Notes:Homology

Definitions

1. Boundary operator: $\partial_n:C_n\rightarrow C_{n-1}$ given by $\partial_n:[a_0,\ldots a_n]\mapsto\sum^n_{i=0}(-1)^i[a_0,\ldots,\hat{a_i},\ldots,a_n]$
2. $\mathbf{n}$-cycles: $Z_n$ (a cycle is defined to have boundary 0, thus $Z_n=\text{Ker}(\partial_n)$ - todo - discussion)
3. $\mathbf{n}$-boundaries: $B_n$ (the image of $\partial_{n+1}$ - all boundaries)
   • Claim: $B_n\le Z_n$ (that is: $B_n$ is a subgroup of $Z_n$)
4. $\mathbf{n}$^th homology group: $H_n:=Z_n/B_n$

Examples 1: $G_1$

$\xymatrix{ & & x \ar@/_.25pc/[dll]_a \\ y \ar@/_.25pc/[drr]_b \\ & & z \ar@/_.5pc/[uu]^c \ar@/_2pc/[uu]_d }$
Our first example, $G_1$

Chain complex: $\xymatrix{ 0 \ar[r]^{\partial_2} & C_1 \ar[r]^{\partial_1} \ar@2{->}[d] & C_0 \ar[r]^{\partial_0=0} \ar@2{->}[d] & 0 \\ & {\langle a,b,c,d\rangle\cong\mathbb{Z}^4} & {\langle x,y,z\rangle\cong\mathbb{Z}^3} }$
$\partial_1:C_1\rightarrow C_0$ morphism:

• We have:
  1. $\partial_1(a)= y-x$,
  2. $\partial_1(b)= z-y$,
  3. $\partial_1(c)= x-z$ and
  4. $\partial_1(d)=x-z$ also
• We extend this to a group homomorphism by defining:
  • $\begin{array}{rcl}\partial_1(\alpha a+\beta b+\gamma c+\delta d)&:=&\alpha\partial_1(a)+\beta\partial_1(b)+\gamma\partial_1(c)+\delta\partial_1(d)\\ & =& \alpha(y-x)+\beta(z-y)+(\gamma+\delta)(x-z)\\& = &(-\alpha+\gamma+\delta)x+(\alpha-\beta)y+(\beta-\gamma-\delta)z\end{array}$, we may write: $\begin{pmatrix}x\\y\\z\end{pmatrix}=\alpha\left(\begin{array}{c}-1\\ 1 \\ 0\end{array}\right)+\beta\begin{pmatrix}0\\-1\\1\end{pmatrix}+\gamma\begin{pmatrix}1\\0\\-1\end{pmatrix}+\delta\begin{pmatrix}1\\0\\-1\end{pmatrix}=\begin{pmatrix}-1 & 0 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & -1\end{pmatrix}\begin{pmatrix}\alpha\\\beta\\\gamma\\\delta\end{pmatrix}$
  • Recall also the rank plus nullity theorem:
    • For $f\in\mathcal{L}(V,W)$ we have $\text{Dim}(\text{Ker}(f))+\text{Dim}(\text{Im}(f))=\text{Dim}(V)$

Computing the homology groups:

• $H_0:=Z_0/B_0=\text{Ker}(\partial_0)/\text{Im}(\partial_1)$
  1. Computing $\text{Ker}(\partial_0)$ (result: $\text{Ker}(\partial_0)=C_0$)
     • By definition, $\partial_0:[a_0]\mapsto 0$, so everything in the domain of $\partial_0$ is in the kernel!
     • Thus $Z_0=C_0$
  2. Computing $\text{Im}(\partial_1)$
     • It is clear from the rank plus nullity theorem mentioned above that we should have $\text{Dim}(\text{Ker}(\partial_1))+\text{Dim}(\text{Im}(\partial_1))=4$ and we'll need to compute the kernel of $\partial_1$ for $H_1$ anyway.
       • See computing the kernel of $\partial_1$ below
       • The dimension of the kernel is $2$ so the dimension of the image is $2$ also!
       • $H_0=\langle x,y,z\rangle/\langle \text{hmm.... thing of dimension 2...}\rangle\cong\mathbb{Z}\ (?)$
• $H_1:=Z_1/B_1:=\text{Ker}(\partial_1)/\text{Im}(\partial 2)$
  1. Computing $\text{Ker}(\partial_1)$ has already been done below
  2. Computing $\text{Im}(\partial_2)$ is easy, it's $0$ - the trivial group
  • Thus:
    • $H_1\cong\text{Ker}(\partial_1)=\langle a+b+c,a+b+d\rangle\cong\mathbb{Z}^2$

Computing the kernel of $\partial_1$

To do this we wish to solve:

• $\begin{pmatrix}-1 & 0 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & -1\end{pmatrix}\begin{pmatrix}\alpha\\\beta\\\gamma\\\delta\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$, which basically amounts to rrefing $\begin{pmatrix}-1 & 0 & 1 & 1 & 0\\ 1 & -1 & 0 & 0 & 0\\ 0 & 1 & -1 & -1 & 0\end{pmatrix}$ giving us $\begin{pmatrix}1 & 0 & -1 & -1 & 0\\ 0 & 1 & -1 & -1 & 0 \\ 0 & 0 & 0 & 0 &0\end{pmatrix}$
  • Yielding: $\alpha=\gamma+\delta$ and $\beta=\gamma+\delta$. Let $\gamma:=s$ and then:
    • $\alpha=s+t$ and $\beta=s+t$, vectorially:
    • If $\begin{pmatrix}\alpha\\\beta\\\gamma\\\delta\end{pmatrix}=s\begin{pmatrix}1\\1\\1\\0 \end{pmatrix}+t\begin{pmatrix}1\\1\\0\\1 \end{pmatrix}$ then $\begin{pmatrix}\alpha\\\beta\\\gamma\\\delta\end{pmatrix}\in\text{Ker}(\partial_1)$
      • This makes perfect sense, it means (informally) $s$ times through $(a\rightarrow b\rightarrow c)$ and $t$ times through $a\rightarrow b\rightarrow d$, which goes $s+t$ times through both $a$ and $b$ all together!
    • Clearly the dimension is 2.

Dealing with generated spaces

Here we see the space $\langle (1,2),(2,1)\rangle$ "inside" $\mathbb{Z}^2$, clearly it is both not, and sort of is "isomorphic" to $\mathbb{Z}^2$. It isn't as there are holes, it is though as it itself is a space of dimension 2...

I don't like being so informal, hence "rings and modules"