orbit space of $\mathbb{Z}_p$ action over complex projective space by permuting the homogeneous coordinates

Question

$Z_p$:=cyclic group of order $p$.

I want to understnd $H_\ast(\mathbb{C} P^{n}/Z_{n+1};Z)$ with $(n+1)$ being a prime number,and the action is given by permuting the homogeneous coordinates.

For example,let's take $n=4$.The fixed point set $(\mathbb{C} P^{4})^{Z_5}$is 5 points.The action is free out of the the fixed point set (since $Z_5$ has no nontrivial subgroups)

My strategy is:

1.The quotient map restrictied to the free part,which is $\mathbb{C} P^{4}-(\mathbb{C} P^{4})^{Z_5}\to \mathbb{C} P^{4}/Z_5-(\mathbb{C} P^{4})^{Z_5}$, is a covering ,so we could compute $H_\ast(\mathbb{C} P^{4}/Z_5-(\mathbb{C} P^{4})^{Z_5})$ using spectral sequence $$H_i(BZ_5,H_j(\mathbb{C} P^{n}-(\mathbb{C} P^{n})^{Z_5}))\Rightarrow H_{i+j}(\mathbb{C} P^{4}/Z_5-(\mathbb{C} P^{4})^{Z_5})$$

2.$\mathbb{C} P^{4}=(\mathbb{C} P^{4}/Z_5-(\mathbb{C} P^{4})^{Z_5})\cup$ 5 copies of cone$(L^7$) ($L^7=S^7/Z_5$ is the lens space) .Use MV sequence 4 times to obtain the homology of the orbit space.

In step 1,there is an extension issue in the spectral sequence.In step 2,similar issue arises when i am in the situation $0\to A\to ?\to B\to 0$.

Is it possible to overcome this extension issue using some geometric observation?What's that.

and

Have people already computed the homology of orbit space of linear cyclic group action over $\mathbb{C} P^{n}$?

where the action is given by $$g\cdot[z_0,\cdots,z_n]:=[\zeta^{a_0}z_0,\cdots,\zeta^{a_n}z_n]$$ where $g$ is the generator of $Z_p$ and $\zeta$ is $p$-th root of unity,and $a_i$ s are some given nonnegative integers.For my problem above,I believe the action is equivalent to $$g\cdot[z_0,z_1,z_2,z_3,z_4]:=[z_0,\zeta^{}z_1,\zeta^{2}z_2,\zeta^{3}z_3,\zeta^{4}z_4]$$ by noticing that the regular representation of $Z_5$ over $\mathbb{C}^5$ is direct sum of the irreducible representations.

Answer 1

I checked some more literature on the computation of homology of orbit space of group action over manifolds,here is my solution:

1.The orbit space $\mathbb{C}P^4/Z_5$ is simply connected,hence $H_1(\mathbb{C}P^4/Z_5;Z)=0$.This is due to a theorem of Armstrong

2.Set $X=\mathbb{C}P^4/Z_5$ and $X_0=(\mathbb{C}P^4)^{Z_5}$,by considering the long exact sequence of homology groups of the pair $(X,X_0)$ ,we know $H_i(X)\cong H_i(X,X_0)$ for $i\geq 2$.But $$H_i(X,X_0)\cong H^{BM}_i(X-X_0)\cong H^{8-i}(X-X_0)$$

where BM means Borel Moore homology.( see Borel-Moore homology in Wikipedia for the 2 isomorphisms above) so we reduced the problem to computing the cohomology and hence the homology of $X-X_0$ by the universal coefficient theorem $$H_i(X)\cong Hom(H_{8-i}(X-X_0),Z)\oplus Ext(H_{7-i}(X-X_0),Z)\quad\text{ for } 2\leq i\leq 8$$ Now use spectral sequence $$H_i(Z_5,H_j(\mathbb{C}P^4-X_0))\Rightarrow H_{i+j}(X-X_0)$$ the $E_{\infty}$ page looks like

To deal with the extension issue for $H_i(X-X_0)$ for $i=3$ or $i=5$,consider the composition of transfer homomorphism and quotient map $$H_i(X-X_0)\to H_i(\mathbb{C} P^4-X_0)\to H_i(X-X_0)$$ It is easy to compute $H_\ast(\mathbb{C} P^4-X_0)$ by MV sequence,which turns out to be a free abelian group,hence the composition should be a 0 map.But by classical theory of transformation groups,this is a $\times 5$ map.this means all possible torsion in $H_\ast(X-X_0)$ should be of order 5.

To deal with the extension issue for $H_7(X-X_0)$,we could actually use identification $$H_0(X,X_0)\cong H^8(X-X_0)\cong Hom(H_8(X-X_0),Z)\oplus Ext(H_7(X-X_0),Z) $$ We know $H_0(X,X_0)\cong 0$,hence there is no torsion elements in $H_7(X-X_0)$.

we conclude that $$H_i(X-X_0)\cong\begin{cases} 0& i=8\\ Z^4& i=7\\ Z& i=6\\ (Z_5)^3& i=5\\ Z& i=4\\ (Z_5)^2& i=3\\ Z& i=2\\ Z_5&i=1\\ Z& i=0 \\ \end{cases}$$

and $$H_i(X)\cong\begin{cases} Z& i=8\\ 0& i=7\\ Z\oplus Z_5& i=6\\ 0& i=5\\ Z\oplus(Z_5)^2& i=4\\ 0& i=3\\ Z\oplus(Z_5)^3& i=2\\ 0&i=1\\ Z& i=0 \\ \end{cases}$$

Answer 2

You can use $Z/5$ - Bredon cohomology of $CP^n$ with given action .Where the coefficient system is constant coeff. system.Then this Bredon cohomology gives the cohomology of the orbit space

Answer 3

The question is not very specific about the coefficients that are being considered. I assume you are mostly interested in coefficients at the prime $p$, or integral coefficients, and actually I can not say much about that. Nevertheless, I will state the obvious here, and give the homology of the orbit spaces $\mathbb{CP}^{p-1}/(\mathbb{Z}/p\mathbb{Z})$ away from the prime $p$. The argument is simple and does not require any geometry or spectral sequence computations. Statements at the prime $p$ will more likely require more careful study of the geometry of the orbit space.

Claim: for any prime $p$ and any field $\mathbb{K}$ of characteristic $\neq p$, there is an isomorphism $$H_\bullet(\mathbb{CP}^{p-1}/(\mathbb{Z}/p\mathbb{Z}),\mathbb{K})\cong H_\bullet(\mathbb{CP}^{p-1},\mathbb{K}).$$

First, $p=2$ is a special case. The orbit space of the $\mathbb{Z}/2$-action on $\mathbb{CP}^1$ is the closed unit disc, with the boundary $S^1$ identified with the complex conjugation involution. This is homeomorphic to $\mathbb{CP}^1$, and the claim follows.

All the other cases are proved uniformly as follows: since we only have a finite group action and the characteristic of the coefficient field $\mathbb{K}$ does not divide the group order, we can use transfer in homology to show that the homology of the orbit space is given by the coinvariants $H_\bullet(\mathbb{CP}^{p-1},\mathbb{K})_{\mathbb{Z}/p\mathbb{Z}}$. So we only need to determine the action of $\mathbb{Z}/p\mathbb{Z}$ on the homology of $\mathbb{CP}^{p-1}$, and the claim follows if this action is trivial. Note that the homology of $\mathbb{CP}^{p-1}$ with $\mathbb{Z}$-coefficients is torsion-free, hence $H_\bullet(\mathbb{CP}^{p-1},\mathbb{K})\cong H_\bullet(\mathbb{CP}^{p-1},\mathbb{Z})\otimes_{\mathbb{Z}}\mathbb{K}$. In particular, we can determine the action of $\mathbb{Z}/p\mathbb{Z}$ on integral homology to get our answer. For any $i$, $H_i(\mathbb{CP}^{p-1},\mathbb{Z})$ is a $0$ or $1$-dimensional representation of $\mathbb{Z}/p\mathbb{Z}$. But $\mathbb{Z}$ does not contain the $p$-th roots of unity if $p$ is odd, so this must always be the trivial representation.