Why don't global coordinates always exist for a manifold?

Question

Let $M$ be a manifold and $(\phi,U)$ a patch. Then $\phi(P)=\bar{x}=\begin{bmatrix} x^1\\ x^2\\ \vdots\\ x^n \end{bmatrix}$ for each $P$ in $U$. But each $P$ in $M$ is in some patch, so this representation must hold for each $P$ in $M$. If $M$ is to represent physical space (or spacetime), then each $P$ in $M$ should be represented by a unique $n$-tuple in $R^n$. All differential geometry texts that I have seen go to great lengths at the outset to claim that, in general, there is no global coordinate system for $M$, but that the best one can do is to find local coordinates.

Granted that finding a global coordinate system by piecing the patches together might be difficult, it seems to me that it must be possible if differential geometry is to be useful in applications.

Could someone straighten me out on this issue? Why not simply postulate a global coordinate system and let the coordinate mapping from $M$ to $R^n$ and the resulting metric tensor describe the geometry?

P.S. By $R^n$ I mean the set of all $n$-tuples with the usual definitions of sum and scalar multiple, not Euclidean $n$-space.

This post imported from StackExchange Mathematics at 2015-01-20 11:58 (UTC), posted by SE-user Heaviside

Comments

I think in cases like this a simple example will be clearer than any deep mathematical arguments: just take the $2$-sphere $S^2$. I think it is quite intuitive on sight that there is no sensible way of assigning two real coordinates two every point. Of course you can for example use $\phi$ and $\theta$ (respectively the angle in the xy-plane, and the angle with the z-axis), but you see that at the north and south pole $\phi$ becomes ill-defined. Basically any such attempt will clearly have continuity issues somewhere (and the conceptual reason for this is due to its non-trivial topology, since topology is basically the study of continuity/`connected'-ness, and the parts of a $2$-sphere are connected differently than parts of $\mathbb R^2$). More exactly you are right that you can always find some bijection to a part of $\mathbb R^n$, but for topologically non-trivial spaces such as $S^2$ these bijections cannot be continuous (that is actually the meaning of `topologically non-trivial :)).

Answer 1

Let's look at your last line:

Why not simply postulate a global coordinate system and let the coordinate mapping from $M$ to $\mathbb{R}^n$ and the resulting metric tensor describe the geometry? P.S. By Rn I mean the set of all n-tuples with the usual definitions of sum and scalar multiple, not Euclidean n-space.

$\mathbb{R}^n$ has a particular, simple and fairly boring global topology - general manifolds have many different, much richer topologies. So the co-ordinate mapping you propose cannot be bijective whilst "capturing" this general, global topology. So now we're needfully talking about a global mapping of the manifold to a subset of $\mathbb{R}^N$, which subset has the same topology as that of $M$ (I'm assuming you still want the mapping to be injective - that's the point of co-ordinates after all to be "labels" - so the only way to break bijectivity is with a non-surjective map).

If you're willing to live with this, then your proposed approach can indeed work in theory, through the Whitney Embedding theorem and, more particularly for Riemannian manifolds, the Nash Embedding Theorem. The latter shows that every Riemannian manifold can be isometrically (i.e. preserving the metric tensor) embedded in a higher dimensional Euclidean space. Or Minkowsky space, if the manifold is pseudo-Riemannian (as with the manifolds that are solutions to the Einstein field equations).

The problem with this approach is that it adds a huge amount of unneedful complexity to the mathematics, and, since the higher dimensions are almost always unphysical(e.g. to the best of our knowledge, the universe is not embedded in a higher dimensional flat space), there are a huge number of unphysical degrees of freedom. The key word in these theorems is higher dimensional: the Whitney theorem's proof shows that you might need as much as to double the dimensions of your manifold to get the dimension of the embedding Euclidean space. The Nash theorem, the more useful one for the present discussion, is much "worse". It assumes that, if $m$ is the dimension of your original manifold, then the embedding Euclidean space may need to have a dimension of up to $m\,(3 m+11)/2$ if the manifold compact manifold, or $m\,(m+1)\,(3m+11)\,/\,2$ if noncompact.

An interesting historical fact here is that, before about the 1930s, the most widespread notion of a manifold was as a subset of a higher dimensional space, a hypersurface defined by some constraint equation of the form $F(X)=0$, where $X$ are the Euclidean co-ordinates. For example, it is natural to think of a 2-sphere in this way as the set of points defined by $X\cdot X = r^2$ in $\mathbb{R}^3$. At about this time though, the simpler, but subtler, modern notion of a manifold as a collection of patches each locally like $\mathbb{R}^N$ began to take root. The main point of the Whitney and Nash theorems is then the proof that these two notions are logically the same and that nothing was lost by taking the simpler, modern patch-and-local approach.

This post imported from StackExchange Mathematics at 2015-01-20 11:58 (UTC), posted by SE-user WetSavannaAnimal aka Rod Vance

Comments

Thanks. Yes, I am aware of the embedding theorems, but it seems to me that there should be a bijective map from the manifold to $R^n$---just from a physical point of view. (After all, we are dealing here with physics, not pure math.) It should be possible to assign to each point in space (or spacetime) a unique $n$-tuple of coordinates. Otherwise, physics simply doesn't make sense. And there seems to be no physical reason for additional coordinates in the sense of the Whitney-Nash theorems. And I think that if one doesn't insist on a global metric the embedding theorems are not operative.

This post imported from StackExchange Mathematics at 2015-01-20 11:58 (UTC), posted by SE-user Heaviside

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@Heaviside Well, of course there can be a bijective map, but it will not preserve structure that is important for descriptions in physics. You can simply label each of the charts for example with a real number weave the $n+1$-tuple into an $n$-tuple using something like a higher dimensional version of Hilbert space filling curves. But I'm guessing this is not what you're meaning? Unless you can at least define a topology on your $\mathbb{R}^N$ that matches that of the manifold of interest, you are doomed. And even if you can do that, the topology you have to bestow on $\mathbb{R}^N$ ....

This post imported from StackExchange Mathematics at 2015-01-20 11:58 (UTC), posted by SE-user WetSavannaAnimal aka Rod Vance

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@Heaviside ... to allow this might be pretty weird and alien. However, you seem to know all this so I must be missing something crucial about your question.

This post imported from StackExchange Mathematics at 2015-01-20 11:58 (UTC), posted by SE-user WetSavannaAnimal aka Rod Vance

Answer 2

First of all, the metric tensor is one additional piece of structure one inserts on a smooth manifold to measure lenghts and angles. The metric is indeed not present in all applications of Differential Geometry to Physics (see e.g. Lagrangian Mechanics). In that case, it is important to know also how to deal with manifolds without metric tensors.

Now, about the coordinate systems the point is that indeed usually manifolds require more than one to be covered. Take the two sphere $S^2$ for example, you need at least two stereographic projections to cover it all. The idea, however is not to piece two coordinate systems to get a global one.

The idea is that given a point, around it there is some coordinate system that works and if that you have any overlapping one, you can be sure that the results and definitions don't depend on which coordinate system you use. One more readily seem example is cartesian and spherical coordinates in $\mathbb{R}^3$: you can use any one of them.

If $(x,U)$ and $(y,V)$ are two coordinate systems, on the overlap $U\cap V$, if you make sure results independ on the coordinate system you can think of them as intrinsic to $M$ and yet use coordinates to carry down calculations.

You can't assume that there is just one coordinate system because if you look at examples you find objects which you certainly want to consider as manifolds which cannot be covered by just one coordinate system.

To clarify those points I recommend you take a look in this two books

1. Modern Differential Geometry for Physicists - C. J. Isham
2. A Comprehensive Introduction to Differential Geometry Vol. 1 - Michael Spivak

Book 2 is more technical and it's for mathematicians, but it's very good. I recommend you first look at 1 and then look at some things in 2 to see some more detailed constructions.

Edit: One counterexample might help you out, so I decided to give one. If $M$ is a smooth manifold and if $(x,U)$ is a coordinate system, then $x : U\subset M\to \mathbb{R}^n$ is a homeomorphism. If there exists one global coordinate system $(x,M)$ then $M$ is homeomorphic to $\mathbb{R}^n$ itself. This is problematic because many manifolds one encounters (not just in Math but in Physics as well) have a more complicated topology.

Let's show that the sphere $S^2$ cannot have a global coordinate system.

1. $S^2$ is compact: in truth, $S^2$ is endowed with the subspace topology and because of that, it suffices to show $S^2$ regarded as a subset of $\mathbb{R}^3$ is closed and bounded. Bounded it's easy, if $p\in S^2$ then $|p|=1$, hence $|p|< 2$ so that $S^2\subset B(0, 2)$ where $B(a,r)$ is the ball centered in $a$ with radius $r$. Closed is also easy: $S^2 = \{(a,b,c)\in \mathbb{R}^3 : a^2+b^2+c^2 = 1\}$ so that if we set $f : \mathbb{R}^3\to \mathbb{R}$ by $f(a,b,c)=a^2+b^2+c^2$ then $S^2 = f^{-1}(1)$, but $f$ is continuous and $\{1\}$ is a closed set so that $S^2$ is closed. Since $S^2$ is closed and bounded, $S^2$ is compact.

2. Suppose now that $S^2$ has a global coordinate system $(x,S^2)$ then $x: S^2\to \mathbb{R}^2$ is a homeomorphism, but since $S^2$ is compact, then $\mathbb{R}^2$ is compact which is obviously wrong. So we are forced to conclude $S^2$ has no global coordinate system.

So your procedure gives one $n$-tuple of numbers for each point of the manifold, but it doesn't respect the topological structure. If we suppose a global coordinate system for the sphere, you get an absurd unless you accept coordinates which are not continuous and those are not really interesting.

Now regarding the metric tensor: I'll say again, the metric is not something you deduce from coordinates. The metric is something you postulate. In GR in particular it is the solution to Einstein's Equations. The formula you say about is just a way to relate coordinate representations of the metric tensor in different coordinate systems, not a way to deduce it from coordinates. For example if $M = \{(a,b)\in \mathbb{R}^2 : b > 0\}$ and if you use cartesian coordinates $(x,y)$ you can define $g$ in coordinates by

$$g = \dfrac{dx\otimes dx + dy\otimes dy}{y^2}$$

This is the coordinate representation of $g$ in this coordinate system. If you choose any other coordinates, $g$ will transform it's representation according to the formula you gave. Also, see that I postulated the metric, instead of deducing it.

This post imported from StackExchange Mathematics at 2015-01-20 11:58 (UTC), posted by SE-user user1620696

Comments

Just a bit more elucidation: The text I referred to in my last edit is, in more detail, Modern Geometry, Methods and Applications: Part I, by Dubrovnin, Fomenko, and Novikov. They begin their exposition by assuming a bijective mapping exists between the fundamental space and $R^n$. They begin, of course, to inject topology pretty soon-but only after establishing the coordinate system. Can anyone clarify the difference between this approach and the abstract manifold one?

This post imported from StackExchange Mathematics at 2015-01-20 11:58 (UTC), posted by SE-user Heaviside

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Oh, come on guys! No takers? I am a retired ekectrical engineering professor, and one of the few things I learned in my long career is how hard it is to answer the simple questions. Classes and texts plunge immediately in to the formalism and crank turning because it is so much easier than dealing with the basics! (And makes homework and exams so much easier!) I am fairly competent at mathematical manipulation, but I am looking for some physical motivation here. Clearly, if you can assign $n$ coordinates at each point you are mapping to $R^n$. But what then?

This post imported from StackExchange Mathematics at 2015-01-20 11:58 (UTC), posted by SE-user Heaviside

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I think I'm missing something about your question. It's not that we assume a bijection between the set and $\mathbb{R}^n$ exists, we define a manifold requiring such bijections to exist locally, but we want them to be continuous. Your procedure makes a global one, the problem is that it is rather odd, because it won't be continuous and all of that. I know you don't want to suppose a manifold to be a topological space, but you need a topology to be able to talk about continuity, convergence, and ultimately do calculus on such space.

This post imported from StackExchange Mathematics at 2015-01-20 11:58 (UTC), posted by SE-user user1620696

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Thanks for your kind response. Please understand that I am not heckling here, but truly am trying to understand. My uncertainty has to do with the motivation for the abstract definition. Start with the bijective mapping. We can surely impose the usual topology of $R^n$. For the sphere embedded in $R^3$, this doesn't suffice because distance on the sphere is different. But this is a metric issue. My question is what motivates restriction of the basic mapping to patches (with an eye tiward applications). Thanks. P.S.: Have a look at the text I mentioned above.

This post imported from StackExchange Mathematics at 2015-01-20 11:58 (UTC), posted by SE-user Heaviside

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It's not just a metric issue, though. The $2$-sphere is not homeomorphic to $\mathbb{R}^2$, even if it's embedded in $\mathbb{R}^3$. That's an issue that arises without reference to metrics. It is homeomorphism (an invertible continuous function) that gives us a space which is nice enough to do calculus with (and, technically, transition functions between patches which are diffeomorphisms, but let's ignore that for the moment). Sure we can make a bijection between $\mathbb{R}^2$ and the $2$-sphere but bijections do not have to be nice at all and there's no guarantee we can develop some useful notion of calculus with a space that just bijects with Euclidean space. Homeomorphisms, from the whole space to all of Euclidean space for some fixed dimension, don't exist in general. That is why we use patches which ARE homeomorphic to Euclidean space. So in a small area, a manifold will look like Euclidean space, but the whole manifold need not look like Euclidean space.

The $2$-sphere, as has been mentioned above, is an example for which the space as a whole simply cannot look like Euclidean space topologically, and it turns out smooth structures are built on topologies so it can't look like it smoothly either. Even if you try not to talk about topology in your definition of a manifold (as, I believe, e.g. Carroll does in his GR text), if you're using a definition that reproduces manifolds as we know them, a topology will be induced.

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Let $r:\mathcal{R}\to\mathcal{M}$ be a mapping from the reals to a general manifold. If a global coordinate system does not exist, what happens to the trajectory at the points where the coordinate transformation including a given initial point fails? If there are two points which cannot be included in one coordinate system, then consider a curve passing through both points---.

This post imported from StackExchange Mathematics at 2015-01-20 11:58 (UTC), posted by SE-user Heaviside

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@Heaviside, I wrote an edit about the implications of a global coordinate system and how you can see the sphere cannot have one. I also wrote about the metric tensor. Now I didn't understand your doubt regarding the curve. The curve is independent of coordinates.

This post imported from StackExchange Mathematics at 2015-01-20 11:58 (UTC), posted by SE-user user1620696