# Finitistic spaces with orbit space a product of projective spaces

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 $G=\mathbb{Z}_{2}$ act freely on a finitistic space $X$. If the mod 2
cohomology of $X$ is isomorphic to the real projective space
$\mathbb{RP}^{2n+1}$ (resp. complex projective space $\mathbb{CP}^{2n+1}$)
then the mod 2 cohomology of orbit spaces of these free actions are
$\mathbb{RP}^{1}\times\mathbb{CP}^{n}$ (resp.
$\mathbb{RP}^{2}\times\mathbb{HP}^{n}$) [7]. In this paper, we have discussed
converse of these result. We have showed that if the mod 2 cohomology of the
orbit space $X/G$ is $\mathbb{RP}^{1}\times\mathbb{CP}^{n}$ (resp.
$\mathbb{RP}^{2}\times\mathbb{HP}^{n}$) then the mod 2 cohomology of $X$ is
$\mathbb{RP}^{2n+1}$ or $\mathbb{S}^{1}\times\mathbb{CP}^{n}$ (resp.
$\mathbb{CP}^{2n+1}$ or $\mathbb{S}^{2}\times\mathbb{HP}^{n}$).

A partial converse of free involutions on the product of projective spaces
$\mathbb{RP}^{n}\times\mathbb{RP}^{2m+1}$ (resp.
$\mathbb{CP}^{n}\times\mathbb{CP}^{2m+1}$) are also discussed.

###### Key words and phrases:

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

###### 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 compact Lie group acting on a space $X$. An interesting problem
is to classify $X$ when $G$ acts freely on $X$ and the orbit space $X/G$ is
known. First, J. C. Su [9] worked in this direction. He proved that if
$G=\mathbb{Z}_{2}$ or $\mathbb{S}^{1}$, acts freely on a space $X$ and the
orbit space $X/G$ is the mod 2 or integral cohomology projective space
$\mathbb{FP}^{n}$, $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$, then the space $X$
is the mod 2 cohomology $n$-sphere $\mathbb{S}^{n}$ or integral cohomology
$(2n+1)$-sphere $\mathbb{S}^{2n+1}$. Su also proved that if
$G=\mathbb{Z}_{q}$, $q$ an odd prime, and $X/G=L_{q}^{2n+1}$ (lens space) then
$X$ is the mod $q$ cohomology $(2n+1)$-sphere $\mathbb{S}^{2n+1}$. It has also
been proved that if $G=\mathbb{S}^{3}$ and $X/G=\mathbb{HP}^{n}$ (quaternion
projective space), then $X$ is either integral cohomology $\mathbb{S}^{4n+3}$
or $\mathbb{S}^{3}\times\mathbb{HP}^{n}$ [4]. Harvey [2] proved that for semi-
free actions of $G=\mathbb{S}^{1}$ when the orbit space $X/G$ is a simply
connected 4-manifold, then the total space is a simply connected 5-manifold.

The orbit spaces of free actions of $G=\mathbb{Z}_{2}$ on projective spaces
$X=\mathbb{FP}^{2n+1}$ and the product of projective spaces
$X=\mathbb{FP}^{n}\times\mathbb{FP}^{m}$, where $\mathbb{F}=\mathbb{R}$ or
$\mathbb{C}$, have been discussed in [7, 8]. For $G=\mathbb{Z}_{2}$ and
$X=\mathbb{RP}^{2n+1}$ (resp. $\mathbb{CP}^{2n+1}$), it is proved that the mod
2 cohomology of the orbit space $X/G$ is
$\mathbb{RP}^{1}\times\mathbb{CP}^{n}$ (resp.
$\mathbb{RP}^{2}\times\mathbb{HP}^{n}$) [7]. We have discussed converse of
these results. We have determined the cohomology classification of a
finitistic space $X$ if the mod 2 cohomology of orbit space $X/G$ is
$\mathbb{RP}^{1}\times\mathbb{CP}^{n}$ or
$\mathbb{RP}^{2}\times\mathbb{HP}^{n}$. For $G=\mathbb{Z}_{2}$ and
$X=\mathbb{RP}^{n}\times\mathbb{RP}^{2m+1}$ (resp.
$\mathbb{CP}^{n}\times\mathbb{CP}^{2m+1}$), one of the possibilities of the
mod 2 cohomology of the orbit space is
$\mathbb{RP}^{1}\times\mathbb{RP}^{n}\times\mathbb{CP}^{m}$ (resp.
$\mathbb{RP}^{2}\times\mathbb{CP}^{n}\times\mathbb{HP}^{m}$) [8]. In this
paper, we have also determined $X$ equipped with a free action of $G$ when the
mod 2 cohomology of the orbit space $X/G$ is
$\mathbb{RP}^{1}\times\mathbb{RP}^{n}\times\mathbb{CP}^{m}$ or
$\mathbb{RP}^{2}\times\mathbb{CP}^{n}\times\mathbb{HP}^{m}$. This gives a
partial converse of results proved in [8].

All spaces are assumed to finitistic connected spaces. Recall that a
paracompact Hausdorff space $X$ is said to be finitistic if every open cover
of $X$ has a finite dimensional open refinement. All compact spaces and all
finite dimensional paracompact spaces are finitistic spaces. We have
considered Čech cohomology with coefficients in $\mathbb{Z}_{2}$. Throughout
the paper, $X\sim_{2}Y$ means $H^{*}(X;\mathbb{Z}_{2})\cong
H^{*}(Y;\mathbb{Z}_{2})$. We have proved the following results:

######  Theorem 1.1.

Let $G=\mathbb{Z}_{2}$ act freely on a finitistic connected space $X$ with
$X/G\sim_{\mathbb{Z}_{2}}\mathbb{RP}^{1}\times\mathbb{CP}^{n}$. Then either
$X\sim_{\mathbb{Z}_{2}}\mathbb{S}^{1}\times\mathbb{CP}^{n}$ or
$X\sim_{\mathbb{Z}_{2}}\mathbb{RP}^{2n+1}$.

######  Theorem 1.2.

Let $G=\mathbb{Z}_{2}$ act freely on a finitistic connected space $X$ with
$X/G\sim_{2}\mathbb{RP}^{2}\times\mathbb{HP}^{n}$. Then
$X\sim_{2}\mathbb{S}^{2}\times\mathbb{HP}^{n}$ or
$X\sim_{2}\mathbb{CP}^{2n+1}$.

######  Theorem 1.3.

Let $G=\mathbb{Z}_{2}$ act freely on a finitistic connected space $X$ with
$X/G\sim_{2}\mathbb{RP}^{1}\times\mathbb{RP}^{n}\times\mathbb{CP}^{m}$ and
$\pi_{1}(B_{G})$ act trivially on $H^{*}(X)$. Then the cohomology algebra of
$X$ is one of the following:

  1. (i)

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

  2. (ii)

$\mathbb{Z}_{2}[x,y,z]/\langle x^{2},y^{m+1},z^{2}+\alpha y^{n}\rangle$, where
$\deg x=1,\deg y=2,\deg z=n$, $\alpha\in\mathbb{Z}_{2}$ and $\alpha=0$ if
$n>m$.

######  Theorem 1.4.

Let $G=\mathbb{Z}_{2}$ act freely on a finitistic connected space $X$ with
$X/G\sim_{2}\mathbb{RP}^{2}\times\mathbb{CP}^{n}\times\mathbb{HP}^{m}$ and
$\pi_{1}(B_{G})$ acts trivially on $H^{*}(X)$. Then the cohomology algebra of
$X$ is isomorphic to

$\mathbb{Z}_{2}[x,y,z]/\langle x^{n+1},y^{m+1},z^{2}+\alpha x^{2}+\beta
y\rangle,$

where $\deg x=2,\deg y=4,\deg z=2$ and $\alpha,\beta\in\mathbb{Z}_{2}$.

##  2\. Preliminaries

In this section, we have recalled some results and definitions which are used
to prove our main results.

Let $G$ be a compact Lie group and $G\hookrightarrow E_{G}\to B_{G}$ 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 $X_{G}=(X\times E_{G})/{G}$. Then the bundle $X\hookrightarrow
X_{G}\to B_{G}$ is called the Borel fibration. We consider the Leray-Serre
spectral sequence for the Borel fibration. If $\pi_{1}(B_{G})$ acts trivially
on $H^{*}(X)$, then the system of local coefficients on $B_{G}$ is simple and
the $E_{2}$-term of the Leray-Serre spectral sequence corresponding to the
Borel fibration is given by

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

This sequence converges to $H^{*}(X_{G})$ as an algebra. We recall some
results which are needed to prove our results:

######  Proposition 2.1.

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

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

and

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

are the homomorphisms

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

For details about spectral sequences, we refer [5]. Let $G$ act freely on $X$
and $h:X_{G}\rightarrow X/G$ be the map induced by $G$-equivariant projection
$X\times E_{G}\rightarrow X$. Then $h$ is a homotopy equivalence [1]. Now, we
recall Gysin sequence of sphere bundles.

######  Proposition 2.2 ([3]).

Let $G=\mathbb{Z}_{2}$ act freely on a finitistic space $X$. The Gysin
sequence of the sphere bundle $G\hookrightarrow X\stackrel{{\scriptstyle
p}}{{\to}}X/G$ is:

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

which start with

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

where $\cup:H^{i}(X/G)\to H^{i+1}(X/G)$ maps $x\to xe$, $e\in H^{1}(X/G)$
denotes the image of $1$ under the map $\cup:H^{0}(X/G)\to H^{1}(X/G)$ and $e$
is called the Euler class of the sphere bundle.

######  Proposition 2.3 ([6]).

Let $G=\mathbb{Z}_{2}$ act freely on a finitistic space $X$. Suppose that
$H^{j}(X)=0$ for all $j>n$, then $H^{j}(X/G)=0$ for all $j>n$.

##  3\. Proof of main Theorems

###### Proof of Theorem 1.1.

As $X/G\sim_{2}\mathbb{RP}^{1}\times\mathbb{CP}^{n}$, we have
$H^{*}(X/G)=\mathbb{Z}_{2}[a,b]/\langle a^{2},b^{n+1}\rangle,$ where $\deg
a=1$ and $\deg b=2$. By connectedness of $X$ and the exactness of Gysin
sequence of $G$-bundle $G\hookrightarrow X\stackrel{{\scriptstyle
p}}{{\to}}X/G$, we get $p^{*}_{0}$ is an isomorphism and the Euler class
$\cup(1)$ is $a$. This implies that for $0\leq i<n$,
$p^{*}_{2i+2}:H^{2i+2}(X/G)\to H^{2i+2}(X)$, $\rho_{2i+1}:H^{2i+1}(X)\to
H^{2i+1}(X/G)$ and $\rho_{2n+1}:H^{2n+1}(X)\to H^{2n+1}(X/G)$ are
isomorphisms. Consequently, $H^{2i+1}(X)\cong H^{2n+1}(X)\cong
H^{2i+2}(X)\cong\mathbb{Z}_{2}$ with bases $\\{u_{2i+1}\\}$, $\\{u_{2n+1}\\}$
and $\\{x^{i+1}\\}$, respectively, where $x=p^{*}_{2}(b)$,
$\rho_{2i+1}(u_{2i+1})=ab^{i}$, $\rho_{2n+1}(u_{2n+1})=ab^{n}$. By Proposition
2.3, $H^{i}(X)=0$ for $i>2n+1$. Now, we compute the cohomology algebra of $X$.
The Leray-Serre spectral sequence associated to the Borel fibration
$X\hookrightarrow X_{G}\to B_{G}$ converges to $H^{*}(X_{G})$. It is easy to
see that $G$ acts trivially on $H^{*}(X)$. Thus, the $E_{2}$-term
$E_{2}^{s,t}$ of the Leray-Serre spectral sequence associated to the Borel
fibration $X\stackrel{{\scriptstyle
i}}{{\hookrightarrow}}X_{G}\stackrel{{\scriptstyle\pi}}{{\to}}B_{G}$ is
$H^{s}(B_{G})\otimes H^{t}(X)$. By the edge homomorphism, we get
$d_{r}(1\otimes x)=0$ for all $r>0$ and $d_{r^{\prime}}(1\otimes u_{1})\not=0$
for some $r^{\prime}>0$. So, $d_{2}(1\otimes u_{1})=t^{2}\otimes 1$. This
gives that $x^{i}u_{1}\not=0$, and hence $u_{2i+1}=x^{i}u_{1}$ for all $0\leq
i\leq n$. Clearly, $x^{n+1}=0$. If $u_{1}^{2}=0$ then
$X\sim_{2}\mathbb{S}^{1}\times\mathbb{CP}^{n}$. If $u_{1}^{2}\not=0$ then
$X\sim_{2}\mathbb{RP}^{2n+1}$. ∎

###### Proof of Theorem 1.2.

As $X/G\sim_{2}\mathbb{RP}^{2}\times\mathbb{HP}^{n}$, we have
$H^{*}(X/G)=\mathbb{Z}_{2}[a,b]/\langle a^{3},b^{n+1}\rangle$, where $\deg
a=1$ and $\deg b=4$. It is easy to see that $\cup:H^{0}(X/G)\to H^{1}(X/G)$
maps 1 to $a$. Since $H^{j}(X/G)=0$ for $j\equiv 3\,(\text{mod 4})$, we get
$p^{*}_{4i}:H^{4i}(X/G)\to H^{4i}(X)$ and $\rho_{4i+2}:H^{4i+2}(X)\to
H^{4i+2}(X/G)$ are isomorphisms for $1\leq i\leq n$. Consequently, for $1\leq
i\leq n$, we get $H^{4i}(X)\cong H^{4i+2}(X)\cong\mathbb{Z}_{2}$ with bases
$\\{x^{i}\\}$ and $\\{u_{4i+2}\\}$, respectively, where $x=p_{4}^{*}(b)$ and
$\rho_{4i+2}(u_{4i+2})=a^{2}b^{i}$. By the exactness of Gysin sequence,
$H^{4i+3}(X)=H^{4i+1}(X)=0$ for all $0\leq i\leq n$. Clearly, $H^{i}(X)=0$ for
all $i>4n+3$. Now, we compute cohomology algebra of $X$. As $\pi_{1}(B_{G})$
acts trivially on $H^{*}(X)$, we get $E^{s,t}_{2}=H^{s}(B_{G})\otimes
H^{t}(X)$. By the edge homomorphism, we get $d_{3}(1\otimes
u_{2})=t^{3}\otimes 1$ and $d_{3}(1\otimes x)=0$. Thus, $x^{i}u_{2}\not=0$,
and hence $u_{4i+2}=x^{i}u_{2}$ for all $0\leq i\leq n$. Clearly, $x^{n+1}=0$.
If $u_{2}^{2}=0$ then $X\sim_{2}\mathbb{S}^{2}\times\mathbb{HP}^{n}$.
Otherwise, we get $X\sim_{2}\mathbb{CP}^{2n+1}$. ∎

###### Proof of Theorem 1.3.

It is given that
$X/G\sim_{2}\mathbb{RP}^{1}\times\mathbb{RP}^{n}\times\mathbb{CP}^{m}$. So, we
get $H^{*}(X/G)=\mathbb{Z}_{2}[a,b,c]/\langle a^{n+1},b^{m+1},c^{2}\rangle$,
where $\deg a=\deg c=1$ and $\deg b=2$. As $X$ is connected,
$H^{0}(X)\cong\mathbb{Z}_{2}$. First, we consider $1\leq n<2m.$ By Proposition
2.3, $H^{i}(X)=0$ for all $i>2m+n+1$. Consider the Gysin sequence of the
sphere bundle $G\hookrightarrow X\stackrel{{\scriptstyle p}}{{\to}}X/G$:

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

which begins with

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

Clearly, the Euler class of the associated principal $G$-bundle
$G\hookrightarrow X\to X/G$ must be nontrivial. We have considered the
following three cases:

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

For $0\leq i<\frac{n}{2}$, we have
$H^{2i}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($2i+1$-copies) and
$H^{2i+1}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($2i+2$-copies) with bases

$\displaystyle\\{b^{i},ab^{i-1}c,a^{2}b^{i-1},a^{3}b^{i-2}c,\cdots,a^{2i-1}c,a^{2i}\\},\text{
and }$
$\displaystyle\\{b^{i}c,ab^{i},a^{2}b^{i-1}c,a^{3}b^{i-1},\cdots,a^{2i}c,a^{2i+1}\\},$

respectively. By the exactness of Gysin sequence, we get $\ker
p_{2i}^{*}=\operatorname{im}\rho_{2i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($i$-copies) with basis $\\{a^{2j+1}b^{i-j-1}c|0\leq j\leq i-1\\}$, and $\ker
p_{2i+1}^{*}=\operatorname{im}\rho_{2i+1}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($i+1$-copies) with basis $\\{a^{2j}b^{i-j}c|0\leq j\leq i\\}$. Consequently,
$\operatorname{im}p^{*}_{2i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($i+1$-copies) with basis $\\{x^{2j}y^{i-j}|0\leq j\leq i\\}$, and
$\operatorname{im}p^{*}_{2i+1}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($i+1$-copies) with basis $\\{x^{2j+1}y^{i-j}|0\leq j\leq i\\}$, where
$p^{*}_{1}(a)=x,p^{*}_{2}(b)=y$. It is easy to see that

$\displaystyle 0\to\operatorname{im}p^{*}_{j}\hookrightarrow
H^{j}(X)\stackrel{{\scriptstyle\rho_{j}}}{{\to}}\operatorname{im}\rho_{j}\to
0,j=2i,2i+1$

are split exact sequences. This gives that

$\displaystyle
H^{2i}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(2i+1\text{-copies}),\text{
and}$ $\displaystyle
H^{2i+1}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(2i+2\text{-copies})$

with bases $\\{x^{2j}y^{i-j},u_{2i}^{2l+1}|0\leq j\leq i,0\leq l\leq i-1\\}$
and $\\{x^{2j+1}y^{i-j},u_{2i+1}^{2j}|0\leq j\leq i\\}$, respectively, where
$\rho_{2i}(u_{2i}^{2l+1})=a^{2l+1}b^{i-l-1}c,0\leq l\leq i-1\text{ and
}\rho_{2i+1}(u_{2i+1}^{2j})=a^{2j}b^{i-j}c,0\leq j\leq i$.

Firstly, suppose that $n=2k$ for some $k$. Then for $\frac{n}{2}\leq i\leq m$,
$H^{2i}(X/G)\cong
H^{2i+1}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($n+1$-copies) with bases

$\displaystyle\\{b^{i},ab^{i-1}c,a^{2}b^{i-1},a^{3}b^{i-2}c,\cdots,a^{n-1}b^{i-k}c,a^{n}b^{i-k}\\},\text{
and }$
$\displaystyle\\{b^{i}c,ab^{i},a^{2}b^{i-1}c,a^{3}b^{i-1},\cdots,a^{n-1}b^{i-k+1},a^{n}b^{i-k}c\\},$

respectively. Again, by the exactness of Gysin sequence, $\ker
p_{2i}^{*}=\operatorname{im}\rho_{2i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k$-copies) and $\ker
p^{*}_{2i+1}=\operatorname{im}\rho_{2i+1}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+1$-copies) with bases $\\{a^{2j+1}b^{i-j-1}c|0\leq j\leq k-1\\}$ and
$\\{a^{2j}b^{i-j}c|0\leq j\leq k\\}$, respectively. This gives that
$\operatorname{im}p^{*}_{2i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+1$-copies) and
$\operatorname{im}p^{*}_{2i+1}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k$-copies) with bases $\\{x^{2j}y^{i-j}|0\leq j\leq k\\}$ and
$\\{x^{2j+1}y^{i-j}|0\leq j\leq k-1\\}$, respectively. We observe that
$H^{j}(X)\cong\operatorname{im}p^{*}_{j}\oplus\operatorname{im}\rho_{j},j=2i,2i+1$.
So, we get

$\displaystyle H^{2i}(X)\cong
H^{2i+1}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(n+1\text{-copies})$

with bases $\\{x^{2j}y^{i-j},u_{2i}^{2l+1}|0\leq j\leq k,0\leq l\leq k-1\\}$
and $\\{x^{2j+1}y^{i-j},u_{2i+1}^{2l}|0\leq j\leq k-1,0\leq l\leq k\\}$,
respectively, where $\rho_{2i}(u_{2i}^{2l+1})=a^{2l+1}b^{i-l-1}c,0\leq l\leq
k-1\text{ and }\rho_{2i+1}(u_{2i+1}^{2l})=a^{2l}b^{i-l}c,0\leq l\leq k$.

For $m<i\leq m+\frac{n}{2}$, we have
$H^{2i}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($n+2m-2i+2$-copies) and
$H^{2i+1}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($n+2m-2i+1$-copies) with bases

$\displaystyle\\{a^{2i-2m-1}b^{m}c,a^{2i-2m}b^{m},a^{2i-2m+1}b^{m-1}c,a^{2i-2m+2}b^{m-1},\cdots,a^{n-1}b^{i-k}c,a^{n}b^{i-k}\\},\text{
and }$
$\displaystyle\\{a^{2i-2m}b^{m}c,a^{2i-2m+1}b^{m},a^{2i-2m+2}b^{m-1}c,a^{2i-2m+3}b^{m-1},\cdots,a^{n-1}b^{i-k+1},a^{n}b^{i-k}c\\},$

respectively. Note that $\ker
p_{2i}^{*}=\operatorname{im}\rho_{2i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+m-i+1$-copies) and $\ker
p^{*}_{2i+1}=\operatorname{im}\rho_{2i+1}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+m-i+1$-copies) with bases $\\{a^{2j+1}b^{i-j-1}c|i-m-1\leq j\leq k-1\\}$
and $\\{a^{2j}b^{i-j}c|i-m\leq j\leq k\\}$, respectively. This implies that
$\operatorname{im}p^{*}_{2i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+m-i+1$-copies) and
$\operatorname{im}p^{*}_{2i+1}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+m-i$-copies) with bases $\\{x^{2j}y^{i-j}|i-m\leq j\leq k\\}$ and
$\\{x^{2j+1}y^{i-j}|i-m\leq j\leq k-1\\}$, respectively. So, we get

$\displaystyle
H^{2i}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(n+2m-2i+2\text{-copies})\text{,
and}$ $\displaystyle
H^{2i+1}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(n+2m-2i+1\text{-copies})$

with bases $\\{x^{2j}y^{i-j},u_{2i}^{2l+1}|i-m\leq j\leq k,i-m-1\leq l\leq
k-1\\}$ and $\\{x^{2j+1}y^{i-j},u_{2i+1}^{2l}|i-m\leq j\leq k-1,i-m\leq l\leq
k\\}$, respectively, where
$\rho_{2i}(u_{2i}^{2l+1})=a^{2l+1}b^{i-l-1}c,i-m-1\leq l\leq k-1$ and
$\rho_{2i+1}(u_{2i+1}^{2l})=a^{2l}b^{i-l}c,i-m\leq l\leq k$.

Now, we compute the cohomology algebra of $X$. Consider the Leray Serre
spectral sequence for the Borel fibration $X\stackrel{{\scriptstyle
i}}{{\hookrightarrow}}X_{G}\stackrel{{\scriptstyle\pi}}{{\rightarrow}}B_{G}$
with $E_{2}^{s,t}=H^{s}(B_{G})\otimes H^{t}(X)$. As $p^{*}_{j}=i^{*}_{j}\circ
h^{*}$ for all $j\geq 0$, we get $d_{r}(1\otimes x)=0=d_{r}(1\otimes y)$ for
all $r\geq 0$ and for all $i,j$; $d_{r_{i,j}}(1\otimes u_{j}^{i})\not=0$ for
some $r_{i,j}>0$. Put $u_{1}^{0}=z$. So, $H^{1}(X)$ is generated by
$\\{x,z\\}$. Trivially, $d_{2}(1\otimes z)=t^{2}\otimes 1$ and so
$d_{2}(1\otimes xz)\not=0$. This implies that $xz=\alpha x^{2}+\beta
y+u_{2}^{1},\alpha,\beta\in\mathbb{Z}_{2}$. So, $H^{2}(X)$ is generated by
$\\{x^{2},y,xz\\}$. By induction, we show that for $1\leq i<\frac{n}{2}$,
$H^{2i-1}(X)$ and $H^{2i}(X)$ are generated by
$\\{xy^{i-1},x^{3}y^{i-2},\cdots,x^{2i-3}y,x^{2i-1},y^{i-1}z,x^{2}y^{i-2}z,\cdots
x^{2i-4}yz,x^{2i-2}z\\}$ and
$\\{y^{i},x^{2}y^{i-1},\cdots,x^{2i-2}y,x^{2i},xy^{i-1}z,x^{3}y^{i-2}z,\cdots,x^{2i-1}z\\}$,
respectively. Now, we prove it for $i+1$. We have $d_{2}(1\otimes
y^{i}z)=t^{2}\otimes y^{i},d_{2}(1\otimes x^{2}y^{i-1}z)=t^{2}\otimes
x^{2}y^{i-1},\cdots,d_{2}(1\otimes x^{2i-2}yz)=t^{2}\otimes
x^{2i-2}y,d_{2}(1\otimes x^{2i}z)=t^{2}\otimes x^{2i}.$ This gives that
$y^{i}z,x^{2}y^{i-1}z,\cdots x^{2i-2}yz,x^{2i}z$ are nonzero elements of
$H^{2i+1}(X)$. Therefore,

$\displaystyle y^{i}z=$
$\displaystyle\alpha_{1}^{0}xy^{i}+\alpha_{3}^{0}x^{3}y^{i-1}+\cdots+\alpha_{2i-1}^{0}x^{2i-1}y+\alpha_{2i+1}^{0}x^{2i+1}$
$\displaystyle+\beta_{0}^{0}u^{0}_{2i+1}+\beta_{2}^{0}u^{2}_{2i+1}+\cdots+\beta^{0}_{2i-2}u_{2i+1}^{2i-2}+\beta_{2i}^{0}u^{2i}_{2i+1}$
$\displaystyle x^{2}y^{i-1}z=$
$\displaystyle\alpha_{1}^{2}xy^{i}+\alpha_{3}^{2}x^{3}y^{i-1}+\cdots+\alpha_{2i-1}^{2}x^{2i-1}y+\alpha_{2i+1}^{2}x^{2i+1}$
$\displaystyle+\beta_{0}^{2}u^{0}_{2i+1}+\beta_{2}^{2}u^{2}_{2i+1}+\cdots+\beta^{2}_{2i-2}u_{2i+1}^{2i-2}+\beta_{2i}^{2}u^{2i}_{2i+1}$
$\displaystyle\vdots$ $\displaystyle x^{2i}z=$
$\displaystyle\alpha_{1}^{2i}xy^{i}+\alpha_{3}^{2i}x^{3}y^{i-1}+\cdots+\alpha_{2i-1}^{2i}x^{2i-1}y+\alpha_{2i+1}^{2i}x^{2i+1}$
$\displaystyle+\beta_{0}^{2i}u^{0}_{2i+1}+\beta_{2}^{2i}u^{2}_{2i+1}+\cdots+\beta^{2i}_{2i-2}u_{2i+1}^{2i-2}+\beta_{2i}^{2i}u^{2i}_{2i+1}$

where $\beta_{j}^{k}\not=0$ for some $j\in\\{0,2,\cdots,2i\\}$ and for all
$k=0,2,\cdots,2i$. From this we get

$\displaystyle t^{2}\otimes y^{i}=$ $\displaystyle\beta_{0}^{0}d_{2}(1\otimes
u^{0}_{2i+1})+\beta_{2}^{0}d_{2}(1\otimes
u^{2}_{2i+1})+\cdots+\beta_{2i}^{0}d_{2}(1\otimes u_{2i+1}^{2i})$
$\displaystyle t^{2}\otimes x^{2}y^{i-1}=$
$\displaystyle\beta_{0}^{2}d_{2}(1\otimes
u^{0}_{2i+1})+\beta_{2}^{2}d_{2}(1\otimes
u^{2}_{2i+1})+\cdots+\beta_{2i}^{2}d_{2}(1\otimes u^{2i}_{2i+1})$
$\displaystyle\vdots$ $\displaystyle t^{2}\otimes x^{2i}=$
$\displaystyle\beta_{0}^{2i}d_{2}(1\otimes
u^{0}_{2i+1})+\beta_{2}^{2i}d_{2}(1\otimes
u^{2}_{2i+1})+\cdots+\beta_{2i}^{2i}d_{2}(1\otimes u^{2i}_{2i+1})$

As the set $\\{y^{i},x^{2}y^{i-1},\cdots,x^{2i-2}y,x^{2i}\\}$ is linearly
independent, the coefficient matrix

$\begin{bmatrix}\beta_{0}^{0}&\beta_{2}^{0}&\cdots&\beta_{2i}^{0}\\\
\beta_{0}^{2}&\beta_{2}^{2}&\cdots&\beta_{2i}^{2}\\\
\vdots&\vdots&\vdots&\vdots\\\
\beta_{0}^{2i}&\beta_{2}^{2i}&\cdots&\beta_{2i}^{2i}\end{bmatrix}$

is invertible. Then $u^{2k}_{2i+1}$ is generated by
$\\{xy^{i},x^{3}y^{i-1},\cdots,x^{2i-1}y,x^{2i+1},y^{i}z,x^{2}y^{i-1}z,\cdots,$
$x^{2i-2}z,x^{2i}z\\}$ for all $0\leq k\leq i$. So, the basis for
$H^{2i+1}(X)$ is $\\{xy^{i},x^{3}y^{i-1},\cdots,x^{2i-3}y^{2},x^{2i-1}y$
$,x^{2i+1},y^{i}z,x^{2}y^{i-1}z,\cdots,x^{2i-2}yz,x^{2i}z\\}$. Also, we have
$d_{2}(1\otimes xy^{i}z)=t^{2}\otimes xy^{i},d_{2}(1\otimes
x^{3}y^{i-1}z)=t^{2}\otimes x^{3}y^{i-1},d_{2}(1\otimes
x^{5}y^{i-2}z)=t^{2}\otimes x^{5}y^{i-2},\cdots,d_{2}(1\otimes
x^{2i+1}z)=t^{2}\otimes x^{2i+1}$. This implies that
$xy^{i}z,x^{3}y^{i-1}z,\cdots,x^{2i+1}z$ are nonzero elements of
$H^{2i+2}(X)$. So, we get

$\displaystyle xy^{i}z=$
$\displaystyle\alpha_{0}^{1}y^{i+1}+\alpha_{2}^{1}x^{2}y^{i}+\cdots+\alpha_{2i-2}^{1}x^{2i-2}y^{2}+\alpha_{2i}^{1}x^{2i}y$
$\displaystyle+\beta_{1}^{1}u^{1}_{2i+2}+\beta_{3}^{1}u^{3}_{2i+2}+\cdots+\beta_{2i+1}^{1}u^{2i+1}_{2i+2}$
$\displaystyle x^{3}y^{i-1}z=$
$\displaystyle\alpha_{0}^{3}y^{i+1}+\alpha_{2}^{3}x^{2}y^{i}+\cdots+\alpha_{2i-2}^{3}x^{2i-2}y^{2}+\alpha_{2i}^{3}x^{2i}y$
$\displaystyle+\beta_{1}^{3}u^{1}_{2i+2}+\beta_{3}^{3}u^{3}_{2i+2}+\cdots+\beta_{2i+1}^{3}u^{2i+1}_{2i+2}$
$\displaystyle\vdots$ $\displaystyle x^{2i+1}z=$
$\displaystyle\alpha_{0}^{2i+1}y^{i+1}+\alpha_{2}^{2i+1}x^{2}y^{i}+\cdots+\alpha_{2i-2}^{2i+1}x^{2i-2}y^{2}+\alpha_{2i}^{2i+1}x^{2i}y$
$\displaystyle+\beta_{1}^{2i+1}u^{1}_{2i+2}+\beta_{3}^{2i+1}u^{3}_{2i+2}+\cdots+\beta_{2i+1}^{2i+1}u^{2i+1}_{2i+2}$

where $\beta_{j}^{k}\not=0$ for some $j\in\\{1,3,\cdots,2i+1\\}$ and for all
$k=1,3,\cdots,2i+1$. Thus, we get

$\displaystyle t^{2}\otimes xy^{i}=$ $\displaystyle\beta_{1}^{1}d_{2}(1\otimes
u^{1}_{2i+2})+\beta_{3}^{1}d_{2}(1\otimes
u^{3}_{2i+2})+\cdots+\beta_{2i+1}^{1}d_{2}(1\otimes u^{2i+1}_{2i+2})$
$\displaystyle t^{2}\otimes x^{3}y^{i-1}=$
$\displaystyle\beta_{1}^{3}d_{2}(1\otimes
u^{1}_{2i+2})+\beta_{3}^{3}d_{2}(1\otimes
u^{3}_{2i+2})+\cdots+\beta_{2i+1}^{3}d_{2}(1\otimes u^{2i+1}_{2i+2})$
$\displaystyle\vdots$ $\displaystyle t^{2}\otimes x^{2i+1}=$
$\displaystyle\beta_{1}^{2i+1}d_{2}(1\otimes
u^{1}_{2i+2})+\beta_{3}^{2i+1}d_{2}(1\otimes
u^{3}_{2i+2})+\cdots+\beta_{2i+1}^{2i+1}d_{2}(1\otimes u^{2i+1}_{2i+2})$

As the set $\\{xy^{i},x^{3}y^{i-1},\cdots,x^{2i+1}\\}$ is linearly
independent, $u^{k}_{2i+2}$ is generated by $\\{y^{i+1},x^{2}y^{i},$
$\cdots,x^{2i}y,xy^{i}z,x^{3}y^{i-1}z,\cdots,x^{2i+1}z\\}$ for all
$k=1,3,\cdots,2i+1$. So, our assumption is true for $i+1$. Similarly, for
$\frac{n}{2}\leq i\leq m$, bases for $H^{2i}(X)$ and $H^{2i+1}(X)$ are
$\\{x^{2j}y^{i-j},x^{2l+1}y^{i-l-1}z$
$|0\leq j\leq k,0\leq l\leq k-1\\}$ and
$\\{x^{2j+1}y^{i-j},x^{2l}y^{i-l}z|0\leq j\leq k-1,0\leq l\leq k\\}$,
respectively, and for $m<i\leq m+\frac{n}{2}$, bases for $H^{2i}(X)$ and
$H^{2i+1}(X)$ are $\\{x^{2j}y^{i-j},x^{2l+1}y^{i-l-1}z|i-m\leq j\leq
k,i-m-1\leq l\leq k-1\\}$ and $\\{x^{2j+1}y^{i-j},x^{2l}y^{i-l}z|i-m\leq j\leq
k-1,i-m\leq l\leq k\\}$, respectively.

Now, suppose that $n=2k+1$ for some $k$. For $n\leq 2i,2i+1\leq 2m$,
$H^{n-1}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($2k+1$-copies) and $H^{2i}(X/G)\cong
H^{2i+1}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($2k+2$-copies) with bases

$\displaystyle\\{b^{k},ab^{k-1}c,a^{2}b^{k-1},a^{3}b^{k-2}c,\cdots,a^{2k-1}c,a^{2k}\\},$
$\displaystyle\\{b^{i},ab^{i-1}c,a^{2}b^{i-1},a^{3}b^{i-2}c,\cdots,a^{n-1}b^{i-k},a^{n}b^{i-k-1}c\\},\text{
and }$
$\displaystyle\\{b^{i}c,ab^{i},a^{2}b^{i-1}c,a^{3}b^{i-1},\cdots,a^{n-1}b^{i-k}c,a^{n}b^{i-k}\\},$

respectively. Then we have $\ker
p^{*}_{n-1}=\operatorname{im}\rho_{n-1}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k$-copies), $\ker
p^{*}_{2i}=\operatorname{im}\rho_{2i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+1$-copies) and $\ker
p^{*}_{2i+1}=\operatorname{im}\rho_{2i+1}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+1$-copies) with bases $\\{a^{2j+1}b^{k-j-1}c|0\leq j\leq k-1\\}$,
$\\{a^{2j+1}b^{i-j-1}c|0\leq j\leq k\\}$ and $\\{a^{2j}b^{i-j}c|0\leq j\leq
k\\}$, respectively. So,
$\operatorname{im}p^{*}_{n-1}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+1$-copies),
$\operatorname{im}p^{*}_{2i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+1$ copies) and
$\operatorname{im}p^{*}_{2i+1}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+1$ copies) with bases $\\{x^{2j}y^{k-j}|0\leq j\leq k\\}$,
$\\{x^{2j}y^{i-j}|0\leq j\leq k\\}$ and $\\{x^{2j+1}y^{i-j}|0\leq i\leq k\\}$,
respectively. As we have done above, we get

$\displaystyle
H^{n-1}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(n\text{-copies}),\text{
and}$ $\displaystyle H^{2i}(X)\cong
H^{2i+1}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(n+1\text{-copies})$

with bases $\\{x^{2j}y^{k-j},u_{2k}^{2l+1}|0\leq j\leq k,0\leq l\leq k-1\\}$,
$\\{x^{2j}y^{i-j},u_{2i}^{2j+1}|0\leq j\leq k\\}$ and
$\\{x^{2j+1}y^{i-j},u_{2i+1}^{2j}|0\leq j\leq k\\}$, respectively, where
$\rho_{2k}(u_{2k}^{2l+1})=a^{2l+1}b^{k-l-1}c,0\leq l\leq k-1$,
$\rho_{2i}(u_{2i}^{2j+1})=a^{2j+1}b^{i-j-1}c,0\leq j\leq k$ and
$\rho_{2i+1}(u_{2i+1}^{2j})=a^{2j}b^{i-j}c,0\leq j\leq k$. Also, for
$2m<2i,2i+1\leq 2m+n+1$, we have
$H^{2i}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($n+2m-2i+2$-copies) and
$H^{2i+1}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($n+2m-2i+1$-copies) with bases

$\displaystyle\\{a^{2i-2m-1}b^{m}c,a^{2i-2m}b^{m},a^{2i-2m+1}b^{m-1}c,a^{2i-2m+2}b^{m-1},\cdots,a^{n-1}b^{i-k},a^{n}b^{i-k-1}c\\},\text{
and }$
$\displaystyle\\{a^{2i-2m}b^{m}c,a^{2i-2m+1}b^{m},a^{2i-2m+2}b^{m-1}c,a^{2i-2m+3}b^{m-1},\cdots,a^{n-1}b^{i-k}c,a^{n}b^{i-k}\\},$

respectively. Note that $\ker
p_{2i}^{*}=\operatorname{im}\rho_{2i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+m-i+2$-copies) and $\ker
p^{*}_{2i+1}=\operatorname{im}\rho_{2i+1}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+m-i+1$-copies) with bases $\\{a^{2j+1}b^{i-j-1}c|i-m-1\leq j\leq k\\}$ and
$\\{a^{2j}b^{i-j}c|i-m\leq j\leq k\\}$, respectively. This implies that
$\operatorname{im}p^{*}_{2i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+m-i+1$-copies) and
$\operatorname{im}p^{*}_{2i+1}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+m-i+1$-copies) with bases $\\{x^{2j}y^{i-j}|i-m\leq j\leq k\\}$ and
$\\{x^{2j+1}y^{i-j}|i-m\leq j\leq k\\}$, respectively. So, we get

$\displaystyle
H^{2i}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(n+2m-2i+2\text{-copies})\text{,
and}$ $\displaystyle
H^{2i+1}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(n+2m-2i+1\text{-copies})$

with bases $\\{x^{2j}y^{i-j},u_{2i}^{2l+1}|i-m\leq j\leq k,i-m-1\leq l\leq
k\\}$ and $\\{x^{2j+1}y^{i-j},u_{2i+1}^{2j}|i-m\leq$
$j\leq k\\}$, respectively, where
$\rho_{2i}(u_{2i}^{2l+1})=a^{2l+1}b^{i-l-1}c,i-m-1\leq l\leq k$ and
$\rho_{2i+1}(u_{2i+1}^{2j})=a^{2j}b^{i-j}c,i-m\leq j\leq k$.

For $1\leq i<n$, $H^{i}(X)$ has the same basis as in the case when $n$ is
even. Again by induction, for $n\leq 2i,2i+1\leq 2m$, bases for $H^{n-1}(X)$,
$H^{2i}(X)$ and $H^{2i+1}(X)$ are $\\{x^{2j}y^{k-j},x^{2l+1}y^{k-l-1}z|0\leq
j\leq k,0\leq l\leq k-1\\}$, $\\{x^{2j}y^{i-j},x^{2l+1}y^{i-l-1}z|0\leq
j,l\leq k\\}$ and $\\{x^{2j+1}y^{i-j},x^{2l}y^{i-l}z|0\leq j,l\leq k\\}$,
respectively, and for $m<i\leq m+\frac{n}{2}$, bases for $H^{2i}(X)$ and
$H^{2i+1}(X)$ are $\\{x^{2j}y^{i-j},x^{2l+1}y^{i-l-1}z|i-m\leq j\leq
k,i-m-1\leq l\leq k\\}$ and $\\{x^{2j+1}y^{i-j},x^{2l}y^{i-l}z|i-m\leq j,l\leq
k\\}$, respectively.

As $d_{2}(1\otimes z^{2})=0$, we get $z^{2}=\alpha x^{2}+\beta y$ for some
$\alpha,\beta\in\mathbb{Z}_{2}$. Thus the cohomology algebra of $X$ is given
by

$\mathbb{Z}_{2}[x,y,z]/\langle x^{n+1},y^{m+1},z^{2}+\alpha x^{2}+\beta
y\rangle,$

where $\deg x=\deg z=1$, $\deg y=2$ and $\alpha,\beta\in\mathbb{Z}_{2}$. This
realizes possibility (i).

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

Clearly, $\rho_{i}:H^{i}(X)\to H^{i}(X/G)$ is trivial for all $1\leq i<n$. For
$0<2i,2i+1<n$, bases for $\ker p^{*}_{2i}$ and $\ker p^{*}_{2i+1}$ are
$\\{a^{2j+1}b^{i-j-1}c,a^{2l}b^{i-l}|0\leq j\leq i-1,1\leq l\leq i\\}$ and
$\\{a^{2j+1}b^{i-j},a^{2l}b^{i-l}c|0\leq j\leq i,1\leq l\leq i\\}$,
respectively. This implies that
$\operatorname{im}p^{*}_{2i}\cong\operatorname{im}p^{*}_{2i+1}\cong\mathbb{Z}_{2}$
with bases $\\{y^{i}\\}$ and $\\{xy^{i}\\}$, respectively, where
$p_{1}^{*}(c)=x$ and $p_{2}^{*}(b)=y$. So, $H^{2i}(X)\cong
H^{2i+1}(X)\cong\mathbb{Z}_{2}$ with bases $\\{y^{i}\\}$ and $\\{xy^{i}\\}$,
respectively.

If $n=2k$ for some $k$, then for $n\leq 2i,2i+1\leq 2m+1$,
$\\{a^{2j+1}b^{i-j-1}c,a^{2l}b^{i-l}|0\leq j\leq k-1,1\leq l\leq k\\}$ and
$\\{a^{2j+1}b^{i-j},a^{2l}b^{i-l}c|0\leq j\leq k-1,1\leq l\leq k\\}$ are bases
for $\ker p^{*}_{2i}$ and $\ker p^{*}_{2i+1}$, respectively. This gives that
$\operatorname{im}p^{*}_{2i}\cong\operatorname{im}p^{*}_{2i+1}\cong\mathbb{Z}_{2}$
with bases $\\{y^{i}\\}$ and $\\{xy^{i}\\}$, respectively, and
$\operatorname{im}\rho_{2i}\cong\operatorname{im}\rho_{2i+1}\cong\mathbb{Z}_{2}$
with bases $\\{a^{n}b^{i-k}\\}$ and $\\{a^{n}b^{i-k}c\\}$, respectively.
Consequently, $H^{2i}(X)\cong
H^{2i+1}(X)\cong\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$ with bases
$\\{y^{i},u^{n}_{2i}\\}$ and $\\{xy^{i},u^{n}_{2i+1}\\}$, respectively, where
$\rho_{2i}(u^{n}_{2i})=a^{n}b^{i-k}$ and
$\rho_{2i+1}(u^{n}_{2i+1})=a^{n}b^{i-k}c$. If $2m+1<2i,2i+1\leq 2m+n+1$ then
bases for $\ker p^{*}_{2i}$ and $\ker p^{*}_{2i+1}$ are
$\\{a^{2j+1}b^{i-j-1}c,a^{2l}b^{i-l}|i-m-1\leq j\leq k-1,i-m\leq l\leq k\\}$
and $\\{a^{2j+1}b^{i-j},a^{2l}b^{i-l}c|i-m\leq j\leq k-1,i-m\leq l\leq k\\}$,
respectively. This implies that $p^{*}_{2i},p^{*}_{2i+1}$ are trivial and
$\operatorname{im}\rho_{2i}\cong\operatorname{im}\rho_{2i+1}\cong\mathbb{Z}_{2}$
with bases $\\{a^{n}b^{i-k}\\}$ and $\\{a^{n}b^{i-k}c\\}$, respectively. Thus,
$H^{2i}(X)\cong H^{2i+1}(X)\cong\mathbb{Z}_{2}$ with bases $\\{u_{2i}^{n}\\}$
and $\\{u^{n}_{2i+1}\\}$ respectively, where
$\rho_{2i}(u^{n}_{2i})=a^{n}b^{i-k}$ and
$\rho_{2i+1}(u^{n}_{2i+1})=a^{n}b^{i-k}c$. Trivially, $x^{2}=y^{m+1}=0$. Put
$u_{2k}^{n}=z$. Then by the edge homomorphism, $d_{r}(1\otimes
x)=d_{r}(1\otimes y)=0$ for all $r>0$ and $d_{r^{\prime}}(1\otimes z)\not=0$
for some $r^{\prime}>0$. As
$X/G\sim_{2}\mathbb{RP}^{1}\times\mathbb{RP}^{n}\times\mathbb{CP}^{m}$,
$r^{\prime}$ must be $n+1$. Note that for all $0\leq i\leq m$, $xy^{i}z$ and
$xy^{i}$ are nonzero elements of $H^{n+2i+1}(X)$ and $H^{n+2i}(X)$,
respectively. Therefore, $xy^{i}z=\alpha xy^{k+m}+\beta u^{n}_{n+2i+1}$ and
$xy^{i}=\alpha^{\prime}y^{k+m}+\beta^{\prime}u^{n}_{n+2i}$ for some
$\alpha,\alpha^{\prime},\beta,\beta^{\prime}\in\mathbb{Z}_{2}$, where $\beta$
and $\beta^{\prime}$ are nonzero. Note that for $n+2i>2m$, we get
$\alpha=\alpha^{\prime}=0$.

As $d_{r}(1\otimes z^{2})=0$ for all $r\geq 0$, we get $z^{2}=\alpha y^{n}$
for some $\alpha\in\mathbb{Z}_{2}$ and $\alpha=0$ for $n>m$. Thus, the
cohomology algebra of $X$ is given by

$\mathbb{Z}_{2}[x,y,z]/\langle x^{2},y^{m+1},z^{2}+\alpha y^{n}\rangle,$

where $\deg x=1,\deg y=2,\deg z=n$ and $\alpha=0$ if $n>m$. This realizes
possibility (ii).

Now, let $n=2k+1$ for some $k\geq 0$. For $n\leq 2i,2i+1\leq 2m+1$, we get
$H^{2i}(X)\cong\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\cong H^{2i+1}(X)$ with bases
$\\{y^{i},u^{n}_{2i}\\}$ and $\\{xy^{i},u^{n}_{2i+1}\\}$, respectively, where
$\rho_{2i}(u^{n}_{2i})=a^{n}b^{i-k-1}c$ and
$\rho_{2i+1}(u^{n}_{2i+1})=a^{n}b^{i-k}$. Also, for $2m+1<2i,2i+1\leq 2m+n+1$,
we have $H^{2i}(X)\cong\mathbb{Z}_{2}\cong H^{2i+1}(X)$ with bases
$\\{u^{n}_{2i}\\}$ and $\\{u^{n}_{2i+1}\\}$, respectively, where
$\rho_{2i}(u^{n}_{2i})=a^{n}b^{i-k-1}c$ and
$\rho_{2i+1}(u^{n}_{2i+1})=a^{n}b^{i-k}$. It is easy to observe that $X$ has
the same cohomology algebra as in the case when $n$ is even.

Case (3): When $\cup:H^{0}(X/G)\to H^{4}(X/G)$ maps $1$ to $a+c$.

Also, in this case $\rho_{i}:H^{i}(X)\to H^{i}(X/G)$ and $p^{*}_{j}$ are
trivial for all $1\leq i<n$ and $2m+1<j\leq 2m+n+1$. Similar calculations show
that for $1\leq i<n$, $H^{2i}(X)\cong H^{2i+1}(X)\cong\mathbb{Z}_{2}$ with
bases $\\{y^{i}\\}$ and $\\{xy^{i}\\}$, respectively, where
$x=p^{*}_{1}(a)=p_{1}^{*}(c),y=p_{2}^{*}(b)$. For $n\leq 2i,2i+1\leq 2m+1$,
$H^{2i}(X)\cong H^{2i+1}(X)\cong\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$ with bases
$\\{y^{i},u_{2i}\\}$, $\\{xy^{i},u_{2i+1}\\}$, respectively; and for
$2m+1<2i,2i+1\leq 2m+n+1$, we have $H^{2i}(X)\cong
H^{2i+1}(X)\cong\mathbb{Z}_{2}$ with bases $\\{u_{2i}\\}$, $\\{u_{2i+1}\\}$,
respectively, where $\rho_{2i}(u_{2i})=a^{n-1}b^{i-k}c+a^{n}b^{i-k}$ and
$\rho_{2i+1}(u_{2i+1})=a^{n}b^{i-k}c$ if $n=2k$; and
$\rho_{2i}(u_{2i})=a^{n}b^{i-k-1}c$ and
$\rho_{2i+1}(u_{2i+1})=a^{n-1}b^{i-k}c+a^{n}b^{i-k}$ if $n=2k+1$.
Consequently, the cohomology algebra of $X$ is given by

$\mathbb{Z}_{2}[x,y,z]/\langle x^{2},y^{m+1},z^{2}+\alpha y^{n}\rangle,$

where $\deg x=1,\deg y=2,\deg z=n$ and $\alpha=0$ if $n>m$. This realizes
possibility (ii).

Similarly, we get the same cohomology algebra for the case when $2m\leq n$.∎

###### Proof of Theorem 1.4.

As $X/G\sim_{2}\mathbb{CP}^{n}\times\mathbb{HP}^{m}\times\mathbb{RP}^{2}$, we
get $H^{*}(X/G)=\mathbb{Z}_{2}[a,b,c]/\langle a^{n+1}$
$,b^{m+1},c^{3}\rangle$, where $\deg a=2,\deg b=4$ and $\deg c=1$. Since $X$
is connected, $H^{0}(X)\cong\mathbb{Z}_{2}$. By Proposition 2.3, $H^{i}(X)=0$
for all $i>4m+2n+2$. Firsty, assume that $1\leq n<2m$. Consider the Gysin
sequence of the sphere bundle $G\hookrightarrow X\stackrel{{\scriptstyle
p}}{{\to}}X/G$:

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

which begins with

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

By the exactness of Gysin sequence, the Euler class of the associated
principal $G$-bundle $G\hookrightarrow X\to X/G$ must be $c$.

For $0\leq i<\frac{n}{2}$, we have
$H^{4i}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($2i+1$-copies),
$H^{4i+1}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($i+1$-copies),
$H^{4i+2}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($2i+2$-copies) and
$H^{4i+3}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($i+1$-copies) with bases

$\displaystyle\\{b^{i},ab^{i-1}c^{2},a^{2}b^{i-1},a^{3}b^{i-2}c^{2},\cdots,a^{2i-1}c^{2},a^{2i}\\},$
$\displaystyle\\{b^{i}c,a^{2}b^{i-1}c,\cdots,a^{2i-2}bc,a^{2i}c\\},$
$\displaystyle\\{b^{i}c^{2},ab^{i},a^{2}b^{i-1}c^{2},a^{3}b^{i-1},\cdots,a^{2i}c^{2},a^{2i+1}\\},\text{
and }$ $\displaystyle\\{ab^{i}c,a^{3}b^{i-1}c,\cdots,a^{2i-1}bc,a^{2i+1}c\\},$

respectively. Again, by the exactness of Gysin sequence, we get $\ker
p_{4i}^{*}=\operatorname{im}\rho_{4i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($i$-copies) with basis $\\{a^{2j+1}b^{i-j-1}c^{2}|0\leq j\leq i-1\\}$, $\ker
p_{4i+1}^{*}=H^{4i+1}(X/G)$, $\ker
p_{4i+2}^{*}=\operatorname{im}\rho_{4i+2}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($i+1$-copies) with basis $\\{a^{2j}b^{i-j}c^{2}|0\leq j\leq i\\}$ and $\ker
p_{4i+3}^{*}=H^{4i+3}(X/G)$. This implies that
$\operatorname{im}p^{*}_{4i}\cong\operatorname{im}p^{*}_{4i+2}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($i+1$-copies) with bases $\\{x^{2j}y^{i-j}|0\leq j\leq i\\}$ and
$\\{x^{2j+1}y^{i-j}|0\leq j\leq i\\}$, respectively, where
$p^{*}_{2}(a)=x,p_{4}^{*}(b)=y$. Clearly,
$\operatorname{im}p^{*}_{4i+j}=\operatorname{im}\rho_{4i+j}=0,j=1,3$. Thus,

$\displaystyle H^{4i+1}(X)=H^{4i+3}(X)=0$ $\displaystyle
H^{4i}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(2i+1\text{-copies}),\text{
and}$ $\displaystyle
H^{4i+2}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(2i+2\text{-copies})$

with bases $\\{x^{2j}y^{i-j},u_{4i}^{2l+1}|0\leq j\leq i,0\leq l\leq i-1\\}$
and $\\{x^{2j+1}y^{i-j},u_{4i+2}^{2j}|0\leq j\leq i\\}$, respectively, where
$\rho_{4i}(u_{4i}^{2l+1})=a^{2l+1}b^{i-l-1}c^{2},0\leq l\leq i-1\text{ and
}\rho_{4i+2}(u_{4i+2}^{2j})=a^{2j}b^{i-j}c^{2},0\leq j\leq i$.

Let us assume that $n=2k$ for some $k>0$. In this case, for $2n\leq 4i+r\leq
4m,0\leq r\leq 3$, we have
$H^{4i}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($2k+1$-copies),
$H^{4i+1}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+1$-copies),
$H^{4i+2}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($2k+1$-copies) and
$H^{4i+3}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k$-copies) with bases

$\displaystyle\\{b^{i},ab^{i-1}c^{2},a^{2}b^{i-1},a^{3}b^{i-2}c^{2},\cdots,a^{2k-1}b^{i-k}c^{2},a^{2k}b^{i-k}\\},$
$\displaystyle\\{b^{i}c,a^{2}b^{i-1}c,\cdots,a^{2k-2}b^{i-k+1}c,a^{2k}b^{i-k}c\\},$
$\displaystyle\\{b^{i}c^{2},ab^{i},a^{2}b^{i-1}c^{2},a^{3}b^{i-1},\cdots,a^{2k-1}b^{i-k+1},a^{2k}b^{i-k}c^{2}\\},\text{
and }$
$\displaystyle\\{ab^{i}c,a^{3}b^{i-1}c,\cdots,a^{2k-3}b^{i-k+2}c,a^{2k-1}b^{i-k+1}c\\},$

respectively. Then by the exactness of Gysin sequence, we get $\ker
p_{4i}^{*}=\operatorname{im}\rho_{4i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k$-copies) with basis $\\{a^{2j+1}b^{i-j-1}c^{2}|0\leq j\leq k-1\\}$, $\ker
p_{4i+2}^{*}=\operatorname{im}\rho_{4i+2}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+1$-copies) with basis $\\{a^{2j}b^{i-j}c^{2}|0\leq j\leq k\\}$ and $\ker
p_{4i+j}^{*}=H^{4i+j}(X/G),j=1,3$. This gives that
$\operatorname{im}p^{*}_{4i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+1$-copies) with basis $\\{x^{2j}y^{i-j}|0\leq j\leq k\\}$ and
$\operatorname{im}p^{*}_{4i+2}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k$-copies) with basis $\\{x^{2j+1}y^{i-j}|0\leq j\leq k-1\\}$. Also,
$\operatorname{im}p^{*}_{4i+j}=\operatorname{im}\rho_{4i+j}=0,j=1$ and $3$,
Therefore,

$\displaystyle H^{4i+1}(X)=H^{4i+3}(X)=0$ $\displaystyle
H^{4i}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(n+1\text{-copies}),\text{
and}$ $\displaystyle
H^{4i+2}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(n+1\text{-copies})$

with bases $\\{x^{2j}y^{i-j},u_{4i}^{2l+1}|0\leq j\leq k,0\leq l\leq k-1\\}$
and $\\{x^{2j+1}y^{i-j},u_{4i+2}^{2l}|0\leq j\leq k-1,0\leq l\leq k\\}$,
respectively, where $\rho_{4i}(u_{4i}^{2l+1})=a^{2l+1}b^{i-l-1}c^{2},0\leq
l\leq k-1\text{ and }\rho_{4i+2}(u_{4i+2}^{2l})=a^{2l}b^{i-l}c^{2},0\leq l\leq
k$. Now, for $4m<4i+r\leq 4m+2n+2,0\leq r\leq 3$; we have
$H^{4i}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($2m+n-2i+2$-copies),
$H^{4i+1}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($m+k-i+1$-copies),
$H^{4i+2}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($2m+n-2i+1$-copies) and
$H^{4i+3}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($m+k-i$-copies) with bases

$\displaystyle\\{a^{2i-2m-1}b^{m}c^{2},a^{2i-2m}b^{m},a^{2i-2m+1}b^{m-1}c^{2},a^{2i-2m+2}b^{m-1},\cdots,a^{2k-1}b^{i-k}c^{2},a^{2k}b^{i-k}\\},$
$\displaystyle\\{a^{2i-2m}b^{m}c,a^{2i-2m+2}b^{m-1}c,\cdots,a^{2k-2}b^{i-k+1}c,a^{2k}b^{i-k}c\\},$
$\displaystyle\\{a^{2i-2m}b^{m}c^{2},a^{2i-2m+1}b^{m},a^{2i-2m+2}b^{m-1}c^{2},a^{2i-2m+3}b^{m-1},\cdots,a^{2k-1}b^{i-k+1},a^{2k}b^{i-k}c^{2}\\},\text{
and }$
$\displaystyle\\{a^{2i-2m+1}b^{m}c,a^{2i-2m+3}b^{m-1}c,\cdots,a^{2k-3}b^{i-k+2}c,a^{2k-1}b^{i-k+1}c\\},$

respectively. Then by the exactness of Gysin sequence, we get $\ker
p_{4i}^{*}=\operatorname{im}\rho_{4i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($m+k-i+1$-copies) with basis $\\{a^{2j+1}b^{i-j-1}c^{2}|i-m-1\leq j\leq
k-1\\}$, $\ker p_{4i+j}^{*}=H^{4i+j}(X/G),j=1,3$; and $\ker
p_{4i+2}^{*}=\operatorname{im}\rho_{4i+2}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($m+k-i+1$-copies) with basis $\\{a^{2j}b^{i-j}c^{2}|i-m\leq j\leq k\\}$. This
gives that
$\operatorname{im}p^{*}_{4i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($m+k-i+1$-copies) with basis $\\{x^{2j}y^{i-j}|i-m\leq j\leq k\\}$,
$\operatorname{im}p^{*}_{4i+j}=\operatorname{im}\rho_{4i+j}=0,j=1,3$; and
$\operatorname{im}p^{*}_{4i+2}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($m+k-i$-copies) with basis $\\{x^{2j+1}y^{i-j}|i-m\leq j\leq k-1\\}$.
Therefore,

$\displaystyle H^{4i+1}(X)=H^{4i+3}(X)=0$ $\displaystyle
H^{4i}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(2m+n-2i+2\text{-copies}),\text{
and}$ $\displaystyle
H^{4i+2}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(2m+n-2i+1\text{-copies})$

with bases $\\{x^{2j}y^{i-j},u_{4i}^{2l+1}|i-m\leq j\leq k,i-m-1\leq l\leq
k-1\\}$ and $\\{x^{2j+1}y^{i-j},u_{4i+2}^{2l}|i-m\leq j\leq k-1,i-m\leq l\leq
k\\}$, respectively, where
$\rho_{4i}(u_{4i}^{2l+1})=a^{2l+1}b^{i-l-1}c^{2},i-m-1\leq l\leq k-1\text{ and
}\rho_{4i+2}(u_{4i+2}^{2l})=a^{2l}b^{i-l}c^{2},i-m\leq l\leq k$.

Next, we compute the cohomology algebra of $X$. Consider the Leray-Serre
spectral sequence for the Borel fibration $X\stackrel{{\scriptstyle
i}}{{\hookrightarrow}}X_{G}\stackrel{{\scriptstyle\pi}}{{\rightarrow}}B_{G}$.
By the edge homomorphisms, we get $d_{r}(1\otimes x)=0=d_{r}(1\otimes y)$ for
all $r>0$ and for all $i,j$; $d_{r_{i,j}}(1\otimes u_{j}^{i})\not=0$ for some
$r_{i,j}>0$. Put $u_{2}^{0}=z$. So, $H^{2}(X)$ is generated by $\\{x,z\\}$.
Clearly, $d_{3}(1\otimes z)=t^{3}\otimes 1$ and so $d_{3}(1\otimes
xz)=t^{3}\otimes x\not=0$. This implies that $xz=\alpha x^{2}+\beta
y+u_{4}^{1},\alpha,\beta\in\mathbb{Z}_{2}$. So, $H^{4}(X)$ is generated by
$\\{x^{2},y,xz\\}$. By the induction, as in Theorem 1.3, we get that for
$1\leq 4i+r<2n,r=0,2$, $H^{4i}(X)$ and $H^{4i+2}(X)$ have bases
$\\{x^{2j}y^{i-j},x^{2l+1}y^{i-l-1}z|0\leq j\leq i,0\leq l\leq i-1\\}$ and
$\\{x^{2j+1}y^{i-j},x^{2l}y^{i-l}z|0\leq j,l\leq i\\}$, respectively.
Similarly, for $2n\leq 4i+r\leq 4m,r=0,2$, bases for $H^{4i}(X)$ and
$H^{4i+2}(X)$ are $\\{x^{2j}y^{i-j},x^{2l+1}y^{i-l-1}z|0\leq j\leq k,0\leq
l\leq k-1\\}$ and $\\{x^{2j+1}y^{i-j},x^{2l}y^{i-l}z|0\leq j\leq k-1,0\leq
l\leq k\\}$, respectively, and for $4m<4i+r\leq 4m+2n+2,r=0,2$, bases for
$H^{4i}(X)$ and $H^{4i+2}(X)$ are $\\{x^{2j}y^{i-j},x^{2l+1}y^{i-l-1}z|i-m\leq
j\leq k,i-m-1\leq l\leq k-1\\}$ and $\\{x^{2j+1}y^{i-j},x^{2l}y^{i-l}z|i-m\leq
j\leq k-1,i-m\leq l\leq k\\}$, respectively.

Now, we take the case when $n=2k+1$ for some $k>0$. In this case, for $2n\leq
4i+r\leq 4m,0\leq r\leq 3$, we have
$H^{4i}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($2k+2$-copies),
$H^{4i+1}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+1$-copies),
$H^{4i+2}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($2k+2$-copies) and
$H^{4i+3}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+1$-copies) with bases

$\displaystyle\\{b^{i},ab^{i-1}c^{2},a^{2}b^{i-1},a^{3}b^{i-2}c^{2},\cdots,a^{2k}b^{i-k},a^{2k+1}b^{i-k-1}c^{2}\\},$
$\displaystyle\\{b^{i}c,a^{2}b^{i-1}c,\cdots,a^{2k-2}b^{i-k+1}c,a^{2k}b^{i-k}c\\},$
$\displaystyle\\{b^{i}c^{2},ab^{i},a^{2}b^{i-1}c^{2},a^{3}b^{i-1},\cdots,a^{2k}b^{i-k}c^{2},a^{2k+1}b^{i-k}\\},\text{
and }$
$\displaystyle\\{ab^{i}c,a^{3}b^{i-1}c,\cdots,a^{2k-1}b^{i-k+1}c,a^{2k+1}b^{i-k}c\\},$

respectively. Again, by the exactness of Gysin sequence, we get $\ker
p_{4i}^{*}=\operatorname{im}\rho_{4i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+1$-copies) with basis $\\{a^{2j+1}b^{i-j-1}c^{2}|0\leq j\leq k\\}$, $\ker
p_{4i+2}^{*}=\operatorname{im}\rho_{4i+2}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+1$-copies) with basis $\\{a^{2j}b^{i-j}c^{2}|0\leq j\leq k\\}$ and $\ker
p_{4i+j}^{*}=H^{4i+j}(X/G),j=1,3$. This gives that
$\operatorname{im}p^{*}_{4i}\cong\operatorname{im}p^{*}_{4i+2}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($k+1$-copies) with bases $\\{x^{2j}y^{i-j}|0\leq j\leq k\\}$ and
$\\{x^{2j+1}y^{i-j}|0\leq j\leq k\\}$, respectively. Note that
$\operatorname{im}p^{*}_{4i+j}=\operatorname{im}\rho_{4i+j}=0,j=1$ and $3$.
Thus,

$\displaystyle H^{4i+1}(X)=H^{4i+3}(X)=0$ $\displaystyle
H^{4i}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(n+1\text{-copies}),\text{
and}$ $\displaystyle
H^{4i+2}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(n+1\text{-copies})$

with bases $\\{x^{2j}y^{i-j},u_{4i}^{2l+1}|0\leq j,l\leq k\\}$ and
$\\{x^{2j+1}y^{i-j},u_{4i+2}^{2l}|0\leq j,l\leq k\\}$, respectively, where
$\rho_{4i}(u_{4i}^{2l+1})=a^{2l+1}b^{i-l-1}c^{2},0\leq l\leq k\text{ and
}\rho_{4i+2}(u_{4i+2}^{2l})=a^{2l}b^{i-l}c^{2},0\leq l\leq k$. Now, for
$4m<4i+r\leq 4m+2n+2,0\leq r\leq 3$; we have
$H^{4i}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($2m+n-2i+2$-copies),
$H^{4i+1}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($m+k-i+1$-copies),
$H^{4i+2}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($2m+n-2i+1$-copies) and
$H^{4i+3}(X/G)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($m+k-i+1$-copies) with bases

$\displaystyle\\{a^{2i-2m-1}b^{m}c^{2},a^{2i-2m}b^{m},a^{2i-2m+1}b^{m-1}c^{2},a^{2i-2m+2}b^{m-1},\cdots,a^{2k}b^{i-k},a^{2k+1}b^{i-k-1}c^{2}\\},$
$\displaystyle\\{a^{2i-2m}b^{m}c,a^{2i-2m+2}b^{m-1}c,\cdots,a^{2k-2}b^{i-k+1}c,a^{2k}b^{i-k}c\\},$
$\displaystyle\\{a^{2i-2m}b^{m}c^{2},a^{2i-2m+1}b^{m},a^{2i-2m+2}b^{m-1}c^{2},a^{2i-2m+3}b^{m-1},\cdots,a^{2k}b^{i-k}c^{2},a^{2k+1}b^{i-k}\\},\text{
and }$
$\displaystyle\\{a^{2i-2m+1}b^{m}c,a^{2i-2m+3}b^{m-1}c,\cdots,a^{2k-1}b^{i-k+1}c,a^{2k+1}b^{i-k}c\\},$

respectively. Now, $\ker
p_{4i}^{*}=\operatorname{im}\rho_{4i}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($m+k-i+2$-copies) with basis $\\{a^{2j+1}b^{i-j-1}c^{2}|i-m-1\leq j\leq
k\\}$, $\ker p_{4i+j}^{*}=H^{4i+j}(X/G),j=1,3$; and $\ker
p_{4i+2}^{*}=\operatorname{im}\rho_{4i+2}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($m+k-i+1$-copies) with basis $\\{a^{2j}b^{i-j}c^{2}|i-m\leq j\leq k\\}$.
Consequently,
$\operatorname{im}p^{*}_{4i}\cong\operatorname{im}p^{*}_{4i+2}\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}$
($m+k-i+1$-copies) with bases $\\{x^{2j}y^{i-j}|i-m\leq j\leq k\\}$ and
$\\{x^{2j+1}y^{i-j}|i-m\leq j\leq k\\}$, respectively. Also,
$\operatorname{im}p^{*}_{4i+j}=\operatorname{im}\rho_{4i+j}=0,j=1,3$.
Therefore,

$\displaystyle H^{4i+1}(X)=H^{4i+3}(X)=0$ $\displaystyle
H^{4i}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(2m+n-2i+2\text{-copies}),\text{
and}$ $\displaystyle
H^{4i+2}(X)\cong\mathbb{Z}_{2}\oplus\cdots\oplus\mathbb{Z}_{2}\,(2m+n-2i+1\text{-copies})$

with bases $\\{x^{2j}y^{i-j},u_{4i}^{2l+1}|i-m\leq j\leq k,i-m-1\leq l\leq
k\\}$ and $\\{x^{2j+1}y^{i-j},u_{4i+2}^{2l}|i-m\leq j,l\leq k\\}$,
respectively, where $\rho_{4i}(u_{4i}^{2l+1})=a^{2l+1}b^{i-l-1}c^{2},i-m-1\leq
l\leq k\text{ and }\rho_{4i+2}(u_{4i+2}^{2l})=a^{2l}b^{i-l}c^{2},i-m\leq l\leq
k$. Now, we discuss the cohomology algebra of $X$. By the repeated use of
induction, for $\frac{n}{2}\leq i\leq m$, bases for $H^{4i}(X)$ and
$H^{4i+2}(X)$ are $\\{x^{2j}y^{i-j},x^{2l+1}y^{i-l-1}z|0\leq j\leq k,0\leq
l\leq k\\}$ and $\\{x^{2j+1}y^{i-j},x^{2l}y^{i-l}z|0\leq j\leq k,0\leq l\leq
k\\}$, respectively, and for $m<i\leq m+\frac{n}{2}$, bases for $H^{4i}(X)$
and $H^{4i+2}(X)$ are $\\{x^{2j}y^{i-j},x^{2l+1}y^{i-l-1}z|i-m\leq j\leq
k,i-m-1\leq l\leq k\\}$ and $\\{x^{2j+1}y^{i-j},x^{2l}y^{i-l}z|i-m\leq j\leq
k,i-m\leq l\leq k\\}$, respectively.

Clearly, $x^{n+1}=y^{m+1}=0$. As $d_{2}(1\otimes z^{2})=0$, $y^{2}=\alpha
x^{2}+\beta y$ for some $\alpha,\beta\in\mathbb{Z}_{2}$. Thus, the cohomology
algebra of $X$ is

$\mathbb{Z}_{2}[x,y,z]/\langle x^{n+1},y^{m+1},z^{2}+\alpha x^{2}+\beta
y\rangle,$

where $\deg x=2,\deg y=4,\deg z=2$ and $\alpha,\beta\in\mathbb{Z}_{2}$.

Similarly, we get the same cohomology algebra for the case when $n\geq 2m.$ ∎

Now, we realize the possibilities of our main Theorems.

######  Example 3.1.

Consider, the antipodal action of $G=\mathbb{Z}_{2}$ on $\mathbb{S}^{1}$ and
trivial action of $G$ on $\mathbb{CP}^{n}$. Then the diagonal action of $G$ on
$X=\mathbb{S}^{1}\times\mathbb{CP}^{n}$ is a free action with the orbit space
$X/G=\mathbb{RP}^{1}\times\mathbb{CP}^{n}$. Also, the map
$[x_{0},x_{1},\cdots,x_{2n},x_{2n+1}]\to[-x_{1},x_{0}\cdots,-x_{2n+1},x_{2n}]$
gives a free action of $G$ on $\mathbb{RP}^{2n+1}$ with the orbit space
$X/G\sim_{2}\mathbb{RP}^{1}\times\mathbb{CP}^{n}$ [7]. These examples realizes
both possibilities of Theorem 1.1.

######  Example 3.2.

Consider, the antipodal action of $G=\mathbb{Z}_{2}$ on $\mathbb{S}^{2}$ and
trivial action of $G$ on $\mathbb{HP}^{n}$. Then the diagonal action of $G$ on
$X=\mathbb{S}^{2}\times\mathbb{HP}^{n}$ is a free action with the orbit space
$X/G=\mathbb{RP}^{2}\times\mathbb{HP}^{n}$. Also, the map
$[z_{0},z_{1},\cdots,z_{n-1},z_{n}]\to[-\overline{z_{1}},\overline{z_{0}}\cdots,-\overline{z_{n}},\overline{z_{n-1}}]$
gives a free action of $G$ on $\mathbb{CP}^{2n+1}$ with the orbit space
$X/G\sim_{2}\mathbb{RP}^{2}\times\mathbb{HP}^{n}$ [7]. These examples realizes
both possibilities of Theorem 1.2.

######  Example 3.3.

Consider, the antipodal action of $G=\mathbb{Z}_{2}$ on $\mathbb{S}^{1}$ and
trivial action on $\mathbb{RP}^{n}\times\mathbb{CP}^{m}$. Then the diagonal
action of $G$ on $\mathbb{S}^{1}\times\mathbb{RP}^{n}\times\mathbb{CP}^{m}$ is
a free action with the orbit space
$\mathbb{RP}^{1}\times\mathbb{RP}^{n}\times\mathbb{CP}^{m}$. This realizes
possibility (i) of Theorem 1.3 when $\alpha=\beta=0$. Now, we consider trival
action of $G$ on $\mathbb{RP}^{n}$ and free action of $G$ on
$\mathbb{RP}^{2m+1}$ as defined in Example 3.1. Then $G$ acts freely on
$X=\mathbb{RP}^{n}\times\mathbb{RP}^{2m+1}$ with the orbit space
$X/G\sim_{2}\mathbb{RP}^{n}\times\mathbb{RP}^{1}\times\mathbb{CP}^{m}$. This
realizes possibility (i) of Theroem 1.3 when $\alpha=0,\beta=1$. Also, if we
take antipodal action of $G$ on $\mathbb{S}^{n}$ and trivial action on
$\mathbb{RP}^{1}\times\mathbb{CP}^{m}$ then $G$ acts freely on
$\mathbb{S}^{n}\times\mathbb{RP}^{1}\times\mathbb{CP}^{m}$ with the orbit
space $\mathbb{RP}^{n}\times\mathbb{RP}^{1}\times\mathbb{CP}^{m}$. This
realizes possibilty (ii) of Theorem 1.3 when $\alpha=0$.

######  Example 3.4.

Consider, the antipodal action of $G=\mathbb{Z}_{2}$ on $\mathbb{S}^{2}$ and
trivial action on $\mathbb{CP}^{n}\times\mathbb{HP}^{m}$. Then the diagonal
action of $G$ on $\mathbb{S}^{2}\times\mathbb{CP}^{n}\times\mathbb{HP}^{m}$ is
a free action with the orbit space
$\mathbb{RP}^{2}\times\mathbb{CP}^{n}\times\mathbb{HP}^{m}$. This realizes
Theorem 1.4 when $\alpha=\beta=0$. Take trival action of $G$ on
$\mathbb{CP}^{n}$ and free action of $G$ on $\mathbb{CP}^{2m+1}$ as defined in
Example 3.2. Then $G$ acts freely on
$X=\mathbb{CP}^{n}\times\mathbb{CP}^{2m+1}$ with the orbit space
$X/G\sim_{2}\mathbb{CP}^{n}\times\mathbb{RP}^{2}\times\mathbb{HP}^{m}$. This
realizes Theorem 1.4 when $\alpha=0,\beta=1$.

## Acknowledgement

We are thankful to the referee for their suggestions to improve the paper,
correcting typos and grammatical errors.

## References

  * [1] T. Tom Dieck,  Transformation groups, De Gruyter Studies in Mathematics, 8, Walter de Gruyter & Co., Berlin, 1987.
  * [2] J. Harvey, M. Kerin and K. Shankar, Semi-free actions with manifold orbit spaces, Doc. Math. 25 (2020), 2085–21214.
  * [3] A. Hatcher, Algebraic topology, Cambridge University Press, Cambridge, 2002.
  * [4] A. Kumari and H. K. Singh, Cohomology classification of spaces with free $\mathbb{S}^{1}$ and $\mathbb{S}^{3}$-actions, Filomat 36 (2022), no. 20, 7021–7026
  * [5] J. McCleary, A user’s guide to spectral sequences, second edition, Cambridge Studies in Advanced Mathematics, 58, Cambridge University Press, Cambridge, 2001.
  * [6] P. L. Q. Pergher, H. K. Singh and T. B. Singh, On $\mathbb{Z}_{2}$ and $\mathbb{S}^{1}$ free actions on spaces of cohomology type $(a,b)$, Houston J. Math. 36 (2010), no. 1, 137–146. MR2610784
  * [7] H. K. Singh and T. B. Singh, Fixed point free involutions on cohomology projective spaces, Indian J. Pure Appl. Math. 39 (2008), no. 3, 285–291.
  * [8] M. Singh, Orbit spaces of free involutions on the product of two projective spaces, Results Math. 57 (2010), no. 1-2, 53–67.
  * [9] J. C. Su, Transformation groups on cohomology projective spaces, Trans. Amer. Math. Soc. 106 (1963), 305–318.


# Does Dataset Complexity Matters for Model Explainers?

1st José Ribeiro  Federal University of Pará - UFPA
Federal Institute of Pará - IFPA
Belém, Brazil
<EMAIL_ADDRESS>2nd Raíssa Silva  IRMB, Montpellier University
La Ligue Contre le Cancer
Montpellier, France
<EMAIL_ADDRESS>3nd Lucas Cardoso  4rd Ronnie Alves  ICEN, Federal
University of Pará
Belém, Brazil
<EMAIL_ADDRESS>Federal University of Pará - UFPA
Vale Institute of Technology - ITV DS
Belém, Brazil
<EMAIL_ADDRESS>
###### Abstract

Strategies based on Explainable Artificial Intelligence - XAI have emerged in
computing to promote a better understanding of predictions made by black box
models. Most XAI measures used today explain these types of models, generating
attribute rankings aimed at explaining the model, that is, the analysis of
Attribute Importance of Model. There is no consensus on which XAI measure
generates an overall explainability rank. For this reason, several proposals
for tools have emerged (Ciu, Dalex, Eli5, Lofo, Shap and Skater). An
experimental benchmark of explainable AI techniques capable of producing
global explainability ranks based on tabular data related to different
problems and ensemble models are presented herein. Seeking to answer questions
such as “Are the explanations generated by the different measures the same,
similar or different?” and “How does data complexity play along model
explainability?” The results from the construction of 82 computational models
and 592 ranks shed some light on the other side of the problem of
explainability: dataset complexity!

###### Index Terms:

Explainable Artificial Intelligence - XAI, Black box model, Dataset
complexity.

##  I Introduction

Recently, technology has increasingly evolved and allowed intelligent
algorithms to be present in our daily lives through solutions to the most
diversified types of problems, thus prompting a need for machine learning
models to solve increasingly complex problems which require characteristics
that justify decision making, in addition to high problem-solving
performance[1][2].

Computational models based on bagging and boosting algorithms, which provide
for high performance and high generalization capacity, are commonly used in
computation to solve regression and classification problems based on tabular
data. However, these models are not considered transparent
algorithms111Transparent Algorithms: Algorithms that generate explanations for
how a given output was produced. Such examples include Decision Tree, Logistic
Regression and K-nearest Neighbors., being considered black box
algorithms222Black Box Algorithms: Machine learning algorithms that have
classification or regression decisions that are hidden from the user. and,
therefore, less used in problems related to sensitive contexts, such as health
and safety [3][4][5].

The limited understanding of black box models requires a search for measures
or tools that can provide information about local explanations — aiming to
predict around an instance through various methods to obtain a local attribute
importance ranking [6] — and global explanations — when it is possible to
understand the logic of all instances of the model by generating a global
attribute importance rank [6][7] — as a means of making decisions
interpretable and, thus, more reliable [8].

Efforts have been made in the Explainable Artificial Intelligence - XAI area
regarding the development of different measures to explain black box models,
even after their training and testing process, called post-hoc analysis [3].
Thus, measures such as Ciu [9], Dalex[10], Eli5[11], Lofo[12], Shap[13], and
Skater[14] emerged to promote the creation of model-agnostic333Model-Agnostic:
means it does not depend on the type of model to be explained. explanations.
Each of these tools is capable of generating explanations using different
techniques, but a fact they have in common is that they all generate global
attribute importance rankings related to the explanation of a model.

The proposal to analyze the ranking of global importance — and not the local —
allows general analysis of how a given model treats and explains the problem
to be solved, and for this reason, it was chosen for this study.

This research raises discussions for the current moment in the XAI area
through two main questions, namely: — Considering the current measures geared
at explaining black box machine learning models, can it be inferred that they
generate global rankings of same, similar, or different explainabilities? —
Following the same idea as in the previous question, are the generations of
equal, similar, or different explainabilities related to specific properties
of a dataset?

Seeking to answer the two hypotheses above, this research emerges as a
comparative analysis of different XAI measures, capable of producing global
explainability ranks based on tabular data related to different problems and
ensemble models.

The main contributions of this research to the machine learning area focused
on XAI are as follows:

  * •

Development of a benchmark that measures the correlations existing between the
explanations generated by the main current XAI measures by comparing them with
each other through different perspectives (properties of the datasets used);

  * •

Generation of comparative results between explainability ranks that lead
researchers in the XAI area to identify which explainability measures to use
against different characteristics of a dataset.

##  II Background

###  II-A Explainable Artificial Intelligence

Recently, there is a growing need to explain models of black box machine
learning in an agnostic way. This includes not only more robust computational
models like Deep Neural Networks but also simpler models like ensembles
[15][16][17][18].

In this way, the herein research identifies in the current literature the
necessity of specific studies on model-agnostic XAI measures for tabular data,
thus enabling explanations of computational models with wide applicability,
such as ensemble trees [17][3].

A bibliographic and practical survey (development) on the main existing XAI
measures was performed through this research, specifically aimed at generating
model-agnostic global explainability ranks that support tabular data. As a
result, a total of six tools were selected, with updated and compatible
libraries with current Python development platforms and Scikit-Learning
algorithms [19]. These tools are: CIU [9], Dalex [20], Eli5 [11], Lofo [12],
SHAP [13] and Skater [14].

Note that all the six measures referenced herein generate explanatory ranks
based on the same machine learning models previously trained, manipulate their
inputs or/and produce new intermediate models (copies). Therefore, a
comparison of the generated explanatory ranks is fair and feasible.

Due to the incompatibility between the current XAI measurement libraries and
the versions of the scikit-lean [19] computational model libraries and
dependencies, only the Random Forest and Gradient Boosting algorithms were
used in this study.

During the initial analyzes and executions, this research found XAI measures
with incompatibilities at the level of libraries and dependencies, which made
it impossible to create uniform tests with the other measures. For this
reason, some XAI means were not used for this research (for example: Alibe-ALE
[21], Lime [22], Ethical ExplainableAI [23], IBM Explainable AI 360 [24], and
Interpreter ML [25]).

Contextual Importance and Utility - CIU is a XAI measure based on Decision
Theory [26] that focuses on serving as a unified measure of model-agnostic
explainability based on the implementation of two different indexes, namely:
Contextual Importance - CI and Contextual Utility - CU, which generally create
equal attribute ranks, but with different values for each index, thus
generating global explanations[9]. Only the CI is used in this research, for
obeying the same CU rank sequence.

Dalex is a set of XAI tools based on the LOCO (leave-one covariate out)
approach and it can generate explainabilities from this approach. In general,
the measure receives the model and the data to be explained; it calculates the
performance of the model, performs new training processes with new generated
data sets, and it does the inversion of each attribute of the data in a
unitary and iterative way, thereby measuring which attributes are most
important to the model based on the performance [27] [10].

Leave One Feature Out - Lofo is a XAI measure that has a similar proposal than
Dalex, but with a main difference regarding the inversion of the attribute
iterativity, because in the Lofo measure the iterative step is the removal of
the attribute to find its global importance to the model based on the
performance [12].

Explain Like I’m Five - Eli5 is a tool that helps explore machine learning
classifiers and explains their predictions by assigning weights to decisions,
as well as exporting decision trees and presenting the importance of the
attributes of the model submitted to the tool [11].

SHapley Additive exPlanations - SHAP is proposed as a unified measure of
attribute importance that explains the prediction of an instance $X$ from the
contribution of an attribute. The contribution of this attribute is calculated
from the game theory of Shapley Value [28]. In this tool, it is possible to
calculate the average of the Shapley Value values by attributes to obtain the
global importance [13][29].

Skater is a set of tools capable of generating rankings of the importance of
model attributes, based on Information Theory [30], through measurements of
entropy in changing predictions, through a perturbation of a certain
attribute. The central idea is that the more a model is dependable upon an
attribute, the greater the change in predictions[14].

A general comparison of the main characteristics of the measures that meet the
requirements of this research is presented in table I.

TABLE I: Main XAI measures surveyed

| XAI
---
meansure
Autor | Algorithm basis |  | Global
---
explanation
(by rank)
| Local
---
explanation
| API
---
compatible
CIU | [9] | Decision Theory | Yes | No | Yes
Dalex | [20] | Leave-one covariate out | Yes | Yes | Yes
Eli5 | [11] | Assigning weights to decisions | Yes | Yes | Yes
Lofo | [12] | Leave One Feature Out | Yes | No | Yes
SHAP/Tree | [13] | Game Theory | Yes | Yes | Yes
Skater | [14] | Information Theory | Yes | Yes | Yes

Figure 1: Visual scheme of all steps and processes performed by the proposed
benchmark.

All the tools described above are capable of generating explainabilities that
go beyond the generation of attribute importance ranks; therefore, the focus
of this article is to compare this most basic unit of model explainability,
i.e., the global importance rank.

With the explanatory ranks generated by the different measures, it is fair to
compare existing correlations between these ranks, even if each measure
generates them based on different algorithms, since the central objective of
each measure is to explain a black box model.

###  II-B Dataset Properties

After being processed through attribute engineering, a dataset is expected to
exhibit characteristics that involve the nature of the problem in a contextual
(problem properties) and technical (dataset properties) manner and which are
generalized by machine learning algorithms to carry out a prediction or a
classification process [8].

Although the contextual side of the problem significantly helps in
interpreting the explainabilities of the models, the herein researchers opted
not to work with these characteristics of the analyzed data, since datasets
utilized refer to different problems with contexts that often go beyond
computing knowledge.

On the other hand, there are properties of datasets which are feasible to be
analyzed, since they are common — even though these datasets represent
different problems. Properties such as dimensionality, number of numerical
attributes, number of binary attributes, the balance between classes, and
entropy are examples of properties that every tabular dataset has and which
directly influence the generalizability of the proposed model [31].

By the time this article was published, research that confronted the inherent
characteristics of different datasets (complexities) and their explanations
through machine learning models was not identified in the literature. In this
sense, this research seeks to fill in this gap in the framework of XAI measure
studies.

##  III Materials and Methods

This research developed a benchmark, Figure 1, which uses tabular data already
known and endorsed by the machine learning community (Figure 1 \- 1),
extracted properties from these datasets (Figure 1 \- 2), clustered these
datasets according to their properties (Figure 1 \- 3), performed the
construction of computational models (Figure 1 \- 4 and 5), generated
explainability ranks for all models (Figure 1 \- 6 and 7), calculated the
correlations obtained from the all rank pairs and analyzed the different
results (Figure 1 \- 8 and 9). All these procedures are explained in detail in
the following topics.

###  III-A Datasets and Preprocess

A total of 41 datasets of binary problems were used, with no data loss, and
selected from the OpenML [32] platform. The most used datasets were selected,
so as to make the results of this research even better for use by the machine
learning community.

The datasets used were as follows: australian, phishing websites, spec,
satellite, analcatdata lawsuit, banknote authentication, blood transfusion
service center, churn, climate model simulation crashes, credit-g, delta
ailerons, diabetes, eeg-eye-state, haberman, heart-statlog, ilpd, ionosphere,
jEdit-4.0-4.2, kc1, kc2, kc3, kr-vs-kp, mc1, monks-problems-1, monks-
problems-2, monks-problems-3, mozilla4, mw1, ozone-level-8hr, pc1, pc2, pc3,
pc4, phoneme, prnn crabs, qsar-biodeg, sonar, spambase, steel-plates-fault,
tic-tac-toe and wdbc.

All datasets went through the following pre-processing steps, as required:
converting categorical attributes to frequency-based ordinals, converting
boolean attributes to integer (0 and 1), and min-max normalization to values
between 0 and 1 [33].

###  III-B Clustering and Multiple Correspondence Analysis

Analyzing different datasets based on their properties allows for comparing
similarities and differences between them, even if these datasets are from
different contexts. For example, we can group a set of datasets based on their
class entropy values and then identify which datasets have more and less
information, thus enabling a dataset complexity impact relationship for future
model explanations.

Based on the proposal laid out above, this research used 15 different
properties (provided by OpenML) extracted from each of the 41 selected
datasets and then used the clustering algorithm k-means [34] to identify
groups of datasets based on their similarities.

The interpretation and validation algorithm between data clusters, called
Silhouettes [35], was used, the K value (clusters) of which ranging between 2
and 20, and K = 3 with Average Silhouette Scores = 0.28 being identified,
Figure 2.

Figure 2: Silhouette coefficients for clustering, using the Kmeans algorithm,
for a $=3$. Distance means ($x$ axis) and label of clusters $0,1$ and $2$
(axis $y$).

In figure 2, it can be seen that for a k=3 the distances between each cluster
(0, 1 and 2) are above the average (red line), so this is an adequate value of
k in relation to the study presented hereby.

The following 15 properties being used included: Number of Features, Number of
Instances, Dimensionality, Percentage of Binary Features, Standard Deviation
Nominal of Attribute Distinct Values, Mean Nominal Attribute Distinct Values,
Class Entropy, Autocorrelation, Number of Numeric Features, Number of Symbolic
Features, Number of Binary Features, Percentage of Symbolic Features,
Percentage of Numeric Features, Majority Class Percentage, and Minority Class
Percentage.

Further information about the values of each property for each dataset can be
found in: https://github.com/josesousaribeiro/XAI-
Benchmark/blob/main/Openml/df_dataset_properties.csv.

In order to identify the different correlations between each dataset cluster
identified with the 15 properties listed above, this research used the
Multiple Correspondence Analysis - MCA [36] in the properties dataset, where
the rows of this table are the observations or individuals $(n)$ concerned —
here are the datasets — and the columns are the different categories of
nominal variables $(p)$ — here are the properties of each dataset.

This research performed the process of binarization in the same table in which
clustering was applied, replaced by $h$ (equal to or above the mean of
attribute values) and $s$ (below the mean of attribute values). For more
information about the new properties dataset, refer to:
https://github.com/josesousaribeiro/XAI-
Benchmark/blob/main/Openml/df_properties_binarized.csv.

This analysis allows for identifying different relationships between dataset
(or even dataset group) and properties (or even property value ranges), as
advocated by the literature [36].

###  III-C Construction of Models

The algorithms used in building each dataset model were Random Forest - RF and
Gradient Boosting - GB, both being tree-based, low-explainability ensembles
[3].

This research tested newer ensemble algorithms such as Light Gradient Boost
[37], CatBoost [38], and Extreme Gradient Boosting [39]; however, not all XAI
metrics that were used supported the output encodings of these new models.
Thus, an option was made to adhere to the Gradient Boosting and Random Forest
algorithms, only.

Notably, the Tuning step was performed through Grid Search [40] with cross-
validation [41] of $fold=3$, thus that the algorithms had a better calibration
on the analyzed data, and a total of 192 candidates through the parameters:
max depth: (1, 10), bootstrap: (True, False), n estimators: (100, 200), min
samples leaf: (1, 10), min samples split:(2,10), max features: (sqrt, log2),
and loss: (deviance, exponential) (the latter only valid for the GB
algorithm).

Finally, each model was trained and tested respectively based on a $70\%-30\%$
ration for each dataset. Then, each model was measured as to their performance
and stability regarding accuracy and precision, recall and Friedman test [42],
so as to characterize each model that was created.

###  III-D XAI Ranks and Correlations

For each of the 41 datasets (divided into 3 different clusters), 2 models were
created, one based on the Random Forest algorithm and the other on Gradient
Boosting, thus generating a total of 82 models. Global explainability ranks
were produced through the 6 different XAI measures for each model, resulting
in a total of 492 ranks.

The Spearman Rank Correlation [43] was used to calculate the correlations
between the ranks that were created. The reason for using this algorithm in
particular, is that it measures the correlation between rank pairs considering
the idea of ranks (positions) in which different values (in this case, dataset
attributes) may appear. In this step, two comparison matrices of rank
correlation pairs are generated for each dataset (one matrix for each
algorithm). Figure 3 shows an example of this matrix created from the Gradient
Boosting - GB model.

Figure 3: Example Spearman correlation matrix (represented in value and color
in each cell) based on explicability ranks generated from Gradient Boosting
model to the dataset $wdbc$. Note that, for this dataset, chosen at random,
the correlations between the rank pairs are low or non-existent (close to
zero).

Correlations such as those shown in Figure 3 are calculated for each dataset
and show how the correlations between different pairs of XAI measures are
presented for each problem and for each computational model being developed.

The results shown in Figure 3 refer to only one dataset and, therefore, are
intermediate results, since they will be summarized in the results of the
other datasets.

###  III-E Benchmark of XAI measures

All the methodology described so far was consolidated and implemented in a
single benchmark. More details about the implementation and execution of the
procedures presented hereby are available in the repository: —
https://github.com/josesousaribeiro/XAI-Benchmark.

##  IV Results and Discussion

The results obtained were computed in groups (2 different computational
models) and clusters (3 different dataset clusters).

The principle for understanding the results of this article is to acknowledge
the different dataset cluster profiles identified in the clustering process
and MCA.

In this sense, according to clustering, based on the 15 different properties
of the 41 datasets that were analyzed, it is possible to assess the existence
of at least 3 different dataset clusters. In quantitative terms, each cluster
has 21 (cluster 0), 17 (cluster 1), and 3 (cluster 2) datasets.

To identify each different dataset profile, an inspection of the value ranges
of each cluster for each property was carried out. This analysis verified the
possible existence of specific property values for datasets of specific
clusters.

At this stage, it was identified that the quantity of datasets in cluster 2
had a small $n$ and, therefore, it was disregarded from these results (the
datasets for this cluster are: kr-vs-kp, Phishing Web sites, and SPECT).
Keeping only clusters 0 and 1.

Intending to consolidate the profile analysis of the clusters, Multiple
Correspondence Analysis - MCA was applied, and the relation between the
datasets and the value ranges of their properties was verified, thus providing
for a better understanding of the complexity of the datasets existing in both
clusters — this practice is already known in the literature [36].

Figure 4: Multiple Correspondence Analysis - MCA with rows (datasets) and
columns (properties) depending on the first and second PCA components.

By inspecting the MCA shown in Figure 4, the relation (shorter distances)
between high quantity of datasets in cluster 0 and above-average values
(symbol $h$) for properties (bigger circles) such as MajorityClassPercentage,
Dimensionality, NumberOfInstances, NumberOfFeatures,
PercentageOfNumericFeatures, and AutoCorrelation was identified. Also, a
relation between a high quantity of datasets of cluster 0 and below-average
values (symbol $s$) for properties such as ClassEntropy,
PercentageOfSymbolicFeatures, and PercentageOfBinaryFeatures was observed.

Also with regard to the MCA shown in Figure 4, by inspection, assessing the
existence of a relation (shorter distances) between a high quantity of
datasets in cluster 1 and above-average values (symbol $h$) is possible for
properties (bigger circles) such as ClassEntropy,
PercentageOfSymbolocFeatures, PercentageOfBinaryFeatures,
NumberOfBinaryFeatures, and MinorityClassPercentage. It is also possible to
identify a relation between a high quantity of datasets in cluster 1 and
below-average values (symbol $s$) for properties sush as
AutoCorrelation,NumberOfFeatures, PercentageOfNumericFeatures,
NumberOfNumericFeatures, NumberOfInstances, Dimensionality, and
MajorityClassPercentage.

It is worth noting that some properties were not mentioned above because they
appear at very close distances for the two dataset clusters.

However, based on the inspection of MCA results, it can be concluded that
cluster 0 is formed by a significant amount of datasets with greater
complexities, whereas cluster 1 is formed by a significant amount of datasets
with minor complexities.

The entire cluster profile identification process can be found in:
https://github.com/josesousaribeiro/XAI-
Benchmark/blob/main/Cluster_Profile.md.

Based on the profile characterization of the two main clusters that were
identified, this research executed the Benchmark for these two sets of
datasets separately.

The benchmark execution for the datasets belonging to cluster 0, Figure 5,
exhibits low or negligible correlations found in most tests, thus showing that
the six XAI measures used hereby generate different explainability ranks for
most of the datasets in this group, regardless of the algorithm (RF or GB) the
computational model is based upon.

Figure 5: Boxplot graphs that summarize all rank pair comparisons ($x$ axis)
and their respective Spearman correlations ($y$ axis) calculated for the
datasets of cluster 0. Results of the model based on Random Forest (left) and
Gradient Boosting (right). The dashed blue lines show the different levels of
correlations. The points identify the positions of the correlation values of
each comparison, and the color of the points refer to the attribute quantities
of each dataset. Note the low variance of most boxplots along with levels of
negligent or low correlations found. Figure 6: Boxplot graphs that summarize
all rank pair comparisons ($x$ axis) and their respective Spearman
correlations ($y$ axis) calculated for the datasets of cluster 1. Results of
the model based on Random Forest (left) and Gradient Boosting (right). The
dashed blue lines show the different levels of correlations. The points
identify the positions of the correlation values of each comparison, and the
color of the points refer to the attribute quantities of each dataset. Note
the high variances of most boxplots along with the high levels of correlations
found.

The results presented by cluster 0, Figure 5, show that datasets with greater
complexity generate more complex models, and in turn, different explanations.
This could justify the low correlations found in the results.

The benchmark execution for the datasets of cluster 1, Figure 6, shows
different correlations from those shown in Figure 5, since there is high
variance in boxplots as well as high positive and negative correlations
between rank pairs in cluster 1.

The results presented in Figure 5 and 6 provide evidence on how to answer the
first hypothesis launched herein: “— Considering the current measures aimed at
explaining black box machine learning models, can it be inferred that they
generate global rankings of same, similar, or different explainabilities?”. It
depends, since the different correlations found demonstrate that the
properties of the datasets directly influence the explainabilities of the
models created from them. Thus, it is possible to make the tools generate
explainabilities with higher correlations with one another (more similar
explainabilities) and lower correlations with one another (less similar
explainabilities).

Noticeably, despite the similar results of the models based on the RF and GB
algorithms, there are important differences in the correlations found, as
shown in Figure 5. For example, the larger variance of some comparisons in the
results of the GB algorithm (shap vs lofo, eli5 vs ci, eli5 vs skater, ci vs
skater and shap vs ci). This proves that the complexity of the algorithm used
in the model (in this case, bagging and boosting tree-based) does have an
influence on explainability generation.

Importantly, the outliers existing in the boxplots of Figures 5 and 6, show
the possibility of other dataset clusters existing amongst those being
analyzed. The colors, used to distinguish the amount of attributes from
different datasets in each boxplot show that the results obtained in the
correlations are not a function of the number of attributes that a dataset
has.

The comparison of the results presented in Figures 5 and 6 demonstrate the
answer to the second hypothesis of this research, namely: “— According to the
same idea as in the previous question, are the generations of equal, similar,
or different explainabilities related to specific properties of a dataset?”.
Yes, since the results of the rank correlations for the two dataset clusters
presented in the benchmark output were different, thus showing that the
complexity properties of the dataset interfere with explainability ranks.

Based on the experiments performed with the 6 different XAI measures and the
resulted obtained therefrom, the following logical reasoning could be
constructed: 1 - If a ensemble model (algorithm and dataset) solves a
classification problem for a low-complexity dataset, there must be few ranks
(or even only one rank) of explainabilities referring to this model, thus
allowing to infer that the different XAI measures have higher correlations
with each other (as seen in Figure 6); 2 - If an ensemble model (algorithm and
dataset) solves a classification problem for a high-complexity dataset, there
must be many explainability ranks for this model, leading the ranks of the
different XAI measures to show lower correlations with each other (as seen in
Figure 5).

Based on the logical reasoning 1 and 2 above, there was identified existence
of evidence about Ensemble-Tree Models with Complex Explainabilities - EMCX
(created from the data of cluster 0) and Ensemble-Tree Models with Simple
Explainabilities - EMSX (created from the data of cluster 1), both terms
proposed hereby. Furthermore, it was identified that the divergence between
the ranks is inherent, primarily from the complexity of the dataset and,
consequently, from the generated model — thus answering the article’s title
question.

The datasets depicted below, which served as the basis for the creation of
EMSX and EMCX, are intended to comply with one of the contributions hereto:

  * •

EMSX Datasets: Ozone level 8hr, Sonar, Spambase, Qsar biodeg, Kc3, Mc1, Pc3,
Mw1, Pc4, Satellite, Pc2, Steel plates fault, Kc2, Pc1, Kc1, Climate model
simulation crashes, and Analcatdata lawsuit;

  * •

EMCX Datasets: Ionosphere, Wdbc, Credit-g, Churn, Australian, Eeg eye state,
Heart statlog, Ilpd, Tic tac toe, JEdit 4.0 4.2, Diabetes, Prnn crabs, Monks
problems 1, Monks-problems 3, Monks problems 2, Delta ailerons, Mozilla4,
Phoneme, Blood transfusion service center, Banknote authentication, and
Haberman;

It is clear that evidence of the existence of EMSX and EMCX was observed from
experiments with ensemble algorithms. For these terms to become generally
applicable (other algorithms), experiments with algorithms of other natures
are necessary.

##  V Final Considerations

Given all the analyzes that were carried out, this research achieves its
objective of being able to observe the impacts of dataset complexity in
explaining black box machine learning models, through the different properties
of the data under analysis. That allowed for finding that the properties of a
dataset — facing the creation of models based on Random Forest and Gradiant
Boosting algorithms — show evidence that EMSX and EMCX do exist. However, many
other properties of the data and of other algorithms still need to be further
explored.

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