Difference between revisions of "0708-1300/Class notes for Thursday, January 17"

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{{0708-1300/Navigation}}
 
{{0708-1300/Navigation}}
  
{{In|
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'''Van-Kampen's Theorem'''
n  = 1 |
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in = <nowiki><< KnotTheory`</nowiki>}}
+
  
<tt>Loading KnotTheory` version of January 13, 2008, 20:30:12.1353.<br>
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Let X be a point pointed topological space such that <math>X = U_1\cup U_2</math> where <math>U_1</math> and <math>U_2</math> are open and the base point b is in the (connected) intersection.
Read more at http://katlas.org/wiki/KnotTheory.</tt>
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{{Graphics|
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Then, <math>\pi_1(X) = \pi_1(U_1)*_{\pi_1(U_1\cap U_2)}\pi_1(U_2)</math>
n  = 2 |
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in = <nowiki>TubePlot[TorusKnot[8, 3]]</nowiki> |
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img= 0708-1300-T83.png |
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out= <nowiki>-Graphics-</nowiki>}}
+
  
{{In|
 
n  = 3 |
 
in = <nowiki>TC[r1_, t1_,r2_,t2_ ] := {
 
  (r1 +r2 Cos[2Pi t2])Cos[2Pi t1],
 
  (r1 +r2 Cos[2Pi t2])Sin[2Pi t1],
 
  r2 Sin[2Pi t2]
 
};</nowiki>}}
 
  
{{In|
 
n  = 4 |
 
in = <nowiki>InflatedTorus[p_, q_, b_] := ParametricPlot3D[
 
  TC[
 
    2, p t - q s,
 
    1 + b(p^2 + q^2)s(1 - (p^2 + q^2)s), q t + p s
 
  ],
 
  {t, 0, 1}, {s, 0, 1/(p^2 + q^2)},
 
  PlotPoints -> {6(p^2 + q^2) + 1, 7},
 
  DisplayFunction -> Identity
 
];</nowiki>}}
 
  
{{Graphics|
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<math>\begin{matrix}
n = 5 |
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&\ \ \ \  U_1&&\\
in = <nowiki>GraphicsArray[{{InflatedTorus[3,8,1], InflatedTorus[3,8,-1]}}]</nowiki> |
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&\nearrow^{i_1}&\searrow^{j_1}&\\
img= 0708-1300-InflatedTori.png |
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U_1\cap U_2&&&U_1\cup U_2 = X\\
out= <nowiki>-GraphicsArray-</nowiki>}}
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  &\searrow_{i_2}&\nearrow^{j_2}&\\
 +
&\ \ \ \  U_2&&\\
 +
\end{matrix}</math>
 +
 
 +
where all the i's and j's are inclusions.
 +
 
 +
 
 +
Lets consider the image of this under the functor <math>\pi_1</math>
 +
 
 +
 
 +
<math>\begin{matrix}
 +
&\ \ \ \  \pi_1(U_1)&&\\
 +
&\nearrow^{i_{1*}}&\searrow^{j_{1*}}&\\
 +
\pi_1(U_1\cap U_2)&&& \pi_1(X)\\
 +
&\searrow_{i_{2*}}&\nearrow^{j_{2*}}&\\
 +
&\ \ \ \  \pi(U_2)&&\\
 +
\end{matrix}</math>
 +
 
 +
 
 +
Now consider the situation as groups:
 +
 
 +
 
 +
<math>\begin{matrix}
 +
&\ \ \ \  G_1&&\\
 +
&\nearrow_{\varphi_1}&\searrow&\\
 +
H&&&G_1*_H G_2\\
 +
&\searrow_{\varphi_2}&\nearrow&\\
 +
&\ \ \ \  G_2&&\\
 +
\end{matrix}</math>
 +
 
 +
 
 +
Where <math>G_1 *_H G_2 = </math>{ words with letters alternating between being in <math>G_1</math> and <math>G_2</math>, ignoring e } / See Later
 +
 
 +
Considering just the set without the identification, we note this is a group with the operation being concatenation of words followed by reduction.
 +
 
 +
Ex: <math>a_1b_1a_2 + a_3b_2a_4 = a_1b_1ab_2a_4</math> where <math>a = a_2a_3</math>
 +
 
 +
 
 +
'''Claim:'''
 +
 
 +
This is really a group.
 +
 
 +
 
 +
So far, we have only defined the "free group of <math>G_1</math> and <math>G_2</math>". We now consider the identification (denoted above by 'See Later') which is
 +
 
 +
<math>\forall h\in H, \phi_1(h) = \phi_2(h</math>)
 +
 
 +
With this identification we have properly defined <math>G_1 *_H G_2</math>
 +
 
 +
 
 +
Note: <math>G_1 *_H G_2</math> is equivalent to { words in <math>G_1\cap G_2\}/ (e_1 = \{\}, e_2 = \{\}, g,h\in G_i, g\cdot h = gh)</math>
 +
 
 +
 
 +
'''Example 0'''
 +
 
 +
<math>\pi_1(S^n) </math> for <math> n\geq 2</math>
 +
 
 +
We can think of <math>S^n</math> as the union of two slightly overlapping open hemispheres which leaves the intersection as a band about the equator. As long as <math>n\geq 2</math> this is connected (but fails for <math>S^1</math>)
 +
 
 +
So, <math>\pi_1(S^n) = \pi_1(U_1)*_{\pi(U_1\cap U_2)}\pi_1(U_2)</math>
 +
 
 +
But, since the hemispheres themselves are contractible, <math>\pi_1(U_1) = \pi_1(U_2) = \{e\}</math>
 +
 
 +
Hence, <math>\pi_1(S^n) = \{e\}</math>
 +
 
 +
 
 +
 
 +
'''Example 1'''
 +
 
 +
Let us consider <math>\pi_1</math> of a a figure eight. Let <math>U_1</math> denote everything above a line slightly beneath the intersection and <math>U_2</math> everything below a line slightly above the intersection point.
 +
 
 +
Now both <math>U_1</math> and <math>U_2</math> are homotopically equivalent to a loop and so <math>\pi_1(U_1) = \pi_2(U_2) = \mathbb{Z}</math>. We can think of these being the groups generated by a loop going around once, I.e., isomorphic to <math><\alpha></math> and <math><\beta></math> respectively.
 +
 
 +
The intersection is an X, contractible to a point and so <math>\pi_1(U_1\cap U_2) = \{e\}</math>
 +
 
 +
So <math>\pi_1</math>(figure 8)<math> = <\alpha>*_{\{\}}<\beta> = F(\alpha,\beta)</math> the free group generated by <math>\alpha</math> and <math>\beta</math>
 +
 
 +
 
 +
This is non abelian
 +
 
 +
 
 +
'''Example 2'''
 +
 
 +
<math>\pi_1(\mathbb{T}^2)</math>
 +
 
 +
We consider <math>\mathbb{T}^2</math> in the normal way as a square with the normal identifications on the sides. We then consider two concentric squares inside this and define <math>U_1</math> as everything inside the larger square and <math>U_2</math> as everything outside the smaller square.
 +
 
 +
Clearly <math>U_1</math> is contractible, and hence <math>\pi_1(U_1) = \{e\}</math>
 +
 
 +
 
 +
Now, the intersection of <math>U_1</math> and <math>U_2</math> is equivalent to an annulus and so <math>\pi_1(U_1\cap U_2) = \mathbb{Z} = <\gamma></math> where <math>\gamma</math> is just a loop in the annulus.
 +
 
 +
Now considering <math>U_2</math>, we note that each of the four outer corner points in the big square are identified, and when we identify edges we are left with something equivalent to a figure 8.
 +
 
 +
Hence <math>\pi_1(U_2) = F(\alpha, \beta)</math> as in example 1
 +
 
 +
 
 +
Hence, <math>\pi_1(\mathbb{T}^2) = \{e\}*F(\alpha,\beta)/(i_{1*}(\gamma) = i_{2*}(\gamma))</math>
 +
 
 +
Now, <math>i_{1*}(\gamma) = e</math>
 +
 
 +
and
 +
 
 +
<math>i_{2*}(\gamma) = \alpha\beta\alpha^{-1}\beta^{-1}</math>
 +
 
 +
I.e., <math>\pi_1(\mathbb{T}^2) = F(\alpha,\beta)/ e = \alpha\beta\alpha^{-1}\beta^{-1}</math>
 +
 
 +
<math> = F(\alpha,\beta)/(\alpha\beta = \beta\alpha)</math>
 +
 
 +
This is just the Free Abelian group on two symbols and,
 +
 
 +
<math>= \{\alpha^n\beta^m\} = \mathbb{Z}^2</math>
 +
 
 +
Hence, <math>\pi_1(\mathbb{T}^2) = \mathbb{Z}^2</math>
 +
 
 +
 
 +
'''Example 3'''
 +
 
 +
 
 +
The two holed torus: <math>\Sigma_2</math>
 +
 
 +
Consider the schematic for this surface, consising of an octagon with edges labeled <math>a_1,b_1,a_1^{-1},b_1^{-1},a_2,b_2,a_2^{-1},b_2^{-1}</math>
 +
 
 +
As in the previous example, consider two concentric circles inside the octagon. Let everything inside the larger circle be <math>U_1</math> and everything outside the smaller circle be <math>U_2</math>.  
 +
 
 +
Clearly <math>\pi_1(U_1) = \{e\}</math> as before.
 +
 
 +
<math>\pi_1(U_1\cap U_2) = <\gamma></math> as before.
 +
 
 +
Now, <math>U_2</math> this times when doing the identifications looks like a clover (4 loops intersecting at one point)
 +
 
 +
Completely analogously to before, we see that <math>\pi_1(U_2) = F(\alpha_1, \beta_1, \alpha_2, \beta_2)</math>
 +
 
 +
Again, <math>i_{1*}(\gamma) = e</math>
 +
 
 +
<math>i_{2*}(\gamma) = \alpha_1\beta_1\alpha_1^{-1}\alpha_2\beta_2\alpha_1^{-1}\beta_2^{-1}</math>
 +
 
 +
 
 +
Therefore,
 +
 
 +
<math>\pi_{\Sigma_2} = F(\alpha_1, \beta_1, \alpha_2, \beta_2)/(e =\alpha_1\beta_1\alpha_1^{-1}\alpha_2\beta_2\alpha_1^{-1}\beta_2^{-1})</math>
 +
 
 +
The ''abelianization'' of this group is
 +
 
 +
<math>\pi_1^{ab}(\Sigma_2) = \pi_1(\Sigma_2)/ gh=hg = F.A.G (\alpha_1,\alpha_2,\beta_1,\beta^2) = \mathbb{Z}^4 \neq \mathbb{Z}^2</math>
 +
 
 +
 
 +
In case someone might want diagrams for the examples above:
 +
 
 +
[[Image:0708-1300_notes_17-01-08c.jpg|200px]]

Latest revision as of 18:16, 6 February 2008

Announcements go here

Van-Kampen's Theorem

Let X be a point pointed topological space such that X = U_1\cup U_2 where U_1 and U_2 are open and the base point b is in the (connected) intersection.

Then, \pi_1(X) = \pi_1(U_1)*_{\pi_1(U_1\cap U_2)}\pi_1(U_2)


\begin{matrix}
&\ \ \ \  U_1&&\\
 &\nearrow^{i_1}&\searrow^{j_1}&\\
U_1\cap U_2&&&U_1\cup U_2 = X\\
 &\searrow_{i_2}&\nearrow^{j_2}&\\
&\ \ \ \  U_2&&\\
\end{matrix}

where all the i's and j's are inclusions.


Lets consider the image of this under the functor \pi_1


\begin{matrix}
&\ \ \ \  \pi_1(U_1)&&\\
 &\nearrow^{i_{1*}}&\searrow^{j_{1*}}&\\
\pi_1(U_1\cap U_2)&&& \pi_1(X)\\
 &\searrow_{i_{2*}}&\nearrow^{j_{2*}}&\\
&\ \ \ \  \pi(U_2)&&\\
\end{matrix}


Now consider the situation as groups:


\begin{matrix}
&\ \ \ \  G_1&&\\
 &\nearrow_{\varphi_1}&\searrow&\\
H&&&G_1*_H G_2\\
 &\searrow_{\varphi_2}&\nearrow&\\
&\ \ \ \  G_2&&\\
\end{matrix}


Where G_1 *_H G_2 = { words with letters alternating between being in G_1 and G_2, ignoring e } / See Later

Considering just the set without the identification, we note this is a group with the operation being concatenation of words followed by reduction.

Ex: a_1b_1a_2 + a_3b_2a_4 = a_1b_1ab_2a_4 where a = a_2a_3


Claim:

This is really a group.


So far, we have only defined the "free group of G_1 and G_2". We now consider the identification (denoted above by 'See Later') which is

\forall h\in H, \phi_1(h) = \phi_2(h)

With this identification we have properly defined G_1 *_H G_2


Note: G_1 *_H G_2 is equivalent to { words in G_1\cap G_2\}/ (e_1 = \{\}, e_2 = \{\}, g,h\in G_i, g\cdot h = gh)


Example 0

\pi_1(S^n) for  n\geq 2

We can think of S^n as the union of two slightly overlapping open hemispheres which leaves the intersection as a band about the equator. As long as n\geq 2 this is connected (but fails for S^1)

So, \pi_1(S^n) = \pi_1(U_1)*_{\pi(U_1\cap U_2)}\pi_1(U_2)

But, since the hemispheres themselves are contractible, \pi_1(U_1) = \pi_1(U_2) = \{e\}

Hence, \pi_1(S^n) = \{e\}


Example 1

Let us consider \pi_1 of a a figure eight. Let U_1 denote everything above a line slightly beneath the intersection and U_2 everything below a line slightly above the intersection point.

Now both U_1 and U_2 are homotopically equivalent to a loop and so \pi_1(U_1) = \pi_2(U_2) = \mathbb{Z}. We can think of these being the groups generated by a loop going around once, I.e., isomorphic to <\alpha> and <\beta> respectively.

The intersection is an X, contractible to a point and so \pi_1(U_1\cap U_2) = \{e\}

So \pi_1(figure 8) = <\alpha>*_{\{\}}<\beta> = F(\alpha,\beta) the free group generated by \alpha and \beta


This is non abelian


Example 2

\pi_1(\mathbb{T}^2)

We consider \mathbb{T}^2 in the normal way as a square with the normal identifications on the sides. We then consider two concentric squares inside this and define U_1 as everything inside the larger square and U_2 as everything outside the smaller square.

Clearly U_1 is contractible, and hence \pi_1(U_1) = \{e\}


Now, the intersection of U_1 and U_2 is equivalent to an annulus and so \pi_1(U_1\cap U_2) = \mathbb{Z} = <\gamma> where \gamma is just a loop in the annulus.

Now considering U_2, we note that each of the four outer corner points in the big square are identified, and when we identify edges we are left with something equivalent to a figure 8.

Hence \pi_1(U_2) = F(\alpha, \beta) as in example 1


Hence, \pi_1(\mathbb{T}^2) = \{e\}*F(\alpha,\beta)/(i_{1*}(\gamma) = i_{2*}(\gamma))

Now, i_{1*}(\gamma) = e

and

i_{2*}(\gamma) = \alpha\beta\alpha^{-1}\beta^{-1}

I.e., \pi_1(\mathbb{T}^2) = F(\alpha,\beta)/ e = \alpha\beta\alpha^{-1}\beta^{-1}

 = F(\alpha,\beta)/(\alpha\beta = \beta\alpha)

This is just the Free Abelian group on two symbols and,

= \{\alpha^n\beta^m\} = \mathbb{Z}^2

Hence, \pi_1(\mathbb{T}^2) = \mathbb{Z}^2


Example 3


The two holed torus: \Sigma_2

Consider the schematic for this surface, consising of an octagon with edges labeled a_1,b_1,a_1^{-1},b_1^{-1},a_2,b_2,a_2^{-1},b_2^{-1}

As in the previous example, consider two concentric circles inside the octagon. Let everything inside the larger circle be U_1 and everything outside the smaller circle be U_2.

Clearly \pi_1(U_1) = \{e\} as before.

\pi_1(U_1\cap U_2) = <\gamma> as before.

Now, U_2 this times when doing the identifications looks like a clover (4 loops intersecting at one point)

Completely analogously to before, we see that \pi_1(U_2) = F(\alpha_1, \beta_1, \alpha_2, \beta_2)

Again, i_{1*}(\gamma) = e

i_{2*}(\gamma) = \alpha_1\beta_1\alpha_1^{-1}\alpha_2\beta_2\alpha_1^{-1}\beta_2^{-1}


Therefore,

\pi_{\Sigma_2} = F(\alpha_1, \beta_1, \alpha_2, \beta_2)/(e =\alpha_1\beta_1\alpha_1^{-1}\alpha_2\beta_2\alpha_1^{-1}\beta_2^{-1})

The abelianization of this group is

\pi_1^{ab}(\Sigma_2) = \pi_1(\Sigma_2)/ gh=hg = F.A.G (\alpha_1,\alpha_2,\beta_1,\beta^2) = \mathbb{Z}^4 \neq \mathbb{Z}^2


In case someone might want diagrams for the examples above:

0708-1300 notes 17-01-08c.jpg