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A Fundamental and Homology groups of Polygons

  1. Dec 13, 2016 #1
    This is an old qual question, and I want to see if I have it right. I had virtually no instruction in homology despite this being about 1/4 of our qualifying exam, so I am feeling a bit stupid and frustrated.

    Anyway,

    I am given a space defined by three polygons with directed edges as follows (a description should suffice).

    1) A triangle with edges ##bca^{-1}##.
    2) A triangle with edges ##ad^{-1}e##
    3) A square with edges ##bdc^{-1}e^{-1}##.

    I am to find a presentation for ## \pi_1(S) ## which I suppose is just what I have given above as labels, so ## < a,b,c,d,e | bca^{-1} = ad^{-1}e = bdc^{-1}e^{-1} = 1 > ##.

    I mean, first of all, is this correct? In most examples I have seen, we are given a single polygonal region, not multiple ones.

    I'm to find ## H_1(S) ## which is equal to ## \pi_1(S)/ [\pi_1(S), \pi_1(S)] ## i.e. the mod of the fundamental group with its commutator subgroup.

    I don't know of the most efficient procedure. I've got 9 commutators ## [a,b], [a,c]...## etc. So that's 27 operations? If I was clever I might know which ones aren't worth doing. For the first element against the first commutator ##[a,b] = aba^{-1}b^{-1} ## I'm getting stuff like:

    ## bcba^{-1}b^{-1} ##
    ## bcca^{-1}c^{-1}##
    ## bcda^{-1}d^{-1}##

    Where I begin to feel like:
    http://i1.kym-cdn.com/photos/images/original/000/234/739/fa5.jpg

    Lastly I'm asked to classify this space from the classification theorem. All I can think to do is glue all the pieces together along their similarly labeled edges, respecting the orientations, etc.

    e.g. ## bca^{-1}ad^{-1}ee^{-1}bdc^{-1} ##

    which reduces to ##bcd^{-1}bdc^{-1} ##

    Which I can relabel ## abcac^{-1}b^{-1}##

    which I see is ## abca(bc)^{-1}## and I can relabel ##bc## (I think) so that I end up with ##abab^{-1}##.

    And darnit, the first time I did this, I swear I ended up with ##aba^{-1}b^{-1}## which is a torus, but I do not know what I have here.

    'elp?

    -Dave K
     
  2. jcsd
  3. Dec 13, 2016 #2

    andrewkirk

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    Are you sure the fourth side of the square is supposed to be ##e^{-1}## and not ##e##? Otherwise ##e## is a closed loop since, if we label as ##X## the vertex at the arrow end of ##b## then the two triangle specifications tell us that ##b## goes into ##X## and ##c,d,e## go out of it. So in the square, when ##c^{-1}## takes us into vertex ##X## we need ##e## to take us out of it, unless it's a loop.

    If the statement is correct as written and ##e## is supposed to be a closed loop, then the shape has only two vertices, which makes it hard to visualise, for me at least. It's still a valid shape, but I think it's just worth checking that's what they meant, since that does make it more difficult. If the fourth side is supposed to be ##e## instead of ##e^{-1}## then the shape has three vertices, and is easier to visualise.

    Also, I don't think what you have written is a presentation, because not all the elements you have written as generators (##a,b,c,d,e##) are in the fundamental group, because they are not loops. As written, the only loops are ##e## and ##b##. If the fourth side is supposed to be ##e## instead of ##e^{-1}## then none of the edges are loops. The loops you need to give as generators are the cycles given in (1), (2) and (3).
     
  4. Dec 14, 2016 #3
    Thanks for your reply. I almost had to go to stackexchange. :eek:

    I am going off my re-write of the question, with the original at home, so I will check. Good point about the presentation though. The problem is I've never really seen a decent amount of examples. Almost every textbook seems to take a "well, you know what we mean" approach. Some of this is a weakness with algebra. (If that is not obvious).
     
  5. Dec 14, 2016 #4
    So, just for clarification here. (And again, I am struggling with the algebra here).

    A presentation of a group consists of the generators and then the relations. For a Torus we have ## <a,b | aba^{-1}b^{-1} = 1 > ##

    So what you are saying are that ##a## and ##b## are loops, but the meaning of ##aba^{-1}b^{-1} = 1 ## is that these loops have a relationship with each other. (In this case, the Torus, which is abelian, we could also write ## ab = ba ##.

    I guess I have been thinking of of ## aba^{-1}b^{-1} = 1 ## as a "loop" since following that path would take me back to the base point. But really it is the relationship between the loops. (Hence called the relations).

    So what I'm calling the relations in my example, you are saying are actually the loops. In order to find the relations I need to do a bit more cutting/pasting or whatever.

    Let me know if I'm on track here. I still need to go home and get the original problem.

    -Dave K
     
    Last edited: Dec 14, 2016
  6. Dec 14, 2016 #5

    fresh_42

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    ... don't you need ##a,b## to be of finite order, too, to speak of a torus? Otherwise you simply have a free Abelian group of rank two. (I'm not sure, just asking out of laziness to look it up.)
     
  7. Dec 14, 2016 #6
    That is the fundamental group of the torus. :smile: (Free Abelian group of rank ##2n## for an ##n##-fold torus).
     
  8. Dec 14, 2016 #7

    andrewkirk

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    Yes that's correct.
     
  9. Dec 14, 2016 #8
    Pretty sure I have the labeling correct.


    16 - 1.jpg
     
  10. Dec 14, 2016 #9
    I have no idea from the original poster's description what space he is trying to find the fundamental group and homology groups of.

    Is he talking about the DISJOINT UNION of two triangles and a square? Is he talking about the plane after the interiors of those three polygons have been removed? Or is he talking about something else entirely? Clarification would be welcome.
     
  11. Dec 14, 2016 #10

    andrewkirk

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    It is neither of those things.

    As I interpret it, the space is the surface obtained by sewing the three polygons together along the matching edges, preserving direction. So, for instance, the two triangles are sewn together along the edge ##a##. I am pretty sure that the surface that is constructed cannot be embedded in 3D space.
     
  12. Dec 14, 2016 #11
  13. Dec 15, 2016 #12
    It is the space defined by the labeling scheme on the three polygonal regions listed (and picture posted).

    Since the polygons share some edge labels, I am assuming I'm supposed to paste the things together. I've never seen anything like it in a textbook, or online, and yet there it is on one of are old qualifiers.
     
  14. Dec 15, 2016 #13
    Yes, you are indeed supposed to paste the things together. Once I caught on (sorry to be so slow on the uptake!), this was lots of fun!!! So I recommend you try it.

    (But in case you are not so inclined, I will tell you a little of what I found way below so you will see it only if you choose to.)










    Hint: First glue the two a's together, making a quadrilateral from the two triangles. Then glue the two d sides together, making a single polygon from the two quadrilaterals. Then notice that the two c sides can "cancel each other out", so glue them together as well. What remains is only a quadrilateral with two b and two e sides. Once the e sides have been now glued together you get a cylinder with its two ends, the former e sides, now each being a circle (i.e., a simple closed curve). When you glue *these* together you must get either a torus or a Klein bottle surface, depending on whether the orientation is preserved or reversed — which is it? I will leave the fun of finding out up to you. Then, then fundamental group and the homology must be that of whatever surface this becomes, which is certainly easy to google and see how to describe with a presentation.
     
  15. Dec 16, 2016 #14
    This is what I have been doing, but I keep ending up with different results. At one point I could swear I ended up with a torus. At another point I got to something I could not seem to reduce any further. Can someone please check my procedure in my first post and let me know either where I went wrong, or if I should have continued further in some direction?
     
  16. Dec 16, 2016 #15

    fresh_42

    Staff: Mentor

    If I combine @zinq's hint (with paper, scissors, spock, uhm ... glue) and the given relations, I end up with ##be^{-1}=e^{-1}b^{-1}##. Does this help?
     
  17. Dec 16, 2016 #16
    Yes, it does. I guess my question is how the choice of cutting and pasting procedure effects the result. I seem like I am coming up with different answers if I start down a different path. I'm admittedly very bad at calculation type stuff, usually due to missing details and copying stuff wrong. But man, I have been looking at this problem for over a week now.
     
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