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

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?

• Are you asking how to construct a $CW$ complex for $\mathbb{R} P^n$? Or do you already know the $CW$ structure and are asking how to compute cellular homology from there? – Yuugi Feb 4 '16 at 17:33
• I know how to construct the CW complex and I also know how to compute the cellular homology except how to exactly derive the incidence number of the differential. I know that it's done via local degree formula but I don't understand how this single step is done. – noctusraid Feb 4 '16 at 17:36
• Do you know how to set up the standard CW structure on $\mathbb{R} P^n$ having one cell $c_i$ in each dimension $i$? Do you know how to set up characteristic maps for each $c_i$? Explicit characteristic maps for $c_i$ and $c_{i-1}$ are a prerequisite for computation of $d_i$. If you know these characteristic maps, then describing them explicitly in your question would be helpful to answerers, who otherwise have to supply the descriptions themselves. – Lee Mosher Feb 5 '16 at 13:39
• ok now I've made an edit, let me know if there is still something missing/unclear. – noctusraid Feb 5 '16 at 14:11

1 Answer

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.

• yes that's what I figured out today too while I was reviewing it. What helped me understand it is to really think of the map the attaching map $f_{\partial \sigma_i}: S^{i-1} \to K^{i-1} \approx \mathbb R P^{i-1}$ being the the two sheeted covering map. – noctusraid Feb 6 '16 at 22:01
• Ok after reading your notes and seeing the picture which I figured out too (the one with the sphere wedge thing with the identity on top and antipodal map on the bottom), I'm pretty sure I got it right, thanks! – noctusraid Feb 6 '16 at 22:04