Cellular homology of the real projective space $\mathbb R P^n$

Question

I've been able to calculate the cellular homology of $\mathbb R P^2$ but I'm struggling to do the same for higher dimensions. My problem is that I don't exactly see how one get to the result $d_i: C_{i}(\mathbb R P^n) \to C_{i-1}(\mathbb R P^n)$ has degree $1+(-1)^i$. I looked this up in Hatcher but our lecture followed Bredon where in chapter 14 they explain it but I do not understand it. In Hatcher they used the local degree, which I know, but I don't understand how the computation works.

I'm interested in seeing how one does the step by step calculation with the local degree as I feel I'm completely lost here.

Edit:

To be more precise: $\mathbb R P^n$ can be identified by $B^n / \tilde{}$ where we identify antipodal points on the boundary. We can also view the real projective space as $S^n/ \tilde{}_{antipodal}$. With these description one sees that we can construct $\mathbb R P^n$ as a CW complex by taking one $\sigma^i$ i-cell from dimension $0$ to $n$. For the attaching maps we do the following: $$ f_{\partial \sigma ^1}: \partial B^1_{\sigma^1} \approx \{0,1\} \to K^{(0)}=\{\sigma ^0\}$$ and for the dimension $1 \leq i \leq n$:

$$f_{\partial \sigma ^i}: \partial B_{\sigma^i}^i \approx S^{i-1} \to K^{(i-1)}$$

where we define this map by antipodal identification. On the chain complex level we get $C_i(\mathbb R P^n)=\mathbb Z \sigma ^i \cong \mathbb Z$. So we have the sequence:

$$0 \stackrel{0}\to \mathbb Z \sigma ^n \stackrel{d_n}\to \mathbb Z \sigma ^{n-1} \stackrel{d_{n-1}}\to \mathbb Z \sigma ^{n-2} \stackrel{d_{n-2}}\to ...\stackrel{d_1}\to \mathbb Z \sigma ^1 \stackrel{0}\to 0$$

We can then compute the cellular homology $H_*(\mathbb R P ^n)=H_*(C_{\cdot}(\mathbb R P^n,d))$ by computing $ker(d_i)/im(d_{i+1})$ where the boundary/differential formula of $d_i$ is:

$$ d_i: \mathbb Z \sigma^i: \to \mathbb Z \sigma^{i-1}$$ defined uniquely by $d_i(\sigma^i)=[\sigma^{i-1}:\sigma^{i}]\sigma^{i-1} \in \mathbb Z \sigma^{i-1}$ where the square brackets denote the incidence number, i.e. $$[\sigma^{i-1}:\sigma^{i}]=deg(p_{\sigma^{i-1}} \circ f_{\partial \sigma^i})$$. And here comes the point where I'm stuck. How do I exactly compute the $d_i$'s?

Answer 1

I myself am struggling too with this but I think I understand it atleast intuitively:

We are looking for the incidence number: $deg(p_{\sigma_{i-1}}\circ f_{\partial \sigma_i})$

Let the attachment map of a point of $\partial B_{\sigma_i}^{i}$ be the map that identifies the point with the antipodal point and maps both on the same point y on $K^{(i-1)}$. Now we go through the process with taking the quotient and projection and get $S^{(i-1)}$. Now each point on this $S^{(i-1)}$ has exactly two points in the preimage of the map $p_{\sigma_{i-1}}\circ f_{\partial \sigma_i}$.

$p_{\sigma_{i-1}}\circ f_{\partial \sigma_i}$ is a local homeomorphism on the neighborhoods of the two points. One of those points has the local degree 1, from the identity (or a map homotopic to the identity) and the other one has local degree $(-1)^i$ (from the antipodal map), and $deg(p_{\sigma_{i-1}}\circ f_{\partial \sigma_i})$ is equal to the sum of the local degrees.

Here it is explained better: http://www.math.wisc.edu/~maxim/Topnotes2.pdf on pages 4-5

But yeah, I am still thinking about it so pls take it with a grain of salt.