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Intuition behind two manifolds being the same.

  1. Aug 25, 2013 #1
    In Sean Carroll's general relativity book he gives a requirement that two (differentiable) manifolds be the same manifold that there exist a diffeomorphism ##\phi## between them; i.e. a one-to-one, invertible and ##C^{\infty}## map.

    Now I wanted to get some intuition why this is the best definition on what we mean by sameness.

    A manifold as I understand it so far is loosely an abstract collection of objects (a set) along with all possible, fully covering, coordinate systems (a maximal atlas). For example, the set of points on the two-sphere ##S^2## along with all different ways of describing all these points by associating them with numbers (obeying certain conditions such as being one-to-one).

    Different choices of atlases only reflects different methods of numbering the different objects in the set, and it's clear that if two sets should be the "same", they would have to exist a one-to-one map between them. But why would we require that this relation be infinitely differentiable?

    I would think that this definition would have to capture the other notions we have for differential manifolds such as curvature etc. But then I would guess there exist a theorem which states that any two (differentiable) manifolds which are diffeomorphic have the same curvature. Does it?

    Any enlightening words are appreciated!
     
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  3. Aug 25, 2013 #2

    Ben Niehoff

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    Because we're talking about differential manifolds. For plain old topological manifolds, all we care about is that the map is continuous.

    And for (pseudo)-Riemannian manifolds, we want the map to be differentiable AND preserve the metric tensor (then it will also preserve the curvature).
     
  4. Aug 25, 2013 #3
    Will a diffeomorphism preserve the metric tensor? If so can you refer me to any theorems on this?
     
  5. Aug 25, 2013 #4

    WannabeNewton

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    Not necessarily. A diffeomorphism preserves the smooth structure. An isometry preserves the metric tensor (Riemannian or pseudo-Riemannian structure). There's no theorem here, it's just a definition. An isometry is a diffeomorphism that preserves the Riemannian of pseudo-Riemannian structure. Just a note on terminology, the term isometry is also used in the context of metric spaces.
     
  6. Aug 25, 2013 #5
    So a diffeomorphism between two manifolds (which per definition implies they being the same manifold) does not imply that the two (differentiable) manifolds have the same curvature?
    In what sense are the two manifolds then the same?

    And how is GR diffeomorphism invariant if the curvature is not invariant under diffeomorphisms?
     
  7. Aug 25, 2013 #6

    WannabeNewton

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    A diffeomorphism just preserves the smooth structure; two diffeomorphic manifolds are the same as far as their smooth structures go just like two homeomorphic topological spaces are the same as far as their topologies and topological properties go.

    GR is diffeomorphism invariant in the sense that the gauge group of GR (under the EFEs) is the diffeomorphism group of the underlying (differentiable) space-time manifold. These are diffeomorphisms that are also endomorphisms i.e. ##\varphi: M\rightarrow M## such that ##\varphi## and ##\varphi^{-1}## are smooth.
     
  8. Aug 25, 2013 #7

    jgens

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    You need to make sure that the inverse is smooth as well.

    There are bunches of ways of thinking about this, but here's one that I like. Basically if two smooth manifolds M and N are diffeomorphic, then the idea is that anything we can prove about M using only its manifold structure (topology, atlas, etc.) should also be provable for N and vice versa. So in some sense they are formally the same object.

    Pretty much. Just want to clarify that your coordinate systems need to be compatible. So the transition from one coordinate system to another should be smooth.

    Yes thinking about coordinates as just applying labels (locally) to the points is a good way to internalize things. Just having a bijective map between the underlying sets of points is not good enough. This map is only guaranteed to prove set-theoretic properties like the size of the set. For example there is a bijective map between R and R2 but clearly the geometry/topology of these spaces are different. Requiring the diffeomorphism to be smooth is just there to require that the manifold properties are preserved.

    You probably need a stronger notion of equivalence to make sure that curvature is preserved. The curvature of a manifold is not intrinsic to the smooth structure, but rather comes from a metric or connection on the manifold. So a stronger equivalence like isometry would do the trick.

    Edit: Ben Niehoff and WannabeNewton beat me to it! I type too slow guyz :'(
     
    Last edited: Aug 25, 2013
  9. Aug 25, 2013 #8

    WannabeNewton

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    *poke*....*poke*
     
  10. Aug 25, 2013 #9
    I would call it a definition of sameness. Essentially, we are defining something called an equivalence relation between differentiable manifolds. There are many other equivalence relations we could define. We can say that they are "the same" if there exists a homeomorphism between them, for example. We can also say that they are "the same" if they both look like the same brand of pancake batter when embedded in ##\mathbb{R}^n## (good luck doing this rigorously). It depends on what you are interested in looking at. In this case, what is interesting about differentiable manifolds?

    As an aside, I think we can exploit the inverse function theorem to obtain this.
     
  11. Aug 25, 2013 #10

    jgens

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    Give R it's usual smooth structure. Then the map f:RR given by f(x) = x3 is smooth and bijective but is not a diffeomorphism.
     
  12. Aug 25, 2013 #11

    I have just studied differential geometry in relation to general relativity (no topology), so that the "smooth structure" of a manifold does not include the concept of curvature seem a bit strange to me. I have to say that I'm a bit disappointed that the mathematicians did not include the concept of curvature into the definition of "sameness". Perhaps I will not be in the near future when I learn more about these things. Anyway; what is the definition of the "smooth structure" of the manifold? Is it just it's level of differentiability?

    With some further thought I guess I could agree with the definition from the fact that a manifold is just a set of objects with no apriori notion of distance between them, and thus no notion of curvature. But then how does one say that a manifold has (for example) a spherical geometry? Can one state that without reference to the metric?
     
  13. Aug 25, 2013 #12

    WannabeNewton

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    A smooth structure (also called maximal smooth atlas) is usually taken to be a maximal collection of smoothly compatible charts on a given topological manifold ##M## (you really don't even need a pre-existing topology, you can construct a smooth atlas on a set and it will naturally induce a topology that is e.g. Hausdorff, second countable, and locally Euclidean).

    What you are overlooking is that there are various types of sameness for a given object. We can look at sameness at the topological level (under homeomorphisms), at the differentiable level (under e.g. diffeomorphisms), and at the level of the metric tensor (under isometries).

    In GR, you have to realize that the underlying smooth structure of space-time is more fundamental than the pseudo-Riemannian structure in the sense of general covariance. If ##g_{ab}## represents a certain space-time geometry then ##\varphi^{*}g_{ab}## will represent the exact same space-time geometry, where ##\varphi \in \text{diff}(M)##
     
    Last edited: Aug 25, 2013
  14. Aug 25, 2013 #13

    jgens

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    The smooth structure is just the maximal smooth atlas. If you want a notion of sameness that includes curvature, then just look at isometry.

    Generally when one makes statements like that there is some metric or connection floating around. I suppose there are some cases where the topology/manifold structure can determine something about what kinds of metric are allowed. For example it's not too difficult to show that S4 does not admit a Lorentzian metric. But for the most part statements like "has a spherical geometry" have a metric lurking somewhere.

    Edit: Beaten again. So many sad faces.
     
    Last edited: Aug 25, 2013
  15. Aug 25, 2013 #14
    Again, sameness can be defined differently. We can define two Riemannian manifolds to be "the same" if there exists an isometric map between them. We can define sameness by looking at what we are interested in.

    I'm speaking of a generalization that says, if ##f:M\to N## is a differentiable map between smooth manifolds and ##df_p:T_pM\to T_{f(p)}N##, with ##p\in M##, is an isomorphism, then there exists an open neighborhood of ##p## such that ##f## restricted to that neighborhood is a diffeomorphism. I was giving the OP a way to confirm a given map is a diffeomorphism, not prove his statement was correct. :tongue:
     
  16. Aug 25, 2013 #15

    WannabeNewton

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    That doesn't tell you if a given map is a diffeomorphism. Taking again my significant other jgens' example, ##f:\mathbb{R}\rightarrow\mathbb{R}## given by ##f(x) = x^3## is differentiable and ##df_p## is an isomorphism hence ##f## satisfies the criteria for the inverse function theorem, but ##f## is not a diffeomorphism. It just tells you that ##f## is a local diffeomorphism.

    EDIT: unless you're talking about a smooth function that fails to satisfy the inverse function theorem in which case yes certainly there is no way it can be a diffeomorphism.
     
    Last edited: Aug 25, 2013
  17. Aug 25, 2013 #16
    So when one talks about the manifold ##M=S^2##, i.e. the two-sphere, one is talking about the set of points on the sphere equipped with the metric ##\sin ^2 \theta d\phi^2 + d\theta^2## (here expressed in spherical coordinates)?
     
  18. Aug 25, 2013 #17

    jgens

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    Yeah unless it's stated otherwise, the standing assumption when you see something like R or S2 is that the usual topology and smooth structures and metrics are taken.
     
  19. Aug 25, 2013 #18

    WannabeNewton

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    It depends entirely on the context. I can simply look at properties of ##S^2## as a topological manifold (i.e.the fact that it's connected, compact, has a trivial fundamental group etc.). I can further look at properties that emerge after introducing the standard smooth structure on ##S^2##. Even further, I can look at properties that emerge after introducing the standard metric tensor ##g_{ab}## on ##S^2## such as the Riemann curvature tensor ##R_{abc}{}{}^{d}\omega_d = 2\nabla_{[a}\nabla_{b]}\omega_{c}## where ##\nabla_a## is associated with ##g_{ab}##. So to reiterate, it depends entirely on the context.
     
  20. Aug 25, 2013 #19
    If ##f## satisfies the given conditions of our inverse function theorem (id est ##f## is a "local diffeomorphism"), for all ##p\in M##, isn't it a diffeomorphism?
     
  21. Aug 25, 2013 #20

    WannabeNewton

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    Oops sorry the above example isn't diffeomorphic on any neighborhood of 0 (I always forget about 0!) but with regards to what you said, if on top of that the map is also bijective then yes. A smooth map between two smooth manifolds is a diffeomorphism if and only if it is a bijective local diffeomorphism.
     
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