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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.

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  • $\begingroup$ Simple example: let $M=S^1$ be a circle. Local coordinates give a continuous, injective map from some piece $U$ of $S^1$ into $\mathbb{R}$. But we need more than coordinate patch, since $U$ can't be all of $S^1$: for continuity, the map must come back to the same point if you go all the way round the circle, so it can't possibly be injective. At best, $U$ can be the circle minus a single point. We need another patch to cover that point. $\endgroup$ – Holographer Jan 21 '15 at 14:42
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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.

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  • $\begingroup$ 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. $\endgroup$ – Heaviside Nov 24 '14 at 4:09
  • $\begingroup$ @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 $\mathbb{R}^N$ .... $\endgroup$ – WetSavannaAnimal Nov 24 '14 at 4:50
  • $\begingroup$ @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. $\endgroup$ – WetSavannaAnimal Nov 24 '14 at 4:51
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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.

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  • $\begingroup$ I do appreciate the references. I do have both. I have looked fairly carefully at Isham, but Spivak's is like trying to drink from a fire hydrant! And I'm not sure you have responded to my question. Physically, each point in space should be mappable to an $n$-tuple. This establishes a coordinate system, doesn't it? The mapping then allows us to compute a metric tensor. This tensor, being symmetric, can be diagonalized at each point, though perhaps not at all points. $\endgroup$ – Heaviside Nov 24 '14 at 3:57
  • $\begingroup$ Problems editing! $\endgroup$ – Heaviside Nov 24 '14 at 4:01
  • $\begingroup$ @Heaviside, the point is that whenever you have a coordinate system $x: U\subset M\to \mathbb{R}^n$, this map $x$ turns out to be a diffeomorphism (a structure preserving map on the category of smooth manifolds). Now, if you have $U = \mathbb{R}^n$ this would imply $M$ to be just $\mathbb{R}^n$ itself. This rules out many interesting topologies and smooth structures. Now, indeed at each point you can map to $\mathbb{R}^n$, but if you create a global map from these maps, the resulting thing won't be structure preserving (although bijective). $\endgroup$ – user1620696 Nov 24 '14 at 13:05
  • $\begingroup$ A coordinate system must be bijective, but it must also preserve the structure present, and in most of cases this patching I understood you wan't to build doesn't preserve it. Now, the metric tensor isn't something you compute from coordinates. It is (like an inner product on a vector space) something you postulate as additional structure. What do you really want to do with it? I think I didn't understand yet. $\endgroup$ – user1620696 Nov 24 '14 at 13:07
  • $\begingroup$ Well, in the purely physical problem of describing spacetime one does not begin with structure on the manifold-but allows it to unfold consequent upon physical postulates. The postulates themselves are phrased in terms of measurable quantities (coordinates). The definition of the metric then arises from the definition of generalized coordinates, say $\bar{y}=\phi(\bar{x})$ as $g_{ij}(\bar{x})=\partial_i\bar(y)\cdot\partial_j\bar(y)$ Problem is reconciling this with the conventional exposition of manifold theory. $\endgroup$ – Heaviside Nov 24 '14 at 15:56
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As a clarifying counterexample, consider any compact smooth surface in $\mathbb{R}^3$, such as sphere.

If there was any global coordinate patch for this surface, then by the continuity of the inverse map, the unit open disk would be compact a compact topological space since continuos maps preserve compactness; a contradiction since unit open ball is not compact.

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