$H_n(\mathbb{R}P^4 \times S^1)$

Question

I have been trying to compute the homology of $\mathbb{R}P^4 \times S^1$ by using cellular homology. Nevertheless, I cannot see what the attaching maps are.

Answer 1

The standard cellular structure on $S^4$ is two cells each of dimensions 0, 1, 2, 3, and 4. The cells of the same dimensions are carried to one another by the antipodal map, so $\mathbb R \mathrm P^4$ has a cellular structure $e^0 \cup e^1 \cup e^2 \cup e^3 \cup e^4$. The antipodal map on $\mathbb R^{2n}$ is orientation-preserving, and the antipodal map on $\mathbb R^{2n+1}$ is orientation-reversing, so the degrees of the attaching maps of the two $e^{2n+1}$ cells add, and the degrees of the attaching maps of the $e^{2n}$ cells cancel.

It follows that $\partial e^1 \to e^0$ and $\partial e^3 \to e^2$ are of degree $0$, while $\partial e^2 \to e^1$ and $\partial e^4 \to e^3$ are of degree $2$.

Now we want to take the cross product of this cell complex with $S^1$, whose standard cell structure is $f^0 \cup f^1$, a 0-cell and a 1-cell, where the degree of $\partial f^1 \to f^0$ is zero.

The product cell structure on $\mathbb R \mathrm P^4 \times S^1$ is a union of cells $e^i \times f^j$. Now, the boundaries of these new cells are $\partial(e^i \times f^j) = (\partial e^i \times f^j) \cup (e^i \times \partial f^j)$, and these attach to the lower-dimensional cells by the maps $(\partial e^i \to e^{i-1}) \times \mathrm{id}_{f^j}$ and $\mathrm{id}_{e^i} \times (\partial f^j \to f^{j-1})$. The degrees of these maps are just the degrees respectively of $\partial e^i \to e^{i-1}$ and of $\partial f^j \to f^{j-1}$.

The associated cellular chain complex is as follows.

$C_5$ is generated by $e^4 \times f^1$.

$C_4$ is generated by $e^3 \times f^1$ and $e^4 \times f^0$.

$C_3$ is generated by $e^2 \times f^1$ and $e^3 \times f^0$.

$C_2$ is generated by $e^1 \times f^1$ and $e^2 \times f^0$.

$C_1$ is generated by $e^0 \times f^1$ and $e^1 \times f^0$.

$C_0$ is generated by $e^0 \times f^0$.

The nonzero boundaries of generators are as follows:

$\partial (e^4 \times f^1) = 2 e^3 \times f^1$,

$\partial (e^4 \times f^0) = 2 e^3 \times f^0$,

$\partial (e^2 \times f^1) = 2 e^1 \times f^1$,

$\partial (e^2 \times f^0) = 2 e^1 \times f^0$.

It follows that

$H_5 =0$,

$H_4 = (\mathbb Z/2)e^3 \times f^1$,

$H_3 = (\mathbb Z/2)e^3 \times f^0$,

$H_2 = (\mathbb Z/2)e^1 \times f^1$,

$H_1 = (\mathbb Z/2)e^1 \times f^0 \oplus \mathbb Z e^0 \times f^1$, and

$H_0 = \mathbb Z e^0 \times f^0$.

Answer 2

Both $\mathbf{R}P^4$ and $S^1$ have standard CW structures, you will find in any book a recipe to define a CW structure on the product space (e.g. in Bredon, or Hatcher), there is a standard way to do this.