Homology Calculation

qwirk · **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 $n \in Z^+$ and set $c_k = \mathbb{Z}\{ x_k,y_k \}$ for $$0 \le k \le n$ or 0 otherwise.

We defined the boundary operator $d_k:c_k \to c_{k-1}$ by

$d(x_k)= \begin{cases} x_{k-1}-y_{k-1}, &k= \mathrm{Odd} \\ x_{k-1}+y_{k-1}, &k= \mathrm{Even} \end{cases}$

$d(y_k)= \begin{cases} y_{k-1}-x_{k-1}, &k= \mathrm{Odd} \\ y_{k-1}+x_{k-1}, &k= \mathrm{Even} \end{cases}$

First question - I presume the notation $\mathbb{Z} \{ a,b \}$ means the free abelian group with generators a,b

?

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

I might as well start by assuming k odd and $0 \le k < n$

Then $\mathbb{Z} \{ x_k,y_k \} \mapsto \mathbb{Z} \{ x_{k-1} - y_{k-1}, y_{k-1}-x_{k-1} \}$

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

Shadow · **Posted:** Fri, 11 Mar 2011 13:21:46 UTC

qwirk wrote:

I am trying to calculate the homology of the chain complex defined as follows

Let $n \in Z^+$ and set $c_k = \mathbb{Z}\{ x_k,y_k \}$ for $$0 \le k \le n$ or 0 otherwise.

We defined the boundary operator $d_k:c_k \to c_{k-1}$ by

$d(x_k)= \begin{cases} x_{k-1}-y_{k-1}, &k= \mathrm{Odd} \\ x_{k-1}+y_{k-1}, &k= \mathrm{Even} \end{cases}$

$d(y_k)= \begin{cases} y_{k-1}-x_{k-1}, &k= \mathrm{Odd} \\ y_{k-1}+x_{k-1}, &k= \mathrm{Even} \end{cases}$

First question - I presume the notation $\mathbb{Z} \{ a,b \}$ means the free abelian group with generators a,b

?

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

I might as well start by assuming k odd and $0 \le k < n$

Then $\mathbb{Z} \{ x_k,y_k \} \mapsto \mathbb{Z} \{ x_{k-1} - y_{k-1}, y_{k-1}-x_{k-1} \}$

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:

$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$

and for the odds is:

$\begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix}$

so clearly the nullspace (kernel) has dimension 1 for each of them, and in the odd case it is obviously spanned by x_k+y_k

and the even case by x_k-y_k

, 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.

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

Shadow wrote:

qwirk wrote:

I am trying to calculate the homology of the chain complex defined as follows

Let $n \in Z^+$ and set $c_k = \mathbb{Z}\{ x_k,y_k \}$ for $$0 \le k \le n$ or 0 otherwise.

We defined the boundary operator $d_k:c_k \to c_{k-1}$ by

$d(x_k)= \begin{cases} x_{k-1}-y_{k-1}, &k= \mathrm{Odd} \\ x_{k-1}+y_{k-1}, &k= \mathrm{Even} \end{cases}$

$d(y_k)= \begin{cases} y_{k-1}-x_{k-1}, &k= \mathrm{Odd} \\ y_{k-1}+x_{k-1}, &k= \mathrm{Even} \end{cases}$

First question - I presume the notation $\mathbb{Z} \{ a,b \}$ means the free abelian group with generators a,b

?

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

I might as well start by assuming k odd and $0 \le k < n$

Then $\mathbb{Z} \{ x_k,y_k \} \mapsto \mathbb{Z} \{ x_{k-1} - y_{k-1}, y_{k-1}-x_{k-1} \}$

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:

$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$

and for the odds is:

$\begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix}$

so clearly the nullspace (kernel) has dimension 1 for each of them, and in the odd case it is obviously spanned by x_k+y_k

and the even case by x_k-y_k

, 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

Shadow · **Posted:** Sat, 12 Mar 2011 07:45:01 UTC

Glad to help. If you have any issues, feel free to post a follow-up.

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

Shadow - so for k odd and $0 \le k+1 \le n$

$h_k = \mathrm{ker} \ d_k / \mathrm{Im} \ d_{k+1}$ gives me $\langle x_{k-1}+y_{k-1} \rangle / \langle x_k+y_k \rangle$ ??

Shadow · **Posted:** Tue, 15 Mar 2011 00:18:50 UTC

qwirk wrote:

Shadow - so for k odd and $0 \le k+1 \le n$

$h_k = \mathrm{ker} \ d_k / \mathrm{Im} \ d_{k+1}$ gives me $\langle x_{k-1}+y_{k-1} \rangle / \langle x_k+y_k \rangle$ ??

NO! Remember, the previous even has $x_{k-1}$ and $y_{k-1}$ which should give you an image of $\langle x_{k}+y_{k}\rangle$

Remember you always need to match where you are, the x_k

and

from different degrees cannot possibly be subgroups of $x_\ell,y_\ell$ for $\ell\ne k$

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

Shadow wrote:

qwirk wrote:

Shadow - so for k odd and $0 \le k+1 \le n$

$h_k = \mathrm{ker} \ d_k / \mathrm{Im} \ d_{k+1}$ gives me $\langle x_{k-1}+y_{k-1} \rangle / \langle x_k+y_k \rangle$ ??

NO! Remember, the previous even has $x_{k-1}$ and $y_{k-1}$ which should give you an image of $\langle x_{k}+y_{k}\rangle$

Remember you always need to match where you are, the x_k

and

from different degrees cannot possibly be subgroups of $x_\ell,y_\ell$ for $\ell\ne k$

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

Shadow · **Posted:** Tue, 15 Mar 2011 02:45:00 UTC

qwirk wrote:

Shadow wrote:

qwirk wrote:

Shadow - so for k odd and $0 \le k+1 \le n$

$h_k = \mathrm{ker} \ d_k / \mathrm{Im} \ d_{k+1}$ gives me $\langle x_{k-1}+y_{k-1} \rangle / \langle x_k+y_k \rangle$ ??

NO! Remember, the previous even has $x_{k-1}$ and $y_{k-1}$ which should give you an image of $\langle x_{k}+y_{k}\rangle$

Remember you always need to match where you are, the x_k

and

from different degrees cannot possibly be subgroups of $x_\ell,y_\ell$ for $\ell\ne k$

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

Looks right.

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