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Topological dimension of the image of a smooth curve in a manifold

  1. Feb 8, 2012 #1
    Here is the situation I am concerned with -

    Consider a smooth curve [itex]g:[0,1] \to M[/itex] where [itex]M [/itex] is a topological manifold (I'd be happy to assume [itex] M [/itex] smooth/finite dimensional if that helps). Let [itex] Im(g) [/itex] be the image of [itex] [0,1] [/itex] under the map [itex] g [/itex]. Give [itex] Im(g) [/itex] the subspace topology induced by [itex] M [/itex].


    The question is this --- as a topological space, does [itex] Im(g) [/itex] always have (topological) dimension [itex] \leq 1 [/itex] ?

    Note that it IS important not to restrict [itex] g [/itex] to be either injective or an immersion - then the result is straightforward. The tricky thing is that [itex] g [/itex] may not be constant rank and also may not be a submanifold of [itex] M [/itex] (or even a manifold at all).

    Also, a reference to a proof is fine, I don't really need to know HOW to prove it, I just need to be certain that it is true. It certainly seems intuitively obvious...

    I've seen a number of statements (without reference or proof) that [itex] Im(g) [/[STRIKE][/STRIKE]itex] must have zero Lebesque measure. I don't immediately see how this would answer the above question, so any references or proofs on this front would be useful as well.

    Thanks!
     
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  3. Feb 8, 2012 #2

    quasar987

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    There are the so-called space-filling curves that are continuous surjective maps [0,1]-->[0,1]^n for any n. In particular, making the n-cube into a torus, you get a continuous curve [0,1]-->T^n whose image has dimension n since it is the whole torus.

    Note that such a curve cannot be injective. Otherwise, it would be a bijection from a compact space into a Hausdorff space, hence a homeomorphism, which would contradict invariance of dimension.

    No differentiable space-filling curve can exist, but there are ones that are smooth almost everywhere.

    These things are discussed in the Dover book "Counter-examples in topology"
     
  4. Feb 8, 2012 #3

    lavinia

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    I think this is your answer.

    For smooth curves around any point where the derivative is not zero there is a neighborhood of the point mapped diffeomorphically onto its image. this i think is the Inverse Function Theorem so the image around such a point is a 1 dimensional manifold.

    Near a point where the derivative is zero this is still true if the point is isolated - not hard to prove. If there is an interval where the derivative is zero then this interval is mapped to a single point. So locally the image looks like a one dimensional manifold or a point.


    Since your interval is compact there can be no points of accumululation that are not in the image. This in any neighborhood of a point where the derivative is not zero the image is an embedded 1 dimensional manifold.
     
  5. Feb 9, 2012 #4

    Bacle2

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  6. Feb 9, 2012 #5

    lavinia

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    I think the Inverse function theorem works because one of the projections onto one of the coordinate axes must have non-zero derivative.
     
  7. Feb 9, 2012 #6

    Bacle2

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    Do you mean when M is embedded in R^n ? I don't know if the OP is assuming this.

    And, my bad: Sard's theorem tells us something about the measure of the image,(since, in the 1xm Jacobian, every point in I=[0,1] is necessarily a critical point) but there is no connection in this respect between measure and topological dimension; e.g., in R^n, any (measurable)subset of dimension less than n will have n-dimensional Lebesgue measure 0, so that does not narrow it down much, definitely not enough.
     
  8. Feb 10, 2012 #7

    lavinia

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    I mean that id the derivative is not zero at some point then the derivative of the composition of the curve with one of the projections must also be non-zero. Smoothness then allows the Inverse Function theorem to guarantee a local diffeomorphism. This seems OK.
     
  9. Feb 12, 2012 #8

    Bacle2

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    A small variant of the problem: if we're looking to preserve Hausdorff dimension--I assume by topological dimension the OP meant Lebesgue covering dimension-- then we can use that continuously-differentiable maps on a compact metric space, i.e., [0,1] , are Lipschitz, so that there is a bound to the scaling of the diameter d of the sets in the cover, and it follows that the Hausdorff dimension is preserved.
     
  10. Feb 13, 2012 #9

    Bacle2

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    Luckily, a topologist/geometer friend of mine helped me to this result:

    http://books.google.com/books?id=Lt...ge&q=Szpilrajn and lebesgue dimension&f=false

    In which Szpilrajn (sp?) showed that inductive (topological) dimension is less than the Haudorff dimension.

    So, for an alternative argument: we start with I=[0,1]. Then we use the fact that

    a C^oo (C^1 is enough, AFAIK) map is Lipschitz, so that f preserves the Hausdorff

    dimension , and then use Szpilrajn's inequality. It then follows, if I did not miss any thing,

    that the image curve has topological dimension <=1.
     
  11. Feb 16, 2012 #10
    Thanks!

    That's definitely what I was looking for - the books I have on dimension theory only talk about the inductive and covering dimensions, and I just found out about the Hausdorff dimension. This definitely does the trick - thanks again.
     
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