# A Why pseudo-Riemannian metric cannot define a topology?

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1. Jan 10, 2016

### victorvmotti

It is not clear for me why a positive definite metric is necessary to define a topology as noted in some textbooks like the one by Carroll.

When we define a manifold we require that it locally looks like Euclidean. But even the Lorentzian metric in SR does not locally looks like Euclidean let alone the pseudo Riemannian metric used in GR.

When, for example, the proper time vanishes as in a null path, the notion of an open ball, distance between events, and open basis cannot be used to define a toplogy.

Does this imply that in cosmology, say through FLRW metric, we can only discuss the topology or global geometry of space, or spatial hypersurface, instead of spacetime?

Also related to this question is that we know there "exists" a coordinate system in which the pseudo-Riemannian metric in GR becomes, locally, a Lorentzian one, thus having canonical signature - + + +.

In FLRW metric we assume an isotropic and homogeneous cosmos based on observation in the, well, "observable" universe.

But how about breaking this assumption and imagine that a global Riemannian metric or coordinates system "exists" for spacetime and only demand that it locally becomes Lorentzian?

Which part of my understanding is correct and which one incorrect?

2. Jan 11, 2016

### robphy

3. Jan 11, 2016

### Orodruin

Staff Emeritus
It is not.it is required to define the topology in terms of open balls in the metric. In general, you do not need a metric at all in a topological space.

There is also a way of defining a topology in terms of a pseudo-Riemannian metric with the basic open sets being the intersections of the interior of the light cones of different points. This is known as the Alexandrov topology.

No.

This is not correct. All that is required is that the manifold is locally homeomorphic to $\mathbb R^n$. the metric does not need to be Euclidean (there does not even have to be a metric!).

4. Jan 11, 2016

### Cruz Martinez

This is true, the topology induced by the lorentzian metric would in general not be locally euclidean. However, the topology that is used in spacetimes is not the one induced by the metric tensor that is in Einstein's field equations, but the topology induced by an atlas of coordinate charts.

You can induce a topology by using the Lorentz metric of SR. This defines a pseudo-metric which you can use to define a basis of open balls for a topology in the usual way. However as I mentioned before this is not the topology we work with in the spacetime of SR. The topology of SR is just the topology of R^4.

5. Jan 11, 2016

### Orodruin

Staff Emeritus
Topology by itself does not care about a metric and calling it euclidean is therefore a misnomer. The topology defined on $\mathbb R^n$ by imposing the topology induced by the standard Euclidean metric is equivalent to the Alexandrov topology you obtain by imposing the Minkowski metric.

6. Jan 11, 2016

### Cruz Martinez

So I guess all the mathematicians out there are using a misnomer? I am not talking about metrics when I say locally euclidean, but about purely topological notions.
Also, I am not talking about the alexandrov topology, but about the topology defined by a basis of open balls as defined by a pseudo-metric.

7. Jan 11, 2016

### micromass

It's not a misnomer. Calling a topology Euclidean or locally Euclidean is perfectly mathematically acceptable.

8. Jan 11, 2016

### ShayanJ

The notion of a metric space that can have a metric induced topology is different from a manifold having a metric tensor. These two metrics aren't the same although although having almost same names.

9. Jan 11, 2016

### Cruz Martinez

Sure, which is why I said: this defines a pseudo-metric. I am defining a pseudo-metric by using the Lorentz metric tensor, not using the metric tensor to define a topology directly.

10. Jan 11, 2016

### ShayanJ

But how can you define a metric tensor on a set that is not even a topological space?

11. Jan 11, 2016

### micromass

You have a Lorentzian manifold, which has a smooth structure and a Lorentz metric tensor. This Lorentz metric tensor defines a pseudo-distance function, which defines a topology.

12. Jan 11, 2016

### Cruz Martinez

Right, which is in general different to the initial topology.

13. Jan 11, 2016

### micromass

And very horribly behaved, I think.

14. Jan 11, 2016

### ShayanJ

Yes, but that manifold is itself a topological space. So you're giving a topology to a space that already has a topology!

15. Jan 11, 2016

### micromass

Ok, so?

16. Jan 11, 2016

### Cruz Martinez

This is not a problem, you can give the same set many different topologies. As a very simple example, you can always give any set the discrete and the trivial topology,

17. Jan 11, 2016

### ShayanJ

Is it mathematically acceptable? A topological space with two different topologies that can be very different? What's the meaning of this?

18. Jan 11, 2016

### micromass

You don't have a topological space with two different topologies. You start with a topological space, and you use it to make a different topological space. See the notion of adjoint,co-adjoint, reflection, etc. in categorical topology.

19. Jan 11, 2016

### stevendaryl

Staff Emeritus
Hi, I'm not sure that your question has been answered, and I'm also not completely sure what your question is. You're certainly right, that for positive-definite metrics, you can define a topology by letting the basic open sets be the "balls" of all points strictly less than a certain distance from some center point. With the Minkowsky metric, that doesn't work, because any two points can be connected by a path of zero distance. So in terms of the Minkowsky metric, every point is nearby every other point.

20. Jan 11, 2016

### PAllen

Which is why (as others noted earlier), if you wanted to discuss topolology induced by the Lorentzian metric, I've always seen it done via light cone diamonds, i.e. the Alexandrov topology. This is not always the same as the manifold topology, but it is for the spacetimes physicists most commonly work with. However, for a super-extremal Kerr solution (for example), they are not the same topology.