Free Actions of Compact Lie Groups on Manifolds

Abstract

If a compact Lie group G acts freely on a manifold X, the resulting orbit space X/G is itself a manifold. This text is concerned with the existence of such actions as well as the cohomological classification of the respective orbit spaces by using some known tools of equivariant cohomology theory and spectral sequences.

1. Introduction

When a topological group $G$ acts on a manifold $X$, we can define the orbit space $X/G,$ that does not necessarily have the structure of a manifold. However, when $G$ is a compact Lie group and we impose the condition that the action be free, which means that the isotropy subgroup $G_x$ contains only the trivial element $1$ of $G$, for any $x\in X,$ then we can construct on $X/G$ a manifold structure.

The general situation above can be illustrated by the construction of the projective spaces ${\mathit{kP}}^n$, for $k=\mathbb{R},\mathbb{C}$ or $\mathbb{H},$ that are orbit spaces of certain free actions of the groups ${\mathbb{Z}}_2,$$S^1$, and $S^3$ on spheres $S^n,$$S^{2n+1}$, and $S^{4n+3},$ respectively. Such spaces, as we know, appear in different contexts and for this reason there is an interest in obtaining certain algebraic and geometric invariants that characterize them, such as their homotopic and cohomological classification.

Thus, given $X$ and $G,$ we can naturally consider the general problem of classifying the space of orbits $X/G,$ through the existence or not of a free action of $G$ on $X,$ which is a typical transformation group problem associated with this data.

The interest in this type of problem, particularly when $G$ is a finite group, has become greater since the publication of work [1], by H. Hopf in 1926, in which one formalizes the purpose of classification of all manifolds whose universal covering is homeomorphic to a sphere $S^n.$ This problem, as we know, is equivalent to the classification of all finite groups that can act freely on $S^n.$

However, we realize that to get a homotopic classification of such spaces can become extremely complicated, even when the space $X$ already has a known classification. For example, let $S^n$ be the $n-$sphere seen as the one-point compactification of euclidean space ${\mathbb{R}}^n,$ which does not have a complete classification of its homotopy groups. As a result, instead of a homotopic classification, we can consider a cohomological classification of these orbit spaces.

In this direction, we realize that the difficulty in computing the cohomology of the orbit space $X/G,$ by direct methods, becomes evident when $X$ has nontrivial cohomology on several levels. On the other hand, many such results have been obtained by using some tools of equivariant cohomology theory. This is due to the fact that as long as $G$ is a compact Lie group acting freely on a space $X,$ there is a homotopy equivalence between the orbit space $X/G$ and the Borel space $X_G,$ so that we can use the so-called Leray-Serre spectral sequence associated with the Borel fibration:

E1

where $B_G$ denotes the classifying space for group $G,$ to investigate the cohomology ring $H^{\ast }\left(X_G;R\right)\cong H^{\ast }\left(X/G;R\right).$

In this text, we briefly deal with these tools and then we present some applications regarding the existence of free actions of certain compact Lie groups in some classes of smooth manifolds, as well as the cohomological classification of the respective orbit spaces.

2. Preliminary concepts

2.1 Group actions and classifying spaces

Let $\mu :G\times X\to X$ be an action of a topological group $G$ on a topological space $X,$ i. e. $\mu$ is a continuous map such that

$$\mu \left(g,\mu \left(h,x\right)\right)=\mu \left(\mathit{gh},x\right),$$E2

$$\mu \left(e,x\right)=x,$$E3

for any $g,h\in G$ and for any $x\in X,$ where $e$ indicates the neutral element of $G$. In this case, we say that $G$ acts on $X$ and $X$ is a $G-$space.

As it is usual, we denote by $\mu \left(g,x\right)=g\left(x\right)$ or simply $\mu \left(g,x\right)=\mathit{gx}$ to indicate the action of the element $g$ of $G$ on $x\in X.$

For each $x\in X,$ the subspace $G\left(x\right)=\left\{\mathit{gx};g\in G\right\}$ is called the orbit of the element $x.$ It is a simple task to show that for any two orbits $G\left(x\right)$ and $G\left(y\right),$ then $G\left(x\right)\cap G\left(y\right)=\varnothing$ or $G\left(x\right)=G\left(y\right).$ Therefore, we can define the orbit space:

$$X/G=\left\{G\left(x\right);x\in X\right\},$$E4

which is provided with the quotient topology induced by the natural map $q:X\to X/G,$ given by $q\left(x\right)=G\left(x\right),$ which is called orbit map.

Example 2.1.1. Any group $G$ acts on itself by multiplication. Precisely, we can define $\mu :G\times G\to G$ by $\mu \left(g,h\right)=\mathit{gh}.$

An action $\mu$ of $G$ on $X$ induces a group homomorphism ${\mathrm{\Gamma }}_{\mu }:G\to \text{Homeo}\left(X\right),$ such that, for each $g\in G,$ we define ${\mathrm{\Gamma }}_{\mu }\left(g\right)=L_g,$ where

$$L_g:X\to X,L_g\left(x\right)=\mathit{gx}.$$E5

The action $\mu$ is called effective when the kernel of the homomorphism ${\mathrm{\Gamma }}_{\mu }$ contains only the trivial element $e\in G$ and is called trivial when $ker{\mathrm{\Gamma }}_{\mu }=G.$

For each $x\in X,$ we call the isotropy subgroup at $x$ the following subgroup of $G$:

$$G_x=\left\{g\in G;\mathit{gx}=x\right\}.$$E6

When $G_x=\left\{e\right\},$ for any point $x\in X$ then the action is called free, and $X$ is said to be a free $G-$space. The set $X^G=\left\{x\in X;\mathit{gx}=x\right\}$ is called the fixed point set of the action.

Remark 2.1.1. If $X$ is a haursdorff space and $G$ is a compact space, it is well known that any action $\mu :G\times X\to X$ is a closed map, according to Theorem 1.2 of [2]. Furthermore, in this case, the subspace $\mu \left(G\times A\right)\subseteq X$ is closed (resp. compact) if $A$ is closed (resp. compact).

Let $X$ and $Y$ be $G-$spaces. If the map $f:X\to Y$ is equivariant, i. e. $f\left(\mathit{gx}\right)=\mathit{gf}\left(x\right),$ for any $g\in G$ and any $x\in X,$ then we can define the map:

$$\overline{f}:X/G\to Y/G,\overline{f}\left(G\left(x\right)\right)=G\left(f\left(x\right)\right).$$E7

Remark 2.1.2. Even though it is possible to investigate actions in arbitrary topological spaces, we are interested in observing certain structures both in $X$ and in the orbit space $X/G,$ so that we will assume, from now on, that $X$ is a manifold (smooth or not) and $G$ is a Lie group.

We recall that a Lie group $G$ is a topological group that is also a (real) finite-dimensional smooth manifold, in which the multiplication operation $\left(g_1,g_2\right)\mapsto g_1{g}_2$ and the inversion map $g\mapsto g^{-1}$ are smooth.

Example 2.1.2. The matrix groups $\mathit{GL}\left(n,\mathbb{R}\right),$$\mathit{GL}\left(n,\mathbb{C}\right),$ of the invertible $n\times n-$matrices with entries in $\mathbb{R}$ or $\mathbb{C},$ respectively, are standard examples of Lie groups, along with respective subgroups (special linear groups) $\mathit{SL}\left(n,\mathbb{R}\right)$ and $\mathit{SL}\left(n,\mathbb{C}\right).$

Example 2.1.3. Let $O\left(n\right)\subset \mathit{GL}\left(n,\mathbb{R}\right)$ be the subgroup of the orthogonal matrices, i. e. those in which ${\mathit{AA}}^t=\mathit{Id}$ and let $U\left(n\right)$ be the subgroup of $\mathit{SL}\left(n,\mathbb{C}\right)$ of the unitary matrices $\overline{A}{A}^t=\mathit{Id}.$ We can define the special orthogonal group by $\mathit{SO}\left(n\right)=O\left(n\right)\cap \mathit{SL}\left(n,\mathbb{R}\right)$ and the special unitary group by $\mathit{SU}\left(n\right)=U\left(n\right)\cap \mathit{SL}\left(n,\mathbb{C}\right).$

We know that $\mathit{SU}\left(2\right)$ is isomorphic to $S^3$ identified as the subgroup of the unitary quaternions. Also we know that the isomorphic groups $U\left(1\right)$ and $\mathit{SO}\left(2\right)$ are isomorphic to the circle group $S^1.$

For each integer $m,$ let ${\mathbb{Z}}_m=\mathbb{Z}/m$ be the group of the integer modulo $m,$ which can be identified with the subgroup of $S^1$ of all $m-$th roots of unity. In particular, we have the following chain of Lie (sub)groups:

$${\mathbb{Z}}_m\subset S^1\cong U\left(1\right)\cong \mathit{SO}\left(2\right)\subset \mathit{SU}\left(2\right)\cong S^3.$$E8

Example 2.1.4. (Free actions on spheres) Let $X$ be the $n-$sphere $S^n\subset {\mathbb{R}}^{n+1}$ and $G$ be the finite group ${\mathbb{Z}}_2.$ Then, $G$ acts freely on $X$ by the antipodal map $\mu \left(1,x\right)=A\left(x\right)=-x,$ and in this case we have $X/G=\mathbb{R}{P}^n$ the real projective space.

For $X=S^{2n-1}\subset {\mathbb{R}}^{2n}\cong {\mathbb{C}}^n$ and $G=S^1$ seen as a subgroup of the complex plane, we can consider the free action induced by complex multiplication:

$$\mu :G\times X\to X,\mu \left(z,\left(z_1,\cdots ,z_n\right)\right)=\left({\mathit{zz}}_1,\cdots ,{\mathit{zz}}_n\right),$$E9

and it follows that $X/G=\mathbb{C}{P}^{n-1}$ is the complex projective space.

Let $X=S^{4n-1}\subset {\mathbb{R}}^{4n}\cong {\mathbb{H}}^n$ and $G=S^3$ identified with the group of the unitary quaternions $\mathit{SU}\left(2\right),$ where $\mathbb{H}$ denotes the quaternion algebra:

$$\mathbb{H}=\left\{\left[\begin{array}{ll}\alpha & -\overline{\beta }\\ \beta & \overline{\alpha }\end{array}\right];\alpha ,\beta \in \mathbb{C}\right\}.$$E10

Similar to the previous case, we can define the free action $\mu :G\times X\to X,$ induced by the multiplication, such that $X/G=\mathbb{H}{P}^{n-1}$ the quaternionic projective space.

Example 2.1.5. (Free involutions on projective spaces) An involution¹ on a space $X$ is a continuous action of the group ${\mathbb{Z}}_2$ on $X.$ Let $\left[x_1,x_2,\cdots ,x_{2n-1},x_{2n}\right]\in \mathbb{R}{P}^{2n-1}=S^{2n-1}/{\mathbb{Z}}_2$ be an arbitrary element. Its easy to see that the map:

$$T\left(\left[x_1,x_2,\cdots ,x_{2n-1},x_{2n}\right]\right)=\left[-x_2,x_1,\cdots ,-x_{2n},x_{2n-1}\right]$$E11

defines a free involution on $\mathbb{R}{P}^{2n-1}.$

Similarly, for $\left[z_1,z_2,\cdots ,z_m,z_{m+1}\right]$ an arbitrary element in $\mathbb{C}{P}^m=S^{2m+1}/S^1,$ if $m>1$ is odd then we can define the free involution:

$$S\left(\left[z_1,z_2,\cdots ,z_m,z_{m+1}\right]\right)=\left[-{\overline{z}}_2,{\overline{z}}_1,\cdots ,-{\overline{z}}_{m+1},{\overline{z}}_m\right]$$E12

$$\mathbb{C}{P}^m.$$E13

When a group $G$ acts on a manifold $X$, in general, we can consider the orbit space $X/G$ with no other additional structure. However, when the action is free and proper, then this orbit space can be seen as a manifold, according to the Quotient Manifold Theorem:

Theorem 2.1.1. (Quotient Manifold Theorem) [4]. Let $G$ be a compact Lie group acting freely (and smoothly) on a smooth manifold $X.$ If the action is free and proper, then $X/G$ is also a smooth manifold of dimension $dim\left(X\right)-dim\left(G\right),$ such that the quotient map $X\to X/G$ is a principal $G-$bundle and a smooth submersion.

Remark 2.1.3. As an immediate consequence of the previous theorem, for every cohomology functor $H^{\ast }$ we have $H^j\left(X/G;R\right)=\left\{0\right\},$ for all $j>dimX-dimG,$ for any commutative ring with unity $R.$

Recall that given any compact Lie group $G$ we can construct the universal $G-$bundle $p_G:E_G\to B_G,$ with fiber space $G,$ where the total space $E_G$ is the $G-$space defined as the join operation² of infinite copies of $G,$ and the base is the quotient space (by diagonal action) $B_G=E_G/G,$ which is called the classifying space for $G$ and $p_G$ is the projection.

Example 2.1.6. (Classifying spaces for ${\mathbb{Z}}_2,S^1$ and $S^3$) For $G={\mathbb{Z}}_2,$ we can see that $B_G=E_G/G\cong \mathbb{R}{P}^{\infty }.$ Consequently, the mod 2 cohomology of the classifying space $B_G$ is given by $H^{\ast }\left(B_G;{\mathbb{Z}}_2\right)={\mathbb{Z}}_2\left[t\right],$ where $degt=1.$ Regarding to $G=S^1,$ since $B_G=E_G/G\cong \mathbb{C}{P}^{\infty },$ then ${\pi }_1\left(B_G\right)=1$ and the mod 2 cohomology is give by $H^{\ast }\left(B_G;{\mathbb{Z}}_2\right)={\mathbb{Z}}_2\left[\tau \right],$ where $deg\tau =2.$ With respect to the group $G=S^3,$ it follows that $B_G=E_G/G\cong \mathbb{H}{P}^{\infty }.$ Since³${\pi }_i\left(B_G\right)\cong {\pi }_{i-1}\left(G\right),$ therefore ${\pi }_1\left(B_G\right)\cong 1$ and the mod 2 cohomology of $B_G$ is given by $H^{\ast }\left(B_G;{\mathbb{Z}}_2\right)={\mathbb{Z}}_2\left[\tau \right],$ where $deg\tau =4.$

2.2 The Leray-Serre spectral sequence of a Borel fibration

Let $X$ be a free $G-$space, $q:X\to X/G$ the orbit map, and $\varpi :E_G\to B_G$ the universal $G-$bundle. The group $G$ acts freely on the product $X\times E_G,$ by the diagonal action $g\left(x,y\right)=\left(\mathit{gx},\mathit{gy}\right),$ where do we get

$$\rho :X\times E_G\to \left(X\times E_G\right)/G=X_G$$E14

the respective orbit map. The quotient space $X_G$ is also know as the Borel space.

Since the projections ${\mathit{\text{proj}}}_1:X\times E_G\to X$ and ${\mathit{\text{proj}}}_2:X\times E_G\to E_G$ are $G-$equivariant, they induce the fibrations $\pi$ and $p,$ respectively, according to the diagram below:

E15

where $\pi$ is called the Borel fibration with fiber $X,$ and $p$ is a principal $G-$bundle.

Moreover, under the above hypothesis, the fiber $E_G$ of $p$ is contractible and therefore $p$ is a homotopy equivalence⁴, which induces a natural isomorphism $p^{\ast }:H^{\ast }\left(X/G;R\right)\to H^{\ast }\left(X_G;R\right),$ for any commutative ring with unit $R.$

By Theorem 5.2 of [9], there is a first quadrant cohomological spectral sequence $\left\{E_r^{\ast ,\ast },d_r\right\}$ converging to $H^{\ast }\left(X_G;R\right)\cong H^{\ast }\left(X/G;R\right),$ as an algebra, such that the $E_2-$page $E_2^{p,q},$ is isomorphic to

$$E_2^{p,q}\cong H^p\left(B_G;{\mathrm{\mathscr{H}}}^q\left(X;R\right)\right),$$E16

where the symbol ${\mathrm{\mathscr{H}}}^q\left(X;R\right)$ indicates a system of local coefficients twisted by the action of the fundamental group ${\pi }_1\left(B_G\right)$ on the cohomology ring of $X.$

When ${\pi }_1\left(B_G\right)$ acts trivially on $H^{\ast }\left(X;R\right),$ the system of local coefficients ${\mathrm{\mathscr{H}}}^q\left(X;R\right)$ is simple and, according to Proposition 5.6 of [9], the $E_2-$page as in (13) takes the form:

$$E_2^{p,q}\cong H^p\left(B_G;R\right){\otimes }_R{H}^q\left(X;R\right),$$E17

what happens, in particular, when ${\pi }_1\left(B_G\right)=1.$

Moreover, by Theorem 5.9 of [9], the homomorphisms

$$H^q\left(B_G;R\right)=E_2^{q,0}\twoheadrightarrow \cdots \twoheadrightarrow E_q^{q,0}\twoheadrightarrow E_{q+1}^{q,0}=E_{\infty }^{q,0}\subseteq H^q\left(X_G;R\right)$$E18

and

$$H^q\left(X_G;R\right)\twoheadrightarrow E_{\infty }^{0,q}=E_{q+1}^{0,q}\subseteq E_q^{0,q}\subseteq \cdots \subseteq E_2^{0,q}=H^q\left(X;R\right)$$E19

coincide with the homomorphisms ${\pi }^{\ast }:H^q\left(B_G;R\right)\to H^q\left(X_G;R\right)$ and $i^{\ast }:H^q\left(X_G;R\right)\to H^q\left(X;R\right),$ respectively.

3. Some applications

Supposing that there is a free action of a group $G$ on $X,$ there is an interest in classifying the orbit space $X/G,$ in the same way as it happens on the real, complex, and quaternionic projective spaces.

In order to do this, we will cite some results that use these tools presented in the previous sections to obtain cohomological classifications for certain known spaces.

However, we observe that in general way, the computation of the cohomology of the orbit space $X/G$ can be a difficult task when the space $X$ is not a sphere or, more generally, when $X$ has nontrivial cohomology $H^j\left(X;R\right)\ne \left\{0\right\}$ on several levels $0<j<dimX$

Remark 3.1. Since all these results use only the cohomological struture of space $X,$ we observe that the same conclusions can be obtained replacing $X$ by any finitistic space that has same $\stackrel{⌣}{C}$ech cohomology algebra. Recall that a finitistic space is a paracompact Hausdorff space whose every open covering has a finite-dimensional open refinement, where the dimension of a covering is one less than the maximum number of members of the covering which intersect nontrivially. It is known [10, 11] that if $G$ is a compact Lie group acting continuously on $X,$ then $X$ is finitistic if and only if the orbit space $X/G$ is finitistic. Therefore, we can consider the problem of cohomology classification of the orbit spaces up to finitistic spaces of isomorphic cohomology to initial space $X.$

3.1 Free actions on spheres and projective spaces

In 1926, Hopf posed the general problem of classifying all groups that present freely in $S^n.$ Posteriorly, in 1957, J. Milnor provided some answers for this problem by showing, among other things, that the symmetric group $S_3$ cannot act freely on $S^n.$

Even considering this classification only on the category of compact Lie groups, this problem still does not have a complete solution for a arbitrary sphere $S^n.$ However, if $n$ is even, the only finite group that acts freely on $S^n$ is the group ${\mathbb{Z}}_2.$

In fact, if $G$ acts freely on $X=S^{2k},$ then the quotient map $X\to X/G$ is a covering projection; therefore, with $\chi \left(\cdot \right)$ the Euler characteristic, it follows that

$$2=\chi \left(S^{2k}\right)=\mid G\mid \cdot \chi \left(S^{2k}/G\right),$$E20

which implies that $\mid G\mid =1$ or $\mid G\mid =2.$ Since the action is free, the only possibility is $\mid G\mid =2$ and then $G={\mathbb{Z}}_2.$ Furthermore, the resulting orbit space $S^n/{\mathbb{Z}}_2$ has the same homotopy type of real projective space $\mathbb{R}{P}^n.$

For $n$ odd, the Section 3.8 of [2] contains a compilation of results related to the existence of free $G-$actions on spheres $S^n.$ In particular, with respect to groups of positive dimension, the Theorem 8.5 of [2] states that a group $G$ that act freely on $S^n$ must be isomorphic to $S^3,$$S^1$_, or $N\left(S^1\right).$

Suppose that $G=S^1$ and let $X=S^n$ and $X_G\to B_G$ the Borel fibration.

For $\left\{E_r^{\ast ,\ast },d_r\right\}$ the associated Leray-Serre spectral sequence, we have

$$E_2^{p,q}=H^p\left(B_G;{\mathbb{Z}}_2\right)\otimes H^q\left(X;{\mathbb{Z}}_2\right)=\left\{\begin{array}{l}{\mathbb{Z}}_2\otimes {\mathbb{Z}}_2,\text{if}p\text{is}\mathrm{odd}\text{and}q=0,n\\ \left\{0\right\},\text{otherwise}.\end{array}\right.$$

Since this sequence converges to $H^{\ast }\left(X/G;{\mathbb{Z}}_2\right)$_, it follows that it cannot collapse on $E_2-$page, which means that there is a nontrivial differential $d_{2k}:E_{2k}^{0,n}\to E_{2k}^{2k,0},$ for $k\in \mathbb{Z}$ such that $n=2k-1.$

Therefore, the sequence collapse on $E_{2k+1}-$page is $E_{\infty }=E_{2k+1},$ whose only nonzero row is $E_{\infty }^{\text{even},0}$ and the total complex is isomorphic to the graded cohomology ring $H^{\ast }\left(\mathbb{C}{P}^{n-1};{\mathbb{Z}}_2\right).$

Proceeding in a similar way, we can show that $H^{\ast }\left(S^{4n-3}/S^3;{\mathbb{Z}}_2\right)=H^{\ast }\left(\mathbb{H}{P}^{n-1};{\mathbb{Z}}_2\right).$

In order to generalize this type of problem, we can consider $X$ a product of spheres. For example, L. W. Cusick [12] showed that if a finite group G acts freely on a product of spheres of even dimensions, $X=S^{2n_1}\times \cdots \times S^{2n_k},$ then then $G$ must be isomorphic to a group of the type ${\mathbb{Z}}_2^r,$ for some $r\le k.$

Concerning on free actions of a finite group ${\mathbb{Z}}_p,$$p$ prime, and the circle group $S^1$ on a product of spheres $S^m\times S^n,$ Dotzel et al. [13] showed the following classification results according to Theorems 3.1.1, 3.1.2, and 3.1.3.

Theorem 3.1.1. Let ${\mathbb{Z}}_p$, $p$ an odd prime, act freely on $X=S^m\times S^n,$$0<m\le n.$ Then, $H^{\ast }\left(X/{\mathbb{Z}}_p;{\mathbb{Z}}_p\right)$ is isomorphic to ${\mathbb{Z}}_p\left[x,y,z\right]/\varphi \left(x,y,z\right),$ as a graded commutative algebra, where $\varphi \left(x,y,z\right)$ is one of the following ideals:

1. $\left(x^2,y^{\left(m+1\right)/2},z^2\right)$$m$ odd, $degx=1$, $y=\beta \left(x\right)$ the Bockstein cohomology operation and $degz=n;$

2. $\left(x^2,y^{\left(m+n+1\right)/2},y^{\left(n-m+1\right)/2}z-{\mathit{ay}}^{\left(n+1\right)/2},z^2-{\mathit{by}}^m\right),$$m$ even, $n$ odd, $degx=1,$$y=\beta \left(x\right)$, $degz=m$, $a,b\in {\mathbb{Z}}_p$, and $a=0$ necessarily when $n<2m;$

3. $\left(x^2,y^{\left(n+1\right)/2},z^2-{\mathit{by}}^m\right),$$n$ odd, $degx=1$, $y=\beta \left(x\right),$$degz=m,$$b\in {\mathbb{Z}}_p,$$b\ne 0$ only when $m$ is even and $2m<n.$

Theorem 3.1.20. Let ${\mathbb{Z}}_2$ act freely on $X=S^m\times S^n,$$0<m\le n.$ Then, $H^{\ast }\left(X/{\mathbb{Z}}_2;{\mathbb{Z}}_2\right)$ is isomorphic to ${\mathbb{Z}}_p\left[y,z\right]/\psi \left(y,z\right),$ as a graded commutative algebra, where $\psi \left(y,z\right)$ is one of the following ideals:

1. $\left(y^{m+2},z^2\right),$$degy=1$_, and $degz=n;$

2. $\left(y^{m+n+1},y^{n-m+1},z,z^2-{\mathit{ay}}^{m}z-{\mathit{by}}^{2m}\right),$$degy=1,$$degz=m,$$a,b\in {\mathbb{Z}}_2$_, and $a=0$ necessarily when $n<2m;$

3. $\left(y^{n+1},z^2-{\mathit{ay}}^{m}z-{\mathit{by}}^{2m}\right),$$degy=1$, $degz=m,$$a,b\in {\mathbb{Z}}_2,$ and $b=0$ necessarirly when $m=n$ or $n<2m.$

Theorem 3.1.3. Let $G=S^1$ act freely on $X=S^m\times S^n,$$0<m\le n.$ Then, $H^{\ast }\left(X/G;\mathbb{Q}\right)$ is isomorphic to $\mathbb{Q}\left[y,z\right]/\psi \left(y,z\right),$ as a graded commutative algebra, where $\psi \left(y,z\right)$ is one of the following ideals:

1. $\left(y^{\left(m+1\right)/2},z^2\right),$$m$ odd, $degy=2$ and $degz=n;$

2. $\left(y^{\left(m+n+1\right)/2},{\mathit{zy}}^{\left(n-m+1\right)/2}-{\mathit{ay}}^{\left(n+1\right)/2},z^2-{\mathit{by}}^m\right),$$m$ even, $n$ odd, $degy=2,$$degz=m,$ and $a=0$ necessarily when $n<2m;$

3. $\left(y^{\left(n+1\right)/2},z^2-{\mathit{by}}^m\right),$$n$ odd, $degy=2$, $degz=m,$ and $b\ne 0$ only when $m$ is even and $2m<n.$

Using the same techniques, it is shown in [14] similar results regarding the action of groups $S^1$ and $S^3$ on the product of spheres, considering both rational and mod 2 coefficients.

Theorem 3.1.4. The group $S^3$ cannot act freely on a $n-$torus $X={\left(S^1\right)}^n.$

Proof. Let $X$ be the $n-$torus ${\left(S^1\right)}^n$ and suppose that $G=S^3$ act freely on $X,$ with $n\ge 3.$ Let $x_1,\cdots ,x_n\in H^1\left(X,{\mathbb{Z}}_2\right)$ be the generators. By Quotient Manifold Theorem, the spectral sequence $\left\{E_r^{\ast ,\ast },d_r\right\}$ associated with the Borel fibration $X_G\to B_G$ does not collapse on the $E_2-$term. Therefore, there must exist some nontrivial differential $d_r^{p,q},$ for a certain $r\ge 2,$ such that

$$E_r^{p,q}\cong E_{r-1}^{p,q}\cong \cdots \cong E_2^{p,q}=H^p\left(B_G;{\mathbb{Z}}_2\right)\otimes H^q\left(X;{\mathbb{Z}}_2\right),$$

and it is clear that this is only possible when $r\ge 4k,$ for some $k\in \mathrm{\mathbb{N}}.$

Let us suppose that $r=4$ and let $y=x_{i_1}{x}_{i_2}{x}_{i_3}\in H^3\left(X;{\mathbb{Z}}_2\right)$ be an element for which $d_4^{0,3}\left(1\otimes y\right)=\tau \otimes 1.$ By dimensional reasons, $d_4^{0,1}\left(1\otimes x_i\right)=0$ for all $1\le i\le n;$ therefore, it follows that

$$\tau \otimes 1=d_4^{0,3}\left(1\otimes y\right)=\left(1\otimes x_{i_1}\right)\left(1\otimes x_{i_2}\right)d_4^{0,1}\left(1\otimes x_{i_3}\right)=0,$$

for $\tau$ the generator of $H^{\ast }\left(B_G;{\mathbb{Z}}_2\right)$, which is a contradiction. Since this argument works for any $r\ge 4$ and for any $y\in H^j\left(X;{\mathbb{Z}}_2\right),$$j\ge 3,$ it follows that $G$ cannot act freely on $X.$ □

Let $p\ge 2$ be a positive integer and $q_1,\cdots ,q_m$ be integers coprime to $p$, where $m\ge 1$. Then the action of ${\mathbb{Z}}_p$ on $S^{2m-1}\subset {\mathbb{C}}^m$ defined by:

$$\left(e^{2\pi {\mathit{iq}}_1/p},\cdots ,e^{2\pi {\mathit{iq}}_m/p}\right)\ast \left(z_1,\cdots ,z_m\right)=\left(e^{2\pi {\mathit{iq}}_1/p}{z}_1,\cdots ,e^{2\pi {\mathit{iq}}_m/p}{z}_m\right)$$

is itself free. Therefore, the resulting orbit space is a compact Hausdorff orientable manifold of dimension $2m-1,$ which is called lens space and it is denoted by:

$$S^{2m-1}/{\mathbb{Z}}_p=L_p^{2m-1}\left(q_1,\cdots ,q_m\right)=L_p^{2m-1}\left(q\right).$$

Theorem 31.5. [15] Let $G={\mathbb{Z}}_2$ act freely on $X=L_p^{2m-1}\left(q\right).$ Then, $H^{\ast }\left(X/G;{\mathbb{Z}}_2\right)$ is isomorphic to one of the following graded commutative algebras:

1. ${\mathbb{Z}}_2\left[x\right]/⟨x^{2m}⟩,$ where $degx=1.$

2. ${\mathbb{Z}}_2\left[x,y\right]/⟨x^2,y^m⟩,$ where $degx=1$ and $degy=2.$

3. ${\mathbb{Z}}_2\left[x,y,z\right]/⟨x^3,y^2,z^{m/2}⟩,$ where $degx=degy=1,$$degz=4$, and $m$ is even.

4. ${\mathbb{Z}}_2\left[x,y,z\right]/⟨x^4,y^2,z^{m/2},x^{2}y⟩,$ where $degx=degy=1,$$degz=4$, and $m$ is even.

5. ${\mathbb{Z}}_2\left[x,y,w,z\right]/⟨x^5,y^2,w^2,z^{m/4},x^{2}y,\mathit{wy}⟩,$ where $degx=degy=1,$$degw=3,$$degz=8$, and $4\mid m.$

Related to actions of ${\mathbb{Z}}_2$ on the product of projective spaces, both real and complex, we can mention the work [16], which provides a list of possible cohomology algebras for the respective orbit spaces. In [17], the authors showed that the group $G=S^3$ cannot act freely on the real projective space of any dimension.

Theorem 3.1.6. The group $G=S^3$ cannot act freely on $X=\mathbb{C}{P}^n,\mathbb{H}{P}^n,$ for any $n>0.$

Proof. Let us suppose that the group $G=S^3$ acts freely on $X=\mathbb{C}{P}^n.$ Then, the spectral sequence $\left\{E_r^{\ast ,\ast },d_r\right\}$ associated with the Borel fibration $X\rightarrowtail X_G\to B_G,$ which has the $E_2-$term given by $E_2^{p,q}=H^p\left(B_G;{\mathbb{Z}}_2\right)\otimes H^q\left(X;{\mathbb{Z}}_2\right),$ converges to $H^{\ast }\left(X_G;{\mathbb{Z}}_2\right)\cong H^{\ast }\left(X/G;{\mathbb{Z}}_2\right),$ as an algebra. By the cohomology structures of $B_G\cong \mathbb{H}{P}^{\infty }$ and $X=\mathbb{C}{P}^n,$ it follows that

$$E_2^{p,q}\cong \left\{\begin{array}{ll}{\mathbb{Z}}_2,& \text{if}p=4i\text{and}q=2j,\text{for}\mathrm{all}i,j\ge 0,\\ \left\{0\right\},& \text{otherwise}.\end{array}\right.$$

Therefore, a differential $d_r^{p,q}:E_r^{p,q}\to E_r^{p+r,q+1-r}$ with bidegree $\left(r,1-r\right),$ is nontrivial only if $p=4i$ and $q=2j\le 2n,$ for some positive integers $i$ and $j.$ In this case, we have the following equality involving the bidegrees: $\left(4i+r,2j+1-r\right)=\left(4k,2l\right),$ for certain integers $k,l>0,$ that is, these numbers must satisfy the linear system:

$$\left\{\begin{array}{l}4i+r=4k,\\ 2j+1-r=2l,\end{array}\right.$$

that clearly has no integer solution; therefore, we conclude that all differentials $d_r^{\ast ,\ast }$ are trivial, for all $r\ge 2.$ This implies that the sequence collapses on its $E_r\cong E_2-$term and contradicts the Quotient Manifold Theorem.

Similarly, let us suppose that the group $S^1$ acts freely on $X=\mathbb{H}{P}^n,$ and let us consider $\left\{E_r^{\ast ,\ast },d_r\right\}$ the spectral sequence associated with the Borel fibration $X_{S^1}\to B_{S^1},$ whose $E_2-$term is given by $E_2^{p,q}\cong H^p\left(B_{S^1};{\mathbb{Z}}_2\right)\otimes H^q\left(X;{\mathbb{Z}}_2\right).$

Let $t$ be the generator of $H^{\ast }\left(\mathbb{C}{P}^{\infty };{\mathbb{Z}}_2\right)\cong H^{\ast }\left(B_{S^1};{\mathbb{Z}}_2\right)$ and $\tau$ be the generator of $H^{\ast }\left(\mathbb{H}{P}^n;{\mathbb{Z}}_2\right).$ Then,

$$E_2^{p,q}\cong \left\{\begin{array}{ll}{\mathbb{Z}}_2,& \text{if}p=2i\text{and}q=4j,i,j\ge 0,\\ \left\{0\right\},& \text{otherwise}.\end{array}\right.$$

By Quotient Manifold Theorem, the spectral sequence does not collapse on it $E_2-$term; therefore, there must exist some nontrivial differential $d_r^{\ast ,\ast }.$ If $r\ge 2$ is the smallest integer for which this happens, so that

$$E_r^{p,q}\cong E_{r-1}^{p,q}\cong \cdots \cong E_2^{p,q},$$

for all $p,q\ge 0,$ we see that this is only possible when the integers $r,i,j$, and $k$ (which are obtained from the equality between the bidegrees involved) satisfy the linear system:

$$\left\{\begin{array}{l}r=2i,\\ 4j+1-r=4k.\end{array}\right.$$

But this system has no integer solution; therefore, the group $S^1$ cannot act freely on $X.$ Since $S^1$ is a subgroup of $S^3,$ then $X$ does not admit any free action of $S^3.$ □

3.2 Free actions on spaces of type $\left(a,b\right)$

Let $X$ be a finite CW complex. We say that $X$ is a space of type $\left(a,b\right),$ characterized by an integer $n>0,$ if

$$H^j\left(X;\mathbb{Z}\right)=\left\{\begin{array}{ll}\mathbb{Z},& \text{if}j=0,n,2n,3n,\\ \left\{0\right\},& \text{otherwise},\end{array}\right.$$E21

whose generators $u_i\in H^{\mathit{\text{in}}}\left(X;\mathbb{Z}\right)$ satisfy the relations ${\mathit{au}}_2=u_1^2$ and ${\mathit{bu}}_3=u_1{u}_2,$ for certain integers $a$ and $b.$ By Universal Coefficient Theorem, the mod 2 cohomology of $X$ is given by $H^{\mathit{\text{in}}}\left(X;{\mathbb{Z}}_2\right)\cong H^{\mathit{\text{in}}}\left(X;\mathbb{Z}\right)\otimes {\mathbb{Z}}_2\cong {\mathbb{Z}}_2,$ for $n=\mathrm{0,1,2,3},$ and the relations come to depend only on the parity of the numbers $a$ and $b.$ In this case, we will use the same symbols to denote the generators, i. e.

$${\mathbb{Z}}_2\cong ⟨u_i⟩\cong H^{\mathit{\text{in}}}\left(X\right)=H^{\mathit{\text{in}}}\left(X;{\mathbb{Z}}_2\right).$$E22

Example 3.2.2. The spaces of type (a, b) were first studied by James [18] and Toda [19]. Note that we can construct examples of these spaces by considering products or unions between certain known spaces, as spheres and projective spaces. Moreover, in Toda’s work, it is shown that it is possible to construct a space of type $\left(a,b\right)$ for any choice of $a$ and $b.$ For example,

1. The product $S^n\times S^{2n}$ is a space of the type $\left(0,1\right),$ characterized by $n.$

2. The one point union $S^n\vee S^{2n}\vee S^{3n}$ is of type $\left(0,0\right),$ characterized by $n.$

3. The one point union $S^6\vee \mathbb{C}{P}^2$ is of type $\left(1,0\right),$ characterized by $n=2.$

4. The projective spaces $\mathbb{R}{P}^3,\mathbb{C}{P}^3,\mathbb{H}{P}^3$ are examples of spaces of type $\left(1,1\right),$ characterized by $n=1,n=2$ and $n=3,$ respectively.

In 2010, Pergher et al. [20] investigated the existence of free actions of the groups ${\mathbb{Z}}_2$ and $S^1$ on spaces of type $\left(a,b\right),$ for $n>1,$ where they concluded that:

Theorem 3.2.1. [20] Let $X$ be a space of type $\left(a,b\right),$ characterized by $n>1.$

1. If $a$ is odd and $b$ is even, then ${\mathbb{Z}}_2$ cannot act freely on $X.$

2. If $a\ne 0,$ then $S^1$ cannot act freely on $X.$

3. If $G={\mathbb{Z}}_2$ act freely on $X$ where both $a$ and $b$ even, then $H^{\ast }\left(X/G;{\mathbb{Z}}_2\right)\cong {\mathbb{Z}}_2\left[x,z\right]/⟨x^{3n+1},z^2,{\mathit{zx}}^{n+1}⟩,$ where $degx=1$ and $degz=n.$

4. If $G=S^1$ act freely on $X,$ then $a=0$ and $H^{\ast }\left(X/G;{\mathbb{Z}}_2\right)$ is isomorphic to one of the following graded commutative algebras

$${\mathbb{Z}}_2\left[x,z\right]/⟨x^{\left(3n+1\right)/2},z^2,{\mathit{zx}}^{\left(n+1\right)/2}⟩,\text{where}degx=2\text{and}degz=n,$$

or

$${\mathbb{Z}}_2\left[x,z\right]/⟨x^{\left(n+1\right)/2},z^2⟩,\text{where}degx=2,degz=2n\text{and}b\text{is}\mathrm{odd}.$$

In [21], Dotzel and Singh constructed a class of examples of free actions of ${\mathbb{Z}}_p,$$p$ prime, on spaces of type $\left(0,0\right),$ by using some known topological operations.

In particular, for $n$ even, it is shown in [22] that the only group that can act freely on $X$ is ${\mathbb{Z}}_2.$ In addiction, the authors construct a example of such action. If $n$ is odd and $X$ is of type $\left(0,1\right)$, then any finite group $G$ which acts freely on $X$ cannot contain the group ${\mathbb{Z}}_p\oplus {\mathbb{Z}}_p,$ for any $p$ odd prime.

Theorem 3.2.2. Let $X$ be a manifold⁵ that is a space of type $\left(0,b\right),$ characterized by $n>1.$ If $G=S^3$ acts freely on $X,$ then $n$ is an odd number of the form $4k-1,$ for some $k\ge 1$ and $b$ is odd. In this case, the cohomology algebra of the orbit space $X/G$ is isomorphic to the graded polynomial algebra ${\mathbb{Z}}_2\left[x,y\right]/⟨x^k,y^2⟩,$ where $degx=4$ and $degy=2n.$

Proof. Let us suppose know that the group $G=S^3$ acts freely on a space $X$ of type $\left(0,b\right)$ and let $\left\{E_r^{\ast ,\ast },d_r\right\}$ be the spectral sequence associated with the Borel fibration $X_G\to B_G,$ with fiber $X,$ such that $E_2^{p,q}=H^p\left(B_G;{\mathbb{Z}}_2\right)\otimes H^q\left(X;{\mathbb{Z}}_2\right),$ which converges to $H^{\ast }\left(X_G;{\mathbb{Z}}_2\right)\cong H^{\ast }\left(X/G;{\mathbb{Z}}_2\right).$

By Quotient Manifold Theorem, it follows that this sequence does not collapse on its $E_2-$term. Then, there must exist some nontrivial differential $d_{r_i}$, for some $r_i\ge 2.$ If $r=min\left\{r_i\right\},$ then

$$E_r^{p,q}\cong E_{r-1}^{p,q}\cong \cdots \cong E_2^{p,q},$$

and this is possible only if $r=4k$ and $n=4k-1,$ for some $k\ge 1.$ This provides the following possibilities for the action of the differentials $d_{4k}^{4l,q},$ for $q=n,2n,3n:$

1. $d_r\left(1\otimes u_1\right)=0,$$d_r\left(1\otimes u_2\right)={\tau }^k\otimes u_1$ and $d_r\left(1\otimes u_3\right)={\tau }^k\otimes u_2,$

2. $d_r\left(1\otimes u_1\right)={\tau }^k\otimes 1$$d_r\left(1\otimes u_2\right)={\tau }^k\otimes u_1$ and $d_r\left(1\otimes u_3\right)=0,$

3. $d_r\left(1\otimes u_1\right)={\tau }^k\otimes 1,$$d_r\left(1\otimes u_2\right)=0$ and $d_r\left(1\otimes u_3\right)={\tau }^k\otimes u_2,$

4. $d_r\left(1\otimes u_1\right)=0,$$d_r\left(1\otimes u_2\right)={\tau }^k\otimes u_1$ and $d_r\left(1\otimes u_3\right)={\tau }^k\otimes u_2,$

5. $d_r\left(1\otimes u_1\right)=0,$$d_r\left(1\otimes u_2\right)=0$ and $d_r\left(1\otimes u_3\right)={\tau }^k\otimes u_2,$

6. $d_r\left(1\otimes u_1\right)=0,$$d_r\left(1\otimes u_2\right)={\tau }^k\otimes u_1$ and $d_r\left(1\otimes u_3\right)=0,$

7. $d_r\left(1\otimes u_1\right)={\tau }^k\otimes 1,$$d_r\left(1\otimes u_2\right)=0$ and $d_r\left(1\otimes u_3\right)=0.$

We will divide the analysis of these cases according to the parity of $b.$

Case$b$odd: In this case, we have the relation $u_1{u}_2=u_3$ and, by the multiplicative properties of the differentials, we have

$$d_r\left(1\otimes u_3\right)=\left(1\otimes u_1\right)d_r\left(1\otimes u_2\right)+\left(1\otimes u_2\right)d_r\left(1\otimes u_1\right).$$

So, if one of the cases $\left(b\right),\left(d\right),\left(e\right)$_, or $\left(g\right)$ occurred, it would lead to the contradiction 0$0={\tau }^k\otimes u_2.$

If case $\left(a\right)$ occurred, then the differentials $d_{4k}^{4i,3n}$ and $d_{4k}^{4i,2n}$ would be isomorphisms, whence it would follow that

$$\mathrm{im}{d}_{4k}^{4i,3n}\cong {\mathbb{Z}}_2⊈\left\{0\right\}=ker{d}_{4k}^{4\left(i+k\right),2n},$$

which is a contradiction.

If case $\left(f\right)$ occurred, then the sequence would collapse on its $E_{4k+1}-$term, with the lines $E_{4k+1}^{\ast ,0}$ and $E_{4k+1}^{\ast ,3n}$ containing an infinite number of nonzero elements. This would contradict the Quotient Manifold Theorem.

Therefore, $\left(c\right)$ is the only possible case, and it produces the following pattern:

$$E_{4k+1}^{p,q}=\left\{\begin{array}{ll}{\mathbb{Z}}_2,& \text{if}p=0,4,\cdots ,4\left(k-1\right)\text{and}q=2n,\\ \left\{0\right\},& \text{otherwise}.\end{array}\right.$$

Then, the sequence collapses on its $E_{4k+1}-$term, and $E_{\infty }^{p,q}\cong E_{4k+1}^{p,q},$ for all $p,q\ge 0.$ So, $H^j\left(X/G;{\mathbb{Z}}_2\right)\cong {\mathrm{Tot}}^j\left(E_{\infty }\right).$

The elements $\tau \otimes 1$ and $1\otimes u_2$ are the only permanent co-cycles, so they determine the nonzero elements $x$ and $y$ in $E_{\infty }^{4,0}$ and $E_{\infty }^{0,2n},$ respectively. By (15), we have ${\pi }^{\ast }\left(\tau \right)=x;$ then, $0={\pi }^{\ast }\left({\tau }^j\right)=x^j$ for all $j\ge k.$ By the structure of the $E_{\infty }-$term, it follows that $y^2=0;$ therefore, $H^{\ast }\left(X/G;{\mathbb{Z}}_2\right)$ is isomorphic to the graded polynomial algebra ${\mathbb{Z}}_2\left[x,y\right]/⟨x^k,y^2⟩,$ where $degx=4$ and $degy=2n.$

Case$b$even: We will show that if $b$ is even, then none of the cases can occur. By the relation $u_1{u}_2=0,$ we have

$$0=\left(1\otimes u_1\right)d_r\left(1\otimes u_2\right)+\left(1\otimes u_2\right)d_r\left(1\otimes u_1\right),$$

and this allows us to eliminate the cases $\left(b\right),\left(c\right)$, and $\left(g\right),$ since they produce the contradiction $0={\tau }^k\otimes u_2.$

By the same reason of the previous case ($b$ odd), we can eliminate case $\left(a\right);$ that is, it implies that

$$\mathrm{im}{d}_{4k}^{4i,3n}⊈ker{d}_{4k}^{4\left(i+k\right),2n}.$$

By a similar reason we can eliminate $\left(d\right),$ since it implies that the differentials $d_{4k}^{4i,2n}$ and $d_{4k}^{4j,3n}$ are isomorphisms.

For case $\left(e\right),$ the sequence would collapse on its $E_{4k+1}-$term, with the lines $E_{4k}^{\ast ,0}$ and $E_{4k}^{\ast ,n}$ containing infinite nonzero elements, which would contradict the Quotient Manifold Theorem. Finally, by the same reason of the previous case, we can eliminate $\left(f\right);$ therefore, when $b$ is even, the space $X$ does not admit any free action of $G.$ □

3.3 Free actions on Dold, Wall, and Milnor manifolds

The Dold manifolds $P\left(m,n\right)$, as they came to be known, were defined by A. Dold [23] as orbit spaces of free actions of ${\mathbb{Z}}_2,$ or equivalently free involutions, on a product of the form $S^m\times \mathbb{C}{P}^n.$ Precisely, for each pair of nonnegative integers $m$ and $n,$$P\left(m,n\right)$ is the orbit space $S^m\times \mathbb{C}{P}^n/T,$ where $T\left(x,\left[z\right]\right)=\left(-x,\left[\overline{z}\right]\right).$

Let $R:S^m\to S^m$ be the involution defined by the reflection of the last coordinate $R\left(x_0,\cdots ,x_m\right)=\left(x_0,\cdots ,x_{m-1},-x_m\right),$ and $1:\mathbb{C}{P}^n\to \mathbb{C}{P}^n$ be the identity map. Since the involution $R\times 1:S^m\times \mathbb{C}{P}^n\to S^m\times \mathbb{C}{P}^n$ commutes with the involution $T$, it induces an involution $S:P\left(m,n\right)\to P\left(m,n\right).$

For each pair of nonnegative integers $m$ and $n,$ the Wall manifold⁶$Q\left(m,n\right)$ is defined as the mapping torus of the homeomorphism $S,$ that is,

$$Q\left(m,n\right)=\frac{P\left(m,n\right)\times \left[0,1\right]}{\left(\left[x,z\right],0\right)\sim \left(S\left[x,z\right],1\right)}.$$E23

Let $m,n$ be integers, such that $0\le n\le m.$ It is called a (real) Milnor manifold⁷ of dimensions $n+m-1$ to the smooth closed submanifold of codimension 1 in $\mathbb{R}{P}^m\times \mathbb{R}{P}^n,$ described in homogeneous coordinates as:

$$\mathbb{R}{H}_{m,n}=\left\{(\left[x_0,\cdots ,x_m\right],[y_0,\cdots ,y_n\left]\right)\in \mathbb{R}{P}^m\times \mathbb{R}{P}^n|x_0{y}_0+\cdots +x_n{y}_n=0\right\},$$E24

which is also denoted by $H\left(m,n\right).$ Equivalently, $\mathbb{R}{H}_{m,n}$ is the total space of the bundle:

E25

The manifolds $P\left(m,n\right),$$Q\left(m,n\right)$_, and $H\left(m,n\right)$ were constructed to provide representatives for generators in odd dimension to the unoriented cobordism ring ${\mathrm{\Re }}_{\ast }$, since we have the projective spaces as representatives in even dimensions. Precisely, the following sets are generator sets for ${\mathrm{\Re }}_{\ast }:$

$$\left\{\left[\mathbb{R}{P}^{2i}\right],\left[P\right(2^r-1,s{2}^r\left)\right];i,r,s\ge 1\right\},$$E26

$$\left\{\left[\mathbb{R}{P}^{2i}\right],\left[Q\right(2^r-2,s{2}^r\left)\right];i,r,s\ge 1\right\},$$E27

and

$$\left\{\left[\mathbb{R}{P}^{2i}\right],\left[H\right(2^k,2t{2}^k\left)\right];i,k,t\ge 1\right\}.$$E28

For these reasons, the analysis of certain structures and algebraic invariants related to the Dold, Milnor, and Wall manifolds is a relevant research topic, as it is done on the works [26, 27, 28] of Mukerjee. On the particular interest of investigating the existence of free actions of compact Lie groups on these spaces and also the cohomology classification of the respective orbit spaces, there are several results in the literature in which we will briefly discuss some of them below.

Regarding the existence of free actions of ${\mathbb{Z}}_2$ on Dold manifolds, Morita et al. [29] partially solved the problem by considering free involutions on $P\left(1,n\right),$ for $n\ge 1$ an odd integer. Later this problem was completely solved by Dey [30], according to the following.

Theorem 3.3.1. [30] If $G={\mathbb{Z}}_2$ acts freely on $X=P\left(m,n\right),$ then $H^{\ast }\left(X/G;{\mathbb{Z}}_2\right)$ is isomorphic to one of the following graded algebras:

1. (i) ${\mathbb{Z}}_2\left[x,y,z\right]/⟨x^2,y^{\left(m+1\right)/2},z^{n+1}⟩,$ where $degx=1$, $degy=degz=2$, and $m$ is odd.

2. ${\mathbb{Z}}_2\left[x,y,z\right]/⟨f,g,z^{\left(n+1\right)/2}+h⟩,$ where $n$ is odd, $degx=degy=1,$$degz=4$, $f=\left(x^{m+1}+{\alpha }_1{x}^{m}y+{\alpha }_2{x}^{m-1}{y}^2\right)$, and $g=\left(y^3+{\beta }_1{\mathit{xy}}^2+{\beta }_2{x}^{2}y\right),$ with ${\alpha }_i,{\beta }_i\in {\mathbb{Z}}_2$, and $h\in {\mathbb{Z}}_2\left[x,y,z\right]$ is either the zero polynomial or it is a homogeneous polynomial of degree $2n+2$ with the highest power of $z$ less than or equal to $\left(n-1\right)/2.$

Concerning on free involutions on Wall manifods $Q\left(m,n\right)$, the work of Khare [31] shows that these manifolds bounds if and only if $n$ is odd or $n=0$ and $m$ odd. By Proposition 3.5 in [32], we can conclude that $X=Q\left(m,\mathrm{odd}\right)$ admit free involutions and about the orbit spaces $X/{\mathbb{Z}}_2$ we have the following result, for some values of $m.$

Theorem 3.3.2. [32, 33] Let $X=Q\left(m,n\right),$ where $n>0$ is odd, equipped with a free action of the group $G={\mathbb{Z}}_2.$

1. If $m=1$ and the induced action of ${\mathbb{Z}}_2$ on the mod 2 cohomology is trivial, then

   $$H^{\ast }\left(X/G;{\mathbb{Z}}_2\right)\cong {\mathbb{Z}}_2\left[x,y,z,w\right]/⟨x^3,y^3,z^2,y^2+\mathit{yw},w^{\left(n+1\right)/2}⟩,$$

   where $degx=degy=degz=1$, and $degw=4.$

2. If $m$ is even and the induced action of ${\mathbb{Z}}_2$ on the mod 2 cohomology is trivial, then $H^{\ast }\left(X/{\mathbb{Z}}_2\right)$ is isomorphic to one of the following graded polynomial algebras:

$${\mathbb{Z}}_2\left[x,y,z,w\right]/⟨x^3,y^2,z^{m+1}+z^{m}y,w^{\left(n+1\right)/2}⟩,$$

where $degx=degy=degz=1$ and $degw=4,$ or

$${\mathbb{Z}}_2\left[x,y,z,w\right]/⟨x^2,y^{m+1},y^2+z,w^{n+1}⟩,$$

where $degx=degy=1$ and $degz=degw=2.$

Example 3.3.1. (Free $S^1-$actions on Dold manifolds) Let $G$ be the group $S^1$ and $m,n$ odd integers, where $m=2k-1,$ for some $k\ge 1.$ Considering $S^m\subseteq {\mathbb{C}}^k,$ we define a free action of $G$ on $S^m\times \mathbb{C}{P}^n$ by:

$$z\ast \left(w,v\right)\mapsto \left(\left({\mathit{zw}}_1,\cdots ,{\mathit{zw}}_k\right),\left[{\mathit{zv}}_0:\cdots :{\mathit{zv}}_n\right]\right),$$E29

where $w=\left(w_1,\cdots ,w_k\right)\in S^m\subseteq {\mathbb{C}}^k$ and $v=\left[v_0:\cdots :v_n\right]\in \mathbb{C}{P}^n.$

Let us consider an arbitrary element $\left[w,v\right]\in P\left(m,n\right)=\left(S^m\times \mathbb{C}{P}^n\right)/T,$ and note that the isotropy subgroup $G_{\left[w,v\right]}$ is trivial. Therefore, $z$ must be equal to $1\in G,$ that is, $G_{\left[w,v\right]}=\left\{1\right\}$, so the induced action on Dold manifold $P\left(m,n\right)$ is free.

If $n$ is odd and $m=2k$ is even, then we can consider

$$S^m=\left\{\left(w,t\right)\in {\mathbb{C}}^k\times \mathbb{R};\parallel w\parallel +|t|=1\right\},$$E30

and the analogous free action of $G$ on $S^m\times \mathbb{C}{P}^n$ is defined by:

$$z\ast \left(\left(w,t\right),v\right)\mapsto \left(\left({\mathit{zw}}_1,\cdots ,{\mathit{zw}}_k,t\right),\left[{\mathit{zv}}_0:\cdots :{\mathit{zv}}_n\right]\right).$$E31

Since this action is free, it induces a free action of $G$ on $P\left(m,n\right),$ as in the previous case.

Theorem 3.3.3. There is no free action of $G=S^1$ on $X=Q\left(m,n\right),$ for any $m,n>0.$

Proof. Recall that⁸$H^{\ast }\left(X;{\mathbb{Z}}_2\right)\cong {\mathbb{Z}}_2\left[x,c,d\right]/⟨x^2,c^{m+1}+c^{m}x,d^{n+1}⟩,$ where $degx=degc=1$ and $degd=2.$ Suppose that there is a free action of $G=S^1$ on $X$ and let $X\hookrightarrow X_G\to B_G$ be the associated Borel fibration, with $\left\{E_r^{\ast ,\ast },d_r\right\},$ and such that $E_2^{p,q}\cong H^p\left(B_G;{\mathbb{Z}}_2\right)\otimes H^q\left(X;{\mathbb{Z}}_2\right),$ that converges, as an algebra, to $H^{\ast }\left(X_G;{\mathbb{Z}}_2\right)\cong H^{\ast }\left(X/G;{\mathbb{Z}}_2\right).$ Since this sequence does not collapse on $E_2-$page, there is some nontrivial differential $d_2^{\ast ,\ast }$, according to the following cases.

Case$m$odd: In this case, we have $d_2^{0,m+1}\left(1\otimes c^{m+1}\right)=0.$ In fact, since $m+1=2r,$ for some $r>0,$ then $d_2^{0,2r}\left(1\otimes c^{2r}\right)=d_2^{0,2r}\left(\left(1\otimes c^r\right)\left(1\otimes c^r\right)\right)=0.$ However, by relation $c^{m+1}=c^{m}x,$ it follows that

$$d_2^{0,m+1}\left(1\otimes c^{m+1}\right)=\left(1\otimes c^m\right)d_2^{0,1}\left(1\otimes x\right)+\left(1\otimes x\right)d_2^{0,m}\left(1\otimes c^m\right).$$

Therefore, $d_2^{0,1}\left(1\otimes x\right)\ne 0$ and $d_2^{0,1}\left(1\otimes c\right)\ne 0,$ cannot occur simultaneously because in this case we will have $0=\tau \otimes c^{m-1}\left(c+x\right),$ which is a contradiction.

Similarly, $d_2^{0,1}\left(1\otimes x\right)=0$ and $d_2^{0,1}\left(1\otimes c\right)\ne 0,$ cannot occur simultaneously. Therefore, it follows there are only the possibilities:

1. $d_2^{0,1}\left(1\otimes c\right)=d_2^{0,1}\left(1\otimes x\right)=0$ and $d_2^{0,2}\left(1\otimes d\right)=\tau \otimes c,$

2. $d_2^{0,1}\left(1\otimes c\right)=d_2^{0,1}\left(1\otimes x\right)=0$ and $d_2^{0,2}\left(1\otimes d\right)=\tau \otimes x.$

We claim that (1) and (2) cannot occur.

If (1) occurs, then for any $j\ge 0,$$k\in \left\{1\cdots ,m\right\}$ and $l\in \left\{1\cdots ,n\right\}$, it follows that

$$d_2^{j,2l}\left({\tau }^j\otimes d^l\right)=\left\{\begin{array}{ll}0,& \text{if}l\text{is\hspace{0.17em}even},\\ {\tau }^{j+1}\otimes {\mathit{cd}}^{l-1},& \text{if}l\text{is}\mathrm{odd},\end{array}\right.$$

$$d_2^{j,k+2l}\left({\tau }^j\otimes c^k{d}^l\right)=\left\{\begin{array}{ll}0,& \text{if}l\text{is\hspace{0.17em}even},\\ {\tau }^{j+1}\otimes c^{k+1}{d}^{l-1},& \text{if}l\text{is}\mathrm{odd},\end{array}\right.$$

$$d_2^{j,2l+1}\left({\tau }^j\otimes {\mathit{xd}}^l\right)=\left\{\begin{array}{ll}0,& \text{if}l\text{is\hspace{0.17em}even},\\ {\tau }^{j+1}\otimes {\mathit{cxd}}^{l-1},& \text{if}l\text{is}\mathrm{odd},\end{array}\right.$$

$$d_2^{j,2l+k+1}\left({\tau }^j\otimes {\mathit{xc}}^k{d}^l\right)=\left\{\begin{array}{ll}0,& \text{if}l\text{is\hspace{0.17em}even},\\ {\tau }^{j+1}\otimes {\mathit{xc}}^{k+1}{d}^{l-1},& \text{if}l\text{is}\mathrm{odd},\end{array}\right.$$

therefore, we will have $E_3^{p,q}\cong \left\{0\right\},$ for all $p$ odd or $q\equiv 2\left(mod4\right)$ and $q>0.$ We can see that the sequence collapses on $E_3-$page, however $E_3^{2r,q}\ne \left\{0\right\},$ for all $q\equiv s\left(mod4\right),$$s=\mathrm{0,2,3},$ and $r\ge 0,$ which contradicts the Quotient Manifold Theorem.

If the case (2) occurs, then for all $j\ge 0,$$k\in \left\{1\cdots ,m\right\}$ and $l\in \left\{1\cdots ,n\right\}$, we have

$$d_2^{j,2l}\left({\tau }^j\otimes d^l\right)=\left\{\begin{array}{ll}0,& \text{if}l\text{is\hspace{0.17em}even},\\ {\tau }^{j+1}\otimes {\mathit{xd}}^{l-1},& \text{if}l\text{is}\mathrm{odd},\end{array}\right.$$

$$d_2^{j,2l+k}\left({\tau }^j\otimes c^k{d}^l\right)=\left\{\begin{array}{ll}0,& \text{if}l\text{is\hspace{0.17em}even},\\ {\tau }^{j+1}\otimes {\mathit{xc}}^k{d}^{l-1},& \text{if}l\text{is}\mathrm{odd},\end{array}\right.$$

while $d_2^{j,2l+1}\left({\tau }^j\otimes {\mathit{xd}}^l\right)=0,$ since $x^2=0.$ Therefore, analogous to case (1), we can conclude that this case is not possible either.

Case$m$even: In this case, we have $d_2^{0,m}\left(1\otimes c^m\right)=0,$ so $d_2^{0,m+1}\left(1\otimes c^{m}x\right)=\left(1\otimes c^m\right)d_2^{0,1}\left(1\otimes x\right),$ while, by relation $c^{m+1}=c^{m}x,$$d_2^{0,m+1}\left(1\otimes c^{m}x\right)=d_2^{0,m+1}\left(1\otimes c^{m+1}\right)=\left(1\otimes c^m\right)d_2^{0,1}\left(1\otimes c\right),$ therefore, we should have necessarily $d_2^{0,1}\left(1\otimes c\right)=d_2^{0,1}\left(1\otimes x\right).$

If $d_2^{0,1}\left(1\otimes c\right)=d_2^{0,1}\left(1\otimes x\right)=\tau \otimes 1,$ then $d_2^{0,2}\left(1\otimes d\right)=0,$ otherwise we will have

$$\mathrm{im}{d}_2^{0,2}=⟨d_2^{0,2}\left(1\otimes d\right)⟩⊈ker{d}_2^{2,1}⟨\tau \otimes \left(c+x\right)⟩,$$

which is a contradiction.

Let us suppose now that $d_2^{0,1}\left(1\otimes c\right)$ and $d_2^{0,1}\left(1\otimes x\right)$ are nontrivial, and $d_2^{0,2}\left(1\otimes d\right)=\tau \otimes c.$ Then, for example,

$$\mathrm{im}{d}_2^{0,4}=⟨\tau \otimes \mathit{cd}⟩\oplus ⟨\tau \otimes c^{2}x⟩⊈ker{d}_2^{2,3}=⟨\tau \otimes {\mathit{xc}}^2⟩\oplus ⟨\tau \otimes c^3⟩,$$

which is a contradiction. Therefore, for $m$ even, we must consider only the cases:

1. $d_2^{0,1}\left(1\otimes c\right)=d_2^{0,1}\left(1\otimes x\right)=\tau \otimes 1$ and $d_2^{0,2}\left(1\otimes d\right)=0;$

   1. $d_2^{0,1}\left(1\otimes c\right)=d_2^{0,1}\left(1\otimes x\right)=0$ and $d_2^{0,2}\left(1\otimes d\right)=\tau \otimes x.$

We will show that both $\left(i\right)$ and $\left(\mathit{ii}\right)$ cannot occur.

If $\left(i\right)$ is true, then for all $j\ge 0,$$k\in \left\{1\cdots ,m\right\}$ and all $l\in \left\{1\cdots ,n\right\}$, we have

$$d_2^{j,k}\left({\tau }^j\otimes c^k\right)=\left\{\begin{array}{ll}0,& \text{if}k\text{is\hspace{0.17em}even},\\ {\tau }^{j+1}\otimes c^{k-1},& \text{if}k\text{is}\mathrm{odd},\end{array}\right.$$

$$d_2^{j,k+1}\left({\tau }^j\otimes c^{k}x\right)=\left\{\begin{array}{ll}{\tau }^{j+1}\otimes c^k,& \text{if}k\text{is\hspace{0.17em}even},\\ {\tau }^{j+1}\otimes \left(c+x\right)c^{k-1},& \text{if}k\text{is}\mathrm{odd},\end{array}\right.$$

$$d_2^{j,2l+k}\left({\tau }^j\otimes c^k{d}^l\right)=\left\{\begin{array}{ll}0,& \text{if}k\text{is\hspace{0.17em}even},\\ {\tau }^{j+1}\otimes c^{k-1}{d}^l,& \text{if}k\text{is}\mathrm{odd},\end{array}\right.$$

and $d_2^{j,2l+1}\left({\tau }^j\otimes {\mathit{xd}}^l\right)={\tau }^{j+1}\otimes d^l.$ Therefore, $E_3^{p,q}\cong \left\{0\right\},$ for all $q\cancel{\equiv }1\left(mod4\right),$ e $p>0.$ However, for $q\equiv 1\left(mod4\right),$$q\ge 5,$ we have

$$E_3^{2j,q}\cong ⟨{\tau }^j\otimes \left(c+x\right)d^{\left(q-1\right)/2}⟩\ne \left\{0\right\}.$$

which contradicts the Quotient Manifold Theorem.

For $\left(\mathit{iii}\right),$ note that it will result in a pattern similar to case (2); therefore, it cannot occur either due to the same arguments. □

In order to investigate the existence of free involutions on a Milnor manifold $X=H\left(m,n\right),$ Dey and Singh [34] showed that if $G={\mathbb{Z}}_2$ acts freely on $X$, with $1<n<m$ and $m\cancel{\equiv }2\left(mod4\right),$ then necessarily $m$ and $n$ must be odd. Furthermore, they construct some examples of such free actions and, in this case, it follows that

$$H^{\ast }\left(X/G;{\mathbb{Z}}_2\right)\cong {\mathbb{Z}}_2\left[x,y,z,w\right]/I,$$E32

where

$$I=⟨z^2,w^2-{\gamma }_1\mathit{zw}-{\gamma }_{2}x-{\gamma }_{3}y,x^{\left(n+1\right)/2}+{\alpha }_0{\mathit{zwx}}^{\left(n-1\right)/2}y+\cdots +{\alpha }_{\frac{n-1}{2}}{\mathit{zwy}}^{\left(n-1\right)/2},$$

$$\left(w+{\beta }_{0}z\right)y^{\left(m-1\right)/2}+\left(w+{\beta }_{1}z\right){\mathit{xy}}^{\left(m-3\right)/2}+\cdots +\left(w+{\beta }_{\frac{n-1}{2}}z\right)x^{\left(n-1\right)/2}{y}^{\left(m-n\right)/2}⟩,$$

with $degx=degy=2,$$degz=degw=1$_, and ${\alpha }_i,{\beta }_i,{\gamma }_i\in {\mathbb{Z}}_2.$

If $G=S^1$ act freely on $X=H\left(m,n\right),$ then $H^{\ast }\left(X/G;{\mathbb{Z}}_2\right)\cong {\mathbb{Z}}_2\left[x,y,w\right]/I,$ where

$$I=⟨x^{\left(n+1\right)/2},{\mathit{wy}}^{\left(m-1\right)/2}+{\mathit{xwy}}^{\left(m-3\right)/2}+\cdots +{\mathit{wx}}^{\left(n-1\right)/2}{y}^{\left(m-n\right)/2},w^2-\mathit{\alpha x}-\mathit{\beta y}⟩,$$

with $degx=degy=2$, $degw=1$, and $\alpha ,\beta \in {\mathbb{Z}}_2.$