Representatives of generators for the homology group of the complex projective space

Question

1. How can I compute the (singular) homology groups of $\Bbb CP^n$ using the long exact sequence of the pair? I can't see what pair do I have to use.

2. Since $\Bbb CP^1$ is homeomorphic to $S^2$, I know that $H_k(\Bbb CP^1)$ is $\Bbb Z$ for $k=0,2$, and is $0$ otherwise. Also, the $H_0(\Bbb CP^1)=\Bbb Z [pt]$, and $H_2(\Bbb CP^1)=\Bbb Z[\Bbb CP^1]$, where $\Bbb CP^1=S^2$ is regarded as a finite $\Delta$-complex (with two $2$-simplices identified by their boundary). More precisely, if we write $S^2=\Delta_1^2\cup\Delta_2^2$ (with $\Delta_1^2$, $\Delta_2^2$ the upper, lower hemispheres, respectively), then $\Delta_1^2-\Delta_2^2$ represents a generator of $H_2(\Bbb CP^1)$.
   As in above, I want to find geometric representatives of generators for the homology groups of $\Bbb CP^n$ for all $n$. Is there a way to do this?

Answer 1

As pointed out in the comment, consider the long exact sequence of the pair $(\mathbb{C}P^n, \mathbb{C}P^{n-1})$ (I'm fleshing the comment out into an answer, mostly for my own practice...). The long exact sequence is

$$\cdots \to H_{k+1}(\mathbb{C}P^n, \mathbb{C}P^{n-1}) \to H_k(\mathbb{C}P^{n-1}) \to H_k(\mathbb{C}P^n) \to H_k(\mathbb{C}P^n, \mathbb{C}P^{n-1}) \to H_{k-1}(\mathbb{C}P^{n-1}) \to \cdots$$

Since the pair is a CW pair, by excision $H_k(\mathbb{C}P^n, \mathbb{C}P^{n-1}) \cong \tilde{H}_k(\mathbb{C}P^n/\mathbb{C}P^{n-1}) \cong \tilde{H}_k(S^{2n})$ because $\mathbb{C}P^n$ is obtained from $\mathbb{C}P^{n-1}$ by attaching a $2n$-cell, so the quotient is homeomorphic to $S^{2n}$.

So now our long exact sequence is

$$\cdots \to \tilde{H}_{k+1}(S^{2n}) \to H_k(\mathbb{C}P^{n-1}) \to H_k(\mathbb{C}P^n) \to \tilde{H}_k(S^{2n}) \to H_{k-1}(\mathbb{C}P^{n-1}) \to \cdots$$

and therefore for $k < 2n-1$ and $k > 2n$, we have $H_k(\mathbb{C}P^n) \cong H_k(\mathbb{C}P^{n-1})$ since $\tilde{H}_k(S^{2n}) \cong 0$ unless $k=2n$ so the exact sequence has zeros flanking $H_k(\mathbb{C}P^{n-1}) \to H_k(\mathbb{C}P^n)$.

Now, induct on $n$, where the claim is that $H_k(\mathbb{C}P^n) \cong \mathbb{Z}$ for $k$ even with $0 \leq k \leq 2n$, and zero otherwise. This is certainly true when $n=1$ since $\mathbb{C}P^1 \cong S^2$.

For the inductive step, suppose we know that $H_k(\mathbb{C}P^{n-1}) \cong \mathbb{Z}$ for even $k$ with $0 \leq k \leq 2n-2$ and zero otherwise. Then by above, we know the homology of $\mathbb{C}P^n$ in degrees $k < 2n-1$ and $k > 2n$.

Go back to the long exact sequence for each of these. For $k= 2n-1$, the inductive hypothesis gives $H_{2n-1}(\mathbb{C}P^{n-1}) \cong 0$ and the exact sequence forces $H_{2n-1}(\mathbb{C}P^n) \cong 0$ since this group is flanked by zeros. For $k=2n$, the portion of the long exact sequence looks like

$$H_{2n}(\mathbb{C}P^{n-1}) \to H_{2n}(\mathbb{C}P^n) \to \tilde{H}_{2n}(S^{2n}) \to H_{2n-1}(\mathbb{C}P^{n-1})$$

but the inductive hypothesis tells us that the two groups on the sides are zeros and we know $\tilde{H}_{2n}(S^{2n}) \cong \mathbb{Z}$, so $H_{2n}(\mathbb{C}P^n) \cong \mathbb{Z}$.

Thus, for all $n$, $H_k(\mathbb{C}P^n) \cong \mathbb{Z}$ for even $k$ with $0 \leq k \leq 2n$ and zero otherwise.