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Following a comment on the answer in this question: Sphere homeomorphic to plane?
I would like to ask: if a sphere(or any surface) is locally homeomorphic to a plane, why doesn't that show us how flat the sphere(or any object) locally is?

Community
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TheQuantumMan
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  • Curvature can come from the embedding into euclidean space, for example there is a flat torus in $\Bbb R^4$. – snulty Aug 08 '16 at 10:38
  • @snulty I don't think I get your point. Why doesn't being locally homeomorphic say anything about the local flatness of the object? – TheQuantumMan Aug 08 '16 at 10:42
  • Well a homeomorphism only says something about continuity, and curvature, like riemann curvature, usually involves derivatives (exterior, partial etc.). You want something more than just the charts, which tell you about local (homeo/diffeo)morphisms. A metric, or a connection (possibly derived from the metric). My point about the torus was, you could have a curved torus or a flat torus depending on how you embed the torus, with the curvature coming from the embedding, in that there's an induced riemannian metric and then curvature or flatness of the levi-civita connection. – snulty Aug 08 '16 at 10:48
  • In either case it's still locally homeomorphic to $\Bbb R^2$, independent of whether it's flat or not. – snulty Aug 08 '16 at 10:50
  • @snulty So, if a sphere with a very big radius(small curvature) says that the sphere locally resembles a plane, then what does the sphere being locally homeomorphic to the plane say about it? – TheQuantumMan Aug 08 '16 at 10:55
  • It says a lot of nice things about it. Firstly it says that it looks two dimensional near every point, so something like this sphere with a hair can't happen. It says locally, that there is a path between 'nearby' points, so for example this comb space won't happen. It also gives you something called locally simply connected, namely that if you have a small loop at a point, you should be able to continuously shrink it to a point $\ldots$ – snulty Aug 08 '16 at 11:06
  • unlike this hawaiian earring space. – snulty Aug 08 '16 at 11:10
  • The point being, even a small sphere with a small radius (large curvature) exhibits these properties. Not just the big sphere, which because it's gaussian curvature is small you expect it to be 'kinda flat' locally. – snulty Aug 08 '16 at 11:13
  • @snulty So, if I were to make an analogy: if we have a differentiable curve y=f(x), the second derivative gives its curvature but the fact that it is differentiable means that locally it resembles a line, right? So, the fact that is locally resembles a line does not give any clues about how "intense" is the curvature of the curve. Is this analogy correct? – TheQuantumMan Aug 08 '16 at 11:22
  • Yeah that's sounds like a nice analogy (invoking the inverse function theorem) which I believe @YannHamdaoui alludes to in their answer. – snulty Aug 08 '16 at 12:30
  • @snulty Yes, I saw the answer afterwards. So, why call it homeomorphic rather than differentiable? – TheQuantumMan Aug 08 '16 at 12:34
  • The comments are getting rather long, and it's suggested we move this to chat, but you would said homeomorphic if you wanted to consider the sphere a topological manifold, diffeomorphic if considering it a differentiable $C^k$ manifold $k=1,2\ldots\infty$. It depends on what context you plan to work with. – snulty Aug 08 '16 at 14:18
  • Let us continue this discussion in chat. – TheQuantumMan Aug 08 '16 at 15:20

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It's because curvature of a surface or a curve isn't conserved under homeomorphisms/diffeomorphisms. If that was the case, then any regular curve or surface would have zero curvature, because they are locally homeomorphic to an interval or ball in $\mathbb R ^2$. Hence it would be a useless measure of curvature.

So in short - when you are considering embedded surfaces into euclidean space the embedding itself defines the curvature of the surface. Any homeomorphic transformation of the embedding may change the curvature.

In the general setting of manifolds its an object called a "connection" that dictates what the curvature is - but that is a whole other story.

Martin
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The point is that being locally homeomorphic is qualitative : you know that it is "locally flat", but you don't know how. The curvature is a quantitative measure : you need to actually exhibit an embedding and do computation on it.

This is a bit like saying that a function $f$ is differentiable tells you nothing about how it actually grows, just that $f$ is locally "linear". You have a whole bunch of differentiable functions that behave very differently locally, from constant ones to exponentials. You need to compute $f$'s derivative to decide if $f$ is more like an horizontal line ($f'(x)=0$), a diagonal line ($f'(x)=1$) or grows even faster ($f'(x) > 1$).

yago
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