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Things that can't happen on a manifold

  1. Aug 29, 2010 #1

    bcrowell

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    In a different thread, JesseM raised ( https://www.physicsforums.com/showpost.php?p=2858281&postcount=37 ) what I thought was an interesting question: can a manifold (over the real numbers) contain points that are infinitely far apart? Since a bare manifold doesn't come equipped with a metric, it's not clear exactly what notion of distance to apply here. It can't be metric distance. It probably can't even be an affine parameter, because I don't think you can define that without a connection, and in any case you can only define an affine parameter along a geodesic, and I think manifolds can contain points that don't have any geodesic connecting them (e.g., events inside the event horizons of two black holes?). I wonder if the mathematically precise way to say this is that the space has to be paracompact: http://en.wikipedia.org/wiki/Paracompact_space Spaces that aren't paracompact don't admit metrics.

    Carroll's text has some good examples of things that aren't metrics: "With all of these examples, the notion of a manifold may seem vacuous; what isn't a manifold? There are plenty of things which are not manifolds, because somewhere they do not look locally like Rn. Examples include a one-dimensional line running into a two-dimensional plane, and two cones stuck together at their vertices. (A single cone is okay; you can imagine smoothing out the vertex.)"

    I think a manifold can admit a metric that changes signature (because the definition of a bare manifold is more basic than manifold+metric), but the standard formulation of GR can't handle changes of signature.

    Anyone got any other illuminating examples?

    I wonder if it's possible/useful to define a manifold over the hyperreals rather than the reals? http://en.wikipedia.org/wiki/Hyperreal_number Then you could certainly have points infinitely far apart.
     
    Last edited by a moderator: Apr 25, 2017
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  3. Aug 29, 2010 #2

    JesseM

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    Note that in my post, I specified I meant either a Riemannian manifold or a pseudo-Riemannian manifold, which means there is a metric defined on the manifold. After posting it occurred to me to think of the Riemann sphere, which is the complex plane distorted onto a sphere with a point "at infinity" at the top--could this surface (including the top point) be turned into a pseudo-Riemannian manifold? If we define distance along any path that doesn't go through the top to just be the distance when we map the path into the regular complex plane and, if the x-axis is the real line and the y-axis is the imaginary line, use the metric ds^2 = dx^2 - dy^2, then we almost have a complete metric on the sphere, the only problem is defining what happens for paths that cross through that pesky point at infinity. I'd be tempted to say the length of any path that passes through it would also be infinite, but what about the path described by y=x in the complex plane (again with x as the real axis and y as the imaginary one)? That would have a length of zero between any two points on the path not at infinity, so would the length of the path be zero even if it went through the top of the sphere? This case might point to some kind of problem in defining a metric on the complete Riemann sphere that satisfies all the requirements of a pseudo-Riemannian metric, I don't know enough about the relevant math though.
     
  4. Aug 29, 2010 #3

    jcsd

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    Seems trivial to me unless I'm missing something. Two spheres form a manifold, but not a metrizable one. Pretty uninteresting example though.

    The long line is also manifold is not metricizable, that's probably more what you're after as it doesn't have the disjoint nature of the p-revious example.

    You can define complex manifolds because the complex numbers are homoemorphic to R^2, the Hyperreals are not homeomorphic to R^n so you can't have hyperreal manifolds.

    The signature is a property of the metric tensor rather than of metrics. Pseudo-Riemannian metric tensors don't define metrics because a pseudo-Riemannian metric is not a metric.
     
    Last edited: Aug 29, 2010
  5. Aug 29, 2010 #4

    atyy

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    There are two not very related definitions of metric.

    One is metric as in a "distance function" http://en.wikipedia.org/wiki/Metric_space

    The other is metric, as in metric tensor which takes vectors as inputs and outputs numbers.
     
  6. Aug 29, 2010 #5

    jcsd

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    They are related metric tensors as they always define a metric (or a function that isn't actually a metric, but does the job of one) on something even if it isn't the whole manifold.

    A pseudo-Riemannian metric as defined by a pseudo-Riemannian metric tensor isn't a metric as it allows distinct points to have null distance. It's more like a function doing the job (or some of the jobs) of a metric.
     
  7. Aug 29, 2010 #6

    JesseM

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    OK, I was confused about the difference between a metric and a metric tensor...but to relate it to my question on the other thread, if we do want to define a full pseudo-Riemannian metric (not just a pseudo-Riemannian metric tensor) on a manifold, can it ever be the case that there will be two points on the manifold that cannot be joined by a curve of finite length? For example, can the points that make up the edge of a Penrose diagram for a Minkoski metric (the ones 'at infinity' in terms of a Minkowski coordiante system) actually be considered part of the spacetime defined by the manifold and metric?
     
  8. Aug 29, 2010 #7

    bcrowell

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    What I meant by a manifold over the hyperreals was simply that you take the standard definition of a manifold, and everywhere it refers to the reals, you change it so it refers to the hyperreals. In other words, it would be an object that can be broken down into things that look like *Rn, where *R is the hyperreals.

    Oops, sorry -- I missed that!

    Re the example of the Riemann sphere, is there even any need to go to two dimensions? I think the one-dimensional analog of the Riemann sphere is just the real projective line: http://en.wikipedia.org/wiki/Real_projective_line Topologically, it's just a circle. You would want to define a line element ds so that when you integrated it between two points, it would be equivalent to subtraction, but I guess there's a problem because [itex]\infty-\infty[/itex] isn't defined. Hmm...this sounds like the situation at a singularity in GR, where the singularity isn't considered to be part of the manifold.
     
  9. Aug 29, 2010 #8

    jcsd

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    Lets say you have a manifold with a metric, you then add some a point or some points to it to get a new manifold, you can't expect to always be able to be able to metricize the new manifold such that the old metric space is a subspace of the new metric space.

    I.e. when you add new points you can't expect to be able to extend the old metric to cover these new points.
     
  10. Aug 29, 2010 #9

    JesseM

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    Maybe this is the case if you try to add the point at infinity to the complex plane to create the Riemann sphere? Anyway my question wasn't really focused on adding new points, it's more about whether it's mathematically possible to have a manifold plus pseudo-Riemannian metric such that you can pick two points which are already part of the manifold, and there is no possible continuous curve joining them (also consisting of solely of points on the manifold) which has a finite length (with 'length' calculated using the metric, of course).
     
  11. Aug 30, 2010 #10

    jcsd

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    Hi Jesse, I'm guessing what you really want to know is can we define a pseudo-Riemannian metric tensor on a non-paracompact path-connected manifold?

    I say metric tensor rather than metric, because the assumption that there exists a metric is the assumption that there is a distance function where all distances are finite. Metrics don't admit infinite or non-defined distances, even when the distance function is a pseudo-Riemannian metric which as I said is only really an impersonation of an actual metric.

    Unfortunately I don't know, I can't see any obvious reason we cannot we could certainly define a metric tensor field on a non-paracompact manifold my trivial example of two spheres demonstartes that (but the two spheres exampel is not path-connected). But it doesn't mean that there isn't one (a reason, obvious or non-obvious that is).

    This is the kind of question that Hurkyl would probably know the answer to.
     
  12. Aug 30, 2010 #11

    George Jones

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    According to Hawking and Ellis, a paracompact manifold admits a Lorentz "metric" if and only if

    (M is not compact) or (M is compact and has Euler number equal to zero).

    Note if M is compact, then M necessarily has closed timelike curves; see

    https://www.physicsforums.com/showthread.php?p=1254758#post1254758.
     
  13. Aug 30, 2010 #12

    jcsd

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    Hi George can you expand on this a bit, i.e. give an example of a paracompact manifold that won't admit a Lorentz metric (especially one that we might've naively thought that it would've).
     
  14. Aug 30, 2010 #13

    George Jones

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    S^4. By the Hairy Ball Theorem, there exists no non-vanishing vector field on all of S^4. If S^4 admitted a Lorentz metric, then the Lorentz metric together with a Riemannian metric (which always exists) could be used to define a non-vanishing vector field on S^4.

    Going to lunch; maybe I'll say more after I get back.
     
  15. Aug 30, 2010 #14

    Hurkyl

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    I'm curious how this works. I know for a Riemannian metric tensor that you can define a metric by taking the distance between two points to be the infimum of the lengths of all curves joining them. I've never heard of a pseudo-Riemannian metric, though, except as an abbreviation for pseudo-Riemannian metric tensor.
     
  16. Aug 30, 2010 #15

    Hurkyl

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    If you're doing nonstandard analysis, then you are certainly interested in the transfer of notions like manifold or metric space. Aside from infinitessimal neighborhoods, the new points would lie between the standard ones and infinity and distances would be either [itex]+\infty[/itex] or hyperfinite (just as for the extended real line).

    I imagine it's useful, but not having studied the subject I don't actually know. I would guess you would like to do things like studying compact sets which include all standard points.
     
    Last edited: Aug 30, 2010
  17. Aug 30, 2010 #16

    jcsd

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    I think you're right I think I may be guilty of no small abuse here.

    But I'd say that the Lorentzian metric would be the length of the globally proper time maximizing (or if no such curve exists proper length minimizing) curve(s) between two events.
     
  18. Aug 30, 2010 #17

    Hurkyl

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    George is the DG expert, not I. :smile: I only know some basics, and probably some esoteric facts related to my interests.
     
  19. Aug 30, 2010 #18

    bcrowell

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    I've tried to construct an example of a manifold in which two points are infinitely far apart. Does this work?

    Let *Z be the set of hyperintegers, http://en.wikipedia.org/wiki/Hyperinteger , and let [itex]M=^*Z\times (0,1)[/itex] be the direct product of the hyperintegers with the open unit interval. Define coordinate charts on M such that for [itex]p=(n,x)\in M[/itex], p's coordinate is x. For all n and for [itex]1/2 < x < 1[/itex], identify (n,x) with (n+1,x-1/2). The idea here is that we have the top half of each chart overlapping with the bottom half of its successor. It seems to me that M is a one-dimensional manifold over the reals according the definition in Wald, p. 12. (That is, Wald's cover and chart family can be defined in obvious ways based on the things I've defined, including making the cover and chart family maximal.)

    There are obvious ways of defining an ordering and a line element ([itex]ds^2=dx^2[/itex]), and this line element corresponds to a certain metric tensor.

    I think the points a=(0,.5) and b=(k,.5), where k is some positive infinite hyperinteger, are infinitely far apart, in the sense that for any real number L, the points p in M such that [itex]|\int_a^p ds| \le L[/itex] do not include b. On the other hand, a and b are connected by a curve [itex]C=\{p|a \le p \le b\}[/itex] that I think is geometrically continuous.

    Anything wrong with this example? Is it a valid example of a manifold, but not a paracompact one?

    I think the interpretation of this is that essentially the hyperreal numbers http://en.wikipedia.org/wiki/Hyperreal_number (or maybe just the hyperreals that differ from some hyperinteger by a real number) are a manifold over the reals. Therefore generalizing from manifolds over the reals to manifolds over the hyperreals doesn't really buy you anything new (or only buys you the existence of infinitesimals?).
     
    Last edited: Aug 30, 2010
  20. Aug 30, 2010 #19

    Hurkyl

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    Your manifold, I think, is the disjoint union of uncountably many copies of R.



    Viewed through standard means, the hyperreals are a horrible topological space. Among other things, I believe it is totally disconnected, but yet it is not discrete. The reason to study nonstandard manifolds is so that you can use the methods of nonstandard analysis.
     
    Last edited: Aug 30, 2010
  21. Aug 30, 2010 #20

    bcrowell

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    I would say that it's the connected (not disjoint) union of uncountably many copies of the unit interval (not R). But I'm prepared to be shown the error of my ways...!?

    Well, my purpose here was just to come up with an example of a manifold that has points infinitely far apart. To me, the methods of NSA are simply the methods of Gauss and Newton, freed of their 20th-century stigma as the technique that dare not speak its name. Do you know of a good book that covers nonstandard manifolds?
     
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