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The definition of manifold I usually see is “A topological space $(X,T)$ in which for every member of the topological space $x \in (X,T)$ there exists a neighborhood $M_x$ such that is homeomorphic to an open set of $\mathbb{R}^n$

The problem with this definition is that it doesn’t seem to imply the intuitive notion that manifolds are locally flat (look flat when you zoom in). All it implies is that there are neighborhoods around points where the topology looks the same, but we know of objects that have the same topology but do not geometrically look the same (.ex a coffee cup and a donut).

It looks like a better definition of a manifold would be a set where for each member of the set there exists a neighborhood where the Riemann curvature tensor is approximately 0. Does this definition work? Is it implied from the 1st definition of a manifold? If so, how?

Thanks

Obama2020
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    what does it mean to look flat? Can you give a formal definition? – Marcos Apr 25 '22 at 22:37
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    Riemannian curvature is not defined for topological manifolds. – pancini Apr 25 '22 at 22:48
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    "looks flat" is a misunderstanding of a manifold. It is often used as an intuition, but it fails to deal with things like the surface of a tetrahedron, which doesn't "look flat" everywhere. Mountains do no make the surface of the earth not a manifold. The notion of curvature of a manifold comes up when you reach Riemannian manifolds. – Thomas Andrews Apr 25 '22 at 22:49
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    Topologically, the surface of a tetrahedron "looks flat." in an abstract sense. There is no way to use local topological features on a tetrahedron to distinguish your space from a plane. But if you are on a geometric tetrahedron in three dimensional space, it doesn't "look flat" at the edges or vertices. – Thomas Andrews Apr 25 '22 at 22:56
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    Your question is jumping too many steps ahead. The Riemann curvature tensor is defined in terms of the connection. To define a connection, you need to first have a smooth manifold (differ – Clemens Bartholdy Apr 25 '22 at 23:02
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    If by flat you mean locally isometric to $\mathbb{R}^n$, by the Killing-Hopf theorem the answer is in general no; e.g. in $2D$ Killing-Hopf says there are only $5$ flat manifolds: plane, cylinder, twisted cylinder, torus, and Klein bottle. – Mo Pol Bol Apr 25 '22 at 23:05
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    The curvature of a Riemannian manifold does not shrink when we consider small pieces. Qualitatively, even a microscopic piece of a unit sphere has Gaussian curvature $1$. – Andrew D. Hwang Apr 25 '22 at 23:53
  • The issue is that a "flat" surface such as a plane in $\mathbb{R}^3$ is topologically indistinguishable from a "bumpy" surface such as the graph of a "rough" function $\mathbb{R}^2\to\mathbb{R}$: They are locally homeomorphic if they have the same dimension. – Kajelad Apr 25 '22 at 23:57
  • @AndrewD.Hwang but when you consider small patches of a Riemannian manifold and move a vector in a loop, the separation vector is small in magnitude. – Obama2020 Apr 26 '22 at 02:53
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    @Marcos by flat I’m talking about the intuitive notion (like an ant on a sphere seeing the surface as flat). I guess it could be made more mathematically rigorous by saying “a region where a vector transported in a loop changes from the initial vector by a small amount”. The question is how the definition of a manifold used captures this. – Obama2020 Apr 26 '22 at 02:55
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    @ThomasAndrews that’s what I was thinking. There are some things said to be manifolds where at certain points they do not look flat, which is part of the reason why I asked the question. So would it be correct to say that a riemannian manifold only looks flat if locally a vector transported along in a loop does not deviate very much from the original vector? – Obama2020 Apr 26 '22 at 02:59
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    Right. The thing is, topologically, a tetrahedron or cube is the same as a sphere. So “flat” doesn’t mean what we normally think it means. – Thomas Andrews Apr 26 '22 at 03:01
  • Sio a geomeometric object which is not flat might still be a manifold. That's because the topology does not reflect that geometry. Just knowing the topology, we can't tell the difference between being on a sphere or being on a tetrahedron. There is no topological difference. – Thomas Andrews Apr 26 '22 at 14:59

1 Answers1

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tl; dr: The definition of a manifold (a topological space having a single local model for its topology) is an asset, not a liability. The problem, as Thomas Andrews identifies, is ambiguity in the qualitative phrasing "locally flat" that we sometimes use when we're trying not to put off Ordinary Civilians.

(It's similarly misleading to speak of homeomorphisms as "continuous deformations," which colloquially and incorrectly suggests isotopies. But I digress.)

The property of looking flat when you zoom in is in the eye of the beholder, and there are advantages in squinting hard, so hard that you can't distinguish anything except outright discontinuity. The universe of topological spaces is beyond human comprehension: Think merely of subsets of Euclidean three-space with the induced topology. One motivation for working with manifolds, as conventionally defined, is to balance generality against tractability. As for generality, a great many physical problems (configuration spaces, vector fields, level sets) can be set naturally on manifolds. Regarding tractability, because any two connected manifolds of the same dimension are locally homeomorphic, dimension is a (topological or smooth) manifold's only local invariant. All else is global: compactness, homology and homotopy groups, etc.

Conversely, what do we change by requiring a manifold to have a local metric structure "nearly the same as" Euclidean space? For one thing, we must decide what "nearly the same as" means. For definiteness, let's work with Riemannian metrics. (These make sense only on smooth manifolds, and so represent a loss of generality.)

Now again: The curvature of a Riemannian metric does not shrink when we restrict attention to small neighborhoods. (Even the holonomy does not shrink, except insofar as the area enclosed by a simple closed loop shrinks with the length of the loop.) If we regard a unit sphere as "not nearly flat" but a microscopic piece of a unit sphere as "nearly flat," then nearly flat is not a local concept, but a global one: After all, locally a sphere and a piece of sphere are congruent, geometrically identical in the strongest Euclidean sense. There is nothing special about a unit sphere: In the seme sense every Riemannian manifold is "nearly flat" if we look at a sufficiently small nighborhood. In other words, a non-flat Riemannian metric can be "nearly flat" only in the sense that a non-zero real number can be "nearly zero," which is "not at all." (Asking for a manifold to admit a flat Riemannian metric is an enormous restriction. Compact flat manifolds are isometrically covered by a Euclidean space, i.e., are tori or spaces covered by tori.)

In an attempt (albeit reasonable-seeming) to capture a non-technical sense of "nearly flat," imposing a geometric condition both complicates the definition and dramatically narrows its scope. That, mathematically speaking, is why we don't.

Andrew D. Hwang
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