# Fixed Point Sets of Involutions on The Product of Three Spheres

Dimpi and Hemant Kumar Singh  Dimpi
Department of Mathematics
University of Delhi
Delhi – 110007, India<EMAIL_ADDRESS>Hemant Kumar Singh
Department of Mathematics
University of Delhi
Delhi – 110007, India<EMAIL_ADDRESS>
###### Abstract.

Let $G=\mathbb{Z}_{2}$ act on a finitistic space $X$ having mod $2$ cohomology
of the product of three spheres
$\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},1\leq n\leq m\leq l$.
In this paper, we have determined the fixed point sets of involutions on $X.$
This generalizes J. C. Su [12] results for involutions on the product of two
sphere $\mathbb{S}^{n}\times\mathbb{S}^{m},n\leq m.$

###### Key words and phrases:

Finitistic space; Fixed point set; Totally nonhomologous to zero; Leray-Serre
spectral sequence

###### 2020 Mathematics Subject Classification:

Primary 57S17; Secondary 55M35

The first author of the paper is supported by SRF of UGC, New Delhi, with
reference no.: 201610039267.

##  1\. Introduction

Let $(G,X)$ be a transformation group, where $G$ is compact Lie group and $X$
is compact Hausdroff space, with the fixed point set $F.$ The study of the
cohomological structure of the fixed point set has been an interesting problem
in transformation groups. J. C. Su [12] gave a conjecture which states that if
$G=\mathbb{Z}_{p},p$ a prime, act on a finitistic space $X$ which satisfies
poincaré duality with respect to Čech cohomology then each component of the
fixed point set also satisfies poincaré duality. Bredon [4] proved Su’s
conjecture, in the case when $X$ is totally nonhomologous to zero in $X_{G}$
(Borel space). In a particular case, Puppe [11] has proved the Bredon’s
conjecture: if $H^{*}(X,\mathbb{Z}_{p})$ generated by $k$ elements as an
algebra, then any component $F_{0}$ of $F,$ $H^{*}(F_{0},\mathbb{Z}_{p})$
generated by at most $k$ elements. Puppe proved that if $X$ is totally
nonhomologous to zero in $X_{G},$ then the number of generators of the
cohomology ring of each component of the fixed point set with
$\mathbb{Z}_{p}$-coefficient is at most the number of generators of
$H^{*}(X,\mathbb{Z}_{p}).$

For the fixed point sets of actions of a group $G$ on the spheres
$\mathbb{S}^{n}$ and on the product of two spheres
$\mathbb{S}^{n}\times\mathbb{S}^{m},$ a lot of work has been done in the
literature. Here, we recall some of the results. The fixed point sets of
$G=\mathbb{Z}_{p},$ $p$ a prime, or $G=\mathbb{T}^{k},k\geq 1$
($\mathbb{T}^{k}$ denotes $k$-dimensional torus) on a finitistic space $X$
having mod $p$ or integral cohomology $n$-sphere are mod $p$ or integral
cohomology $r$-sphere, where $-1\leq r\leq n$ and $n-r$ is even for
$G=\mathbb{Z}_{p},p>2$ and $G=\mathbb{T}^{k}$ (see Chapter III, [3]).
According to this well known result, the fixed point sets of transformation
group $(G,X_{1}\times X_{2})$ (with diagonal action), where $X_{1}$ and
$X_{2}$ has mod $p$ or integral cohomology of spheres, has the same mod $p$ or
integral cohomology algebra of the product of two spheres. Swan [8] gave
sufficient conditions which insure that the fixed point set of
$G=\mathbb{Z}_{p},~{}p$ a prime, actions on mod $p$ cohomology product of two
even dimensional spheres have mod $p$ cohomology of product of two even
dimensional spheres of lower dimensions. Su [12] determined the fixed point
sets of $G=\mathbb{Z}_{p}$ actions on a space $X$ with mod $p$ cohomology
algebra product of spheres $\mathbb{S}^{m}\times\mathbb{S}^{n}.$ Further,
Chang et al. [6] and David [7] discussed the fixed point sets of circle
actions and torus actions on the product of even dimensional spheres and the
product of odd dimensional spheres, respectively. Allday [1] discussed the
fixed point sets of torus actions on cohomology product of three odd
dimensional spheres. In continuation, it seems to be an interesting problem to
determine the fixed point sets of involutions on a finitistic space $X$ having
mod $2$ cohomology of the arbitrary product of three spheres
$\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},$ $1\leq n\leq m\leq
l.$

In this paper, we have determined the possibilities of the fixed point sets of
involutions on a finitistic space $X$ having mod $2$ cohomology of the
arbitrary product of three spheres
$\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},$ $1\leq n\leq m\leq
l.$

##  2\. Preliminaries

We recall some known facts that will be used in this paper. A paracompact
Hausdroff space is called finitistic if every open covering has a finite
dimension refinement. For example: (1) Compact Hausdroff spaces are finitistic
spaces. (2) $\prod_{n\geq 1}(\mathbb{S}^{n}\times\mathbb{R}^{k})$ is
noncompact finitistic space. (3) Paracompact spaces with finite covering
dimension are finitistic spaces. Let $G$ be a finite cyclic group acting on a
finitistic space $X$ with the fixed point set $F.$ Note that $G$ act freely on
contractible space $E_{G}$ (an infinite join of $G$ with itself) and the orbit
map $E_{G}\rightarrow B_{G}$ is the universal G-bundle. The associated
fibration $X\stackrel{{\scriptstyle
i}}{{\hookrightarrow}}X_{G}\stackrel{{\scriptstyle\pi}}{{\rightarrow}}B_{G}$
is called the Borel fibration, where $X_{G}=(X\times E_{G})/G$ (Borel space)
and $B_{G}=E_{G}/G$ (classifying space). If $F$ is nonempty and $x\in F,$ then
$\eta_{x}:B_{G}\hookrightarrow X_{G}$ is cross section of $\pi,$ where
$B_{G}\approx\\{x\\}\times_{G}E_{G},$ and we get
$H^{*}(X_{G})=ker~{}\eta_{x}^{*}\oplus im~{}\pi^{*}.$ The induced homomorphism
$\eta_{x}^{*}$ depends on the component $F_{0}$ of $F$ in which $x$ lies. If
$\alpha\in H^{n}(X_{G})$ such that $\alpha\in Ker~{}\eta_{x}^{*},$ then the
image of $\alpha$ under the restriction of $j:F_{G}\hookrightarrow X_{G}$ on
$(F_{0})_{G}$ does not involve the term of $H^{n}(B_{G})\otimes H^{0}(F_{0}).$
For details, we refer [3, 4].
A space $X$ is said to be totally nonhomologous to zero (TNHZ) in $X_{G}$ with
respect to the ring $R$ if the inclusion map $i:X\hookrightarrow X_{G}$
induces a surjection in the cohomology $i^{*}:H^{*}(X_{G};R)\rightarrow
H^{*}(X;R).$ So, if $X$ is TNHZ in $X_{G},$ then for $a\in H^{n}(X)$ there
exist $\alpha^{\prime}\in H^{n}(X_{G})$ such that $i^{*}(\alpha^{\prime})=a.$
Define $\alpha=\alpha^{\prime}-\pi^{*}\eta_{x}^{*}(\alpha^{\prime}).$ Then it
is easy to see that $i^{*}(\alpha)=a$ and $\eta_{x}^{*}(\alpha)=0.$ We have
used the following propositions to determine the fixed point sets of product
three of spheres.

######  Proposition 2.1.

([3]) Let $G=\mathbb{Z}_{2}$ act on finitistic space $X$ and $\sum$ rk
$H^{i}(X,\mathbb{Z}_{2})<\infty,$ then the following statements are
equivalent:
(a) $X$ is TNHZ in $X_{G}.$
(b) $\sum_{i\geq 0}$ rk $H^{i}(F,\mathbb{Z}_{2})=\sum_{i\geq 0}$ rk
$H^{i}(X,\mathbb{Z}_{2}).$
(c) $\pi_{1}(B_{G})$ acts trivially on $H^{*}(X;\mathbb{Z}_{2})$ and spectral
sequence $E^{r,q}_{2}$ of fibration $X\hookrightarrow X_{G}\rightarrow B_{G}$
degenerates.

######  Proposition 2.2.

([3]) Let $G=\mathbb{Z}_{2}$ act on the finitistic space $X.$ Suppose that $A$
be a closed and invariant subset of $X,$ and $H^{i}(X,A;\mathbb{Z}_{2})=0$ for
$i>n$. Then $j^{*}:H^{k}(X_{G},A_{G};\mathbb{Z}_{2})\rightarrow
H^{k}(F_{G},A_{G}\cap F_{G};\mathbb{Z}_{2})$ is an isomorphism for $k>n$. If
$(X,A)$ is TNHZ in $(X_{G},A_{G})$, then $j^{*}$ is a monomorphism for all
$k.$

######  Proposition 2.3.

([5]) Let $X$ be TNHZ in $X_{G}$ and $\\{\gamma_{j}\\}$ be a set of
homogeneous elements in $H^{*}(X_{G};\mathbb{Z}_{2})$ such that
$\\{i^{*}(\gamma_{j})\\}$ forms $\mathbb{Z}_{2}$-basis of
$H^{*}(X;\mathbb{Z}_{2}).$ Then, $H^{*}(X_{G};\mathbb{Z}_{2})$ is the free
$H^{*}(B_{G})$-module generated by $\\{\gamma_{j}\\}.$

######  Proposition 2.4.

([3]) For a finitistic $G$-space $X,$ where $G$ is cyclic group of prime order
$p,$ we have $\sum_{i\geq j}$ rk $H^{i}(F;\mathbb{Z}_{2})\leq\sum_{i\geq j}$
rk $H^{i}(X;\mathbb{Z}_{2}),$ for each $j.$

######  Definition 2.5.

[3] A space $X$ is said to be poincaré duality space over a field $k$ of
formal dimension $r$ if the following conditions are satisfied:
(a) $H^{*}(X;k)$ is finitely generated.
(b) $H^{i}(X;k)=0$ for $i>r$ and $H^{r}(X;k)\approx k.$
(c) An element $u\in H^{i}(X;k)$ is nonzero if and only if there exists an
element $u^{\prime}\in H^{r-i}(X;k)$ called poincaré dual of $u,$ such that
the cup product $0\neq uu^{\prime}\in H^{r}(X;k).$

Recall that the poincaré duals of basis elements are unique [10].

######  Proposition 2.6.

([3]) Let $X$ be a finitistic poincaré duality space over $\mathbb{Z}_{2}$ of
formal dimension $n$, where $p$ is prime. Let $G=\mathbb{Z}_{2}$ act on $X,$
where $X$ is TNHZ in $X_{G}.$ Then, for each component $F_{0}$ of $F$ is a
poincaré duality space of formal dimension $r\leq n.$ If $r=n,$ then $F=F_{0}$
is connected and the restriction $H^{*}(X;\mathbb{Z}_{2})\rightarrow
H^{*}(F;\mathbb{Z}_{2})$ are isomorphisms.

######  Proposition 2.7.

[11] Let $G=\mathbb{Z}_{2}$ act on a finitistic space $X,$ where $X$ is TNHZ
in $X_{G}.$ If $H^{*}(X;\mathbb{Z}_{2})$ is generated by $k$ elements as an
algebra, then each component of the fixed point set is generated by at most
$k$ elements.

######  Proposition 2.8.

([3]) Let $G=\mathbb{Z}_{2}$ act on a finitistic space $X$ and
$\pi_{1}(B_{G})$ acts nontrivially on $H^{*}(X).$ Then, the $E_{2}$ term of
the Leray-Serre spectral sequence of the fibration $X\stackrel{{\scriptstyle
i}}{{\hookrightarrow}}X_{G}\stackrel{{\scriptstyle\pi}}{{\rightarrow}}B_{G}$
is given by

$E_{2}^{k,i}=\begin{cases}\text{ker $\tau$}&\mbox{for}~{}~{}k=0,\\\ \text{ker
$\tau$/im $\sigma$}&\text{for }k>0,\\\ \end{cases}$

where $\tau=\sigma=1+g^{*},$ $g^{*}$ is induced by a generator $g$ of $G.$

######  Proposition 2.9.

([3]) Let $G=\mathbb{Z}_{2}$ act on a finitistic space $X.$ Then, for any
$a\in H^{n}(X),$ the element $ag^{*}(a)$ is permanent cocycle in spectral
sequence of $X_{G}\rightarrow B_{G}.$

Recall that
$H^{*}(\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l};\mathbb{Z}_{2})=\mathbb{Z}_{2}[a,b,c]/<{a^{2},b^{2},c^{2}}>,$
where deg $a=n,$ deg $b=m$ and deg $c=l,~{}1\leq n\leq m\leq l.$
Throughout the paper, $H^{*}(X)$ will denote the Čech cohomology of a space
$X$ with coefficient group $G=\mathbb{Z}_{2},$ and $X\sim_{2}Y,$ means
$H^{*}(X;\mathbb{Z}_{2})\cong H^{*}(Y;\mathbb{Z}_{2}).$

##  3\. Fixed point Sets of involutions on the product of three spheres

Let $G=\mathbb{Z}_{2}$ act on a finitistic space
$X\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},$ $1\leq
n\leq m\leq l.$ If the fixed point set is empty, then the cohomology ring of
the orbit space $X/G$ has been discussed in [9]. In this section, we determine
the possibilities of nonempty fixed points sets of involutions on $X.$ First,
we determine the possibilities of nonempty connected fixed point sets of
involutions on $X.$

###  3.1. Connected fixed point sets of involutions on $X$

Suppose that $X$ is TNHZ in $X_{\mathbb{Z}_{2}},$ then we have $\sum$ rk
$H^{i}(F)=\sum$ rk $H^{i}(X)=8.$ By Proposition 2.6, we get $F$ is poincaré
duality space of formal dimension $r\leq n+m+l.$ Let $0<q_{1}\leq q_{2}\leq
q_{3}\leq q_{4}\leq q_{5}\leq q_{6}<r$ be the non vanishing dimensions of
$H^{*}(F),$ and ${u_{i}}$ is generator of $q_{i}^{th}$ cohomology group of
$F,1\leq i\leq 6$ and $v$ is generator of $H^{r}(F).$ Assume that $\alpha\in
H^{n}(X_{G}),\beta\in H^{m}(X_{G})$ and $\gamma\in H^{l}(X_{G})$ represents
generators $a,b$ and $c$ of $H^{*}(X,x),$ respectively, such that
$\eta_{x}(\alpha)=\eta_{x}(\beta)=\eta_{x}(\gamma)=0.$ By Proposition 2.3, we
get
$\\{1,\alpha,\beta,\gamma,\alpha\beta,\alpha\gamma,\beta\gamma,\alpha\beta\gamma\\}$
forms a basis for $H^{*}(X_{G})$ over $H^{*}(B_{G})$-module. Note that the
images of generators of $H^{*}(X_{G})$ under the homomorphism
$j^{*}:H^{*}(X_{G})\rightarrow H^{*}(F_{G})$ involves generators of
$H^{*}(F).$ Since $a^{2}=b^{2}=c^{2}=0,$ we get $\alpha^{2},\beta^{2}$ and
$\gamma^{2}$ may be zero or nonzero, and hence $u_{i}u_{j},1\leq i,j\leq 6,$
may be zero or nonzero. Further, if each generator $u_{i}$ is poincaré dual of
itself or any two generators are poincaré dual of itself or any four
generators are poincaré dual of itself, then we get $H^{*}(F)$ must have more
than three generators, which contradicts Proposition 2.7. As $q_{i}\leq
q_{i+1},1\leq i\leq 5,$ we get $u_{1},u_{2}$ and $u_{3}$ are poincaré duals of
$u_{6},u_{5}$ and $u_{4},$ respectively. So, we have
$u_{1}u_{6}=u_{2}u_{5}=u_{3}u_{4}=v\in H^{r}(F).$ Clearly,
$q_{4}=r-q_{3},q_{5}=r-q_{2}$ and $q_{6}=r-q_{1},$ and
$u_{2}u_{6}=u_{3}u_{5}=u_{3}u_{6}=u_{4}u_{5}=u_{4}u_{6}=u_{5}u_{6}=v^{2}=0~{}\&~{}u_{i}v=0,1\leq
i\leq 6.$ Thus, the possibilities of the fixed point set will depend on the
cup products $u_{1}u_{2},u_{1}u_{3},u_{1}u_{4},u_{1}u_{5},u_{2}u_{3}$ and
$u_{2}u_{4}$ which may be zero or nonzero. So, we have considered the cases
depending upon how many cup products out of these six are simultaneously
nonzero.

######  Theorem 3.1.

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},$ where
$1\leq n\leq m\leq l$ and $X$ is TNHZ in $X_{G}.$ If $F$ is connected and none
of the cup products $u_{1}u_{2},u_{1}u_{3},u_{1}u_{4},u_{1}u_{5},u_{2}u_{3}$
and $u_{2}u_{4}$ of generators of $H^{*}(F)$ is nonzero, then
$F\sim_{2}\underset{3}{\\#}\mathbb{P}^{3}(q),q\leq n$ and $q=1,2,4,8.$

###### Proof.

As discussed above, we have $H^{*}(X_{G})$ is the free $H^{*}(B_{G})$-module
with basis
$\\{1,\alpha,\beta,\gamma,\alpha\beta,\alpha\gamma,\beta\gamma,\alpha\beta\gamma\\}.$
Then we write

$j^{*}(\alpha)=A_{1}t^{n-q_{1}}\otimes
u_{1}+A_{2}t^{n-q_{2}}\otimes{u_{2}}+A_{3}t^{n-q_{3}}\otimes{u_{3}}+A_{4}t^{n-q_{4}}\otimes{u_{4}}+$
$A_{5}t^{n-q_{5}}\otimes{u_{5}}+A_{6}t^{n-q_{6}}\otimes{u_{6}}+A_{7}t^{n-r}\otimes{v},$
$j^{*}(\beta)=A_{1}^{{}^{\prime}}t^{m-q_{1}}\otimes
u_{1}+A_{2}^{{}^{\prime}}t^{m-q_{2}}\otimes{u_{2}}+A_{3}^{{}^{\prime}}t^{m-q_{3}}\otimes{u_{3}}+A_{4}^{{}^{\prime}}t^{m-q_{4}}\otimes{u_{4}}+$
$A_{5}^{{}^{\prime}}t^{m-q_{5}}\otimes{u_{5}}+A_{6}^{{}^{\prime}}t^{m-q_{6}}\otimes{u_{6}}+A_{7}^{{}^{\prime}}t^{m-r}\otimes{v},$
$j^{*}(\gamma)=A_{1}^{{}^{\prime\prime}}t^{l-q_{1}}\otimes
u_{1}+A_{2}^{{}^{\prime\prime}}t^{l-q_{2}}\otimes{u_{2}}+A_{3}^{{}^{\prime\prime}}t^{l-q_{3}}\otimes{u_{3}}+A_{4}^{{}^{\prime\prime}}t^{l-q_{4}}\otimes{u_{4}}+$
$A_{5}^{{}^{\prime\prime}}t^{l-q_{5}}\otimes{u_{5}}+A_{6}^{{}^{\prime\prime}}t^{l-q_{6}}\otimes{u_{6}}+A_{7}^{{}^{\prime\prime}}t^{l-r}\otimes{v},$

where all
$A_{i},A^{\prime}_{i},A^{\prime\prime}_{i}\in\mathbb{Z}_{2}~{}\forall~{}1\leq
i\leq 7.$ As all cup products are zero, we get
$u_{4}^{2}=u_{5}^{2}=u_{6}^{2}=0.$ So, we have

$j^{*}(\alpha\beta\gamma)=A_{1}A_{1}^{{}^{\prime}}A_{1}^{{}^{\prime\prime}}t^{l-q_{1}}\otimes
u_{1}^{3}+A_{2}A_{2}^{{}^{\prime}}A_{2}^{{}^{\prime\prime}}t^{l-q_{2}}\otimes{u_{2}^{3}}+A_{3}A_{3}^{{}^{\prime}}A_{3}^{{}^{\prime\prime}}t^{l-q_{3}}\otimes
u_{3}^{3}.$

As $j^{*}$ is injective and $\alpha\beta\gamma\neq 0,$ we get
$j^{*}(\alpha\beta\gamma)\neq 0.$ Thus, at least one of $u_{1}^{3},u_{2}^{3}$
and $u_{3}^{3}$ must be nonzero. Now, we consider following cases:
Case (i) Any one of $u_{1}^{3},u_{2}^{3}$ and $u_{3}^{3}$ is nonzero.
Assume $u_{1}^{3}\neq 0$ and $u_{2}^{3}=u_{3}^{3}=0.$ We must have
$u_{2}^{2}=u_{3}^{2}=0$ and $u_{6}=u_{1}^{2}.$ Thus, we get
$u_{1}^{3}=u_{3}u_{4}=u_{2}u_{5}=v$ is generator of $H^{r}(F).$ Clearly, the
cohomology algebra of the fixed point set $F$ has five generators which
contradicts Proposition 2.7. So, this case not possible.
Case (ii) Any two of $u_{1}^{3},u_{2}^{3}$ and $u_{3}^{3}$ are nonzero.
Assume $u_{1}^{3}\neq 0,u_{2}^{3}\neq 0$ and $u_{3}^{3}=0.$ Then, we must have
$u_{6}=u_{1}^{2},u_{5}=u_{2}^{2},$ $u_{3}^{2}=0$ and
$u_{1}^{3}=u_{2}^{3}=u_{3}u_{4}=v$ generator of $H^{r}(F).$ Clearly,
$q_{1}=q_{2}$ and $q_{5}=q_{6}.$ It is easy to observe that the cohomological
algebra of the fixed point set $F$ has four generators, a contradiction. So,
this case is also not possible.
Case(iii) $u_{1}^{3},u_{2}^{3}$ and $u_{3}^{3}$ are nonzero.
Clearly, $u_{1}^{2},u_{2}^{2}$ and $u_{3}^{2}$ are all nonzero, and we must
have $u_{4}=u_{3}^{2},u_{5}=u_{2}^{2}$ and $u_{6}=u_{1}^{2}.$ Therefore,
$u_{1}^{3}=u_{2}^{3}=u_{3}^{3}$ is generator of $H^{r}(F).$ In this case, we
must have $q_{1}=q_{2}=q_{3}.$ So, we get $u_{1},u_{2},u_{3}$ are generator of
degree $q,$ and $u_{1}^{2},u_{2}^{2},u_{3}^{2}$ are generators of degree $2q.$
Thus, the cohomological algebra of fixed point set $F$ is given by

${\mathbb{Z}_{2}[u_{1},u_{2},u_{3}]}/{<u_{1}^{4},u_{2}^{4},u_{3}^{4},u_{1}^{3}+u_{2}^{3},u_{2}^{3}+u_{3}^{3},u_{1}^{3}+u_{3}^{3},u_{1}u_{2},u_{2}u_{3},u_{1}u_{3}>},$

where deg $u_{1}=$ deg $u_{2}=$ deg $u_{3}=q.$ Thus,
$F\sim_{2}\underset{3}{\\#}\mathbb{P}^{3}(q).$ By Proposition 2.4, we get
$q\leq n,$ where $q=1,2,3,4$ [2]. ∎

######  Theorem 3.2.

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},$ where
$1\leq n\leq m\leq l$ and $X$ is TNHZ in $X_{G}.$ If $F$ is connected and one
cup product from $u_{1}u_{2},u_{1}u_{3},u_{1}u_{4},u_{1}u_{5},u_{2}u_{3}$ and
$u_{2}u_{4}$ of generators of $H^{*}(F)$ is nonzero, then
$F\sim_{2}\mathbb{P}^{3}(r_{1})\\#(\mathbb{P}^{2}(r_{2})\times\mathbb{S}^{r_{3}}),$
where $min~{}\\{r_{1},r_{2}\\}\leq n$ and $r_{1},r_{2}=1,2,4,8.$

###### Proof.

First, assume that $u_{1}u_{2}\neq 0$ and
$u_{1}u_{3}=u_{1}u_{4}=u_{1}u_{5}=u_{2}u_{3}=u_{2}u_{4}=0.$
It is easy to see that either $u_{1}u_{2}=u_{5}$ or $u_{1}u_{2}=u_{6}.$
Suppose that $u_{1}u_{2}=u_{5}.$ Then, we have
$u_{1}u_{6}=u_{2}^{2}u_{1}=u_{3}u_{4}$ is generator of $H^{r}(F).$ As all the
cup products $u_{1}u_{4},u_{1}u_{5},u_{2}u_{3}$ and $u_{2}u_{4}$ are zero, we
must have $u_{2}^{2}=u_{6}$ and $u_{4}^{2}=0.$ Now, we consider four cases:
(i) $u_{1}^{2}=u_{3}^{2}=0$ (ii) $u_{1}^{2}\neq 0,u_{3}^{2}=0$ (iii)
$u_{1}^{2}=0,u_{3}^{2}\neq 0,$ and (iv) $u_{1}^{2}\neq 0,u_{3}^{2}\neq 0.$
Case (i): $u_{1}^{2}=u_{3}^{2}=0.$
In this case, the cohomology algebra of the fixed point set has four
generators, namely, $u_{1},u_{2},u_{3}$ and $u_{4},$ which contradicts
Proposition 2.7.
Case (ii): $u_{1}^{2}\neq 0$ and $u_{3}^{2}=0.$
Since, $u_{1}u_{3}=u_{1}u_{4}=u_{1}u_{5}=0,$ we must have $u_{6}=u_{1}^{2}.$
So, $u_{2}^{2}u_{1}=u_{1}^{3}=u_{3}u_{4}$ is generator of $H^{r}(F).$ Thus,
the cohomological algebra of the fixed point set is generated by four
generators $u_{1},u_{2},u_{3}$ and $u_{4},$ a contradiction.
Case (iii): $u_{1}^{2}=0$ and $u_{3}^{2}\neq 0.$
In this case, we get $u_{4}=u_{3}^{2}$ and $q_{1}=q_{2}=q_{3}.$ Thus,
$H^{2q_{1}}(F)$ is generated by $u_{3}^{2},u_{1}u_{2},u_{2}^{2},$ and
$u_{2}^{2}u_{1}=u_{3}^{3}$ generates $H^{3q_{1}}(F).$ Hence, the cohomological
algebra of the fixed point set is given by

${\mathbb{Z}_{2}[u_{1},u_{2},u_{3}]}/{<u_{1}^{2},u_{2}^{3},u_{3}^{4},u_{3}^{3}+u_{2}^{2}u_{1},u_{1}u_{3},u_{2}u_{3}>},$

where deg $u_{1}=$ deg $u_{2}=$ deg $u_{3}=q_{1}.$ Thus,
$F\sim_{2}\mathbb{P}^{3}(q_{1})\\#(\mathbb{P}^{2}(q_{1})\times\mathbb{S}^{q_{1}}).$
By Proposition 2.4, we get $q_{1}\leq n$ and $q_{1}=1,2,4,8$ [2].
Case (iv): $u_{1}^{2}\neq 0$ and $u_{3}^{2}\neq 0.$
In this case, we get $u_{4}=u_{3}^{2}$ and $u_{6}=u_{1}^{2}=u_{2}^{2}.$ So,
$u_{2}^{2}u_{1}=u_{1}^{3}=u_{3}^{3}$ is generator of $H^{r}(F),$ and
$q_{1}=q_{2}=q_{3}.$ By the change of basis, $d=u_{1}+u_{2},$ we get
$d^{2}=0,$ $u_{2}d=u_{2}u_{1}+u_{2}^{2}\neq 0,u_{6}d=u_{1}u_{6}\neq 0.$
Clearly, $u_{3}d=u_{4}d=u_{5}d=0$ and $u_{2}^{2}d=u_{3}^{2}$ is generator of
$H^{3q_{1}}(F).$ Hence,
$F\sim_{2}\mathbb{P}^{3}(q_{1})\\#(\mathbb{P}^{2}(q_{1})\times\mathbb{S}^{q_{1}}).$
Now, we suppose $u_{1}u_{2}=u_{6}.$ Then,
$u_{1}^{2}u_{2}=u_{2}u_{5}=u_{3}u_{4}$ generates $H^{r}(F).$ Clearly,
$u_{5}=u_{1}^{2}$ and $u_{4}^{2}=0.$ Now, $u_{2}^{2}$ and $u_{3}^{2}$ are
either zero or nonzero. Similarly, as discussed above, if $\\{u_{2}^{2}=0$ and
$u_{3}^{2}=0\\}$ or $\\{u_{2}^{2}\neq 0$ and $u_{3}^{2}=0\\},$ then the
cohomology ring $H^{*}(F)$ has four generators, which is not possible. Now, if
$u_{2}^{2}=0$ and $u_{3}^{2}\neq 0,$ then we must have $u_{4}=u_{3}^{2}.$
Thus, $q_{1}=q_{2}=q_{3},$ and $u_{1}^{2}u_{2}=u_{3}^{3}$ generates
$H^{3q_{1}}(F).$ Hence, the cohomological algebra of the fixed point set is
given by

${\mathbb{Z}_{2}[u_{1},u_{2},u_{3}]}/{<u_{1}^{3},u_{2}^{2},u_{3}^{4},u_{3}^{3}+u_{1}^{2}u_{2},u_{1}u_{3},u_{2}u_{3}>},$

where deg $u_{1}=$ deg $u_{2}=$ deg $u_{3}=q_{1}.$ Thus,
$F\sim_{2}\mathbb{P}^{3}(q_{1})\\#(\mathbb{P}^{2}(q_{1})\times\mathbb{S}^{q_{1}}),$
$q_{1}\leq n.$ Finally, if $u_{2}^{2}$ and $u_{3}^{2}$ both are nonzero, then
we get $u_{4}=u_{3}^{2}$ and $u_{5}=u_{1}^{2}=u_{2}^{2}.$ By the change of
basis, $d=u_{1}+u_{2},$ we get
$F\sim_{2}\mathbb{P}^{3}(q_{1})\\#(\mathbb{P}^{2}(q_{1})\times\mathbb{S}^{q_{1}}),q_{1}\leq
n.$

Similarly, if $u_{1}u_{3}\neq 0$ or $u_{2}u_{3}\neq 0,$ then
$F\sim_{2}\mathbb{P}^{3}(q_{1})\\#(\mathbb{P}^{2}(q_{1})\times\mathbb{S}^{q_{1}}).$
Next, assume that $u_{1}u_{4}\neq 0.$ Here, we must have $u_{1}u_{4}=u_{6},$
and $u_{1}^{2}u_{4}=u_{2}u_{5}=u_{3}u_{4}$ is generator of $H^{r}(F).$
Clearly, $u_{4}^{2}=u_{5}^{2}=0$ and $u_{3}=u_{1}^{2}.$ If $u_{2}^{2}=0,$ then
cohomology algebra of the fixed point set has four generators, which is not
possible. If $u_{2}^{2}\neq 0,$ then $u_{5}=u_{2}^{2}$ and we get
$F\sim_{2}\mathbb{P}^{3}(q_{2})\\#(\mathbb{P}^{2}(q_{1})\times\mathbb{S}^{q_{4}}),$
$q_{1}\leq n$ and $q_{1},q_{2}=1,2,4,8.$

Similarly, if $u_{1}u_{5}\neq 0,$ then
$F\sim_{2}\mathbb{P}^{3}(q_{2})\\#(\mathbb{P}^{2}(q_{1})\times\mathbb{S}^{q_{5}}),$
and if $u_{2}u_{4}\neq 0,$ then
$F\sim_{2}\mathbb{P}^{3}(q_{1})\\#(\mathbb{P}^{2}(q_{2})\times\mathbb{S}^{q_{4}}).$
This completes the proof. ∎

######  Theorem 3.3.

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},$ where
$1\leq n\leq m\leq l$ and $X$ is TNHZ in $X_{G}.$ If $F$ is connected and two
cup products from $u_{1}u_{2},u_{1}u_{3},u_{1}u_{4},u_{1}u_{5},u_{2}u_{3}$ and
$u_{2}u_{4}$ of generators of $H^{*}(F)$ are nonzero, then $F$ must be one of
the following:

  1. (1)

$F\sim_{2}(\mathbb{P}^{5}(r_{1})\\#({\mathbb{S}^{r_{2}}\times\mathbb{S}^{r_{3}}}),$
where $r_{1}=1,2,4,8$ and $min\\{r_{1},r_{2},r_{3}\\}\leq n.$

  2. (2)

$F\sim_{2}(\mathbb{P}^{2}(r_{1})\\#\mathbb{P}^{2}(r_{1}))\times\mathbb{S}^{r_{2}},$
where $r_{1}=1,2,4,8$ and $min\\{r_{1},r_{2}\\}\leq n.$

  3. (3)

$F\sim_{2}{\mathbb{Z}_{2}[x,y,z]}/{<x^{3},u^{3},v^{2},xv+u^{2},yz>},$ where
$(u,v)\in\\{(y,z),(z,y)\\}$ and
$(\mbox{deg~{}}x,\mbox{deg~{}}y,\mbox{deg~{}}z)\in\\{(q_{1},q_{2},q_{3}),(q_{1},q_{2},q_{4}),(q_{2},q_{1},q_{3}),(q_{1},q_{3},q_{5}),(q_{3},q_{1},q_{2})\\}.$

  4. (4)

$F\sim_{2}{\mathbb{Z}_{2}[x,y,z]}/{<x^{2},y^{3},z^{3},y^{2}+z^{2}+ux,yz>},$
where $u\in\\{y,z\\}$ and $(\mbox{deg~{}}x,\\\
\mbox{deg~{}}y,\mbox{deg~{}}z)\in\\{(q_{1},q_{2},q_{3}),(q_{2},q_{1},q_{3}),(q_{3},q_{1},q_{2})\\}.$

###### Proof.

Clearly, we have $6\choose 2$$=15$ cases depending on pairs of nonzero cup
products. It is easy to see that out of these fifteen cases, the following six
nonzero pairs are not possible: (1) $u_{1}u_{2},u_{2}u_{4}$ (2)
$u_{1}u_{3},u_{2}u_{4}$ (3) $u_{1}u_{4},u_{1}u_{5}$ (4)
$u_{1}u_{4},u_{2}u_{3}$ (5) $u_{1}u_{5},u_{2}u_{3},$ and (6)
$u_{1}u_{5},u_{2}u_{4}.$ A contradiction of these cases occurs from the
generators whose cup products are zero. Now, we consider remaining nonzero
pairs of cup products:
Case (7): $u_{1}u_{2}\neq 0~{}\&~{}u_{1}u_{5}\neq 0.$
As $u_{1}u_{3}=u_{1}u_{4}=u_{2}u_{3}=u_{2}u_{4}=0,$ we must have either
$u_{1}u_{5}=u_{1}u_{5}=u_{6}$ or {$u_{1}u_{2}=u_{5}$ and $u_{1}u_{5}=u_{6}$}.
First, suppose that $u_{1}u_{2}=u_{1}u_{5}=u_{6}.$ Then,
$u_{1}^{2}u_{2}=u_{1}^{2}u_{5}=u_{2}u_{5}=u_{3}u_{4}$ is generator of
$H^{r}(F).$ So, $u_{1}^{2}\neq 0.$ As $u_{1}u_{3}=u_{1}u_{4}=0,$ we have
either $u_{1}^{2}=u_{2}$ or $u_{1}^{2}=u_{5}.$ Further, in either cases,
$u_{2}$ or $u_{5}$ have two poincaré duals, a contradiction.
Next, suppose that $u_{1}u_{2}=u_{5}$ and $u_{1}u_{5}=u_{6}.$ Then, we have
$u_{1}u_{5}=u_{1}^{2}u_{2}$ and $u_{1}^{3}u_{2}=u_{2}^{2}u_{1}=u_{3}u_{4}$ is
generator of $H^{r}(F).$ Since, $u_{1}u_{3}=u_{1}u_{4}=0,$ we have
$u_{2}=u_{1}^{2}.$ Thus, $u_{1}u_{2}=u_{1}^{3},u_{1}^{2}u_{2}=u_{1}^{4}$ and
$u_{1}^{5}=u_{3}u_{4}$ are generators of $H^{q_{5}}(F),H^{q_{6}}(F)$ and
$H^{r}(F),$ respectively. Clearly, $u_{4}^{2}=0.$ If $u_{3}^{2}\neq 0,$ then
we must have $u_{4}=u_{3}^{2},$ and hence
$F\sim_{2}\mathbb{P}^{5}(q_{1})\\#\mathbb{P}^{3}(q_{3}),$ where
$5q_{1}=3q_{3}.$ By Adam [2], this is not possible. If $u_{3}^{2}=0,$ then we
get
$F\sim_{2}\mathbb{P}^{5}(q_{1})\\#(\mathbb{S}^{q_{3}}\times\mathbb{S}^{q_{4}}).$
By Proposition 2.4, $q_{1}\leq n.$ This realizes possibility (1).

Similarly, for Case (8) $u_{1}u_{3}\neq 0~{}\&~{}u_{1}u_{4}\neq 0,$ we get
$F\sim_{2}\mathbb{P}^{5}(q_{1})\\#(\mathbb{S}^{q_{2}}\times\mathbb{S}^{q_{5}}),$
and for Case (9) $u_{2}u_{3}\neq 0~{}\&~{}u_{2}u_{4}\neq 0,$ we get
$F\sim_{2}\mathbb{P}^{5}(q_{6})\\#(\mathbb{S}^{q_{1}}\times\mathbb{S}^{q_{6}}).$
So, these cases also realizes possibility (1).
Case (10): $u_{1}u_{4}\neq 0~{}\&~{}u_{2}u_{4}\neq 0.$
In this case, we must have $u_{2}u_{4}=u_{5}$ and $u_{1}u_{4}=u_{6}.$ Thus, we
have $u_{1}^{2}u_{4}=u_{2}^{2}u_{4}=u_{3}u_{4}$ is a generator of $H^{r}(F).$
As $u_{1}u_{2}=u_{1}u_{3}=u_{1}u_{5}=u_{2}u_{3}=0,$ we get
$u_{1}^{2}=u_{2}^{2}=u_{3}.$ Clearly, $q_{1}=q_{2}$ and $u_{4}^{2}=0.$ Hence,
$F\sim_{2}(\mathbb{P}^{2}(q_{1})\\#\mathbb{P}^{2}(q_{1}))\times\mathbb{S}^{q_{4}},q_{1}\leq
n.$ This realizes possibility (2).
Case (11): $u_{1}u_{2}\neq 0$ and $u_{1}u_{4}\neq 0.$
In this case, we get either $u_{1}u_{4}=u_{1}u_{5}=u_{6}$ or
{$u_{1}u_{2}=u_{5}$ and $u_{1}u_{4}=u_{6}$}. If $u_{1}u_{2}=u_{1}u_{4}=u_{6},$
then by uniquesness of poincaré dual, this is not possible.
Next, Suppose that $u_{1}u_{2}=u_{5}$ and $u_{1}u_{4}=u_{6}.$ Then, we have
$u_{1}^{2}u_{4}=u_{2}^{2}u_{1}=u_{3}u_{4}$ generates of $H^{r}(F).$ Clearly,
$u_{1}^{2}=u_{3},$ $u_{2}^{2}=u_{6}$ and $u_{4}^{2}=0.$ Consequently,
$u_{1}^{3}=u_{2}^{3}=u_{4}^{2}=0.$ Hence, the cohomology ring of the fixed
point set is isomorphic to

${\mathbb{Z}_{2}[u_{1},u_{2},u_{4}]}/{<u_{1}^{3},u_{2}^{3},u_{4}^{2},u_{1}u_{4}+u_{2}^{2},u_{2}u_{4}>}$

where deg $u_{1}=q_{1},$ deg $u_{2}=q_{2}$ and deg $u_{4}=q_{4}.$ This
realizes possibility (3) for $(u,v)=(y,z),$ $x=u_{1},y=u_{2}$ and $z=u_{4}.$

Similarly, for Case (12) $u_{1}u_{3}\neq 0~{}\&~{}u_{1}u_{5}\neq 0,$ the
cohomology ring of the fixed point set is isomorphic to

${\mathbb{Z}_{2}[u_{1},u_{3},u_{5}]}/{<u_{1}^{3},u_{3}^{3},u_{5}^{2},u_{1}u_{5}+u_{3}^{2},u_{3}u_{5}>}$

where deg $u_{1}=q_{1},$ deg $u_{3}=q_{3}$ and deg $u_{5}=q_{5}.$ This also
realizes possibility (3) for $(u,v)=(y,z),$ $x=u_{1},y=u_{3}$ and $z=u_{5}.$
Case (13): $u_{1}u_{2}\neq 0$ and $u_{1}u_{3}\neq 0.$
In this case, we have following subcases (i) $u_{1}u_{2}=u_{1}u_{3}=u_{6}$
(ii) $u_{1}u_{2}=u_{5}~{}\&~{}u_{1}u_{3}=u_{4}$ (iii)
$u_{1}u_{2}=u_{5}~{}\&~{}u_{1}u_{3}=u_{6},$ and (iv)
$u_{1}u_{2}=u_{6}~{}\&~{}u_{1}u_{3}=u_{4}.$ Note that subcase (i) is not
possible.
Subcase (ii): $u_{1}u_{2}=u_{5}$ and $u_{1}u_{3}=u_{4}.$ Then, we have
$u_{2}^{2}u_{1}=u_{3}^{2}u_{1}=u_{1}u_{6}$ generates $H^{r}(F).$ It is easy to
see that $u_{6}=u_{3}^{2}=u_{2}^{2}$ is generator of $H^{q_{6}}(F).$ Then,
$q_{2}=q_{3}.$ Now, if $u_{1}^{2}=0,$ then
$F\sim_{2}(\mathbb{P}^{3}(q_{2})\\#\mathbb{P}^{3}(q_{2}))\times\mathbb{S}^{q_{1}}.$
If $u_{1}^{2}\neq 0,$ then we must have $u_{1}^{2}=u_{6}.$ So, we get
$q_{1}=q_{2}=q_{3}.$ By the change of basis $d=u_{1}+u_{2},$ we get
$F\sim_{2}(\mathbb{P}^{3}(q_{1})\\#\mathbb{P}^{3}(q_{1}))\times\mathbb{S}^{q_{1}}.$
These realizes possibility (2).
Subcase (iii): $u_{1}u_{2}=u_{5}$ and $u_{1}u_{3}=u_{6}.$ Then,
$u_{1}^{2}u_{3}=u_{2}^{2}u_{1}=u_{3}u_{4}$ generates $H^{r}(F).$ Clearly,
$u_{4}=u_{1}^{2}$ and $u_{1}u_{3}=u_{2}^{2}.$ Consequently,
$u_{1}^{3}=u_{2}^{3}=0.$ If $u_{3}^{2}=0,$ then the cohomology ring of the
fixed point set is isomorphic to

${\mathbb{Z}_{2}[u_{1},u_{2},u_{3}]}/{<u_{1}^{3},u_{2}^{3},u_{3}^{2},u_{1}u_{3}+u_{2}^{2},u_{2}u_{3}>},$

where deg $u_{1}=q_{1},$ deg $u_{2}=q_{2}$ and deg $u_{3}=q_{3}.$ This
realizes possibility (3) for $(u,v)=(y,z),$ $x=u_{1},y=u_{2}$ and $z=u_{3}.$
If $u_{3}^{2}\neq 0,$ then we get $u_{4}=u_{3}^{2}=u_{1}^{2}.$ Thus,
$q_{1}=q_{2}=q_{3},$ and by change of basis $d=u_{1}+u_{3},$ we get
$\\{d,u_{2},u_{3}\\}$ generates $H^{q_{1}}(F),$
$\\{u_{3}d,u_{2}d,u_{2}^{2}=u_{3}^{2}+u_{3}d\\}$ generates $H^{2q_{1}}(F)$ and
$u_{2}^{2}d=u_{3}^{2}d$ generates $H^{3q_{1}}(F).$ Thus, the cohomology ring
$H^{*}(F)$ is isomorphic to

${\mathbb{Z}_{2}[d,u_{2},u_{3}]}/{<d^{2},u_{2}^{3},u_{3}^{3},u_{3}d+u_{2}^{2}+u_{3}^{2},u_{2}u_{3}>},$

where deg $d=$deg $u_{2}=$deg $u_{3}=q_{1}.$ This realizes possibility (4) for
$u=z,$ $x=d,y=u_{2}$ and $z=u_{3}.$
Subcase (iv): $u_{1}u_{2}=u_{6}$ and $u_{1}u_{3}=u_{4}.$ Then,
$u_{1}^{2}u_{2}=u_{3}^{2}u_{1}$ generates $H^{r}(F).$ It is easy to see that
$u_{5}=u_{1}^{2}$ and $u_{6}=u_{3}^{2}.$ If $u_{2}^{2}=0,$ then the cohomology
ring of the fixed point set is isomorphic to

${\mathbb{Z}_{2}[u_{1},u_{2},u_{3}]}/{<u_{1}^{3},u_{2}^{2},u_{3}^{3},u_{1}u_{2}+u_{3}^{2},u_{2}u_{3}>},$

where deg $u_{1}=q_{1},$ deg $u_{2}=q_{2}$ and deg $u_{3}=q_{3}.$ This
realizes possibility (3) for $(u,v)=(z,y)$ and
$x=u_{1},y=u_{2}~{}\&~{}z=u_{3}.$ Now, if $u_{2}^{2}\neq 0,$ then
$u_{5}=u_{1}^{2}=u_{2}^{2},$ and again by the change of basis $d=u_{1}+u_{2},$
we get $H^{*}(F)$ is isomorphic to

${\mathbb{Z}_{2}[d,u_{2},u_{3}]}/{<d^{2},u_{2}^{3},u_{3}^{3},u_{2}d+u_{2}^{2}+u_{3}^{2},u_{2}u_{3}>},$

where deg $d=$ deg $u_{2}=$ deg $u_{3}=q_{1}.$ This realizes possibility (4)
for $u=y,$ $x=d,y=u_{2}$ and $z=u_{3}.$

Similarly, for Case (14) $u_{1}u_{3}\neq 0~{}\&~{}u_{2}u_{3}\neq 0,$ we
realize possibility (2), possibility (3) for $(u,v)=(y,z)~{}\&~{}(z,y);$
$x=u_{2},y=u_{1}~{}\&~{}z=u_{3},$ and possibility (3) for $u=y~{}\&~{}z;$
$x=u_{2},y=d~{}\&~{}z=u_{3},$ and for Case (15) $u_{1}u_{2}\neq
0~{}\&~{}u_{2}u_{3}\neq 0,$ we realize possibility (2), possibility (3) for
$(u,v)=(y,z)~{}\&~{}(z,y);$ $x=u_{3},y=u_{2}~{}\&~{}z=u_{1},$ and possibility
(3) for $u=y~{}\&~{}z;$ $x=u_{1},y=u_{2}~{}\&~{}z=d.$ ∎

######  Theorem 3.4.

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},$ where
$1\leq n\leq m\leq l$ and $X$ is TNHZ in $X_{G}.$ If $F$ is connected and
three cup products from
$u_{1}u_{2},u_{1}u_{3},u_{1}u_{4},u_{1}u_{5},u_{2}u_{3}$ and $u_{2}u_{4}$ of
generators of $H^{*}(F)$ are nonzero. Then, $F$ must be one of the following:

  1. (1)

$F\sim_{2}{\mathbb{Z}_{2}[x,y]}/{<x^{6},y^{3},x^{4}+y^{2}>},$ where deg
$x=q_{1}$ and deg $y\in\\{q_{2},q_{3}\\}.$

  2. (2)

$F\sim_{2}{\mathbb{Z}_{2}[x,y,z]}/{<x^{2+i},y^{2+j},z^{2+k},a_{i}(yz+x^{i}),a_{j}(xz+y^{j}),a_{k}(xy+z^{k})>},$
where $i,j,k\in\\{0,2\\}$ or $i=j=k=1,a_{0}=0,a_{1}=a_{2}=1$ and deg
$x=q_{1},$ deg $y=q_{2}~{}\&$ deg $z\in\\{q_{3},q_{4}\\}.$ In particular, for
$i=j=k=0,$
$F\sim_{2}\mathbb{S}^{r_{1}}\times\mathbb{S}^{r_{2}}\times\mathbb{S}^{r_{3}},$
where $r_{1}\leq n,r_{2}\leq m$ and $r_{3}\leq l.$

###### Proof.

Here, we have ${6\choose 3}=20$ cases depending on which three nonzero cup
products are taken from
$u_{1}u_{2},u_{1}u_{3},u_{1}u_{4},u_{1}u_{5},u_{2}u_{3}$ and $u_{2}u_{4}.$ Out
of these twenty cases, the following four cases are possible: (1)
$u_{1}u_{2}\neq 0,u_{1}u_{3}\neq 0~{}\&~{}u_{1}u_{4}\neq 0,$ (2)
$u_{1}u_{2}\neq 0,u_{1}u_{3}\neq 0~{}\&~{}u_{1}u_{5}\neq 0,$ (3)
$u_{1}u_{2}\neq 0,u_{1}u_{3}\neq 0~{}\&~{}u_{2}u_{3}\neq 0,$ and (4)
$u_{1}u_{2}\neq 0,u_{1}u_{4}\neq 0~{}\&~{}u_{2}u_{4}\neq 0.$
Case (1) $u_{1}u_{2}\neq 0,u_{1}u_{3}\neq 0$ and $u_{1}u_{4}\neq 0.$
In this case, we must have $u_{1}u_{3}=u_{4},u_{1}u_{2}=u_{5},$ and
$u_{1}u_{4}=u_{6}.$ Thus, $u_{1}u_{4}=u_{1}^{2}u_{3}$ and
$u_{1}u_{2}^{2}=u_{1}^{3}u_{3}=u_{1}u_{3}^{2}$ is the generator of $H^{r}(F).$
So, $u_{3}^{2},u_{2}^{2}$ and $u_{1}^{3}$ are nonzero. Since, $u_{1}u_{5}=0,$
we must get $u_{3}=u_{1}^{2}.$ Thus, $u_{4}=u_{1}^{3},$ $u_{6}=u_{1}^{4},$ and
$u_{1}^{5}=u_{2}^{2}u_{1}.$ Hence, the cohomology ring $H^{*}(F)$ is
isomorphic to

${\mathbb{Z}_{2}[u_{1},u_{2}]}/{<u_{1}^{6},u_{2}^{3},u_{2}^{2}+u_{1}^{4}>,}$

where deg $u_{1}=q_{1}$ and deg $u_{2}=q_{2}.$ This realize possibility (1)
for $x=u_{1},y=u_{2}$ and deg $y=q_{2}.$
Similarly, for Case (2) $u_{1}u_{2}\neq 0,u_{1}u_{3}\neq 0,$ $u_{1}u_{5}\neq
0,$ we realizes possibility (1) for $x=u_{1},y=u_{3}$ and deg $y=q_{3}.$
Case(3) $u_{1}u_{2}\neq 0,u_{1}u_{3}\neq 0$ and $u_{2}u_{3}\neq 0.$
As the cup products $u_{1}u_{4},u_{1}u_{5}$ and $u_{2}u_{3}$ are zero, only
the following subcases are possible: (i)
$u_{1}u_{2}=u_{4},u_{1}u_{3}=u_{5}~{}\&~{}u_{2}u_{3}=u_{6},$ (ii)
$u_{1}u_{2}=u_{5},u_{1}u_{3}=u_{6}~{}\&~{}u_{2}u_{3}=u_{4},$ and (iii)
$u_{1}u_{2}=u_{6},u_{1}u_{3}=u_{4}~{}\&~{}u_{2}u_{3}=u_{5}.$
Subcase (i) $u_{1}u_{2}=u_{4},u_{1}u_{3}=u_{5}$ and $u_{2}u_{3}=u_{6}.$ Then,
$u_{1}u_{2}u_{3}$ is the generator of $H^{r}(F).$ Now, if
$u_{1}^{2},u_{2}^{2}$ and $u_{3}^{2}$ are all zero, then
$F\sim_{2}\mathbb{S}^{q_{1}}\times\mathbb{S}^{q_{2}}\times\mathbb{S}^{q_{3}}.$
By Proposition 2.4, we get $r_{1}\leq n,r_{2}\leq m$ and $r_{3}\leq l.$ This
realizes possibility (2) for $i=j=k=0.$

Suppose that $u_{1}^{2},u_{2}^{2}$ and $u_{3}^{2}$ are all nonzero. Then, we
must get $u_{4}=u_{3}^{2},u_{5}=u_{2}^{2}$ and $u_{6}=u_{3}^{2}.$ Clearly,
$u_{1}^{3}=u_{2}^{3}=u_{3}^{3}=u_{1}u_{2}u_{3}$ and
$u_{1}^{4}=u_{2}^{4}=u_{3}^{4}=0.$ Thus, the cohomology ring $H^{*}(F)$ is
isomorphic to

${\mathbb{Z}_{2}[u_{1},u_{2},u_{3}]}/{<u_{1}^{4},u_{2}^{4},u_{3}^{4},u_{1}u_{2}+u_{3}^{2},u_{1}u_{3}+u_{2}^{2},u_{2}u_{3}+u_{1}^{2}>},$

where deg $u_{1}=q_{1},$ deg $u_{2}=q_{2}$ and deg $u_{3}=q_{3}.$ This
realizes possibility (2) for $i=j=k=2.$

Suppose that any two of $u_{1}^{2},u_{2}^{2}$ and $u_{3}^{2}$ are nonzero.
Assume $u_{1}^{2}$ and $u_{2}^{2}$ are nonzero and $u_{3}^{2}=0.$ Then,
$u_{5}=u_{2}^{2}$ and $u_{6}=u_{1}^{2}.$ So, the cohomology ring $H^{*}(F)$ is
isomorphic to

${\mathbb{Z}_{2}[u_{1},u_{2},u_{3}]}/{<u_{1}^{4},u_{2}^{4},u_{3}^{2},u_{1}u_{3}+u_{2}^{2},u_{2}u_{3}+u_{1}^{2}>},$

where deg $u_{1}=q_{1},$ deg $u_{2}=q_{2}$ and deg $u_{3}=q_{3}.$ This
realizes possibility (2) for $i=j=2~{}\&~{}k=0.$

Finally, suppose that one of $u_{1}^{2},u_{2}^{2}$ and $u_{3}^{2}$ is nonzero.
Assume $u_{1}^{2}\neq 0$ and $u_{2}^{2}=u_{3}^{2}=0.$ Then, $u_{6}=u_{1}^{2}.$
So, the cohomology ring $H^{*}(F)$ is isomorphic to

${\mathbb{Z}_{2}[u_{1},u_{2},u_{3}]}/{<u_{1}^{4},u_{2}^{2},u_{3}^{2},u_{2}u_{3}+u_{1}^{2}>},$

where deg $u_{1}=q_{1},$ deg $u_{2}=q_{2}$ and deg $u_{3}=q_{3}.$ This
realizes possibility (2) for $i=2~{}\&~{}j=k=0.$
Subcase (ii) $u_{2}u_{3}=u_{4},u_{1}u_{2}=u_{5}$ and $u_{1}u_{3}=u_{6}.$ Then,
$u_{1}^{2}u_{3}=u_{2}^{2}u_{1}=u_{3}^{2}u_{2}$ generates $H^{r}(F).$ Clearly,
$u_{1}^{2}=u_{4},u_{3}^{2}=u_{5}$ and $u_{2}^{2}=u_{6}.$ Thus, the cohomology
ring $H^{*}(F)$ is isomorphic to

${\mathbb{Z}_{2}[u_{1},u_{2},u_{3}]}/{<u_{1}^{3},u_{2}^{3},u_{3}^{3},u_{2}u_{3}+u_{1}^{2},u_{1}u_{2}+u_{3}^{2},u_{1}u_{3}+u_{2}^{2}>},$

where deg $u_{1}=q_{1},$ deg $u_{2}=q_{2}$ and deg $u_{3}=q_{3}.$ This
realizes possibility (2) for $i=j=k=1.$
Subcase (iii) $u_{2}u_{3}=u_{5},u_{1}u_{2}=u_{6}$ and $u_{1}u_{3}=u_{4}.$ In
this subcase, we get same cohomology ring $H^{*}(F)$ as in subcase (ii).
Case (4) $u_{1}u_{2}\neq 0,u_{1}u_{4}\neq 0,u_{2}u_{4}\neq 0.$
In this case, we must have $u_{1}u_{2}=u_{3},u_{1}u_{4}=u_{5}$ and
$u_{2}u_{4}=u_{6}.$ So, $u_{1}u_{2}u_{4}$ is generator of $H^{r}(F).$ If
$u_{4}^{2}\neq 0,$ then either $u_{4}^{2}=u_{5}$ or $u_{6},$ it fails the
uniqueness of poincaré dual of a generating element. So, $u_{4}^{2}=0.$ Now,
if $u_{1}^{2}=u_{2}^{2}=0,$ then
$F\sim_{2}\mathbb{S}^{q_{1}}\times\mathbb{S}^{q_{2}}\times\mathbb{S}^{q_{3}}.$
This realizes possibility (2) for $i=j=k=0.$ If both $u_{1}^{2}$ and
$u_{2}^{2}$ are nonzero, then we get $u_{6}=u_{1}^{2}$ and $u_{5}=u_{2}^{2}.$
Thus, the cohomology ring $H^{*}(F)$ is isomorphic to

${\mathbb{Z}_{2}[u_{1},u_{2},u_{4}]}/{<u_{1}^{4},u_{2}^{4},u_{4}^{2},u_{1}u_{4}+u_{2}^{2},u_{2}u_{4}+u_{1}^{2}>},$

where deg $u_{1}=q_{1},$ deg $u_{2}=q_{2}$ and deg $u_{4}=q_{4}.$ This
realizes possibility (2) for $i=j=2~{}\&~{}k=0.$
Finally, if any one of $u_{1}^{2}$ and $u_{2}^{2}$ is nonzero, say
$u_{1}^{2}\neq 0,$ then $u_{6}=u_{1}^{2}.$ Thus, the cohomology ring
$H^{*}(F)$ is isomorphic to

${\mathbb{Z}_{2}[u_{1},u_{2},u_{4}]}/{<u_{1}^{4},u_{2}^{2},u_{4}^{2},u_{2}u_{4}+u_{1}^{2}>},$

where deg $u_{1}=q_{1},$ deg $u_{2}=q_{2}$ and deg $u_{4}=q_{4}.$ This
realizes possibility (2) for $i=2~{}\&~{}j=k=0.$
It is easy to observe that the remaining sixteen cases can be discarded by the
zero cup products or by the uniqueness of poincaré duals. ∎

######  Theorem 3.5.

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},$ where
$1\leq n\leq m\leq l$ and $X$ is TNHZ in $X_{G}.$ If $F$ is connected and any
four cup products from
$u_{1}u_{2},u_{1}u_{3},u_{1}u_{4},u_{1}u_{5},u_{2}u_{3}$ and $u_{2}u_{4}$ of
generators of $H^{*}(F)$ are nonzero. Then, $F$ must be one of the following:

  1. (1)

$F\sim_{2}\mathbb{P}^{3}(r_{1})\times\mathbb{S}^{r_{2}},$ where
$r_{1}=1,2,4,8.$

  2. (2)

$F\sim_{2}{\mathbb{Z}_{2}[x,y]}/{<x^{4},y^{4},x^{3}+y^{2},y^{2}x>},$ where deg
$x=q_{1}$ and deg $y=q_{2}.$

###### Proof.

We consider $6\choose 4$$=15$ cases, depending on which pair of cup products
$u_{1}u_{2},u_{1}u_{3},\\\ u_{1}u_{4},u_{1}u_{5},u_{2}u_{3},u_{2}u_{4}$ are
zero and rest four cup products are nonzero. Out of these fifteen cases, the
following six cases are possible: (i) $u_{1}u_{3}=u_{2}u_{3}=0,$ (ii)
$u_{1}u_{3}=u_{1}u_{5}=0,$ (iii) $u_{1}u_{4}=u_{1}u_{5}=0,$ (iv)
$u_{1}u_{4}=u_{2}u_{4}=0,$ (v) $u_{1}u_{5}=u_{2}u_{3}=0,$ and (vi)
$u_{1}u_{5}=u_{2}u_{4}=0.$
Case (i) $u_{1}u_{3}=u_{2}u_{3}=0.$
We have $u_{1}u_{2},u_{1}u_{4},u_{1}u_{5}$ and $u_{2}u_{4}$ are nonzero cup
products. Clearly, $u_{1}u_{5}=u_{6}.$ By the uniqueness of poincaré dual of
$u_{1}^{2},$ we must have $u_{1}u_{4}=u_{5}.$ Similarly, we get
$u_{1}u_{2}=u_{3}$ and $u_{1}u_{5}=u_{2}u_{4}=u_{6}.$ Then,
$u_{1}^{3}u_{4}=u_{1}u_{2}u_{4}$ is the generator of $H^{r}(F).$ It is easy to
see that, $u_{2}=u_{1}^{2}$ and $u_{4}^{2}=0.$ Thus, $0=u_{1}u_{3}=u_{1}^{4},$
and hence $F\sim_{2}\mathbb{P}^{3}(q_{1})\times\mathbb{S}^{q_{4}}.$ Note that
$q_{1}=1,2,4,8.$ This realize possibility (1)
Similarly, for Case (ii) & (iii), we get
$F\sim_{2}\mathbb{P}^{3}(q_{2})\times\mathbb{S}^{q_{1}},$ for Case (iv), we
get $F\sim_{2}\mathbb{P}^{3}(q_{1})\times\mathbb{S}^{q_{3}},$ and for Case
(v), we get $F\sim_{2}\mathbb{P}^{3}(q_{1})\times\mathbb{S}^{q_{2}}.$ This
also realizes possibility (1).
Case (vi) $u_{1}u_{5}=u_{2}u_{4}=0.$
As $u_{1}u_{5}=u_{2}u_{4}=0,$ we get $u_{1}u_{4}=u_{6},$
$u_{1}u_{2}=u_{4},u_{1}u_{3}=u_{5}$ and $u_{1}u_{4}=u_{2}u_{3}=u_{6}.$ Thus,
$u_{1}^{3}u_{2}=u_{1}u_{2}u_{3}$ generates $H^{r}(F).$ Clearly,
$u_{1}^{2}=u_{3}.$ So, if $u_{2}^{2}=0,$ then
$F\sim_{2}\mathbb{P}^{3}(q_{1})\times\mathbb{S}^{q_{2}}.$ This realizes
posibility (1). If $u_{2}^{2}\neq 0,$ then we must have
$u_{5}=u_{1}^{3}=u_{2}^{2}$ and $u_{1}^{3}u_{2}=u_{2}^{3}$ generates
$H^{r}(F).$ We have $0=u_{1}u_{5}=u_{1}^{4}$ and
$0=u_{2}u_{4}=u_{2}^{2}u_{1}.$ Hence, the cohomology ring $H^{*}(F)$ is
isomorphic to

${\mathbb{Z}_{2}[u_{1},u_{2}]}/{<u_{1}^{4},u_{2}^{4},u_{1}^{3}+u_{2}^{2},u_{2}^{2}u_{1}>},$

where deg $u_{1}=q_{1}$ and deg $u_{2}=q_{2}.$ This realizes possibility (2).
It is easy to observe that the remaining nine cases are not possible. ∎

######  Theorem 3.6.

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},$ where
$1\leq n\leq m\leq l$ and $X$ is TNHZ in $X_{G}.$ If $F$ is connected and any
five cup products from
$u_{1}u_{2},u_{1}u_{3},u_{1}u_{4},u_{1}u_{5},u_{2}u_{3}$ and $u_{2}u_{4}$ of
generators of $H^{*}(F)$ are nonzero. Then,
$F\sim_{2}\mathbb{P}^{3}(r_{1})\times\mathbb{S}^{r_{1}},$ where
$r_{1}=1,2,4,8.$

###### Proof.

Here, we have six cases depending on which cup product out of
$u_{1}u_{2},u_{1}u_{3},u_{1}u_{4},\\\ u_{1}u_{5},u_{2}u_{3}$ and $u_{2}u_{4}$
is zero, and rest five cup products are nonzero. Clearly, the only possible
case is $u_{1}u_{5}=0.$ We have either $u_{2}u_{4}=u_{6}$ or
$u_{2}u_{4}=u_{5}.$
If $u_{2}u_{4}=u_{6},$ then we must have
$u_{1}u_{3}=u_{6},u_{1}u_{4}=u_{2}u_{3}=u_{5}$ and $u_{1}u_{2}=u_{3}.$ Thus,
$u_{1}^{3}u_{2}=u_{2}^{3}u_{1}$ generates $H^{r}(F).$ Clearly,
$u_{1}^{2}=u_{2}^{2}=u_{4}.$ So, we get $u_{1}^{4}=u_{2}^{4}=0.$ By the change
of basis $d=u_{1}+u_{2},$ we get $\\{d,u_{2}\\}$ generates $H^{q_{1}}(F),$
$\\{u_{2}d,u_{2}^{2}\\}$ generates $H^{q_{3}}(F),$
$\\{u_{2}^{2}d,u_{2}^{3}\\}$ generates $H^{q_{5}}(F),$ and $u_{2}^{3}d$
generates $H^{r}(F).$ Hence,
$F\sim_{2}\mathbb{P}^{3}(q_{1})\times\mathbb{S}^{q_{1}}.$
Now, if $u_{2}u_{4}=u_{5},$ then we must have
$u_{1}u_{3}=u_{5},u_{1}u_{4}=u_{2}u_{3}=u_{6},$ and $u_{1}u_{2}=u_{4}.$ So,
$u_{1}^{3}u_{2}=u_{2}^{3}u_{1}$ generates $H^{r}(F).$ Simiarly, by the change
of basis, we get $F\sim_{2}\mathbb{P}^{3}(q_{1})\times\mathbb{S}^{q_{1}}.$
Hence, our claim. ∎

If all cup products $u_{1}u_{2},u_{1}u_{3},u_{1}u_{4},u_{1}u_{5},u_{2}u_{3}$
and $u_{2}u_{4}$ are nonzero, then we get
$u_{1}u_{2}=u_{3},u_{1}u_{3}=u_{4},u_{1}u_{4}=u_{2}u_{3}=u_{5},u_{1}u_{5}=u_{2}u_{4}=u_{6}.$
So, we must have $u_{1}^{2}=u_{2},$ and hence $u_{1}^{7}=0.$ Thus, we have:

######  Theorem 3.7.

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},$ where
$1\leq n\leq m\leq l$ and $X$ is TNHZ in $X_{G}.$ If $F$ is connected and all
cup products $u_{1}u_{2},u_{1}u_{3},u_{1}u_{4},u_{1}u_{5},u_{2}u_{3}$ and
$u_{2}u_{4}$ of generators of $H^{*}(F)$ are nonzero, then
$F\sim_{2}\mathbb{P}^{7}(q),~{}~{}~{}q=1,2,4,8.$

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n_{1}}\times\mathbb{S}^{n_{2}}\times\cdots\mathbb{S}^{n_{k}},$
where $1\leq n_{1}\leq n_{2}\leq\cdots\leq n_{k}$ and $X$ is TNHZ in $X_{G}.$
Then, rk$H^{*}(F)=2^{k}.$ Suppose that $F$ is connected. Let $0<q_{1}\leq
q_{2}\leq\cdots\leq q_{2^{k}-2}<r$ be non vanising dimensions of $H^{*}(F)$
and $u_{i}$ is generator of $q_{i}^{th}$ cohomology group, and $v$ is
generator of $H^{r}(F).$ WLOG, we may assume that $u_{1},\cdots,u_{2^{k-1}-1}$
has poincaré duals $u_{2^{k}-2},\cdots,u_{2^{k-1}},$ respectively. Thus,
$v=u_{1}u_{2^{k}-2}=\cdots=u_{2^{k-1}-1}u_{2^{k-1}}.$ Clearly, $u_{i}u_{j}=0$
if $i+j\geq 2^{k}.$ It is easy to derive the following results:

######  Theorem 3.8.

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n_{1}}\times\mathbb{S}^{n_{2}}\times\cdots\times\mathbb{S}^{n_{k}},$
where $1\leq n_{1}\leq n_{2}\leq\cdots\leq n_{k}$ and $X$ is TNHZ in $X_{G}.$
If $F$ is connected and all the cup products of generators of $H^{*}(F)$ other
than generator of $r^{th}$ cohomology group are zero, then
$F\sim_{2}\underset{k}{\\#}\mathbb{P}^{k}(q),q\leq n_{1}$ and $q=1,2,4,8.$

######  Theorem 3.9.

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n_{1}}\times\mathbb{S}^{n_{2}}\times\cdots\times\mathbb{S}^{n_{k}},$
where $1\leq n_{1}\leq n_{2}\leq\cdots\leq n_{k}$ and $X$ is TNHZ in $X_{G}.$
If $F$ is connected and all the cup products $u_{i}u_{j},$ where $i+j\leq
2^{k}-2,i\neq j,$ of generators of $H^{*}(F)$ other than $r^{th}$ cohomology
group are nonzero, then $F\sim_{2}\mathbb{P}^{2^{k}-1}(q),~{}~{}~{}q=1,2,4,8.$

Next, suppose that $X$ is not TNHZ in $X_{G}$ and $F$ is nonempty connected
fixed point set of involutions on $X.$ So, we have $0<$ rk $H^{*}(F)<8.$ By
the Floyd’s formula [3], we get rk $H^{*}(F)=2,4$ or $6.$ If $\pi_{1}(B_{G})$
acts nontrivially on $H^{*}(X),$ then we have proved the following theorem:

######  Theorem 3.10.

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},1\leq n\leq
m\leq l$ and $X$ is not TNHZ in $X_{G}.$ If $F$ is nonempty connected fixed
point set and $\pi_{1}(B_{G})$ acts nontrivially on $H^{*}(X),$ then $F$ must
be the one of the following:

  1. (1)

$F\sim_{2}\mathbb{S}^{r_{1}},0<r_{1}\leq n+m+l.$

  2. (2)

$F\sim_{2}\mathbb{S}^{r_{1}}\times\mathbb{S}^{r_{2}},0<r_{1}\leq
min\\{2n,l\\}~{}or~{}n,0<q_{2}\leq~{}max\\{2n,l\\}~{}or~{}2m.$

  3. (3)

$F\sim_{2}\mathbb{P}^{3}(r_{1}),r_{1}\leq~{}min\\{2n,l\\}~{}or~{}n,r_{1}=1,2,4,8.$

  4. (4)

$F\sim_{2}\mathbb{P}^{2}(r_{1})\\#\mathbb{P}^{2}(r_{1}),r_{1}\leq~{}min\\{2n,l\\}~{}or~{}n,r_{1}=1,2,4,8.$

  5. (5)

$F\sim_{2}\mathbb{P}^{2}(r_{1})\vee\mathbb{S}^{r_{2}},r_{1}\leq~{}min\\{2n,l\\}~{}or~{}n,0<r_{2}\leq
2n+l~{}or~{}n+2m,r_{1}=1,2,4,8.$

###### Proof.

As $\pi_{1}(B_{G})=G$ acts nontrivially on $H^{*}(X),$ we must have one of the
following conditions: $(i)~{}n=m<l$ $(ii)~{}n<m=l$ $(iii)~{}n=m=l.$
If $n=m<l,$ then $G$ acts nontrivially on $H^{n}(X).$ Let $g$ be generator of
$G.$ We have exactly three possibilities of nontrivial actions: (1)
$g^{*}(a)=b~{}\&$ $g^{*}(b)=a,$ (2) $g^{*}(a)=a~{}\&$ $g^{*}(b)=a+b$ and (3)
$g^{*}(a)=a+b~{}\&$ $g^{*}(b)=b.$ First, we consider nontrivial action defined
by $g^{*}(a)=b~{}\&~{}g^{*}(b)=a.$ By the natuality of cup product, we get $G$
also acts nontrivially on $H^{n+l}(X).$ Note that $\sigma=\tau=1+g^{*}.$ By
Proposition 2.8, we get $E_{2}^{0,i}\cong\mathbb{Z}_{2}$ and
$E_{2}^{k,i}=0~{}~{}~{}\forall~{}~{}k>0,~{}~{}i=n,n+l.$ For $i\neq n,n+l$,
$\pi_{1}(B_{G})$ acts trivially on $H^{i}(X),$ and hence, $E_{2}^{k,i}\cong
H^{k}(B_{G})\otimes H^{i}(X)\cong\mathbb{Z}_{2}~{}~{}~{}\forall~{}~{}k\geq
0,~{}~{}i=0,l,2n,2n+l.$ By Proposition 2.9, $1\otimes ab$ is a permanent
cocycle.

First, suppose that $E_{2}^{*,*}\neq E_{\infty}^{*,*}.$ As $F\neq\emptyset,$
we must have $d_{l+1}(1\otimes c)=0.$ If $d_{l-2n+1}$ is nontrivial, then
$d_{2n+l+1}$ must be trivial. Thus rk $H^{k}(X_{G})=2$ for large value of $k,$
and by Proposition 2.2, we get rk $H^{*}(F)=2.$ So, $F\sim_{2}\mathbb{S}^{q},$
and by Proposition 2.4, we get $0<q\leq n+m+l.$ Similarly, if $d_{l-2n+1}$ is
trivial, then $d_{2n+l+1}$ must be nontrivial. Clearly,
$F\sim_{2}\mathbb{S}^{q},0<q\leq n+m+l.$ This realizes possibility (1).

Next, supppose that $E_{2}^{*,*}=E_{\infty}^{*,*}.$ Then rk $H^{*}(F)=4.$
Assume that $x\in F,$ and $\alpha\in H^{*}(X_{G},x_{G})$ and $\beta\in
H^{*}(X_{G},x_{G})$ represent $1\otimes ab\in E_{2}^{0,2n}$ and $1\otimes c\in
E_{2}^{0,l},$ respectively. Clearly,
$\\{t^{i}\alpha,t^{i}\beta,t^{i}\alpha\beta\\}$ forms a basis for
$H^{k}(X_{G},x_{G})$ over $\mathbb{Z}_{2}$-module. Let $0<q_{1}\leq q_{2}\leq
q_{3}$ be the non vanishing dimensions of $H^{*}(F)$ and $u_{i}$ generates
${q_{i}}^{th}$ cohomology group of $F,1\leq i\leq 3.$ Then we write

$j^{*}(\alpha)=A_{1}t^{2n-q_{1}}\otimes u_{1}+A_{2}t^{2n-q_{2}}\otimes
u_{2}+A_{3}t^{2n-q_{3}}\otimes u_{3},$

$j^{*}(\beta)=B_{1}t^{l-q_{1}}\otimes u_{1}+B_{2}t^{l-q_{2}}\otimes
u_{2}+B_{3}t^{l-q_{3}}\otimes u_{3},$

So, we get

$j^{*}(\alpha\beta)=A_{1}B_{1}t^{2n-2q_{1}}\otimes
u_{1}^{2}+(A_{1}B_{2}+A_{2}B_{1})t^{2n+l-q_{1}-q_{2}}\otimes
u_{1}u_{2}+A_{2}B_{2}t^{2n+l-2q_{2}}\otimes u_{2}^{2},$

where $A_{i}~{}\&~{}B_{i}$ are in $\mathbb{Z}_{2}.$ Clearly,
$u_{1}u_{3}=u_{2}u_{3}=u_{3}^{2}=0.$ As $\alpha\beta\neq 0,$
$j^{*}(\alpha\beta)$ is one of the following: (1) $t^{2n+l-2q_{1}}\otimes
u_{1}^{2},$ (2) $t^{2n+l-2q_{2}}\otimes u_{2}^{2},$ (3)
$t^{2n+l-q_{1}-q_{2}}\otimes u_{1}u_{2},$ (4) $t^{2n+l-2q_{1}}\otimes
u_{1}^{2}+t^{2n+l-q_{1}-q_{2}}\otimes u_{1}u_{2},$ (5) $t^{2n+l-2q_{2}}\otimes
u_{2}^{2}+t^{2n+l-q_{1}-q_{2}}\otimes u_{1}u_{2},$ and (6)
$t^{2n+l-2q_{1}}\otimes u_{1}^{2}+t^{2n+l-2q_{2}}\otimes u_{2}^{2}.$
It is easy to observe that the cases (5) and (6) are not possible. If
$j^{*}(\alpha\beta)=t^{2n+l-2q_{1}}\otimes u_{1}^{2}+t^{2n+l-2q_{2}}\otimes
u_{2}^{2},$ then either rk $H^{*}(F)>4$ or $j^{*}(\alpha\beta)=0,$ a
contradiction. And, if $j^{*}(\alpha\beta)=t^{2n+l-2q_{2}}\otimes
u_{2}^{2}+t^{2n+l-q_{1}-q_{2}}\otimes u_{1}u_{2},$ then
$j^{*}(\alpha\beta)=0,$ again a contradiction.
Now, suppose that $j^{*}(\alpha\beta)=t^{2n+l-2q_{1}}\otimes
u_{1}^{2}+t^{2n+l-q_{1}-q_{2}}\otimes u_{1}u_{2}.$ Then, if
$u_{1}u_{2}=u_{1}^{2},$ then $j^{*}(\alpha\beta)=0,$ a contradiction. So,
$u_{2}=u_{1}^{2},$ and hence $F\sim_{2}\mathbb{P}^{3}(q_{1}),q_{1}<2n$ or
$q_{1}<l,$ according as $2n<l$ or $l<2n,$ respectively. Note that, if
$q_{1}=2n$ or $q_{1}=l,$ then $u_{1}=a|F,$ and we get $u_{1}^{2}=0,$ a
contradiction. This realizes possibility (3).
Next, suppose that $j^{*}(\alpha\beta)=t^{2n+l-q_{1}-q_{2}}\otimes
u_{1}u_{2}.$ In this case, we have four conditions
$(i)~{}u_{1}^{2}=0~{}\&~{}u_{2}^{2}=0,~{}(ii)~{}u_{1}^{2}\neq
0~{}\&~{}u_{2}^{2}=0,~{}(iii)~{}u_{1}^{2}=0~{}\&~{}u_{2}^{2}\neq 0,$ and
$~{}(iv)~{}u_{1}^{2}\neq 0~{}\&~{}u_{2}^{2}\neq 0.$ Clearly, if
$u_{1}^{2}=0~{}\&~{}u_{2}^{2}=0,$ then
$F\sim_{2}\mathbb{S}^{q_{1}}\times\mathbb{S}^{q_{2}}.$ Now, if $u_{1}^{2}\neq
0~{}\&~{}u_{2}^{2}=0,$ then for $u_{2}=u_{1}^{2},$ we have
$F\sim_{2}\mathbb{P}^{3}(q),$ and for $u_{1}u_{2}=u_{1}^{2},$ we get
$q_{1}=q_{2}$ and by the change of basis $d=u_{1}+u_{2},$ we have
$F\sim_{2}\mathbb{P}^{2}(q)\\#\mathbb{P}^{2}(q).$ Next, if
$u_{1}^{2}=0~{}\&~{}u_{2}^{2}\neq 0,$ then $u_{1}u_{2}=u_{2}^{2},$ and we get
$F\sim_{2}\mathbb{P}^{2}(q)\\#\mathbb{P}^{2}(q).$ Finally, if $u_{1}^{2}\neq
0~{}\&~{}u_{2}^{2}\neq 0,$ then we must have $u_{1}u_{2}=u_{2}^{2}=u_{1}^{2},$
and we get $F\sim_{2}\mathbb{P}^{2}(q)\vee\mathbb{S}^{q}.$
Now, suppose that $j^{*}(\alpha\beta)=t^{2n+l-2q_{1}}\otimes u_{1}^{2},$ then
as above we have four conditions:
$(i)~{}u_{2}^{2}=0~{}\&~{}u_{1}u_{2}=0,~{}(ii)~{}u_{2}^{2}\neq
0~{}\&~{}u_{1}u_{2}=0,~{}(iii)~{}u_{2}^{2}=0~{}\&~{}u_{1}u_{2}\neq 0,$ and
$~{}(iv)~{}u_{2}^{2}\neq 0~{}\&~{}u_{1}u_{2}\neq 0.$ If
$u_{2}^{2}=0~{}\&~{}u_{1}u_{2}=0,$ then
$F\sim_{2}\mathbb{P}^{2}(q_{1})\vee\mathbb{S}^{q_{2}}$ or
$F\sim_{2}\mathbb{P}^{2}(q_{1})\vee\mathbb{S}^{q_{3}},$ accordingly
$u_{1}^{2}=u_{3}$ or $u_{1}^{2}=u_{2},$ respectively. If $u_{2}^{2}\neq
0~{}\&~{}u_{1}u_{2}=0,$ then we must have $u_{1}^{2}=u_{2}^{2},$ and hence
$F\sim_{2}\mathbb{P}^{2}(q)\\#\mathbb{P}^{2}(q).$ If
$u_{2}^{2}=0~{}\&~{}u_{1}u_{2}\neq 0,$ then for $u_{2}=u_{1}^{2}$ we get
$F\sim_{2}\mathbb{P}^{3}(q),$ and for $u_{1}u_{2}=u_{1}^{2},$ we get
$q_{1}=q_{2}$ and by the change of basis $d=u_{1}+u_{2},$ we have
$F\sim_{2}\mathbb{P}^{2}(q)\\#\mathbb{P}^{2}(q).$ If $u_{2}^{2}\neq
0~{}\&~{}u_{1}u_{2}\neq 0,$ then we must have
$u_{1}u_{2}=u_{2}^{2}=u_{1}^{2},$ and we get
$F\sim_{2}\mathbb{P}^{2}(q)\vee\mathbb{S}^{q}.$ Clearly, all the inequalities
hold by Propostion 2.4.
Finally, suppose that $j^{*}(\alpha\beta)=t^{2n+l-2q_{1}}\otimes u_{1}^{2}.$
This case is similar to the case when
$j^{*}(\alpha\beta)=t^{2n+l-2q_{1}}\otimes u_{1}^{2}.$
For the other two nontrivial actions of $G$ on $H^{n}(X),$ we get the same
result.
Next, if $n<m=l$ or $n=m=l,$ then we get the same possibilities as in the case
when $n=m<l.$ ∎

######  Remark 3.11.

If $\pi_{1}(B_{G})$ acts trivially on $H^{*}(X)$ and rk $H^{*}(F)=2$ or $4,$
then we get same possibilities of nonempty connected fixed point sets as in
Theorem 3.10. Note that for the smooth manifolds in Theorem 3.10, we get
possibilities (1)-(4) only.

###  3.2. Possibilities of disconnected fixed point sets

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},1\leq n\leq
m\leq l.$ In this section, we determine the possibilities of a disconnected
fixed point set $F$ when $X$ is TNHZ in $X_{G}.$ We have $\sum$ rk
$H^{i}(F)=8.$ Clearly, $F$ has at most eight components. By Proposition 2.6,
each component $F_{0}$ of $F$ is a poincaré duality space of formal dimension,
say $r\leq n+m+l.$ If rk $H^{*}(F_{0})=1$ or $2$ or $3,$ then $F_{0}$ have the
mod $2$ cohomology of a point or a sphere or a projective space of height two,
respectively. Now, if rk $H^{*}(F_{0})=4,$ then $F_{0}$ have the mod $2$
cohomology of either a product of spheres
$\mathbb{S}^{q_{1}}\times\mathbb{S}^{q_{2}},$ or a projective space of height
three $\mathbb{P}^{3}(q)$ or a connected sum of two projective spaces of
height two $\mathbb{P}^{2}(q)\\#\mathbb{P}^{2}(q)$ [4]. For any component
$F_{0}$ of $F$ with rk $H^{*}(F_{0})\leq 4,$ we have following theorem:

######  Theorem 3.12.

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},$ where
$1\leq n\leq m\leq l$ and $X$ is TNHZ in $X_{G}.$ If $F$ is disconnected and
the rank of cohomology ring of a component of $F$ is at most four, then $F$
must be one of the following:

  1. (1)

$F\sim_{2}\underset{i=0}{\overset{3}{\sqcup}}F_{i},$ where
$F_{i}\sim_{2}\mathbb{S}^{r_{i}},0\leq i\leq 3.$

  2. (2)

$F\sim_{2}\underset{i=0}{\overset{2}{\sqcup}}F_{i},$ where
$F_{0}\sim_{2}\mathbb{P}^{2}(r_{1})$ or $\mathbb{S}^{r_{2}}+pt,$
$F_{1}\sim_{2}\mathbb{P}^{2}(r_{3})$ and $F_{2}\sim_{2}\mathbb{S}^{r_{4}}.$

  3. (3)

$F\sim_{2}\underset{i=0}{\overset{2}{\sqcup}}F_{i},$ where
$F_{0}\sim_{2}\mathbb{S}^{r_{1}}\times\mathbb{S}^{r_{2}}$ or
$\mathbb{P}^{3}(r_{1})$ or $\mathbb{P}^{2}(r_{1})\\#\mathbb{P}^{2}(r_{2})$ and
$F_{i}\sim_{2}\mathbb{S}^{r_{i}},i=1,2.$

  4. (4)

$F\sim_{2}\underset{i=0}{\overset{2}{\sqcup}}F_{i},$ where
$F_{0}\sim_{2}\mathbb{S}^{r_{1}}\times\mathbb{S}^{r_{2}}$ or
$\mathbb{P}^{3}(r_{1})$ or $\mathbb{P}^{2}(r_{1})\\#\mathbb{P}^{2}(r_{1}),$
$F_{1}\sim_{2}\mathbb{P}^{2}(r_{2})$ and $F_{2}\sim_{2}pt.$

  5. (5)

$F\sim_{2}\underset{i=0}{\overset{1}{\sqcup}}F_{i},$ where
$F_{i}\sim_{2}\mathbb{S}^{r_{1}}\times\mathbb{S}^{r_{2}}$ or
$\mathbb{P}^{3}(r_{1})$ or
$\mathbb{P}^{2}(r_{1})\\#\mathbb{P}^{2}(r_{1}),i=0,1.$

Further, if rk $H^{*}(F_{0})>4,$ where $F_{0}$ is a component of $F,$ then we
have following theorems:

######  Theorem 3.13.

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},$ where
$1\leq n\leq m\leq l$ and $X$ is TNHZ in $X_{G}.$ If $F$ is disconnected and
the rank of cohomology ring of a component of $F$ is five, then $F$ must be
one of the following:

  1. (1)

$F\sim_{2}\underset{i=0}{\overset{1}{\sqcup}}F_{i},$ where
$F_{0}\sim_{2}\underset{3}{\\#}\mathbb{P}^{2}(r_{1})$ or
$\mathbb{P}^{2}(r_{1})\\#({\mathbb{S}^{r_{2}}\times\mathbb{S}^{r_{3}}})$ or
$\mathbb{P}^{4}(r_{1})$ and $F_{1}\sim_{2}\mathbb{P}^{2}(r_{1}).$

  2. (2)

$F\sim_{2}\underset{i=0}{\overset{2}{\sqcup}}F_{i},$ where
$F_{0}\sim_{2}\underset{3}{\\#}\mathbb{P}^{2}(r_{1})$ or
$\mathbb{P}^{2}(r_{1})\\#({\mathbb{S}^{r_{2}}\times\mathbb{S}^{r_{3}}})$ or
$\mathbb{P}^{4}(r_{1})$ and $F_{1}\sim_{2}\mathbb{S}^{r_{4}}$ or
$F_{2}\sim_{2}pt.$

###### Proof.

Let $F_{0}$ be the component of $F$ such that rk $H^{*}(F_{0})=5.$ Suppose
that $0<q_{1}\leq q_{2}\leq q_{3}<r$ are non vanishing dimensions of
$H^{*}(F_{0}),$ and $u_{i}$ generates $H^{q_{i}}(F_{0}),1\leq i\leq 3$ and $v$
generates $H^{r}(F_{0}).$ Clearly, $u_{i}v=0,1\leq i\leq 3.$
First, we consider that each generator $u_{i},1\leq i\leq 3,$ is poincaré dual
of itself. In this case, we get $u_{1}^{2}=u_{2}^{2}=u_{3}^{2}$ generates
$H^{r}(F),$ and hence $F_{0}\sim_{2}\underset{3}{\\#}\mathbb{P}^{2}(q_{1}).$
Now, we consider that exactly one generator is poincaré dual of itself. Let
$u_{2}$ be poincaré dual of itself. So, we have $u_{3}$ is the poincaré dual
of $u_{1}.$ Then, $u_{1}u_{3}=u_{2}^{2}$ generates $H^{r}(F_{0}).$ Clearly,
$u_{2}u_{3}=u_{3}^{2}=0.$ Suppose that $u_{1}u_{2}=0.$ Further, if
$u_{1}^{2}\neq 0,$ then we must have $u_{3}=u_{1}^{2},$ and hence
$F_{0}\sim_{2}\mathbb{P}^{3}(q_{1})\\#\mathbb{P}^{2}(q_{2}),$ which is not
possible. If $u_{1}^{2}=0,$ then
$F_{0}\sim_{2}\mathbb{P}^{2}(q_{2})\\#({\mathbb{S}^{q_{1}}\times\mathbb{S}^{q_{3}}}).$
Now, suppose $u_{1}u_{2}\neq 0.$ Then, we get $u_{3}=u_{1}u_{2}.$ So,
$u_{1}^{2}u_{2}=u_{2}^{2}$ generates $H^{r}(F_{0})$ and we must have
$u_{2}=u_{1}^{2}.$ Hence, $F_{0}\sim_{2}\mathbb{P}^{4}(q_{1}),$ where
$q_{1}=1,2,4,8.$ For the remaining cases, when $u_{1}$ or $u_{3}$ is poincaré
dual of itself, we only get
$F_{0}\sim_{2}\mathbb{P}^{2}(q)\\#({\mathbb{S}^{q}\times\mathbb{S}^{q}}).$
Since rk $H^{*}(F)=8$ and rk $H^{*}(F_{0})=5,$ then $F$ have at most four
components. Clearly, the components other than $F_{0}$ have the mod $2$
cohomology of a point or a sphere or a projective space of height two. Hence,
our claim. ∎

######  Theorem 3.14.

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},$ where
$1\leq n\leq m\leq l$ and $X$ is TNHZ in $X_{G}.$ If $F$ is disconnected and
one of the rank of cohomology ring of a component of $F$ is six, then
$F\sim_{2}\underset{i=1}{\overset{2}{\sqcup}}F_{i},$ where
$F_{1}\sim_{2}\underset{2}{\\#}\mathbb{P}^{3}(r_{1})$ or
$\mathbb{P}^{3}(r_{1})\\#(\mathbb{S}^{r_{2}}\times\mathbb{S}^{r_{3}})$ or
$\mathbb{P}^{2}(r_{1})\times\mathbb{S}^{r_{2}}$ or $\mathbb{P}^{5}(r_{1})$ and
$F_{2}\sim_{2}\mathbb{S}^{r_{2}}.$

###### Proof.

Let $F_{0}$ be the component of $F$ with rk $H^{*}(F_{0})=6.$ Assume that
$0<q_{1}\leq q_{2}\leq q_{3}\leq q_{4}<r$ are non vanishing dimensions of
$H^{*}(F_{0}),$ and $u_{i}$ generates $H^{q_{i}}(F_{0}),1\leq i\leq 4$ and $v$
generates $H^{r}(F_{0}).$ If each generator $u_{i}$ is poincaré dual of itself
or any two generators are poincaré dual of itself, then this contradicts
Proposition 2.7. As $q_{i}\leq q_{i+1},1\leq i\leq 3,$ we must have $u_{4}$
and $u_{3}$ are poincaré duals of $u_{1}$ and $u_{2},$ respectively. So,
$u_{1}u_{4}=u_{2}u_{3}$ generates $H^{r}(F_{0}).$ Clearly,
$u_{2}u_{4}=u_{3}u_{4}=0.$ So, the cup products $u_{1}u_{2}$ and $u_{1}u_{3}$
are either zero or nonzero.

First, Suppose that $u_{1}u_{2}=u_{1}u_{3}=0.$ Clearly,
$u_{3}^{2}=u_{4}^{2}=0.$ Now, we have $u_{1}^{2}$ and $u_{2}^{2}$ are either
zero or nonzero. If $u_{1}^{2}=u_{2}^{2}=0,$ then
$F\sim_{2}(\mathbb{S}^{q_{2}}\times\mathbb{S}^{q_{3}})\\#(\mathbb{S}^{q_{1}}\times\mathbb{S}^{q_{4}}),$
which contradicts Proposition 2.7. If $u_{1}^{2}\neq 0$ and $u_{2}^{2}\neq 0,$
then we get $u_{4}=u_{1}^{2}$ and $u_{3}=u_{2}^{2}.$ Thus,
$F\sim_{2}\mathbb{P}^{3}(q_{1})\\#\mathbb{P}^{3}(q_{2}).$ If $u_{1}^{2}\neq 0$
and $u_{2}^{2}=0,$ then we get
$F\sim_{2}\mathbb{P}^{3}(q_{1})\\#(\mathbb{S}^{q_{2}}\times\mathbb{S}^{q_{3}}).$
Similarly, if $u_{1}^{2}=0$ and $u_{2}^{2}\neq 0,$ then
$F\sim_{2}\mathbb{P}^{3}(q_{1})\\#(\mathbb{S}^{q_{1}}\times\mathbb{S}^{q_{4}}).$

Now, suppose that $u_{1}u_{2}\neq 0$ and $u_{1}u_{3}=0.$ Then, we have either
$u_{1}u_{2}=u_{3}$ or $u_{1}u_{2}=u_{4}.$ If $u_{1}u_{2}=u_{3},$ then
$u_{1}u_{4}=u_{1}u_{2}^{2}$ generates $H^{r}(F_{0}).$ So, $u_{2}^{2}\neq 0,$
and we get $u_{4}=u_{2}^{2}.$ Further, if $u_{1}^{2}=0,$ then we get
$F\sim_{2}\mathbb{P}^{2}(q_{1})\times\mathbb{S}^{q_{2}}.$ For $u_{1}^{2}\neq
0,$ we get $u_{1}^{2}=u_{2}^{2}.$ By the change of basis $d=u_{1}+u_{2},$ we
get $F\sim_{2}\mathbb{P}^{2}(q_{1})\times\mathbb{S}^{q_{1}}.$ Similarly, if
$u_{1}u_{2}=u_{4},$ then
$F\sim_{2}\mathbb{P}^{2}(q_{1})\times\mathbb{S}^{q_{2}}.$

Now, suppose that $u_{1}u_{2}=0$ and $u_{1}u_{3}\neq 0.$ Then,
$u_{1}u_{3}=u_{4}$ and $u_{1}^{2}u_{3}$ generates $H^{r}(F_{0}).$ Clearly,
$u_{3}^{2}=0$ and $u_{2}=u_{1}^{2}.$ Thus,
$F\sim_{2}\mathbb{P}^{2}(q_{1})\times\mathbb{S}^{q_{3}}.$

Finally, suppose that both $u_{1}u_{2}$ and $u_{1}u_{3}$ are nonzero. In this
case, we get $u_{3}=u_{1}u_{2}$ and $u_{4}=u_{1}u_{3}.$ Clearly,
$u_{2}=u_{1}^{2},$ and $F\sim_{2}\mathbb{P}^{5}(q_{1}),$ where
$q_{1}=1,2,4,8.$
Note that, the component other than $F_{0}$ have sphere cohomology. Hence, our
claim. ∎

######  Theorem 3.15.

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},$ where
$1\leq n\leq m\leq l$ and $X$ is TNHZ in $X_{G}.$ If $F$ is disconnected and
one of the rank of cohomology ring of component of $F$ is seven, then $F$ must
be one of the following:

  1. (1)

$F\sim_{2}\underset{i=1}{\overset{2}{\sqcup}}F_{i},$ where
$F_{1}\sim_{2}(\mathbb{P}^{2}(r_{1})\times\mathbb{S}^{r_{2}})\\#\mathbb{P}^{2}(r_{3})$
or $\mathbb{P}^{4}(r_{1})\\#(\mathbb{S}^{r_{2}}\times\mathbb{S}^{r_{3}})$ or
$\mathbb{P}^{6}(r_{1})$ and $F_{2}\sim_{2}pt.$

  2. (2)

$F\sim_{2}{\mathbb{Z}_{2}[u_{1},u_{2}]}/{<u_{1}^{5},u_{2}^{3},u_{1}^{3}+u_{2}^{2},u_{1}^{2}u_{2}>}+pt,$
where deg $u_{1}=r_{1}$ and deg $u_{2}=r_{2}.$

###### Proof.

Let $F_{0}$ be the component of $F$ such that rk $H^{*}(F_{0})=7.$ Clearly,
the component of $F$ other than $F_{0}$ have point cohomology. Assume that
$0<q_{1}\leq q_{2}\leq q_{3}\leq q_{4}\leq q_{5}<r$ are the non vanishing
dimensions of $H^{*}(F_{0})$ with $u_{i},1\leq i\leq 5$ is generator of
$H^{q_{i}}(F_{0})$ and $v$ generates $H^{r}(F_{0}).$ If each generator $u_{i}$
is poincaré dual of itself or any three genertors are poincaré duals of
itself, then we get a contradiction of Proposition 2.7. Clearly, $u_{3}$ is
poincaré dual of itself and $u_{5}$ and $u_{4}$ are poincaré duals of $u_{1}$
and $u_{2},$ respectively. Therefore, $u_{1}u_{5}=u_{2}u_{4}=u_{3}^{2}$
generates $H^{r}(F_{0}).$ Note that the cup products
$v^{2}=u_{2}u_{5}=u_{3}u_{4}=u_{3}u_{5}=u_{4}^{2}=u_{5}^{2}=u_{i}v=0,1\leq
i\leq 5.$ The remaining cup products $u_{1}u_{2},u_{1}u_{3},u_{1}u_{4},$ and
$u_{2}u_{3}$ may be zero or nonzero. Now, we considered cases accordingly to
how many cup products are nonzero.
It is easy to observe that, if any three cup products are nonzero or none of
the cup products are nonzero then these cases are not possible. So, we
consider the following cases:
Case(1): If any one cup product from
$u_{1}u_{2},u_{1}u_{3},u_{1}u_{4},u_{2}u_{3}$ is nonzero.
First, Suppose that $u_{1}u_{2}\neq 0,$ then either $u_{1}u_{2}=u_{4}$ or
$u_{1}u_{2}=u_{5}.$ Let $u_{1}u_{2}=u_{4}.$ Then,
$v=u_{1}u_{5}=u_{2}^{2}u_{1}=u_{3}^{2}.$ If $u_{2}^{2}=u_{3},$ then
$u_{3}^{2}=u_{2}^{4}\neq 0$ but $u_{2}u_{3}=u_{2}^{3}=0,$ a contradiction. By
the uniqueness of poincaré dual of $u_{2},$ we get $u_{2}^{2}=u_{5}.$ If
$u_{1}^{2}=0,$ then we get
$F\sim_{2}(\mathbb{P}^{2}(q_{1})\times\mathbb{S}^{q_{2}})\\#\mathbb{P}^{2}(q_{3}),$
and $q_{1},q_{3}\in\\{1,2,4,8\\}.$ If $u_{1}^{2}\neq 0,$ then we must have
$u_{2}^{2}=u_{1}^{2},$ and by the change of basis $d=u_{1}+u_{2},$ we get
$F\sim_{2}(\mathbb{P}^{2}(q_{1})\times\mathbb{S}^{q_{1}})\\#\mathbb{P}^{2}(q_{3}).$
Similarly, if $u_{1}u_{2}=u_{5},$ then
$F\sim_{2}(\mathbb{P}^{2}(q_{1})\times\mathbb{S}^{q_{2}})\\#\mathbb{P}^{2}(q_{3}).$

Suppose that $u_{1}u_{3}\neq 0,$ then $u_{1}u_{3}=u_{5}.$ So,
$u_{1}^{2}u_{3}=u_{2}u_{4}=u_{3}^{2}=v.$ Clearly, $u_{1}^{2}=u_{3}$ and
$u_{2}^{2}=0.$ Thus,
$F\sim_{2}\mathbb{P}^{4}(q_{1})\\#(\mathbb{S}^{q_{2}}\times\mathbb{S}^{q_{2}}),q_{1}=1,2,4,8.$
Similarly, if $u_{1}u_{4}\neq 0,$ then we get
$F\sim_{2}(\mathbb{P}^{2}(q_{1})\times\mathbb{S}^{q_{4}})\\#\mathbb{P}^{2}(q_{3}),$
and if $u_{2}u_{3}\neq 0,$ then we get
$F\sim_{2}\mathbb{P}^{4}(q_{2})\\#(\mathbb{S}^{q_{1}}\times\mathbb{S}^{q_{5}}).$
These realizes possibility (1).
Case(2): If any two cup products from
$u_{1}u_{2},u_{1}u_{3},u_{1}u_{4},u_{2}u_{3}$ are nonzero.
It is easy observe that the only possible nonzero pair is
$u_{1}u_{2}~{}\&~{}u_{1}u_{3}.$ We must have $u_{1}u_{3}=u_{6}$ and
$u_{1}u_{2}=u_{4}$. Thus, $u_{1}^{2}u_{3}=u_{2}^{2}u_{1}=u_{3}^{2}=v.$
Clearly, $u_{3}=u_{1}^{2}$ and $u_{1}u_{3}=u_{2}^{2}.$ Thus, the cohomology
ring of $H^{*}(F_{0})$ isomorphic to

${\mathbb{Z}_{2}[u_{1},u_{2}]}/{<u_{1}^{5},u_{2}^{3},u_{1}^{3}+u_{2}^{2},u_{1}^{2}u_{2}>},$

where deg $u_{1}=q_{1}$ and deg $u_{2}=q_{2}.$ This realizes possibility (2).
Case(3): If all cup products $u_{1}u_{2},u_{1}u_{3},u_{1}u_{4},u_{2}u_{3}$ are
nonzero.
In this case, we must have $u_{1}u_{2}=u_{3},$ $u_{1}u_{3}=u_{4},$
$u_{1}u_{4}=u_{2}u_{3}=u_{5}$ and $u_{1}^{2}=u_{2}.$ Hence,
$F_{0}\sim_{2}\mathbb{P}^{6}(q),$ where $q=1,2,4,8.$ This realizes possibility
(1). ∎

Next, suppose that $X$ is not TNHZ in $X_{G}$ and $F$ is disconnected fixed
point set of involutions on $X.$ A proof of the following theorem is similar
to proof of Theorem 3.10.

######  Theorem 3.16.

Let $G=\mathbb{Z}_{2}$ act on
$X\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},1\leq n\leq
m\leq l$ and $X$ is not TNHZ in $X_{G}.$ If $F$ is disconnected fixed point
set and $\pi_{1}(B_{G})$ acts nontrivially on $H^{*}(X),$ then $F$ must be the
one of the following:

  1. (1)

$F\sim_{2}\mathbb{S}^{0}.$

  2. (2)

$F\sim_{2}\mathbb{S}^{r_{1}}+\mathbb{S}^{r_{2}},0\leq r_{1},r_{2}\leq n+m+l.$

  3. (3)

$F\sim_{2}\mathbb{P}^{2}(r_{1})+pt,r_{1}=1,2,4,8,r_{1}\leq n+m+l.$

  4. (4)

$F\sim_{2}\mathbb{S}^{r_{1}}\vee\mathbb{S}^{r_{2}}+pt,0\leq r_{1},r_{2}\leq
n+m+l.$

Note that if $\pi_{1}(B_{G})$ acts trivially on $H^{*}(X)$ and rk $H^{*}(F)=2$
or $4,$ then we get same possibilities of disconnected fixed point sets as in
Theorem 3.16. For the smooth manifolds possibilities (3) and (4) will not
occur.

Next, we give some examples to realizes above Theorems:

######  Example 3.17.

Let $G=\mathbb{Z}_{2}$ act on
$Y\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m},n\leq m.$ If $Y$ is TNHZ in
$Y_{G},$ then the possibilities of mod $2$ cohomology of fixed point sets are
$\mathbb{S}^{r_{1}}\times\mathbb{S}^{r_{2}},\mathbb{P}^{2}(r_{1})\\#\mathbb{P}^{2}(r_{1}),\mathbb{P}^{3}(r_{1}),\\\
~{}\mathbb{P}^{2}(r_{1})+pt$ and $\mathbb{S}^{r_{1}}+\mathbb{S}^{r_{2}},$ and
if $Y$ is not TNHZ in $Y_{G},$ then $F\sim_{2}\mathbb{S}^{r_{1}}$ [4]. Recall
that if $G$ acts on $Z\sim_{2}\mathbb{S}^{l},l>0,$ then mod $2$ cohomology of
the fixed point set is $\mathbb{S}^{r_{3}},r_{3}\geq 0.$ After taking the
diagonal action of $G$ on $X=Y\times
Z\sim_{2}\mathbb{S}^{n}\times\mathbb{S}^{m}\times\mathbb{S}^{l},$ we get the
possibilities of mod $2$ cohomology of fixed point sets are
$\mathbb{S}^{r_{1}}\times\mathbb{S}^{r_{2}}\times\mathbb{S}^{r_{3}},(\mathbb{P}^{2}(r_{1})\\#\mathbb{P}^{2}(r_{1}))\times\mathbb{S}^{r_{3}},\mathbb{P}^{3}(r_{1})\times\mathbb{S}^{r_{3}},(\mathbb{P}^{2}(r_{1})+pt)\times\mathbb{S}^{r_{3}}$
and $(\mathbb{S}^{r_{1}}+\mathbb{S}^{r_{2}})\times\mathbb{S}^{r_{3}},$ in the
case when $X$ is TNHZ in $X_{G}$ and
$F\sim_{2}\mathbb{S}^{r_{1}}\times\mathbb{S}^{r_{3}},$ in the case when $X$ is
not TNHZ in $X_{G}.$ This realizes possibility (2) of Theorems $\ref{3.3},$
3.4 and $\ref{3.10},$ possibility (1) of Theorems 3.5 and $\ref{3.6},$
possibilities (1) and (5) of Theorem $\ref{3.12},$ and Theorem $\ref{3.14}.$

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