Compute the homology groups \(H_i(\RP^n/\RP^m;\ZZ)\) of the quotient space \(\RP^n/\RP^m\), for \(m<n\). Use cellular homology, using the standard CW structure on \(\RP^n\) with \(\RP^m\) as its \(m\)-skeleton.
Solution
1. \(\RP^n\) has a CW structure with one cell \(e^k\) in each dimension \(0 \le k \le n\), and \(\RP^m\) is its \(m\)-skeleton.
This is the standard CW structure: \(\RP^k\) is obtained from \(\RP^{k-1}\) by attaching one \(k\)-cell along the double cover \(S^{k-1}\to\RP^{k-1}\).
2. The quotient \(X = \RP^n/\RP^m\) has one \(0\)-cell (the collapsed \(\RP^m\)) and one cell in each dimension \(m+1 \le k \le n\).
Collapsing the subcomplex \(\RP^m\) to a point replaces its cells by one \(0\)-cell and keeps the cells \(e^k\) with \(k>m\).
3. In positive degrees the cellular chain groups of \(X\) are\[ C_k(X) \cong \begin{cases} \ZZ,&m+1\le k\le n,\\ 0,&\text{otherwise}, \end{cases} \]and \(C_0(X)\cong\ZZ\).
The cellular chain groups are free on the cells listed in 2.
4. \(d_{m+1} = 0\).
The boundary of the \((m+1)\)-cell is attached to the \(m\)-skeleton \(\RP^m\), which is collapsed to a point, so the attaching map is constant.
5. For \(m+2 \le k \le n\), \(d_k = 1 + (-1)^k\) (i.e. \(d_k = 2\) for \(k\) even and \(d_k = 0\) for \(k\) odd).
The cellular boundary of the \(k\)-cell of \(\RP^n\) has degree \(1 + (-1)^k\), and these cells are unaffected by the collapse.
6. Hence\[ H_0(X)=\ZZ, \qquad H_i(X)=0\quad(0<i\le m). \]
The quotient is connected, so \(H_0(X)\cong\ZZ\). There are no positive-dimensional cells below degree \(m+1\), hence \(H_i(X)=0\) for \(0<i\le m\).
7. The bottom positive-degree group is\[ H_{m+1}(X) \cong \begin{cases} \ZZ,&n=m+1,\\ \ZZ,&n\ge m+2\text{ and }m+1\text{ is even},\\ \ZZ/2,&n\ge m+2\text{ and }m+1\text{ is odd}. \end{cases} \]
By 4, \[ H_{m+1}(X) = \ZZ/\operatorname{im}d_{m+2}. \] If \(n=m+1\), there is no \(d_{m+2}\), so this group is \(\ZZ\). If \(n\ge m+2\), then 5 gives \[ d_{m+2} = \begin{cases} 2,&m+2\text{ is even},\\ 0,&m+2\text{ is odd}. \end{cases} \] Since \(m+2\) has the opposite parity from \(m+1\), the displayed cases follow.
8. For \(m+2 \le k \le n-1\), \(H_k(X) = \ZZ/2\) if \(k\) is odd and \(H_k(X) = 0\) if \(k\) is even.
If \(k\) is even then \(d_k = 2\) is injective so \(\ker d_k = 0\); if \(k\) is odd then \(d_k = 0\) so \(\ker d_k = \ZZ\), and \(d_{k+1} = 2\) (since \(k+1\) is even) so \(\operatorname{im} d_{k+1} = 2\ZZ\), giving \(H_k = \ZZ/2\).
9. If \(n\ge m+2\), then\[ H_n(X) \cong \begin{cases} \ZZ,&n\text{ odd},\\ 0,&n\text{ even}. \end{cases} \]
There is no \(d_{n+1}\), so \(H_n=\ker d_n\). Under the hypothesis \(n\ge m+2\), 5 applies: \(d_n=0\) for odd \(n\) and \(d_n=2\) for even \(n\).
10. Thus, if \(n=m+1\), then\[ \boxed{ H_i(X;\ZZ) \cong \begin{cases} \ZZ,&i=0,n,\\ 0,&\text{otherwise}. \end{cases}} \]If \(n\ge m+2\), then\[ \boxed{ H_i(X;\ZZ) \cong \begin{cases} \ZZ,&i=0,\\ \ZZ,&i=m+1\text{ and }i\text{ is even},\\ \ZZ,&i=n\text{ and }n\text{ is odd},\\ \ZZ/2,&m+1\le i<n\text{ and }i\text{ is odd},\\ 0,&\text{otherwise}. \end{cases}} \]
Combine <1>6–<1>9. In the case \(n=m+1\), 4 gives \(d_n=0\) even when \(n\) is even; equivalently, collapsing the entire boundary skeleton of the sole remaining \(n\)-cell gives \[ \RP^n/\RP^{n-1}\cong D^n/S^{n-1}\cong S^n. \]