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Approaching the concept of a manifold from a physicists perspective there is one particular question that I have not been able to answer myself:

Is a manifold an intrinsically geometric object, or is the geometry purely specified by the additional structure of a metric?

For example, consider the unit sphere $S^{2}$. Without specifying a metric it is simple a set, along with a topology. Can one say that this is a geometric object because it is locally homeomorphic to $\mathbb{R}^{2}$? From an intuitive point of view, I can see that a geometric object such as a sphere can exist without the need to specify any sort of metric defined on it, but I'm not sure whether this intuition carries over to manifolds in general - is it even correct to say that $S^{2}$ is a unit sphere without introducing a metric on it?!

Will
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1 Answers1

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The question is, what you mean by $S^2$. If it is only a set, then you don't have much structure and its only property is the cardinality $c$.

If you can see the topology, then you know what are neighbourhoods, limits, homology and homotopy groups, Euler characteristic and some kind of "dimension" (although this is not trivial).

A manifold structure is even a richer structure (for example, the notion of dimension is much cleaner and more canonical). However, very often people think of smooth manifolds in which case you additionally get an idea of differentiability, tangent spaces and tangent vectors: this can still be done without a metric. All the field of differential topology deals with smooth structures and differential invariants that are, sometimes, defined via a metric, but are independent on the choice of the metric. By the way, on the manifold $S^2$, there is only one smooth structure up to isomorphism, which is not necessarily the case for higher-dimensional spheres.

But if you don't have a metric, you can not define distances and angles between tangent vectors in a canonical way. If this is "geometry", then no, you don't have it.

Peter Franek
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  • So in general, a manifold does not have an intrinsic sense of geometry then - a metric must be defined on it for it to possess a geometrical structure?! I've always thought of geometry as the measurement of distance, angles etc. but is there a more general definition then? – Will Dec 07 '16 at 11:46
  • No, no metric is defined by the manifold structure. But there are some properties that any metric must possess -- topological / differential properties (such as that if you integrate the curvature of a metric on a surface, you get the Euler characteristic, for example) What is geometry? is a good question. You may be interested also in this discussion http://math.stackexchange.com/questions/192055/what-is-a-geometry – Peter Franek Dec 07 '16 at 11:47
  • Note that if you embed $S^2$ in $\Bbb R^3$ in some way, and measure everything on $S^2$ by looking at it as something really happening in $\Bbb R^3$ with the standard, euclidean metric and geometry, then that will uniquely define a metric and a geometry on $S^2$ (although different embeddings give different results). We say that this structure is "inherited" from $\Bbb R^3$, or in more fancy (and intrinsic) terms, "pulled back over the embedding". – Arthur Dec 07 '16 at 12:08
  • @PeterFranek Ah ok. Thanks for the link. What exactly do you mean by the metric being defined by the manifold structure? Do you mean that the choice of manifold, i.e. whether it is Riemannian, Pseudo-Riemannian, etc. determines the geometric structure? – Will Dec 07 '16 at 12:21
  • @Will I said "No metric is defined by the manifold strucure" I mean "(No metric) (is defined by) (the manifold structure)". In other words, if somebody gives you a smooth atlas, then there is no canonical metric. There always is some, however. A pseudo-Riemannian metric doesn't even need to exists. – Peter Franek Dec 07 '16 at 12:45
  • @PeterFranek Sorry, I misunderstood what you wrote before. So would it be correct at all to say that a manifold is not intrinsically a geometric object, and one must give additional structure to it, such as a metric, in order for a geometry to be defined on it? For example, if one takes Euclidean space, if one introduces the standard Pythagorean metric on it, then one arrives at Euclidean geometry, however, without this additional structure it is simply isomorphic to $\mathbb{R}^{n}$?! – Will Dec 07 '16 at 12:53
  • @Will There is no consensus on an exact definition of "geometry". So it is better to avoid this word, if you want to ask a formal and precise question. I can just repeat what I said before, smooth structure on a manifold does not determine a metric. – Peter Franek Dec 07 '16 at 13:00
  • @PeterFranek I see, fair enough. I wasn't so concerned before, but what got me thinking about it was reading Terry Tao's blog post "What is a gauge" (https://terrytao.wordpress.com/2008/09/27/what-is-a-gauge/). At the end of the section on coordinate systems he says "...More generally, a coordinate system $\Phi$ can be viewed as an isomorphism $\Phi: A \to G$ between a given geometric (or combinatorial) object A in some class (e.g. a circle), and a standard object G in that class (e.g. the standard unit circle)... – Will Dec 07 '16 at 13:37
  • @PeterFranek ...(To be pedantic, this is what a global coordinate system is; a local coordinate system, such as the coordinate charts on a manifold, is an isomorphism between a local piece of a geometric or combinatorial object in a class, and a local piece of a standard object in that class.)..." ... It is this last part in particular that got me a bit confused about the subject. – Will Dec 07 '16 at 13:37
  • @Will I haven't read the blog so I don't know what exactly is he trying to say. However, if I can guess, "A geometric or combinatorial object in a class" sounds like a very high-level language, without caring too much about a precise setting (like whether or not we use a particular metric). – Peter Franek Dec 07 '16 at 20:07
  • @PeterFranek Ok, so is the statement implicitly assuming that the the object (such as a manifold) as a loose definition of geometry, such as having a metric defined on it, however not specifying the nature of it at all (not specifying the nature of the metric or additional structure)?! – Will Dec 08 '16 at 15:18