Difference between revisions of "Notes:Homology"
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+ | * [[/Real projective plane]] | ||
+ | * [[/Torus]] | ||
==Definitions== | ==Definitions== | ||
# '''Boundary operator: ''' {{M|\partial_n:C_n\rightarrow C_{n-1} }} given by {{M|1=\partial_n:[a_0,\ldots a_n]\mapsto\sum^n_{i=0}(-1)^i[a_0,\ldots,\hat{a_i},\ldots,a_n]}} | # '''Boundary operator: ''' {{M|\partial_n:C_n\rightarrow C_{n-1} }} given by {{M|1=\partial_n:[a_0,\ldots a_n]\mapsto\sum^n_{i=0}(-1)^i[a_0,\ldots,\hat{a_i},\ldots,a_n]}} | ||
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*# {{M|1=\partial_1(d)=x-z}} also | *# {{M|1=\partial_1(d)=x-z}} also | ||
* We extend this to a [[group homomorphism]] by defining: | * We extend this to a [[group homomorphism]] by defining: | ||
− | ** {{M|1=\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)\\<!-- | + | ** <span style="font-size:0.8em;">{{M|1=\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+(-\gamma-\delta)z\end{array} }}, we may write: <span style="font-size:0.8em;">{{M|1=(xyz)=\alpha\left(−110\right)+\beta(0−11)+\gamma\begin{pmatrix} | + | -->& = &(-\alpha+\gamma+\delta)x+(\alpha-\beta)y+(\beta-\gamma-\delta)z\end{array} }}</span>, we may write: <span style="font-size:0.8em;">{{M|1=(xyz)=\alpha\left(−110\right)+\beta(0−11)+\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}(αβγδ) }}</span> |
** Recall also the [[rank plus nullity theorem]]: | ** Recall also the [[rank plus nullity theorem]]: | ||
*** For {{M|f\in\mathcal{L}(V,W)}} we have {{M|1=\text{Dim}(\text{Ker}(f))+\text{Dim}(\text{Im}(f))=\text{Dim}(V)}} | *** For {{M|f\in\mathcal{L}(V,W)}} we have {{M|1=\text{Dim}(\text{Ker}(f))+\text{Dim}(\text{Im}(f))=\text{Dim}(V)}} | ||
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*#* It is clear from the [[rank plus nullity theorem]] mentioned above that we should have {{M|1=\text{Dim}(\text{Ker}(\partial_1))+\text{Dim}(\text{Im}(\partial_1))=4}} and we'll need to compute the kernel of {{M|\partial_1}} for {{M|H_1}} anyway. | *#* It is clear from the [[rank plus nullity theorem]] mentioned above that we should have {{M|1=\text{Dim}(\text{Ker}(\partial_1))+\text{Dim}(\text{Im}(\partial_1))=4}} and we'll need to compute the kernel of {{M|\partial_1}} for {{M|H_1}} anyway. | ||
*#** See ''computing the kernel of {{M|\partial_1}}'' below | *#** See ''computing the kernel of {{M|\partial_1}}'' below | ||
+ | *#** The dimension of the kernel is {{M|2}} so the dimension of the image is {{M|2}} also! | ||
+ | *#** {{M|1=H_0=\langle x,y,z\rangle/\langle y-x, z-y\rangle\ (\cong\mathbb{Z}\ ?)}} (although surely there are other choices for {{M|1=\langle y-x, z-y\rangle}}) | ||
+ | * {{M|1=H_1:=Z_1/B_1:=\text{Ker}(\partial_1)/\text{Im}(\partial 2)}} | ||
+ | *# Computing {{M|\text{Ker}(\partial_1)}} has already been done below | ||
+ | *# Computing {{M|\text{Im}(\partial_2)}} is easy, it's {{M|0}} - the [[trivial group]] | ||
+ | ** Thus: | ||
+ | *** {{M|1=H_1\cong\text{Ker}(\partial_1)=\langle a+b+c,a+b+d\rangle\cong\mathbb{Z}^2}} | ||
===Computing the kernel of {{M|\partial_1}}=== | ===Computing the kernel of {{M|\partial_1}}=== | ||
To do this we wish to solve: | To do this we wish to solve: | ||
− | * {{M|1=\begin{pmatrix}-1 & 0 & | + | * {{M|1=\begin{pmatrix}-1 & 0 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & -1\end{pmatrix}(αβγδ)=(000) }}, which basically amounts to [[rrefing]] {{M|1=\begin{pmatrix}-1 & 0 & 1 & 1 & 0\\ 1 & -1 & 0 & 0 & 0\\ 0 & 1 & -1 & -1 & 0\end{pmatrix} }} giving us {{M|1=\begin{pmatrix}1 & 0 & -1 & -1 & 0\\ 0 & 1 & -1 & -1 & 0 \\ 0 & 0 & 0 & 0 &0\end{pmatrix} }} |
− | ** | + | ** Yielding: {{M|1=\alpha=\gamma+\delta}} and {{M|1=\beta=\gamma+\delta}}. Let {{M|1=\gamma:=s}} and {{M|\delta:=t}} then: |
+ | *** {{M|1=\alpha=s+t}} and {{M|1=\beta=s+t}}, vectorially: | ||
+ | *** If {{M|1=\begin{pmatrix}\alpha\\\beta\\\gamma\\\delta\end{pmatrix}=s\begin{pmatrix}1\\1\\1\\0 \end{pmatrix}+t(1101) }} then {{M|(αβγδ)\in\text{Ker}(\partial_1)}} | ||
+ | **** This makes perfect sense, it means (''informally'') {{M|s}} times through {{M|(a\rightarrow b\rightarrow c)}} and {{M|t}} times through {{M|a\rightarrow b\rightarrow d}}, which goes {{M|s+t}} times through both {{M|a}} and {{M|b}} all together! | ||
+ | *** Clearly the dimension is 2. | ||
+ | ===Computing the image of {{M|\partial_1}}=== | ||
+ | Take the following system: | ||
+ | * {{M|1=(−10111−10001−1−1)\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 {{M|\mathbb{Z}^3}} spanned by: | ||
+ | * {{M|\langle y-x,z-y\rangle}} | ||
+ | ==Dealing with generated spaces== | ||
+ | [[File:NonSurjectiveInvertable"Lin"Map.JPG|thumb|Here we see the space {{M|\langle (1,2),(2,1)\rangle}} "inside" {{M|\mathbb{Z}^2}}, clearly it is both not, and sort of is "isomorphic" to {{M|\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" |
Latest revision as of 07:30, 15 October 2016
Contents
[hide]Definitions
- Boundary operator: ∂n:Cn→Cn−1 given by ∂n:[a0,…an]↦∑ni=0(−1)i[a0,…,^ai,…,an]
- n-cycles: Zn (a cycle is defined to have boundary 0, thus Zn=Ker(∂n) - todo - discussion)
- n-boundaries: Bn (the image of ∂n+1 - all boundaries)
- Claim: Bn≤Zn (that is: Bn is a subgroup of Zn)
- nth homology group: Hn:=Zn/Bn
Examples 1: G1
Chain complex:
∂1:C1→C0 morphism:
- We have:
- ∂1(a)=y−x,
- ∂1(b)=z−y,
- ∂1(c)=x−z and
- ∂1(d)=x−z also
- We extend this to a group homomorphism by defining:
- ∂1(αa+βb+γc+δd):=α∂1(a)+β∂1(b)+γ∂1(c)+δ∂1(d)=α(y−x)+β(z−y)+(γ+δ)(x−z)=(−α+γ+δ)x+(α−β)y+(β−γ−δ)z, we may write: (xyz)=α(−110)+β(0−11)+γ(10−1)+δ(10−1)=(−10111−10001−1−1)(αβγδ)
- Recall also the rank plus nullity theorem:
- For f∈L(V,W) we have Dim(Ker(f))+Dim(Im(f))=Dim(V)
Computing the homology groups:
- H0:=Z0/B0=Ker(∂0)/Im(∂1)
- Computing Ker(∂0) (result: Ker(∂0)=C0)
- By definition, ∂0:[a0]↦0, so everything in the domain of ∂0 is in the kernel!
- Thus Z0=C0
- Computing Im(∂1)
- It is clear from the rank plus nullity theorem mentioned above that we should have Dim(Ker(∂1))+Dim(Im(∂1))=4 and we'll need to compute the kernel of ∂1 for H1 anyway.
- See computing the kernel of ∂1 below
- The dimension of the kernel is 2 so the dimension of the image is 2 also!
- H0=⟨x,y,z⟩/⟨y−x,z−y⟩ (≅Z ?) (although surely there are other choices for ⟨y−x,z−y⟩)
- It is clear from the rank plus nullity theorem mentioned above that we should have Dim(Ker(∂1))+Dim(Im(∂1))=4 and we'll need to compute the kernel of ∂1 for H1 anyway.
- Computing Ker(∂0) (result: Ker(∂0)=C0)
- H1:=Z1/B1:=Ker(∂1)/Im(∂2)
- Computing Ker(∂1) has already been done below
- Computing Im(∂2) is easy, it's 0 - the trivial group
- Thus:
- H1≅Ker(∂1)=⟨a+b+c,a+b+d⟩≅Z2
Computing the kernel of ∂1
To do this we wish to solve:
- (−10111−10001−1−1)(αβγδ)=(000), which basically amounts to rrefing (−101101−100001−1−10) giving us (10−1−1001−1−1000000)
- Yielding: α=γ+δ and β=γ+δ. Let γ:=s and then:
- α=s+t and β=s+t, vectorially:
- If (αβγδ)=s(1110)+t(1101) then (αβγδ)∈Ker(∂1)
- This makes perfect sense, it means (informally) s times through (a→b→c) and t times through a→b→d, which goes s+t times through both a and b all together!
- Clearly the dimension is 2.
- Yielding: α=γ+δ and β=γ+δ. Let γ:=s and then:
Computing the image of ∂1
Take the following system:
- (−10111−10001−1−1)(αβγδ)=(xyz)
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 Z3 spanned by:
- ⟨y−x,z−y⟩
Dealing with generated spaces
I don't like being so informal, hence "rings and modules"