# homology groups of a product-quotient of spheres

In many textbooks the integral homology groups of an $$n$$-dimensional sphere $$S^n$$ and of its quotient, the real projective space $$\mathbb{R}\mathbb{P}^n=S^n/\{\pm 1\}$$ are explained. As further examples, I want to work out the integral homology groups of the product-quotient $$X_{m,n}:= (S^m\times S^n)/\{\pm 1\}$$, which is doubly covered by $$S^m\times S^n$$ and doubly covers $$\mathbb{R}\mathbb{P}^m \times \mathbb{R}\mathbb{P}^n$$. But, for general $$m$$ and $$n$$, I encountered difficulty in computing especially the groups of intermediate degrees (see below). I have tried some cases:

1. If $$n=1$$ and $$m\geq 2$$, then the projection $$X_{m,1}\rightarrow S^1/\{\pm 1\}$$ exhibits $$X_{m,1}$$ as a sphere bundle over a circle whose geometric monodromy acts a fiber by $$-1$$. By the Mayer-Vietoris sequence, I could compute the homology groups as $$\text{if m is even,}\qquad H_i(X_{m,1},\mathbb{Z})= \begin{cases} &\mathbb{Z}\quad (i=0,1)\\ &\mathbb{Z}/2 \ (i=m)\\ &0\qquad \text{otherwise.} \end{cases}$$ $$\text{if m is odd and >1,}\quad H_i(X_{m,1},\mathbb{Z})= \begin{cases} &\mathbb{Z}\quad (i=0,1,m,m+1)\\ &0\qquad \text{otherwise.} \end{cases}$$ This case looks complete.

2. For general $$m,n$$, using Kuenneth formula and [Hatcher's algebraic topology, Proposition 3G.1] I see the Betti numbers as the number of $$(-1)$$-invariant cohomology classes on the product. In particular, I see the rational homology groups.

3. For $$m=n=2$$: By 2. above we see $$H_0=H_4=\mathbb{Z}$$ and $$H_1=\mathbb{Z}/2$$ since its fundamental group is of order 2. The remaining degree $$2,3$$ can be obtained using Poincare duality and the universal coefficient formula as $$H_2= \mathbb{Z}/2$$ and $$H_3=0$$. This case can be understood.

4. For $$m=2,n=4$$: The same method as above computes $$H_0,H_1,H_4,H_5,H_6$$ but the remaining two groups (both torsion) seems not to be reachable by my ad hoc method.

Question: How can we compute the full integral homology of $$X_{m,n}$$ ?

Maybe a good CW-structure should be introduced, but I couldn't figure out. Any help is appreciated. Also, please point and correct my computations above if there are missing stuffs. Thank you in advance.

• There is a CW structure compatible with the antipodal action on the sphere, use it. Commented Dec 9, 2022 at 15:50
• For this specific case, writing down the CW-structure (even a $\Delta$-complex structure) and then computing cellular/simplicial homology is probably the most accessible. However, there are general techniques to study the homology of quotients by (free) involutions. a) If $G$ is a finite group acting freely and continuously on a Hausdorff space $X$ and $R$ is a ring s.t. $|G|\in R^{\times}$, then the canonical projection induces isomorphisms $H_i(X;R)_G\rightarrow H_i(X/G;R)$, where the notation means coinvariants. Commented Dec 9, 2022 at 17:26
• b) if $G=\mathbb{Z}/2\mathbb{Z}$ specifically, there is a LES $\dotsc\rightarrow H_k(X/G;\mathbb{Z}/2\mathbb{Z})\rightarrow H_k(X;\mathbb{Z}/2\mathbb{Z})\rightarrow H_k(X/G;\mathbb{Z}/2\mathbb{Z})\rightarrow\dotsc$, where the first map is given by taking a simplex to the sum of its two lifts and the second map is induced by the canonical projection. In total, this can allow you to compute $H_k(X/G;R)$ for $R=\mathbb{F}_p$, $p$ prime and $R=\mathbb{Q}$. Then, using UCT/Bockstein, you can try reconstructing $H_k(X;\mathbb{Z})$ from that. Commented Dec 9, 2022 at 17:26
• Thanks for the comments. I once tried CW-decomposition, but I couldn't compute the boundary (differential) of the complex correctly ($\partial \circ \partial$ didn't vanish!). Commented Dec 9, 2022 at 18:24
• So, I want more hints..... Commented Dec 9, 2022 at 18:25

Let $$t$$ denote the antipodal map on $$S^n$$. There is a simple CW structure for $$S^n$$ compatible with $$t$$, which we define inductively by $$S^0=\{*,*t\}$$, for a point $$*$$, and $$S^n=S^{n-1}\cup_\phi e_n\cup_\psi e_nt,$$ where $$e_n$$ is an $$n$$-cell and $$\phi,\psi$$ are identifications of the boundaries of $$e_n$$ and $$e_nt$$ with $$S^{n-1}$$.

Intuitively, we regard each sphere as the equator of the next, and glue on a couple of hemispheres.

The boundary maps $$d$$ are given by $$de_r=e_{r-1}(1+(-1)^rt),\qquad de_0=0.$$

To understand the sign $$(-1)^r$$ here, note if $$r$$ is odd then $$t$$ preserves the orientation of $$S^{r-2}$$, so the boundary of $$e_r$$ must be the difference $$e_{r-1}-e_{r-1}t$$ in order to be closed. Conversely if $$r$$ is even, then $$t$$ reverses the orientation of $$S^{r-2}$$, so the boundary of $$e_r$$ must be the sum $$e_{r-1}+e_{r-1}t$$ in order to be closed.

We may take the tensor product of the resulting chain complexes for $$S^n$$ and $$S^m$$, to obtain a chain complex for $$S^n\times S^m$$.

Here the degree $$r$$ chain group is generated by $$\begin{eqnarray*}e_a\otimes e_{r-a},\\ e_at\otimes e_{r-a},\\ e_a\otimes e_{r-a}t,\\ e_at\otimes e_{r-a}t.\end{eqnarray*}$$ for $$a=0,\cdots, r|a\leq n, r-a\leq m$$. The differential is given by $$\begin{eqnarray*}d(e_i\otimes e_j)&=&e_{i-1} \otimes e_j\\ + &(-1)^i& e_{i-1}t \otimes e_j\\+&(-1)^i& e_{i} \otimes e_{j-1}\\&(-1)^{i+j}& e_{i} \otimes e_{j-1}t. \end{eqnarray*}$$

That is we use Leibniz's rule for tensor products:$$d(a\otimes b)=da\otimes b+ (-1)^{{\rm deg}(a)} a\otimes db.$$

Finally, we obtain a chain complex for $$X=(S^n\times S^m)/\langle t\rangle,$$ by taking our chain complex and identifying $$e_i\otimes e_j\sim e_it\otimes e_jt,\qquad e_i\otimes e_jt\sim e_it\otimes e_j.$$

We may write the boundary maps $$d$$ as matrices, with respect to the basis $$\ldots e_a\otimes e_b, e_at\otimes e_b, e_{a+1}\otimes e_{b-1},\ldots$$.

The homology groups of $$X$$ may then be computed as an exercise in linear algebra. For example, if $$0< and $$n,m$$ both odd I get that the non-zero homology groups are:

$$\begin{eqnarray*} H_0(X)\cong H_{n+m}(X)&\cong& \mathbb{Z},\\ H_m(X)\cong H_n(X)&\cong&\mathbb{Z},\\ H_{2i-1}(X)\cong H_{n+m-2i}&\cong& \mathbb{Z}/2\mathbb{Z}, \qquad{\rm for} \,\,i=1,\cdots,(m-1)/2.\end{eqnarray*}$$

I suggest that you do this calculation yourself, as it is quite fiddly and I may have made an error. However the above answer has the symmetries one would expect. If you require help, I can give you the matrices.

• I'm grateful to your great contribution! I will try to follow the computation. Commented Jan 15, 2023 at 17:13
• I found the general answer for arbitrary $m,n$ will be too complicated to describe, at present. Nevertheless, I think I understood the method, thanks again. I will retry in future, hopefully... Commented Mar 10, 2023 at 5:59