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Question about differentiable structures

  1. Nov 11, 2008 #1

    quasar987

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    If M is a topological manifold, a smooth structure A (or maximal atlas) on M is a set of smoothly compatible charts of M that is maximal in the sense that if we consider any chart that is not in A, then there is some chart in A with whom it is not smoothly compatible.

    Now, it is a fact that some topological manifolds admit no smooth structure and others admit many distinct ones. One could ask how many distinct smooth structures exists on a given manifold, but differential topologists choose to classify the smooth structures on a manifold up to diffeomorphism (meaning that two smooth structures A, A' are diffeomorphic if there exists a diffeomorphism F:(M,A)-->(M,A')). Could someone explain this choice to me?

    Because what is a smooth structure if not a way to make sense of differentiability of map on a topological space? So at first I thought that diffeomorphic smooth structures admit the same smooth maps (In the sense that for any smooth manifold N, a map G:(M,A)-->N is smooth if and only if G(M,A')-->N is smooth). But this is not the case, for the following reason:

    Lemma: Suppose that A, A' are two smooth differentiable structures on M that admit the same smooth maps. Then A = A'.

    Proof: To see this, consider (U,f) a chart on (M,A). Then f must be smooth as a map on (M,A'). This means that for any chart (V,g) in (M,A'), the coordinate representation of f, namely f o g^-1, is smooth. Inversely, given a chart (V,g) in (M,A'), g must be smooth on (M,A), so for any chart (U,f) in (M,A), the coordinate representation of g, namely g o f^-1, is smooth. So every pair of charts (U,f), (V,g) in A and A' respectively are smoothly compatible. This means that A=A'. QED

    If it were true that two diffeomorphic smooth structures admit the same smooth maps, then there would not exist distinct diffeomorphic smooth structures, which is absurd (see 2nd paragraph).
     
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  3. Nov 12, 2008 #2
    I think the first thing you want do is the third problem in Lee's introduction to smooth manifolds. it says that given any topological manifold of dim > 0 with a smooth atlas, one can construct uncountably many distinct smooth structures. with this in mind, it seems that the equivalence relation of having the same smooth structure is way too restrictive. So, a diffeomorphism is a less strict equivalence relation: first you have that the manifolds don't have to be equal and the topological data is preserved & second you have various things about the smooth structures are preserved. later you can be less restrictive by considering the equivalence relation of spaces being smoothly homotopic.
     
  4. Nov 12, 2008 #3
    If it were true that two diffeomorphic smooth structures admit the same smooth maps, then there would not exist distinct diffeomorphic smooth structures, which is absurd (see 2nd paragraph).

    there most certainly does exist two distinct diffeomorphic smooth structures. (R,{identity}) and (R, f(x)=x^3) is an example. in this case, it is easy to see that they do not admit the same smooth maps. for example, the map f(x)=x^1/3 is smooth on the second smooth structure, but not smooth on the first. so, your conclusion doesn't follow from your hypothesis.
     
  5. Nov 12, 2008 #4

    Hurkyl

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    It's the same thing you do with any mathematical structure. Diffeomorphisms are to differentiable manifolds as bijections are to sets, isomorphisms are to vector spaces, homeomorphisms are to topological spaces, isomorphisms are to groups, et cetera.
     
  6. Nov 12, 2008 #5
    BTW the problem with your "proof" is the second sentence above, which is decidedly not true in general.

    I think you are mixing up the terms "smooth" and "homeomorphic" in much of your narrative.
     
  7. Nov 12, 2008 #6

    quasar987

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    I am aware of this problem from Lee and also that the charts (R,id) and (R, x^3) are contained in two distinct but diffeomorphic smooth structures of R. In fact, these were the examples I had in mind when I wrote the first line of paragraph 2, "Now, it is a fact that some topological manifolds admit no smooth structure and others admit many distinct ones."

    As for the rest of your post, I do not understand its meaning.

    Did you miss the word "not" up there? "If it were true that two diffeomorphic smooth structures admit the same smooth maps, then there would not exist distinct diffeomorphic smooth structures, which is absurd (see 2nd paragraph)." What is absurd is the conclusion that there would not be manifolds with many distinct smooth structures.

    Don't you find it strange that two distinct but diffeomorphic smooth structures on a manifold give rise to different smooth maps. Under this circumstance, I would be hesitant to call these smooth manifolds "the same" !

    But it's true here by hypothesis. I defined what it means for 2 smooth structures to have the same smooth maps like so:

    Dfn: If M is a topological manifold and A, A' two smooth structures on M. I say that A and A' admit the same smooth map if for any smooth manifold N, a map G:(M,A)-->N is smooth if and only if G(M,A')-->N is smooth.

    In the proof, since the coordinate maps of A are smooth with respect to A, they must be smooth maps with respect to A'.
     
  8. Nov 12, 2008 #7
    Sorry, I misread the lemma. So, now I'm confused as to what the confusion is: diffeomorphic differentiable structures are essentially the same differentiable structure and will share the same differentiable functions -- up to, of course, the amount of differentiability of the diffeomorphism, C^1, C^2, ..., C^infinity,....
     
  9. Nov 12, 2008 #8

    quasar987

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    How do you get to the conclusion that diffeomorphic differentiable structures share the same differentiable functions?

    The point of the lemma is to allow for the following corollary (which constitutes the last sentence of my original post):

    Corollary: If two smooth structures A, A' on a topological manifold M are distinct, then they do not admit the same smooth maps.

    Proof: If they did, then they would be equal, by the lemma.

    In particular, even if two smooth structures are diffeomorphic, as soon as they are distinct, they do not admit the same smooth maps.

    So with that in mind, I question the reason for classifying smooth structures on a manifold only up to diffeomorphism, since arguably, two diffeomorphic smooth structure could hardly be called "the same" if they do not allow for the same smooth maps!
     
  10. Nov 12, 2008 #9

    Hurkyl

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    Nope. Their respective classes of smooth maps shold be bijective, not equal. The bijection is given by composing with the diffeomorphism.

    Compare: the following two Abelian group structures on {0, 1} are isomorphic:
    Structure 1: 0 + 0 = 0, 0 + 1 = 1, 1 + 1 = 0
    Structure 2: 0 @ 0 = 1, 0 @ 1 = 0, 1 @ 1 = 1

    Now, what group homomorphisms have these two groups as its source?
     
  11. Nov 13, 2008 #10
    I really think you should re-investigate what definition of diffeomorphism you are thinking of here, because as far as I use the term, diffeomorphisms by definition preserve the differentiable structures of the manifolds involved. Certainly, this is true if you are assuming that diffeomorphisms are smooth, i.e., the induced mappings on the coordinate patches of the manifolds are C^\infinity.

    In particular, if you have a differentiable function defined on one of the manifolds, you can use the diffeomorphism or its inverse to induce a differentiable function on the other manifold.
     
  12. Nov 13, 2008 #11

    All I am saying is the obvious fact that there exists distinct diffeomorphic smooth structures. the statement you wrote is just plain false. another way of saying what you said is that you can conclude that the smooth structures are not distinct when your manifolds are diffeomorphic and give rise to the same smooth maps. but that does not necessarily mean that there are no distinct diffeomorphic smooth structures. in fact, i provided a counter example to your claim (and noted that in that case the smooth structures do not give rise to the same maps).

    but, fine. believe something entirely false: i really don't care.
     
  13. Nov 13, 2008 #12

    quasar987

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    Mmh! Seen like that it doen't seem so strange anymore.

    Thanks for the insight.
     
  14. Sep 28, 2010 #13
    How do we prove this statment
    "the third problem in Lee's introduction to smooth manifolds. it says that given any topological manifold of dim > 0 with a smooth atlas, one can construct uncountably many distinct smooth structures."

    Thank you!
     
  15. Oct 2, 2010 #14
    In this exercise, it's critical to understand how differentiable structures may be distinct even if they are the same "up to diffeomorphism."

    As usual, it's best to start with the simplest example of a topological manifold of dim > 0, namely, the real line. Can you think of a differentiable structure on the real line that is different from the usual differentiable structure?

    Hint: pathological examples are unnecessary here. You should be familiar with homeomorphisms of the real line whose inverses aren't everywhere differentiable from calc I. So choose one, then use it to construct a differentiable structure that is diffeomorphic to, but distinct from, the usual differentiable structure.

    By definition, all topological manifolds of dim > 0 contain neighborhoods homeomorphic to open subsets of products of copies of the real line. As such, there are always uncountably many points where things could "go wrong." Use these facts to complete the proof.
     
  16. Oct 2, 2010 #15
    Thank you jasomill!
    Actually, I understood the examples you mentioned on the real line. However I do not know how to write a formal proof for the statment. I tried the following :
    Let A={gi:ui----->R^n} be an atlas on M. Let f: B^n ----> B^n be a homeomorphism s.t. not differentiable at 0 e.g. f(x)=( x1^(1/3), ....,xn^(1/3) ). Now define the "new strucure" to be the same ui's in A but the maps are f composed with gi (fgi) . This strucure is different from A because fgg^-1=f and f is not differentiable by assumption. However, I could not prove that this "strucuture" is really an atlas ( I canot decide if f g1 g2^-1 f^-1
    is smooth or not.)
    Looking to hear from you soon.
    Thank you.
     
  17. Oct 2, 2010 #16

    quasar987

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    So you have a problem with the transition functions. In the example of R, this problem does not arise because the queer atlas has only one chart.

    Choose any point P in the manifold of interest. Show that you can modify the atlas A to an atlas A' which is smoothly compatible with A (i.e. determines the same smooth structure as A) such that there exists a nbhd V of P which intersects a unique chart (U,g) of A'. Moreover, g(U) = R^n, g(P)=0 and g(V) = B^n. Then for every r in (-1,1), set fr:R^n-->R^n, fr(x):=((x1-r)1/3, x2,...,xn), and (U,gr) where gr=fr o g.

    Call A'r the atlas A' where the chart (U,g) is replaced with the chart (U,gr). The pathology happens in V and since no other chart of A'r intersects V, the transition function problem does not arise. And each atlas A'r determines a different smooth structure since for each r,s, gr o gs-1 is not differentiable at (r1/3-s,0,...,0).
     
  18. Oct 3, 2010 #17
    That's a great idea!
    Thank you for the help.
     
  19. Oct 8, 2011 #18
    Sorry quasar, but how can sure that you can modify the atlas A to A' as you describe before?. In my opinion, we can write function f: B^n---> B^n s.t x--->x.absvalue(x)^-2/3, this map is homeomoprhism and it is not differentiable at 0. Now, if we compose it by g which is taken at first, we get an atlas( to do this in fact we must modify our first atlas) but how?
     
  20. Oct 8, 2011 #19

    quasar987

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    Pick a chart (U,g) s.t. g(U)=R^n, and set V:=g-1(Bn). If (W,h) is any other chart in A around P, replace it with (W',h'), where W'=W - cl(V), and h' = h|W'.
     
  21. Oct 8, 2011 #20

    lavinia

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    In differential topology diffeomorphic manifolds have all of the same properties. That is why diffeomorphism is important.
     
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