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Are C^1 knots Orientable?

  1. Jan 4, 2014 #1

    WWGD

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    Hi, hope this is not too simple:

    Are ##S^1 ## -knots (meaning homeomorphisms of ##S^1 ## into ## R^3## or ## S^3## , classified under isotopy) that are just C1 - knots orientable? If we were working " in the smooth category " , then we could just say that we can pushfoward the orientation form ##dx## by a diffeomorphism, and the diffeomorphism would give us a nowhere-zero form, which would be positive or negative depending on whether the diffeo. is orientation-preserving or not. But if we don't know if the embedding of the knot is a smooth embedding , can we still guarantee orientability? I think we can say that if ##S^1## is embedded, then it is a submanifold, and submanifolds admit tubular neighborhoods, meaning the normal bundle of the knot is trivial , which implies orientability. What if we only know that the embedding is a homeomorphism, but we do not address issues of degree of smoothness, i.e., we do not know whether the embedding is even ##C^1## , can we determine orientability?

    Also: outside of the "smooth category", where we cannot work with differential forms, do we define orientability in terms of the fundamental class (i.e., the generator of the top homology class )? Then I guess we would have to decide whether the map induced in top homology preserves this class ?

    Thanks.
     
  2. jcsd
  3. Jan 6, 2014 #2

    lavinia

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    I think C^1 suffices because it provides a continuous tangent vector field along the curve.
     
    Last edited: Jan 6, 2014
  4. Jan 6, 2014 #3

    WWGD

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    Thanks, lavinia. Do you have a nice definition of orientable bundle other than having a reduction of the structure group to O(n)? This definition makes sense, in that O(n) preserves orientation, but it does not seem very geometrically understandable.
     
  5. Jan 6, 2014 #4

    lavinia

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    Sorry about my typos WWGD. I just fixed them.

    One definition of orientation for a k dimensional vector bundle is that the bundle of k-frames has a non-zero section. If the bundle is smooth over a manifold then this means that there is a non-zero k form on the bundle, just like in the case of an orientation for the tangent bundle.

    This definition just says that one can pick an orientation for each fiber and that these orientations are consistent, that is that locally they vary continuously. The rigorous definition is that in an open ball,U, around any point there should be a bundle isomorphism from [itex]U \times E [/itex] into [itex]\pi^{-1}U[/itex] that is orientation preserving .

    Another way to think of orientability of a fiber is to think of its unit sphere as being oriented. If you do not have a Riemannian metric then one can define an orientation as the orientation of a k- simplex that does not intersect the zero section. This is equivalent to choosing a generator of the cohomology group, [itex]H^{k}(E,E-0)[/itex].

    Then the compaibility condition is that there is a cohomology class in [itex]H^{k}(\pi^{-1}U,\pi^{-1}U-0)[/itex]
    that restricts to the chosen generator of [itex]H^{k}(E,E-0)[/itex] for each fiber above points in [itex]U[/itex].

    The advantage of this way of defining it is that the cohomology groups can have different coefficients. With some coefficients the bundle may be orientable while with others it may not. For instance every vector bundle is orientable with Z2 coefficients. The usual way of defining orientability is equivalent to using Z coefficients.
     
    Last edited: Jan 6, 2014
  6. Jan 6, 2014 #5

    WWGD

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    Thanks. lavinia; I was thinking too, about a result that any closed curve without self-intersection living in a manifold is embedded. It seems little can go wrong with a one-dimensional subspace that does not intersect itself. But I can't think of a proof right now. Of course, C^1 does not guarantee that there is no self-intersection either.
     
    Last edited: Jan 6, 2014
  7. Jan 6, 2014 #6

    lavinia

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    I am not sure what embedding means but whenever you have a continuous bijection of a compact set onto a Hausdorff space it is a homeomorphism.The closed curve without intersection is a continuous bijection of the circle into the manifold which is a Hausdorff space and so is a homeomorphism onto its image( with the subspace topology). Is that what you mean?

    BTW: What are you studying?
     
  8. Jan 6, 2014 #7

    WWGD

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    Thanks; good idea, the continuous bijection between compact+Hausdorff pulls you out of a lot of difficult places.

    I'm doing some work on contact submanifolds of 4-manifolds, and some Lefschetz fibrations with open books.

    I'm hoping to put it together towards a thesis soon. Thanks for your feedback; it's helped me clarify my ideas.
     
  9. Jan 9, 2014 #8

    WWGD

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    Hey, lavinia, I think this works, tho it is not geometrically too nice: I think a homeomorphism (the knot being a homeomorphic copy of ## S^1 ##) would send the top homology class of an orientable manifold ( the fundamental class ) , to the fundamental class. Doesn't a homeomorphism induce an isomorphism in homology?
     
  10. Jan 9, 2014 #9

    jgens

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    Yes. There are actually much weaker conditions (like weak equivalence) that induce isomorphisms on homology.
     
  11. Jan 9, 2014 #10

    jgens

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    This is definitely false in the topological and homotopy categories. The problem is that weak equivalences in general have no homotopy inverse (although in the CW category they do). Yet they certainly induce isomorphisms on homology.
     
    Last edited: Jan 9, 2014
  12. Jan 9, 2014 #11

    WWGD

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    Yes, my bad; I was thinking of the subgroup of the mapping class group that induces the identity
    on homology; not the ones that induce isomorphisms. I deleted my previous answer after you saw it, sorry.
     
  13. Jan 9, 2014 #12

    jgens

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    No worries. We all make mistakes sometimes and quickly realize the error.
     
  14. Jan 9, 2014 #13

    lavinia

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    It is true that a homeomorphism induces an isomorphism of homology groups but when you embed one manifold in another the induced map on homology may be zero. For instance, if you embed a circle as the boundary of the closed disk. Any knot in Euclidean space is homologous to zero because Euclidean space has zero homology except in dimension zero.

    In your case, there is a homeomorphism of the circle onto a submanifold of Euclidean space. In Euclidean space theis submanifold's homology classes are all zero.
     
  15. Jan 9, 2014 #14

    WWGD

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    .

    But is this also the case for the top homology (when embedding ## S^1 ## in Euclidean n-space)? I agree the circle embedded as the boundary of
    a disk is (n-1)- homologous to zero (i.e., its (n-1)-st homology is zero), since it bounds the disk. But I think this is not so for the top homology, when embedding in Euclidean n-space since an n-knot cannot bound an (n+1)-submanifold. Still, there are no ## S^1## knots in ## \mathbb R^n ## beyond ## \mathbb R^3 ## ; there is enough space to uknot everything , tho I don't have a good proof of this last (only for n=2, we have Schoenflies, but Schoenflies does not hold for n>2, I'm pretty sure; for n=3, there is the Alexander Horned Sphere).

    But it's a good point. I guess, e.g., on a torus you can have meridians or parallels as homologically non-trivial embeddings of ##S^1 ## , and non-meridional , and in general loops that are neither homologous to merdians nor homologous to parallels ( whose removal disconnects the torus) as cases where the inclusion induces the zero map in homology.
     
    Last edited: Jan 9, 2014
  16. Jan 9, 2014 #15

    WWGD

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    I wonder if the definition of submanifold I'm familiar with implies that a submanifold is oriented -- ##N^n ## is a submanifold of ##M^M## ; (## n \leq m ##) if N is embedded in M like ## \mathbb R^n ## is "standardly-embedded" in ##\mathbb R^M## , (meaning ## (x_!,x_2 ,..,x_n) \rightarrow (x_1,x_2,..,x_n,0,0,..,0)##) i.e., there are subspace charts ## (U_i, \phi_i) ## for ## N^n ## with ## \phi_i(U \cap N)=(x_1,x_2,..,x_n,0,..,0)## . But I may be wrong, since this seems to be almost precisely the way the projective spaces are embedded in higher-dimensional ones; no cells beyond a certain dimension.
     
  17. Jan 9, 2014 #16

    jgens

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    Nah. Your definition (that of an embedded submanifold) does not imply orientability. Another standard definition (that of an immersed submanifold) is strictly weaker and the primary difference is that the topology on your submanifold may actually be finer than the subspace topology.
     
  18. Jan 9, 2014 #17

    WWGD

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    Well, yes, I guess we first need to make sure the submanifold is embedded to talk about orientability, so that the manifold is a subspace.
     
  19. Jan 10, 2014 #18

    lavinia

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    Euclidean space has zero homology in all dimensions greater than zero. So the top homology class of an embedded submanifold is zero.

    Not every manifold can be oriented. But every manifold (compact) can be embedded as a submanifold of Euclidian space. For instance, the Klein bottle and the projective plane can both be embedded in Euclidean 4 space but neither can be oriented.
     
  20. Jan 10, 2014 #19

    lavinia

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    What does this have to do with orientability of the manifold?
     
  21. Jan 10, 2014 #20

    lavinia

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    All that you need is an immersion to talk about orientability.

    Suppose f: M ->N is an immersion and that N is oriented. Let E denote the tangent bundle of N

    Then the induced bundle, f*(E), over M splits into a Whitney sum of the tangent bundle of
    M and the normal bundle to the immersion. Suppose that the normal bundle can be oriented - which for instance will be true if it is a trivial bundle. Then M can also be oriented using the orientation of N and the orientation of the normal bundle.

    Both the orientability of N and the normal bundle to the immersion are required. A counterexample would be any unorientable n manifold that is the boundary of an n+1 dimensional manifold. For example, the Klein bottle is the boundary of a 3 manifold. The normal bundle of the bounding manifold is always orientable because it is always a trivial line bundle. This is true because on the boundary, the idea of an inward pointing direction is well defined.
    So the 3 manifold that the Klein bottle bounds is itself, non-orientable.
     
    Last edited: Jan 10, 2014
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