Homology Groups of the 2D Torus

  • #1
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Main Question or Discussion Point

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:

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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].

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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 \(\displaystyle p \gamma \), 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
 

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  • #2
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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##.
 
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  • #3
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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
 
  • #4
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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
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##.
 
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  • #5
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Thanks so much ... that clears up that matter and gives me the confidence to go on ...

Thanks again,

Peter
 

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