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Week of...

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Fall Semester

1

Sep 10

About, Tue, Thu

2

Sep 17

Tue, HW1, Thu

3

Sep 24

Tue, Photo, Thu

4

Oct 1

Questionnaire, Tue, HW2, Thu

5

Oct 8

Thanksgiving, Tue, Thu

6

Oct 15

Tue, HW3, Thu

7

Oct 22

Tue, Thu

8

Oct 29

Tue, HW4, Thu, Hilbert sphere

9

Nov 5

Tue,Thu, TE1

10

Nov 12

Tue, Thu

11

Nov 19

Tue, Thu, HW5

12

Nov 26

Tue, Thu

13

Dec 3

Tue, Thu, HW6

Spring Semester

14

Jan 7

Tue, Thu, HW7

15

Jan 14

Tue, Thu

16

Jan 21

Tue, Thu, HW8

17

Jan 28

Tue, Thu

18

Feb 4

Tue

19

Feb 11

TE2, Tue, HW9, Thu, Feb 17: last chance to drop class

R

Feb 18

Reading week

20

Feb 25

Tue, Thu, HW10

21

Mar 3

Tue, Thu

22

Mar 10

Tue, Thu, HW11

23

Mar 17

Tue, Thu

24

Mar 24

Tue, HW12, Thu

25

Mar 31

Referendum,Tue, Thu

26

Apr 7

Tue, Thu

R

Apr 14

Office hours

R

Apr 21

Office hours

F

Apr 28

Office hours, Final (Fri, May 2)

Register of Good Deeds

Errata to Bredon's Book


Announcements go here
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 $\partial :C_{n}^{CW}\rightarrow C_{n1}^{CW}$ by $\partial \sigma =\sum _{\tau \in K_{n1}}[\tau :\sigma ]\tau$
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 $int(D_{\tau }^{n})\mapsto B^{n}$ 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 $[\tau :\sigma ]=deg(p_{\tau }\circ f_{\partial \sigma }:S^{n1}\rightarrow S^{n1})$
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
$f_{\partial \sigma }:S^{n1}\rightarrow \infty$
Hence $C_{n}^{CW}=<\sigma >,C_{0}^{CW}=<\infty >$ 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, $C_{0}^{CW}=\{p\},C_{1}^{CW}$ is generated by the figure 8 with one loop labeled a and the other labeled b, and $C_{2}^{CW}$ 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) $\mathbb {R} P^{n}=D^{n}\cup \mathbb {R} P^{n1}=D^{n}\cup D^{n1}\cup \cdots \cup D^{0}$
So, $C_{*}^{CW}(\mathbb {R} P^{n})is<D^{n}>\rightarrow <D^{n1}>\rightarrow \cdots \rightarrow <D^{0}>$
The boundary map alternates between 0 and 2 where it is 2 for $<D^{i}>\rightarrow <D^{i1}>$ 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 $z\in \mathbb {C} ^{n+1}$, $\alpha \in \mathbb {C} \{0\}$
$\mathbb {C} P^{n}=\{[z_{0},\cdots ,z_{n}]:z_{n}\neq 0\}\cup \{[\ldots ,0]\}=\{[z_{0},\cdots ,z_{n1},1]\}\cup \mathbb {C} P^{n1}$
=$\mathbb {C} ^{n}\cup \mathbb {C} P^{n1}=\mathbb {R} ^{2n}\cup \mathbb {C} P^{n1}=B^{2n}\cup \mathbb {C} P^{n1}$
I.e., $\mathbb {C} P^{n}=D^{2n}\cup D^{2n1}\cup \cdots$
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)