$Z_p$:=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.


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.


3 Answers 3


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

enter image description here

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 enter image description here

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


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


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.

  • $\begingroup$ Matthias,Thanks for your comment.Definitely,i am interested in the homology with integral coefficients. $\endgroup$
    – user2015
    Mar 28, 2015 at 18:49
  • $\begingroup$ @user2015: I thought as much. You might also want to consider $\mathbb{F}_p$-coefficients, at least this will get rid of the extension problems in the spectral sequence. It may be necessary to write out a full equivariant cell structure to get the computation at $p$. The case $p=2$ in my comment actually gives the integral statement. In this case, the homology is not so terribly complicated, but then again the prime $2$ case is not terribly good as a starting point for generalizations. $\endgroup$ Mar 28, 2015 at 19:27
  • $\begingroup$ @Matthias.I agree.And it is usually not so easy to write down a G-CW structure explicitly. $\endgroup$
    – user2015
    Mar 28, 2015 at 19:33

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