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

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'''Van-Kampen's Theorem'''
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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.
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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>
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<math>\begin{matrix}
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&\ \ \ \  U_1&&\\
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&\nearrow^{i_1}&\searrow^{j_1}&\\
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U_1\cap U_2&&&U_1\cup U_2 = X\\
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&\searrow_{i_2}&\nearrow^{j_2}&\\
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&\ \ \ \  U_2&&\\
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\end{matrix}</math>
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where all the i's and j's are inclusions.
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Lets consider the image of this under the functor <math>\pi_1</math>
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<math>\begin{matrix}
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&\ \ \ \  \pi_1(U_1)&&\\
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&\nearrow^{i_{1*}}&\searrow^{j_{1*}}&\\
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\pi_1(U_1\cap U_2)&&& \pi_1(X)\\
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&\searrow_{i_{2*}}&\nearrow^{j_{2*}}&\\
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&\ \ \ \  \pi(U_2)&&\\
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\end{matrix}</math>
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Now consider the situation as groups:
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<math>\begin{matrix}
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&\ \ \ \  G_1&&\\
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&\nearrow_{\varphi_1}&\searrow&\\
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H&&&G_1*_H G_2\\
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&\searrow_{\varphi_2}&\nearrow&\\
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&\ \ \ \  G_2&&\\
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\end{matrix}</math>
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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
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Considering just the set without the identification, we note this is a group with the operation being concatenation of words followed by reduction.
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Ex: <math>a_1b_1a_2 + a_3b_2a_4 = a_1b_1ab_2a_4</math> where <math>a = a_2a_3</math>
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'''Claim:'''
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This is really a group.
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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
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<math>\forall h\in H, \phi_1(h) = \phi_2(h</math>)
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With this identification we have properly defined <math>G_1 *_H G_2</math>
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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>
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'''Example 0'''
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<math>\pi_1(S^n) </math> for <math> n\geq 2</math>
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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>)
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So, <math>\pi_1(S^n) = \pi_1(U_1)*_{\pi(U_1\cap U_2)}\pi_1(U_2)</math>
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But, since the hemispheres themselves are contractible, <math>\pi_1(U_1) = \pi_1(U_2) = \{e\}</math>
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Hence, <math>\pi_1(S^n) = \{e\}</math>
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'''Example 1'''
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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.
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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.
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The intersection is an X, contractible to a point and so <math>\pi_1(U_1\cap U_2) = \{e\}</math>
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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>
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This is non abelian
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'''Example 2'''
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<math>\pi_1(\mathbb{T}^2)</math>
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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.
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Clearly <math>U_1</math> is contractible, and hence <math>\pi_1(U_1) = \{e\}</math>
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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.
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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.
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Hence <math>\pi_1(U_2) = F(\alpha, \beta)</math> as in example 1
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Hence, <math>\pi_1(\mathbb{T}^2) = \{e\}*F(\alpha,\beta)/(i_{1*}(\gamma) = i_{2*}(\gamma))</math>
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Now, <math>i_{1*}(\gamma) = e</math>
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and
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<math>i_{2*}(\gamma) = \alpha\beta\alpha^{-1}\beta^{-1}</math>
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I.e., <math>\pi_1(\mathbb{T}^2) = F(\alpha,\beta)/ e = \alpha\beta\alpha^{-1}\beta^{-1}</math>
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<math> = F(\alpha,\beta)/(\alpha\beta = \beta\alpha)</math>
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This is just the Free Abelian group on two symbols and,
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<math>= \{\alpha^n\beta^m\} = \mathbb{Z}^2</math>
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Hence, <math>\pi_1(\mathbb{T}^2) = \mathbb{Z}^2</math>
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'''Example 3'''
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The two holed torus: <math>\Sigma_2</math>
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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>
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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>.
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Clearly <math>\pi_1(U_1) = \{e\}</math> as before.
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<math>\pi_1(U_1\cap U_2) = <\gamma></math> as before.
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Now, <math>U_2</math> this times when doing the identifications looks like a clover (4 loops intersecting at one point)
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Completely analogously to before, we see that <math>\pi_1(U_2) = F(\alpha_1, \beta_1, \alpha_2, \beta_2)</math>
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Again, <math>i_{1*}(\gamma) = e</math>
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<math>i_{2*}(\gamma) = \alpha_1\beta_1\alpha_1^{-1}\alpha_2\beta_2\alpha_1^{-1}\beta_2^{-1}</math>
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Therefore,
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<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>
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The ''abelianization'' of this group is
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<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>
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In case someone might want diagrams for the examples above:
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[[Image:0708-1300_notes_17-01-08c.jpg|200px]]

Latest revision as of 19:16, 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