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A Torus with two Möbius Strips

  1. Dec 16, 2016 #1
    Another zany homology question:

    Just let me know if I have my labeling scheme right so far.

    I have a torus. I have cut two holes into it and am attaching Möbius strips around the holes. (Clearly we are not in 3 dimensions?)

    My Torus is represented by a polygon with the labeling scheme ## a b a^{-1}b^{-1} ## and my strips are labeled ## a b c b ## as follows (I'll subscript them after pasting)

    Torusmobius.jpg

    I'll subscript the torus with 1s, and the 2 mobius strips will be subscripted with 2s and 3s.

    I think I cut a hole in the Torus just by cutting a corner off of the polygon. I'll attach the strips to attain the following monstrosity:

    torusmobius2.jpg


    I end up with a long labeling scheme:

    ##a_1 b_1 c_1 b_2 b_1 b_3 c_3 b_3 a_1^{-1} b_1^{-1} ##

    Before I say any further, is this what cutting a hole in a Torus and gluing a Möbius band even would look like? Obviously I can't picture this outside the polygons here. Thanks for your patience.

    -Dave K
     
  2. jcsd
  3. Dec 16, 2016 #2
    "Before I say any further, is this what cutting a hole in a Torus and gluing a Möbius band even would look like?"

    Do you mean "a" Möbius band, or "two" of them?
     
  4. Dec 16, 2016 #3
    Two.
     
  5. Dec 19, 2016 #4
    Sorry, that's not what I end up with. I get:


    ##a_1 b_2 c_2 b_2 b_1 b_3 c_3 b_3 a_1^{-1} b_1^{-1} ##

    Still, if anyone can let me know if I have even this preliminary step right I would appreciate it.

    Might as well keep going.

    When I relabel I get:

    ## a b c d a^{-1}b^{-1} ##

    Which still isn't something I know from the classification theorem.
     
  6. Dec 20, 2016 #5

    lavinia

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    I don't see how you diagram accomplishes what you want. For instance, once you cut the corner s off the identification of ##b1## no longer makes sense. Can you explain this better?
     
  7. Dec 22, 2016 #6
    Sorry, I didn't realize there'd been an update to this thread. Thought I was going to have to go to stackexchange. ::shudder::

    I admit that I am confused as to how to properly make this hole with regards to the polygonal region. I will revisit.

    -Dave K
     
  8. Dec 22, 2016 #7
    Nope. Don't know how to do it. I am thinking that cutting a hole in a torus equates to cutting a corner off the torus, which is really like adding a side. i.e. when Munkres glues two tori together:

    upload_2016-12-22_14-1-8.png

    So why doesn't my ## b_1 ## make sense?
     
  9. Dec 22, 2016 #8

    lavinia

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    It might make sense. I found it confusing because you cut off part of one ##b_{1}## that is identified to the other. So where does the identification go?
     
  10. Dec 22, 2016 #9
    Of course in a regular torus, the ##b_1## would go back to the other ##b_1##, The ##a_1## similarly. The Mobius strips would be joined up in their usual way, though I don't know how this effects the overall "shape". I certainly can't visualize it! But at least according to the question, this thing should be able to be classified, so it's an n-fold torus, n-fold projective plane or a sphere, although it certainly looks like something much weirder to me.

    -Dave K
     
  11. Dec 23, 2016 #10

    lavinia

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    If you mean the diagram for a torus with a hole in it that you showed here in post 7 then nothing is wrong with ##b_1##.

    The curve ##c## is a closed loop and you have attached a Mobius band to it using a half the boundary of the band which is not a closed loop. So the band gets folded and is no longer a surface with a boundary. Is that what you want?

    One thing that can be done is to attach the entire boundary of the Mobius band to the curve ##c##. This lines up two boundary circles so that the resulting glued together space is a closed surface without boundary. Can you redraw this space to make a polygon with edges identified?

    I think by Van Kampen's Theorem that the relation will be ##aba^{-1}b^{-1}c^{-1}##.
     
    Last edited: Jul 12, 2017
  12. Jul 4, 2017 #11
    If you are willing to use the theorem that classifies all "closed surfaces"* that says that one needs just two pieces of information to determine which surface one has: the Euler characteristic** and the orientability. Since the surface described here — let's call it M — contains a Möbius band, it must be non-orientable. The Euler characteristic of a torus is 0, so that of a torus with two holes is 0 - 2 = -2. The Euler characteristic of a Möbius band is 0 as is that of a circle (like the boundary of a Möbius band), so it follows that the closed surface M has Euler characteristic -2.

    The only closed surface that is non-orientable and has Euler characteristic = -2 is the connected sum of 4 projective planes — or what is the same thing, of 2 Klein bottles. It's fairly standard to determine its homology groups as they are listed in https://topospaces.subwiki.org/wiki/Homology_of_compact_non-orientable_surfaces.

    —————
    * A closed surface is a compact surface without boundary.

    ** The Euler characteristic of an abstract polyhedral surface N is V - E + F, where V, E, F are the numbers of vertices, edges, and faces of the polyhedron. It is remarkable that this integer depends only on the topological type of the surface N.
     
  13. Jul 11, 2017 #12

    mathwonk

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    as just suggested, the euler number is very useful for these questions. note that every surface is obtained from a sphere by adding handles and/or cross caps (mobius bands). i.e. by taking connected sum with a torus and/or a projective plane. Moreover adding a handle makes the euler number go down by 2 while adding a cross cap makes it go down by 1. So as stated the surface here has euler number -2. Moreover there are exactly two surfaces having each even euler number ≤ 0. In particular, adding 4 crosscaps to a sphere gives the same result as adding one handle and two crosscaps. thus although it is not true in general that 2 crosscaps are the same as one handle, they are in the presence of another crosscap. Thus every surface is obtained by adding a certain number ≥ 0 of handles and either 0,1, or 2 crosscaps. There is only one surface with euler number 2, and one for each odd negative number, the latter necessarily being non orientable. This is nicely explained in the book by Massey, Alg Top, an introduction, using triangulations. I hope I got this right. I only learned it a couple years ago by teaching it to a group of brilliant youngsters at Epsilon camp.
    http://epsiloncamp.org/
     
    Last edited: Jul 13, 2017
  14. Jul 12, 2017 #13

    lavinia

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    I think the OP's question is how to represent the torus with the two Mobius bands attached as a polygon with edge identifications.
     
    Last edited: Jul 14, 2017
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