Definition of a smooth manifold.

1. Dec 13, 2005

vanesch

Staff Emeritus
Is it correct that the definition of a smooth manifold is an equivalence class (under diffeomorphism) of atlasses ?

(this discussion is related to a discussion I try to start in general relativity concerning the hole argument).

2. Dec 14, 2005

Hurkyl

Staff Emeritus
Nope. It's just the pair of a topological space M and a (maximal, $C^{\infty}$) atlas on M.

3. Dec 15, 2005

vanesch

Staff Emeritus
Ok, first a question: if two topological spaces are homeomorphic, they are in fact the same topological space, right ?

Of course you can have two different constructions, which are ALSO satisfying the definition of a topological space (but which have additional structure). If there exists then a homeomorphism between both, that doesn't mean that the two spaces are identical, but that's because of the additional structure. But, an abstract topological space (M1,T1) and another abstract topological space (M2,T2), when they are homeomorphic, I think one can say that this is the SAME topological space, or am I wrong here ?

Now, imagine we have an atlas on some abstract set M. Do we need to talk about the set M, or can we construct the set M from the atlas itself (including the charts, overlapping parts and transition functions) ? I thought the latter. It is sufficient to have an atlas, with all transition functions between overlapping parts of charts, and then the set M is the equivalence class of the points of the different charts, over the transition functions ; meaning: if a point in a part of a chart does NOT belong to an overlapping part, then it is, by itself, a point of M (an equivalence class consisting of just one point), and if a point of a chart does belong to an overlapping part, then all the equivalent points in other charts form one equivalence class and hence this is one point of M.
I would think that the topology of the charts induces a topology over M, and as such, turns M into a topological space.

So doesn't just specifying an atlas (as I said, including overlapping regions and transition functions between the overlapping regions) give us a smooth manifold, without even having a topological space as starters, but is generated by the atlas ?

And once we are there, doesn't a diffeomorphism just "add more charts and overlap functions" to the atlas ?

Or am I totally off here ?

4. Dec 15, 2005

matt grime

"Same" is not a formal mathematical term. Homeomorphic objects are isomorphic in the category Top, but are not in general equal.

An isomorphism class of topological spaces with smooth atlases is exactly that: an isomorphism class of smooth manifolds.

It is not a smooth manifold itself.

In fact it is not even close: it isn't even a set but a proper class of objects.

Last edited: Dec 15, 2005
5. Dec 15, 2005

vanesch

Staff Emeritus
Ok, let me put my ramblings in context. The context is the so-called "hole argument" of Einstein in general relativity. I started out in the relativity forum, then - because the question sounded more mathematical - I came over here, and now I find myself again switching back to the relativity case :grumpy:

The entire problem seems to be that in general relativity, "spacetime" could be a smooth manifold M1 with a metric, but could also be a smooth manifold M2 with a metric, where there is a diffeomorphism mapping M1 onto M2 and taking along the metric, and that seemed to be (and still seems to be) a problem, because then "we don't know what manifold general relativity tells us is out there".
Now, my point was that the only structure we know of the smooth manifold, is nothing else but an atlas (with a metric): we just know coordinate patches and their metric, which are locally solutions to the Einstein equations (all this is in patches of R^4). So this atlas is the defining property of the manifold (at least, that's how I understand things here, please correct me if I'm wrong) ; it is not that we had a pre-existing topological space (constructed in some way or other, and hence possessing extra structure) on which we constructed also an atlas. So my idea was that it makes simply no sense to talk about two _different_ such manifolds which are related by a diffeomorphism, as they both originate from one and the same atlas, and that by the very nature of such a manifold, it will *always* be defined "up to a diffeomorphism" (as this is the relationship that conserves the entire structure).

As such, I wondered if, in the definition of a manifold, it even made sense to start with an abstract topological space ; and that the atlas that goes with it is not, by itself, sufficient.

You can make up from what I write that I'm not a mathematician, so probably you can teach me a lot here (I'm all ears).

Let me take a (to me) more familiar example to try to make clear the problem that I'm trying to make clear for myself. Consider 2-dimensional real vectorspaces V1 and V2.
Now, to me, there is only ONE 2-dimensional real vector space (as a mathematical idea). If people would tell me that there are several, but that they are all connected with a linear invertible transformation (the isomorphism of vector spaces), then that doesn't make much sense to me, in the following way:
Imagine that someone objects to Euclidean geometry (with an origin) in 2 dimensions, that the relationship between "the physical plane" (somewhere in Egypt or so) and a 2-dimensional vector space is ambiguous, because I can always think of a linear transformation of the 2-dimensional vector space, and as such, that the "laws of Euclidean geometry" do not define uniquely the 2-dimensional vector space (because it doesn't say *which* one, we can always use a linear transformation and obtain the same laws), then that wouldn't be a very serious objection to saying that the mathematical structure that represents the "physical Euclidean plane" is THE two-dimensional real vectorspace, would it ?

The reason why there is, in this sense, only ONE two-dimensional real vectorspace, is that you always end up working with R^2, and that all two-dimensional real vector spaces are (concerning the vector space structure) isomorphic to R^2.

Now, back to general relativity. Imagine that, through the Einstein equations, we arrive at a set of charts (as solutions on patches of R^4, solving the Einstein equations). That is all we know of our "manifold". But isn't that sufficient ? Isn't that completely defining the manifold we're after ? So what does it mean to say that we "only defined a manifold up to a general diffeomorphism" ?
Isn't that similar to saying that, when working with R^2, we only defined the 2-dimensional real vector space up to a linear transformation ?

EDIT: all this is just part of an effort on my part to wrap my mind around this "hole argument" in general relativity...

Last edited: Dec 15, 2005
6. Dec 15, 2005

George Jones

Staff Emeritus
I would like to get more involved in this thread, but right now I'm goofing off and procrastinating, so I'm just going to make a few disjoint comments.

1) A manifold is pair consisting of a topological space and either a maximal atlas or an equivalence class of atlases. This is because 2 atlases are equivalent if and only if they determine the same maximal atlas.

2)
I think you're thinking of topology, not set. A manifold can be defined as a set M and a maximal atlas. An atlas consists of a bunch of charts, and a chart maps a subset of M into a subset of R^n. Without the set, I don't see how charts (and thus an atlas) can be defined. Given a set M and an atlas, I do think that the atlas can be used to generate a topology on M.

3) Does homeomorphism mean the same?

A category is, roughly, a bunch of sest that all have the same structure imposed on them, together with a bunch of mappings that preserve the stucture. Examples: 1) category of sets with the mapping being mappings; 2) category of groups with the mappings being homomorphisms; category of vectory spaces with linear mappings; 4) category of topological spaces with continuous mappings. If, within a particular category, a mapping is bijective and preserves structure both ways, then it is an isomorphism. Example: homeomorphisms and the category of topological spaces.

If there is an isomorphism between 2 members of a give category, are they the same? Mathematicians like Hurkyl and Matt probably don't find this question to be particularly meaningful.

Let give another, more detailed example. R is set that also is: 1) (with its standard topology) a topological space; 2) an abelian group, with group "multiplication" being addition, and the group identity being 0; 3) a topological group, since the group operations are continuous. Similarly, R+, the set of all positive real numbers, is a topological group, with group multiplication being (numeric) mutiplication and the group identity 1.

These 2 topological groups don't look that similar - the group operations and identities look very different - but they are ismorphic as topological groups. Define f : R -> R+ by f(x) = e^x for every x in R. Then f(a+b) = f(a)f(b), and f is a homomorphism. f has all the other requires properties as well.

Regards,
George

7. Dec 15, 2005

matt grime

If you have a metric then you are talking about more than a mere smooth manifold entirely.

On another point, you only refer to THE something or other if any two realizations of it are canonically isomorphic, and your examples fail to be uniquely determined since you are taking bases.

It also makes perfect sense to talk about two different, yet diffeomorphic smooth manifolds. Consider the following spaces:

The space of lines in R^(n+1)\{0}, and the sphere S^n with antipodal points identified. These are smooth projective varieties, obviously not the equal but isomorphic; they originate from distinct spaces and atlases.

And one final thing, the underlying manifold as a topological construct is not sufficient to fix the Riemannian metric. There are an uncountable number of topologically equivalent Riemannian tori.

Last edited: Dec 15, 2005
8. Dec 16, 2005

vanesch

Staff Emeritus
These are the typical examples one finds in texts on manifolds, but I find this highly disturbing, because the approach is different, from what you have to do when you DON'T have an a priori structure from which you extract your manifold. The lines in R^(n+1)\{0} is a very rich construction ; it is much more than a smooth manifold ; it only happens to _also_ fit the defining axioms of a smooth manifold, but it is much more ; in the same way, S^n is much more than a smooth manifold. And it is this extra structure which allows you to distinguish them. But if you limit yourself to their "manifold" structure, then I'd say that these are two things are the same.

Imagine, to make the link to the "Einstein hole argument", that we have a physical theory that, when made certain choices, concludes that "each electron corresponds to a line in the set of lines in R^(n+1)\{0}". Now, imagine that if you make other choices, that in that theory you come to "each electron corresponds to a point of S^n".
Now, should one go and complain about the essential incompleteness of such a theory, because it cannot determine if this electron over here is going to be a "line in R^(n+1)\{0}", or whether it will be "a point in S^n" ? (which, to me, is the essence of the hole argument) And that, as such, the theory is not capable of giving us an ontology of what an electron is ? line or point of S^n ?
Or is it more reasonable to say that the theory tells us that the electrons are points of a smooth manifold, which can, for instance, be represented by a construction like S^n, or if we like, by R^(n+1)\{0}, but that all this is just different representations of the same abstract manifold M ? And that as such, the mathematical representation of an electron is an element of M, and that the theory determines that precisely ?

In GR, you don't have an a priori construction (such as "lines in R^(n+1)") or so. You only have an atlas, from which you're supposed to construct the manifold behind it. That's a totally different approach than when you START from a construction (such as S^n) and you are going to explore this construction from a "manifold" point of view.

9. Dec 16, 2005

vanesch

Staff Emeritus
Ok.

Ah, but you can define the set in the way I indicated (at least, I think! Just to be shot down by a mathematician maybe).

Imagine I tell you:

I have an atlas consisting of two charts (I just talk, for the moment about the image of the chart), U1 and U2, where U1 = ]0,1[ and U2 = ]0,1[.
Now, they have two domains of overlap:
]0,0.1[ of U1 overlaps with ]0.9,1[ of U2 and the mapping is u1 -> u2=u1+0.9

]0.9,1[ of U1 overlaps with ]0,0.1[ of U2 and the mapping is u1-> u2=u1-0.9

This is what I call "my atlas" (the images of the charts, the definition of the overlapping regions of the images of the charts, and the transition functions between them).

Now, does this, or doesn't this, define a smooth manifold ?

I would think it does !
In the following way:
the set M consists of all points of U1 and U2 which don't belong to an overlap region, together with the equivalence classes under the overlapping functions, of the overlapping regions.
So in our case, M consists of the points of ]0.1,0.9[ of U1, the points ]0.1,0.9[ of U2, and then the set of pairs {u1,u1+0.9} for u1 in ]0,0.1[ and the set of pairs {u1,u1-0.9} for u1 in ]0.9,1[.
The topology induced on M is the standard one on U1 and U2 ; as such, M becomes a topological space. And the chart (as a function) is also evident (the mapping from a part of M into U1 and U2).

As such, we constructed a manifold M, with an atlas.

And one can see that this manifold is diffeomorphic (unless I made a mistake) to S^1.

I somehow understand that. But when a physicist uses a mathematical structure which is supposed to represent the ontology of some physical system, I think "isomorphism" does mean "the same", in the sense that one shouldn't complain about being able to obtain isomorphic, but different structures, and not to know which one is now "the real one".

And I have the impression that this confusion is what the hole argument is all about. Look at my example of a constructed M. I could now construct a different M1, where I apply a smooth function to each of the elements of M. But is M1 "something different" ?

Yes, but that's because R is a specific construction, which is then observed to ALSO satisfy the axioms of different mathematical ideas, such as "topological space", "abelian group", ...

But let me give another example: consider the set of decimal expansions (with the necessary identifications) R_blue, and consider the set of dedekind cuts R_red... are these now DIFFERENT real number systems ? Or is there just one real number system, and these are different constructions of the same idea ?

Well, to me, these two groups, as groups, are the same of course. We've just found two different constructions to represent the group.
The constructions have extra structure and are different in their extra structure.

10. Dec 16, 2005

matt grime

I really don't get the problem; as you yourself have explained you can only recover the underlying manifold upto diffeomorphism from its atlas. What is bad with that? It is (a modern belief that it is) important in maths to distinguish between things that are equal and things that are isomorphic.

And agian you're using 'the same' as if it were meaningful in a rigorous sense! If you want to be rigrous either use equal or isomorphic.

11. Dec 16, 2005

vanesch

Staff Emeritus
Well, it was such an important problem to Einstein, that he refrained from considering generally covariant equations for two years and even got convinced that such a theory didn't make sense (the "hole argument"). And the argument still holds somehow.

http://plato.stanford.edu/entries/spacetime-holearg/

Well, I don't understand the difference. Maybe you can help me.

I can understand that two CONSTRUCTIONS are different (simply because they are constructed differently, as your example of lines in R^(n+1)\{0} and S^n indicated), but I cannot understand that two abstract structures, when all of the structure is equivalent between the two, are somehow different. But, as I said, I'm not a mathematician, so there must be a good reason why mathematicians insist upon the difference. Nevertheless, it gives interpretational problems in physics.

I guess if I learned some category theory that this would be clearer to me, but I never got around doing that...

12. Dec 16, 2005

matt grime

From reading that article the problems are nothing to do with category theory and entirely to do with physics and physical interpretation.

But in the unrelated (I think) notion of category theory, owing to another thread I came up with this idea of why equal and isomoprhic are important.

Consider the n-gon for n even, eg square or hexagon, and consider its symmetry group. Then the stabilizer of a vertex x is equal to the stabilizer of its oppositve vertex ie it is genuinely the 'same', but these are not equal to the stabilizers of any other vertices, though they are isomorphic as abstract groups.

Last edited: Dec 16, 2005
13. Dec 16, 2005

Hurkyl

Staff Emeritus
I'm glad you think that way vanesch! So, you won't mind trading your set of ten \$1 bills for my set of ten pennies? (After all, they're isomorphic sets, so they cannot be different... )

14. Dec 16, 2005

mathwonk

geez...you guys have infinite patience; to me there is no mathematical discussion here at all.

this is like asking if words are collections of symbols.... wellllll... yeah, but that's really not the point of them!

a manifold is a geometric object, and there is a rule for deciding if two of them are the same. try to understand some examples of equivalent ones, and what the equivalence relation means.

charts are just B.S. i.e. a manifold is a geometric object with a concept of distance, (it might as well have one for beginners), and the ability to take derivatives, and if your eyesight isn't too good, it looks like you are just in euclidean space. that's all.

Last edited: Dec 16, 2005
15. Dec 17, 2005

matt grime

Having had time to digest that link I think I have the idea it is trying to say something about the suitability of differential geometry as a model of space-time *if* we make some assumptions about the existence (substantiation) of the universe.

16. Dec 17, 2005

vanesch

Staff Emeritus
Nah, there's extra structure to it

17. Dec 17, 2005

matt grime

Yes. and your manifolds need to have extra structure if they are to be used for space-time stuff.

18. Dec 17, 2005

Hurkyl

Staff Emeritus
There's an analogy here.

The two sets are isomorphic, but there's clearly a difference between them. Now, can you extend the analogy to topological spaces?

19. Dec 17, 2005

mathwonk

here are some examples: as metric spaces a circle and an ellipse are not isoimorpohic, because the distances change, but as smooth manifolds they are the same, because the projection of an ellipse inscribed in a cxircle, out onto the circle is a smooth isomorphism. I.e. although not distance preserving, it is differentiable with a differentiablke inverse.
A circle inscribed in a square is not equivalent to the square however, since the projection fron the circle to the square does not make the tangent line to the square correspond linearly to the tangent line to the circle. Indeed the square has no unique tangent line, so is not really a smooth manifold at all.
the curve y = x^3 is a smooth manifold and is smoothly equivalent to the x axis via projection, hence also to the y axis, by the map taking (0,y) to (y,0) and then projecting up to the curve. But the projection of y=x^3 onto the y axis is not a smooth equivalence because it smushes the tangent line of the curve at (0,0) to one point, hence the smooth map carrying one space to the other does not induce linear isomorphisms of every approximating linear space.
In (ideal) spacetime, presumably (I am a novice here) there is not just a smooth structure but also a metric structure given by the lorenz metric, and equivalences should preserve that too.
but for the purposes of study one can ask wkjetehr differing models for spacetime are continuously equivalent, smoothly equivalent, or even smoothly equivalent while also preserving distance.
so as others have said, a mathematical model for a space consists of a set of points plus various notions associated to those points when considered as a whole, like "nearness", "local linearity", actual distance, angles between curves, straightness ("minimal length" curves).
These various concepts are measrued and quantified by introducing coordinates locally, i.e. numbers, which are called "charts".
Of course a chart works only locally so it is needed to use several to cover the whole space, and the collection is called an atlas. Different charts overlap near each point, so they must be able to give equival;ent answers to the same questions asked near that p[oint.
hence the two sets of coordinartes obtained near the same point must themselves be related continuously, or smoothly, or metrically, or conformally, or etc.....

It is meaningless to ask which mathematical model is the "real" spacetime, since all comprise only a tiny fraction of the proeprties of real spacetime, and those too they extend to meaningless levels of precision, where quantum models are more apropriate.
i hope this is useful.

Last edited: Dec 17, 2005
20. Jan 1, 2006

bchui

The definition of "manifold" should not required any atlas $$\Phi$$ at all.

A well know fact is that "any curves on the plane which intersect itself is not a manifold", say for example $$M=\{ (x,y)| (x^2+y^2)^2 =x^2-y^2\}$$

Closed intervals $$$a,b$$$ are also non-manifolds since the neighborhoods of the end points cannot be homeomorphic to any open subsets of the real line.

The sphere [tex]S^n[tex] is a manifold due to the existence of an atlas

A "manifold" or the "C^r topology" are only sets with the property:
a class of "compatible and maximal atlas" can be defined on it but has nothing to do with the atlas itself.