Compute \(H_i(\mathbb{RP}^n/\mathbb{RP}^m)\) for \(m < n\) by cellular homology, using the standard CW structure on \(\mathbb{RP}^n\) with \(\mathbb{RP}^m\) as its \(m\) skeleton.
Solution
Let \[ Y=\mathbb{RP}^n/\mathbb{RP}^m, \qquad m<n. \] The quotient CW structure has one \(0\)-cell and one cell in each dimension \[ m+1,m+2,\dots,n. \] Thus its reduced cellular chain groups are \[ \widetilde C_i(Y)\cong \begin{cases} \mathbb Z,&m<i\le n,\\ 0,&\text{otherwise}. \end{cases} \] For \(i>m+1\), the cellular boundary is the usual projective-space boundary \[ d_i= \begin{cases} 2,&i\text{ even},\\ 0,&i\text{ odd}, \end{cases} \] while \(d_{m+1}=0\) because its target chain group has been collapsed away.
1. If \(n=m+1\), then\[ \widetilde H_n(Y)\cong\mathbb Z \]and all other reduced homology groups vanish.
There is a single nonzero reduced cellular chain group, in degree \(n\), and both adjacent boundary maps are zero.
Assume henceforth that \(n\ge m+2\).
2. In the bottom degree \(m+1\),\[ \widetilde H_{m+1}(Y)\cong \begin{cases} \mathbb Z,&m+1\text{ even},\\ \mathbb Z_2,&m+1\text{ odd}. \end{cases} \]
The outgoing differential is zero. The incoming differential is \(d_{m+2}\), which is zero when \(m+2\) is odd and multiplication by \(2\) when \(m+2\) is even. These are exactly the two displayed cases.
3. For \(m+1<i<n\),\[ \widetilde H_i(Y)\cong \begin{cases} \mathbb Z_2,&i\text{ odd},\\ 0,&i\text{ even}. \end{cases} \]
If \(i\) is even, \(d_i=2\) is injective, so there are no cycles. If \(i\) is odd, \(d_i=0\) and \(d_{i+1}=2\), giving \[ H_i\cong\mathbb Z/2\mathbb Z. \]
4. In the top degree \(n\),\[ \widetilde H_n(Y)\cong \begin{cases} \mathbb Z,&n\text{ odd},\\ 0,&n\text{ even}, \end{cases} \qquad(n>m+1). \]
There is no incoming boundary from degree \(n+1\). Thus top homology is \(\ker d_n\), which is \(\mathbb Z\) for odd \(n\) and \(0\) for even \(n\).
Finally \(Y\) is connected, so \[ H_0(Y)\cong\mathbb Z. \] Together <1>1–<1>4 give all homology groups of \(\mathbb{RP}^n/\mathbb{RP}^m\).