Register to reply

The topology of spacetimes

by kevinferreira
Tags: spacetimes, topology
Share this thread:
kevinferreira
#1
Jan1-13, 10:49 AM
P: 123
Mod note: This thread contains an off-topic discussion from the thread http://www.physicsforums.com/showthread.php?p=4216768

The radius of the largest ball about the origin in TpM that can be mapped diffeomorphically...
So a notion of distance is used... I wonder how.
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
dextercioby
#2
Jan1-13, 11:24 AM
Sci Advisor
HW Helper
P: 11,928
Quote Quote by kevinferreira
But that's troubling, as in the wikipedia article of the exponential map it is said explicitly

<The radius of the largest ball about the origin in TpM that can be mapped diffeomorphically...>

So a notion of distance is used... I wonder how.
The manifold is riemannian, it has a metric tensor which creates a metric from the topological point of view. See http://en.wikipedia.org/wiki/Riemannian_manifold particularly this part <The tangent bundle of a smooth manifold M assigns to each fixed point of M a vector space called the tangent space, and each tangent space can be equipped with an inner product.> The inner product (which creates a norm, hence a metric topology) is induced by the geometrical metric (tensor).
WannabeNewton
#3
Jan1-13, 11:28 AM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,638
Quote Quote by kevinferreira View Post
So a notion of distance is used... I wonder how.
The riemannian metric / metric tensor defines an inner product on Tp(M). An inner product induces a norm.
EDIT: dextercioby beat me to it while I was typing =D

kevinferreira
#4
Jan1-13, 12:07 PM
P: 123
The topology of spacetimes

Of course, thank you both.
But then, aha, the tangent space is indeed a topological vector space with a topology induced by the notion of distance induced by the norm induced by the inner product, itself defined by the metric of the Riemannian manifold!
Naty1
#5
Jan1-13, 01:35 PM
P: 5,632
why I do not regret leaving mathematical interpretations to others! {LOL}

But then, aha, the tangent space is indeed a topological vector space with a topology induced by the notion of distance induced by the norm induced by the inner product, itself defined by the metric of the Riemannian manifold!
pervect
#6
Jan1-13, 02:38 PM
Emeritus
Sci Advisor
P: 7,634
Quote Quote by kevinferreira View Post
Of course, thank you both.
But then, aha, the tangent space is indeed a topological vector space
I'd agree - at least off the top of my head.

with a topology induced by the notion of distance induced by the norm induced by the inner product, itself defined by the metric of the Riemannian manifold!
No, you need to define the topological "neighborhood" (open balls) by something that's always greater than zero, like x^2 + y^2 + z^2 + t^2, not x^2 + y^2 + x^2 - t^2, which would be the inner product.

But I don't think there's anything that prevents you from doing that if you want.
kevinferreira
#7
Jan1-13, 02:59 PM
P: 123
Quote Quote by pervect View Post
No, you need to define the topological "neighborhood" (open balls) by something that's always greater than zero, like x^2 + y^2 + z^2 + t^2, not x^2 + y^2 + x^2 - t^2, which would be the inner product.

But I don't think there's anything that prevents you from doing that if you want.
Hmm, indeed. It is true that for two vectors [itex]X,Y\in T_p(M)[/itex] we may have [itex]g(X,Y)≤0[/itex], so that distance is not positively defined, and therefore it isn't really a metric space. This is very problematic. Maybe we can just take the absolute value of
[itex]g(X,Y)[/itex], but then we have to check that the triangular inequality is still true.
Hmm, [itex]g[/itex] doesn't even define a distance in M, as M is pseudo-Riemannian, the distance is usually defined as
[tex] \int_a^b ds=\int_a^b\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}[/tex]
and this is always positive.
DaleSpam
#8
Jan1-13, 04:03 PM
Mentor
P: 17,340
Quote Quote by kevinferreira View Post
Hmm, indeed. It is true that for two vectors [itex]X,Y\in T_p(M)[/itex] we may have [itex]g(X,Y)≤0[/itex], so that distance is not positively defined, and therefore it isn't really a metric space. This is very problematic.
I think that a pseudo-Reimannian manifold is probably not a metric space. Because of path dependency issues there is probably not a unique measure of distance between two points in every manifold. E.g. some manifolds may have two events where there are two geodesics with different lengths that connect them.

In any case, even if that can be overcome the topology of a pseudo-Riemannian manifold is inherited from the underlying manifold, not from the metric nor from the inner product of vectors in the tangent spaces.
George Jones
#9
Jan1-13, 04:33 PM
Mentor
George Jones's Avatar
P: 6,246
Quote Quote by kevinferreira View Post
Hmm, indeed. It is true that for two vectors [itex]X,Y\in T_p(M)[/itex] we may have [itex]g(X,Y)≤0[/itex], so that distance is not positively defined, and therefore it isn't really a metric space.
There isn't any problem.
Quote Quote by pervect View Post
But I don't think there's anything that prevents you from doing that if you want.
Take any basis for [itex]T_p(M)[/itex]. This basis can contain any combination of timelight, lightlike, and spacelike vectors. Use this basis to set up a bijection between [itex]T_p(M)[/itex] and [itex]\mathbb{R}^4[/itex] in the standard way. Call a subset of [itex]T_p(M)[/itex] open if the corresponding subset of [itex]\mathbb{R}^4[/itex] is open in its standard topology. This turns [itex]T_p(M)[/itex] into a topological vector space that is homeomorphic to [itex]\mathbb{R}^4[/itex]. The toploogy (class of open sets) of [itex]T_p(M)[/itex] arrived at in this way is independent of the original basis used.

This is equivalent to introducing a (positive-definite) norm on [itex]T_p(M)[/itex].
kevinferreira
#10
Jan1-13, 04:34 PM
P: 123
Quote Quote by DaleSpam View Post
I think that a pseudo-Reimannian manifold is probably not a metric space. Because of path dependency issues there is probably not a unique measure of distance between two points in every manifold. E.g. some manifolds may have two events where there are two geodesics with different lengths that connect them.

In any case, even if that can be overcome the topology of a pseudo-Riemannian manifold is inherited from the underlying manifold, not from the metric nor from the inner product of vectors in the tangent spaces.
Even if there isn't a unique way of measuring path lengths, it's enough to have one with distance properties in order to have a metric space, I think.
Yes, that is true, a manifold is a topological space before any other thing you may define on it. Can that be used on the tangent bundle?
micromass
#11
Jan1-13, 06:27 PM
Mentor
micromass's Avatar
P: 18,334
Quote Quote by DaleSpam View Post
I think that a pseudo-Reimannian manifold is probably not a metric space.
Every manifold is a metric space by Whitney's embedding theorem.
dextercioby
#12
Jan2-13, 03:41 AM
Sci Advisor
HW Helper
P: 11,928
It's high time I threw in the reference for the one interested

Naber G.L. - The geometry of Minkowski spacetime (Springer, 1992)(271p)
pervect
#13
Jan2-13, 04:17 AM
Emeritus
Sci Advisor
P: 7,634
Quote Quote by kevinferreira View Post
Hmm, indeed. It is true that for two vectors [itex]X,Y\in T_p(M)[/itex] we may have [itex]g(X,Y)≤0[/itex], so that distance is not positively defined, and therefore it isn't really a metric space.
Maybe my point was too simple. We know that a manifold is a topological space. But we also know that the Minkowskii notion of distance isn't always positive. So it's not suitiable for defining "open balls".

So we ask - do we actually use the Minkowskii notion of distance to define the topology of our 4-d space time? (Forget about the tangent space, for the moment, I'm talking about how space-time is a 4-d manifold).

For instance, if we are now observing a distant event in the andromeda galaxy, and it's Lorentz interval is zero, does that mean it's close to us, in our neighborhood?

THe answer is no, and no.
TrickyDicky
#14
Jan2-13, 05:44 PM
P: 3,043
Quote Quote by George Jones View Post
There isn't any problem.


Take any basis for [itex]T_p(M)[/itex]. This basis can contain any combination of timelight, lightlike, and spacelike vectors. Use this basis to set up a bijection between [itex]T_p(M)[/itex] and [itex]\mathbb{R}^4[/itex] in the standard way. Call a subset of [itex]T_p(M)[/itex] open if the corresponding subset of [itex]\mathbb{R}^4[/itex] is open in its standard topology. This turns [itex]T_p(M)[/itex] into a topological vector space that is homeomorphic to [itex]\mathbb{R}^4[/itex]. The toploogy (class of open sets) of [itex]T_p(M)[/itex] arrived at in this way is independent of the original basis used.

This is equivalent to introducing a (positive-definite) norm on [itex]T_p(M)[/itex].
Yes, what I always have a hard time understanding is that if both the topology and the distance function is the same in a pseudo-Riemannian manifold as in any Riemannian manifold as pervect and micromass also point out(metric space per Whitney theorem, R^4 topology, etc), then what is the deal with the different kind of vectors (timelike, lightlike, spacelike) of the Lorentzian metric tensor, how can they give rise to so much physics if mathematically it is all equivalent to using a positive-definite inner product wrt the integrated distance function and the topology?
IOW, if the Lorentzian metric tensor only has a local significance, why are its pseudometric features extended to determine the global features of the manifold?
DaleSpam
#15
Jan2-13, 06:37 PM
Mentor
P: 17,340
Quote Quote by micromass View Post
Every manifold is a metric space by Whitney's embedding theorem.
OK, I guess they must just use the length of the shortest path, regardless of whether or not there are multiple geodesics.
WannabeNewton
#16
Jan2-13, 07:03 PM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,638
Quote Quote by DaleSpam View Post
OK, I guess they must just use the length of the shortest path, regardless of whether or not there are multiple geodesics.
Any topological manifold is metrizable. As the requirement is a topological manifold, this is done before a riemannian or pseudo riemannian metric is even equipped to the manifold.
kevinferreira
#17
Jan2-13, 07:12 PM
P: 123
Quote Quote by TrickyDicky View Post
Yes, what I always have a hard time understanding is that if both the topology and the distance function is the same in a pseudo-Riemannian manifold as in any Riemannian manifold as pervect and micromass also point out(metric space per Whitney theorem, R^4 topology, etc), then what is the deal with the different kind of vectors (timelike, lightlike, spacelike) of the Lorentzian metric tensor, how can they give rise to so much physics
Topologies and notions of neighbourhood do not determine uniquely all the properties we want spacetime to have. The notion of manifold is introduced in order to have local coordinates, and we make furthermore the assumption that it is a Riemannian manifold, i.e. that a line element on the manifold is given by
[tex]
ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}
[/tex]
It is the form of [itex]g_{\mu\nu}[/itex] that will therefore determine all the odd properties of the manifold that we're used to see as nice in euclidean space.

Quote Quote by TrickyDicky View Post
if mathematically it is all equivalent to using a positive-definite inner product wrt the integrated distance function and the topology?
IOW, if the Lorentzian metric tensor only has a local significance, why are its pseudometric features extended to determine the global features of the manifold?
I don't think it is 'equivalent', the metric tensor defines a inner product on each tangent space, while the integrated distance function is defined on the manifold itself. You may define a line on your manifold with only one vector, for example by
[tex]
\frac{d\gamma}{ds}(0) = X
[/tex]
and therefore the path [itex]\gamma(s)[/itex] will give you a line 'along' X, on the manifold.
The metric tensor only has local significance, but then you can imagine that its local features compose the global properties as seen locally, so that in the end the whole ensemble of local features will 'add up' to form the manifold. You can find a lot of these things in mathematics, e.g. the Cantor set, which is just a set defined by very simple rules on euclidean space with the usual distance, but then in the end you get an ensemble which has completely different properties and very strange ones indeed.
You may also think about Minkowski space (flat spacetime), and about how you may define this light cone lat an event, and this is not just a drawing on paper, space itself gets different properties on different regions. You may want to view this as only locally defined, but there's nothing that contradicts the fact that you can extend this (relational) properties to the whole space.
DaleSpam
#18
Jan2-13, 09:17 PM
Mentor
P: 17,340
Quote Quote by WannabeNewton View Post
Any topological manifold is metrizable. As the requirement is a topological manifold, this is done before a riemannian or pseudo riemannian metric is even equipped to the manifold.
I don't see how that works if you have a manifold without a metric. A metric space needs to have a unique notion of distance defined; how can you do that without defining a metric and making it a (pseudo-) Riemannian manifold?

Distances on a coffee cup and a donut are different even though they are the same topologically. Do you have an explanation how that works?


Register to reply

Related Discussions
The usual topology is the smallest topology containing the upper and lower topology Topology and Analysis 2
The usual topology is the smallest topology containing the upper and lower topology Calculus & Beyond Homework 0
When are two spacetimes the same Special & General Relativity 1
A democracy of spacetimes? Beyond the Standard Model 10
Stringy spacetimes Beyond the Standard Model 0