Link btw manifolds and space-time

[robphy]

Any topological model that fully describes SR and GR must actually remove the Hausdorff constraint. The problem is that Riemannian geometry requires a manifold to be Hausdorff, a fact, by the way, that was unknown at the time SR and GR was developed.

Why? Are you claiming that Riemannian requires [and not merely chooses for convenience] the Hausdorff condition? Can you provide a reference?

Historically, lots of things were unknown at the time SR and GR were being developed. For a long time, the emphasis was on systems-of-PDEs in coordinate patches and not the global structure underlying the modern formulations of spacetime [including causal structures].

One of the requirements for a metric on a Riemannian manifold is that it the triangle inequality must hold. The Minkowski "metric" which has a negative definite signature is obviously not a metric by that definition.

The triangle-inequality has some nice properties... which is needed for certain properties of Riemannian geometry. But, the clock effect tells us that that triangle-inequality is not satisfied by triangle with timelike-vectors... we have instead the reverse-triangle inequality.

FYI, Minkowski is NOT http://mathworld.wolfram.com/NegativeDefiniteMatrix.html" . It has signature (-+++) or (+---), depending on your convention.

In any case, as Riemannian generalizes Euclidean, Lorentzian (or more generally semi-riemannian) generalizes Riemannian. One can further generalize to Finslerian manifolds, complex-manifolds, non-metric manifolds, non-manifold topological spaces, etc... In the mathematical hierarchy, is there is anything so sacred about Riemannian geometry with its positive-definite metric?

The point for modeling the physical world is: which best models spacetime... [with hopefully each mathematical structure in model having some physical interpretation]?

(By the way, from a projective-geometric viewpoint [in the spirit of Felix Klein], one can see that Minkowskian geometry is a completely consistent geometric theory [as one does with Elliptic and Hyperbolic geometries].)

Now we could simply put the term "pseudo" in front of anything and be done with it but that does not make the real mathematical issue go away. It is a pseudo solution.

This sounds like the comments that regard the complex numbers as "strange" number systems.

There is another way around it, by defining an "Einstein algebra", but then we cannot any longer think in terms of a topological model of space-time.

Geroch's Einstein Algebra? Using that does NOT say that you "cannot any longer think in terms of a topological model of space-time"... rather, you don't have to think in that way. You can [and do] use the Einstein Algebra for ordinary spacetime. (By the way, some noncommutative-geometric approaches use a similar approach.)

So in short, this is not an SR or GR problem, it is a mathematics problem.
Nevertheless and obviously, mathematics can still supply a workable framework for SR and GR, but clearly not an mathematically complete one.

It may be a mathematics problem... but don't lose sight of the physicist's goal: find the best model for spacetime [sufficient for the problem under study].

 

[MeJennifer]

FYI, Minkowski is NOT http://mathworld.wolfram.com/NegativeDefiniteMatrix.html" . It has signature (-+++) or (+---), depending on your convention.

Ops, of course you are correct, it can be positive, negative and 0. I should have written it is not a positive-definite.

It may be a mathematics problem... but don't lose sight of the physicist's goal: find the best model for spacetime [sufficient for the problem under study].

I fully agree!
And clearly a pseudo-Riemannian manifold is currently the best we have. To me, and perhaps of my lack of knowledge in this area, it is a workable but not a completely satisfactory solution to model SR and GR on.

 

[AlphaNumeric]

Sorry, my error. Not all manifolds are Hausdorff but all Riemannian ones are because they are differentiable and you need a notion of 'neighbourhoods' around each point in the manifold to create the differentiable structure (at least that seems to be my applied maths understanding of it).

A pseudo-Riemannian manifold still satisfies the Hausdorff requirement, since you can pick open charts which don't intersect around any two points you like, which gives you the Hausdorff property.

Besides, weren't you just saying that (pseudo)Riemannian manifolds aren't Hausdorff, particularly those used in GR?! You didn't answer where the non-Hausdorff points in Minkowski space-time are.

Nakahara has the following requirements for a Riemannian manifold :

1. It's differentiable
2. It's metric g is defined at every point and has the property g_{p}(V,U) = g_{p}(U,V)
3a. g_{p}(U,U) \geq 0 for all U and equality iff U=0.

He then extends this to a pseudo-Riemannian manifold by changing 3a. to the following :

3b. if g_{p}(U,V)=0 for any U in the tangent space, then V=0.

It's not quite the same, but does share some of the properties of 3a. It's not that Pseudo-Riemannian are totally invalid, it's that their metric isn't held to the same restriction as the Riemannian ones.

The Minkowski "metric" which has a negative definite signature is obviously not a metric by that definition.?

No, it's obviously not a Riemannian metric. It's still a metric if you slacken your constraints somewhat.

There's nothing wrong with doing that provided you're consistent. Take SUSY for instance. By slackening the restrictions on Lie algebras to allow graded Lie algebras you develop a new, rich algebraic system.

Yes, it's important you're consistent with your usage of Riemannian and Pseudo-R. results, even Hawking once got them confused and tried to publish a paper using Riemannian results but applied to GR manifolds (at least so said my old GR lecturer).

 

[cesiumfrog]

Any topological model that fully describes SR and GR must actually remove the Hausdorff constraint.

Why? (It seems like you didn't answer the question: What physics are you neglecting if you constrain the model to be Hausdorff? Personally, seems like a non-Hausdorff space would be the one behaving unphysically.)

 

[pervect]

BTW, does anyone have an example of a manifold (which must by definition be covered by charts diffeomorphic to the familiar R^n) that is not Hausdorff?

I gather it's supposed to be possible, but I've never seen an example.

Wikipedia says that any metric space is Hausdorff, for whatever it's worth, so presumably our non-Hausdorff manifold won't have a metric (unless the wiki is wrong).

I don't understand all the excitement over the word "psuedo" in "Psuedo-Riemannian manifold" - are we being trolled?

 

[Doodle Bob]

Wikipedia says that any metric space is Hausdorff, for whatever it's worth, so presumably our non-Hausdorff manifold won't have a metric (unless the wiki is wrong).

i believe that the proof for this goes something like: Let x and y be distinct points in the metric space. then d=dist(x,y)>0. Let U={z: dist(x,z)<d/3} and V={z: dist(z,y)<d/3}, both open sets by the metric topology. By the triangle inequality, U and V are disjoint.(oops, i probably just gave away a homework problem there...)

So, if a manifold has a Riemannian metric that defines a distance on it, then it is necessarily Hausdorff. However, i do believe there are examples of topological manifolds (maybe even differentiable manifolds) that do not admit a partition of unity and hence do not have even a well-defined global Riemannian metric.

 

[quasar987]

BTW, does anyone have an example of a manifold (which must by definition be covered by charts diffeomorphic to the familiar R^n) that is not Hausdorff?

I gather it's supposed to be possible, but I've never seen an example.

Wikipedia says that any metric space is Hausdorff, for whatever it's worth, so presumably our non-Hausdorff manifold won't have a metric (unless the wiki is wrong).

The key word is 'non-metrizability'. See

http://en.wikipedia.org/wiki/Metrization_theorem

for a plenitude of metrizability theorems and also an example of a non metrizable (and hence, not Hausdorff) manifold:

An example of a space that is not metrizable is the real line with the lower limit topology.

 

[Hurkyl]

Manifolds are explicitly stated to be Hausdorff. The reason is that it is useful. A space such as the line with a double point at the origin

------:------

satisfies all the other conditions of being a manifold, and can even have a differential structure. But the extra point at the origin essentially contributes nothing beyond its identity: any continuous function, vector field, tensor field, or whatever has exactly the same value at both of the points at the origin.


Metrizability implies Hausdorff. Nonmetrizability does not imply non-Hausdorff. The famous example of a nonmetrizable manifold, the long line, is clearly a Hausdorff space.

Incidentally, some authors require that manifolds must also be metrizable, so the long line wouldn't be a manifold.


In pseudoRiemannian geometry, the word "metric" refers to the metric tensor, rather than the metric one would study in the context of metric spaces.

 

[cesiumfrog]

does anyone have an example of a manifold (which must by definition be covered by charts diffeomorphic to the familiar R^n) that is not Hausdorff? I gather it's supposed to be possible, but I've never seen an example.

It seems trivial that Euclidean space satisfies the Hausdorff axioms, right? And by the definition of a manifold, around every point in a manifold is a region "topologically equivalent" to Euclidean space. Sound's like a Diff. Geom. homework question now, "prove homeomorphisms preserve the Hausdorff condition".

if a manifold has a Riemannian metric[..]

From above, I think "manifold" implies Hausdorff, which is stronger (ie. doesn't depend on there actually being a metric, let alone on whether it is Riemannian).

an example of a non metrizable (and hence, not Hausdorff) manifold

I think that example of a topological space is not a manifold.

 

[Hurkyl]

It seems trivial that Euclidean space satisfies the Hausdorff axioms, right? And by the definition of a manifold, around every point in a manifold is a region "topologically equivalent" to Euclidean space. Sound's like a Diff. Geom. homework question now, "prove homeomorphisms preserve the Hausdorff condition".

It would, if being Hausdorff was a local property. (It's not)

Again, I cite the line with two origins:

-----:-----

This space is locally homeomorphic to Euclidean space. Here's a neighborhood of the top point at the origin that is homeomorphic to an interval:

--[/color]---'---[/color]--[/color]

Here is a neighborhood of the bottom point at the origin that is homeomorphic to an interval:

--[/color]---.---[/color]--[/color]


And, of course, any other point can be covered with an interval that misses the origin entirely.

 

[pervect]

Manifolds are explicitly stated to be Hausdorff. The reason is that it is useful. A space such as the line with a double point at the origin

------:------

satisfies all the other conditions of being a manifold, and can even have a differential structure. But the extra point at the origin essentially contributes nothing beyond its identity: any continuous function, vector field, tensor field, or whatever has exactly the same value at both of the points at the origin.


Metrizability implies Hausdorff. Nonmetrizability does not imply non-Hausdorff. The famous example of a nonmetrizable manifold, the long line, is clearly a Hausdorff space.

Incidentally, some authors require that manifolds must also be metrizable, so the long line wouldn't be a manifold.


In pseudoRiemannian geometry, the word "metric" refers to the metric tensor, rather than the metric one would study in the context of metric spaces.

OK, thanks, this is just the sort of info I was looking for. I don't believe that Wald, for instance, defines a manifold as necessarily being "Hausdorff" in "General Relativity".

What Wald does say is "Viewed as topological spaces, we shall consider in this book only manifolds which are Hausdorff and paracompact; these terms are defined in Appendix A". So he basically avoids the issue without discussing the details.

As far as a "metric space" goes - for the most part in GR I'm interested in spaces with a metric tensor. Is it correct, mathematically, to view a space with a metric tensor as a specific example of a more general "metric space"?

 

[Hurkyl]

The definition I'm going by for manifold came from Spivak, and Wikipedia agrees too; maybe some authors permit an even more definition of manifold that permit them to be non-Hausdorff, but I haven't seen it.

A Riemannian metric tensor (on a sufficiently "small" manifold) can be used to create a metric space structure, and I imagine the metric tensor can be recovered from the metric space structure.

But I don't know of a similar thing that can be said for a non-Riemannian metric tensor. (Of course, that doesn't mean such a thing doesn't exist; I just don't know of it)

 

[vanesch]

Maybe this is a moot point, but I had the impression that MeJennifer's claim comes from the following: the metric on a pseudo-Riemannian manifold is not a positive-definite metric, and as such, cannot be used to define a generating set of open balls which generates a topology. The topology on space time, however, is not THIS one. It is not the one that "follows from the pseudo-riemanian metric", but another one, that follows from the local equivalence to an Euclidean space R^4 (a chart).
The topology on R^4 comes from the Euclidean metric on R^4, and not from any pseudo-riemanian metric induced on R^4.

At least, that's how I understand it.

 

[quasar987]

The topology on space time, however, is not THIS one. It is not the one that "follows from the pseudo-riemanian metric", but another one, that follows from the local equivalence to an Euclidean space R^4 (a chart).

Can you say that again please? The topology on space-time is not the open ball aka metric topology induced by the metric tensor, instead it is...what? What is this topology "that follows from the local equivalence to an Euclidean space R^4 (a chart)"?

 

[vanesch]

Can you say that again please? The topology on space-time is not the open ball aka metric topology induced by the metric tensor, instead it is...what? What is this topology "that follows from the local equivalence to an Euclidean space R^4 (a chart)"?

Well, a smooth manifold is defined by an atlas, which contains diffeomorphisms between open sets in R^n (with the usual Euclidean metric + topology) and the abstract manifold M, which is, in order to even be able to define "continuous", also equipped with a topology. The charts are hence local "isomorphisms of topology" (homeomorphisms is the name I think). Now, once we have the manifold, we can define a symmetric tensor over it which will give us the semi-Riemanian metric. However, the "balls" of the semi-riemanian metric are not a generator for the topology on the manifold which was used to define the charts!

Indeed, look at simple Minkowski space M. There is one single chart in an atlas, which is a mapping from M to R^4: c: p->(x0,x1,x2,x3).

If we take as "fundamental balls" with radius epsilon in the Minkowski metric, we have, for instance,

(x0-y0)^2 - (x1-y1)^2 - (x2-y2)^2 - (x3-y3)^2 < epsilon

as its image through the chart in R^4. But that's not even a compact set in R^4 under its usual topology! It means that points which are (nearly) lightlike connected, are in "close neighbourhood". But, using an intermediate point, ANY TWO POINTS can be lightlike connected! So this means that any two points are in the neighbourhood of each other. This means that continuous functions must be constant over M, and even the coordinate functions are not continuous functions.

 

[MeJennifer]

Maybe this is a moot point, but I had the impression that MeJennifer's claim comes from the following: the metric on a pseudo-Riemannian manifold is not a positive-definite metric, and as such, cannot be used to define a generating set of open balls which generates a topology.

Correct, furthermore sequences do not neccesarily converge to one single point in non-Hausdorff spaces.

The topology on space time, however, is not THIS one. It is not the one that "follows from the pseudo-riemanian metric", but another one, that follows from the local equivalence to an Euclidean space R^4 (a chart).
The topology on R^4 comes from the Euclidean metric on R^4, and not from any pseudo-riemanian metric induced on R^4.

At least, that's how I understand it.

I know what you mean, but there is absolutely nothing in SR and GR that implies that! Forgive me for taking the liberty to think that that "solution" is simply "mathematical spielerei" to avoid, rather than adress, the issue.

 

[vanesch]

I know what you mean, but there is absolutely nothing in SR and GR that implies that!

I think there is, but it is implicit. The point is that in GR, one requires "smooth" transition functions between "coordinates", but in order to say what "smooth" is, one uses the standard 4 real variable into 4 real variable diffeomorphisms, as in calculus. That uses implicitly the standard topology on R^4 which is generated by the open balls of the Euclidean metric, but one doesn't say so explicitly. It is essentially the topology used to decide which transition functions are diffeomorphisms which defines the topology on the manifold, and the "accepted" diffeomorphisms in GR (the "changes of coordinates") are smooth functions between R^4 -> R^4 with standard (Euclidean) topology.
One doesn't use any bizarre "Minkowski topology" (whatever that may mean!) to find out whether coordinate transformations are "smooth".

EDIT: in fact, when limiting oneself to topology, one should just use "continuous", but I had the impression that in relativity, one wants to use smooth manifolds, and not just topological manifolds.

 

[MeJennifer]

but I had the impression that in relativity, one wants to use smooth manifolds, and not just topological manifolds.

Sure, how otherwise are you going to use differential calculus.

But, however, how can a smooth manifold ever have singularities if only smooth deformations are allowed?

 

[coalquay404]

The mathematical model of special and general relativity is most certainly incomplete.
Space-time, as modeled by a Riemann manifold, is not Hausdorff.

This is technically correct, but only technically. If you want to talk about spacetime solutions to the EFEs it is customary to utter the following incantation before one begins:

Spacetime is a four-dimensional paracompact, connected smooth Hausdorff manifold without boundary.

Non-Hausdorff spacetimes don't arise directly in the context of the EFEs. Where they do appear is when one looks at certain quotient spaces in general relativity. The example with which I am most familiar is that of the quotient spaces involved in the maximally extended covering spaces for Taub-NUT space. It's a reasonably interesting area (if you like that sort of thing) but not a particularly fruitful one. Petr Hajicek wrote a nice paper on causality in non-Hausdorff spaces in the seventies, when such things were popular, but I'm not aware of any more recent work.

In fact the concept of a manifold being Hausdorff did not even exist when SR and GR were developed and no-one, not even Cartan, has resolved it since. Space-time, as modeled by a Riemann manifold, does not even have a valid metric.

There are several confusing things here. Firstly, what is the problem you think needs to be resolved? Secondly, there may perhaps be some difficulty with your terminology. Don't refer to things such as Riemannian or pseudo-Riemannian manifolds; it will only confuse people when you try to talk to somebody who's not familiar with the subject area. It's better in the context of general relativity to talk simply of a differentiable manifold (with the explicit degree of differentiability being largely unimportant unless you're interested in studying reductions of the EFEs in the context of, say, weighted Sobolev spaces), and then to specify the type of metric structure that you're working with on that manifold. For example, classical general relativity (I'm talking about Einstein's way of viewing it) deals with differentiable manifolds which are endowed with a pseudo-Riemannian metric structure, i.e., a space of indefinite metric forms. Modern GR deals with topological identifications between a spacetime and a foliation by a three-dimensional manifold endowed with positive definite metric structure.

Sure some like to shove it under the rug by simply placing the term pseudo in front of everything and then claiming that all is well. But that obviously won't do anything for those who like to think exact!
For them it is like someone saying "Well, admittedly it is not true but for sure it is pseudo-true".

I can assure you that nobody who knows what they're talking about "shoves [anything] under the rug". In the field, when people use pseudo-Riemannian or Riemannian, their usage is quite clear.

 

[vanesch]

Sure, how otherwise are you going to use differential calculus.

Right, but differential calculus is based upon standard R^4 topology.

But, however, how can a smooth manifold ever have singularities if only smooth deformations are allowed?

I'm not 100% sure that the manifold itself becomes non-smooth at GR singularities. Rather, the METRIC becomes singular, no ? But I'm no expert, maybe someone better versed in this can correct me.

 

[quasar987]

Right, but differential calculus is based upon standard R^4 topology.
I'm not 100% sure that the manifold itself becomes non-smooth at GR singularities. Rather, the METRIC becomes singular, no ? But I'm no expert, maybe someone better versed in this can correct me.

I of course am not that "better versed" person, but I was under the impression that there were two types of singularities: real ones and those that can be made to disappear under a proper change of coordinates. For example, when one changes from spherical coordinates to the Kruskal-Szekeres coordinates in the black hole solution, the singularity at the horizon disappears but not the one at the center of the BH.

 

[vanesch]

I of course am not that "better versed" person, but I was under the impression that there were two types of singularities: real ones and those that can be made to disappear under a proper change of coordinates. For example, when one changes from spherical coordinates to the Kruskal-Szekeres coordinates in the black hole solution, the singularity at the horizon disappears but not the one at the center of the BH.

Yes, but that is a singularity in the metric. That is, some part of the metric tensor blows up (and if it blows up in one coordinate system then it blows up in all). But the metric is just a 2-tensor defined over the manifold. This is like having, in Euclidean space, say, a vector field which takes on the form:

v(p) = (x+y) 1_x + (z+x) 1_y + (z^3 + y^3)/(x^2+y^2) 1_z

Now, this vector field blows up on the z-axis, and this is not a coordinate system problem: in any coordinate system, this vector field blows up. However, does this mean that the Euclidean space itself, over which the vector field is defined, blows up ?

In the same way, it is not because a 2-tensor (which happens to be called a pseudo-riemanian metric) blows up at a certain point of a manifold, that the manifold itself has a problem there.

Look for instance at the Kerr BH, in the Kerr form. The metric goes singular for x^2 + y^2 = a^2 ; z = 0. But the coordinate mapping is still smooth there.

Again, caveat: I'm no expert at all in this!

 

[George Jones]

Maximally extended Schwarzschild (and Kerr) are both timelike geodeically incomplete. There exist worldlines of freely falling observers that end after a finite amount of proper time - these freely falling observers "fall off of spacetime" after a finite amounts of time have elapsed on their watches.

Some spacetimes exhibit this type of behaviour without having curvature invariants blow up. Other spacetimes that are geodesically complete are classified as singular because they have inextenable worldlines that have bounded 4-acceleration - observers in rockets that have finite thrusts "fall off the edge of spacetime".

Of course all of this - geodesics, 4-acceleration, etc. - is determined by the metric.

 

[vanesch]

Maximally extended Schwarzschild (and Kerr) are both timelike geodeically incomplete. There exist worldlines of freely falling observers that end after a finite amount of proper time - these freely falling observers "fall off of spacetime" after a finite amounts of time have elapsed on their watches.

Some spacetimes exhibit this type of behaviour without having curvature invariants blow up. Other spacetimes that are geodesically complete are classified as singular because they have inextenable worldlines that have bounded 4-acceleration - observers in rockets that have finite thrusts "fall off the edge of spacetime".

Of course all of this - geodesics, 4-acceleration, etc. - is determined by the metric.

Could we picture this "falling off spacetime" as just a quirk of the metric, as in the following sense:
picture the Euclidean plane as a manifold (on which we can define a traditional (x,y) chart). Now define the metric tensor ds^2 = 1/(1+x^2 + y^2)^2(dx^2 + dy^2). Clearly, any radial line going "all the way" on the manifold will just have a total accumulated length of pi/2 (from the top of my head, if I'm not mistaking). So the whole Euclidean plane just "looks like a disk" from the metric PoV (of course, in this case, we could probably extend it).

 

[MeJennifer]

Firstly, what is the problem you think needs to be resolved?

Well it must be my lack of intelligence to see how the following is consistent:

1. Gravity is curvature of space-time
2. Space-time is a Riemannian manifold with a positive definite metric
3. The shape of this manifold is determined by the EFE in terms of a metric tensor that is not a valid metric.

I suppose the consistency of the emperor's fabric is beyond my level.

 

[cesiumfrog]

1. Gravity is curvature of space-time
2. Space-time is a Riemannian manifold with a positive definite metric
3. The shape of this manifold is determined by the EFE in terms of a metric tensor that is not a valid metric.

I suppose the consistency of the emperor's fabric is beyond my level.

(2) is mistaken. Space-time is a not a Riemannian manifold (instead it is a different particular type of manifold, specifically the type named psuedo-Riemannian), with a metric that has a Lorentzian signature (hence is not positive definite).

I thought you'd seen this earlier along the thread, but nonetheless.. Now that you know the 2nd fibre that the emperor really uses, do you still have any issue with the fabric's consistency?

 

[coalquay404]

Well it must be my lack of intelligence to see how the following is consistent:

1. Gravity is curvature of space-time
2. Space-time is a Riemannian manifold with a positive definite metric
3. The shape of this manifold is determined by the EFE in terms of a metric tensor that is not a valid metric.

I suppose the consistency of the emperor's fabric is beyond my level.

I wouldn't say you lack intelligence, but you're definitely confused. (2) is incorrect. It should read "Spacetime (M,g) is a four-dimensional paracompact, connected smooth Hausdorff manifold M without boundary, and with an indefinite (or, if you like, pseudo-Riemannian) metric structure g."

(3) is also completely incorrect.

 

[George Jones]

Could we picture this "falling off spacetime" as just a quirk of the metric, as in the following sense:
picture the Euclidean plane as a manifold (on which we can define a traditional (x,y) chart). Now define the metric tensor ds^2 = 1/(1+x^2 + y^2)^2(dx^2 + dy^2). Clearly, any radial line going "all the way" on the manifold will just have a total accumulated length of pi/2 (from the top of my head, if I'm not mistaking). So the whole Euclidean plane just "looks like a disk" from the metric PoV (of course, in this case, we could probably extend it).

What is the metric when expressed in terms of coordinates rho and theta, where r = tan(rho), and r and theta are related to x and y in the standard way?

 

[pervect]

I don't know if this is the source of the confusion, but I hope we can all agree that points connected by null curves, are not necessarily neighbors in the sense used by topological spaces.

Thus if I am looking out at the night sky at Andromeda, I see some particular event happening that is not "close" to me, even though it is connected to me by a light ray which ideally has a Lorentz interval of zero.

What makes the Lorentz interval interesting is not that it defines points which are "neighbors", "close" or "nearby", (it doesn't do that function) - rather, it is interesting because it is an invariant of space-time, something that different observers can agree on.

 

[vanesch]

What makes the Lorentz interval interesting is not that it defines points which are "neighbors", "close" or "nearby", (it doesn't do that function) - rather, it is interesting because it is an invariant of space-time, something that different observers can agree on.

yes, that was my point. The topology over the spacetime manifold is NOT given by the lorentz metric, but rather by the standard topology over R^4 (which is, in itself, given by the Euclidean metric over R^4). This is because we consider "coordinate transformations" (which are nothing else but transition maps between charts) to be continuous, when they are continuous in the standard calculus sense.