# Counter-examples to the Hasse Principle among the twists of the Klein
Quartic

Elisa Lorenzo García  Elisa Lorenzo García: Institut de Mathématiques,
Université de Neuchâtel, Rue Emile-Argand 11, 2000, Neuchâtel, Switzerland —
Laboratoire IRMAR, Office 602, Université de Rennes 1, Campus de Beaulieu,
35042, Rennes Cedex, France<EMAIL_ADDRESS>elisa.lorenzogarcia@univ-
rennes1.fr Michaël Vullers<EMAIL_ADDRESS>and  Michaël Vullers
Dedicated to Jaap Top on the occasion of his 62th birthday.

###### Abstract.

In this paper we inspect from closer the local and global points of the twists
of the Klein quartic. For the local ones we use geometric arguments, while for
the global ones we strongly use the modular interpretation of the twists. The
main result is providing families with (conjecturally infinitely many) twists
of the Klein quartic that at counter-examples to the Hasse Principle.

##  1\. Introduction

The Klein quartic $\mathcal{C}$ is the curve in the projective plane given by
the equation

$X^{3}Y+Y^{3}Z+Z^{3}X=0$

which is isomorphic to the modular curve $X(7)$ over the cyclotomic field
$\mathbb{Q}(\zeta_{7})$. It has a number of interesting properties which are
deeply studied in [2]. For instance it is a Hurwitz curve, meaning that it has
the maximum number of automorphisms possible for its genus. In fact it is the
smallest genus example of such a curve.

Given a curve $C$ over a field $k$, we say that a curve $C^{\prime}/k$ is a
twist of $C$ if over an algebraic closure $\overline{k}$ of $k$ the curves $C$
and $C^{\prime}$ are isomorphic. Geometrically speaking both curves are equal,
however they may have very different arithmetic properties. In this paper we
focus our attention on the following question: Suppose that $C^{\prime}$ is a
twist of $\mathcal{C}$, _does $C^{\prime}$ violate the Hasse principle?_ That
is, does $C^{\prime}$ have a point over every completion of $\mathbb{Q}$ but
not over $\mathbb{Q}$ itself?

In [10] and [11] a similar question was asked and answered for the modular
curves $X_{0}(N)$. The modularity of these curves is strongly used there to
prove the existence of local and global points. In this paper we start by
using a classification of all the twists of the Klein quartic [8], we use
geometric arguments as well as Hensel’s Lemma to deal with the existence of
local points and we only use modular reasonings to study rational points.
Those differ by the ones used in the aforementioned papers. Ther are based on
the central result in [3].

Our main result is the following:

######  Theorem 1.1.

There are (conjecturally infinetely many) twists of the Klein quartic yielding
counterexamples to the Hasse principle.

##  2\. Non-Archimedean local points

In this section we study the existence of $\mathbb{Q}_{p}$-points for primes
$p$ not dividing the discriminant or the order of the automorphism group of
the twists of the Klein quartic.

######  Proposition 2.1.

Let $p$ be a prime different from $2$, $3$ or $7$ and let
$C^{\prime}/\mathbb{F}_{p}$ be a twist of the Klein quartic such that $p$ does
not divide its discriminant. Then $C^{\prime}(\mathbb{F}_{p})\neq\emptyset$.

###### Proof.

If $p\geq 33$, the result follow immediately from the Hasse-Weil bound, so
assume $p<33$. Let $\pi$ denote the Frobenius on $\mathrm{Jac}(C^{\prime})$.
Since

$\\#C^{\prime}(\mathbb{F}_{p})=p+1-\mathrm{tr}(\pi),$

it suffices to show that $\mathrm{tr}(\pi)\neq p+1$. In their paper [9]
Meagher and Top show that

$\mathrm{tr}(\pi)=\left\\{\begin{array}[]{ll}0&\text{if }p\equiv
3,5,6\,\,\mathrm{mod}\,\,7\\\
3\mathrm{tr}(\pi_{E}),0,-\mathrm{tr}(\pi_{E}),\mathrm{tr}(\alpha\pi_{E})\text{
or }\mathrm{tr}(\pi_{E})&\text{if }p\equiv
1,2,4\,\,\mathrm{mod}\,\,7\end{array}\right.$

where $\pi_{E}$ is the Frobenius of the elliptic curve $E/\mathbb{F}_{p}$
given by $y^{2}+xy=x^{3}+5x^{2}+7x$ and $\alpha\in\mathrm{End}(E)$ satisfies
$\alpha^{2}+\alpha+2=0$.

It follows immediately that if $p\equiv 3,5,6\,\,\mathrm{mod}\,\,7$ then there
is nothing to prove, so assume $p\equiv 1,2,4\,\,\mathrm{mod}\,\,7$. This
gives us only the cases $p=11$, $23$ or $29$. For which the elliptic curve $E$
is ordinary and
$\mathrm{End}(E)\cong\mathbb{Z}[\alpha]\subset\mathbb{Q}(\sqrt{-7})$ after
making $\alpha=\frac{-1+\sqrt{-7}}{2}$. Moreover, after this identification,
we get $\pi_{E}=\pm 2\pm\sqrt{-7}$, $\pm 4\pm\sqrt{-7}$, $\pm 1\pm 2\sqrt{-7}$
for $p=11$, $23$, $29$ respectively.

Straightforward substitution now yields $\mathrm{tr}(\pi)\neq p+1$ hence
$C(\mathbb{F}_{p})\neq\emptyset$. ∎

######  Corollary 2.2.

Let $p\neq 2,3,7$ be a prime. If $C^{\prime}$ is a twist of the Klein quartic
with $p$ not dividing its discriminant, then
$C^{\prime}(\mathbb{Q}_{p})\neq\emptyset$.

###### Proof.

Let $\overline{C^{\prime}}/\mathbb{F}_{p}$ be the reduction of $C^{\prime}$,
then $\overline{C^{\prime}}$ is a twist of the Klein quartic over
$\mathbb{F}_{p}$, in particular it is a smooth curve. From Proposition 2.1 one
has $\overline{C^{\prime}}(\mathbb{F}_{p})\neq\emptyset$ hence by Hensel’s
Lemma it follows that $C^{\prime}(\mathbb{Q}_{p})\neq\emptyset$. ∎

##  3\. Real points

In this section, and before proving that all twists $C^{\prime}$ of the Klein
quartic have real points, we introduce a special twist:

$C_{0}:x^{4}+y^{4}+z^{4}+6(xy^{3}+yz^{3}+zx^{3})-3(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})+3xyz(x+y+z)=0.$

We fix the isomorphism $\phi_{0}:\,C_{0}\rightarrow\mathcal{C}$ as in [8, Sec.
6]. The special property of this twist is that among all twists of the Klein
quartic, the Klein quartic itself included, it is the one with the smallest
field of definition of its automorphism group, namely $\mathbb{Q}(\sqrt{-7})$.
This is why, in [8] instead of directly computing the twists of $\mathcal{C}$,
the twists of $C_{0}$ are computed. Of course, both sets are equal, but the
algorithm developed in [7] runs faster while inputing $C_{0}$. Moreover,
$C_{0}$ is isomorphic over $\mathbb{Q}$ to $X(7)$.

######  Proposition 3.1.

Let $C^{\prime}$ be a twist of the Klein quartic $\mathcal{C}$. Then
$C^{\prime}(\mathbb{R})\neq\emptyset$.

###### Proof.

By the classification of twists of the Klein quartic in [8] when
$K=\mathbb{R}$ we get that there are only two twists of the Klein quartic up
to $\mathbb{R}$-isomorphism, the Klein quartic itself and the twist $C_{0}$.
The Klein quartic has for instance the rational point $(1:0:0)$. Plugging
$(x:1:1)$ into $C_{0}$ yields a degree $4$ polynomial in $x$ over $\mathbb{R}$
which has real solutions, hence $C_{0}(\mathbb{R})\neq\emptyset$. ∎

##  4\. Rational points

In order to study the rational points of the twists of the Klein quartic, we
start by recalling some definitions and rephrasing the main theorem of [3].

Let $\chi_{p}$ be the quadratic character of $G_{\mathbb{Q}}$ obtained from
the $p$-th cyclotomic character. The Galois action on the $p$-torsion points
of an elliptic curve $E/\mathbb{Q}$ produces a representation
$\overline{\rho}_{E,p}:\,G_{k}\rightarrow\operatorname{PGL}_{2}(\mathbb{F}_{p})$,
determined up to conjugation by the isomorphism class of $E$ unless the
$j$-invariant of $E$ is $0$ or $1728$, and with determinant $\chi_{p}$. Going
the other way, we say that a representation
$\rho:\,G_{\mathbb{Q}}\rightarrow\operatorname{PGL}_{2}(\mathbb{F}_{p})$ is
realized by an elliptic curve $E/\mathbb{Q}$ if $\overline{\rho}_{E,p}=\rho$,
where this last equality is considered up to conjugation inside
$\operatorname{PGL}_{2}(\mathbb{F}_{p})$.

The automorphism group of $X(p)$ is naturally (via the modular interpretation)
isomorphic to $\operatorname{PSL}_{2}(\mathbb{F}_{p})$ and it is defined over
the only quadratic field $k_{p}$ contained in the cyclotomic field
$\mathbb{Q}(\zeta_{p})$. Fix $V:=\begin{pmatrix}0&-v\\\ 1&0\end{pmatrix}$ for
$v$ a non-square in $\mathbb{F}_{p}^{\times}$. Then, by [3, Prop. 2.1] we have
that the action of the non-trivial element $\tau$ of
$\operatorname{Gal}(k_{p}/\mathbb{Q})$ on an element
$g\in\operatorname{Aut}(X(p))\simeq\operatorname{PSL}_{2}(\mathbb{F}_{p})$ is
given by $g\mapsto VgV$.

A cocycle
$\xi\in\operatorname{H}^{1}(G_{\mathbb{Q}},\operatorname{Aut}(X(p)))$ defining
a twist of $X(p)$ can be seen as a representation
$\rho:\,G_{\mathbb{Q}}\rightarrow\operatorname{Aut}(X(p))\rtimes\operatorname{Gal}(k_{p}/\mathbb{Q})\simeq\operatorname{PGL}_{2}(\mathbb{F}_{p}):\,\sigma\mapsto(\xi(\sigma),\overline{\sigma})$,
where the last isomorphism is given by sending $(g,1)$ to $g$ and $(1,\tau)$
to $V$. Going the other way around, starting by a representation
$\rho:\,G_{\mathbb{Q}}\rightarrow\operatorname{PGL}_{2}(\mathbb{F}_{p})$ whose
determinant is the quadratic character $\chi_{p}$, we can define the cocycle
$\xi$ given by the first projection of $\rho$ by using the previous
isomorphism, i.e. setting $\xi(\sigma)=\pi_{1}(\rho(\sigma))$. This cocycle
produces a twist $X_{\rho}(p)$. We can as well define a cocycle
$\xi^{\prime}(\sigma)=\pi_{1}(\sigma^{-1})^{t}$ defining a twist
$X^{\prime}_{\rho}(p)$. These two twists, defined over the same field, are
equivalent under the conditions in [3, Lemma 2.1].

######  Theorem 4.1.

([3, Thm. 2.1]) The map $X(p)\rightarrow X(1)$ define a surjective map from
the set of non-cuspidal rational points with $j\neq 0,1728$ on the curves
$X_{\rho}(p)$ and $X^{\prime}_{\rho}(p)$ to the set of isomorphism classes of
elliptic curves defined over $\mathbb{Q}$ realizing $\rho$.

It is important to note that not every twist of $X(7)$ arises as
$X_{E}(7):=X_{\overline{\rho}_{E,7}}(7)$ or
$X_{E}^{-}(7):=X^{\prime}_{\overline{\rho}_{E,7}}(7)$ for some elliptic curve
$E$, in particular $\mathcal{C}$ itself does not arise from an elliptic curve.
This type of twists have brought the attention of the community: explicit
equations for $X_{E}(7)$ are given in [4], and the study of their rational
points for some particular examples are in [12], which is used to solve a
particularly difficult diophantine equation.

In the case in which the elliptic curve $E$ does not have CM, we follow [14,
Thm. 1.5] to determine the possible images of the representations
$\overline{\rho}_{E,7}$.

######  Theorem 4.2.

Let $E$ be an elliptic curve defined over $\mathbb{Q}$. If $E$ does not have
CM, then the image of $\overline{\rho}_{E,7}$ is isomorphic to $S_{3}$ (then
$j(E)=3^{3}\cdot 5\cdot 7^{5}/2^{7}$) or has cardinality $14$, $42$ or $336$.
If $E$ has CM then $\overline{\rho}_{E,7}$ has cardinality $7$ or larger than
$12$.

###### Proof.

The aforementioned reference classify the possibilities for
$\rho_{E,7}:\,G_{\mathbb{Q}}\rightarrow\operatorname{GL}_{2}(\mathbb{F}_{7})$
when $E/\mathbb{Q}$ does not have CM.

There are only a finite number of $E/\mathbb{Q}$ with CM up to
$\overline{\mathbb{Q}}-$isomorphism. When $7$ does not divide the discriminant
of the CM-field and up to quadratic twists, the corresponding representation
is maximal according to the LMFDB [6]. When $7$ divides the discriminant then
the CM-field is $\mathbb{Q}(\sqrt{-7})$ and again according to the LMFDB, and
up to quadratic twist, $\overline{\rho}_{E,7}$ has cardinality $14$. ∎

Twists not arising from elliptic curves do not have rational points apart from
eventually the cusps.

######  Proposition 4.3.

Let $\phi\colon C^{\prime}\rightarrow C_{0}$ be a twist defined over a field
$L$. If $L\cap\mathbb{Q}(\zeta_{7})=\mathbb{Q}(\sqrt{-7})$, then $C^{\prime}$
has no rational cusps.

###### Proof.

The cusps of $C_{0}$ are defined over $\mathbb{Q}(\zeta_{7})$ but not over
$\mathbb{Q}(\sqrt{-7})$. An isomorphism $\phi\colon C^{\prime}\rightarrow
C_{0}$ maps $C^{\prime}(\mathbb{Q})$ to $C_{0}(L)$. If $C_{0}$ has no cusps
over $L$, then $C^{\prime}$ has no cusps over $\mathbb{Q}$. ∎

##  5\. Counter-examples to the Hasse Principle

In the classification of twists of the Klein quartic in [8] there are $11$
different families (or cases) of twists accordingly to the isomorphism class
of the Galois group of the field of definition of the twists. The families 6.
and 10. do not appear when the Klein quartic is considered to be defined over
$\mathbb{Q}$. And the family $11.$ contains all the twists $X_{E}(7)$ when
$\overline{\rho}_{E,7}$ is a big as possible which is the generic case: all
those twists have then rational points according to Theorem 4.1. We hence
study the other cases in order to find counter-examples to the Hasse
Principle.

######  Proposition 5.1.

Twists from cases $1$. and $2$. satisfy $C^{\prime}(\mathbb{Q}_{2})=\emptyset$

###### Proof.

Case $1$. corresponds to the curve $C_{0}$. It is straightforward to check
that $\overline{C_{0}}(\mathbb{F}_{2})=\emptyset$. For case $2$. we
distinguish the cases $m$ even or odd, where $m$ is the parameter of the
family given in [8, Thm. 6.1]. This time
$\overline{{C}^{\prime}}(\mathbb{F}_{2})\neq\emptyset$, but
$\overline{{C}^{\prime}}(\mathbb{Z}/4\mathbb{Z})=\emptyset$. ∎

######  Proposition 5.2.

Twists in case $7$. satisfy $C^{\prime}(\mathbb{Q}_{7})=\emptyset$.

###### Proof.

Independently of the valuations at $7$ of the parameters defining this family
we always obtain, after simplification, an equation of the form $Q^{2}+7G=0$
with $Q,G\in\mathbb{Z}[x,y,z]$ homogeneous of degree 2 and 4 respectively, and
not divisible by $7$. This model of the twists clearly shows the hyperelliptic
reduction behavior of the Klein quartic at $7$, see [5, Prop. 1.2]. While
trying to construct $\mathbb{Q}_{7}$-points we find an obstruction. We do find
points over $\mathbb{F}_{7}$ in the conic, but they are double points of the
quartic and they cannot even be lifted to $\mathbb{Z}/7^{2}\mathbb{Z}$. ∎

######  Proposition 5.3.

For infinitely many twists in cases $5$., $9$. and $11$. we have
$C^{\prime}(\mathbb{Q})\neq\emptyset$.

###### Proof.

They are infinitely many non-CM elliptic curves giving Galois representations
with $\overline{\rho}_{E,7}$ isomorphic to the corresponding groups of order
$14$, $42$ and $336$ respectively, see [14, Thm. 1.5]. ∎

We look next to families that do provide counter-examples to the Hasse
Principle.

###  5.1. Cases 3., 4. and 8

For those twists the field of definition is $L=K(\alpha,\beta,\gamma)$ with
$K=\mathbb{Q}(\sqrt{-7})$ and $\alpha,\beta,\gamma$ are the roots of a degree
$3$ irreducible polynomial $f=x^{3}+Ax^{2}+Bx+C$ with coefficients in
$\mathbb{Q}$. For case 3 we have that the splitting field of $f$ over
$\mathbb{Q}$ has discriminant $\Delta=-7q^{2}$, for case 4. $\Delta=q^{2}$,
and for case 8 we have that the splitting field of $f$ has discriminant
$\Delta$ not of the form $q^{2}$ or $-7q^{2}$.

Let $\phi\colon C^{\prime}\rightarrow C_{0}$ be the twist associated with $L$.
Then $\phi$ is given by

$\left(\begin{array}[]{ccc}d&-3\alpha+2\beta+\gamma&\alpha\beta-3\beta\gamma+2\alpha\gamma\\\
d&\alpha-3\beta+2\gamma&2\alpha\beta+\beta\gamma-3\alpha\gamma\\\
d&2\alpha+\beta-3\gamma&-3\alpha\beta+2\beta\gamma+\alpha\gamma\\\
\end{array}\right)$

which has determinant equal to $-21d\sqrt{\Delta}$. Taking
$(d,\Delta)=(\sqrt{-7},-7q^{2})$ gives us case 3, $(d,\Delta)=(1,q^{2})$ gives
us case 4, and taking $(d,\Delta)=(\sqrt{\Delta},\Delta)$ yields case 8.

Therefore if $p$ is a prime such that $p\neq 2,3,7$ and $p\nmid\Delta$, then
$p\nmid\det(M)$ hence $C^{\prime}$ has good reduction at $p$. From Proposition
2.1, and its corollaries, it follows that for such primes
$C^{\prime}(\mathbb{Q}_{p})\neq\emptyset$.

We can assume the coefficient $A=0$. Hence, accordingly to [8] the twist
$C^{\prime}$ is given by the equation

$\displaystyle F=$ $\displaystyle\,\,3d^{4}x^{4}-63d^{2}Bx^{2}y^{2}+9\cdot
21d^{2}Cx^{2}yz+21d^{2}B^{2}x^{2}z^{2}+147d\sqrt{\Delta}xy^{3}$
$\displaystyle+147d\sqrt{\Delta}Bxyz^{2}-147d\sqrt{\Delta}Cxz^{3}-9\cdot
49B^{2}y^{4}+98\cdot 27BCy^{3}z$
$\displaystyle+147(2B^{3}-27C^{2})y^{2}z^{2}-9\cdot
98B^{2}Cyz^{3}-49B^{4}z^{4}=0.$

With $\Delta=-4B^{3}-27C^{2}$.

We study now the $\mathbb{Q}_{2}$-points, $\mathbb{Q}_{3}$-points and
$\mathbb{Q}_{7}$-points.

######  Lemma 5.4.

Let $f(x)=x^{3}+Bx+C$ and $C^{\prime}$ as above, if $2\nmid C$ then
$C^{\prime}(\mathbb{Q}_{2})\neq\emptyset$.

###### Proof.

If $2\nmid C$ then $2\nmid\Delta$ and $F$ reduces modulo $2$ as follows

$\displaystyle x^{4}+x^{2}yz+xy^{3}+xz^{3}+y^{2}z^{2}$ $\displaystyle\text{if
}2\mid B$ $\displaystyle
x^{4}+x^{2}y^{2}+x^{2}yz+x^{2}z^{2}+xy^{3}+xyz^{2}+xz^{3}+y^{4}+y^{2}z^{2}+z^{4}$
$\displaystyle\text{if }2\nmid B$

Note that if $2\mid B$, then the point $(0:1:0)$ is a solution over
$\mathbb{F}_{2}$, If $2\nmid B$, then $(1:1:1)$ is a solution over
$\mathbb{F}_{2}$. Both of which can be lifted with Hensel’s Lemma since the
derivative with respect to $x$ does not vanish. ∎

######  Lemma 5.5.

Let $f(x)=x^{3}+Bx+C$ and $C^{\prime}$ as above. If $3\nmid B$ then
$C^{\prime}(\mathbb{Q}_{3})\neq\emptyset$.

###### Proof.

If $3\nmid B$, then since $\Delta=-4B^{3}-27C^{2}$ it follows that
$3\nmid\Delta$ and therefore $3\nmid d\sqrt{\Delta}$ and $3\nmid d$. After
making the change of variable $z\mapsto 3z$ and dividing by $3$ the equation,
we obtain modulo $3$ the non-singular point $(0:1:0)$, that lifts to a point
in $\mathbb{Q}_{3}$ by Hensel’s Lemma. ∎

######  Proposition 5.6.

Let $f(x)=x^{3}+Bx+C$ and $C^{\prime}$ as above. Assume $7\mid B$ and
$7\mid\mid C$. Then $C^{\prime}(\mathbb{Q}_{7})\neq\emptyset$.

###### Proof.

We check with Magma [1] that after writing $B=7b$ and $C=7c$ we get non-
singular points over $\mathbb{F}_{7}$ for all the values of
$b,c\in\mathbb{F}_{7}$ unless $c=0$.

∎

For primes $p$ dividing the discriminant we prove the existence of
$\mathbb{Q}_{p}$-points under the following conditions.

######  Proposition 5.7.

Let $p\neq 2,3,7$ be a prime number. If $p\mid B$ and $p\mid\mid C$ then
$C^{\prime}(\mathbb{Q}_{p})\neq\emptyset$.

###### Proof.

One can check with Magma [1] that writing $B=pb$ and $C=pc$ we get that
$F\equiv y^{2}(-3969c^{2}xy-441b^{2}y^{2}+2646bcyz-3969c^{2}z^{2})\text{ mod
}p$ which contains non-singular point over $\mathbb{F}_{p}$ if $c\neq 0$. ∎

Let us address now the question about rational points.

######  Proposition 5.8.

In cases 4. and 8. $C^{\prime}(\mathbb{Q})=\emptyset$.

###### Proof.

From Theorem 4.2 we know that these twists do not come from elliptic curves,
i.e. they are not of the form $X_{E}(7)$. Then Theorem 4.1 tells us that those
twists have no non-cuspidal rational point. Finally, Proposition 4.3 gives us
the non existence of cuspidal rational points. ∎

######  Proposition 5.9.

In case 3., if $C^{\prime}\not\simeq_{\mathbb{Q}}\mathcal{C}$ and the field of
definition is not the field $6.0.214375.1$ (with the [6] notation) then
$C^{\prime}(\mathbb{Q})=\emptyset$.

###### Proof.

The same proof than in previous proposition applies except that we have to
remove the special case in which the elliptic curve has $j$-invariant
$3^{3}\cdot 5\cdot 7^{5}/2^{7}$, in this case the 7-torsion is defined over
the field $6.0.214375.1$ or a quadratic extension of it and we need to remove
the single case in which the conditions of Proposition 4.3 are not hold,
namely for the Klein quartic that it does have rational cuspidal points. ∎

As a consequence of the previous results, we conclude:

######  Theorem 5.10.

Let $f(x)=x^{3}+Bx+C\in\mathbb{Z}[x]$ be an irreducible polynomial such that
all of the following conditions hold

  1. (i)

$2\nmid C$,

  2. (ii)

$3\nmid B$,

  3. (iii)

$7\mid B$ and $7\mid\mid C$,

  4. (iv)

$p\mid B$ and $p\mid\mid C$ for any prime $p\neq 2,3,7$ dividing
$-4B^{3}-27C^{2}$.

  5. (v)

The splitting field $\mathbb{Q}_{f}\not\simeq\mathbb{Q}(\zeta_{7})$ nor the
field $6.0.214375.1$.

Then $f$ gives a twist of the Klein quartic (constructed as in Subsection 5.1)
that is a counterexample to the Hasse principle.

######  Remark 5.11.

Notice that the condition (iii) in Theorem 5.10 implies that $f$ is
irreducible because it is Eisenstein at $7$.

In order to get polynomials $f$ satisfying all the conditions above we need
$B=\pm 2^{a}7^{b}\prod p_{i}^{e_{i}}$ and $C=\pm 3^{c}7\prod p_{i}$ with
$e_{i},b\geq 1$, $a,c\geq 0$, the $p_{i}\neq 2,3,7$ prime and such that the
discriminant is only divisible by the primes $7$ and $p_{i}$. This last
condition yields an equality of the form: $3^{3+2c}\pm
1=2^{3a+2}7^{3b-2}p_{i}^{3e_{i}-2}$. In order to get the left hand size of
this equality to be $0$ modulo $4$ we need the first sign to be positive, and
in order to be $0$ modulo $7$ we need $3\mid c$. Actually, if $c=3c^{\prime}$
with $c^{\prime}\neq 3\text{ mod }7$ then $4\mid\mid 3^{3+6c^{\prime}}+1$ and
$7\mid\mid 3^{3+6c^{\prime}}+1$.

######  Conjecture 5.12.

There are infinitely many values of $c^{\prime}\neq 3\text{ mod }7$ such that
all the primes different from $2$ and $7$ in the factorization of
$3^{3+6c^{\prime}}+1$ appear with an exponent that is $1\text{ mod }3$.

Any such $c^{\prime}$ produces a factorization
$3^{3+6c^{\prime}}+1=2^{2}7p_{i}^{3e_{i}-2}$ giving $B=-7\prod p_{i}^{e_{i}}$
and $C=\pm 3^{3c^{\prime}}7\prod p_{i}$ satisfying conditions $(i)-(iv)$ in
Theorem 5.10.

######  Example 5.13.

After checking condition $(v)$ in Theorem 5.10, we find that the polynomials
$x^{3}-7x\pm 7$, $x^{3}-7\cdot 19\cdot 37x\pm 3^{3}\cdot 7\cdot 19\cdot 37$
and $x^{3}-7\cdot 31\cdot 61\cdot 271x\pm 3^{6}\cdot 7\cdot 31\cdot 61\cdot
271$ yield different twists of the Klein quartic that are counterexamples to
the Hasse principle.

Accordingly to the folklore conjecture that for curves, under the assumption
that $Sha(J)$ is finite, the Brauer-Manin obstruction is the only obstruction
to the Hasse principle, we should have that the obstruction to have rational
points of the counter-examples given in Example 5.13 is due to this
phenomenon. If we could find a degree 1 rational divisor in their jacobian, we
could indeed prove it under the finite $Sha$ assumption [13, p. 37].
Unfortunately, we were not able to find such a divisor.

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# On Rings of Baire one functions

A. Deb Ray  Department of Pure Mathematics, University of Calcutta, 35,
Ballygunge Circular Road, Kolkata - 700019, INDIA<EMAIL_ADDRESS>and
Atanu Mondal  Department of Pure Mathematics, University of Calcutta, 35,
Ballygunge Circular Road, Kolkata - 700019, INDIA<EMAIL_ADDRESS>
###### Abstract.

This paper introduces the ring of all real valued Baire one functions, denoted
by $B_{1}(X)$ and also the ring of all real valued bounded Baire one
functions, denoted by $B_{1}^{*}(X)$. Though the resemblance between $C(X)$
and $B_{1}(X)$ is the focal theme of this paper, it is observed that unlike
$C(X)$ and $C^{*}(X)$ (real valued bounded continuous functions),
$B_{1}^{*}(X)$ is a proper subclass of $B_{1}(X)$ in almost every non-trivial
situation. Introducing $B_{1}$-embedding and $B_{1}^{*}$-embedding, several
analogous results, especially, an analogue of Urysohn’s extension theorem is
established.

###### Key words and phrases:

$B_{1}(X)$, $B_{1}^{*}(X)$, zero set of a Baire one function, completely
separated by $B_{1}(X)$, $B_{1}$-embedded, $B_{1}^{*}$-embedded.

###### 2010 Mathematics Subject Classification:

26A21, 54C30, 54C45, 54C50

The second author is supported by Council of Scientific and Industrial
Research, HRDG, India. Sanction No.- 09/028(0998)/2017-EMR-1

##  1\. Introduction

The real valued Baire class one functions on a real variable is very important
and intensively studied in Real analysis. Several characterizations of Baire
one functions defined on metric spaces were obtained by different
mathematicians in [1], [8] etc. Lee, Tang and Zhao [5] characterized Baire
class one functions in terms of the usual $\epsilon$-$\delta$ formulation as
in the case of continuous functions under the assumptions that X and Y are
complete separable metric spaces. In [7], Baire one functions on normal
topological spaces have been investigated and characterized by observing that
pull-backs of open sets are $F_{\sigma}$.

The study of rings of continuous functions was initiated long back and the
theory was enriched by publication of several outstanding results, cited in
[2], [3], [4], [6], etc., established by various well known mathematicians
around the first half of the twentieth century. The class of all real valued
Baire one functions defined on any topological space is of course a superset
of the class of all real valued continuous functions on the same space and
therefore, it is a natural query whether one can extend the study on the class
of Baire one functions. This paper puts forward some basic results which came
out of such investigation.

In Section 2, we introduce two rings, the ring $B_{1}(X)$ of all the real
valued Baire one functions and the ring $B_{1}^{*}(X)$ of all the real valued
bounded Baire one functions. We observe that $B_{1}(X)$ is a lattice ordered
commutative ring with unity and the other one, i.e., $B_{1}^{*}(X)$ is a
commutative subring with unity (and a sublattice) of $B_{1}(X)$. The rest of
this section is devoted to build the basics that are required to carry on
research in this field.

Though the focal theme of this work is to observe the similarities of
$B_{1}(X)$ with $C(X)$, we find in section 3 that the scenario is quite
different. It is well known that $C(X)=C^{*}(X)$ defines a pseudocompact space
$X$ and there are plenty of examples, including compact spaces, which are
pseudocompact. But in case of $B_{1}(X)$ and $B_{1}^{*}(X)$, the equality
occurs very rarely.

Section 4 introduces zero sets of $B_{1}(X)$ and discusses several algebraic
equalities involving unions and intersections of zero sets of Baire one
functions. Also, the relationships between zero set of modulus (as well as,
the power) of a Baire one function $f$ with the zero set of $f$ are
established. The sets which are completely separated by $B_{1}(X)$ is
characterized via zero sets of functions from $B_{1}(X)$. However, the well
known result that two sets $A$ and $B$ are completely separated by $C(X)$ if
and only if $\overline{A}$ and $\overline{B}$ are completely separated by
$C(X)$ becomes one sided in the context of $B_{1}(X)$. That it is indeed one-
sided is supported by an example.

The final section of this paper is devoted for developing the idea of
$B_{1}$-embedding and $B_{1}^{*}$-embedding of a set and thereby obtaining an
analogue of Urysohn’s Extension theorem in this new context.

##  2\. $B_{1}(X)$ and $B_{1}^{*}(X)$

Let $X$ be any arbitrary topological space and $C(X)$ be the collection of all
real valued continuous functions from $X$ to $\mathbb{R}$. We define
$B_{1}(X)$ as the collection of all pointwise limit functions of sequnces in
$C(X)$. So, $B_{1}(X)=\\{f:X\rightarrow\mathbb{R}$ $:$ $\exists$
$\\{f_{n}\\}\subseteq C(X)$, for which $\\{f_{n}(x)\\}$ pointwise converges to
$f(x)$ for all $x\in$ $X$ $\\}$, is called the set of Baire class one
functions or Baire one functions. It is clear that, $C(X)\subseteq B_{1}(X)$.

Let $f$ and $g$ be two functions in $B_{1}(X)$. There exist two sequences of
continuous functions $\\{f_{n}\\}$ and $\\{g_{n}\\}$ such that, $\\{f_{n}\\}$
converges pointwise to $f$ and $\\{g_{n}\\}$ converges pointwise to $g$ on
$X$. Then

  * •

$\\{f_{n}+g_{n}\\}$ pointwise converges to $f+g$.

  * •

$\\{-f_{n}\\}$ pointwise converges to $-f$.

  * •

$\\{f_{n}.g_{n}\\}$ converges pointwise to $f.g$.

  * •

$\\{|f_{n}|\\}$ converges pointwise to $|f|$.

In view of the above observations, it is easy to see that $(B_{1}(X),+,.)$ is
a commutative ring with unity 1 (where $\textbf{1}:X\rightarrow\mathbb{R}$ is
given by $\textbf{1}(x)=1$, $\forall x\in X$) with respect to usual pointwise
addition and multiplication.
In [7], Baire one functions are described in terms of pull-backs of open sets
imposing conditions on domain and co-domain of functions.

######  Theorem 2.1.

[7] (i) For any topological space $X$ and any metric space $Y$, $B_{1}(X,Y)$
$\subseteq\mathscr{F_{\sigma}}(X,Y)$, where $B_{1}(X,Y)$ denotes the
collection of Baire one functions from $X$ to $Y$ and
$\mathscr{F_{\sigma}}(X,Y)=\\{f:X\rightarrow Y:f^{-1}(G)$ is an $F_{\sigma}$
set, for any open set $G\subseteq Y$}.
(ii) For a normal topological space $X$, $B_{1}(X,\mathbb{R})$
$=\mathscr{F_{\sigma}}(X,\mathbb{R})$, where $B_{1}(X,\mathbb{R})$ denotes the
collection of Baire one functions from $X$ to $\mathbb{R}$ and
$\mathscr{F_{\sigma}}(X,\mathbb{R})=\\{f:X\rightarrow\mathbb{R}:f^{-1}(G)$ is
an $F_{\sigma}$ set, for any open set $G\subseteq\mathbb{R}$}.

Define a partial order ‘$\leq$’ on $B_{1}(X)$ by $f\leq g$ iff $f(x)\leq
g(x)$, $\forall x\in X$. It is clear that $(B_{1}(X),\leq)$ is a lattice,
where $sup\\{f,g\\}=f\vee g=\frac{1}{2}\\{(f+g)+|f-g|\\}$ and
$inf\\{f,g\\}=f\wedge g=\frac{1}{2}\\{(f+g)-|f-g|\\}$ both are in $B_{1}(X)$.
Also for any $f,g,h\in B_{1}(X)$

  * •

$f\leq g\implies f+h\leq g+h$.

  * •

$f\geq 0$ and $g\geq 0\implies f.g\geq 0.$

So, $B_{1}(X)$ is a commutative lattice ordered ring with unity. Moreover
$C(X)$ is a commutative subring with unity and also a sublattice of
$B_{1}(X)$.
If a Baire one function $f$ on $X$ is a unit in the ring $B_{1}(X)$ then
$\\{x\in X\ :\ f(x)=0\\}=\emptyset$. The following result shows that in case
of a normal space, this condition is also sufficient.

######  Theorem 2.2.

For a normal space $X$, $f\in B_{1}(X)$ is a unit in $B_{1}(X)$ if and only if
$Z(f)=\\{x\in X\ :\ f(x)=0\\}=\emptyset$.

###### Proof.

If $f\in B_{1}(X)$ is a unit then clearly the condition holds. Let $f\in
B_{1}(X)$ be such that $Z(f)=\emptyset$. Define
$\frac{1}{f}(x)=\frac{1}{f(x)}$, for all $x\in X$. To show that
$\frac{1}{f}\in B_{1}(X)$. Let $U=(a,b)$ be any open interval in $\mathbb{R}$.
It is enough to show that $\left(\frac{1}{f}\right)^{-1}(U)$ is $F_{\sigma}$.
$\left(\frac{1}{f}\right)^{-1}(U)=\\{x\in X\ :\ a<\frac{1}{f(x)}<b\\}$.
Case 1 : Suppose $0\notin U$. Then $\\{x\in X\ :\
a<\frac{1}{f(x)}<b\\}=\\{x\in X\ :\ \frac{1}{b}<f(x)<\frac{1}{a}\\}$
$=f^{-1}\left(\frac{1}{b},\frac{1}{a}\right)$, when $a\neq 0,b\neq 0$ and
$\left(\frac{1}{f}\right)^{-1}(U)=f^{-1}\left(\frac{1}{b},\infty\right)$ or
$f^{-1}\left(-\infty,\frac{1}{a}\right)$ according as $a=0$ or $b=0$ . In any
case the resultant set is an $F_{\sigma}$ set.
Case 2 : Let $0\in U$. Since $Z(f)=\emptyset$, $f(x)\neq 0$ and hence
$\frac{1}{f(x)}\neq 0$,
$\left(\frac{1}{f}\right)^{-1}(U)=\left(\frac{1}{f}\right)^{-1}\left((a,0)\right)\cup\left(\frac{1}{f}\right)^{-1}\left((0,b)\right)$.
Then $\left(\frac{1}{f}\right)^{-1}(U)=\\{x\in X\ :\
a<\frac{1}{f(x)}<0\\}\bigcup\\{x\in X\ :\ 0<\frac{1}{f(x)}<b\\}=$ $\\{x\in X\
:\ -\infty<f(x)<\frac{1}{a}\\}\bigcup\\{x\in X\ :\ \frac{1}{b}<f(x)<\infty\\}$
$=f^{-1}\left(-\infty,\frac{1}{a}\right)\cup
f^{-1}\left(\frac{1}{b},\infty\right)$, which is an $F_{\sigma}$ set.
Hence, $\frac{1}{f}\in B_{1}(X)$. ∎

However, the following theorem provides an useful sufficient criterion to
determine units of $B_{1}(X)$, where $X$ is any topological space.

######  Theorem 2.3.

Let $X$ be a topological space and $f\in B_{1}(X)$ be such that $f(x)>0$ _(_
or $f(x)<0$_)_ , $\forall x\in X$, then $\frac{1}{f}$ exists and belongs to
$B_{1}(X)$.

###### Proof.

Without loss of generality let, $f\in B_{1}(X)$ and $f(x)>0,\forall x\in X$.
Then there exists a sequence of continuous functions $\\{f_{n}\\}$ such that
$\\{f_{n}\\}$ converges pointwise to $f$. Now we construct a sequence of
continuous functions $\\{g_{n}\\}$, where $g_{n}(x)=|f_{n}(x)|+\frac{1}{n}$,
$\forall n\in\mathbb{N}$ and $\forall x\in X$. Clearly $g_{n}(x)>0$, $\forall
n\in\mathbb{N}$ and $\forall x\in X$.
We now show that, $\\{g_{n}(x)\\}$ converges to $f(x)$, for each $x\in X$.
For each $x\in X$, $\\{f_{n}(x)\\}$ converges to $f(x)$ and
$\\{\frac{1}{n}\\}$ converges to $0$ imply that $\\{g_{n}(x)\\}$ converges to
$|f(x)|=f(x)$, $\forall x\in X$.
Consider the function $g:\mathbb{R}-\\{0\\}\rightarrow\mathbb{R}$ defined by
$g(y)=\frac{1}{y}$, $g$ being continuous $\\{g\circ g_{n}\\}$ is a sequence of
continuous functions from $X$ to $\mathbb{R}$. Our claim is $\\{g\circ
g_{n}\\}$ converges to $g\circ f$ on $X$.
Let, $\epsilon>0$. Then there exists a $\delta>0$ such that $|(g\circ
g_{n})(x)-(g\circ f)(x)|$ $=|g(g_{n}(x))-g(f(x))|<\epsilon$, whenever
$|g_{n}(x)-f(x)|<\delta$ (By using continuity of $g$).
Since $\\{g_{n}$} converges pointwise to $f$, for $\delta>0$, $\exists$
$K\in\mathbb{N}$ such that, $|g_{n}(x)-f(x)|<\delta$, whenever $n\geq K$. So,
$|(g\circ g_{n})(x)-(g\circ f)(x)|<\epsilon$ whenever $n\geq K$.
Hence, $\\{(g\circ g_{n})\\}$ converges pointwise to $g\circ f$, i.e. $g\circ
f\in B_{1}(X)$. Now $(g\circ f)(x)=\frac{1}{f(x)}$ shows that $\frac{1}{f}$
belongs to $B_{1}(X)$. ∎

In the last theorem, we have shown that composition of a typical continuous
function $g:\mathbb{R}-\\{0\\}\rightarrow\mathbb{R}$ given by
$g(x)=\frac{1}{x}$ and a typical Baire one function $f:X\rightarrow\mathbb{R}$
produces a Baire one function $g\circ f$. In the next theorem we further
generalize this.

######  Theorem 2.4.

Let $f$ be any Baire one function on $X$ and
$g:\mathbb{R\rightarrow\mathbb{R}}$ be a continuous function then the
composition function $g\circ f$ is also a Baire one function.

###### Proof.

Since $f\in B_{1}(X)$, there exists a sequence of continuous functions
$\\{f_{n}\\}$ which converges pointwise to $f$. The functions $g\circ f_{n}$
are all defined and continuous functions on $X$, $\forall n\in\mathbb{N}$. Let
$\epsilon>0$ be any arbitrary positive real number and $x\in X$. By continuity
of $g$ there exists a positive $\delta$ depending on $\epsilon$ such that,
$|(g\circ f_{n})(x)-(g\circ f)(x)|=|g(f_{n}(x))-g(f(x))|<\epsilon$, whenever
$|f_{n}(x)-f(x)|<\delta$. Again by using pointwise convergence of
$\\{f_{n}\\}$ we can find a natural number $K$ such that,
$|f_{n}(x)-f(x)|<\delta,\forall n\geq K$. So, $|(g\circ f_{n})(x)-(g\circ
f)(x)|<\epsilon,\forall$ $n\geq K$.
Hence, $g\circ f$ is a Baire one function on $X$. ∎

We introduce another subcollection of $B_{1}(X)$, called bounded Baire one
functions, denoted by $B_{1}^{*}(X)$ consisting of all real valued bounded
Baire one functions on $X$. i.e., $B_{1}^{*}(X)=\\{f\in B_{1}(X):f$ is bounded
on $X\\}$. $B_{1}^{*}(X)$ also forms a commutative lattice ordered ring with
unity $1$, which is a subring and sublattice of $B_{1}(X).$

##  3\. Is $B_{1}^{*}(X)$ always a proper subring of $B_{1}(X)$?

In case of rings of continuous functions we have seen that there are spaces
for which $C^{*}(X)$ coincides with $C(X)$, where $C(X)$ and $C^{*}(X)$ denote
respectively the collection of all real valued continuous functions and the
collection of all real valued bounded continuous functions on $X$. For
example, if $X$ is compact then $C(X)=C^{*}(X)$. In fact, the spaces for which
$C(X)=C^{*}(X)$ are known as pseudocompact spaces. But for Baire one
functions, the scenario is quite different. We show in the next theorem that
for most of the spaces unbounded Baire one functions do exist.

######  Theorem 3.1.

Let $X$ be any topological space. If $f\in C(X)$ is such that $0\leq f(x)\leq
1$, $\forall x\in X$ and $0$ is a limit point of the range set $f(X)$, then
there exists an unbounded Baire one function on $X$. (i.e., $B_{1}^{*}(X)$ is
a proper subset of $B_{1}(X)$)

###### Proof.

For each $n\in\mathbb{N}$, define $g_{n}:X\rightarrow\mathbb{R}$ by

$g_{n}(x)=\begin{cases}n^{2}f(x)&if\ x\in f^{-1}([0,\frac{1}{n}])\\\
\frac{1}{f(x)}&\ if\ x\in f^{-1}([\frac{1}{n},1])\end{cases}$

Each $g_{n}$ is continuous and it is clear that $\\{g_{n}(x)\\}$ converges
pointwise to the function $g:X\rightarrow\mathbb{R}$ defined by

$g(x)=\begin{cases}0&\ if\ x\in Z(f)\\\ \frac{1}{f(x)}&\ if\ x\notin
Z(f)\end{cases}$

So, $g$ is a Baire one function on $X$. Since $0$ is a limit point of $f(X)$,
$g$ is unbounded Baire one function. ∎

######  Remark 3.2.

If $B_{1}(X)=B_{1}^{*}(X)$, then for every $f\in C(X)$, $0$ cannot be a limit
point of $f(X)$. In fact, we can say more, if $B_{1}(X)=B_{1}^{*}(X)$, then
for every $f\in C(X)$, $r$ is not a limit point of $f(X)$, where $r$ is any
real number. This follows from the fact that if $r$ is a limit point of the
range set of the continuous function $f$, then $0$ becomes a limit point of
the set $g(X)$, where $g=f-r$ and we can apply the previous theorem to the
function $(0\vee|g|)\wedge 1$ to get an unbounded function.

######  Remark 3.3.

One may observe that whenever $B_{1}(X)=B_{1}^{*}(X)$ and $X$ possesses
atleast one non-constant continuous function then $X$ cannot be connected,
because in that case $f(X)$ must be an interval and hence possesses a limit
point. Also, it follows easily from $B_{1}(X)=B_{1}^{*}(X)$ that $X$ is
pseudocompact. Therefore, it is natural to ask, Which class of spaces is
precisely determined by $B_{1}(X)=B_{1}^{*}(X)$?

As a consequence of Remark 3.2 we obtain,

######  Theorem 3.4.

A completely Hausdorff space $X$ _(_ i.e., where any two distinct points are
completely separated by continuous function _)_ is totally disconnected if
$B_{1}(X)=B_{1}^{*}(X)$.

###### Proof.

Since $X$ is completely Hausdorff, it possesses a non-constant continuous
function. So, by previous remark $B_{1}(X)=B_{1}^{*}(X)$ implies $X$ is
disconnected. Now assume that $C$ be any component of $X$ and $x,y$ be two
distinct points in $C$. Since $X$ is completely Hausdorff, there exists a
continuous function $f$ on $X$ such that $f(x)=0$ and $f(y)=1.$ Then the
function $g=f\big{|}_{C}$ is a continuous function on $C$ and $C$ being
connected $g(C)$ must be an interval. So, $f(X)$ has a limit point as $g$ is
the restriction function of $f$ on $C$. This contradicts to the fact that
$B_{1}(X)=B_{1}^{*}(X)$. So, $C$ cannot contain more than one point. Since $C$
is arbitrary component of $X$, every component of $X$ contains single point.
Hence, $X$ is totally disconnected. ∎

######  Remark 3.5.

Converse of the theorem is not true. $\mathbb{Q}$, the set of all rational
numbers is both completely Hausdorff and totally disconnected but $\mathbb{Q}$
possesses an unbounded real valued continuous function, hence an unbounded
Baire one function.

In the context of ring homomorphism we may further get results similar to the
known results about homomorphism from $C(Y)$ (or $C^{*}(Y)$) to $C(X)$.

######  Theorem 3.6.

Every ring homomorphism $t$ from $B_{1}(Y)$ (or $B_{1}^{*}(Y)$) to $B_{1}(X)$
is a lattice homomorphism.

######  Theorem 3.7.

Every ring homomorphism $t$ from $B_{1}(Y)$ (or $B_{1}^{*}(Y)$) to $B_{1}(X)$
takes bounded functions to bounded functions.

######  Corollary 3.8.

If a completely Hausdorff space $X$ is not totally disconnected, then
$B_{1}(X)$ cannot be a homomorphic image of $B_{1}^{*}(Y)$, for any $Y$.

######  Corollary 3.9.

$B_{1}(X)$ and $B_{1}^{*}(X)$ are isomorphic if and only if they are
identical.

######  Theorem 3.10.

Let, $t$ be a homomorphism from $B_{1}(Y)$ into $B_{1}(X)$, such that
$B_{1}^{*}(X)\subseteq t(B_{1}(Y))$. Then $t$ maps $B_{1}^{*}(Y)$ onto
$B_{1}^{*}(X)$.

##  4\. Zero sets in $B_{1}(X)$

The zero set $Z(f)$ of a function $f\in B_{1}(X)$ is defined by $Z(f)=\\{x\in
X:f(x)=0\\}$ and the collection of all zero sets in $B_{1}(X)$ is denoted by
$Z(B_{1}(X))$. We say a subset $E$ of $X$ is a zero set in $B_{1}(X)$ if
$E=Z(f)$, for some $f\in B_{1}(X)$.
We call a set to be a cozero set in $B_{1}(X)$ if it is the complement of a
zero set in $B_{1}(X)$. $Z(B_{1}(X))$ is closed under finite union and finite
intersection as $Z(f)\cup Z(g)=Z(f.g)$ and $Z(f)\cap
Z(g)=Z(f^{2}+g^{2})=Z(|f|+|g|)$.
It is evident that, $Z(f)=Z(|f|)=Z(f^{n})$, for all $f\in B_{1}(X)$ and for
all $n\in$ $\mathbb{N}$, $Z(\textbf{0})=X$ and $Z(\textbf{1})=\emptyset$. Here
0 and 1 denote the constant functions whose values are $0$ and $1$ on $X$.

Examples of zero and cozero sets in $B_{1}(X):$

  * •

Every zero set of a continuous function is also a zero set of a Baire one
function. Every clopen set $K$ of $X$ is in $Z(B_{1}(X))$, as it is a zero set
of a continuous function.

  * •

For any $f\in B_{1}(X),\\{x\in:f(x)\geq 0\\}$ and $\\{x\in:f(x)\leq 0\\}$
belong to $Z(B_{1}(X))$ and they are the zero sets of the functions $f-|f|$
and $f+|f|$ respectively.

  * •

For any real number $r\in\mathbb{R}$, $\\{x\in X:f(x)\leq r\\}$ and $\\{x\in
X:f(x)\geq r\\}$ are also in $Z(B_{1}(X))$.

  * •

For any real number $r\in\mathbb{R}$, $\\{x\in X:f(x)<r\\}$ and $\\{x\in
X:f(x)>r\\}$ are cozero sets in $B_{1}(X)$.

It is easy to observe that, for any $f\in B_{1}(X)$ there exists $g\in
B_{1}^{*}(X)$ given by $g=f\wedge\textbf{1}$ such that $Z(f)=Z(g)$. Hence
$Z(B_{1}(X))$ and $Z(B_{1}^{*}(X))$ produce same family of zero sets.

######  Theorem 4.1.

For any $f\in B_{1}(X),Z(f)$ is a $G_{\delta}$ set.

###### Proof.

Since $\mathbb{R}$ is a metric space, by Theorem 2.1, $f^{-1}(G)$ is an
$F_{\sigma}$ set, for every open set $G$ of $\mathbb{R}$. Therefore in
particular $f^{-1}(\mathbb{R}-\\{0\\})$ is an $F_{\sigma}$ set. i.e.
$Z(f)=f^{-1}(\\{0\\})$ is a $G_{\delta}$ set. ∎

######  Corollary 4.2.

Every cozero set in $Z(B_{1}(X))$ is $F_{\sigma}$.

Observation: Countable union of zero sets in $B_{1}(X)$ need not be a zero set
in $B_{1}(X)$. For example, $\mathbb{Q}$ can be written as a countable union
of singleton sets and each singleton set is a zero set in $B_{1}(\mathbb{R})$,
but $\mathbb{Q}$ is not a zero set in $B_{1}(\mathbb{R})$, as it is not a
$G_{\delta}$ set in $\mathbb{R}$. However it can be proved that, $Z(B_{1}(X))$
is closed under countable intersection. To establish this result we need two
important lemmas, which are already proved for the functions of a real
variable. We generalize these results here for any arbitrary topological
spaces.

######  Lemma 4.3.

If $f:X\rightarrow\mathbb{R}$ is a bounded Baire one function with bound $M$,
where $X$ is any topological space, then there exists a sequence of continuous
function $\\{f_{n}\\}$ such that, each $f_{n}$ has the same bound $M$ and
$\\{f_{n}\\}$ converges pointwise to $f$ on $X$.

###### Proof.

Let $\\{g_{n}\\}$ be a sequence of continuous functions that converges
pointwise to $f$ on $X$. Suppose also that $|f(x)|\leq M$, $\forall x\in X$.
Define $\\{f_{n}\\}$ by
$f_{n}(x)=$ $\begin{cases}-M&if$ $g_{n}(x)\leq-M\\\ g_{n}(x)&if$ $-M\leq
g_{n}(x)\leq M\\\ M&if$ $g_{n}(x)\geq M.\end{cases}$
Each $f_{n}$ of the sequence of functions $\\{f_{n}(x)\\}$ is continuous and
has bound $M$. Also $\\{f_{n}(x)\\}$ converges pointwise to $f$ on $X$ with
bound $M$. ∎

######  Lemma 4.4.

Let $\\{f_{k}\\}$ be a sequence of Baire one functions defined on a
topological space $X$ and let $\sum\limits_{k=1}^{\infty}M_{k}<\infty$, where
each $M_{k}>0$. If $|f_{k}(x)|\leq M_{k}$, $\forall k\in\mathbb{N}$ and
$\forall x\in X$, then the function $f(x)=\sum\limits_{k=1}^{\infty}f_{k}(x)$
is also a Baire one function on $X$.

###### Proof.

Since each $f_{k}$ is a Baire one function, for each positive integer $k$
there exists a sequence of continuous functions $\\{g_{ki}\\}_{i=1}^{\infty}$
on $X$ such that $\\{g_{ki}\\}_{i=1}^{\infty}$ converges pointwise to the
function $f_{k}$ on $X$ and by lemma 4.3 we can choose
$\\{g_{ki}\\}_{i=1}^{\infty}$ in such a way that, $|g_{ki}|\leq M_{k},\forall
i\in\mathbb{N}$.
For each $n\in\mathbb{N}$, let $h_{n}=g_{1n}+g_{2n}+\ldots+g_{nn}$.
It is easy to verify that each $h_{n}$ is continuous on $X$. We will show that
$\\{h_{n}\\}$ converges pointwise to $f$ on $X$.
Fix a point $x\in X$ and let $\epsilon>0$ be an arbitrary positive real
number. Since $\sum\limits_{k=1}^{\infty}M_{k}<\infty$, we can find
$K\in\mathbb{N}$ so that $\sum\limits_{k=K+1}^{\infty}M_{k}<\epsilon$.
Now choose an integer $N>K$ such that
$|g_{ki}(x)-f_{k}(x)|<\frac{\epsilon}{K}$ for $1\leq k\leq K$ and $\forall$
$i\geq N$.
For any $n\geq N$ we have,
$|h_{n}(x)-f(x)|=|\sum\limits_{k=1}^{n}g_{kn}(x)-\sum\limits_{k=1}^{\infty}f_{k}(x)|\leq|\sum\limits_{k=1}^{n}(g_{kn}(x)-f_{k}(x))|+|\sum\limits_{k=n+1}^{\infty}f_{k}(x)|$
$\leq|\sum\limits_{k=1}^{K}(g_{kn}(x)-f_{k}(x))|+\sum\limits_{k=K+1}^{n}|g_{kn}(x)|+\sum\limits_{k=K+1}^{\infty}|f_{k}(x)|<\sum\limits_{k=1}^{K}\frac{\epsilon}{K}+2\sum\limits_{k=K+1}^{\infty}M_{k}<3\epsilon$.
Since $\epsilon$ is arbitrary, it follows that $\\{h_{n}(x)\\}$ converges
pointwise to $f$ on $X$. Hence $f(x)=\sum\limits_{k=1}^{\infty}f_{k}(x)$ is a
Baire one function on $X$. ∎

######  Theorem 4.5.

$Z((B_{1}(X)))$ is closed under countable intersection.

###### Proof.

Let $Z(f_{n})\in Z(B_{1}(X)),\forall n\in\mathbb{N}$.
We define $g_{n}(x)=|f_{n}(x)|\wedge\frac{1}{2^{n}}$, $\forall x\in X$ and
$\forall n\in\mathbb{N}$ and let $g(x)=\sum\limits_{n=1}^{\infty}g_{n}(x)$.
Here, $|g_{n}(x)|\leq\frac{1}{2^{n}},\forall n\in\mathbb{N}$ and $\forall x\in
X$. Also $\sum\limits_{n=1}^{\infty}\frac{1}{2^{n}}<\infty$. So, by lemma 4.4
$g\in B_{1}(X)$ and
$Z(g)=\bigcap\limits_{n=1}^{\infty}Z(g_{n})=\bigcap\limits_{n=1}^{\infty}Z(f_{n})$.
∎

######  Definition 4.6.

Two subsets $A$ and $B$ are said to be completely separated in $X$ by
$B_{1}(X)$, if there exists a function $f\in B_{1}^{*}(X)$ such that $f(A)=r$
and $f(B)=s$ with $r<s$ and $r\leq f\leq s$, $\forall x\in X$.

It is enough to say that $A$ and $B$ are completely separated by $B_{1}(X)$,
if we get a function $g$ in $B_{1}(X)$ satisfying $g(x)\leq r,\forall x\in A$
and $g(x)\geq s,\forall x\in B$, for then, $(r\vee g)\wedge s$ has the
required property. Moreover we can replace $r$ and $s$ by $0$ and $1$, as
there always exists a continuous bijection $h$ from $[r,s]$ to $[0,1]$ and
$h\circ g$ is a Baire one function with the desired property.
It is well known that two sets $A$ and $B$ are completely separated by $C(X)$
if and only if $\overline{A}$ and $\overline{B}$ are completely separated by
$C(X)$. But in case of completely separated by $B_{1}(X)$, this result is one-
sided. Certainly, $\overline{A}$ and $\overline{B}$ are completely separated
by $B_{1}(X)$ implies that $A$ and $B$ are completely separated by $B_{1}(X)$.
But the converse is not true, as seen in the following example.

######  Example 4.7.

In $[0,1]$, the sets $A=[0,1)$ and $B=\\{1\\}$ are completely separated by
$B_{1}([0,1])$, because $\\{f_{n}\\}\subseteq C[0,1]$ defined by
$f_{n}(x)=x^{n}$, $\forall x\in[0,1]$ and $\forall n\in\mathbb{N}$ converges
to the function $f(x)$ defined by

$f(x)=$ $\begin{cases}0&if$ $0\leq x<1\\\ 1&if$ $x=1.\\\ \end{cases}$
So, $f$ belongs to $B_{1}^{*}[0,1]$ and $f(A)=0,f(B)=1$. But $\overline{A}$,
$\overline{B}$ are not disjoint and therefore are not completely separated by
$B_{1}([0,1])$.

######  Theorem 4.8.

Two subsets of $X$ are completely separated by $B_{1}(X)$ if and only if they
are contained in disjoint zero sets in $Z(B_{1}(X))$.

###### Proof.

Let $Z(f)$ and $Z(g)$ are two members of $Z(B_{1}(X))$ such that $A\subseteq
Z(f)$ and $B\subseteq Z(g)$ with $Z(f)\cap Z(g)=\emptyset$. Clearly, the zero
set of $|f|+|g|$ is empty and we may define
$h(x)=\frac{|f(x)|}{|f(x)|+|g(x)|}$. Here $(|f(x)|+|g(x)|)>0$ on $X$ and
$|f|+|g|\in B_{1}(X)$, so by Theorem 2.3 $\frac{1}{|f(x)|+|g(x)|}\in B_{1}(X)$
which implies $h(x)=\frac{|f(x)|}{|f(x)|+|g(x)|}\in B_{1}(X)$. Now,
$h(Z(f))=0$ and $h(Z(g))=1$. So, $Z(f)$ and $Z(g)$ are completely separated by
$B_{1}(X)$. Therefore $A$ and $B$ are completely separated by $B_{1}(X)$.
Conversely, let $A$ and $A^{\prime}$ be completely separated by $B_{1}(X)$.
So, there exists $f\in B_{1}(X)$ such that $f(A)={0}$ and $f(A^{\prime})={1}$.
The disjoint sets $F=\\{x:f(x)\leq\frac{1}{3}\\}$ and
$F^{\prime}=\\{x:f(x)\geq\frac{2}{3}\\}$ belong to $Z(B_{1}(X))$ and
$A\subseteq F,A^{\prime}\subseteq F^{\prime}$. ∎

##  5\. $B_{1}$-embedded and $B_{1}^{*}$-embedded

In this section we introduce the analogues of $C$-embedding and
$C^{*}$-embedding, called $B_{1}$-embedding and $B_{1}^{*}$-embedding in
connection with the extensions of Baire one functions.

######  Definition 5.1.

A subset $Y$ of a topological space $X$ is called $B_{1}$-embedded in $X$, if
each $f\in B_{1}(Y)$ has an entension to a $g\in B_{1}(X)$. i.e. $\exists$
$g\in B_{1}(X)$ such that $g\big{|}_{Y}=f$.
Similarly, $Y$ is called $B_{1}^{*}$-embedded in $X$, if each $f\in
B_{1}^{*}(Y)$ has an entension to a $g\in B_{1}^{*}(X)$.

######  Theorem 5.2.

Any $B_{1}$-embedded subset is $B_{1}^{*}$-embedded.

###### Proof.

Let $Y$ be any $B_{1}$-embedded subset of $X$ and $f\in B_{1}^{*}(Y)$. There
is a natural number $n$ such that $|f(y)|\leq n$, $\forall y\in Y$. Since $Y$
is $B_{1}$-embedded, it has an extension to $g\in B_{1}(X)$. Taking $h=(-n\vee
g)\wedge n$. We get $h\in B_{1}^{*}(X)$ and $g(y)=h(y),\forall$ $y\in Y$. So,
$h$ is an extension of $f$ and $h$ is bounded. Hence $Y$ is
$B_{1}^{*}$-embedded in $X$. ∎

We establish an analogue of Urysohn’s extension theorem for continuous
functions which gives a necessary and sufficient condition for a subspace to
be $B_{1}^{*}$-embedded in a topological space $X$. Before proving this
theorem we need a lemma that ensures that the uniform limit of a sequence of
Baire one functions is a Baire one function. It is to be noted that a special
case of this particular result when functions are on real line has already
been established. We prove the result in a more general setting.

######  Lemma 5.3.

Let $\\{f_{n}\\}$ be a sequence of Baire one functions on $X$. If
$\\{f_{n}\\}$ converges uniformly to $f$ on $X$ then $f$ is a Baire one
function on $X$.

###### Proof.

Let $\\{f_{n}\\}$ be a sequence of Baire one functions converging uniformly to
$f$ on $X$. By definition of uniform convergence, for each $k\in\mathbb{N}$,
there exists a subsequence $\\{f_{n_{k}}\\}$ such that
$|f_{n_{k}}(x)-f(x)|<\frac{1}{2^{k}},\forall x\in X$.
Consider the sequence $\\{f_{n_{k+1}}-f_{n_{k}}\\}$. Then
$|f_{n_{k+1}}(x)-f_{n_{k}}(x)|\leq|f_{n_{k+1}}(x)-f(x)|+|f_{n_{k}}(x)-f(x)|\leq\frac{1}{2^{k+1}}+\frac{1}{2^{k}}=(\frac{3}{2})2^{-k}$.
Let $M_{k}=(\frac{3}{2})2^{-k}$ and note that,
$|f_{n_{k+1}}(x)-f_{n_{k}}(x)|\leq M_{k},\forall x\in X$ and
$\sum\limits_{k=1}^{\infty}M_{k}<\infty$, where $M_{k}>0,\forall
k\in\mathbb{N}$. So, $\sum_{k=1}^{\infty}[f_{n_{k+1}}(x)-f_{n_{k}}(x)]$ is a
convergent series and so by lemma 4.4 the sum function
$\sum_{k=1}^{\infty}[f_{n_{k+1}}(x)-f_{n_{k}}(x)]$ is a Baire one function on
$X$. Now,
$\sum_{k=1}^{\infty}[f_{n_{k+1}}(x)-f_{n_{k}}(x)]=\lim\limits_{N\to\infty}\sum_{k=1}^{N}[f_{n_{k+1}}(x)-f_{n_{k}}(x)]=f(x)-f_{n_{1}}(x)$.
Since, $f_{n_{1}}$ is a Baire one function, $f$ is also a Baire one function.
∎

######  Theorem 5.4.

A subset $Y$ of a topological space $X$ is $B_{1}^{*}$-embedded in $X$ iff any
pair of subsets of $Y$ which are completely separated in $Y$ by $B_{1}(Y)$ are
also completely separated in $X$ by $B_{1}(X)$.

###### Proof.

Let $Y$ be a $B_{1}^{*}$-embedded subspace of $X$ and $P$, $Q$ be subsets of
$Y$ completely separated in $Y$ by $B_{1}(Y)$. So, there exists $f\in
B_{1}^{*}(Y)$ such that, $f(P)=0$ and $f(Q)=1$. Since $Y$ is
$B_{1}^{*}$-embedded in $X$, $f$ has an extension to $g\in B_{1}^{*}(X)$ such
that $g(P)=0$ and $g(Q)=1$. Hence, $P$, $Q$ are completely separated in $X$ by
$B_{1}(X)$.
Conversely, assume that any two completely separated sets in $Y$ by $B_{1}(Y)$
are also completely separated in $X$ by $B_{1}(X$). Let $f_{1}\in
B_{1}^{*}(Y)$. Then there exists a natural number $m$ such that,
$|f_{1}(y)|\leq m$, $\forall y\in Y$.
For $n\in\mathbb{N}$, let $r_{n}=(\frac{m}{2})({\frac{2}{3}})^{n}$. Then
$3r_{n+1}=2r_{n},\forall n$ and $|f_{1}|\leq 3r_{1}$.
Let $A_{1}=\\{y\in Y:f_{1}(y)\leq-r_{1}\\}$ and $B_{1}=\\{y\in Y:f_{1}(y)\geq
r_{1}\\}$. Clearly, $A_{1}$ and $B_{1}$ are disjoint members in $Z(B_{1}(Y))$
and therefore, are completely separated in $Y$ by $B_{1}(Y)$. By our
hypothesis $A_{1}$, $B_{1}$ are completely separated in $X$ by $B_{1}(X)$. So,
there exists $g_{1}\in B_{1}^{*}(X)$ such that, $g_{1}(A_{1})=-r_{1}$ and
$g_{1}(B_{1})=r_{1}$ also $-r_{1}\leq g_{1}\leq r_{1}$.
Let $f_{2}=f_{1}-g_{1}\big{|}_{Y}$, then $f_{2}\in B_{1}^{*}(Y)$ and
$|f_{2}|\leq 2r_{1}$ on $Y$.
So, $|f_{2}|\leq 3r_{2}$.
Let $A_{2}=\\{y\in Y:f_{2}(y)\leq-r_{2}\\}$ and $B_{2}=\\{y\in Y:f_{2}(y)\geq
r_{2}\\}$. Then $A_{2}$ and $B_{2}$ are completely separated in $X$ by member
of $B_{1}(X)$. So, there exists $g_{2}:X\rightarrow[-r_{2},r_{2}]$, such that,
$g_{2}\in B_{1}^{*}(X)$ and $g_{2}(A_{2})=-r_{2},$ $g_{2}(B_{2})=r_{2}$.
Taking $f_{3}=f_{2}-g_{2}\big{|}_{Y}$ we get $f_{3}\in B_{1}^{*}(Y)$ with
$|f_{3}|\leq 2r_{2}=3r_{3}$.
If we continue this process and use principle of mathematical induction then
we get two sequences $\\{f_{n}\\}\subseteq B_{1}^{*}(Y)$ and
$\\{g_{n}\\}\subseteq B_{1}^{*}(X)$, with the following properties:
$|f_{n}|\leq 3r_{n}$ and $f_{n+1}=f_{n}-g_{n}\big{|}_{Y},\forall
n\in\mathbb{N}$.
Let $g(x)=\sum\limits_{n=1}^{\infty}g_{n}(x),\forall x\in X$. It is clear
that, $|g_{n}(x)|\leq r_{n},\forall n$ and $\forall x\in\mathbb{X}$, $g_{n}\in
B_{1}(X),\forall n\in\mathbb{N}$. Also,
$\sum\limits_{n=1}^{\infty}r_{n}<\infty$, where $r_{n}>0,\forall
n\in\mathbb{N}$.
So, by lemma 4.4 $g$ is a Baire one function on $X$.
We claim that, $g(y)=f_{1}(y),\forall y\in Y$.
$g(y)=\lim\limits_{n\to\infty}[g_{1}(y)+g_{2}(y)+...+g_{n}(y)]=\lim\limits_{n\to\infty}[f_{1}(y)-f_{2}(y)+f_{2}(y)-f_{3}(y)+...+f_{n}(y)]=f_{1}(y)-\lim\limits_{n\to\infty}f_{n+1}(y)=f_{1}(y)$
(as $|f_{n}|\leq 3r_{n}$ and $\sum\limits_{n=1}^{\infty}r_{n}$ is a geometric
series with common ratio less than $1$). So, $f\in B_{1}^{*}(Y)$ has an
extension $g\in B_{1}(X)$ and we know that if a Baire one function $f\in
B_{1}^{*}(Y)$ has an extension in $B_{1}(X)$, then it has an extension in
$B_{1}^{*}(X)$. Hence, $Y$ is $B_{1}^{*}$-embedded. ∎

######  Theorem 5.5.

If $Y$ is any $B_{1}$-embedded subspace of a topological space $X$ then $Y$ is
completely separated in $X$ by $B_{1}(X)$, from any zero set of $Z(B_{1}(X))$
disjoint from $Y$.

###### Proof.

Let $Y$ be a $B_{1}$-embedded subspace of $X$ and $Z(f)$ be a zero set in
$Z(B_{1}(X))$, disjoint from $Y$.
Let $h:Y\rightarrow\mathbb{R}$ be a map defined as, $h(y)=\frac{1}{|f(y)|}$.
Here, $|f|\bigg{|}_{Y}\in B_{1}(Y)$ and $|f(y)|>0$ on $Y$ as $Z(f)\cap
Y=\emptyset$. So, by Theorem 2.3, $h\in B_{1}(Y)$. As $Y$ is $B_{1}$-embedded,
$h$ has an extension $g\in B_{1}(X)$. Then $|f|.g\in B_{1}(X)$ and
$(|f|.g)(Z(f))=0$, $(|f|.g)(y)=1,\forall$ $y\in Y$. It follows that $Y$ and
$Z(f)$ are completely separated in $X$ by $B_{1}(X)$. ∎

######  Theorem 5.6.

A $B_{1}^{*}$-embedded subset $Y$ of $X$ is a $B_{1}$-embedded subset in $X$
if and only if it is completely separated in $X$ by $B_{1}(X)$ from any
disjoint zero set in $Z(B_{1}(X))$.

###### Proof.

Let $B_{1}^{*}$-embedded subset $Y$ of $X$ is $B_{1}$-embedded then the result
follows from Theorem 5.5.
Conversely, let $Y$ be completely separated in $X$ by $B_{1}(X)$ from any zero
set in $Z(B_{1}(X))$ disjoint from $Y$. Let $f\in B_{1}(Y)$. We consider the
function $tan^{-1}:\mathbb{R}\rightarrow(-\frac{\pi}{2},\frac{\pi}{2})$. By
Theorem 2.4 the function $tan^{-1}\circ f$ is a bounded Baire one function. As
$Y$ is $B_{1}^{*}$-embedded in $X$, $tan^{-1}\circ f$ has an extension to a
function $g\in B_{1}(X)$.
Let $Z=\\{x\in X:|g(x)|\geq\frac{\pi}{2}\\}$, then $Z\in Z(B_{1}(X))$ and
$Z\cap Y=\emptyset$. So, by hypothesis, there exists a Baire one function
$h:X\rightarrow[0,1]$ such that $h(Z)=0$ and $h(Y)=1$. We see that, $g.h\in
B_{1}(X)$ and $\forall x\in X,|(g.h)(x)|<\frac{\pi}{2}$. So, $tan(g.h)\in
B_{1}(X)$ and $\forall$ $y\in Y$, $tan(g.h)(y)=f(y)$. Hence, $Y$ is
$B_{1}$-embedded. ∎

######  Corollary 5.7.

For any topological space $X$, a zero set $Z\in Z(B_{1}(X))$ is
$B_{1}^{*}$-embedded if and only if it is $B_{1}$-embedded.

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