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

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?

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 $$\{g^ie_j:0\leq i give an equivariant cell structure on $$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,

• Wow! This is fantastic! Did not expect such an explicit description!
– pfw
May 26, 2022 at 12:45

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.

• By the way, each of the 3-ball inside $S^3$ looks a bit like a lense and thus it makes sense to pick the term "lens space" for the quotients. May 26, 2022 at 12:11
• Thank you, Ian! :)
– pfw
May 26, 2022 at 12:44