Special cell decomposition for spheres with free $\mathbb{Z}/p\mathbb{Z}$-action by orthogonal transformations?

Question

Consider the unit sphere $S^d$ in $\mathbb{R}^{d+1}$ with the antipodal action $\nu \colon x\mapsto -x$. This turns $S^d$ into a free $\mathbb{Z}/2\mathbb{Z}$-space.

Construct a CW-complex structure for $S^d$ with 2 cells in each dimension (which we think of as hemispheres) as follows: start with two vertices/$0$-cells. Attach to segments to get a circle, glue in two disks to get a $2$-sphere etc.

$\nu$ induces an action on chains. Write $\mathrm{id}$ for the identity map on chains and let $\theta = \mathrm{id}+\nu$. The cell structure described above gives rise to an interesting family of chains $h_i\in C_i(S^d;\mathbb{Z}/2\mathbb{Z})$, given by one of the two hemispheres, such that $h_0$ is an elementary $0$-chain, $\theta h_d$ is the fundamental cycle of $S^d$ and $\partial h_i = \theta h_{i-1}$ for all $i\geq 1$.

Some handwavy argument tells me that this should generalize to free $\mathbb{Z}/p\mathbb{Z}$-actions on $S^d$ as follows: Assume the cyclic group $G=\mathbb{Z}/p\mathbb{Z}$ of prime order $p$ acts freely on $S^d$ by linear orthogonal transformations (so we are considering the unit sphere of a $(d+1)$-dimensional linear orthogonal representation $V$ of $G$ such that the induced action on that sphere is free). Let $\nu\colon S^d\rightarrow S^d$ be a generator for this action. Consider the two special elements $s= \mathrm{id}-\nu$ and $t=\mathrm{id}+\nu+\dots+\nu^{p-1}$ acting on chains of $S^d$. Can we find chains $h_i\in C_i(S^d;\mathbb{Z}/p\mathbb{Z})$ such that $h_0$ is an elementary $0$-chain, $t h_d$ is the fundamental cycle of $S^d$ and $\partial h_i=s h_{i-1}$ if $i$ is odd and $\partial h_i=t h_{i-1}$ if $i$ is even? Is there an analogous CW-complex structure like the hemispheres in the case of the antipodal action on $S^d$ from which we can read of the chains $h_i$? How to make this precise?

Answer 1

This is a more explicit version of Ian's answer.

From the representation theory of $\mathbb{Z}/p$, we can assume that $V=\mathbb{C}^{m+1}$ with the generator $g$ of $\mathbb{Z}/p$ acting as $g.z=(\omega_0z_0,\dotsc,\omega_mz_m)$ for some primitive $p$-th roots of unity $\omega_0,\dotsc,\omega_m$. Put $I=[0,1]$ and $W=\{re^{i\theta}:0\leq r\leq 1,\;0\leq\theta\leq 2\pi/p\}$. Then put \begin{align*} e_{2k} &= \{z\in S^{2m+1}:z_k\in I,\; z_j=0\text{ for } j>k\} \\ e_{2k+1} &= \{z\in S^{2m+1}:z_k\in W,\; z_j=0\text{ for } j>k\}. \end{align*} There is a homeomorphism $f_{2k}\colon B^{2k}=B(\mathbb{C}^k)\to e_{2k}$ given by $$ f_{2k}(z) = (z_0,\dotsc,z_{k-1},\sqrt{1-\|z\|^2},0,\dotsc,0) $$ There are continuous surjections $p_k\colon B(\mathbb{C}^k)\times[0,1]\to e_{2k+1}$ and $q_k\colon B(\mathbb{C}^k)\times[0,1]\to B(\mathbb{C}^k\oplus\mathbb{R})=B^{2k+1}$ given by \begin{align*} p_k(z,t) &= (z_0,\dotsc,z_{k-1},\sqrt{1-\|z\|^2}\,e^{2\pi it/p},0,\dotsc,0) \\ q_k(z,t) &= (z,\sqrt{1-\|z\|^2}\,(2t-1)) \end{align*} One checks that $$ p_k(z,t)=p_k(z',t') \Leftrightarrow (z=z' \wedge (t=t' \vee \|z\|=1)) \Leftrightarrow q_k(z,t)=q_k(z',t'). $$ It follows that there is a unique map $f_{2k+1}\colon B^{2k+1}\to e_{2k+1}$ with $f_{2k+1}\circ q_k=p_k$, and that this is a homeomorphism.

One can now check that the cells $\{g^ie_j:0\leq i<p,\;0\leq j\leq 2m+1\}$ give an equivariant cell structure on $S^{2m+1}$. The cellular boundary operator is $\partial(e_{2k})=\sum_ig^ie_{2k-1}$ and $\partial(e_{2k+1})=g^{u_k}e_{2k}-e_{2k}$, where $u_k$ is determined by $\omega_k^{u_k}=e^{2\pi i/p}$. In particular, in the basic case where $\omega_k=e^{2\pi i/p}$ for all $k$ we have $\partial(e_{2k+1})=g.e_{2k}-e_{2k}$.

Most of this is in Section V.5 of the 1962 book Cohomology operations by Steenrod and Epstein,

Answer 2

Yes, there is such a decomposition. The spheres on which $\mathbb{Z}/p\mathbb{Z}$ acts freely are necessarily odd dimensional for $p>2$. View each $S^{2m-1}$ as the unit sphere in $\mathbb{C}^m$, with $\mathbb{C}^{m-1}$ viewed as a subspace of $\mathbb{C}^m$ and invariant under the action of $\mathbb{Z}/p\mathbb{Z}$. In the case $m=1$, you have the CW-structure on $S^1$ coming from the boundary of the $p$-gon. Inside $S^3$, this copy of $S^1$ bounds a disc. The $p$-distinct images of this disc under the action of $\mathbb{Z}/p\mathbb{Z}$ are your 2-cells, and their complement is $p$ disjoint open 3-balls. The same sort of argument works to get from $S^{2m-1}$ to $S^{2m+1}$: you have a CW-structure on $S^{2m-1}$ with exactly $p$ cells of each (relevant) dimension. Inside $S^{2m+1}$ you pick a $2m$-disc that bounds your $S^{2m-1}$ and the images of this disc under the elements of $\mathbb{Z}/p\mathbb{Z}$ form the $2m$-cells, while the closures of the complementary regions form the $(2m+1)$-cells.