# 0708-1300/Class notes for Tuesday, March 25

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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:

${\displaystyle C_{n}^{CW}(K):=}$

and the boundary maps via ${\displaystyle \partial :C_{n}^{CW}\rightarrow C_{n-1}^{CW}}$ by ${\displaystyle \partial \sigma =\sum _{\tau \in K_{n-1}}[\tau :\sigma ]\tau }$

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

Let's make this precise.

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

Example:

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

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

Theorem

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

Examples:

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

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

Hence ${\displaystyle C_{n}^{CW}=<\sigma >,C_{0}^{CW}=<\infty >}$ and all the rest are zero. Hence, ${\displaystyle H_{p}(S^{n})=\mathbb {Z} }$ for p = 0 or n and is zero otherwise.

2) Consider the torus thought of as a square with the usual identifications and ${\displaystyle \sigma }$ is the interior. Hence, ${\displaystyle C_{0}^{CW}=\{p\},C_{1}^{CW}}$ is generated by the figure 8 with one loop labeled a and the other labeled b, and ${\displaystyle C_{2}^{CW}}$ is generated by the entire torus.

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

Now, ${\displaystyle \partial a=[p:a]p}$

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

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

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

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

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

I.e., ${\displaystyle <\sigma >\mapsto \mapsto ^{0}

}$

${\displaystyle \partial \sigma =[a:\sigma ]a+b:\sigma b=0_{a}+2b}$

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

The kernal of ${\displaystyle \rightarrow

}$

is everything, so the homology is ${\displaystyle H_{1}(K)=/2b=0\cong \mathbb {Z} \oplus \mathbb {Z} /2}$, ${\displaystyle H_{0}(K)=\mathbb {Z} }$ and ${\displaystyle H_{2}(K)=0}$.

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

So, get ${\displaystyle <\sigma >\rightarrow ^{0}\rightarrow ^{0}

}$

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

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

So, ${\displaystyle C_{*}^{CW}(\mathbb {R} P^{n})is\rightarrow \rightarrow \cdots \rightarrow }$

The boundary map alternates between 0 and 2 where it is 2 for ${\displaystyle \rightarrow }$ if i is even and 0 if it is odd.

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

### Second Hour

${\displaystyle \mathbb {C} P^{n}:\{[z_{0},\cdots ,z_{n}]:z_{i}\in \mathbb {C} \ not\ all\ zero\}=\mathbb {C} ^{n+1}/z\sim \alpha z}$ for ${\displaystyle z\in \mathbb {C} ^{n+1}}$, ${\displaystyle \alpha \in \mathbb {C} -\{0\}}$

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

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

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