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Homology Groups of the 2D Torus

  1. Apr 19, 2014 #1
    I am reading James Munkres' book, Elements of Algebraic Topology.

    Theorem 6.2 on page 35 concerns the homology groups of the 2-dimensional torus.

    Munkres shows that [itex] H_1 (T) \simeq \mathbb{Z} \oplus \mathbb{Z} [/itex] and [itex] H_2 (T) \simeq \mathbb{Z} [/itex].

    After some work I now (just!) follow the proof that [itex] H_1 (T) \simeq \mathbb{Z} \oplus \mathbb{Z} [/itex] but I need some help to understand a point in the proof of [itex] H_2 (T) \simeq \mathbb{Z} [/itex].

    Munkres' argument to show [itex] H_2 (T) \simeq \mathbb{Z} [/itex] is as follows:

    -----------------------------------------------------------------------------

    To compute [itex] H_2 (T) [/itex], note that by (2) any 2-cycle d of T must be of the form [itex] p \gamma [/itex] for some p. Each such 2-chain is in fact a cycle,by (4) , and there are no 3-chains for it to bound. We conclude that

    [itex] H_2 (T) \simeq \mathbb{Z} [/itex]

    and this group has as generator the 2-cycle [itex] \gamma [/itex].

    ------------------------------------------------------------------------------------------------


    I do not fully understand why any 2-cycle d of T must be of the form [TEX] p \gamma [/TEX] for some p.

    Can someone please explain exactly why this follows?

    Would appreciate some help.




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    To give members of the Physics Forums the context of the post above, the text of Theorem 6.2 and its proof follow:

    attachment.php?attachmentid=68801&stc=1&d=1397891920.jpg
    attachment.php?attachmentid=68802&stc=1&d=1397891920.jpg


    Some of my thoughts ... ...

    Basically, to show that any 2-cyclce of L (i.e. T) os of the form [MATH] p \gamma [/MATH], we have to show the following:

    If [itex] d = \sum_i n_i \sigma_i [/itex] where [itex] \partial d = 0 [/itex] then [itex] d = p \gamma [/itex].

    We have, of course that [itex] \gamma = \sum_i \sigma_i [/itex]

    Note that we have that if d is a 2-chain of L and if [itex] \partial d [/itex] is carried by A then d is a multiple of [itex] \gamma [/itex].

    Munkres defines 'carried by' in the following text taken from page 31:


    attachment.php?attachmentid=68805&stc=1&d=1397894545.jpg


    Hope someone can help.

    Peter
     
    Last edited: Apr 19, 2014
  2. jcsd
  3. Apr 19, 2014 #2

    micromass

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    Take a ##2##-cycle ##d##. Then obviously ##d## is a ##2##-chain. And also by definition of ##2##-cycle, we have ##\partial d = 0##. In particular, ##\partial d## is carried by ##L##. Thus ##(2)## implies that ##d## is a multiple of ##\gamma##, which means by definition that ##d=p\gamma## for some ##p##.
     
  4. Apr 19, 2014 #3
    Thanks micromass ...

    OK so we take a 2-cycle d ... but how do we know that an arbitrary 2-cycle is carried by L?

    BTW ...Do you mean carried by L or A? If as I suspect you mean A ... then ... how do we know that an arbitrary cycle is carried by A ... this is exactly my problem ...

    Again, thanks for you prompt help

    Peter
     
  5. Apr 19, 2014 #4

    micromass

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    I meant ##A##. And we don't need to show that an arbitrary cycle is carried by ##A## (I doubt it's even true). To apply ##(2)##, we need to show that for any cycle ##d##, we have that ##\partial d## is carried by ##A##. This is what we need to show. But ##\partial d = 0## by definition of a cycle. And of course ##0## is carried by ##A##.
     
  6. Apr 19, 2014 #5
    Thanks so much ... that clears up that matter and gives me the confidence to go on ...

    Thanks again,

    Peter
     
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