Trefor at 19:15, 11 April 2008

2008-04-11T19:15:23Z

← Older revision		Revision as of 15:15, 11 April 2008
Line 25:		Line 25:
	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.		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.

Trefor at 18:01, 25 March 2008

2008-03-25T18:01:31Z

New page

{{0708-1300/Navigation}}

==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:

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

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

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

Let's make this precise.

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

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.

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

'''Theorem'''

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

'''Examples:'''

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

<math>f_{\partial\sigma}: S^{n-1}\rightarrow \infty</math>

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

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

Ie we get <math>\mathbb{Z}\rightarrow\mathbb{Z}^2\rightarrow\mathbb{Z}</math>

Now, <math>\partial a = [p:a]p</math>

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

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

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

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

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

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

<math>\partial\sigma = [a:\sigma]a+b:\sigma b = 0_a + 2b</math>

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

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

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

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

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

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

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

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

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

===Second Hour===

<math>\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</math> for <math>z\in\mathbb{C}^{n+1}</math>, <math>\alpha\in\mathbb{C}-\{0\}</math>

<math>\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}</math>
=<math> \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}</math>

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

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

0708-1300/Class notes for Tuesday, March 25 - Revision history

Trefor at 19:15, 11 April 2008

Trefor at 18:01, 25 March 2008