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In the context of GTR spacetime, I'm trying to get the basic idea of a Riemannian manifold clear in my mind. Apologies for the longwindedness.

Question 1. Is this a reasonable, simplified summary of the steps needed to define such a manifold?

  1. Start with an amorphous collection of points.

  2. Find the minimum number of parameters needed to locate each point in space. That number will be the dimension of the manifold.

  3. Construct a coordinate system to locate each point in space. I may need more than one coordinate system to to cover the whole manifold. I need a “good” coordinate system so that each point is uniquely described.

  4. Decide on a metric that gives the infinitesimal distance between two adjacent point. The metric need not be the same all over the manifold (as $g_{\mu\upsilon}$ isn't in spacetime). In that case I would have a metric tensor field defined over the manifold.

  5. The manifold has to be continuous and differentiable (not quite sure what this means, but I assume it's the obvious - I have to be able to differentiate it everywhere).

  6. I can now start constructing scalar fields, covariant and contravariant vectors, tensors etc.

Question 2. Assuming I do all the above with a flat plane (a table top for example), using Cartesian coordinates. Can I impose any kind of metric to my new “table top” manifold? Could I just make a weird one up: $$\left(\begin{array}{cc} x^{2} & \sin xy\\ x+y & 13y \end{array}\right)$$

for example, instead of the Euclidean $$\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)$$

Thank you.

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Hint: Since the manifold should be Riemannian, one has to make sure that the tensor $g_{\mu\nu}$ is symmetric and positive definite. –  Qmechanic Oct 21 '11 at 16:14
    
But beware that sometimes people casually use the word "Riemannian" to mean "smooth" rather than "of positive definite signature", which, as Qmechanic pointed out it really should be. Re point 5, continuous and differentiable refers to the way your coordinate patches are sewed together. –  twistor59 Oct 21 '11 at 16:38
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1a. as well as a set of points, you need a set of "open sets", to make a topological space. You also need maps to and from your parameter space to the set of points, to make your 2. possible. There's a sense in which you can't take the sort of short cuts you're trying to take here. There are established short cuts, but not just any short cut is OK. I suggest you find a copy of Nakahara, "Geometry, Topology, and Physics", amazon.com/Geometry-Topology-Physics-Graduate-Student/dp/…, or something similar, and become familiar with the beast. –  Peter Morgan Oct 21 '11 at 17:13
    
@Peter4075: FWIW, there's a book I can't recommend highly enough. Sam Lilley "Discovering Relativity for yourself", Cambridge University Press, 1981. –  Mike Dunlavey Oct 21 '11 at 20:58
    
Just my two cents, but I think the comparison of the two answers below, my own and Ron's, is the perfect example of how precise mathematical language can aid in understanding. In what took me 9 lines to define, using "naive" language, it took Ron 6 steps to define, one of which was just as long as my entire definition! Learning the math is not an easy task, but once learned, it will make things so much easier. By the way, I wouldn't be so intimidated by thigns like "Hausdorff", "second-countable", and "maximal". It sounds intimidating, but once you learn what they mean, it's not so bad:) –  Jonathan Gleason Oct 21 '11 at 22:15
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3 Answers

up vote 3 down vote accepted

My recommendation would be to take the time and seriously learn the definition of a manifold. It is so crucial to modern mathematics and physics, especially GR, to learn such a fundamental thing in an improper, even if in a more efficient way in the beginning, in the long run, it will harm you.

First you should learn some basic point-set topology so that you can understand the definition of a manifold.

Definition of a chart: Let $M$ be a topological space. Then, we say that $(U,\phi )$ is a chart on $M$ iff $U\subseteq M$ is open and $\phi :U\rightarrow \mathbb{R}^n$ is a homeomorphism onto an open subsetof $\mathbb{R}^n$.

Definition of an atlas: Let $M$ be a topological space. Then, we say that $\left\{ (U_i,\phi _i)|\, i\in I\right\}$, $I$ an index set, is an atlas iff each $(U_i,\phi _i)$ is a chart on $M$, and for $U_i\cap U_j$ nonempty, the natural maps from $\phi _i(U_i\cap U_j)$ to $\phi _j(U_i\cap U_j)$ and vice versa are diffeomorphisms. Note that each chart is required to map into euclidean space of the same dimension.

Definition of a manifold: A smooth manifold is a second countable Hausdroff topological space $M$ equipped with a maximal atlas.

Intuitively, the charts given you coordinates (because they map points in the manifold to points in $\mathbb{R}^n$, which can naturally be thought of as coordiantes). An atlas is a collection of charts in which the cooridnates "mesh smoothly". As any atlas is contained in a unique maximal atlas, the maximal condition is somewhat of a technicality.

But you're interested in pseudo-Riemannian manifolds. I say pseudo-Riemannian because, as we're on the Physics stackexchange, you're probably interested in relativity. To talk about pseudo-Riemannian manifolds, we need an additional piece of structure: the metric. To formally define that, you'd have to define tangent spaces and vector fields, at a minimum, and this is really too much content for a single answer.

In any case, I do stress that you learn the underlying mathematics. It will be more difficult than more "naive" approaches, but it will help you in the long run.

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Thanks, but that's (a) a counsel of perfection and (b) pitched way over my head ("second countable Hausdroff topological space M equipped with a maximal atlas"!). I was hoping for more of a dumbed down answer to my dumbed down, "naive" question, which actually (he say's plaintively) took me ages to formulate. I'm not a physics or mathematics graduate. I'm a manual worker trying, for the fun of it, to learn the basics of GTR. Schutz ("A first course in general relativity") defines a manifold as "essentially a continuous space which looks locally like a Euclidean space". That's about my level! –  Peter4075 Oct 21 '11 at 20:08
    
@Peter4075 Well, since you're doing this for fun, which is great, instead of making a living at this stuff, I won't stress the "learning it the right way" approach. I still do feel it would give you a deeper insight into the physics. I would say your description of manifold is not bad as far as intuition goes. As for Question 2, you can make up any metric you want, but, depending on what you do, it might not be a metric. For it to be called a metric, it has to be symmetric, nondegenerate, and smooth. This essentially means that your matrix has to be symmetric and invertible. –  Jonathan Gleason Oct 21 '11 at 20:26
    
@Jonathan: the definition of manifolds does not require a maximal atlas, which is a philosophically dubious and completely unnecessary object, and it is as straightforward as giving local charts and transformations. –  Ron Maimon Oct 21 '11 at 21:27
    
@RonMaimon Well, a manifold requires an atlas, and some basic results of manifold theory require the atlas to be maximal. It doesn't matter in the end whether in the definition you require your atlas to be maximal or not, because every atlas uniquely determines a maximal atlas. That being said, if this weren't the case, you would need to require the atlas to be maximal. And it certainly is not "philosophically dubious". –  Jonathan Gleason Oct 21 '11 at 22:04
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@Jonathan Gleason: It is not my opinion, it is a mathematical fact. If you wish to learn about this and make up your own mind, there is Paul Cohen's classic book "Set Theory and the Continuum Hypothesis", where the foundations of set theory were given their modern form. The results there show that you can give ordinal representations for the real numbers larger than any uncountable size, and this isn't fixed by better axioms. The idea that the set of real numbers have definite properties, like the set of integers does, is busted. –  Ron Maimon Oct 23 '11 at 6:53
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I believe naive approaches are extrememly important, because mathematics is a social process, which selects certain ideas as important according to the whims of humans, while physics is about nature, and selects ideas by criteria of natural processes. Why should the two processes agree?

Well, they often do agree! That's Wigner's "unreasonable efficacy of mathematics in the natural sciences". It might be expected that the notions of mathematics, chosen as they are for their appeal to the human intellect, will not coincide very will with the notions of physics, which are chosen for their match to nature. But they coincide more often than not, and in rare cases when they don't, like quantum mechanics, the mathematics of the new structure is often very beautiful and appealing by classical standards of mathematical beauty. Wigner considered that a deep surprising property of nature. I'm not sure I agree, but it is interesting.

But it always pays to check that the social construct of the mathematical object is doing a good job matching nature, and not trust that the social construct got the thing right, because half the time it doesn't. In the case of manifolds, the mathematicians didn't do that bad, the definition is more or less ok (although a plurality of mathematicians probably don't agree, only they can't think of a replacement!).

  1. The philosophical idea of imposing more and more structure on some sort of "amorphous set of points" is a reasonable and popular one, although I do not personally like it without further specification about the collection of points. The naive collection you are thinking about is uncountable, and there is not much point in thinking of it as a well defined structureless collection, because there are always intrinsically undecidable questions about these. But I will reinterpret your statement to mean that you start with an amorphous countable collection of points, some sort of random lattice, and you make the lattice limit finer and finer to define the points on the manifold. Then it is ok.
  2. The first problem you have is that there is no way to assign a dimension to a set of points based on the "number of parameters". If you allow functions which are discontinuous at a countable set of positions, there is the Cantor function from 2 dimensions to 1 (interleaving the decimal sequences). There are also continuous function from 1 dimension to 2, the space-filling curves. In order to define a notion of dimension, you need to go straight to point 5, and define continuous maps.
  3. (no, not three, 5) The notion of a differential structure, or a continuous structure, is telling you which functions are continuous. You can do this on the countable lattice of points directly, by giving a topological structure to these--- for example by defining a topological metric. Then you can say a function on these points is continuous if the limit as the distance approaches zero of the value at one point is equal to the value at the neighbor. This property can be used to define the function values on the uncountable set of cauchy-sequences by the metric, and will give an uncountable topological space.
  4. But most topological spaces are going to be crazy wild--- for a mundane example, consider a branching graph--- this is mostly one dimensional, but it has branch points. You can also consider a Sierpinski triangle for a one-dimensional object which branches everywhere, or any other sufficiently complex fractal. The notion of manifold is designed to exclude objects which branch. So you impose the condition that there is a continuous 1-1 map from $R^n$ to the space, and this is a coordinate chart. You also impose the condition that where two coordinate charts overlap, you can transform between the two charts, and the transformation is continuous. This recovers the mathematicians definition of a continuous manifold.
  5. Before you make a metric, you need to have a differentiable manifold. If you are mystified by the notion of a differentiable manifold--- consider the surface of a polyhedron in three dimensional space. If you have a function which is differentiable in three dimensions, and you consider its restriction to the polyhedron, it is no longer differentiable, so the polyhedron is not a differentiable manifold (or rather it is not a differentially embedded manifold--- so it doesn't get a differentiable structure from the embedding), but it is a continuous manifold, because any continuous function restricts to a continuous function. The intrinsic definition is in terms of the transition functions from the different coordinate charts which cover the space--- these should be differentiable for a differentiable manifold, analytic for an analytic manifold, and so on.
  6. Once you have a differentiable manifold can define a tangent space, tensor spaces, and the metric tensor, however you like to do it. Just remember that the physicist's metric tensor is not related to the topologists "metric function", because it is not positive definite, and it gives a zero distance for objects which are topologically separated, two points separated by a light ray. The topologists metric is just an easy concrete tool to specify an object as unnaturally huge as a topology.

The process is dependent on a preexisting structure, namely $R^n$. You define $R^n$ by Cauchy sequences, and then you define manifolds from $R^n$. This makes many mathematicians unhappy, because they feel that something as intrinsic as a manifold should be defined without reference to an external structure and coordinate charts to this external space. Perhaps you can deny a special role to $R^n$ by using a suitable condition on the discrete lattice which guarantees that it reproduces a manifold in the limit, but nobody has thought up the right condition which includes all manifolds and excludes wilder things with branches and so on.

The definition at the end is too restrictive too! The modern notion of a space has to include singularities of a certain kind, at the very least orbifolds (mirror-reflection points/sheets). So you have to keep the process by which it was built in the back of the head, for modification. Orbifolds are, in my opinion, the biggest modification of GR predicted by string theory.

As far as your last question, yes, you can make a metric up, but you have to make sure that

  • It is symmetric! You example is not symmetric, and to make it a metric, you need to take the symmetric part, average it with its transpose.
  • The metric shouldn't change type--- it should always have two positive eigenvalues everywhere, so that it stays Euclidean, and doesn't turn Lorentzian somewhere. This is also violated by your example, which is singular along the x and y axes, and changes type some places there (after taking its symmetric part).
  • If it is a Lorentzian metric, the curvature should satisfy a positive energy condition of some type. This condition is restrictive enough to give certain physical results already, like singularity theorems.

This is a physical restriction, there might be some mathematical sense in studying geomtries that turn from Euclidean to Lorentzian along a surface. For your example, the metric tensor is singular along the x-axis and y-axis.

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more unphysical considerations about uncountability. -1 Nor is the rest of the answer pedagogically good enough to avoid that downvote –  joseph f. johnson Jan 17 '12 at 23:17
    
@joseph f. johnson: You are free to downvote, but the uncountability business is important in ways you do not yet appreciate. It would be good for you to read "Set Theory and the Continuum Hypothesis" or another treatment of forcing, so that you can appreciate how fungible uncountable collections really are. –  Ron Maimon Jan 18 '12 at 2:44
    
I knew Paul Cohen personally, I understand the issues better than most people. –  joseph f. johnson Jan 18 '12 at 3:18
    
@joseph f. johnson: If you knew him, why not talk to him? I don't think you follow his stuff very deeply, because of the comments you make regarding this stuff. By the way, sorry for your loss, he was a truly great man. –  Ron Maimon Jan 18 '12 at 3:47
    
Thankyou. We talked about number theory, Lie Groups and unitary representations since he was, as always, working on the Riemann hypothesis and wanted to know about the Selberg Zeta Function and the Trace Formula. –  joseph f. johnson Jan 18 '12 at 4:22
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Basically you are fine except for one major ingredient: you need what amounts to an entire new step in between your steps 3 and 4. You need to have a notion of tangent space and connection between infinitesimally near points before you can proceed to step 4 for an important reason: mathematically, this reason takes the form: the notion of tensor isn't defined on nearby points, but on what a mathematician calls vectors in the tangent space, and physically: because the distance between two nearby points depends, at first glance, on the path you take between them. The point is that logically one starts with the notion of a quadratic differential form, and to go from that to distance you take a line integral over a path from one nearby point to the other. Since the conceptions are logically distinct, and physically distinct, you need to have this intermediate step that I am about to explain:

Step 3 and a half. Tangent vectors. Note: you have to do step 5 before you do step 4 or my new step 3 and a half.

You have to get the notion of a flat vector starting at any point $P$ of your set but heading straight off the set, into space if necessary and you think of your set as embedded in space, or living in a whole world of its own which only has $P$ as a point of contact with your set, if you do not think of your set as embedded in a larger flat space. This notion can be provided by the partial derivatives in different directions of your coordinates, I omit the details unless you want them. Hence for each coordinate function $x^i$ around $P$ you have constructed $\partial \over \partial x^i$, which is a tangent vector, and there are $n$ of these where $n$ is the dimension of your set or space.

The $\partial \over \partial x^i$ for $i$ from one to $n$ span a vector space, the tangent space, denoted $T_P$, since they can be combined to form other linear differential operators on the space of differentiable function on your set.

Only now can you define the metric tensor field: it is an association (for each $P$) of a positive definite quadratic form on each $T_P,$ which is differentiable as $P$ varies.

I.e., before you can, as you put it,

Decide on a metric that gives the infinitesimal distance between two adjacent points.

you have to say that two adjacent points determine one of these tangent vectors that give the direction you would head off in in order to get from one of them, $P$, to the other one. I.e., you really have to replace the notion of « adjacent points » by « vector in the tangent space,» and « infinitesimal distance » by « Euclidean distance inside the tangent space.»

And by now, perhaps you see that at least part of your step 6 really has to be done before you can pick a metric tensor field since it is, after all, a tensor field...

But except for getting things in the wrong order and leaving out an essential step, you were fine.

==Question 2==

Well, as already pointed out by somebody, your matrix had better always be positive definite symmetric. But what I want to further point out is that, technically,

Assuming I do all the above with a flat plane (a table top for example),
using Cartesian coordinates.

well, umm, all the above included picking a metric tensor field, right? so you already picked it. Not to be picky, but what you really want to ask is, alright, assuming you have all the notions of differentiable manifold, coordinates, tangent vectors, tensors, vector fields, tensor fields, now what conditions do four functions of two variables have to satisfy so that they give the components in Cartesian coordinates of a metric tensor field? And as pointed out, you need that they define a symmetric positive definite matrix for all values of $x,y$ when you arrange them in a matrix like you did.

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