Difference between revisions of "0708-1300/Class notes for Tuesday, March 25"

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Example:
 
Example:
  
If you just have the segment consisting of two endpoints and the line connecting them, and call this <math>\tau</math>, then <math>p_{\tau}</math> takes the two end points to <math>\infty</math> and the rest (the open interval) gets mapped to B^1. Hence, we get a circle.  
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If you just have the segment consisting of two endpoints and the line connecting them, and call this <math>\tau</math>, then <math>p_{\tau}</math> takes the two end points to <math>\infty</math> and the rest (the open interval) gets mapped to <math>B^1</math>. Hence, we get a circle.  
  
  

Latest revision as of 14:15, 11 April 2008

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Typed Notes

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

First Hour

Definition

We define the CW chain complex via:

C_n^{CW}(K):=<K_n>

and the boundary maps via \partial:C_n^{CW}\rightarrow C^{CW}_{n-1} by \partial\sigma=\sum_{\tau\in K_{n-1}}[\tau:\sigma]\tau

where [\tau:\sigma] is roughly the number of times that \partial\sigma covers \tau.


Let's make this precise.

For f_{\sigma}:D_{\sigma}^n\rightarrow K (not quite an embedding) this restricts to a map f_{\partial\sigma}:S_{\sigma}^n\rightarrow K. Given \tau\in K_m let p_{\tau}:K^n\rightarrow S^n = B^n\cup\{\infty\} such that int(D^n_{\tau})\mapsto B^n and the rest maps to the point \infty.


Example:

If you just have the segment consisting of two endpoints and the line connecting them, and call this \tau, then p_{\tau} takes the two end points to \infty and the rest (the open interval) gets mapped to B^1. Hence, we get a circle.


We thus can now formally define [\tau:\sigma]= deg(p_{\tau}\circ f_{\partial\sigma}:S^{n-1}\rightarrow S^{n-1})


Theorem


(C^{CW}_*,\partial) is a chain complex; \partial^2 = 0 and H_*^{CW}(K) = H_*(K)


Examples:


1) S^n = \{\infty\}\cup D^n for n>1

f_{\partial\sigma}: S^{n-1}\rightarrow \infty

Hence C^{CW}_n = <\sigma>, C^{CW}_0 = <\infty> and all the rest are zero. Hence, H_p(S^n) = \mathbb{Z} for p = 0 or n and is zero otherwise.


2) Consider the torus thought of as a square with the usual identifications and \sigma is the interior. Hence, C^{CW}_0 =\{p\}, C^{CW}_1 is generated by the figure 8 with one loop labeled a and the other labeled b, and C^{CW}_2 is generated by the entire torus.

Ie we get \mathbb{Z}\rightarrow\mathbb{Z}^2\rightarrow\mathbb{Z}

Now, \partial a = [p:a]p

but \partial a takes the two endpoints of a (both p) and maps them to p. Neither point is mapped to \infty. Hence, deg\partial a:S^0\rightarrow S^0 = 0

Note: This ought to be checked from the definition of degree but was just stated in class


Now, [\sigma:a] = the degree of the map that takes the square to the figure 8...and hence is \pm 1\mp 1 = 0.

Hence the boundary map is zero at all places, so H_n(\mathbb{T}^2) = \mathbb{Z} if n = 0 or 2, \mathbb{Z}^2 if n = 1 and is zero otherwise.


3) Consider the Klein bottle thought of as a square with the usual identifications. Under p_b\circ f_{\partial\sigma} takes this to a circle with side labled b.

I.e., <\sigma>\mapsto<a,b>\mapsto^0<p>

\partial\sigma = [a:\sigma]a+b:\sigma b = 0_a + 2b

Where the sign may be negative. Or more eloquantly put: "2b or -2b, that is the question"

The kernal of <a,b>\rightarrow<p> is everything, so the homology is H_1(K) = <a,b>/2b=0\cong\mathbb{Z}\oplus\mathbb{Z}/2, H_0(K)=\mathbb{Z} and H_2(K) = 0.


4) \Sigma_g the "g holed torus" or "surface of genus g" is formed by the normal diagram with edges identified in sets of 4 such as aba^{-1}b^{-1}

So, get <\sigma>\rightarrow^0<a_1,\cdots, a_g, b_1,\cdots,b_g>\rightarrow^0<p>

Therefore, H_p = \mathbb{Z} if p = 2 or 0, \mathbb{Z}^{2g} if p = 1 and zero elsewhere.


5) \mathbb{R}P^n= D^n\cup \mathbb{R}P^{n-1} = D^n\cup D^{n-1}\cup\cdots\cup D^0

So, C_*^{CW}(\mathbb{R}P^n) is <D^n>\rightarrow<D^{n-1}>\rightarrow\cdots\rightarrow <D^0>

The boundary map alternates between 0 and 2 where it is 2 for <D^i>\rightarrow<D^{i-1}> if i is even and 0 if it is odd.

Hence the homology alternates between \mathbb{Z}/2 for odd p and 0 for even p.


Second Hour

\mathbb{C}P^n:\{[z_0,\cdots,z_n]:z_i\in\mathbb{C}\ not\ all\ zero\}=\mathbb{C}^{n+1}/z\sim\alpha z for z\in\mathbb{C}^{n+1}, \alpha\in\mathbb{C}-\{0\}

\mathbb{C}P^n=\{[z_0,\cdots,z_n]:z_n\neq 0\}\cup\{[\ldots,0]\} = \{[z_0,\cdots,z_{n-1},1]\}\cup\mathbb{C}P^{n-1} = \mathbb{C}^n\cup\mathbb{C}P^{n-1} = \mathbb{R}^{2n}\cup\mathbb{C}P^{n-1} = B^{2n}\cup\mathbb{C}P^{n-1}


I.e., \mathbb{C}P^n = D^{2n}\cup D^{2n-1}\cup\cdots


So, C^{CW}_*(\mathbb{C}P^n) alternates between 0 (for odd p) and \mathbb{Z} for even p and thus as this also as the homology. (and clearly trivial greater than p)