Some doubts concerning the mathematical bases of GR

[TrickyDicky]

The normal definition involving second countable hausdorff assumption is the one used 99% of the time.

Maybe even 99.9%, but that doesn't prove anything mathematically.

A pseudometric space requires a global function of two points satsifying some axioms. I have explained that (unlike the Riemannian -> metric space case) there is no unique natural definition of this function at all.

Maybe, but that would mean even less chance to be Hausdorff, as we get even farther from the metric space structure.

 

[TrickyDicky]

Let's stick to the definition of Hausdorff space.
"Points x and y in a topological space Xcan be separated by neighbourhoods if there exists a neighbourhood Uof x and a neighbourhood V of y such that Uand V are disjoint (U∩ V = ∅). X is a Hausdorff space if any two distinct points of X can be separated by
neighborhoods. "
How are distinct points on the light cone of a Lorentzian manifold separated by neighbourhoods if their distance is zero?

 

[PAllen]

Maybe even 99.9%, but that doesn't prove anything mathematically.Maybe, but that would mean even less chance to be Hausdorff, as we get even farther from the metric space structure.

A definition is not a proof. The definition of pseudo-riemannian manifold used by essentiallly all authors is based on the normal mathematical definition of manifold, which includes the requirement of 'second countable hausdorff'.

Your second point is total nonsense. If something is part of the definition, there is no 'chance not to be hausdorff'. If I define natural numbers as integers greater than zero, what is the chance of natural number < 0?

 

[PAllen]

Let's stick to the definition of Hausdorff space.
"Points x and y in a topological space Xcan be separated by neighbourhoods if there exists a neighbourhood Uof x and a neighbourhood V of y such that Uand V are disjoint (U∩ V = ∅). X is a Hausdorff space if any two distinct points of X can be separated by
neighborhoods. "
How are distinct points on the light cone of a Lorentzian manifold separated by neighbourhoods if their distance is zero?

A pseudo-riemannian manifold is first of all a manifold. The normal definition of manifold requires the hausdorff property before even defining a metric. This is getting ridiculous. The sequence of definitions is:

topological space -> Hausdorff -> manifold-> semi-riemannian manifold.

The open sets and Hausdorff property are defined without any reference to the metric, which is not even defined yet.

This is getting just ridiculous as you ignore everything matterwave, myself, and George Jones are saying.

 

[Dale]

Let's stick to the definition of Hausdorff space.
"Points x and y in a topological space Xcan be separated by neighbourhoods if there exists a neighbourhood Uof x and a neighbourhood V of y such that Uand V are disjoint (U∩ V = ∅). X is a Hausdorff space if any two distinct points of X can be separated by
neighborhoods. "
How are distinct points on the light cone of a Lorentzian manifold separated by neighbourhoods if their distance is zero?

This is the definition I was referring to above in post 4. Note that nowhere in that definition does it refer to distance, and that distances are not required to define a neighborhood in topological spaces. The facts that some points in the neighborhood have zero distance and some points outside of the neighborhood also have zero distance is not relevant.

 

[Haelfix]

How are distinct points on the light cone of a Lorentzian manifold separated by neighbourhoods if their distance is zero?

As others have explained, you are taking 3 separate mathematical structures and trying to mix them outside of their domains of definition in a single sentence.

'neighborhoods' refer to a concept in point set topology.

'Distance' typically involves concepts at the level of a metric (distance function) but in fact, can mean several different things even when that is understood.

A 'light cone' is yet even more structure, involving a specific choice of observer and a causal structure.

 

[DrGreg]

My understanding is that the underlying topology of a manifold as used in GR is nothing to do with the metric tensor, but is simply the topology inherited via the coordinate maps from Euclidean $\mathbb{R}^4$ space. I.e. an open set in the manifold is just the coordinate image of an open set in Euclidean $\mathbb{R}^4$ coordinate space.

 

[TrickyDicky]

My understanding is that the underlying topology of a manifold as used in GR is nothing to do with the metric tensor, but is simply the topology inherited via the coordinate maps from Euclidean $\mathbb{R}^4$ space. I.e. an open set in the manifold is just the coordinate image of an open set in Euclidean $\mathbb{R}^4$ coordinate space.

Yes, I know that is the usual assumption, the OP was asking if this assumption is rigorously backed mathematically, not only based on authority amd physical convenience.
We know the toplogy of manifold in GR is not necessarily R^4.
And on the other hand metrics induce topologies, a fact that is being ignored by most posters.

 

[TrickyDicky]

As others have explained, you are taking 3 separate mathematical structures and trying to mix them outside of their domains of definition in a single sentence.

'neighborhoods' refer to a concept in point set topology.

'Distance' typically involves concepts at the level of a metric (distance function) but in fact, can mean several different things even when that is understood.

A 'light cone' is yet even more structure, involving a specific choice of observer and a causal structure.

You don't think metrics can induce topological structures in pseudo-Riemannian manifolds?

 

[TrickyDicky]

This is the definition I was referring to above in post 4. Note that nowhere in that definition does it refer to distance, and that distances are not required to define a neighborhood in topological spaces. The facts that some points in the neighborhood have zero distance and some points outside of the neighborhood also have zero distance is not relevant.

We are not talking about topological spaces in general but about pseudo-Riemannian manifolds that are by definition endowed with a metric, that induces a certain topology on the manifold that imply distance.

 

[TrickyDicky]

A pseudo-riemannian manifold is first of all a manifold. The normal definition of manifold requires the hausdorff property before even defining a metric. This is getting ridiculous. The sequence of definitions is:

topological space -> Hausdorff -> manifold-> semi-riemannian manifold.

The open sets and Hausdorff property are defined without any reference to the metric, which is not even defined yet.

This is getting just ridiculous as you ignore everything matterwave, myself, and George Jones are saying.

You have the sequence wrong, at least according to Hawking and Ellis, whom I trust more than the wikipedia.
All your arguments are based on authority like percentages of authors and definitions without mathematical proof, that is ok in itself but ignores that in the OP I was asking for mathematical rigor rather than authority or physical convenience arguments.

 

[TrickyDicky]

This excerpt might help clarify some confusion that keeps being posted about metrics and metric tensors:

From the wikipedia entry on Metric:
" Important cases of generalized metrics In differential geometry, one considers metric tensors, which can be thought of as "infinitesimal" metric functions. They are defined as inner products on the tangent space with an appropriate differentiability requirement. While these are not metric functions as defined in this article, they induce metric functions by integration. A manifold with a metric tensor is called a Riemannian manifold. If one drops the positive definiteness requirement of inner product spaces, then one obtains a pseudo-Riemannian metric tensor, which integrates to a pseudo-semimetric . These are used in the geometric study of the theory of relativity, where the tensor is also called the "invariant distance"."

 

[TrickyDicky]

The topology of a manifold is determined by the local mappings to a Euclidean space and the standard topology in the Euclidean space (which is Hausdorff).

Therefore, all (topological) manifolds (of which differentiable manifolds are a subsection) are Hausdorff.

This has nothing to do with the metric.

Sorry to answer this late.
What you are saying applies to general differentiable manifolds rather than general manifolds, (see Hawking and Ellis "The large scale structure of spacetime").
And precisely what is being asked for in this thread is a rigorous mathematical proof (or a reference to it), that a pseudo-Riemannian manifold keeps that property, or if it looses it due to its pseudometricity.

 

[TrickyDicky]

A possible route to understand this is that as I commented usually (see singularity theory I am wikipedia)manifolds are defined as spaces without singularities(discontinuities), in the case of GR they occur thru degeneration of the manifold structure, precisely due to its pseudometricity.

 

[Dale]

We are not talking about topological spaces in general but about pseudo-Riemannian manifolds that are by definition endowed with a metric, that induces a certain topology on the manifold that imply distance.

That is not how I learned it. The metric does not induce the topology, the manifold does. At least, that was the way I learned GR, and it appears to be better than your approach for exactly the reason that you are running into here.

 

[TrickyDicky]

Yes, a topological manifold has a topological structure just by being a manifold, this in no way contradicts the fact that when a manifold has a metric, this metric induces the topology.

You are right that the issues I'm bringing up are not mentioned in the usual GR texts, that is why I'm asking for some mathematical proofs, not just convenient assumptions.

 

[Dale]

Yes, a topological manifold has a topological structure just by being a manifold, this in no way contradicts the fact that when a manifold has a metric, this metric induces the topology.

It sounds like a contradiction to me. If the manifold already has a topology then if you introduce a metric which has a topology then it seems that you could get situations where the manifold topology and the metric topology contradict each other. In fact, I think that is exactly this contradiction that is causing your concerns. The topology you are trying to induce via the metric includes all points on the light cone as indistinguishable from a given point, while the topolgy of the manifold distinguishes them. The latter defines the topology of a pseudo-Riemannian manifold, not the former.

On the other hand, if you believe (which I don't) that both the manifold and the metric induce separate topologies and that there is no contradiction in that fact, then simply use the manifold topology rather than the metric topology for defining the manifold as Hausdorff. Presumably, if you have both then you can use either wherever convenient.

 

[PAllen]

You have the sequence wrong, at least according to Hawking and Ellis, whom I trust more than the wikipedia.
All your arguments are based on authority like percentages of authors and definitions without mathematical proof, that is ok in itself but ignores that in the OP I was asking for mathematical rigor rather than authority or physical convenience arguments.

Proof is not relevant at this point. These are definitions. If one defines natural numbers as integers greater than zero, there is no such thing as proving natural numbers are positive.

It is true that there are two definitions of manifold, the common one and the more general one. Most books on GR base it on the common one. It is a definitional choice.

I gave wikipedia links because they are easy to find. However, my GR books that use manifolds all start from the common definition.

This link clarifies some things, and mentions Hawking and Ellis less common usage:

http://mathworld.wolfram.com/TopologicalManifold.html

This link clarifies the usual usage (e.g. that manifold assumes T2-space = hausdorff space property). See the description 'all manifolds'. This clearly means 'under the common definition', otherwise it would be wrong (as opposed to just being shorthand).

http://mathworld.wolfram.com/ParacompactSpace.html

Obviously, for their investigations, Hawking and Ellis have chosen to start from less common definitions. They deliberately start from a manifold that is not necessarily a topological manifold. As I don't have their book, I can't say much more on this.

So, trying to rephrase what the OP is possibly getting at:

If one uses the definitional scheme of Hawking and Ellis, it is then meaningful to ask about proving the Hausdorff property under some particular conditions. Other questions which are true by definition in the common definitional framework also become interesting.

At this point, having clarified that the OP specifically refers the Hawking and Ellis sheme, it would useful for a re-statement of the specific questions the OP wants to discuss.

Unfortunately, I can't contribute further, as I have only studied the more common framework and don't have a copy of Hawking and Ellis.

 

[PAllen]

This excerpt might help clarify some confusion that keeps being posted about metrics and metric tensors:

From the wikipedia entry on Metric:
" Important cases of generalized metrics In differential geometry, one considers metric tensors, which can be thought of as "infinitesimal" metric functions. They are defined as inner products on the tangent space with an appropriate differentiability requirement. While these are not metric functions as defined in this article, they induce metric functions by integration. A manifold with a metric tensor is called a Riemannian manifold. If one drops the positive definiteness requirement of inner product spaces, then one obtains a pseudo-Riemannian metric tensor, which integrates to a pseudo-semimetric . These are used in the geometric study of the theory of relativity, where the tensor is also called the "invariant distance"."

Note that this talks about integrating the pseudo-riemannian metric into a pseudo-semimetric not a pseudometric. Semi-metric is defined in the same article, and pseudo-semimetric is obvious by context. This validates my argument that a pseudometric structure cannot (generally) be imposed (by integration) on pseudo-riemannian manifold, because the pseudo-metric tensor does not integrate into pseudometric. Even for integrating to a pseudo-semimetric, one would have to add some definitions to specify the integration (e.g. minimum interval (any type - timelike, spacelike, or null) over all geodesics [parallel transport definition] connecting two points). It is clear this definition (which is not unique) would, indeed, produce a pseuo-semimetric but not either a semimetric or a pseudometric.

Thus, it is true, as I claimed, that there simply no connection between pseudo-Riemannian manifolds and psuedometric spaces. This is contrast to the fact that any connected Riemannian manifold can be treated as a metric space via natural, unique, integration.

 

[Matterwave]

The topology on a manifold should be induced from the mappings to the Euclidean space. I.e. open sets in the Euclidean space are mapped to open sets in the manifold.

I can always introduce a metric (or more precisely a distance function) to this manifold (since it's a set of points after all). I can also always introduce a distance function that would induce a topology that contradicts the topology induced by the mappings. I can introduce the trivial distance function d(m,n)=0 for all m and n, and this will certainly induce a non-hausdorff topology; however, this is not how we define a topology on a manifold. At least, I have always seen the manifold topology defined the first way and not this second, more convoluted way.

 

[micromass]

I don't see how a pseudometric could ever come into play here.

Every Riemannian manifold or pseudo-Riemannian manifold is by definition a manifold. A manifold is almost always taken to be a locally Euclidean, second countable Hausdorff space.

By Whitney's embedding theorem, a smooth manifold can always be embedded in $\mathbb{R}^n$ for a suitable n. As such, a smooth manifold is metrizable. Thus every smooth manifold can be given the structure of a metric space.

If we define a pseudometric on a smooth manifold, then this pseudometric is always a metric. Indeed, a pseudometric space is a metric space if and only if it is Hausdorff.

It is true that there are non-Hausdorff manifolds, these are topological spaces that are locally Euclidean and second countable. But these are usually not regarded as topological manifolds.

Note that a non-Hausdorff manifold is not even necessarily pseudometrizable. For example, the line with two origins is perhaps the most famous example of a non-Hausdorff manifold. But this line with two origins is $T_1$ and not Hausdorff, therefore it cannot be pseudometrizable.

We can even go further, a non-Hausdorff manifold is always a $T_1$. Indeed: given two point x and y in our manifold M. Take a Euclidean neighborhood U of x. If y is not in our neighborhood, then U is a neighborhood of x that does not contain y and as such the $T_1$ axiom is satisfied. If y is in our neighborhood, then (since our neighborhood is locally Euclidean), we can find a smaller neighborhood around x which does not contain y. Again, then $T_1$-axiom is satisfied.

So a non-Hausdorff manifold is always $T_1$ and non-Hausdorff. As such, a non-Hausdorff manifold is never pseudo-metrizable. If it were pseudometrizable, then it would be either Hausdorff or not $T_1$.

So talking about pseudometrizable non-Hausdorff manifolds is useless, since there are no such things.

 

[TrickyDicky]

Thanks Dalespam, Pallen an matterwave for your interesting contributions, they're surely helpful.
Micromass, that is a great , really informative post, thanks , I value it even more coming from a mathematician rather than a physicist or relativist
that might be contaminated by old habits in their thinking about GR.

 

[TrickyDicky]

Note that this talks about integrating the pseudo-riemannian metric into a pseudo-semimetric not a pseudometric. Semi-metric is defined in the same article, and pseudo-semimetric is obvious by context. This validates my argument that a pseudometric structure cannot (generally) be imposed (by integration) on pseudo-riemannian manifold, because the pseudo-metric tensor does not integrate into pseudometric. Even for integrating to a pseudo-semimetric, one would have to add some definitions to specify the integration (e.g. minimum interval (any type - timelike, spacelike, or null) over all geodesics [parallel transport definition] connecting two points). It is clear this definition (which is not unique) would, indeed, produce a pseuo-semimetric but not either a semimetric or a pseudometric.

This is right, the non-positive definite metric tensor integrates to give a both pseudo- and semi-metric, this means it doesn't only relax the point separation axiom of metric spaces, but the triangle inequality axiom. Pseudosemimetric spaces are also tipically non-Hausdorff, but see the post by micromass.
We must bear in mind that the change of signature of the metric tensor (and therefore its distance function upon integration) is the only difference between a Riemannian and a Pseudo-Riemannian manifold. So anything that eliminates that difference makes them indistinguishable.

This is contrast to the fact that any connected Riemannian manifold can be treated as a metric space via natural, unique, integration.

See micromass answer.

 

[TrickyDicky]

...every smooth manifold can be given the structure of a metric space.

If we define a pseudometric on a smooth manifold, then this pseudometric is always a metric. Indeed, a pseudometric space is a metric space if and only if it is Hausdorff.
So talking about pseudometrizable non-Hausdorff manifolds is useless, since there are no such things.

These are both great points.
In the interest of rigor and as PAllen points out let's call the distance function in question a pseudosemimetric for the Lorentzian manifold used in GR case.
Now if defining a pseudosemimetric in a smooth manifold we actually define a metric, I guess because the smooth manifold topology is the one that rules, mathematically (at least topologically) how do we make a distinction between Riemannian and Pseudo-Riemannian manifolds if their only difference is in the metric tensor that in one case integrates to a metric and in the other to a pseudosemimetric?

 

[Ben Niehoff]

how do we make a distinction between Riemannian and Pseudo-Riemannian manifolds if their only difference is in the metric tensor that in one case integrates to a metric and in the other to a pseudosemimetric?

That is the only distinction. They are both manifolds. One has been given a positive-definite metric tensor; the other has been given an indefinite metric tensor.

Given any (connected) manifold, it is always possible to put a positive-definite metric tensor on it, which can always be integrated to a metric (taking the infimum of multiple geodesics, if need be).

Given any manifold, it is not always possible to put an indefinite-signature metric tensor on it. There are topological obstructions. For example, a torus can have a Lorentzian metric tensor, but a sphere cannot.

 

[atyy]

Given any manifold, it is not always possible to put an indefinite-signature metric tensor on it. There are topological obstructions. For example, a torus can have a Lorentzian metric tensor, but a sphere cannot.

So when people give an example of AdS/CFT and say eg. it's string theory on AdS5 X S5, there isn't a Lorentzian metric on S5?

 

[PAllen]

So when people give an example of AdS/CFT and say eg. it's string theory on AdS5 X S5, there isn't a Lorentzian metric on S5?

I am guessing Ben meant a 2-sphere, and that for higher dimensions, answer is different.

A little searching turns up the following claim to a proof that an n-sphere admits a Lorentzian metric if n is odd:

http://mathoverflow.net/questions/4...ossible-that-sn-can-have-a-lorentz-metric-why

See the checked answer.

 

[Ben Niehoff]

So when people give an example of AdS/CFT and say eg. it's string theory on AdS5 X S5, there isn't a Lorentzian metric on S5?

No, the S5 part of AdS5 x S5 has a Riemannian metric on it (i.e., the timelike part of the total metric lies entirely within the AdS5 factor).

However, as Pallen points out, the odd-dimensional spheres can be given Lorentzian metrics. This is because S^(2n+1) can be decomposed as an S^1 fibered over CP^n. This S^1 can then be chosen as the timelike direction.

 

[atyy]

@Ben Niehoff & PAllen, got it - thanks!

 

[TrickyDicky]

That is the only distinction.

Yes, that's my point. That is the only difference and the one that gives rise to many of the properties of GR, like the different type of geodesics (null, timelike..), the causal structure, the geodesic incompleteness and the singularities, etc...
But if that difference is obviated by the smooth manifold structure that turns pseudosemimetrics spaces into metric spaces how do we justify mathematically all those properties associated to the pseudosemimetricity that are such important features of GR?

 

[Ben Niehoff]

Yes, that's my point. That is the only difference and the one that gives rise to many of the properties of GR, like the different type of geodesics (null, timelike..), the causal structure, the geodesic incompleteness and the singularities, etc...
But if that difference is obviated by the smooth manifold structure that turns pseudosemimetrics spaces into metric spaces how do we justify mathematically all those properties associated to the pseudosemimetricity that are such important features of GR?

I'm not really sure what you're getting at. None of this causes a problem for GR. GR only cares about the metric tensor; whether it integrates into any global structure is bonus.

When we say some space is a "manifold", all we mean is that it has certain topological properties. Topology cares about how the various points in a set are connected. It is completely agnostic as to the concept of those points having any "distance" defined between them.

A manifold is just some space that can be covered by open sets, each of which looks just like an open set of R^n. By "looks just like", I mean that the points in the open set U on the manifold are connected to each other the same way as the points in the open set V of R^n. At no point do I care about the Euclidean distance which is possible to define in R^n; it's irrelevant.

Now, an added fact is that I can use the natural Euclidean distance in R^n to define a notion of distance on the manifold. One uses the usual tangent space construction to define a Riemannian metric tensor (turning our manifold into a Riemannian manifold). This metric tensor can be integrated to obtain a global distance function, turning our Riemannian manifold into a metric space. (One can also define metric spaces which are not manifolds, so in fact this object is both a Riemannian manifold and a metric space, those being independent properties).

Metric spaces have the additional property that the distance function can be used to define a topology. That is, open sets can be defined as the interiors of metric balls. It happens that when we do this to a Riemannian manifold, the topology induced by the metric structure agrees with the topology we already had from the manifold structure. This is not hard to prove.

But these concepts do not carry over to the pseudo-Riemannian case. As you have pointed out, a pseudo-Riemannian metric tensor does not integrate to a distance function, for one. Whatever object a pseudo-Riemannian metric integrates to, it must fail to satisfy the distinguishability axiom,

$$d(x,y) = 0 \; \text{iff} \; x = y,$$
and hence, the topology induced by such a distance function will not agree with the topology we already have from the manifold structure. This is easy to see in flat Minkowski space, whose topology is that of R^4.

There is no reason to expect a pseudo-Riemannian metric tensor to induce a topology that agrees with the one already present, because the theorem in the Riemannian case relies upon the details of all the axioms. Just because "pseudo-Riemannian metric tensor" contains the words "Riemannian metric tensor" does not mean you can borrow theorems and expect them to still be true.

 

[TrickyDicky]

I'm not really sure what you're getting at. None of this causes a problem for GR. GR only cares about the metric tensor; whether it integrates into any global structure is bonus.

When we say some space is a "manifold", all we mean is that it has certain topological properties. Topology cares about how the various points in a set are connected. It is completely agnostic as to the concept of those points having any "distance" defined between them.

But these concepts do not carry over to the pseudo-Riemannian case. As you have pointed out, a pseudo-Riemannian metric tensor does not integrate to a distance function, for one. Whatever object a pseudo-Riemannian metric integrates to, it must fail to satisfy the distinguishability axiom,

$$d(x,y) = 0 \; \text{iff} \; x = y,$$
and hence, the topology induced by such a distance function will not agree with the topology we already have from the manifold structure.

Ok, so from this I interpret that Riemannian and Pseudo-Riemannian manifolds are topologically indistinguishable, no?

OTOH, when you say that GR only cares about the metric tensor, I am not sure how to make this statement compatible with the fact that in GR preserving lengths, that is, distance, is fundamental, as it is the Levi-Civita connection that being torsion-free assures integrability of the metric tensor.

What I'm getting at is that the GR features derived from its metric tensor indefiniteness seem to be overridden by the smooth manifold topology, so that at least at the large scale (not at the infinitesimal level of the metric tensor) there seems to be no difference between Riemannian and Pseudo-Riemannian manifolds, it could only have R^4 topology in the case of GR.

And all lengths in Pseudo-Riemannian manifolds must equal the Riemannian manifold case since the pseudo-Riemannian metric tensor can only integrate to a metric in a smooth manifold.

 

[Ben Niehoff]

Spacetimes are not pseudometric spaces. They are manifolds on which we have defined a pseudo-Riemannian metric tensor.

Do you agree that various points along a null geodesic are distinct points?

 

[TrickyDicky]

Spacetimes are not pseudometric spaces. They are manifolds on which we have defined a pseudo-Riemannian metric tensor.

Right.

Do you agree that various points along a null geodesic are distinct points?

Yes.

I asked a related question before that was left unanswered. How can we separate distinct points on the null light cone if their distance can be zero? Now, given the fact that due to the fact that the smooth manifold structure underlying Pseudo-Riemannian manifolds assures the spacetime is not pseudosemimetric but metric, we should be able to adjudicate a non-zero distance to distinct points on the null cone but I don't know how when their metric tensor is vanishing, ds=0.

 

[Ben Niehoff]

I asked a related question before that was left unanswered. How can we separate distinct points on the null light cone if their distance can be zero?

"Distance" is not part of the subject of topology, period.

Points along a null geodesic are separate because they have different values of affine parameter. This is the whole point of saying the manifold can be covered by open sets that map continuously to open sets of R^n.

A pseudo-Riemannian metric tensor does not induce a topology that agrees with the underlying manifold structure. We do not use, nor care, what topology the pseudo-Riemannian metric tensor does induce, precisely because it disagrees with the underlying manifold structure.

Just because a structure can be defined does not mean that it is useful or physically reasonable. GR uses differential geometry, which is done on manifolds, and hence it is the manifold structure we require. Physically, it is reasonable that null geodesics be a series of distinct points, rather than a single point, because it is our physical observation that light rays travel.

Note also, that on any manifold, we can define a Riemannian metric tensor. But in GR we choose not to, because a Riemannian metric tensor is incompatible with the physical requirement of local Lorentz symmetry. So this is another example of a structure that is possible to define, but is left unused in the context of GR.