# Finitistic Spaces with the orbit space $\mathbb{FP}^{n}\times\mathbb{S}^{m}$

Anju Kumari and Hemant Kumar Singh  Anju Kumari, 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 $G=\mathbb{S}^{d}$, $d=0,1$ or $3$, act freely on a finitistic connected
space $X$. This paper gives the cohomology classification of $X$ if a mod 2 or
rational cohomology of the orbit space $X/G$ is isomorphic to the product of a
projective space and sphere $\mathbb{FP}^{n}\times\mathbb{S}^{m}$, where
$\mathbb{F}=\mathbb{R},\mathbb{C}$ or $\mathbb{H}$, respectively. For a free
involution on $X$, a lower bound of covering dimension of the coincidence set
of a continuous map $f:X\to\mathbb{R}^{k}$ is also determined.

###### Key words and phrases:

Free action; Finitistic space; Leray-Serre spectral sequence; Gysin sequence;
Euler class

###### 2010 Mathematics Subject Classification:

Primary 55T10; Secondary 57S99

This paper is supported by the Science and Engineering Research Board
(Department of Science and Technology, Government of India) with reference
number- EMR/2017/002192

##  1\. Introduction

Let $G$ be a group acting on a compact Hausdorff space $X$. It has been an
interesting problem to classify the orbit space $X/G$ for a given space $X$
when $G$ acts freely on $X$. It is well known that the projective spaces
$\mathbb{FP}^{n}$, where $\mathbb{F}=\mathbb{R}$, $\mathbb{C}$ or
$\mathbb{H}$, are the orbit spaces of the standard free actions of
$G=\mathbb{S}^{d}$ on $\mathbb{S}^{(d+1)n+d}$, where $d=0,1$ and $3$, and the
lens space $L_{p}^{2n+1}$ is the orbit space of the standard free action of
$G=\mathbb{Z}_{p}$, $p$ an odd prime, on $\mathbb{S}^{2n+1}$. A lot of work
have been done in the literature in this direction for various spaces $X$ and
different groups $G$. For example: (i) $X=L_{p}^{2n+1}$ or $\mathbb{FP}^{n}$
or $\mathbb{FP}^{n}\times\mathbb{FP}^{m}$, where $\mathbb{F}=\mathbb{R}$ or
$\mathbb{C}$ and $G=\mathbb{Z}_{2}$ [16, 18, 19], (ii) $X=\mathbb{F}H_{r,s}$,
$\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$ and $0\leq s\leq r$, (the Real or
complex Milnor manifold) and $G=\mathbb{Z}_{2}$ or $\mathbb{S}^{1}$ [3], (iii)
$X=\mathbb{S}^{n}\times\mathbb{S}^{m},1\leq n\leq m$ and $G=\mathbb{Z}_{p},p$
a prime, or $\mathbb{S}^{1}$ or $\mathbb{S}^{3}$ [5, 10], and (iv) $X=P(1,n)$
(Dold manifold) and $G=\mathbb{Z}_{2}$ [13]. Another thread of research is to
classify a space $X$ for a given orbit space $X/G$, where $G$ acts on $X$. For
semi-free actions of $G=\mathbb{S}^{1}$ with the orbit space $X/G$ is a simply
connected 4-manifold, it has been proved that the total space is a simply
connected 5-manifold [7]. Duan [6] shows that every 1-connected 4-manifold $M$
is the quotient of some regular circle action on the space
${\\#}_{r-1}\mathbb{S}^{2}\times\mathbb{S}^{3}$, where $r=\text{rank
}H^{2}(M,\mathbb{Z})$. If $G$ acts freely on $X$, then the cohomology
classification of $X$ is discussed for (i) $X/G=L_{p}^{2n+1}$ or
$\mathbb{CP}^{n}$ and $G=\mathbb{Z}_{p},p$ a prime or $\mathbb{S}^{1}$ [20],
and for $X/G=\mathbb{HP}^{n}$ and $G=\mathbb{S}^{3}$ [11]. We contribute to
this question by determining the cohomology classification of a finitistic
space $X$ equipped with free action of $G=\mathbb{Z}_{2},\mathbb{S}^{1}$ or
$\mathbb{S}^{3}$ and the orbit space $X/G$ is a mod $2$ or rational cohomology
$\mathbb{FP}^{n}\times\mathbb{S}^{m},$ where
$\mathbb{F}=\mathbb{R},\mathbb{C}$ or $\mathbb{H}$, respectively. These
results describes the converse of some results for $G=\mathbb{Z}_{2}$ or
$\mathbb{S}^{1}$ actions proved by Dotzel et. al. [5], and for
$G=\mathbb{S}^{3}$ actions proved in [10].

For free $\mathbb{Z}_{2}$-space $X$ and any space $Y$, the coincidence set of
a continuous map $f:X\to Y$ is defined as $A(f)=\\{x\in X|f(x)=f(gx)\text{ for
each }g\in\mathbb{Z}_{2}\\}$. The classical Borsuk-Ulam Theorem states that
for a continuous map $f:\mathbb{S}^{n}\to\mathbb{R}^{k}$, the coincidence set
$A(f)=\\{x\in\mathbb{S}^{n}|f(x)=f(-x)\\}$ is nonempty, where $\mathbb{S}^{n}$
is equipped with antipodal action. Munkholm [14] shows that for a map
$f:\mathbb{S}^{n}\to M$, topological dimension of coincidence set $A(f)$ is
greater than or equal to $n-k$, where $M$ is a compact $k$-dimensional
topological manifold, $n>k$. Biasi et al. [1] extends this result for
generalized manifolds. In this paper, we determine a lower bound of the
covering dimension of coincidence set $A(f)$ of a continuous map
$f:X\to\mathbb{R}^{k}$ if
$H^{*}(X/G;\mathbb{Z}_{2})=\mathbb{Z}_{2}[a,b]/\langle a^{n+1},b^{2}\rangle$,
where $\deg a=1$ and $\deg b=m$.

##  2\. Preliminaries

Let $G$ be a compact Lie group and $G\to E_{G}\to B_{G}$ be the universal
principal $G$-bundle, where $B_{G}$ is the classifying space. Suppose $G$ acts
freely on a finitistic space $X$. The associated bundle
$X\hookrightarrow(X\times E_{G})/{G}\to B_{G}$ is a fibre bundle with fibre
$X$. Put $X_{G}=(X\times E_{G})/{G}$. The bundle $X\hookrightarrow X_{G}\to
B_{G}$ is called the Borel fibration. Then there exists the Leray-Serre
spectral sequence for the Borel fibration $X\stackrel{{\scriptstyle
i}}{{\hookrightarrow}}X_{G}\stackrel{{\scriptstyle\pi}}{{\rightarrow}}B_{G}$
which converges to $H^{*}(X_{G})$ as an algebra with
$E_{2}^{k,l}=H^{k}(B_{G};\mathcal{H}^{l}(X;R))$ . If $B_{G}$ is simply
connected then the system of local coefficients on $B_{G}$ is simple then the
$E_{2}$-term becomes

$E_{2}^{k,l}=H^{k}(B_{G};R)\otimes H^{l}(X;R).$

We recall some results which are needed to prove our results:

######  Proposition 2.1.

Let $X\stackrel{{\scriptstyle
i}}{{\hookrightarrow}}X_{G}\stackrel{{\scriptstyle\pi}}{{\rightarrow}}B_{G}$
be the Borel fibration. Suppose that the system of local coefficients on
$B_{G}$ is simple, then the edge homomorphisms

$H^{k}(B_{G})\cong E_{2}^{k,0}\longrightarrow
E_{3}^{k,0}\longrightarrow\cdots\longrightarrow E_{k}^{k,0}\longrightarrow
E_{k+1}^{k,0}=E_{\infty}^{k,0}\subset H^{k}(X_{G}),\text{ and }$

$H^{l}(X_{G})\longrightarrow E_{\infty}^{0,l}=E_{l+1}^{0,l}\subset
E_{l}^{0,l}\subset\cdots\subset E_{2}^{0,l}\cong H^{l}(X)$

are the homomorphisms

$\pi^{*}:H^{k}(B_{G})\to
H^{k}(X_{G})~{}~{}~{}\textrm{and}~{}~{}~{}i^{*}:H^{l}(X_{G})\to H^{l}(X).$

For details about spectral sequences, we refer the reader to [12]. Let
$h:X_{G}\rightarrow X/G$ be the map induced by the $G$-equivariant projection
$X\times E_{G}\rightarrow X$. Then h is a homotopy equivalence [4]. All the
cohomologies are Čech cohomology with coefficients in $R$, where
$R=\mathbb{Q}$ or $\mathbb{Z}_{2}$. Note that $X\sim_{R}Y$ means
$H^{*}(X;R)\cong H^{*}(Y;R)$. For $R=\mathbb{Q}$ and for $G=\mathbb{S}^{1}$ or
$\mathbb{S}^{3}$, we assume that the associated sphere bundle
$G\hookrightarrow X\to X/G$ is orientable.

######  Proposition 2.2.

([17, 9]) Let $G=\mathbb{S}^{1}$ or $\mathbb{S}^{3}$, act freely on a
finitistic space $X$. If $H^{i}(X;R)=0$ for all $i>n$, then $H^{i}(X/G;R)=0$
for all $i>n$.

Now, we recall Gysin sequence of sphere bundles.

######  Proposition 2.3.

Let $G=\mathbb{S}^{d}$, $d=0,1$ or $3$, act freely on a finitistic space $X$.
The Gysin sequence of the sphere bundle $G\hookrightarrow
X\stackrel{{\scriptstyle p}}{{\to}}X/G$ is:

$\displaystyle\cdots\rightarrow H^{i}(X/G)\stackrel{{\scriptstyle
p^{*}_{i}}}{{\longrightarrow}}H^{i}(X)\stackrel{{\scriptstyle\rho_{i}}}{{\longrightarrow}}H^{i-d}(X/G)\stackrel{{\scriptstyle\cup}}{{\longrightarrow}}H^{i+1}(X/G)\stackrel{{\scriptstyle
p_{i+1}^{*}}}{{\longrightarrow}}H^{i+1}(X)\rightarrow\cdots$

which start with

$\displaystyle 0\longrightarrow$ $\displaystyle
H^{d}(X/G)\stackrel{{\scriptstyle
p^{*}_{d}}}{{\longrightarrow}}H^{d}(X)\stackrel{{\scriptstyle\rho_{d}}}{{\longrightarrow}}H^{0}(X/G)\stackrel{{\scriptstyle\cup}}{{\longrightarrow}}H^{d+1}(X/G)\stackrel{{\scriptstyle
p^{*}_{d+1}}}{{\longrightarrow}}H^{d+1}(X)\longrightarrow\cdots$

where $\cup:H^{i}(X/G)\to H^{i+d+1}(X/G)$ maps $x\to x\cup u$ and $u\in
H^{d+1}(X/G)$ denotes the Euler class of the sphere bundle. It is easy to
observe that $p^{*}_{i}$ is an isomorphism for $0\leq i\leq d-1$ for $d=1$ or
2.

In this paper, we have considered finitistic spaces. These spaces were
introduced by R. G. Swan [21] and have been discovered as relevant spaces for
the study of cohomological aspects of tranformation groups [2]. Recall that a
space is said to be finitistic if it is paracompact, Hausdorff and every open
cover of it have a finite dimensional open refinement. All compact spaces, all
finite dimensional and all finite-dimensional paracompact spaces are some
examples of finitistic spaces. Note that the space
$X=\prod\limits_{n=1}^{\infty}\mathbb{S}^{n}\times\mathbb{R}^{k}$ is an
example of finitistic space which is neither compact nor has finite covering
dimension. Recall that
$H^{*}(\mathbb{RP}^{n}\times\mathbb{S}^{m};R)=R[a,b]/\langle
a^{n+1},b^{2}\rangle$, where $\deg a=1$ and $\deg b=m$.

##  3\. Main Theorems

Let $G=\mathbb{S}^{d}$, $d=0,1$ or $3$, act freely on a finitistic space $X$
having the mod $2$ or rational cohomology the product of spheres
$\mathbb{S}^{(d+1)n+d}\times\mathbb{S}^{m}$. It has been proved that one of
the possibilities of the orbit space $X/G$ are the mod $2$ or rational
cohomology $\mathbb{FP}^{n}\times\mathbb{S}^{m}$, where
$\mathbb{F}=\mathbb{R},\mathbb{C}$ or $\mathbb{H}$, respectively. Using
techniques of the Gysin sequence of sphere bundles and the Leray-Serre
spectral sequence of Borel fibration, we discuss converse of these statements.
These results also describes the cohomology classification of a finitistic
connected free $G$-space $X$ with the orbit space a product of projective
space and sphere. First, we have discussed for free actions of
$G=\mathbb{S}^{3}$.

######  Theorem 3.1.

Let $G=\mathbb{S}^{3}$ act freely on a finitistic connected space $X$ with
$X/G\sim_{R}\mathbb{HP}^{n}\times\mathbb{S}^{m}$, where $R$ is $\mathbb{Q}$ or
$\mathbb{Z}_{2}$. Then the cohomology algebra of $X$ with coefficients in $R$
is isomorphic to the cohomology algebra of one of the following:

  1. (i)

$\mathbb{S}^{m}\times\mathbb{S}^{4n+3}$;

  2. (ii)

$\mathbb{S}^{3}\times\mathbb{S}^{m}\times\mathbb{HP}^{n}$;

  3. (iii)

$\mathbb{HP}^{n}\times\mathbb{S}^{7}$ and $m=4$;

  4. (iv)

$R[x,y]/\langle x^{n+1},y^{4}\rangle$, where $\deg x=4$, $\deg y=3$, $m=6$,
$\beta(y)=x$ and $R=\mathbb{Z}_{2}$, where $\beta:H^{3}(X/G)\to H^{4}(X/G)$
denotes the Bockstein homomorphism associated with the coefficient sequence
$0\to\mathbb{Z}_{2}\to Z_{4}\to\mathbb{Z}_{2}\to 0$.

###### Proof.

It is clear that $H^{i}(X/G)\cong H^{i}(X)$ for $i=0,1$ and 2, and
$H^{i}(X)=0$ for all $i>m+4n+3$. We consider different possibilities of the
Euler class in the Gysin sequence:

Case(a): When $\cup:H^{0}(X/G)\to H^{4}(X/G)$ is trivial.

First, assume that $m>4n$. In this case, $\cup:H^{i}(X/G)\to H^{i+4}(X/G)$ is
trivial for all $i\geq 0$. For $0\leq i<n$ and $k=0,m$, we get $\rho_{k+4i+3}$
and $p^{*}_{k+4i+4}$ are isomorphisms. This implies that $H^{k+4i+3}(X)\cong
H^{k+4i+4}(X)\cong R$. Suppose that $\\{u_{4i+3}\\}$ and $\\{x^{i+1}\\}$
denotes bases for $H^{4i+3}(X)$ and $H^{4i+4}(X)$, respectively, where
$\rho_{4i+3}(u_{4i+3})=a^{i}$ and $p^{*}_{4}(a)=x$. Note that $H^{i}(X)=0$ for
$i\not=4n+3$, and $4n<i<m$. If $m\not=4n+3$, then by the exactness of the
Gysin sequence, $H^{4n+3}(X)\cong H^{m}(X)\cong R$ with bases $\\{u_{4n+3}\\}$
and $\\{y\\}$, respectively; and if $m=4n+3$ then $H^{4n+3}(X)\cong R\oplus R$
with basis $\\{u_{4n+3},y\\}$, where $\rho_{4n+3}(u_{4n+3})=a^{n}$ and
$p^{*}_{m}(b)=y$. Also, $H^{m+4n+3}(X)\cong R$. Let $\\{v_{m+4i+3}\\}$ be
basis for $H^{m+4i+3}(X)$ with $\rho_{m+4i+3}(v_{m+4i+3})=a^{i}b$ for all
$0\leq i\leq n$. Clearly, $H^{j}(X)=0$ for all $k\leq j\equiv k+1\text{ or
}k+2(\text{mod }4)<k+4n$; and $k=0,m$; and $j\not=4n+3$. Note that
$u_{3}^{2}=0$.

Now we assume that $m\leq 4n$. We consider four cases $m\equiv j(\text{mod
}4),0\leq j\leq 3$.
m $\equiv$ 0(mod 4): Let $m=4i_{0}$ for some $i_{0}\leq n$. In this case,
$H^{i}(X)=0$ for $i\equiv 1(\text{mod }4)$ or $i\equiv 2(\text{mod }4)$. Also,
$\rho_{4i-1}$ and $p^{*}_{4i}$ are isomorphisms for all $i\geq 0$. This
implies that for each $0\leq i<i_{0}$, we have $H^{4i}(X)\cong R$ and
$H^{4i+3}(X)\cong R$ with bases $\\{x^{i}\\}$ and $\\{u_{4i+3}\\}$,
respectively, and for $i_{0}\leq i\leq n$, $H^{4i}(X)\cong R\oplus R$ and
$H^{4i+3}(X)\cong R\oplus R$ with bases $\\{x^{i},x^{i-i_{0}}y\\}$ and
$\\{u_{4i+3},v_{4i+3}\\}$, respectively, where
$p^{*}_{m}(b)=y,p^{*}_{4}(a)=x,\rho_{4i+3}(u_{4i+3})=a^{i}$ and
$\rho_{4i+3}(v_{4i+3})=a^{i-i_{0}}b$. Also, for $1\leq i\leq i_{0}$,
$H^{4n+4i}(X)\cong R$ and $H^{4n+4i+3}(X)\cong R$ with bases
$\\{x^{n+i-i_{0}}y\\}$ and $\\{v_{4n+4i+3}\\}$, respectively, where
$\rho_{4n+4i+3}(v_{4n+4i+3})=a^{n+i-i_{0}}b$. Clearly, $u_{3}^{2}=0$. Thus,

$H^{i}(X)=\begin{cases}R&\text{ if }j\leq i\equiv 0\text{ or }3(\text{mod
}4)\leq m-1+j,j=0\text{ or }4n+4\\\ R\oplus R&\text{ if }m\leq i\equiv 0\text{
or }3(\text{mod }4)\leq 4n+m+3\\\ 0&\text{ otherwise.}\end{cases}$

m $\equiv$ 1(mod 4): Let $m=4{i_{0}}+1$ for some $i_{0}<n$. For
$i\not=i_{0}+j,1\leq j\leq n+1$, we get $p^{*}_{4i}$ are isomorphisms. This
implies that $H^{4i}(X)\cong R$ with basis $\\{x^{i}\\}$ for $0\leq i\leq
i_{0}$, where $p^{*}_{4}(a)=x$. For $i_{0}+1\leq i\leq n$, $H^{4i}(X)\cong
R\oplus R$ with basis $\\{x^{i},v_{4i}\\}$; and for $n+1\leq i\leq i_{0}+n+1$,
$H^{4i}(X)\cong R$ with basis $\\{v_{4i}\\}$, where
$\rho_{4i}(v_{4i})=a^{i-i_{0}-1}b$. It is clear that $H^{4i+2}(X)=0$ for all
$i\geq 0$; $H^{4i+3}(X)\cong R\cong H^{m+4i}(X)$, for all $0\leq i\leq n$,
with bases $\\{u_{4i+3}\\}$ and $\\{x^{i}y\\}$, where
$\rho_{4i+3}(u_{4i+3})=a^{i}$ and $p^{*}_{m}(b)=y$; and
$H^{4i+3}(X)=H^{4i+1}(X)=0$ otherwise. Obviously, $u_{3}^{2}=0$. We have

$H^{i}(X)=\begin{cases}R&\text{ if }j\leq i\equiv 0(\text{mod }4)\leq
j+m-1,j=0\text{ or }4n+4,\text{ or }\\\ &\;\;\;\;j\leq i\equiv j(\text{mod
}4)\leq j+4n,j=3\text{ or }m\\\ R\oplus R&\text{ if }m<i\equiv 0(\text{mod
}4)\leq 4n\\\ 0&\text{ otherwise.}\end{cases}$

m $\equiv$ 2(mod 4): Let $m=4i_{0}+2$ for some $i_{0}<n$. We get $\rho_{4i+3}$
and $p^{*}_{4i}$ are isomorphisms, for all $i\geq 0$. Therefore, for $0\leq
i\leq n$, $H^{4i+3}(X)\cong H^{4i}(X)\cong R$ with bases $\\{u_{4i+3}\\}$ and
$\\{x^{i}\\}$, respectively, and for $i>n$, $H^{4i+3}(X)=H^{4i}(X)=0$, where
$\rho_{4i+3}(u_{4i+3})=a^{i}$ and $p^{*}_{4}(a)=x$. Also, we have
$H^{4i+1}(X)=0$, for $i\not=i_{0}+j,1\leq j\leq n+1$, and for $i_{0}+1\leq
i\leq i_{0}+n+1$, $H^{4i+1}(X)\cong H^{4i-2}(X)\cong R$ with bases
$\\{v_{4i+1}\\}$ and $\\{x^{i-i_{0}-1}y\\}$, where
$\rho_{4i+1}(v_{4i+1})=a^{i-i_{0}-1}b$ and $p^{*}_{m}(b)=y$. Consequently,

$H^{i}(X)=\begin{cases}R&\text{ if }j\leq i\equiv j\text{ or }j+3(\text{mod
}4)\leq j+4n+3,j=0\text{ or }m\\\ 0&\text{ otherwise.}\end{cases}$

m $\equiv$ 3(mod 4): Let $m=4i_{0}+3$ for some $i_{0}<n$. Then $H^{4i}(X)=0$
for $i>n,$ and $H^{4i}(X)\cong R$ with basis $\\{x^{i}\\},$ for $0\leq i\leq
n$, where $p^{*}_{4}(a)=x$. Also, $H^{4i+1}(X)=0$ for all $i\geq 0$. By the
exactness of the Gysin sequence, for $i_{0}+1\leq i\leq i_{0}+n+1$,
$H^{4i+2}(X)\cong R$ with basis $\\{v_{4i+2}\\}$, where
$\rho_{4i+2}(v_{4i+2})=a^{i-i_{0}-1}b$, and $H^{4i+2}(X)=0$, otherwise. Now,
for $0\leq i<i_{0}$, $H^{4i+3}(X)\cong R$ with basis $\\{u_{4i+3}\\}$; for
$i_{0}\leq i\leq n$, $H^{4i+3}(X)\cong R\oplus R$ with basis
$\\{u_{4i+3},x^{i-i_{0}}y\\}$; and for $n<i\leq i+i_{0}$, $H^{4i+3}(X)\cong R$
with basis $\\{x^{i-i_{0}}y\\}$, where $\rho_{4i+3}(u_{4i+3})=a^{i}$ and
$p^{*}_{m}(b)=y$. Clearly, $u_{3}^{2}=0$. We have

$H^{i}(X)=\begin{cases}R&\text{ if }j\leq i\equiv j(\text{mod }4)\leq
j+4n,j=0\text{ or }m+3,\text{ or }\\\ &\;\;\;\;j+3\leq i\equiv 3(\text{mod
}4)\leq j+m-4,j=0\text{ or }4n+4\\\ R\oplus R&\text{ if }m\leq i\equiv
3(\text{mod }4)\leq 4n+3\\\ 0&\text{ otherwise.}\end{cases}$

Consider the Leray-Serre spectral sequence for the Borel fibration
$X\stackrel{{\scriptstyle
i}}{{\hookrightarrow}}X_{G}\stackrel{{\scriptstyle\pi}}{{\rightarrow}}B_{G}$
which converges to $H^{*}(X_{G})$ as an algebra and
$E_{2}^{k,l}=H^{k}(B_{G};R)\otimes H^{l}(X;R)$ for all $k,l\geq 0$. The
possible nontrivial differentials in the Leray-Serre spectral sequence for the
Borel fibration $X\stackrel{{\scriptstyle
i}}{{\hookrightarrow}}X_{G}\stackrel{{\scriptstyle\pi}}{{\rightarrow}}B_{G}$
are $\\{d_{4r}\\}_{r\geq 1}$. By the edge homomorphisms and the fact that
$p^{*}_{j}=i^{*}\circ h^{*}$ for all $j\geq 0$, we get $d_{r}(1\otimes
x)=0=d_{r}(1\otimes y)$ for all $r\geq 0$. Note that $u_{4i+3}$ and
$v_{m+4i+3}$ are not in image of $p^{*}_{4i+3}$ and $p^{*}_{m+4i+3}$,
respectively, so their image must be nonzero under some differential for all
$0\leq i\leq n$. Consequently, $x^{i}u_{3}\not=0$ and $x^{i}u_{3}y\not=0$, for
all $i$. Obviously, $x^{n+1}=y^{2}=0$.

For $m>4n$, it is clear that $u_{4i+3}=\alpha_{i}x^{i}u_{3}$ and
$v_{m+4i+3}=\beta_{i}x^{i}u_{3}y$ for some nonzero elements
$\alpha_{i},\beta_{i}$ in $R$ and $0\leq i\leq n$. In particular, for
$m=4n+3$, $x^{n}u_{3}$ can not be equal to any multiple of $y$ and so
$u_{4n+3}$ is generated by $\\{y,x^{n}u_{3}\\}$.

Now, suppose $m\leq 4n$. If $m=4i_{0}$, for some $i_{0}\leq n$ then
$u_{4i+3}=\alpha_{i}x^{i}u_{3}$ for $0\leq i<i_{0}$ and
$v_{4i+3+m}=\beta_{i}x^{i}u_{3}y$ for $n+1-i_{0}\leq i\leq n$, where
$\alpha_{i}$’s and $\beta_{i}$’s are nonzero elements in $R$. We observe that
$x^{i}u_{3}$ can not be equal to any multiple of $x^{i-i_{0}}u_{3}y$, for all
$i_{0}\leq i\leq n$. Therefore, the elements $u_{4i+3}$ and $v_{4i+3}$ are
generated by $\\{x^{i}u_{3},x^{i-i_{0}}u_{3}y\\}$ for each $i_{0}\leq i\leq
n$. If $m=4i_{0}+1$ for some $i_{0}<n$ then the elements $x^{i}u_{3}y$ cannot
be equal to any multiple of $x^{4i+3+m}$ for each $0\leq i\leq n-i_{0}+1$.
Consequently, $v_{4i+m+3}$ is generated by $\\{x^{i}u_{3}y,x^{4i+m+3}\\}$ and
$u_{4i+3}$ is generated by $\\{x^{i}u_{3}\\}$ for all $0\leq i\leq n$. If
$m=4i_{0}+2$ for some $i_{0}<n$ then we have $u_{4i+3}=\alpha_{i}x^{i}u_{3}$
and $v_{m+4i+3}=\beta_{i}x^{i}u_{3}y$ for some nonzero elements
$\alpha_{i},\beta_{i}$ in $R$ and $0\leq i\leq n$. In this case, $u_{3}^{2}$
may be both zero or nonzero. If $m=4i_{0}+3$ then
$v_{m+4i+3}=\alpha_{i}x^{i}u_{3}y$, where $\alpha_{i}\in R$, for all $0\leq
i\leq n$. As $x^{i}u_{3}$ can not be equal to $x^{i-i_{0}}y$, for all
$i_{0}\leq i\leq n$, therefore, $u_{4i+3}$ is generated by
$\\{x^{i}u_{3},x^{i-i_{0}}y\\}$, for all $0\leq i\leq n$. For $m\not=3$,
obviously $u_{3}^{2}=0$. If $u_{3}^{2}\not=0$, for $m=3$, then $d_{4}(1\otimes
v_{6})=0$, a contradiction. Therefore, $u_{3}^{2}=0$. The cohomology algebra
of $X$ is given by

$H^{*}(X)=R[x,u_{3},y]/\langle x^{n+1},u_{3}^{2},y^{2}\rangle,$

where $\deg x=4$, $\deg u_{3}=3$ and $\deg y=m$. This realizes the possibility
(ii).

If $u_{3}^{2}\not=0$ then $m$ must be 6 and $u^{2}_{3}=\alpha y$ for some
$\alpha$ nonzero in $R$. By the commutativity of cup product, we get
$2u_{3}^{2}=0$. This implies that $R$ cannot be $\mathbb{Q}$. Therefore,

$H^{*}(X)\cong R[x,u_{3}]/\langle x^{n+1},u_{3}^{4}\rangle,$

$\deg x=4$ and $\deg u_{3}=3$ only when $R=\mathbb{Z}_{2}$. By the properties
of Steenrod squares $Sq^{3}(y)=y^{2}\not=0$. As $3$ is not a power of 2, we
get $Sq^{3}=Sq^{2}\circ Sq^{1}+Sq^{1}\circ Sq^{2}$. This gives that
$Sq^{1}(y)=x$. Note that $Sq^{1}$ is the Bockstein homomorphism associated
with the coefficient sequence $0\to\mathbb{Z}_{2}\to Z_{4}\to\mathbb{Z}_{2}\to
0$. This realizes the possibility (iv) of the Theorem.

Case(b): When $\cup:H^{0}(X/G)\to H^{4}(X/G)$ maps $1$ to $ca$ for some
$c\not=0$ in $R$.

First, we suppose that $m>4n$. For $0\leq i<n$ and $j=0,m$, we get
$\cup:H^{4i+j}(X/G)\to H^{4i+4+j}(X/G)$ are isomorphisms; $\rho_{4i+3+j}$ and
$p^{*}_{4i+4+j}$ are trivial homomorphisms. Consequently,
$H^{4i+3+j}(X)=H^{4i+4+j}(X)=0$. Also, $H^{4i+1+j}(X)=H^{4i+2+j}(X)=0$ except
for $H^{4n+3}(X)$. Note that $H^{i}(X)=0$ if $(4n<i<m\text{ and }i\not=4n+3)$
and $H^{m+4n+3}(X)\cong R$ with basis $\\{u_{m+4n+3}\\}$, where
$\rho_{m+4n+3}(u_{m+4n+3})=a^{n}b$. For $m\not=4n+3$, $\rho_{4n+3}$ and
$p^{*}_{m}$ are isomorphisms. Consequently, $H^{4n+3}(X)\cong R\cong H^{m}(X)$
with bases $\\{u_{4n+3}\\}$ and $\\{y\\}$, respectively, where
$p^{*}_{m}(b)=y$ and $\rho_{4n+3}(u_{4n+3})=a^{n}$. If $m=4n+3$ then
$H^{4n+3}(X)\cong R\oplus R$ with basis $\\{u_{4n+3},y\\}$, where
$\rho_{4n+3}(u_{4n+3})=a^{n}$ and $p^{*}_{m}(b)=y$. Also,
$H^{m+4n+1}(X)=H^{m+4n+2}(X)=0$. For $m\not=4n+3$, we get

$H^{i}(X)=\begin{cases}R&\text{ if }i=0,4n+3,m,m+4n+3\\\ 0&\text{
otherwise}\end{cases}$

and for $m=4n+3$, we get

$H^{i}(X)=\begin{cases}R\oplus R&\text{ if }i=4n+3\\\ R&\text{ if
}i=0,m+4n+3\\\ 0&\text{ otherwise.}\end{cases}$

For $m\leq 4n$, we get the same cohomology groups as in $m>4n$.

Now, we compute the cohomology ring structure of $X$. In the Leray-Serre
spectral sequence for the Borel fibration $X\hookrightarrow
X_{G}\stackrel{{\scriptstyle\pi}}{{\to}}B_{G}$, $d_{4r^{\prime}}(1\otimes
u_{4n+3})\not=0$ for some $r^{\prime}>0$ and $d_{4r}(1\otimes y)=0$ for all
$r\geq 0$. Firstly, suppose that $m=4i_{0}$ and $r^{\prime}=n-i_{0}+1$, where
$1\leq i_{0}\leq n$, then $E_{4n-m+5}^{k,4n+3}=0,E_{4n-m+5}^{k,q}=E_{2}^{k,q}$
for all $k\geq 0$ and $q=0,m+4n+3$. Also, $E_{4n-m+5}^{k,m}=E_{2}^{k,m}$ for
all $k\leq 4n-m$ and trivial otherwise. Since $G$ acts freely on $X$,
therefore, $d_{m+4n+4}(1\otimes u_{m+4n+3})=ct^{n+i_{0}+1}\otimes 1$ for some
$c\not=0$ in $R$. Then $E_{\infty}^{4k,q}=R$ for all ($4k\leq m+4n,q=0$),
($4k\leq 4n-m,q=m$) and trivial otherwise. This implies that $t\otimes 1\in
E_{2}^{4,0}$ and $1\otimes y\in E_{2}^{0,m}$ are permanent cocycles. Then by
edge homomorphism there exist $u\in E_{\infty}^{4,0}$ and $w\in
E_{\infty}^{0,m}$ corresponding $t\otimes 1$ and $1\otimes y$ respectively
with $\pi^{*}(t)=u$. We have $w^{2}=u^{n+i_{0}+1}=u^{n-i_{0}+1}w=0$. This
implies that

$\displaystyle\textnormal{Tot}E_{\infty}^{*,*}\cong\mathbb{R}[u,w]/\langle
w^{2},u^{n+i_{0}+1},u^{n-i_{0}+1}w\rangle,$

where $\deg u=4\text{ and }\deg w=m.$ Then there exist an element $v\in
H^{m}(X_{G})$ corresponding to $w\in E_{\infty}^{0,m}$ such that $i^{*}(v)=y$.
We have $u^{n-i_{0}+1}v=\alpha u^{n+1}$ and $v^{2}=\beta u^{2i_{0}}+\gamma
u^{i_{0}}v$, where $\alpha,\beta,\gamma\in R$ and $\beta=0$ if $m>4n$. So, the
ring cohomology of $X/G$ is given by $H^{*}(X_{G})\cong R[u,v]/\langle
u^{n-i_{0}+1}v-\alpha u^{n+1},v^{2}-\beta u^{2i_{0}}-\gamma
u^{i_{0}}v,u^{n+i_{0}+1}\rangle$ where $\deg u=4,\deg v=m$ and
$\alpha,\beta,\gamma\in R$, $\beta=0$ if $m>4n$, which is a contradiction. So,
$r^{\prime}$ must be $n+1$. This implies that $d_{4n+4}(1\otimes
yu_{4n+3})\not=0$. Consequently, $u_{4n+m+3}=\alpha yu_{4n+3}$ for some
$\alpha\not=0$ in $R$. Obviously, $y^{2}=0$, and $u_{4n+3}^{2}=0$ for
$m\not\in\\{4n+3,8n+6\\}$. If $m=4n+3$ then $u_{4n+3}^{2}\not=\beta
u_{4n+m+3}$ for any $\beta$ in $R$ and we get $u_{4n+3}^{2}=0$. If $m=8n+6$
then $u_{4n+3}^{2}$ may be both zero or nonzero. Thus,

$H^{*}(X,R)=R[y,u_{4n+3}]/\langle y^{2},u_{4n+3}^{2}\rangle,$

where $\deg y=m$ and $\deg u_{4n+3}=4n+3$. This realizes possibility (i). If
$u_{4n+3}^{2}\not=0$ then $u_{4n+3}=\alpha y$ for some nonzero element
$\alpha\in R$. Thus, the cohomology algebra of $X$ is given by
$R[u_{4n+3}]/\langle u_{4n+3}^{4}\rangle,$ where $\deg u_{4n+3}=4n+3$. By [8,
Theorem 4L.9], this cohomology algebra is not possible for $R=\mathbb{Z}_{2}$
and by the commutativity of cup product this cohomology algebra is also not
possible for $R=\mathbb{Q}$.

Case(c): When $\cup:H^{0}(X/G)\to H^{4}(X/G)$ maps $1$ to $cb$, where
$c\not=0$ in $R$.

In this case $m$ must be 4, and we get $H^{4i}(X)\cong R$ with basis
$\\{x^{i}\\}$, where $p^{*}_{4}(a)=x$. Clearly,
$H^{3}(X)=H^{4i+1}(X)=H^{4i+2}(X)=0$ for all $i\geq 0$. As
$\ker\cup:H^{4i}(X/G)\to H^{4i+4}(X/G)$ is generated by $a^{i-1}b$, we have
$H^{4i+3}(X)\cong R$ with basis $\\{u_{4i+3}\\}$, where
$\rho_{4i+3}(u_{4i+3})=a^{i-1}b$ for all $1\leq i\leq n+1$. We must have
$d_{4}(1\otimes u_{7})=0$ and $d_{8}(1\otimes u_{7})\not=0$. Consequently,
$u_{4i+7}=c_{i}x^{i}b_{7}$ for some nonzero $c_{i}\in R$, for all $1\leq i\leq
n$. Therefore, $H^{*}(X)\cong R[x,u_{7}]/\langle u_{7}^{2},x^{n+1}\rangle$,
where $\deg x=4$ and $\deg u_{7}=7$. This realizes the possibility (iii).

Case(d): When $\cup:H^{0}(X/G)\to H^{4}(X/G)$ maps $1$ to $ca+c^{\prime}b$,
where $c,c^{\prime}\not=0$ in $R$.

In this case also $m$ must be 4 and $H^{4}(X)\cong R$ with basis $\\{x\\}$,
where $p^{*}_{4}(a)=x$. By the exactness of the Gysin sequence, we get
$H^{3}(X)=H^{4i+1}(X)=H^{4i+2}(X)=0$ for all $i\geq 0$. As
$\cup:H^{4i}(X/G)\to H^{4i+4}(X/G)$ is an isomorphism, we get
$H^{4i+3}(X)=H^{4i+4}(X)=0$ for all $0<i<n$. Note that $H^{4n+4}(X)=0$ and
$\ker\cup:H^{4n}(X/G)\to H^{4n+4}(X/G)$ is generated by
$\\{a^{n-1}b-\frac{c}{c^{\prime}}a^{n}\\}$. This implies that
$H^{4n+3}(X)\cong R$ with basis $\\{u_{4n+3}\\}$, where
$\rho_{4n+3}(u_{4n+3})=a^{n-1}b-\frac{c}{c^{\prime}}a^{n}$. Obviously,
$H^{4n+7}(X)\cong R$ with basis $\\{u_{4n+7}\\}$, where
$\rho_{4n+7}(u_{4n+7})=a^{n}b$. Using similar arguments as in case(b),
$H^{*}(X)\cong R[x,u_{4n+3}]/\langle u_{4n+3}^{2},x^{2}\rangle$, where $\deg
x=4$ and $\deg u_{4n+3}=4n+3$. This realizes the possibility (i). ∎

Next, we discuss similar result for circle actions with the orbit space
product of a complex projective space and sphere:

######  Theorem 3.2.

Let $G=\mathbb{S}^{1}$ act freely on a finitistic connected space $X$ with
$X/G\sim_{R}\mathbb{CP}^{n}\times\mathbb{S}^{m}$, where $R$ is $\mathbb{Q}$ or
$\mathbb{Z}_{2}$. The cohomology algebra of $X$ with coefficients in $R$ is
isomorphic to the cohomology algebra of one of the following:

  1. (i)

$\mathbb{S}^{m}\times\mathbb{S}^{2n+1}$;

  2. (ii)

$\mathbb{S}^{1}\times\mathbb{S}^{m}\times\mathbb{CP}^{n};$

  3. (iii)

$\mathbb{RP}^{2n+1}\times\mathbb{S}^{m};$

  4. (iv)

$R[x,y]/\langle x^{n+1},y^{2}+\alpha x^{3}\rangle$, where $\deg x=2,\deg
y=3,m=2$, $\alpha\in R$ and $\alpha=0$ for $R=\mathbb{Q}$.

###### Proof.

As $X$ is connected, $H^{0}(X/G)\cong H^{0}(X)$, and clearly, $H^{i}(X)=0$ for
all $i>m+2n+1$. We consider the following cases:

Case(a): When $\cup:H^{0}(X/G)\to H^{2}(X/G)$ is trivial.
First assume that $m>2n$. In this case, $\cup:H^{i}(X/G)\to H^{i+2}(X/G)$ is
trivial for all $i\geq 0$. Then $\rho_{k+2i+1}$ and $p^{*}_{k+2i+2}$ are
isomorphisms for all $0\leq i<n$ and $k=0,m$. This implies that
$H^{k+2i+1}(X)\cong H^{k+2i+2}(X)\cong R$. Suppose that $\\{u_{2i+1}\\}$ and
$\\{x^{i+1}\\}$ denotes the bases for $H^{2i+1}(X)$ and $H^{2i+2}(X)$,
respectively, where $\rho_{2i+1}(u_{2i+1})=a^{i}$ and $p^{*}_{2}(a)=x$. Also,
$H^{m+2n+1}(X)\cong R$. Let $\\{v_{m+2i+1}\\}$ be basis for $H^{m+2i+1}(X)$
with $\rho_{m+2i+1}(v_{m+2i+1})=a^{i}b$ for all $0\leq i\leq n$. Note that
$H^{i}(X)=0$, for all $2n+1<i<m$. If $m\not=2n+1$, then by the exactness of
Gysin sequence, $H^{2n+1}(X)\cong H^{m}(X)\cong R$ with bases $\\{u_{2n+1}\\}$
and $\\{y\\}$, respectively; and if $m=2n+1$ then $H^{2n+1}(X)\cong R\oplus R$
with basis $\\{u_{2n+1},y\\}$, where $\rho_{2n+1}(u_{2n+1})=a^{n}$ and
$p^{*}_{m}(b)=y$.

Now, suppose that $m\leq 2n$. We consider two cases when $m$ is even or odd:

m is odd: Let $m=2i_{0}+1$ for some $i_{0}<n$. In this case,
$\cup:H^{2i}(X/G)\to H^{2i+2}(X/G)$ is trivial for all $i\geq 0.$ Then
$\rho_{2i-1}$ and $p^{*}_{2i}$ are isomorphisms for all $0\leq i\leq i_{0}$.
This implies that $H^{2i-1}(X)\cong H^{2i}(X)\cong R$ with bases
$\\{u_{2i-1}\\}$ and $\\{x^{i}\\}$, respectively, where
$\rho_{2i-1}(u_{2i-1})=a^{i-1}$ and $p^{*}_{2}(a)=x$. For $i_{0}+1\leq i\leq
n$, $H^{2i}(X)\cong H^{2i-1}(X)\cong H^{2n+1}(X)\cong R\oplus R$ with bases
$\\{x^{i},v_{2i}\\}$, $\\{yx^{i-i_{0}-1},u_{2i-1}\\}$ and
$\\{yx^{n-i_{0}},u_{2n+1}\\}$, respectively; and for $n<i\leq i_{0}+n$, we
have $H^{2i}(X)\cong H^{2n+m+1}(X)\cong H^{2i+1}(X)\cong R$ with bases
$\\{v_{2i}\\}$, $\\{v_{2n+m+1}\\}$ and $\\{yx^{i-i_{0}}\\}$, respectively,
where $\rho_{2i}(v_{2i})=a^{i-i_{0}-1}b$, $\rho_{2i+1}(u_{2i+1})=a^{i}$ and
$p^{*}_{m}(b)=y$.

m is even: Let $m=2i_{0}$ for some $i_{0}\leq n$. We get that $\rho_{2i-1}$
and $p^{*}_{2i}$ are isomorphisms for all $i\geq 0$. This implies that for
each $0\leq i<i_{0}$, $H^{2i}(X)\cong H^{2i+1}(X)\cong R$ with bases
$\\{x^{i}\\}$ and $\\{u_{2i+1}\\}$, respectively, and for $i_{0}\leq i\leq n$,
we get $H^{2i}(X)\cong H^{2i+1}(X)\cong R\oplus R$ with bases
$\\{x^{i},x^{i-i_{0}}y\\}$ and $\\{u_{2i+1},v_{2i+1}\\}$, respectively, where
$p^{*}_{2}(a)=x,p^{*}_{m}(b)=y,\rho_{2i+1}(u_{2i+1})=a^{i}$ and
$\rho_{2i+1}(v_{2i+1})=a^{i-i_{0}}b$. Also, for $1\leq i\leq i_{0}$,
$H^{2n+2i}(X)\cong H^{2n+2i+1}(X)\cong R$ with bases $\\{x^{n+i-i_{0}}y\\}$
and $\\{v_{2n+2i+1}\\}$, respectively, where
$\rho_{2n+2i+1}(v_{2n+2i+1})=a^{n+i-i_{0}}b$.

Finally, for all $m\leq 2n$, we have

$H^{i}(X)=\begin{cases}R&\text{ if }j\leq i\leq m+j-1,j=0\text{ or }2n+2,\\\
R\oplus R&\text{ if }m\leq i\leq 2n+1\\\ 0&\text{ otherwise.}\end{cases}$

Now, we compute the cohomology ring structure of $X$. In the Leray Serre
spectral sequence for the Borel fibration $X\stackrel{{\scriptstyle
i}}{{\hookrightarrow}}X_{G}\stackrel{{\scriptstyle\pi}}{{\rightarrow}}B_{G}$,
it is easy to observe that $d_{r}(1\otimes x)=0=d_{r}(1\otimes y)$ for all
$r\geq 0$, and the images of $u_{2i+1}$ and $v_{m+2i+1}$ must be nonzero under
some differential, for all $0\leq i\leq n$. Consequently, $x^{i}u_{1}\not=0$
and $x^{i}u_{1}y\not=0$. Obviously, $x^{n+1}=y^{2}=0$.

For $m>2n$, it is clear that for $0\leq i\leq n$,
$u_{2i+1}=\alpha_{i}x^{i}u_{1}$ and $v_{m+2i+1}=\beta_{i}x^{i}u_{1}y$ for some
nonzero elements $\alpha_{i},\beta_{i}$ in $R$. In particular, for $m=2n+1$,
$x^{n}u_{1}$ can not be equal to any multiple of $y$ and so $u_{2n+1}$ is
generated by $\\{y,x^{n}u_{1}\\}$.

Now, suppose $m\leq 2n$. If $m=2i_{0}+1$ for some $i_{0}<n$ then
$u_{2i+1}=\alpha_{i}x^{i}u_{1}$ for $0\leq i<i_{0}$ and
$v_{m+2i+1}=\beta_{i}x^{i}u_{1}y$ for $n-i_{0}\leq i\leq n$, where
$\alpha_{i}$’s and $\beta_{i}$’s are nonzero elements in $R$. As $x^{i}u_{1}$
can not be equal to $x^{i-i_{0}}y$, for all $i_{0}\leq i\leq n$, therefore,
$u_{2i+1}$ is generated by $\\{x^{i}u_{1},x^{i-i_{0}}y\\}$, for all $i_{0}\leq
i\leq n$. Also, $x^{i}u_{1}y$ can not be equal to any multiple of
$x^{i+i_{0}+1}$, so $v_{m+2i+1}$ is generated by
$\\{x^{i+i_{0}+1},x^{i}u_{1}y\\}$, for all $0\leq i<n-i_{0}$. If $m=2i_{0}$,
for some $i_{0}\leq n$ then $u_{2i+1}=\alpha_{i}x^{i}u_{1}$ for $0\leq
i<i_{0}$ and $v_{2i+m+1}=\beta_{i}x^{i}u_{1}y$ for $n-i_{0}+1\leq i\leq n$,
where $\alpha_{i}$’s and $\beta_{i}$’s are nonzero elements in $R$. We observe
that $x^{i}u_{1}$ can not be equal to any multiple of $x^{i-i_{0}}u_{1}y$, for
all $i_{0}\leq i\leq n$. Thus, the elements $u_{2i+1}$ and $v_{2i+1}$ are
generated by $\\{x^{i}u_{1},x^{i-i_{0}}u_{1}y\\}$. Note that $u_{1}^{2}$ may
be both zero or nonzero. If $u_{1}^{2}=0$ then
$X\sim_{R}\mathbb{S}^{1}\times\mathbb{CP}^{n}\times\mathbb{S}^{m}$. If
$u_{1}^{2}\not=0$, then by the commutativity of cup product,
$R=\mathbb{Z}_{2}$ and
$X\sim_{\mathbb{Z}_{2}}\mathbb{RP}^{2n+1}\times\mathbb{S}^{m}$. This realizes
possibilities (ii) and (iii) of the theorem.

Case(b): When $\cup:H^{0}(X/G)\to H^{2}(X/G)$ maps $1$ to $ca$ for some
$c\not=0$ in $R$.

First, suppose that $m>2n$. In this case, $\rho_{2i+1+j}$ and $p^{*}_{2i+2+j}$
are trivial for all $0\leq i<n$ and $j=0$ or $m$. Consequently,
$H^{2i+1+j}(X)=H^{2i+2+j}(X)=0$. Note that $H^{i}(X)=0$ for $2n+1<i<m$ and
$H^{m+2n+1}(X)\cong R$ with basis $\\{u_{m+2n+1}\\}$, where
$\rho_{m+2n+1}(u_{m+2n+1})=a^{n}b$. Thus, for $m\not=2n+1$, $\rho_{2n+1}$ and
$p^{*}_{m}$ are isomorphisms. This implies that $H^{2n+1}(X)\cong
H^{m}(X)\cong R$ with bases $\\{u_{2n+1}\\}$ and $\\{y\\}$ respectively. If
$m=2n+1$ then $H^{2n+1}(X)\cong R\oplus R$ with basis $\\{u_{2n+1},y\\}$,
where $\rho_{2n+1}(u_{2n+1})=a^{n}$ and $p^{*}_{m}(b)=y$.

It is easy to see that, for $m\leq 2n$, the cohomology groups and generators
are the same as above.

In the Leray-Serre spectral sequence for the Borel fibration $X\hookrightarrow
X_{G}\stackrel{{\scriptstyle\pi}}{{\to}}B_{G}$, $d_{2r^{\prime}}(1\otimes
u_{2n+1})\not=0$ for some $r^{\prime}>0$ and $d_{2r}(1\otimes y)=0$ for all
$r\geq 0$. Let if possible $m=2i_{0}$ and $r^{\prime}=n-i_{0}+1$, where $1\leq
i_{0}\leq n$. Then as done in previous theorem, we get $H^{*}(X_{G})\cong
R[u,v]/\langle u^{n-i_{0}+1}v-\alpha u^{n+1},v^{2}-\beta u^{2i_{0}}-\gamma
u^{i_{0}}v,u^{n+i_{0}+1}\rangle$ where $\deg u=2,\deg v=m$ and
$\alpha,\beta,\gamma\in R$, $\beta=0$ if $m>2n$, which is a contradiction.
Therefore, $r^{\prime}$ must be $n+1$. This implies that $d_{2n+2}(1\otimes
yu_{2n+1})\not=0$. Consequently, $u_{m+2n+1}=\alpha yu_{2n+1}$ for some
$\alpha\not=0$ in $R$. Obviously, $y^{2}=0$ and $u_{2n+1}^{2}=0$ for
$m\not\in\\{2n+1,4n+2\\}$. If $m=2n+1$ then
$u_{2n+1}^{2}\not=\alpha^{\prime}u_{m+2n+1}$ for any $\alpha^{\prime}$ in $R$
and so $u_{2n+1}^{2}=0$. If $m=4n+2$ then $u_{2n+1}^{2}$ may be both zero or
nonzero. If $u_{2n+1}^{2}=0$ then $H^{*}(X)=R[y,u_{2n+1}]/\langle
y^{2},u_{2n+1}^{2}\rangle$, where $\deg y=m$ and $\deg u_{2n+1}=2n+1$. This
realizes possibility (i) of the theorem. If $u_{2n+1}^{2}\not=0$ then
$u_{2n+1}=\beta y$ for some nonzero element $\beta$ in $R$. So, we get
$H^{*}(X)=R[u_{2n+1}]/\langle u_{2n+1}^{4}\rangle$, where $\deg u_{2n+1}=2n+1$
and $m=4n+2$. By the commutativity of cup product this is not possible for
$R=\mathbb{Q}$. Also, by [8, Theorem 4L.9], it is not possible for
$R=\mathbb{Z}_{2}$.

Case(c): When $\cup:H^{0}(X/G)\to H^{2}(X/G)$ maps $1$ to $cb$, where
$c\not=0$ in $R$.

In this case, $m$ must be 2 and $H^{1}(X)=0$. As $H^{2i-1}(X/G)=0$ for all
$i\geq 0$, therefore, we have $H^{2i-2}(X)\cong\operatorname{im}p^{*}_{2i-2}$
and $H^{2i+1}(X)\cong\ker\\{\cup:H^{2i}(X/G)\to H^{2i+2}(X/G)\\}$ with bases
$\\{x^{i}\\}$ and $\\{u_{2i+1}\\}$, respectively, where
$\rho_{2i+1}(u_{2i+1})=a^{i-1}b$ and $p^{*}_{2}(a)=x$ for all $1\leq i\leq
n+1$. In the Leray-Serre spectral sequence, $d_{2r}(1\otimes x)=0$ for all
$r\geq 0$ and $d_{2r^{\prime}}(1\otimes u_{3})\not=0$ for some $r^{\prime}>0$.
If $r^{\prime}=1$ then $H^{*}(X_{G})\cong R[u,v]/\langle
u^{n+2},v^{n+1},uv\rangle$, where $\deg u=\deg v=2$, a contradiction.
Therefore, $r^{\prime}$ must be $2$, and we get $x^{i}u_{3}\not=0$ for all
$1\leq i\leq n$. This implies that $u_{2i+3}=\alpha_{i}x^{i}u_{3}$ for some
nonzero $\alpha_{i}\in R$. We have $u_{3}^{2}=\alpha x^{3}$ for some
$\alpha\in R$. By the commutativity of cup product, we get $2u_{3}^{2}=0$.
Therefore, for $R=\mathbb{Q}$, $\alpha$ must be zero. Therefore,
$H^{*}(X)\cong R[x,u_{3}]/\langle x^{n+1},u_{3}^{2}-\alpha x^{3}\rangle$,
where $\deg x=2,$ $\deg u_{3}=3$ and $\alpha=0$ for $R=\mathbb{Q}$. This
realizes possibility (iv).

Case(d): When $\cup:H^{0}(X/G)\to H^{2}(X/G)$ maps $1$ to $ca+c^{\prime}b$,
where $c,c^{\prime}\not=0$ in $R$.

We have $m=2$ and $\cup:H^{2i}(X/G)\to H^{2i+2}(X/G)$ is an isomorphism, for
all $0<i<n$. By the exactness of Gysin sequence, $H^{2i-1}(X)=H^{2i+2}(X)=0$
for all $0<i\leq n$; and $H^{2}(X)\cong H^{2n+1}(X)\cong H^{2n+3}(X)\cong R$
with bases $\\{x\\}$, $\\{u_{2n+1}\\}$ and $\\{u_{2n+3}\\}$, respectively,
where $p^{*}_{2}(a)=x$,
$\rho_{2n+1}(u_{2n+1})=a^{n-1}b-\frac{c}{c^{\prime}}a^{n}$ and
$\rho_{2n+3}(u_{2n+3})=a^{n}b$. Hence, $H^{*}(X)\cong R[x,u_{2n+1}]/\langle
u_{2n+1}^{2},x^{2}\rangle$, where $\deg x=2$ and $\deg u_{2n+1}=2n+1$. This
realizes possibility (i). ∎

Finally, we classify a finitistic space $X$ equipped with a free involution
and the orbit space a product of real projective space and sphere:

######  Theorem 3.3.

Let $G=\mathbb{Z}_{2}$ act freely on a finitistic connected space $X$ with
$X/G\sim_{\mathbb{Z}_{2}}\mathbb{RP}^{n}\times\mathbb{S}^{m}$. Then the
cohomology algebra of $X$ is one of the following:

  1. (i)

$\mathbb{S}^{m}\times\mathbb{S}^{n}$;

  2. (ii)

$\mathbb{Z}_{2}[x,y]/\langle x^{n+1},y^{2}+\alpha x^{2}\rangle$, where $\deg
x=\deg y=1,m=1$ and $\alpha\in\mathbb{Z}_{2}$;

  3. (iii)

$\mathbb{Z}_{2}[y]/\langle y^{4}\rangle,$ where $\deg y=n$, $m=2n$ and $n=1,2$
or 4.

###### Proof.

Clearly, $H^{0}(X/G)\cong H^{0}(X)$ and $H^{i}(X)=0$ for all $i>m+n$. It is
easy to see that Euler class of the 0-sphere bundle $X\to X/G$ must be
nontrivial. Now, we consider the following cases:

Case(a): When $\cup:H^{0}(X/G)\to H^{1}(X/G)$ maps $1$ to $a$.

First, we suppose that $m\geq n$. For $m\not=n$, we get $\cup:H^{k+i}(X/G)\to
H^{k+i+1}(X/G)$ is an isomorphism for all $1\leq i<n$ and $k=0$ or $m$. This
implies that $\rho_{k+i}$ and $p^{*}_{k+i+1}$ are trivial homomorphism.
Consequently, $H^{k+i}(X)=0$ and $H^{n}(X)\cong H^{m}(X)\cong R$ with bases
$\\{u_{n}\\}$ and $\\{y\\}$ respectively, where $\rho_{n}(u_{n})=a^{n}$ and
$p^{*}_{m}(b)=y$. Note that $H^{i}(X)=0$ for $n<i<m$, $H^{m+n}(X)\cong R$ with
basis $\\{u_{m+n}\\}$, where $\rho_{m+n}(u_{m+n})=a^{n}b$. For $n=m$,
$H^{n}(X)\cong R\oplus R$ with basis $\\{u_{n},y\\}$, where
$\rho_{n}(u_{n})=a^{n}$ and $p^{*}_{m}(b)=y$.

Next, suppose that $m<n$. We get $\cup:H^{i}(X/G)\to H^{i+1}(X/G)$ is an
isomorphism for all $1<i<n+m$ and $i\not=m-1,n$. Note that the map
$\cup:H^{m-1}(X/G)\to H^{m}(X/G)$ is injective with $\operatorname{im}\cong R$
with basis $\\{a^{m}\\}$, and for $i=n$ it is surjective with $\ker\cong R$
with basis $\\{a_{n}\\}$, where $\rho_{n}(u_{n})=a^{n}$. This implies that
$H^{i}(X)=0$ for $1<i<n+m$ and $i\not=m,n$, and $H^{m}(X)\cong H^{n}(X)\cong
R$ with basis $\\{y\\}$ and $\\{u_{n}\\}$, respectively, where
$p^{*}_{m}(b)=y$ and $\rho_{n}(u_{n})=a^{n}$. Also, $H^{n+m}\cong R$ with
basis $\\{u_{n+m}\\}$, where $\rho_{n+m}(u_{n+m})=a^{n}b$.

Now, we compute the cohomology ring structure of $X$. In the Leray-Serre
spectral sequence for the Borel fibration $X\hookrightarrow
X_{G}\stackrel{{\scriptstyle\pi}}{{\to}}B_{G}$, $d_{r^{\prime}}(1\otimes
u_{n})\not=0$ for some $r^{\prime}>0$ and $d_{r}(1\otimes y)=0$ for all $r\geq
0$. Let it possible for $n>m$, $r^{\prime}=n-m+1$. Then we get
$H^{*}(X_{G})\cong R[u,v]/\langle u^{n-m+1}v-\alpha u^{n+1},v^{2}-\beta
u^{2m}-\gamma u^{m}v,u^{n+i_{0}+1}\rangle$ where $\deg u=1,\deg v=m$ and
$\alpha,\beta,\gamma\in R$, $\beta=0$ if $m>n$, which is a contradiction.
Therefore, $r^{\prime}$ must be $n+1$. This implies that $d_{n+1}(1\otimes
yu_{n})\not=0$. Consequently, $u_{m+n}=yu_{n}$. Obviously, $y^{2}=0$, and
$u_{n}^{2}=0$ for $m\not\in\\{n,2n\\}$. If $m=n$ then $u_{n}^{2}\not=u_{m+n}$
and so $u_{n}^{2}=0$. If $m=2n$ then $u_{n}^{2}$ may be both zero or nonzero.
If $u_{n}^{2}=0$ then $H^{*}(X)=\mathbb{Z}_{2}[y,u_{n}]/\langle
y^{2},u_{n}^{2}\rangle$, where $\deg y=m$ and $\deg u_{n}=n$. If
$u_{n}^{2}\not=0$ then the cohomology algebra of $X$ is
$\mathbb{Z}_{2}[u_{n}]/\langle u_{n}^{4}\rangle$, where $\deg u_{n}=n$ and
$m=2n$. By [8, Theorem 4L.9], $n=1,2,4$ or 8. This realizes possibility (i)
and (iii).

Case(b): When $\cup:H^{0}(X/G)\to H^{2}(X/G)$ maps $1$ to $b$.

In this case, $m$ must be 1 and $\operatorname{im}p^{*}_{1}=\mathbb{Z}_{2}$
with basis $\\{p^{*}_{1}(a)\\}$. Also,
$\operatorname{im}p^{*}_{i}\cong\operatorname{im}\rho_{i}\cong\mathbb{Z}_{2}$
with basis $\\{p^{*}_{i}(a^{i})\\}$ and $\\{a^{i-1}b\\},$ respectively, for
all $0<i\leq n$. Consequently,
$H^{i}(X)\cong\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$ with basis
$\\{x^{i},u_{i}\\}$, where $\rho_{i}(u_{i})=a^{i-1}b$ and $p^{*}_{1}(a)=x$.
Obviously, $H^{n+1}(X)\cong\mathbb{Z}_{2}$ with basis $\\{u_{n+1}\\}$, where
$\rho_{n+1}(u_{n+1})=a^{n}b$. In the Leray-Serre spectral sequence,
$d_{r}(1\otimes x)=0$ for all $r\geq 0$ and $d_{2}(1\otimes u_{1})\not=0$.
This implies that for $1\leq i\leq n$, $x^{i}u_{1}\not=0$, and hence
$u_{i+1}=x^{i}u_{1}+\alpha_{i}x^{i+1}$ for some $\alpha_{i}\in\mathbb{Z}_{2}$.
As $d_{2}(1\otimes xu_{1})\not=0$, we have $u_{1}^{2}=\alpha x^{2}$ for some
$\alpha\in\mathbb{Z}_{2}$. Therefore,
$H^{*}(X)\cong\mathbb{Z}_{2}[x,u_{1}]/\langle x^{n+1},u_{1}^{2}+\alpha
x^{2}\rangle$, where $\deg x=\deg u_{1}=1$. This realizes possibility (ii).

Case(c): When $\cup:H^{0}(X/G)\to H^{2}(X/G)$ maps $1$ to $a+b$.

In this case also $m$ must be 1 and $\cup:H^{i}(X/G)\to H^{i+1}(X/G)$ is an
isomorphism for all $0<i\leq n$. Consequently, $H^{i}(X)=0$ for $1<i<n$. Note
that $\ker p^{*}_{1}\cong\operatorname{im}\rho_{n}\cong\mathbb{Z}_{2}$ with
bases $\\{a+b\\}$ and $\\{a^{n}+a^{n-1}b\\}$, respectively. This implies that
for $n\not=1$, $H^{1}(X)\cong H^{n}(X)\cong\mathbb{Z}_{2}$ with bases
$\\{x\\}$ and $\\{u_{n}\\}$, respectively, where $p^{*}_{1}(a)=x$ and
$\rho_{n}(u_{n})=a^{n}+a^{n-1}b$. Obviously, $H^{n+1}(X)\cong\mathbb{Z}_{2}$
with basis $\\{u_{n+1}\\}$, where $\rho_{n+1}(u_{n+1})=a^{n}b$. In particular
for $n=1$, $H^{1}(X)\cong\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$,
$H^{2}(X)\cong\mathbb{Z}_{2}$ with bases $\\{x,u_{1}\\}$ and $\\{u_{2}\\}$,
respectively, where $p^{*}_{1}(a)=x,\rho_{1}(u_{1})=a+b$ and
$\rho_{2}(u_{2})=ab$. We must have $u_{n+1}=xu_{n}$ and $u_{n}^{2}=0$. Thus,
$X\sim_{\mathbb{Z}_{2}}\mathbb{S}^{1}\times\mathbb{S}^{n}$. This realizes
possibility (i). ∎

Now, we determine covering dimension of coincidence set $A(f)$ of continuous
maps $f:X\to\mathbb{R}^{k}$, where $X$ is a finitistic space equipped with
free involution and
$X/G\sim_{\mathbb{Z}_{2}}\mathbb{RP}^{n}\times\mathbb{S}^{m}$.

######  Theorem 3.4.

Let $G=\mathbb{Z}_{2}$ act freely on a finitistic space $X$ with
$X/G\sim_{\mathbb{Z}_{2}}\mathbb{RP}^{n}\times\mathbb{S}^{m},$ where
$n>4,m>1$. If $f:X\to\mathbb{R}^{k}$ be any continuous map, then
$cov.dim(A(f))\geq n-k$ for $k\leq n$.

###### Proof.

By Theorem 3.3, the Volovikov’s index $in(X)$ [22] is $n$. Note that
$in(A(f))\geq in(X)-k$ [23], so we have $in(A(f))\geq n-k$. As $G$ acts freely
on $X$, it induces a free action on $A(f)$. By the definition of $in(X)$,
$H^{n-k}(B_{G})\to H^{n-k}(A(f)_{G})$ is injective. This implies that
$cohom.dim(A(f)_{G})\geq n-k$, where $cohom.dim$ denotes the cohomological
dimension of a space. As $A(f)/G$ and $A(f)_{G}$ are homotopy equivalent,
$cohom.dim(A(f)/G)\geq n-k$. Consequently, $cohom.dim(A(f))\geq n-k$ [15,
Proposition A.11]. The result follows from the fact that $cov.dim(A(f))\geq
cohom.dim(A(f))$. ∎

##  4\. Examples

Consider the standard free actions of $G=\mathbb{S}^{d}$ on
$\mathbb{S}^{(d+1)n+d}$, for $d=0,1$ or 3, and trivial action on
$\mathbb{S}^{m}$, then under the diagonal action $G$ acts freely on
$\mathbb{S}^{(d+1)n+d}\times\mathbb{S}^{m}$ with the orbit space
$\mathbb{FP}^{n}\times\mathbb{S}^{m}$, where
$\mathbb{F}=\mathbb{R},\mathbb{C}$ or $\mathbb{H}$, respectively. This
realizes the possibility (i) of Theorems 3.1, 3.2 and 3.3. Similarly, if we
take free actions of $G=\mathbb{S}^{d}$ on $\mathbb{S}^{2d+1}$ and trivial
action on $\mathbb{FP}^{n}$ then $G$ acts freely on
$\mathbb{FP}^{n}\times\mathbb{S}^{2d+1}$ with the orbit space
$\mathbb{FP}^{n}\times\mathbb{FP}^{1}$, respectively. This realizes the
possibility (iii) of Theorem 3.1, when $\alpha=0$, the possibility (iv) of
Theorem 3.2 and the possibility (ii) of Theorem 3.3. If we take free action of
$G=\mathbb{S}^{d}$ on itself, for $d=1$ or 3, and trivial action on
$\mathbb{FP}^{n}\times\mathbb{S}^{m}$, where $\mathbb{F}=\mathbb{C}$ or
$\mathbb{H}$, respectively then $G$ acts freely on
$\mathbb{S}^{d}\times\mathbb{FP}^{n}\times\mathbb{S}^{m}$ with the orbit space
$\mathbb{FP}^{n}\times\mathbb{S}^{m}$. This realizes the possibility (ii) of
Theorems 3.1 and 3.2.

Now, consider free action of $\mathbb{Z}_{4}$ on
$\mathbb{S}^{3}\subseteq\mathbb{C}^{2}$ defined by
$(z_{1},z_{2})\mapsto(z_{1}e^{2\pi i/4},z_{2}e^{2\pi i/4})$. This induces a
free involution on $\mathbb{RP}^{3}$ with the orbit space $\text{L}^{3}(4,1)$.
We know that
$\text{L}^{3}(4,1)\sim_{\mathbb{Z}_{2}}\mathbb{RP}^{1}\times\mathbb{S}^{2}$.
Recall that if $G=\mathbb{Z}_{2}$ acts freely on a finitistic space $X$ with
the mod 2 cohomology $\mathbb{CP}^{3}$, then the orbit spaces $X/G$ is the mod
2 cohomology $\mathbb{RP}^{2}\times\mathbb{S}^{4}$ [16]. These examples
realizes the possibility (iii) of Theorem 3.3 for $n=1$ and $n=2$,
respectively.

Remark. Let $X$ be a connected Hausdorff space equipped with free action of
$G=\mathbb{Z}_{2}$. If the orbit space
$X/G=\mathbb{RP}^{n}\times\mathbb{S}^{m}$ then $X$ is homeomorphic to
$\mathbb{S}^{n}\times\mathbb{S}^{m}$. The significance of Theorem 3.3 lies
with the fact that if
$X/G\sim_{\mathbb{Z}_{2}}\mathbb{RP}^{n}\times\mathbb{S}^{m}$ then $X$ may
have the mod 2 cohomology isomorphic to $\mathbb{RP}^{n}\times\mathbb{S}^{1}$
or $\mathbb{S}^{n}\times\mathbb{S}^{1}$ for $m=1$; and
$\mathbb{S}^{4}\times\mathbb{S}^{2}$ or $\mathbb{CP}^{3}$ for $m=4$ and $n=2$.

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# Fuzzy Signed Measure

Jun Tanaka  University of California, Riverside, USA<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
(Date: January, 10, 2008)

###### Key words and phrases:

fuzzy measure, signed measure, Hahn decomposition, fuzzy sets

###### 2000 Mathematics Subject Classification:

Primary: 28A12, 28E10

##  1\. Introduction

It is well known that one could obtain the Caratheodory Extension Theorem on
fuzzy measurable space. Now a fuzzy measure is defined on $\sigma$-algebras.
As one already saw, Fuzzy measure is a classical measure, provided that fuzzy
sets are restricted to classical sets. As the classical measure theory goes,
we will define a fuzzy signed measure on $\sigma$-algebras, as well as
positive and negative sets. Herein, we will show that the Fuzzy Hahn
Decomposition Theorem, which is a generalization of the classical Hahn
Decomposition Theorem, decompose any space X into a positive set A and a
negative set B such that A+B=X and the signed measure of $A\wedge B$ is 0.

##  2\. Preliminaries

In this section, we shall briefly review the well know facts about lattice
theory (e.g. Birkhoff [1], Iwamura), propose an extension lattice, and
investigate its properties. Later in this section, we shall review
Caratheodory Extension Theorem on fuzzy measurable sets. (L,$\wedge$,$\vee$)
or simply L, under closed operations $\wedge$ $\vee$, is called a lattice. If
it satisfies in addition to the distributive law is called a lattice. For two
lattices L and L’, a bijection from L to L’, which preserves lattice
operations are called a lattice isomorphism, or simply an isomorphism. If
there is an isomorphism from L to L’, then L is called lattice-isomorphic with
L’, and we write L $\cong$ L’. We write x $\leq$ y if x $\wedge$ y = x or,
equivalently, if x $\wedge$ y = y. L is called complete, if any subset A of L
includes the supremum $\vee$A, and infimum $\wedge$A, with respect to the
above order. A complete lattice L includes the maximum and minimum elements,
which are denoted by I and O, or 1 and 0, respectively [1].

######  Definition 1.

Unless otherwise stated, X is a space and $\mu_{\Box}$ is a membership
function of any fuzzy set $\Box$. If a family $\sigma$ of membership functions
on X satisfies the following conditions, then it is called fuzzy
$\sigma$-algebra;

(1) $\forall\alpha\in$ [0,1], $\alpha$ is constant; $\alpha\in\sigma$.

(2) $\forall\mu\in\sigma$, 1-$\mu\in\sigma$.

(3) if $(\mu_{n})_{n\in N}\in\sigma^{N}$, then $\sup\mu_{n}\in\sigma$.

######  Definition 2.

If m : $\sigma$ $\mapsto$ $\mathbb{R}\cup\\{\infty\\}$ satisfying the
following properties, then m is called a fuzzy measure.

(1) m $(\emptyset)$ = m $(0)$ = 0.

(2) $\forall\mu,\eta\in\sigma$ s.t. $m(\mu),m(\eta)\geq$ 0 : $\mu\leq\eta$
$\Rightarrow$ m $(\mu)\leq$ m $(\eta)$.

(3) $\forall\mu,\eta\in\sigma$ : m$(\mu\vee\eta)+$ m$(\mu\wedge\eta)=$
m$(\mu)+$ m$(\eta)$.

(4) $(\mu_{n})_{n\in N}$ $\subset\sigma^{N}$ such that
$\mu_{1}\leq\mu_{2}\leq\cdots\leq\mu_{n}\leq\cdots$ : $\sup\mu_{n}=\mu$
$\Rightarrow$ m$(\mu)=\lim$ m$(\mu_{n})$.

######  Definition 3.

By an outer fuzzy measure $m^{\ast}$, we mean an extended real-value set
function defined on [0,1]X, having the following properties:

(1) $m^{\ast}$(0) = 0

(2) $m^{\ast}$ $(\mu)\leq$ $m^{\ast}$ $(\eta)$ for $\mu\leq\eta$

(3) $m^{\ast}(\vee_{i=1}^{\infty}\mu_{E_{i}})$ $\leq$
$\vee_{i=1}^{\infty}m^{\ast}(\mu_{E_{i}})$.

######  Example 1.

Suppose

$m^{\ast}(\mu_{E})=\begin{cases}0\quad&\mu_{E}=0\\\ 1\quad&\mu_{E}\neq
0\end{cases}$

Then $m^{\ast}$ is an outer fuzzy measure which is not fuzzy measure on
[0,1]X, if X has at least two points.

######  proposition 1.

Let F be a class of fuzzy subsets of X containing empty set such that for
every $\mu_{A}\leq\mu_{X}$, there exists a sequence
$(\mu_{B_{n}})^{\infty}_{n=1}$, $\mu_{B_{n}}\in F$ such that $\mu_{A}\leq$
$\sup(\mu_{B_{n}})^{\infty}_{n=1}$. Let $\tau$ be an extended real-valued
function on F such that $\tau(0)$ = 0 and $\tau(\mu_{A})\geq$ 0 for
$\mu_{A}\in$ F. Then $m^{\ast}$ defined on [0,1]X by $m^{\ast}(\mu_{A})$ =
$\inf\\{\tau(\vee^{\infty}_{i=1}\mu_{B_{n}}):\mu_{B_{n}}\in F$ s.t.
$\mu_{A}\leq\vee^{\infty}_{i=1}\mu_{B_{n}}\\}$ is a outer fuzzy measure.

###### Proof.

Clearly, $m^{\ast}(\emptyset)$ = 0. Secondly, if $\mu_{A_{1}}\leq\mu_{A_{2}}$
and $\mu_{A_{2}}\leq\vee^{\infty}_{i=1}\mu_{B_{n}}$, then
$\mu_{A_{1}}\leq\vee^{\infty}_{i=1}\mu_{B_{n}}$. Thus
$m^{\ast}(\mu_{A_{1}})\leq m^{\ast}(\mu_{A_{2}})$.

Finally, let $\mu_{E_{n}}\leq\mu_{X}$ for each natural number n. Then
$m^{\ast}$($\mu_{E_{n}}$) = $\infty$ for some n,
$m^{\ast}$($\vee_{n=1}^{\infty}\mu_{E_{n}}$) $\leq$
$\vee_{n=1}^{\infty}m^{\ast}\mu_{E_{n}}$.

∎

The following theorem is an extension of the above proposition.

######  Theorem 2.

The class B of $m^{\ast}$-fuzzy measurable sets is a $\sigma$-algebra. Also
$\overline{m}$, the restriction of $m^{\ast}$ to B, is a measure.

######  Theorem 3.

Let m be a fuzzy measure on a $\sigma$-algebra $\sigma\subseteq$ [0,1]X.

Suppose for $\mu_{E}\leq\mu_{X}$

$m^{\ast}(\mu_{E})=\inf\\{m(\vee^{\infty}_{i=1}\mu_{E_{n}}):\mu_{E_{n}}\in\sigma
s.t.\mu_{E}\leq\vee^{\infty}_{i=1}\mu_{E_{n}}\\}$

Then the following properties hold:

(i) $m^{\ast}$ is an outer fuzzy measure.

(ii) $\mu_{E}\in\sigma$ implies m($\mu_{E}$) = $m^{\ast}$($\mu_{E}$)

(iii) $\mu_{E}\in\sigma$ implies $\mu_{E}$ is $m^{\ast}$-fuzzy measurable.

(iv) The restriction $\overline{m}$ of $m^{\ast}$ to the $m^{\ast}$-fuzzy
measurable sets in an extension of m to a fuzzy measure on a fuzzy
$\sigma$-algebra containing $\sigma$.

(v) If m is fuzzy $\sigma$-finite, then $\overline{m}$ is the only fuzzy
measure (on the smallest fuzzy $\sigma$-algebra containing $\sigma$) that is
an extension of m.

##  3\. Fuzzy Signed Measure

Let $m_{1}$ and $m_{2}$ be fuzzy measures defined on the same $\sigma$-algebra
$\sigma$. If one of them is finite, the set function
$m(\mu_{E})=m_{1}(\mu_{E})-m_{2}(\mu_{E})$ , $\mu_{E}\in\sigma$ is well
defined and countably additive on $\sigma$. However, it is not necessarily
nonnegative; it is called a signed measure.

######  Definition 4.

By a fuzzy signed measure on the fuzzy measurable space (X, $\sigma$) we mean
$\nu$ : $\sigma$ $\mapsto$ $\mathbb{R}\cup\\{\infty\\}$ or
$\mathbb{R}\cup\\{-\infty\\}$, satisfying the following property:

(1) $\nu(\emptyset)$ = $\nu(0)$ = 0.

(2) $\forall\mu,\eta\in\sigma$ s.t. $\nu(\mu),\nu(\eta)\geq$ 0 : $\mu\leq\eta$
$\Rightarrow$ $\nu(\mu)\leq\nu(\eta)$.

$\forall\mu,\eta\in\sigma$ s.t. $\nu(\mu),\nu(\eta)\leq$ 0 : $\mu\leq\eta$
$\Rightarrow$ $\nu(\eta)\leq\nu(\mu)$.

(3) $\forall\mu,\eta\in\sigma$ :
$\nu(\mu\vee\eta)+\nu(\mu\wedge\eta)=\nu(\mu)+\nu(\eta)$.

(4) $(\mu_{n})_{n\in N}$ $\subset\sigma^{N}$ such that
$\mu_{1}\leq\mu_{2}\leq\cdots\leq\mu_{n}\leq\cdots$ : $\sup\mu_{n}=\mu$
$\Rightarrow$ $\nu(\mu)=\lim\nu(\mu_{n})$.

This is meant in the sense that if the left-hand side is finite, the series on
the right-hand side is convergent, and if the left-hand side is $\pm\infty$,
then the series on the right-hand side diverges accordingly.

######  Remark 1.

The Fuzzy signed measure is a fuzzy measure when it takes only positive value.
Thus, the fuzzy signed measure is a generalization of fuzzy measure.

######  Definition 5.

A is a positive fuzzy set if for any fuzzy measurable set E in A,
$\nu(\mu_{E})\geq 0$. Similarly, B is a negative fuzzy set if for any fuzzy
measurable set E in B, $\nu(\mu_{E})\leq 0$.

######  Lemma 1.

Every fuzzy subset of a positive fuzzy set is a positive fuzzy set and any
countable union of positive fuzzy sets is a positive fuzzy set.

###### Proof.

The first claim is clear. Before we show the second claim, we need to show
that every union of positive sets is a positive set. Let A, B be fuzzy
positive sets and E $\leq$ A $\vee$ B be a fuzzy measurable set. By (2) in
Definition 4, 0 $\leq$ $\nu(\mu_{B}\wedge\mu_{E})$ \-
$\nu(\mu_{A}\wedge\mu_{B}\wedge\mu_{E})$. By (3), $\nu(\mu_{E})\geq$ 0\. Now
by induction, every finite union of fuzzy positive sets is a positive fuzzy
set. Let $A_{n}$ be a positive fuzzy set for all n and E $\leq$ $\vee$ $A_{n}$
be a fuzzy measurable set. Then $\mu_{E_{m}}$ :=
$\mu_{E}\wedge\vee_{n=1}^{m}\mu_{A_{n}}$ =
$\vee_{n=1}^{m}\mu_{E}\wedge\mu_{A_{n}}$. Then $E_{m}$ is a fuzzy measurable
set and a positive fuzzy set. In particular, $\mu_{E_{m}}\leq\mu_{E_{m+1}}$
for all n and $\mu_{E}$ = $\vee_{n=1}^{\infty}\mu_{E_{m}}$. Thus 0
$\leq\lim\nu(\mu_{E_{m}})$ = $\nu(\mu_{E})$. Therefore $\vee$ $A_{n}$ is a
positive set.

∎

######  Lemma 2.

Let E be a fuzzy measurable set such that 0 $<\nu(\mu_{E})<\infty$. Then there
is a positive fuzzy set A $\leq E$ with $\nu(\mu_{A})>$ 0.

###### Proof.

If E is a positive fuzzy set, we take A=E. Otherwise, E contains a set of
negative measure. Let $n_{1}$ be the smallest positive integer such that there
is a measurable set $E_{1}\subset$ E with $\nu(\mu_{E_{1}})<-\frac{1}{n_{1}}$.
Proceeding inductively, if E$\wedge\wedge_{j=1}^{k-1}E_{j}^{C}$ is not already
a positive set, let $n_{k}$ be the smallest positive integer for which there
is a fuzzy measurable set $E_{k}$ such that $E_{k}\leq
E\wedge\wedge_{j=1}^{k-1}E_{j}$ and $\nu(\mu_{E_{k}})<-\frac{1}{n_{k}}$.

Let A = $(\vee E_{k})^{C}$.

Then $\nu(\mu_{E})$ =
$\nu(\mu_{E}\wedge\mu_{A})+\nu(\mu_{E}\wedge\vee\mu_{E_{k}})$ =
$\nu(\mu_{E}\wedge\mu_{A})+\nu(\vee\mu_{E_{k}})$. Since $\nu(\mu_{E})$ is
finite, $\lim_{n\rightarrow\infty}\nu(\vee^{n}\mu_{E_{k}})$ is finite and
$\nu(\vee\mu_{E_{k}})\leq$ 0\. Since $\nu(\mu_{E})>$ 0 and
$\nu(\vee\mu_{E_{k}})\leq$ 0, $\nu(\mu_{E}\wedge\mu_{A})>$ 0.

We will show that A is a positive set. Let $\epsilon>$ 0\. Since
$\frac{1}{n_{k}}\rightarrow$ 0, we may choose k such that
$-\frac{1}{n_{k}-1}$, which is greater than $-\epsilon$. Thus A contains no
fuzzy measurable sets of measure less than $-\epsilon$. Since $\epsilon$ was
arbitrary positive number, it follows that A can contain no sets of negative
measure and so must be a positive fuzzy set.

∎

##  4\. Fuzzy Hahn Decomposition

Without loss of generality, let’s omit + $\infty$ value of $\nu$. Let
$\lambda$ = $\sup\\{\nu(\mu_{A}):A$ is a fuzzy positive set $\\}$.

Then $\lambda\geq$ 0 since $\nu(\emptyset)$ = 0.

Let $A_{i}$ be a sequence of positive fuzzy sets such that $\lambda$ =
$\lim\nu(\mu_{A_{i}})$ and $\mu_{A}$ = $\vee\mu_{A_{i}}$. By Theorem 1, A is a
positive fuzzy set and $\lambda\geq\nu(\mu_{A})$.

$\vee^{n}\mu_{A_{i}}\leq$ $\mu_{A}$ for any n implies
$\nu(\vee^{n}\mu_{A_{i}})\geq$ 0 for any n. Thus $\lambda$ =
$\lim\nu(\mu_{A_{i}})$ = $\nu(\mu_{A})$ = 0.

Let $E\leq A^{C}$ be a positive set. Then $\nu(\mu_{E})\geq$ 0 and $A\vee E$
is a positive fuzzy set. Thus $\lambda\geq$
$\nu(\mu_{A}\vee\mu_{E})=\nu(\mu_{A})+\nu(\mu_{E})-\nu(\mu_{A}\wedge\mu_{E})$
= $\lambda+\nu(\mu_{E})-\nu(\mu_{A}\wedge\mu_{E})$. Thus
$\nu(\mu_{E})=\nu(\mu_{A}\wedge\mu_{E})$. $\nu(\mu_{E})$ = 0 since
$\mu_{A}\wedge\mu_{E}\leq\mu_{A}\wedge\mu_{A}^{C}$ and
$\nu(\mu_{A}\wedge\mu_{A}^{C})$ = 0.

Thus, $A^{C}$ contains no fuzzy positive subsets of positive measure and hence
no subsets of positive measure by Lemma 2. Consequently, $A^{C}$ is a negative
fuzzy set.

##  5\. conclusion

Let X be a space. Then by the previous theorem, we find such a positive fuzzy
set A and a negative fuzzy set B (= $A^{C}$). By the fuzzy measurability of
$\nu$, $\nu(\mu_{A}\wedge\mu_{A}^{C})$ = 0. $A+A^{C}$ = X in the sense that
$\mu_{A}+\mu_{A^{C}}$ = 1. These characteristics provides X = $A\cup B$ and
$A\cap B=\emptyset$ in the classical set sense.

## References

  * [1] 1\. G. Birkhoff, Lattice Theory 3rd ed., AMS Colloquim Publications, Providence, RI, 1967
  * [2] 2 T. Iwamura, Sokuran, Kyoritu Shuppan, Tokyo, 1966.
  * [3] 3\. Z. Wang G.J. Klir, Fuzzy Measure Theory, Springer, 1993.
  * [4] 4\. L.A. Zadeh, Fuzzy sets, Inform and Control 8 (1965), 338-353
  * [5] 5\. L.A. Zadeh, Calculus of Fuzzy restrictions, Fuzzy Sets and their Applications to Cognitive and Decision Processes, L.A. Zadeh et al., eds., Academic Press, New York, 1975, pp. 1-39.
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