? That is, fields equipped with an absolute value. Well, I mean, topological fields are kind of a generalization of them, right? And of ordered fields as well, but I don't know so much about those in topological terms. Anyway, some time ago I was wondering if some of the purely topological facts I either knew or had read somewhere about valued fields still held for general topological fields.
Having now actually read some of this stuff and briefly looked up more of it, I can now say that no, topological fields are not very nice in general. (Apparently much of the specialness of valued fields comes from them being "locally retrobounded"[8]?) Let me give some examples![9]
(Note: The following being a brief summary, I may not distinguish between things I actually know and things I just read somewhere, or the intermediate category of "things I skimmed", or things I intend to read, or... you get the idea. Sources on valued fields: Coppel's "Number Theory", Wiȩsław's "Topological Fields". Sources on more general topological fields: Wiȩsław, also a bit of Warner's "Topological Fields". Sources on other stuff: The internet.)
1. Completion. Every abelian (or more generally, every balanced[0]) group can be completed to a complete abelian (more generally, balanced) topological group. (This can be proved just via abstract nonsense.) Every topological ring can be completed to a complete topological ring. (This unfortunately I don't think can be). Every topological field can be completed to... a complete topological ring. Yup, the completion doesn't have to be a field. Nor even a domain. If it's a field, it's automatically a topological field, for what that's worth. Valued fields complete to fields, we all know, but even what Wiȩsław calls "normed fields"[4] don't necessarily. (I know this because I tried to prove they do, and quickly found that they don't -- take two absolute values on Q and take their maximum; this is a norm, and the completion will not be a domain. (OK, if one of them is the usual absolute value I've only checked that it's not a field, but I'd bet it's not a domain either. In fact I think I recall reading that it'll be the product of the respective completions? Need to check about that.))
2. On the subject of completion: If you have an algebraically closed valued field, and you complete it, the result is again algebraically closed[5]. For a general topological field... will it even be a field? Hell if I know. Maybe if it's a field it'll be algebraically closed? I have no idea. Honestly, this seems like it should be true, but the potential lack of fieldness scares me.
3. Independence: If two nontrivial absolute on a field induce distinct topologies, they induce independent topologies, in the sense that if we let Fi be F with the i'th topology (i=1,2), then the diagonal is dense in F1×F2. (I.e., given any x1,x2∈F, you can find a sequence in F that converges to x1 wrt the first absolute value and x2 wrt the second. Hence, "independence". Obviously sequences aren't enough in general but hey, metric.)[6] So in particular they're necessarily incomparable! But more general topologies on a field (staying compatible with the field structure, of course) form a complete lattice! (I guess we have to include the discrete and indiscrete topologies for that to work.) So they certainly need not be incomparable, and even if incomparable they can still be dependent.
4. Finite dimensional topological vector spaces: It's a well-known fact that given a complete valued field K, the only Hausdorff topology on K^n making it a topological vector space over K is the product topology. In the one-dimensional case, you don't even need completeness. Well, the latter statement is definitely false about more general topological fields, because if you have a strictly coarser nontrivial field topology, it'll make K a topological vector space over the original topology. I haven't yet found a counterexample which is complete, but I'd bet there is one. (Terminology note: A field satisfying the one-dimensional version of this is known as "strictly minimal" (since as noted above it must be minimal) or "straight".) Note: This also works for ordered fields, or fields with an "absolute value" taking values in the nonnegative elements of some ordered field that need not be R. Also I think I read it works for locally retrobounded fields? Maybe such fields as I described *are* locally retrobounded? That would surprise me, from what little I know of this notion.
5. Schmidt's theorem: Schmidt proved that if a field is complete with respect to two independent absolute values, it must be algebraically closed. More precisely, a field F is complete wrt two independent absolute values iff it is algebraically closed and |F|^ℵ0=|F|. (This has the nice corollary that any algebraic automorphism of a complete but not algebraically closed field is automatically a continuous automorphism.) For two independent field topologies generally...? No idea. (There's a version also for a different special type of topological fields -- also among the "locally retrobounded" -- but I'm not talking about that here, because I don't much care. Also it's not exactly the same I don't think?)
6. Probably lots more but this is what comes to mind right now.
Incidentally, one of the most important facts about topological fields is that any locally compact field comes from an absolute value, and then once you know that you can use various techniques to go ahead and classify them. The latter I've seen proved (in Coppel), the former I hadn't. Wiȩsław also gives a proof of this which I haven't read yet but seems like it should actually be understandable, and very unlike the more "natural" proof that Wikipedia hints at. Well, assuming you can prove it that way; I assume you can. What WP suggests is, the field is locally compact, so take a Haar measure; then multiplication by a nonzero element just scales the Haar measure, so define |x| to be the scaling factor (and |0|=0). Then |x|=0 iff x=0, and the map is obviously multiplicative... however it isn't necessarily an absolute value, just a power of one. (Of course, if we already know the classification of local fields, we know the only case where it isn't is C, where it's the square of one.) Well, somehow you show that fact, and that it induces the original topology. But those two steps sound scary[7].
Conclusion: Now when I see a purely topological/algebraic fact about valued fields, I know that "Does this generalize to topological fields?" is not necessarily a good question.
-Harry
[0]By this, I mean that the right and left uniformities are the same. Apparently there are unbalanced groups that cannot be completed to a topological group (though I guess we have to ask, completed in which of the 3 uniformities[3]? I no longer remember or much care).
[3]I suppose there's a fourth if it's locally compact, but I get the idea nobody cares about that uniformity, and, more relevantly, in that case it's already complete!
[4]By a "norm" on a field F, Wiȩsław means |•|:F→[0,∞) satisfying:
- |x|=0 iff x=0
- |x+y|≤|x|+|y|
- |xy|≤|x||y|
- |-x|=|x|
[5]Worth noting: Wiȩsław gives a single proof for this that works for any valued algebraically closed field, whereas (though I hadn't seen an actual proof before) from what descriptions I've heard it seemed like everyone else does it by doing it one way for the non-Archimedean case, and presumably handling the Archimedean case by just noting that the only complete Archimedean fields are R and C.
[6]This is known as the "weak approximation theorem" or the "Artin-Whaples approximation theorem". Well, really, that's for any finite number of them, not just two, but I didn't want to bother stating that.
[7]OK, it's a general fact that when considering absolute values on a field, the triangle inequality can actually be replaced by the weaker requirement that |x+y|≤2max(|x|,|y|). (See this earlier post for a proof.) So if you have |x+y|≤C*max(|x|,|y|), raising |•| to a power just does the same to C, hence some power of |•| is an absolute value. So I assume the way to do it is to show that this inequality holds for some C, but again, no idea how you'd do that. It'd be funny if it just worked by abstractly showing that there is some C, because if we already know the classification of local fields, we know the highest C that will occur is 4!)
[8]A.k.a. "of type V".
[9]OK, so some of these are just examples of "I haven't seen it stated to be true so if it is I doubt anyone's proved it", rather than examples of "It's false", but...
