General relativity and curvilinear coordinates

["Don't panic!"]

But when you say "the geometry is Euclidean in the local neighborhood around each point", that sounds like it's different from "a manifold that is non-Euclidean at every point". So I'm not sure what you're trying to say.

Sorry, I didn't word this very well. What I meant really was that on a curved manifold, such as a 2-sphere, in a text that I've read it said that Cartesian coordinates can only be constructed in an infinitesimal neighbourhood of each point on the manifold. Is it possible then with more general coordinates to describe larger (finite) neighbourhoods around each point on the manifold? (using spherical polar coordinates one can cover a 2-sphere with two coordinate charts, right?)

 

[PeterDonis]

in a text that I've read it said that Cartesian coordinates can only be constructed in an infinitesimal neighbourhood of each point on the manifold.

Correct. How small "infinitesimal" is, in practice, depends on how accurate your measurements are.

Is it possible then with more general coordinates to describe larger (finite) neighbourhoods around each point on the manifold

Of course. Isn't that what we've been saying all along?

(using spherical polar coordinates one can cover a 2-sphere with two coordinate charts, right?)

Yes.

 

["Don't panic!"]

Of course. Isn't that what we've been saying all along?

I think what has confused me earlier is that we discussed how points on a manifold could be locally described by n-tuples in \mathbb{R}^{n} and we can do this because the manifold is locally homeomorphic to Euclidean space, but to me this implies that the manifold is locally flat and so Cartesian coordinates are applicable. However, if we do consider larger patches that are curved (in the sense that the geometry over them is non-Euclidean) how can we still describe the points in terms of n-tuples of coordinates in \mathbb{R}^{n}?

 

[stevendaryl]

I think what has confused me earlier is that we discussed how points on a manifold could be locally described by n-tuples in \mathbb{R}^{n} and we can do this because the manifold is locally homeomorphic to Euclidean space, but to me this implies that the manifold is locally flat and so Cartesian coordinates are applicable. However, if we do consider larger patches that are curved (in the sense that the geometry over them is non-Euclidean) how can we still describe the points in terms of n-tuples of coordinates in \mathbb{R}^{n}?

Well, take the example of the surface of the Earth. The "patch" that is the entire Earth except for the north and south poles can be described by the pair of coordinates:

(latitude, longitude).​

So it's a 2-D manifold described by two coordinates. It's certainly non-Euclidean, though.

Here is perhaps a way to think about it: You have a two-dimensional space. That means that the points on that space can be described by two real numbers: (x,y). For simplicity, let me assume that it is Riemannian, which just means that the relevant notion of "length" of a curve through space is always positive. Let \mathcal{P}_0 and \mathcal{P}_1 be two points that are close enough together that there is a unique minimal-distance path connecting the two. Let the coordinate of the two point be written as:

\mathcal{P}_0 = (x, y)
\mathcal{P}_1 = (x+\delta x, y + \delta y)

Let \delta s be the distance between the points (as measured along the unique minimal-distance path).

\delta s^2 can be expressed as a double power series in \delta x and \delta y as follows:

\delta s^2 = A_{xx} \delta x^2 + A_{xy} \delta x \delta y + A_{yx} \delta y \delta x + A_{yy} \delta y^2
+ B_{xxx} \delta x^3 + B_{xxy} \delta x^2 \delta y + ...
+ C_{xxxx} \delta x^4 + C_{xxxy} \delta x^3 \delta y + ...
+ ...

The first line groups together the terms that are second-order in the differences \delta x, \delta y. The second line are the terms of 3rd order, the third line are the terms of 4th order, etc.

So far, that's true for any 2D patch, and for any coordinate system on that patch. We can define the coordinate system to be "locally cartesian at point \mathcal{P}_0" if

1. A_{xx} = A_{yy} = 1
2. A_{xy} = A_{yx} = 0
3. For all i, j, k, B_{ijk} = 0

For every 2-D patch, and for every point on that patch, there is a coordinate system that is locally cartesian at that point.

For the coordinate system to be globally cartesian within the patch, it would have to be the case that all the terms are zero except for the first line (only 2nd order terms).

 

["Don't panic!"]

Well, take the example of the surface of the Earth. The "patch" that is the entire Earth except for the north and south poles can be described by the pair of coordinates:
(latitude, longitude).So it's a 2-D manifold described by two coordinates. It's certainly non-Euclidean, though.

Thanks for your detailed explanation.
I think my confusion is, if the manifold is non-Euclidean, and we consider a patch that is large enough that it's geometry is non-Euclidean, when talks of assigning coordinates to points in this patch, is it that we can map the points to n-tuples in \mathbb{R}^{n} with respect to some coordinate map (defining a coordinate system) and not necessarily that the geometry has to Euclidean in order to do this?

 

[PeterDonis]

if the manifold is non-Euclidean, and we consider a patch that is large enough that it's geometry is non-Euclidean, when talks of assigning coordinates to points in this patch, is it that we can map the points to $n$-tuples in $\mathbb{R}^{n}$ with respect to some coordinate map (defining a coordinate system) and not necessarily that the geometry has to Euclidean in order to do this?

Correct. Euclidean geometry means the coordinate n-tuples plus a particular metric, the Euclidean one. Non-Euclidean geometry means the coordinate n-tuples plus a different metric, a non-Euclidean one.

 

["Don't panic!"]

Correct. Euclidean geometry means the coordinate n-tuples plus a particular metric, the Euclidean one. Non-Euclidean geometry means the coordinate n-tuples plus a different metric, a non-Euclidean one.

So in saying that a manifold is locally homeomorphic to \mathbb{R}^{n} is this the statement that each point on a manifold can be represented as an n-tuple in \mathbb{R}^{n}, however in general it is not possible to construct coordinate maps in such a way that they cover the whole of a manifold, but can cover a patch of it (hence are local). The coordinate maps that map each point within a given patch on the manifold to an n-tuple in \mathbb{R}^{n} will in general be non-Cartesian as the geometry within the patch that they are describing will be generally non-Euclidean. Despite this, the geometry in a sufficiently small neighbourhood around each point in a given patch will be Euclidean and thus one can construct local Cartesian coordinates within such a neighbourhood.

 

[PeterDonis]

So in saying that a manifold is locally homeomorphic to $\mathbb{R}^{n}$

"Homeomorphic" is a topological term, not a geometric term; it doesn't include the metric. A manifold meets a stronger condition: the metric is locally Euclidean.

is this the statement that each point on a manifold can be represented as an $n$-tuple in $\mathbb{R}^{n}$

Not exactly. "Locally homeomorphic" means that any open neighborhood of any point can be assigned coordinates in this way, i.e., that we can set up a one-to-one mapping between the points in any open neighborhood and $n$-tuples of real numbers. But this in itself says nothing about the metric; a metric doesn't even have to exist. Manifolds in physics always have metrics, so they meet stronger conditions than just this one, as above.

however in general it is not possible to construct coordinate maps in such a way that they cover the whole of a manifold, but can cover a patch of it (hence are local).

This is because, in general, there might not be an open neighborhood of every point that covers the entire manifold. For example, there is no open neighborhood of any point on a 2-sphere that covers the entire 2-sphere (there can't be, since the 2-sphere as a whole is closed).

The coordinate maps that map each point within a given patch on the manifold to an $n$-tuple in $\mathbb{R}^{n}$ will in general be non-Cartesian as the geometry within the patch that they are describing will be generally non-Euclidean.

Yes, but, as above, this is a stronger condition than just homeomorphism, because it involves the metric. A 2-sphere and an irregular blob with a closed 2-dimensional surface are homeomorphic (and both are locally homeomorphic to $\mathbb{R}^n$ ), but they do not have the same metric (i.e., geometry).

 

["Don't panic!"]

Ah ok, I think I'm starting to get it a bit more now. So the homeomorphism guarantees that we can assign n-tuples of real numbers to each point and then the metric on a given patch determines the type of coordinate maps we can construct from that patch to \mathbb{R}^{n} (by this I mean that if the patch is small enough that the metric is Euclidean then we can use Cartesian coordinate maps, as well as others. However, if the patch is such that the metric is non-Euclidean within it then it will not be able to construct Cartesian coordinate maps (apart from infinitesimally close to each point), however other (more general) coordinate maps will be possible).

 

[Ibix]

I'm not sure if this will help (maybe I'm teaching my grandmother this), but...

One can cover (nearly) the whole of the surface of a cylinder with coordinates (φ,z). One can then simply perform x=φ, y=z and draw a chart on the x-y plane. There is a one-to-one correspondence between the points (φ,z) on the cylinder and the points (x,y) on the chart (there's a bit of a problem over how to represent φ=0 or 2π, but we'll ignore that for the time being). Furthermore, distances are preserved by the map - if I move in a straight line from (φ₀,z₀) to (φ₀+dφ,z₀+dz) then (assuming I was careful about the choice of scale for z) the distance $\sqrt{d\phi^2+dz^2}$ on the cylinder and the distance $\sqrt{dx^2+dy^2}$ on the chart are equal (and that holds as d → Δ).

One can also cover (nearly) the whole surface of a 2-sphere with coordinates (φ,z). Once again one can simply perform x=φ, y=z and draw a chart on the x-y plane (this is not quite a Mercator projection - not sure if it has a proper name). There is a one-to-one correspondence between the points (φ,z) on the sphere and the points (x,y) on the chart (although there are now problems at the poles as well as φ=0 or 2π, but again we'll ignore that). However, in this case distances are not preserved. In general, $\sqrt{d\phi^2+dz^2} \neq \sqrt{dx^2+dy^2}$ (however careful I was about the choice of z). You can easily see this by considering a half circle centered on the north pole (a full circle isn't legit since we'd have to cross the coordinate singularity in our naive choice of map). The path length on the sphere gets arbitrarily small as z tends towards 1; on the map it is π whatever the value of y.

The difference between the two cases is because the cylinder comes equipped with a Euclidean metric while the sphere does not. The cylinder has no intrinsic curvature while the sphere does.

However - look again at the sphere on the equator. Distances up the lines of constant longitude are pretty much correct close to the equator (they'd be perfect if we'd picked a Mercator projection) and distances along the equator are spot on, since the total length of the equator is 2πR in both the map and the sphere. So locally we have a very close approximation to a Euclidean metric. As long as we don't stray too far from the equator the map we've drawn is fine. It's only if we go too far away that we start to notice distortion.

For the manifolds we work with in GR we can always find a way to project the manifold onto ℝⁿ so that there is a "small" region like the equatorial belt in the sphere example around the point of interest. In the sphere example, all we need to do is move the pole to 90° away from our point of interest and hey presto - this is effectively what a street atlas of your town has done. How small is small depends on how strongly curved the manifold is and how precise we need to be. But if there is curvature, inevitably there comes a point where "let's pretend it's Euclidean" will have your GPS driving you into walls...

 

[PeterDonis]

So the homeomorphism guarantees that we can assign n-tuples of real numbers to each point and then the metric on a given patch determines the type of coordinate maps we can construct from that patch to $\mathbb{R}^{n}$

Not quite. An assignment of n-tuples of real numbers to each point is a coordinate map from the patch to $\mathbb{R}^{n}$. The geometry of the manifold determines the form of the line element $ds^2$ when expressed in terms of that coordinate map.

 

[Cruz Martinez]

So in saying that a manifold is locally homeomorphic to \mathbb{R}^{n} is this the statement that each point on a manifold can be represented as an n-tuple in \mathbb{R}^{n}, however in general it is not possible to construct coordinate maps in such a way that they cover the whole of a manifold, but can cover a patch of it (hence are local). The coordinate maps that map each point within a given patch on the manifold to an n-tuple in \mathbb{R}^{n} will in general be non-Cartesian as the geometry within the patch that they are describing will be generally non-Euclidean. Despite this, the geometry in a sufficiently small neighbourhood around each point in a given patch will be Euclidean and thus one can construct local Cartesian coordinates within such a neighbourhood.

This is entirely correct, except I wouldn't say "represented", I would say "mapped" or "labeled". Also the points in the manifold are not mapped to n-tuples of R^n, but to single elements of R^n, which are n-tuples of numbers, this is just a small correction to what you said.

 

[Cruz Martinez]

Not exactly. "Locally homeomorphic" means that any open neighborhood of any point can be assigned coordinates in this way, i.e., that we can set up a one-to-one mapping between the points in any open neighborhood and $n$-tuples of real numbers. But this in itself says nothing about the metric; a metric doesn't even have to exist. Manifolds in physics always have metrics, so they meet stronger conditions than just this one, as above.

Locally homeomorphic to R^n means that around each point we can find a neighborhood of the point and a homeomorphism from the neighborhood to R^n. This doesn't imply we can do this for any neighborhood in the manifold, only that we can always find one such neighborhood around each point.
Also, manifolds in physics don't always have a metric, what about phase-space in classical mechanics?

This is because, in general, there might not be an open neighborhood of every point that covers the entire manifold. For example, there is no open neighborhood of any point on a 2-sphere that covers the entire 2-sphere (there can't be, since the 2-sphere as a whole is closed).

The whole 2-sphere is also open in the subspace topology, in fact it is clopen. The fact that a set is closed in some topology doesn't mean it isn't open as well.

 

[PeterDonis]

This doesn't imply we can do this for any neighborhood in the manifold

Can you give an example of a manifold that has an open neighborhoods which doesn't have a homeomorphism to $\mathbb{R}^n$?

manifolds in physics don't always have a metric, what about phase-space in classical mechanics?

Hm, good point. You can define a metric on phase space, but it won't be unique, and AFAIK it isn't really used for anything. There is, however, a unique volume form on phase space, so the notion of "phase space volume" has a well-defined meaning, even though "phase space distance" doesn't.

The whole 2-sphere is also open in the subspace topology, in fact it is clopen.

The subspace topology on a 2-sphere isn't the relevant topology for this discussion.

 

[micromass]

Can you give an example of a manifold that has an open neighborhoods which doesn't have a homeomorphism to $\mathbb{R}^n$?

Manifold: the circle $S^1$. The open neighborhood: the entire circle $S^1$.

The subspace topology on a 2-sphere isn't the relevant topology for this discussion.

Then which one is, because your statement confused me.

 

[Cruz Martinez]

Can you give an example of a manifold that has an open neighborhoods which doesn't have a homeomorphism to $\mathbb{R}^n$?

Yeah, in fact we are talking about one, the 2-sphere, all of it is a neighborhood of any of its points, which doesn't have a homeomorphim to R^2. The whole 2-sphere is a subset of R^3 which has the subspace topology, ths topology is locally euclidean, this is what makes it a 2-manifold. The other two conditions, second countability and the Hausdorff axiom are true a fortiori by virtue of it being a subset of R^n.

The subspace topology on a 2-sphere isn't the relevant topology for this discussion.

It actually is the topology which makes the study of the 2-sphere as a 2-manifold relevant, as implied
above.

EDIT: I mean locally euclidean as in locally homeomorphic to R^n.

 

[PeterDonis]

the 2-sphere, all of it is a neighborhood of any of its points

But not an open neighborhood. I specified open neighborhoods, since that's what is specified in the definition of a manifold.

The whole 2-sphere is a subset of R^3 which has the subspace topology, ths topology is locally euclidean, this is what makes it a 2-manifold.

No, it isn't. What makes it a 2-manifold is that there is a homeomorphism from any open neighborhood of the 2-sphere to $\mathbb{R}^2$. That fact is independent of any embedding of the 2-sphere into $\mathbb{R}^3$ or any other space; you don't need to use or even know about the embeddings to construct the homeomorphisms.

It actually is the topology which makes the study of the 2-sphere as a 2-manifold relevant, as implied
above.

No; what is "implied above" is incorrect, as I said earlier in this post. The relevant topology for the study of the 2-sphere as a 2-manifold, at least in the context of this discussion (see below for why), is the intrinsic topology induced by the usual metric on the 2-sphere, which is given by the line element $ds^2 = R^2 \left( d\theta^2 + \sin^2 \theta d\phi^2 \right)$ in the usual spherical coordinates. An open set in this topology is any set of points whose distance $s$ from a chosen point, as given by the above metric, is less than some chosen value $D$, with the additional condition that $D$ must be less than $\pi R$, where $R$ is the number appearing in the line element (this is to ensure that the "antipodal point" of the chosen point is not included in any open set).

The reason why the intrinsic topology I just described is the one that's relevant for this discussion is that, as I said in a previous post, when we start talking about the geometry of spacetime, which was what originally started this thread, we don't have the option of viewing it extrinsically, because we don't have any way of going "outside" spacetime and observing how it is embedded in any higher dimensional space (as far as we can tell, no such space even exists). The only way we can study the geometry of spacetime is intrinsically, using measurements made within the manifold. So if we're going to use the 2-sphere as a "warmup" for doing this, we need to make sure we don't use any extrinsic facts about it like its embedding into $\mathbb{R}^3$, since we can't do that for spacetime.

 

[micromass]

But not an open neighborhood. I specified open neighborhoods, since that's what is specified in the definition of a manifold.

The entire space is always open. It's an axiom of topology.

No; what is "implied above" is incorrect, as I said earlier in this post. The relevant topology for the study of the 2-sphere as a 2-manifold, at least in the context of this discussion (see below for why), is the intrinsic topology induced by the usual metric on the 2-sphere, which is given by the line element $ds^2 = R^2 \left( d\theta^2 + \sin^2 \theta d\phi^2 \right)$ in the usual spherical coordinates. An open set in this topology is any set of points whose distance $s$ from a chosen point, as given by the above metric, is less than some chosen value $D$, with the additional condition that $D$ must be less than $\pi R$, where $R$ is the number appearing in the line element (this is to ensure that the "antipodal point" of the chosen point is not included in any open set).

And in this intrinsic topology, the entire 2-sphere is open and closed. By the way, this intrinsic topology is the same as the topology by looking at it as a subspace of $\mathbb{R}^3$.

 

[PeterDonis]

The entire space is always open. It's an axiom of topology.

Drat, you're right. And in fact, the entire space is always clopen, correct? (Since the empty set is also always open, and the complement of any open set is closed. So the empty set is also always clopen.)

in this intrinsic topology, the entire 2-sphere is open and closed. By the way, this intrinsic topology is the same as the topology by looking at it as a subspace of $\mathbb{R}^3$.

I see I'll have to spend some time working through this. But I still think it should be emphasized that the embedding of the 2-sphere in $\mathbb{R}^3$ is not relevant for this discussion, for the reasons given in my previous posts.

 

[Cruz Martinez]

But not an open neighborhood. I specified open neighborhoods, since that's what is specified in the definition of a manifold.

The whole sphere is an open neighborhood.

No, it isn't. What makes it a 2-manifold is that there is a homeomorphism from any open neighborhood of the 2-sphere to $\mathbb{R}^2$. That fact is independent of any embedding of the 2-sphere into $\mathbb{R}^3$ or any other space; you don't need to use or even know about the embeddings to construct the homeomorphisms.

Indeed, you don't need to know about the embedding, however the definition of the 2-sphere uses an embedding. In fact the topologies would be identical anyway.

 

[PeterDonis]

the definition of the 2-sphere uses an embedding

The usual definition does, yes. But you could construct a definition that did not; just use the one I gave in my previous post (but without the restriction $D < \pi R$, which, as you and micromass have established, is not needed since the entire space is an open set).

the topologies would be identical anyway.

Yes, agreed. I was confused about that before.

 

[&quot;Don&#039;t panic!&quot;]

I've been reading Schutz's book "Geometrical methods of mathematical physics" in which he states that a manifold is such that it's local topology is identical to \mathbb{R}^{n} which allows as to locally map points on the manifold to points in \mathbb{R}^{n} in a one-to-one fashion. Obviously we don't want to restrict the global topology to \mathbb{R}^{n}, as this would effectively restrict us to only Euclidean geometry. I'm not to confident in my understanding of topology, so when he talks of the local topology being the same as \mathbb{R}^{n} is he simply meaning that the open subsets are "constructed" in the same way locally?
Before we introduce a measure of distance and curvature then there is no notion of shape on the manifold and so is it correct to say that we can choose any coordinates we like, regardless of the size of the patch we're considering to label points within that patch? Is it only when one introduces the notion of distance, through a metric, that curvature comes into play (through the curvature of the metric - Riemann curvature tensor), and in doing so introduces non-Euclidean geometry to the manifold thus restricting when we can use Cartesian coordinates (i.e. only when the geometry is locally Euclidean), and if the geometry on the patch is non-Euclidean then Cartesian coordinate maps simply don't exist and we must use other more general coordinate systems?

 

[micromass]

I've been reading Schutz's book "Geometrical methods of mathematical physics" in which he states that a manifold is such that it's local topology is identical to \mathbb{R}^{n} which allows as to locally map points on the manifold to points in \mathbb{R}^{n} in a one-to-one fashion. Obviously we don't want to restrict the global topology to \mathbb{R}^{n}, as this would effectively restrict us to only Euclidean geometry. I'm not to confident in my understanding of topology, so when he talks of the local topology being the same as \mathbb{R}^{n} is he simply meaning that the open subsets are "constructed" in the same way locally?

It means that around every point in the manifold, there is some open set which is homeomorphic to $\mathbb{R}^n$. We usually want smooth manifolds, which means that we additionally have that the homeomorphisms are smoothly compatible.

Before we introduce a measure of distance and curvature then there is no notion of shape on the manifold and so is it correct to say that we can choose any coordinates we like, regardless of the size of the patch we're considering to label points within that patch?

The shape of the manifold is partially already determined by the topology and the smooth structure. But this shape can vary wildly. For example, in topology, a donut is the same as a coffee cup. So in some sense the shape is preserved in the sense that there is always one "hole". But since we have no metric, the actual distances won't be preserved. If we add the notion of a metric, then the distances are preserved.

Is it only when one introduces the notion of distance, through a metric, that curvature comes into play (through the curvature of the metric - Riemann curvature tensor), and in doing so introduces non-Euclidean geometry to the manifold thus restricting when we can use Cartesian coordinates (i.e. only when the geometry is locally Euclidean), and if the geometry on the patch is non-Euclidean then Cartesian coordinate maps simply don't exist and we must use other more general coordinate systems?

Non-Euclidean geometry is essentially a property of the metric. So we can't have non-Euclidean geometry without a metric.

Also, strictly speaking, the curvature is not a property of the metric, but of the connection. But of course we can always choose a good connection associated with the metric (the Levi-Civita connection).

 

[&quot;Don&#039;t panic!&quot;]

It means that around every point in the manifold, there is some open set which is homeomorphic to Rn\mathbb{R}^n. We usually want smooth manifolds, which means that we additionally have that the homeomorphisms are smoothly compatible.

What exactly is the topology of \mathbb{R}^{n}?

Non-Euclidean geometry is essentially a property of the metric. So we can't have non-Euclidean geometry without a metric.

I was really trying to give a good motivation as to when one would want to/ be forced to use non-Cartesian coordinate systems to map points on a patch of the manifold to points in \mathbb{R}^{n}? Would what I put in the post before yours be correct? Sorry to harp on so much about this point, but I just feel it's a rather crucial stepping stone towards me gaining a better understanding of the subject.

 

[micromass]

What exactly is the topology of \mathbb{R}^{n}?

The collection of open subsets of $\mathbb{R}^n$. A subset $G$ of $\mathbb{R}^n$ is open if every point $p\in G$ contains an open ball $B(p,\varepsilon)\subseteq G$. The open ball is defined as $B(p,\varepsilon) = \{x\in \mathbb{R}^n~\vert~d(x,p)<\varepsilon\}$.

I was really trying to give a good motivation as to when one would want to/ be forced to use non-Cartesian coordinate systems to map points on a patch of the manifold to points in \mathbb{R}^{n}? Would what I put in the post before yours be correct? Sorry to harp on so much about this point, but I just feel it's a rather crucial stepping stone towards me gaining a better understanding of the subject.

We use non-Cartesian coordinate system because sometimes Cartesian coordinate systems do not exist on the manifold.

 

[&quot;Don&#039;t panic!&quot;]

The collection of open subsets of Rn\mathbb{R}^n. A subset GG of Rn\mathbb{R}^n is open if every point p∈Gp\in G contains an open ball B(p,ε)⊆GB(p,\varepsilon)\subseteq G. The open ball is defined as

Doesn't this rely on the notion of a distance between points though?

We use non-Cartesian coordinate system because sometimes Cartesian coordinate systems do not exist on the manifold.

Why exactly is this the case though?
Is this to do with the geometry being non-Euclidean, or just that the topology is such that Cartesian coordinate maps are not possible?

 

[micromass]

Doesn't this rely on the notion of a distance between points though?

Yes, the distance function on $\mathbb{R}^n$ induces the topology. Many other distance functions also induces that same topology.

Why exactly is this the case though?
Is this to do with the geometry being non-Euclidean, or just that the topology is such that Cartesian coordinate maps are not possible?

It has to do with the geometry being non-Euclidean.

 

[PeterDonis]

Obviously we don't want to restrict the global topology to $\mathbb{R}^{n}$, as this would effectively restrict us to only Euclidean geometry.

Not quite; there are other geometries with the same underlying topology, at least in the 4-d spacetime case--for example, a flat or open FRW spacetime. But there are many other geometries that are studied in GR that have different underlying topology.

 

[PeterDonis]

the distance function on $\mathbb{R}^n$ induces the topology.

But note that this "distance function", in the case of spacetime, may not be the same as the one given by the metric that is used to determine the spacetime geometry. For example, the topology on spacetime distinguishes distinct points along a null worldline; it is induced by a distance function that gives different, nonzero values for different pairs of points. But the metric that determines the spacetime geometry gives zero "spacetime distance" along a null worldline.

 

[micromass]

But note that this "distance function", in the case of spacetime, may not be the same as the one given by the metric that is used to determine the spacetime geometry. For example, the topology on spacetime distinguishes distinct points along a null worldline; it is induced by a distance function that gives different, nonzero values for different pairs of points. But the metric that determines the spacetime geometry gives zero "spacetime distance" along a null worldline.

Right, but this is a bit complicated. The spacetime has an a priori topology, and as such the spacetime is covered with patches that are homeomorphic to $\mathbb{R}^4$ with the usual Euclidean topology. The tangent spaces of the spacetime on the other hand are $\mathbb{R}^4$ with the Minkowski-inner product (and thus with a special topology). This induces a second topology on the spacetime which is not the same as the a priori topology.