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.

  • 3
    $\begingroup$ There is a CW structure compatible with the antipodal action on the sphere, use it. $\endgroup$ Commented Dec 9, 2022 at 15:50
  • 1
    $\begingroup$ 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. $\endgroup$
    – Thorgott
    Commented Dec 9, 2022 at 17:26
  • 1
    $\begingroup$ 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. $\endgroup$
    – Thorgott
    Commented Dec 9, 2022 at 17:26
  • $\begingroup$ 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!). $\endgroup$
    – Ayaka
    Commented Dec 9, 2022 at 18:24
  • $\begingroup$ So, I want more hints..... $\endgroup$
    – Ayaka
    Commented Dec 9, 2022 at 18:25

1 Answer 1


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 $0<<m<<n$ 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.

  • $\begingroup$ I'm grateful to your great contribution! I will try to follow the computation. $\endgroup$
    – Ayaka
    Commented Jan 15, 2023 at 17:13
  • $\begingroup$ 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... $\endgroup$
    – Ayaka
    Commented Mar 10, 2023 at 5:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .