Difference between revisions of "Exercises:Saul - Algebraic Topology - 7/Exercise 7.6"

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(Added first part of question, saving work)
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====Solutions====
 
====Solutions====
 
=====Part I=====
 
=====Part I=====
 +
<div style="float:right;margin:0px;margin-left:0.2em;">
 +
{{:Exercises:Saul - Algebraic Topology - 7/Exercise 7.6/X Complex}}
 +
</div>The homology groups of the {{M|2}}-torus have been computed in previous assignments, the result was:
 +
* {{M|H_0^\Delta(T^2)\cong\mathbb{Z} }}
 +
* {{M|H_1^\Delta(T^2)\cong\mathbb{Z}\oplus\mathbb{Z}\eq:\mathbb{Z}^2}}
 +
* {{M|H_2^\Delta(T^2)\cong\mathbb{Z} }}
 +
* {{M|H_n^\Delta(T^2)\cong 0}}<ref group="Note">Here {{M|0}} denotes the [[trivial group]]</ref> for {{M|n\in\{3,\ldots\}\subset\mathbb{N} }}
 +
To calculate the homology groups of {{M|X}} we must start by computing the boundary of the generators (which are extended to group homomorphisms on free Abelian group by the characteristic property of free Abelian groups):
 +
* {{M|2}}-simplices:
 +
*: {{M|\partial_2(A)\eq c}}, {{M|\partial_2(B)\eq c}}
 +
* {{M|1}}-simplices:
 +
*: {{M|\partial_1(\alpha)\eq v_1-v_0}}, {{M|\partial_1(\beta)\eq v_2-v_0}}, {{M|\partial_1(\gamma)\eq v_2-v_1}}
 +
*: {{M|\partial_1(a)\eq v_2-v_3}}, {{M|\partial_1(b)\eq v_4-v_3}}, {{M|\partial_1(c)\eq 0}}
 +
*: {{M|\partial_1(x)\eq v_4-v_5}}, {{M|\partial_1(y)\eq v_4-v_6}}, {{M|\partial_1(z)\eq v_6-v_5}}
 +
* Note that {{M|\partial_0:(\text{anything})\mapsto 0}}, so we do not mention in here, it is also obvious that the entire of the domain is the kernel of {{M|\partial_0}} from this definition.
 +
Now the images and kernels:
 +
* {{M|\text{Im}(\partial_2)\eq \langle c\rangle}} - by inspection
 +
* {{M|\text{Ker}(\partial_2)\eq \langle A-B \rangle}} - by inspection
 +
* {{M|\text{Im}(\partial_1)\eq \langle v_0-v_1,\ v_0-v_2,\ v_2-v_3,\ v_3-v_4,\ v_4-v_5,\ v_6-v_4\rangle}}
 +
** Calculated by starting with just the first vector, {{M|\partial_1(\alpha)}}, going over each of the remaining vectors and adding it to this linearly-independent set if said vector is not a linear combination of what we have in this set.
 +
** {{M|\partial_1(\gamma)\eq\partial_1(\beta)-\partial_1(\beta)}}, then we see {{M|\partial_1(c)}} is the identity element, {{M|0}}, so cannot be in a basis set for obvious reasons, and {{M|\partial_1(z)\eq\partial_1(x)-\partial_1(y)}}, hence the chosen basis consists of all but these.
 +
* {{M|\text{Ker}(\partial_1)\eq\langle\alpha-\beta+\gamma, c, x-y-z\rangle}}
 +
** Computed by "''RReffing in {{M|\mathbb{Z} }}''", I may upload a picture of these matrices, but as it is {{M|10}} columns by {{M|7}} rows I am not eager to type both the starting matrix and it's reduced form out.
 +
* {{M|\text{Ker}(\partial_0)\eq\langle v_0,v_1,v_2,v_3,v_4,v_5,v_6\rangle}}, as discussed above and from the definition of {{M|\partial_0}}
 +
Note that for any higher values:
 +
* {{M|\text{Ker}(\partial_n)\eq 0}} for {{M|n\ge 3}}, {{M|n\in\mathbb{N} }}, as these groups are trivial groups, and a group homomorphism must map the identity to the identity, couple this with the domain of {{M|\partial_n}} has one element and the result follows.
 +
* {{M|\text{Im}(\partial_n)\eq 0}} for {{M|n\ge 3}}, {{M|n\in\mathbb{N} }}, as the image of a trivial group must be the identity element of the co-domain group.
 +
======Homology groups of {{M|X}}======
 +
* {{MM|H_2^\Delta(X):\eq\frac{\text{Ker}(\partial_2)}{\text{Im}(\partial_3)}\eq\frac{\langle A-B \rangle}{0}\cong \langle A-B \rangle\cong\mathbb{Z} }}
 +
* {{MM|H_1^\Delta(X):\eq\frac{\text{Ker}(\partial_1)}{\text{Im}(\partial_2)}\eq\frac{\langle\alpha-\beta+\gamma, c, x-y-z\rangle}{\langle c\rangle}\cong\langle\alpha-\beta+\gamma,x-y-z\rangle\cong\mathbb{Z}^2 }}
 +
* {{MM|H_0^\Delta(X):\eq\frac{\text{Ker}(\partial_0)}{\text{Im}(\partial_1)}\eq\frac{\langle v_0,v_1,v_2,v_3,v_4,v_5,v_6\rangle}{\langle v_0-v_1,\ v_0-v_2,\ v_2-v_3,\ v_3-v_4,\ v_4-v_5,\ v_6-v_4\rangle} }}
 
=====Part II=====
 
=====Part II=====
 
<noinclude>
 
<noinclude>

Revision as of 17:30, 28 February 2017

Exercises

Exercises 7.6

Question

  1. Compute the singular homology groups of [ilmath]T^2:\eq\mathbb{S}^1\times\mathbb{S}^1[/ilmath] and of [ilmath]X:\eq\mathbb{S}^1\vee\mathbb{S}^1\vee\mathbb{S}^2[/ilmath]
  2. Prove that [ilmath]T^2[/ilmath] and [ilmath]X[/ilmath] are not homotopy equivalent spaces

Solutions

Part I
[ilmath]\Delta[/ilmath]-complex for [ilmath]X[/ilmath]
[ilmath]\xymatrix{ \bullet_{v_0} \ar[dd]_\alpha \ar[dr]^\beta & & \bullet_{v_3} \ar@2{->}[dr]^b \ar@2{->}[dl]_a & & \bullet_{v_5} \ar[dd]^z \\ & \bullet_{v_2} \ar@2{<-}[dr]_a \ar@{}[rr]|(.3){\LARGE{\mathbf{A} } }|(.7){\LARGE{\mathbf{ B} } } & & \bullet_{v_4} \ar@{<-}[ur]^x \\ \bullet_{v_1} \ar[ur]_\gamma & & \bullet_{v_3} \ar@2{->}[ur]_b \ar@2{->}[uu]^c_c & & \bullet_{v_6} \ar[ul]^y }[/ilmath]
[ilmath]0[/ilmath]-simplices: [ilmath]v_0,v_1,v_2,v_3,v_4,v_5,v_6[/ilmath] (7)
[ilmath]1[/ilmath]-simplices: [ilmath]\alpha,\beta,\gamma,a,b,c,x,y,z[/ilmath] (9)
[ilmath]2[/ilmath]-simplices: [ilmath]A,B[/ilmath] (2)
The homology groups of the [ilmath]2[/ilmath]-torus have been computed in previous assignments, the result was:
  • [ilmath]H_0^\Delta(T^2)\cong\mathbb{Z} [/ilmath]
  • [ilmath]H_1^\Delta(T^2)\cong\mathbb{Z}\oplus\mathbb{Z}\eq:\mathbb{Z}^2[/ilmath]
  • [ilmath]H_2^\Delta(T^2)\cong\mathbb{Z} [/ilmath]
  • [ilmath]H_n^\Delta(T^2)\cong 0[/ilmath][Note 1] for [ilmath]n\in\{3,\ldots\}\subset\mathbb{N} [/ilmath]

To calculate the homology groups of [ilmath]X[/ilmath] we must start by computing the boundary of the generators (which are extended to group homomorphisms on free Abelian group by the characteristic property of free Abelian groups):

  • [ilmath]2[/ilmath]-simplices:
    [ilmath]\partial_2(A)\eq c[/ilmath], [ilmath]\partial_2(B)\eq c[/ilmath]
  • [ilmath]1[/ilmath]-simplices:
    [ilmath]\partial_1(\alpha)\eq v_1-v_0[/ilmath], [ilmath]\partial_1(\beta)\eq v_2-v_0[/ilmath], [ilmath]\partial_1(\gamma)\eq v_2-v_1[/ilmath]
    [ilmath]\partial_1(a)\eq v_2-v_3[/ilmath], [ilmath]\partial_1(b)\eq v_4-v_3[/ilmath], [ilmath]\partial_1(c)\eq 0[/ilmath]
    [ilmath]\partial_1(x)\eq v_4-v_5[/ilmath], [ilmath]\partial_1(y)\eq v_4-v_6[/ilmath], [ilmath]\partial_1(z)\eq v_6-v_5[/ilmath]
  • Note that [ilmath]\partial_0:(\text{anything})\mapsto 0[/ilmath], so we do not mention in here, it is also obvious that the entire of the domain is the kernel of [ilmath]\partial_0[/ilmath] from this definition.

Now the images and kernels:

  • [ilmath]\text{Im}(\partial_2)\eq \langle c\rangle[/ilmath] - by inspection
  • [ilmath]\text{Ker}(\partial_2)\eq \langle A-B \rangle[/ilmath] - by inspection
  • [ilmath]\text{Im}(\partial_1)\eq \langle v_0-v_1,\ v_0-v_2,\ v_2-v_3,\ v_3-v_4,\ v_4-v_5,\ v_6-v_4\rangle[/ilmath]
    • Calculated by starting with just the first vector, [ilmath]\partial_1(\alpha)[/ilmath], going over each of the remaining vectors and adding it to this linearly-independent set if said vector is not a linear combination of what we have in this set.
    • [ilmath]\partial_1(\gamma)\eq\partial_1(\beta)-\partial_1(\beta)[/ilmath], then we see [ilmath]\partial_1(c)[/ilmath] is the identity element, [ilmath]0[/ilmath], so cannot be in a basis set for obvious reasons, and [ilmath]\partial_1(z)\eq\partial_1(x)-\partial_1(y)[/ilmath], hence the chosen basis consists of all but these.
  • [ilmath]\text{Ker}(\partial_1)\eq\langle\alpha-\beta+\gamma, c, x-y-z\rangle[/ilmath]
    • Computed by "RReffing in [ilmath]\mathbb{Z} [/ilmath]", I may upload a picture of these matrices, but as it is [ilmath]10[/ilmath] columns by [ilmath]7[/ilmath] rows I am not eager to type both the starting matrix and it's reduced form out.
  • [ilmath]\text{Ker}(\partial_0)\eq\langle v_0,v_1,v_2,v_3,v_4,v_5,v_6\rangle[/ilmath], as discussed above and from the definition of [ilmath]\partial_0[/ilmath]

Note that for any higher values:

  • [ilmath]\text{Ker}(\partial_n)\eq 0[/ilmath] for [ilmath]n\ge 3[/ilmath], [ilmath]n\in\mathbb{N} [/ilmath], as these groups are trivial groups, and a group homomorphism must map the identity to the identity, couple this with the domain of [ilmath]\partial_n[/ilmath] has one element and the result follows.
  • [ilmath]\text{Im}(\partial_n)\eq 0[/ilmath] for [ilmath]n\ge 3[/ilmath], [ilmath]n\in\mathbb{N} [/ilmath], as the image of a trivial group must be the identity element of the co-domain group.
Homology groups of [ilmath]X[/ilmath]
  • [math]H_2^\Delta(X):\eq\frac{\text{Ker}(\partial_2)}{\text{Im}(\partial_3)}\eq\frac{\langle A-B \rangle}{0}\cong \langle A-B \rangle\cong\mathbb{Z} [/math]
  • [math]H_1^\Delta(X):\eq\frac{\text{Ker}(\partial_1)}{\text{Im}(\partial_2)}\eq\frac{\langle\alpha-\beta+\gamma, c, x-y-z\rangle}{\langle c\rangle}\cong\langle\alpha-\beta+\gamma,x-y-z\rangle\cong\mathbb{Z}^2 [/math]
  • [math]H_0^\Delta(X):\eq\frac{\text{Ker}(\partial_0)}{\text{Im}(\partial_1)}\eq\frac{\langle v_0,v_1,v_2,v_3,v_4,v_5,v_6\rangle}{\langle v_0-v_1,\ v_0-v_2,\ v_2-v_3,\ v_3-v_4,\ v_4-v_5,\ v_6-v_4\rangle} [/math]
Part II

Notes

  1. ↑ Here [ilmath]0[/ilmath] denotes the trivial group

References


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