Difference between revisions of "Notes:Homology"

Latest revision as of 07:30, 15 October 2016

• /Real projective plane
• /Torus

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: $G_1$

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!
       • $H_0=\langle x,y,z\rangle/\langle y-x, z-y\rangle\ (\cong\mathbb{Z}\ ?)$ (although surely there are other choices for $\langle y-x, z-y\rangle$)
• $H_1:=Z_1/B_1:=\text{Ker}(\partial_1)/\text{Im}(\partial 2)$
  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.

Computing the image of $\partial_1$

Take the following system:

• $\begin{pmatrix}-1 & 0 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & -1\end{pmatrix}\begin{pmatrix}\alpha\\\beta\\\gamma\\\delta\end{pmatrix}=\begin{pmatrix}x\\y\\z\end{pmatrix}$

Looking at it we see that the first column add the second column is minus the third, so the colspan is clearly 3. We can write this as the subset of $\mathbb{Z}^3$ spanned by:

• $\langle y-x,z-y\rangle$

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"