# Levi-Civita connection and pseudoRiemannian metric

1. Mar 4, 2014

### TrickyDicky

One of the properties of the unique Levi-Civita connection is that it preserves the metric tensor at each point's tangent space, allowing the definition of invariant intervals between points in the manifold. I'd be interested in clarifying: when the metric preserved by the L-C connection is a PseudoRiemannian metric, the PseudoRiemannian manifold is a:
-pseudometric space
-metric space
-none of the above

2. Mar 4, 2014

### micromass

Staff Emeritus
Every manifold (with or without pseudo-Riemannian metric or connections) is a metric space. So on every manifold you can define a metric such that the topology coincide with the metric topology.

This has nothing to do with the pseudo-Riemannian metric.

But maybe you have something else in mind?

3. Mar 4, 2014

### TrickyDicky

Ok, but the fact is that thru the unique connection in GR spacetimes a particular structure of distance functions between events is developed specially for null curves and geodesics, based on the light cone structure defined by the pseudoriemmanian metric in the space tangent at each event, very different from the distance functions in metric spaces and from wich very important physical consequences are derived.

4. Mar 4, 2014

### WannabeNewton

Invariant intervals have nothing to do with the Levi-Civita connection. This is a trivial property of the metric tensor in and of itself. The Levi-civita connection preserves inner products along arbitrarily extended curves when the vectors in the inner product are parallel transported, which is much stronger. Note the inner product is not preserved along arbitrary curves if the transport is done differently e.g. Lie transport will not preserve the inner product, even if we substitute in the Levi-Civita connection, unless the curve is an integral curve of a Killing field.

All topological manifolds are metric spaces as has already been stated. This has nothing to do with your previous statements.

Last edited: Mar 4, 2014
5. Mar 5, 2014

### TrickyDicky

I know every manifold is metrizable, you are missing my point.
I'm trying to get an understabding on how Exactly the distance functions in spacetime are compatible with the metric space structure.
How for instance a lightlike interval, or an arbitrary null Curve in spacetime M fits in the requurements of the distance functions of metric spaces?

6. Mar 5, 2014

### TrickyDicky

To be more specific, how is the distance in a lightlike geodesic that can be zero between 2 distinct points can be defined if the spacetime is a metric space in wich the distance between 2 distinct points cannot be zero?

7. Mar 5, 2014

### TrickyDicky

You need a connection preserving the metric tensor if you want to use properties of the metric(trivial or not) between points in a general spacetime, don't you?

I mean, the lightcone structure is only defined locally at a point's tangent space or is also defined globally in a curved lorentzian manifold ?(I leave out Minkowski spacetime since it is itself a vector space).

8. Mar 5, 2014

### micromass

Staff Emeritus
Do you mean to define a distance function

$$d(x,y) = \textrm{inf}\{\textrm{Length}(\gamma)~\vert~\gamma:[a,b]\rightarrow M~\text{smooth and}~\gamma(a) = x,~\gamma(b)=y\}$$

for $x,y\in M$? And whether $M$ becomes a (pseudo-)metric space under this $d$?

The definition and properties of $d$ still don't require a connection though.

9. Mar 5, 2014

### TrickyDicky

I'm asking if the pseudoRiemannian metric defines some distance function in a pseudoRiemannian manifold. I'm only mentioning the connection because it preserves the metric tensor.

10. Mar 5, 2014

### micromass

Staff Emeritus
Can you tell us the distance function you're considering?

11. Mar 5, 2014

### DrGreg

I think you are confusing "metric space" and "metric tensor". They are separate concepts which just happen to share the same name "metric". As I understand it, the metric space structure of a manifold is the topology inherited via coordinate charts from Euclidean $\mathbb{R}^4$. It's needed to define "continuity", "differentiability", "smoothness", etc, but isn't used for distance measurements via the metric tensor.

12. Mar 5, 2014

### micromass

Staff Emeritus
You mean the right thing, but I want to be pedantic here. A manifold has (a priori) only a topological structure (and a smooth structure which is not interesting here). Not every topological space induces a metric space, but in a manifold that is the case. So we can show that there exists a distance function such that the topology generated by that distance function (meaning, the topology generated by the open balls $\{x\in M~\vert~d(x,a)<r\}$) is the original manifold topology.

I can already see this thread is going to be confusing because the word metric is used in two senses. So, I propose the following terminology:

1) A metric always means a metric tensor.
2) A distance function on a set $X$ is a function $d:X\times X\rightarrow \mathbb{R}$ such that $d(x,y)=0$ if and only if $x=y$, $d(x,y) = d(y,x)$ and $d(x,z)\leq d(x,y) + d(y,x)$. (Analogous definitions for pseudo-distance function)
3) A distance space is a pair $(X,d)$ where $X$ is a set and $d$ is a distance function on $X$. (Analogous definition for pseudo-distance space).

So I propose never to use the word "metric space" in this thread, or never to use the word "metric" when it actually means a distance function.

So the OP asks us whether a pseudo-Riemannian manifold is
a) A distance space
b) A pseudo-distance space
c) Neither

The question that I want to ask for the OP is which distance function $d$ he is considering concretely.

13. Mar 5, 2014

### TrickyDicky

I said it $ds^2=g_{\mu\nu} dx_\mu dx_\nu$ in a curved pseudoRiemannian manifold with $g_{\mu\nu}$ a pseudoRiemannian metric tensor, allowing distances along curves to be determined through integration, in the specific case of a null curve.

I'm just trying to clear up the differences wrt the Riemannian case of the pseudoriemannian generalization and how it would affect if at all the metric space structure of the Riemannian case, because when M is any connected Riemannian manifold, then we can turn M into a metric space by defining the distance of two points as the infimum of the lengths of the paths (continuously differentiable curves) connecting them, but I wonder how can we define this for null curves in a curved pseudoRiemannian manifold.

14. Mar 5, 2014

### micromass

Staff Emeritus
OK, so you don't have a specific distance function in mind. Your question is whether we can define a suitable (pseudo)-distance function from the metric tensor in the way that we can do this in Riemannian manifolds. Right?

15. Mar 5, 2014

### TrickyDicky

Exactly.

16. Mar 5, 2014

### WannabeNewton

No. The metric is defined at all points of space-time and is completely independent of a connection. A connection (in this setting defined on sections of tangent bundles) is a way to transport vectors from one point to another along curves and is independent of a metric. They have two completely unrelated roles.

It's defined globally based on how it varies from point to point as the metric varies from point to point. The metric is a field it isn't just defined at a single point. It defines a different light cone at each point and if the space-time is time orientable then the light cone variation will be smooth. The local light cone structure is trivial so physics wouldn't be interesting in curved space-times if all we had was local light cone structure.

17. Mar 5, 2014

### TrickyDicky

I'm only concerned here with the unique L-C connection, that is related to the metric only in the sense that it is the only one compatible with it for torsionless manifolds. But you are right, in this thread the connection has basically been a distracting element. I shouldn't have even mentioned it. Sorry about that.

Right. And my question was how do the distances obtained globally with the pseudoriemannian metric, specifically in the case of null curves get along with the manifold being a distance(metric) space?

18. Mar 5, 2014

### micromass

Staff Emeritus
Given a piecewise smooth curve $\alpha:[a,b]\rightarrow M$. We can define the length of the curve as

$$\int_a^b \sqrt{|g_{\alpha(t)}(\dot{\alpha}(t),\dot{\alpha}(t)|}dt$$

See for example O'Neill "Semi-Riemannian Geometry". Thing is that this will violate the triangle inequality.

19. Mar 5, 2014

### TrickyDicky

And in the case of the null curve the identity of indescirnibles is also violated.

20. Mar 5, 2014

### micromass

Staff Emeritus
Right. But that is something I feel is a minor problem. The triangle inequality is much worse.