# Coordinate patches on curved manifolds

1. Feb 16, 2013

### joda80

Dear All,

I've been studying differential geometry for some time, but there is one thing I keep failing to understand. Perhaps you can help out (I think the question is quite simple):

Can I use Cartesian coordinates to cover a curved manifold? I.e., is there an atlas that only contains Cartesian coordinate patches (I envision small squares covering a sphere, which should work if the squares overlap).

On the one hand I think this should be possible as the charts are allowed (and indeed, supposed) to overlap.

On the other hand, in classical diff.-geo. text books (not using manifolds), it says that no Cartesian coordinates are possible on curved manifolds.

How can these two ideas be reconciled?

Thanks a lot and best regards,

Johannes

2. Feb 16, 2013

### WannabeNewton

In general, a smooth manifold M has no single global chart i.e. a one element atlas $\left \{ (M,\varphi ) \right \}$ such that $\varphi :M\rightarrow \mathbb{R}^{n}$ is a homeomorphism. The classic example is $S^{2}$ which needs at least two charts (e.g. stereographic projection). However, given an arbitrary smooth manifold M, a maximal smooth atlas, and a smooth coordinate chart $(U,\varphi )$ in the atlas, a cartesian coordinate representation $\varphi (p) = (x^{1},...,x^{n})\in \mathbb{R}^{n}, p\in U$ is certainly a possibility.

3. Feb 16, 2013

### TrickyDicky

If you specifically refer to curved manifolds, they certainly cannot be covered globally with one set of Cartesian coordinates, but that's different from covering the curved manifold with local overlapping charts(not necessarily Cartesian) wich is what defines all smooth manifolds.
See the difference global versus local?

4. Feb 16, 2013

### joda80

Thanks for the responses! I understand. The motivation for the question was the role of "general coordinate transformations" in GR. The way I understand it is as follows:

We can use the local Cartesian coordinates to define tangent vectors, based on which we define the tangent space and tensors living in this space. The Cartesian patch and the associated tangent space represent the local inertial system. To obtain the covariant equations, we merely need to change the coordinates within this tangent space to something where the metric tensor is coordinate dependent. This includes coordinate systems that are accelerated and, by the equivalence principle, contain the effects of gravity. Now we take the SR equations and make sure they have the desired transformation properties. If they do, we have the laws that include the effect of gravity. Spacetime curvature isn't explicitly considered here at that stage. All these considerations could also be done in the Minkowski space (I think). The question is then, how does one infer a curved spacetime in the presence of gravity? (I thought it was because on a curved manifold the metric tensor is *also* coordinate-dependent, just as in the accelerated frames -- I think somewhere here might be my misunderstanding).

5. Feb 16, 2013

### WannabeNewton

Well to start with, tangent vectors and tangent spaces are certainly not defined by using cartesian coordinates. They are coordinate independent constructions (of course). Let $M$ be a smooth manifold and $p\in M$. A derivation at $p$ is a linear map $X:C^{\infty }(M)\rightarrow \mathbb{R}$ that satisfies $X(fg) = f(p)Xg + g(p)Xf,\forall f,g\in C^{\infty }(M)$. We then define the tangent space $T_{p}(M)$ to be the set of all such derivations at $p$ and its elements as tangent vectors. Tensors are more general constructions that are still certainly not constructed explicitly from cartesian coordinates. What do you mean by the statement that making sure SR equations behave covariantly implies they take gravity into account? SR doesn't talk about gravity at all - we work in a globally flat space -time (Minkowski space). In SR we want to make sure equations are lorentz covariant; GR generalizes this vastly of course. If gravitational tidal effects are present then you can take two neighboring geodesics, connect them with a so called connecting / separation vector and see how they converge or diverge from one another by measuring the relative acceleration of one of the neighboring geodesics with respect to the other. This will allow you to measure the curvature by virtue of the equation of geodesic deviation.

Last edited: Feb 16, 2013
6. Feb 17, 2013

### joda80

I don’t seem to formulate very well where the problem lies.

So let me back up a bit: The way I understood the gist of general relativity (see e.g., Weinberg's approach in Gravitation and Cosmology) was that we start out on in flat space, where we know the equations of (SRT) are true. As we seek the equations in curved space (with gravity), we need to construct equations that transform like tensors under general coordinate transformations and yet reduce to the known SRT equations if the metric is flat. These general coordinates are needed because in curved space, Cartesian coordinates are impossible. E.g., we could take the equation of motion in Minkowski space and simply by using the covariant (instead of the ordinary) derivative, obtain the equation that is valid also in the presence of gravity (the geodesic equation). This viewpoint seems to be taken in Weinberg's Gravitation and Cosmology (e.g., pp. 105-106). Or do I misinterpret that?

If I didn't, then how do we embed these ideas in the context of manifolds? If I'm free to choose either Cartesian or any other coordinate system in my tangent spaces, then mathematically what is the difference between the local inertial system (=tangent space), an accelerated system (=tangent space with "general" coordinates, as in a rotating frame), and a system on a curved manifold with "true" gravity (which we can analyze only locally, i.e., also in a flat tangent space)? So if we always stay in flat tangent spaces with arbitrary coordinates, the role of general coordinate transformations isn't quite clear to me (it would be, if we could go from flat to curved space, but as we're stuck with tangent spaces, this isn't really possible ... or is it?).

It's amazing (and a little frustrating) how much you can read and presumably learn about this stuff and still miss some very basic notions... :-\$

7. Feb 17, 2013

### WannabeNewton

My first advice would be to not think of everything from the basis of cartesian coordinates. Yes what Weinberg is saying is correct of course: we want equations that are made up of covariant derivatives and tensor quantities so that if a physically relevant expression equals zero in one coordinate system, it equals zero in all coordinate systems. It is a mathematical way of stating that coordinates are nothing more than computational tools and that geometric objects exist independent of what coordinate basis they are represented in. So we have (1) general covariance and (2) the statement that if the lorentz metric is flat then such equations should reduce to equations satisfied in SR. You are also mixing up tangent spaces with coordinate charts. They are not the same thing. A coordinate chart consists of an open subset of the manifold and a homeomorphism to an open subset of euclidean space whereas a tangent space at a point is as defined in the previous post. Also I'm totally lost on your last few sentences because you are using tangent spaces in a wrong way (e.g. "flat" tangent space?). If you could clarify that last paragraph I would be much indebted. Thanks.

8. Feb 17, 2013

### WannabeNewton

Just for some clarification, is your confusion one stemming from the physics or the math? If it is the latter as I suspect it is then I would suggest taking a look at proper math textbooks to learn the various differential geometry concepts because physics books tend to butcher these kinds of things and leave the reader confused.

9. Feb 17, 2013

### joda80

Indeed, I'm working with a couple of pure math books aside from physics books ... though mainly with Wald's "General Relativity", which seems to be rather clean/rigorous regarding the mathematical treatment. I think my problem is mostly of mathematical nature. If you're no bored yet, I'll give it another try:

Isn't one (standard?) way to construct a tangent space to use the coordinate lines given by the chart? This is how I interpret Wald's (General Relativity) exposition (p. 17): "Note that the above coordinate basis vector $X_{\mu}$ associated with a chart $\phi$ is the tangent to the curves on $M$ obtained by keeping all coordinate values except $x_{\mu}$ constant"). So the basis of the tangent space is related to the coordinate system. If we were to choose local Cartesian coordinates on the manifold, we would have a Cartesian basis in our tangent space, if we chose some curvilinear coordinates, the basis would no longer be Cartesian (which is fine, of course). But the tangent space is still Euclidean, irrespective of the coordinates used to construct the basis, isn't it? And I thought that "Euclidean space" is equivalent to a "space in which there is no gravity". So if indeed the tangent space is Euclidean, then how can we describe effects of gravity in this space.

10. Feb 17, 2013

### WannabeNewton

Ahhh ok I see your confusion now mate. I first want to refer you to Lee's text on smooth manifolds to clear up later confusions (I absolutely love Wald to death but there are better ways to construct certain things). Let's take it from the top shall we? Let $M$ be a Hausdorff, second countable topological space (second countable so we can employ paracompactness which Wald does himself on occasion as you will see). We say that $M$ is locally euclidean of dimension n if $\forall p\in M$, $\exists U\subseteq M$ open and a map $\varphi :U\rightarrow \mathbb{R}^{n}$ such that $\varphi$ is a homeomorphism. We call $(U,\varphi)$ a coordinate chart and the set of all coordinate charts covering $M$ an atlas. Furthermore, we shall now require that $M$ be a smooth manifold which means the atlas is now required to be a maximal smooth atlas i.e. $\forall (U,\varphi ), (V,\psi )$ in the atlas, $\psi \circ \varphi ^{-1}:\varphi (U \cap V)\rightarrow \psi (U\cap V)$ must be a diffeomorphism (these are called smooth transition maps and the charts are said to be smoothly compatible) and any chart smoothly compatible with all charts in the atlas is already in the atlas (the maximal part). You can certainly define the tangent space the way Wald does it but the way Lee does it is more instructive and makes you see you shouldn't care about the coordinates at all until computations come along.

Now, using notation more along the lines of Wald, for a given metric tensor $g_{ab}$ on our smooth manifold, there exists a unique torsion - free derivative operator $\triangledown _{a}$ such that $\triangledown _{c}g_{ab} = 0$ (you can do a similar thing for derivative operators with torsion c.f. problem 1 of chapter 3). Let $\omega _{c}$ be any dual vector field then we define the riemann curvature tensor by $\triangledown _{a}\triangledown _{b}\omega _{c} - \triangledown _{b}\triangledown _{a}\omega _{c} = R_{abc}^{d}\omega _{d}$. Now let $(M,g_{ab})$ be a space - time. We can describe what the curvature tensor does by showing it is related to how a vector parallel transported around a closed loop fails to return to its initial value and of course for this we need some region on the space - time (the pseudo - riemannian manifold). We do not detect it using nothing but $T_{p}(M)$! What do we mean when we say the space - time is flat i.e. there is no presence of gravity? We simply mean that $R_{abc}^{d} = 0$ identically i.e. there is no curvature. This has nothing to do with the fact that $T_{p}(M)$ is isomorphic to $\mathbb{R}^{n}$. Your only confusion is you are looking at the vector space - structure of the tangent space but when talking about curvature we are concerned with the pseudo - riemannian and smooth structure of the space - time.

11. Feb 17, 2013

### TrickyDicky

To expand on the clear explanations by WN and since I had typed this that tries to clear other possible misconceptions of the OP:
The tangent space is associated with each point of the manifold, what kind of effects of gravity do you want to describe at a point? There aren't any.
Maybe in this phrase from another of your posts lies most of your confusion: " This includes coordinate systems that are accelerated and, by the equivalence principle, contain the effects of gravity."
Actually coordinate systems don't contain the effects of gravity, because they don't contain any physical effect at all, they're just labels that can always be used locally due to how manifolds are mathematically defined.
Also it is not strictly correct to call tangent spaces Euclidean, in (pseudo)Riemannian manifolds like it is the case in GR, tangent spaces associated to the points of the manifold have an inner product that is not Euclidean in general (it is not in the case of GR). Don't mix this with the property that this space is flat to first order approximation and therefore locally isometric to the manifold's point neighbourhood and to R^n. Note depending on the level of the structure I want to highlight I coud use different terms here, if thinking only of the vector space structure of the tangent space, I'd say isomorphic to R^n, if I was stressing the smooth structure I'd say diffeomorphic to R^n, since I'm stressing the pseudoRiemannian case (you are interested in GR) I say locally isometric to R^n.

Last edited: Feb 17, 2013
12. Feb 18, 2013

### joda80

Thanks, guys! I think I understand (the metric tensor and the curvature are properties of the manifold, not of the tangent spaces; accordingly, the coordinates chosen to describe the tensors merely determine their components, not the tensors themselves, which of course is the purpose of tensors).

Why can’t gravity be defined at a point on a manifold – after all, we’re dealing with differential equations that are valid (only) at a point (the curvature tensor is also a *local* property of the manifold – or isn’t it?).

The statement about the coordinate systems, and in fact, the role of coordinate transformations still confuse me. I understand that physics is independent of the coordinate systems, but isn't it true that an observer in a freely falling elevator in a gravity field observes no gravity, while an observer in an accelerated system in free space does feel gravity (this is what Einstein called the 'happiest thought of his life'). I would suspect that it depends on what quantities we’re considering. A Riemann tensor in spacetime is independent of the coordinates, but the acceleration in ℝ$^3$ is not. This is why we’re interested in Riemann-tensor formulations, after all, because only they have the desired objectivity.

I still have some difficulty conceptualizing the roles of the coordinate transformations and general covariance in the manifold perspective. Per Wald’s Appendix C1, these coordinate transformations are equivalent to maps between two manifolds. I could think of two interpretations:

a) We are in free space, where there is no gravity. Accordingly, the laws of SR hold. Now we do a transformation to an accelerated coordinate system and construct equations that transform in the desired way. The fact that we can associate the coordinate transformation with the effect of gravity is based on the equivalence principle. This is the ‘passive’ interpretation of the local diffeomorphism in Wald’s App. C1.

b) Alternatively, we can take an ‘active’ perspective and map the tensors from a manifold with vanishing curvature to one that is curved, and again expect the tensors to transform in the desired manner (however, we would have to require that the tensors don’t change during the process).

Do any of these interpretations make any sense? (If not, how do you look at these things?).

13. Feb 18, 2013

### WannabeNewton

(b) is wrong; we are concerned with diffeomorphisms and you cannot make the curvature of a riemannian manifold vanish identically with a diffeomorphism. As for your first point, this goes way back to simple calculus in $\mathbb{R}^n$. Let $f:U\rightarrow \mathbb{R}^m$ be a map where $U\subseteq \mathbb{R}^n$ is open. Let $a\in U$ then we say $f$ is differentiable at $a$ if there exists a linear map $D$ such that $lim_{h\rightarrow 0}\frac{f(a + h) - f(a) - Df(a)}{\left \| h \right \|} = 0$. This is just the familiar calculus definition. Do you remember your epsilon delta definition of limits of functions? We required the use of open balls. You can't use nothing but a point and no extra information whatsoever to talk about these things. Manifolds are no different in spirit; local is not exactly a mathematically defined term but it is usually used when talking about properties true for neighborhoods of points e.g. local compactness, local connectedness / path connectedness, local continuity, locally euclidean etc.

EDIT: Read chapter 4 of Wald, it should clear things up for you.

Last edited: Feb 18, 2013
14. Apr 4, 2013

### joda80

Just to follow up: I think I found out what my misunderstanding was. Somehow I thought that the coordinate systems on a manifold were always the Riemann normal coordinates (i.e., valid only in a small neighborhood of a given point). Accordingly, I thought all coordinate systems were describing local inertial frames, and doing transformations from one inertial frame to another didn't seem like what we want in general relativity. But I've now convinced myself, hopefully not in error, that the global* curvilinear coordinates compatible with the curved manifold are just as valid a choice for the coordinates (the definition of the manifold only says that there need to exist open subsets of M that can homeomorphically be mapped into R^n, not that these subsets need to cover only infinitesimal neighborhoods). This should solve my problem of how to fit the global/curvilinear coordinates into the manifold picture.

Cheers ...

* I assume a well-behaved, albeit "warped" spacetime structure, like a warped 2D surface for which there should always be one global curvilinear coordinate system that can be mapped via a homeomorphism into R^2, right? (Well, my point is merely that it doesn't need to be valid only in a small neighborhood of a point.)

15. Apr 4, 2013

### Ben Niehoff

You still need to amend your thinking a bit, to envision coordinate patches that are bigger than infinitesimal neighborhoods, but still smaller than "global".

Consider what it would mean to put global coordinates on some surface. Since the coordinate patch is homeomorphic to R^2, and the coordinate patch covers the entire surface, then the whole surface must be homeomorphic to R^2. This only works if the surface is R^2. If the surface is, say, a sphere, then you'll have to cut a hole in it somewhere to map it onto R^2 (and this violates the definition of what makes a homeomorphism).

But taking the sphere example, the standard longitude and lattitude coordinates cover most of the sphere, except for the north and south poles, and the international date line. These are all points where the definition of "homeomorphism" is violated, because multiple different coordinate values correspond to the same point on the manifold.

So here's a case in which there is a coordinate chart that covers almost all of a manifold, but is still not global. In general, there does not need to exist any global chart. In fact, the existence of a global chart imples that the whole manifold is homeomorphic to some R^n. Hence if the manifold is not homeomorphic to R^n, there cannot be a global chart.

16. Apr 4, 2013

### WannabeNewton

Well first off note that "infinitesimal" neighborhood (by which I assume you mean $\epsilon$ radii open balls) have no meaning on general topological manifolds, there is no notion of a metric unless you impose one. Secondly, a curved manifold will NOT in general have a global chart (i.e. a one element atlas). Even a 2 - sphere, as Ben pointed out, requires AT LEAST two charts (Stereographic projection).

17. Apr 5, 2013

### joda80

Thanks ... I knew I was going to get in trouble for using the word "global" ;) Point taken! Just as a check, though: Let's assume a 2D surface that looks like this:

http://www.bu.edu/tech/files/images/surface.jpg

(from www.bu.edu)

And let's also assume that this image repeats itself periodically in space in x- and y-directions. Wouldn't a curved surface like this have a single curvilinear global system (I think it can uniquely be mapped to R^2).

EDIT: Updated the URL (the original ones didn't wotk for some reason)

Last edited: Apr 5, 2013
18. Apr 5, 2013

### micromass

Staff Emeritus

19. Apr 5, 2013

### WannabeNewton

It doesn't work for me either but I have a good idea of what you are asking and the answer is no it is not that simple. Forget surfaces for a second and just focus on $\mathbb{R}$. Consider a countable infinity of copies of $S^{1}$. Your argument is if we have a way of "gluing" these copies together then it should be homeomorphic to $\mathbb{R}$. One method of going about this "gluing" is by using the wedge product. Pick a base point $p\in S^{1}$ from each copy of $S^{1}$. Then, the wedge sum of the countably infinitely many copies of $S^{1}$ will be the quotient space obtained by collapsing all those base points to a single point (i.e. identifying them all with each other). It can be shown that this quotient space is homeomorphic to the quotient space $\mathbb{R}/\mathbb{Z}$ so no it cannot simply be be homeomorphic to $\mathbb{R}$.

Last edited: Apr 5, 2013
20. Apr 5, 2013

### joda80

I updated the URL in my previous post, sorry it didn't work. I'm not sure if we mean the same thing regarding gluing the surfaces together (though we might -- I never stumbled upon the wegde sum before). This was just my way of saying that the above pattern (see URL) was supposed to repeat itself infinitely many times. So what if we didn't glue anything together, but in 1D simply defined a manifold as

M = sin(x), -infty < x < infty

This manifold is curved, and should be homeomorphic to R. (Likewise, a smooth but wrinkly/rippled 2D surface with infinite extent should be globally homeomorphic to R^2, no?)

Last edited: Apr 5, 2013