# Are there spaces that are globally Euclidean, but locally not?

I was reading this question, and I idly wondered if there could be spaces that were globally Euclidean, but locally not, and couldn't think of any circumstances in which that might be the case.

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What does it mean to be globally Euclidean? – Qiaochu Yuan Dec 9 '10 at 8:47
Please define globally Euclidean. – Alex B. Dec 9 '10 at 8:49
Qiaochu beat me to it :-) Anyway, once you have (somehow) defined globally Euclidean, define locally Euclidean. Chances are that that will settle your question. – Alex B. Dec 9 '10 at 8:50

As the comments have suggested, the question is a bit under-determined, but let me supply some definitions in a plausible way and answer it based on those definitions.

Let $P$ be a property of topological spaces. (Formally speaking this could be any "predicate", but it would only be asking for trouble if $P$ were not a topological property of topological spaces -- i.e., such that if $X$ and $Y$ are homeomorphic topological spaces, then $X$ has property $P$ iff $Y$ has property $P$.)

Then we say that $X$ is locally P if for every point of $x$ there is a neighborhood base $\{U_x\}$ each of whose elements $U_x$ has property $P$ in the subspace topology.

We say that $X$ is globally P if it has property $P$. Why would we say this? Well, it's a matter of emphasis, presumably by way of comparison to something which locally has property $P$.

Now define a space to be Euclidean if it is homeomorphic to $\mathbb{R}^n$ for some $n \in \mathbb{N}$.

Now I claim that any space which is globally Euclidean is also locally Euclidean. In other words, I claim that for all $n \in \mathbb{N}$, every $x \in \mathbb{R}^n$ admits a neighborhood base, each of whose members is homeomorphic to Euclidean $m$-space for some $m$. But indeed, for any $\epsilon > 0$, the open $\epsilon$-ball around $x$ is isomorphic to $\mathbb{R}^n$.

Note that in general "globally P" need not imply "locally P". For instance a connected space need not be locally connected, and a quasi-compact (non-Hausdorff!) space need not be locally quasi-compact.

In fairness I should say that some mathematicians don't like the idea that globally $P$ need not imply locally $P$, so they weaken the definition of locally $P$ to "each point of $X$ admits some neighborhood which has property $P$". I would call this weakly locally P and then it is trivial that globally $P$ implies weakly globally $P$: for each $x$, take $X$ itself as a neighborhood which has property $P$.

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Some similar issue: Does weakly locally euclidean imply locally euclidean? – Freeze_S Apr 8 '14 at 12:40

Yes, $\mathbb{Z}$ is one.

(Note: As pointed out in the comments, $\mathbb{Z}$ is technically locally Euclidean as it is discrete. If you are bothered by such things, replace $\mathbb{Z}$ by something like $\mathbb{Q}$, or the union of copies of the Cantor set. I think that, despite this flaw, $\mathbb{Z}$ is a better example of the point I wish to make so I'm going to stick with it.)

Of course, as Qiaochu and Alex effectively say in the comments, this all depends on what you mean by "globally". But before I say what that means, let me point out that "locally X" is not so straightforward. In topology, we say that a space is "locally X" if whenever we have $p \in U$ with $U$ open (or just a neighbourhood of $p$) then there is some neighbourhood $V$ of $p$ with $p \in V \subseteq U$ such that $V$ has property X.

This works fine with lots of properties such as connected, path connected, compact, and similar. When we put the word "Euclidean" in it, something is a bit weird. What does it mean to say that a space is Euclidean? We don't tend to define that. Rather we define "locally Euclidean" all in one go. It's like the definition of the limit of a sequence. We write $\lim_{n \to \infty} s_n = s$ but we never decompose that funny symbol "$\lim_{n \to \infty}$" and talk about $n$ actually going to $\infty$ in the same way that we talk about $x$ going to $0$ in the expression $\lim_{x \to 0} \sin(x) = 0$.

Well, except that we do. There is a way to make sense of $n$ going to $\infty$, but it comes second, not first. We could happily deal with limits and the like without ever knowing how to really make sense of $n$ going to $\infty$. But to know what $n$ going to $\infty$ means, we have to know about limits.

So it is with "locally Euclidean". We can decompose this, but it comes second. And the decomposition, into "locally" + "Euclidean", is almost determined by what "locally Euclidean" means. I say almost, and this is the point of the ramble, because there are actually two things that it could mean which both give the same answer for "locally Euclidean".

1. A space is "Euclidean" if it is homeomorphic to $\mathbb{R}^n$ for some $n$.
2. A space is "Euclidean" if it is homeomorphic to an open subset of $\mathbb{R}^n$ for some $n$.

For "locally Euclidean", both of these give the same answer because any open subset of $\mathbb{R}^n$ contains lots of small open subsets that are homeomorphic to $\mathbb{R}^n$. But if we replace "Euclidean" by some other space, this equivalence will break.

And so back to the original question. Why is $\mathbb{Z}$ globally Euclidean but not locally Euclidean? Again, we need to know what everything means. But now on a slightly higher level. What are we doing when we talk about "locally X"? We're saying that we can get out our microscopes and examine our space near $p$ and find that property X holds. By doing this, we effectively ignore a huge part of our space. In fact, we can ignore almost as much as we like (within reason) and focus down only on a tiny, tiny bit near $p$.

So what happens if we reverse this process? Instead of taking a microscope and examining our space really close up, let's put it in a rocket, shoot it off to the Moon and look at it through a telescope. Then, not only do we not see any of the small detail, we can't really tell if all that detail is there in the first place. The resolution of our telescope means that we cannot examine our space very closely at all. Indeed, we can't even tell if it is all there or not.

So when put on the Moon, $\mathbb{R}$ and $\mathbb{Z}$ are indistinguishable. So $\mathbb{Z}$ is Euclidean in the opposite sense to "locally Euclidean" but is (clearly) not locally Euclidean.

If you look for this concept in the literature, you won't find it under the heading of "globally X". That's because that would be too confusing a term. Rather, what I'm talking about is coarse geometry. And it's a lot of fun, and I wish I knew much, much more about it than I do.

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Isn't $\mathbb{Z}$ locally $\mathbb{R}^0$? – Joe Johnson 126 Dec 9 '10 at 15:25
I find this answer to be written somewhat misleadingly. Clearly rockets, telescopes and the moon are out of place in a formal mathematical definition. At the end of your post, you confirm (to a sufficiently knowledgeable reader) that what you actually mean by "Z is globally Euclidean" is that "The metric space Z is quasi-isometric to the metric space R". But I don't know anyone who refers to this property as "globally Euclidean", whereas in general if someone said "X is globally P", I would take that to mean "X has property P" and the word "globally" is there for emphasis. – Pete L. Clark Dec 9 '10 at 15:44
The idea of stepping further and further away from a metric space is exactly how Gromov explains his ultrafilter limit construction. It's not precise, but still conveys the idea very forcefully. Similarly, in describing the Fulton-MacPherson compactification, Kontsevich talks about looking at configurations with a microscope.I don't think these metaphors are a problem at all. – Grumpy Parsnip Dec 9 '10 at 16:16
@Jim: we know the mathematics behind what Andrew is talking about, so we suppose that's what he means. I wish he would be more explicit, or at least say earlier in his message that he's talking about a metric concept, whereas the OP used the tag general-topology. I also think that given that the OP is missing a precise definition, such a definition (rather than a metaphorical description) should be part of an answer. These are just my opinions of course: I have disagreed with Andrew before, and I know that he can take it cheerfully. – Pete L. Clark Dec 9 '10 at 16:58
@Pete: Your last sentence is completely correct. I'm not going to respond otherwise because it's the sort of thing that (as with our other disagreements) would take us 5 minutes in a pub to decide that we actually agreed but expressed ourselves differently. If we ever end up at the same conference, there'd better be a good pub nearby. – Loop Space Dec 9 '10 at 19:15

(disclaimer: I'm not writing anything deep in this answer, but it might be useful for people starting to learn manifolds/topology)

I guess it would be safe to say no, globally euclidean always implies locally euclidean, although of course it all depends on how you define these concepts.

A (not very useful) definition could be:

X is globally euclidean if it is homeomorphic to $\mathbb{R}^n$ for some n.

X is euclidean at x $\in X$ if there exists an open neighbourhood of x homeomorphic to an open subset of a globally euclidean space.

X is locally euclidean if it is euclidean at every point $x \in X$.

From this (tautological) definition then it is obvious that

If X is globally euclidean then X is locally euclidean.

The converse is of course false: take the real line $\mathbb{R}$ and the circle $S^1$, the first is globally euclidean, the second is locally euclidean but not globally. If $S^1$ were globally euclidean then (by a general result of algebraic topology "invariance of dimension") it would be the one-dimensional euclidean space $\mathbb{R}$. But $\mathbb{R}$ is contractible and $S^1$ is not (or $S^1$ is compact and $\mathbb{R}$ is not) (or if you remove a point from $S^1$ it stays connected but if you remove a point from $\mathbb{R}$ it splits in two).

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Just another attempt could be:
Define sth is locally adjective if it admits a neighborhood base consisting of adjective neighborhoods at every point and sth is globally adjective if it admits a basis for the topology consisting of adjective opens...

However, that would be again just seeking for some consistency pressing in some technical dress, so hmm I think best is still to stick with the luckily almost widely accepted aggreement presented by Pete Clark - though I like Andrew's view on these things although it's kind of vague.

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