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Represntative Surfaces/Curves in homology. Examples.?

  1. Apr 5, 2010 #1
    Hi, everyone:
    A couple of questions, please:

    1) Examples of representative surfaces or curves:

    Please let me know if this is a correct definition of a surface representing

    H_2(M;Z):


    Let M be an orientable m-manifold, Z the integers; m>2 . Let S be an orientable

    surface embedded in M. Then H_2(S;Z)=Z . We then say that S represents

    H_2(M;Z)~Z if the homomorphism h: Z-->H_2(M;Z) sends 1 --as a generator of Z --

    to the homology class of S. specifically, if we have h(1)=alpha ; alpha a homology

    class, then there is an embedding i:S-->M , with [i(S)] =alpha.


    If this is correct. Anyone know of examples of representative curves or surfaces.?




    2)An argument for why non-orientable manifolds have top homology zero, and

    for why orientable manifolds have top homology class Z.?.


    I have no clue on this one. I know homology zero means that all cycles

    are boundaries, but I don't see how this is equivalent to not being orientable.


    Thanks in Advance.
     
  2. jcsd
  3. Apr 7, 2010 #2

    quasar987

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    How about this for 2: Specialize to simplicial homology (and thus, to triangulable manifolds). It is more or less obvious that in an n-manifold M the only n-chain in M that has a chance of being a cycle will be some integer multiple of the chain induced by a triangulation of M (in the sense of my reply to your question 1). Now, when does the n-chain induced by a triangulation is a cycle? Precisely when M is orientable. To get the flavor of this, consider the 2-sphere (orientable) and pick a simple triangulation. It is easy to see that the vertices of the triangles can be ordered so that this 2-chain is cycle. The same procedure works for the 2-torus. Ok, so it works in this case. Next consider the Mobius bundle (i.e. the Mobius strip, but infinitely long in order to remove the boundary... more like a "Mobius cylinder") Triangulate it in a simple way and start ordering the vertices of the triangles in order to make the induced 2-chain into a cycle. You will see that this is impossible. Exactly the same procedure can be applied to the Klein bottle and the projective plane.

    So, if M is orientable, then in the language and notation of my reply to your question 1, the only nontrivial cycle is represented by M itself, and H_n(M;Z)={k[M]:k\in Z}. (Clearly [M] is not a boundary as there are no (n+1)-chains in M for dimensional reasons.) If M is not orientable, then the only n-cycle that has a chance of being a cycle is not a cycle and thus H_n(M;Z)=0.
     
    Last edited: Apr 7, 2010
  4. Apr 7, 2010 #3
    Why must H_2(M;Z) be isomorphic to Z? Once this is false, the rest doesn't make sense.
     
  5. Apr 7, 2010 #4
    Zhentil wrote:

    "Why must H_2(M;Z) be isomorphic to Z? Once this is false, the rest doesn't make sense. "

    If M is orientable, its homology over Z is Z. I don't know if this is a definition or a result,

    but I know it to be true.
     
  6. Apr 7, 2010 #5
    Its top homology is Z. You specified that the dimension was greater than 2.
     
  7. Apr 7, 2010 #6
    Btw, it shouldn't be hard to see that H_2(S^3,Z)=0, since the image of any two-simplex that isn't surjective lands in R^3, and hence retracts to a point.
     
  8. Apr 8, 2010 #7
    Zhentil wrote, in part:

    "Its top homology is Z. You specified that the dimension was greater than 2. "

    Your right. Now I am confused; I don't know if this is an intrinsic orientation of

    the surface, or if it is some sort of orientation relative to, and compatible with,

    that of the space in which the surface is embedded.
     
  9. Apr 8, 2010 #8
    Continuing my previous, and using the helpful comment by Quasar --correct me if
    I am wrong , Q * -- if the embedded surface S were orientable (leaving aside for
    now issues of the relation between the orientability of S, and that of M) , then
    it would be possible to triangulate it ( I think we only need C^1 to be able to
    triangulate a manifold, sure that smoothness is way more than enough), and
    to triangulate it and orient it in such a way that it is a cycle, and , within S,
    it would be the only 2-cycle, again, within S, it would not be a boundary, since
    there are no 3-chains in dimension 2. Then the cycle S is a 2-cycle which is
    not a boundary, and the only cycles in dimension 2 are integral multiples of Z,
    (and, again repeating Q's argument.) none of these cycles would be a boundary,
    so that H_2(S;Z)=Z.
     
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