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 Post subject: Homology CalculationPosted: Fri, 11 Mar 2011 07:18:12 UTC
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Joined: Fri, 3 Sep 2010 09:29:45 UTC
Posts: 140
I am trying to calculate the homology of the chain complex defined as follows

Let and set for or 0 otherwise.

We defined the boundary operator by

First question - I presume the notation means the free abelian group with generators ?

Second - I don't really know where to go with this question.

I might as well start by assuming k odd and

Then

I am not really sure how to calculate the image/kernel of these maps? Any hints appreciated

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 Post subject: Re: Homology CalculationPosted: Fri, 11 Mar 2011 13:21:46 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
qwirk wrote:
I am trying to calculate the homology of the chain complex defined as follows

Let and set for or 0 otherwise.

We defined the boundary operator by

First question - I presume the notation means the free abelian group with generators ?

Second - I don't really know where to go with this question.

I might as well start by assuming k odd and

Then

I am not really sure how to calculate the image/kernel of these maps? Any hints appreciated

First of all, yes, it is the free abelian group on two generators. The point of the presentation you're given is to calculate the kernal using linear algebra.

Notice that the matrix for the even ones is:

and for the odds is:

so clearly the nullspace (kernel) has dimension 1 for each of them, and in the odd case it is obviously spanned by and the even case by , since those are sent to zero clearly from the description. As the matrix has rank 1, that's the entire kernel. By the rank nullity theorem you should be able to deduce that the image has dimension 1 as well, and go from there.

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 Post subject: Re: Homology CalculationPosted: Sat, 12 Mar 2011 03:52:30 UTC
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Joined: Fri, 3 Sep 2010 09:29:45 UTC
Posts: 140
qwirk wrote:
I am trying to calculate the homology of the chain complex defined as follows

Let and set for or 0 otherwise.

We defined the boundary operator by

First question - I presume the notation means the free abelian group with generators ?

Second - I don't really know where to go with this question.

I might as well start by assuming k odd and

Then

I am not really sure how to calculate the image/kernel of these maps? Any hints appreciated

First of all, yes, it is the free abelian group on two generators. The point of the presentation you're given is to calculate the kernal using linear algebra.

Notice that the matrix for the even ones is:

and for the odds is:

so clearly the nullspace (kernel) has dimension 1 for each of them, and in the odd case it is obviously spanned by and the even case by , since those are sent to zero clearly from the description. As the matrix has rank 1, that's the entire kernel. By the rank nullity theorem you should be able to deduce that the image has dimension 1 as well, and go from there.

Thanks Shadow - I'll have a digest of this and see how I go

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 Post subject: Posted: Sat, 12 Mar 2011 07:45:01 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
Glad to help. If you have any issues, feel free to post a follow-up.

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 Post subject: Posted: Tue, 15 Mar 2011 00:12:59 UTC
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Joined: Fri, 3 Sep 2010 09:29:45 UTC
Posts: 140
Shadow - so for k odd and

gives me ??

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 Post subject: Posted: Tue, 15 Mar 2011 00:18:50 UTC
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Location: Austin, TX
qwirk wrote:
Shadow - so for k odd and

gives me ??

NO! Remember, the previous even has and which should give you an image of

Remember you always need to match where you are, the and from different degrees cannot possibly be subgroups of for

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 Post subject: Posted: Tue, 15 Mar 2011 00:54:39 UTC
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Joined: Fri, 3 Sep 2010 09:29:45 UTC
Posts: 140
qwirk wrote:
Shadow - so for k odd and

gives me ??

NO! Remember, the previous even has and which should give you an image of

Remember you always need to match where you are, the and from different degrees cannot possibly be subgroups of for

Sorry, you are right. So the homology groups are zero (except for the cases $k=n$)?

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 Post subject: Posted: Tue, 15 Mar 2011 02:45:00 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
qwirk wrote:
qwirk wrote:
Shadow - so for k odd and

gives me ??

NO! Remember, the previous even has and which should give you an image of

Remember you always need to match where you are, the and from different degrees cannot possibly be subgroups of for

Sorry, you are right. So the homology groups are zero (except for the cases $k=n$)?

Looks right.

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