0708-1300/Class notes for Thursday, January 17

From Drorbn
Jump to: navigation, search
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)

&\ \ \ \  U_1&&\\
U_1\cap U_2&&&U_1\cup U_2 = X\\
&\ \ \ \  U_2&&\\

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

Lets consider the image of this under the functor \pi_1

&\ \ \ \  \pi_1(U_1)&&\\
\pi_1(U_1\cap U_2)&&& \pi_1(X)\\
&\ \ \ \  \pi(U_2)&&\\

Now consider the situation as groups:

&\ \ \ \  G_1&&\\
H&&&G_1*_H G_2\\
&\ \ \ \  G_2&&\\

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


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


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


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}


\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