# From actions of an abelian group on itself to left braces

A. Ballester-Bolinches Departament de Matemàtiques, Universitat de València;
Dr. Moliner, 50; 46100 Burjassot, València, Spain<EMAIL_ADDRESS>ORCID 0000-0002-2051-9075 R. Esteban-Romero Departament de Matemàtiques,
Universitat de València; Dr. Moliner, 50; 46100 Burjassot, València, Spain;
<EMAIL_ADDRESS>ORCID 0000-0002-2321-8139 L. A. Kurdachenko Department of
Algebra and Geometry, Oles Honchar Dnipro National University, Dnipro 49010,
Ukraine<EMAIL_ADDRESS>ORCID 0000-0002-6368-7319Departament de
Matemàtiques, Universitat de València; Dr. Moliner, 50; 46100 Burjassot,
València, Spain V. Pérez-Calabuig Departament de Matemàtiques, Universitat de
València; Dr. Moliner, 50; 46100 Burjassot, València, Spain; Vicent.Perez-
<EMAIL_ADDRESS>ORCID 0000-0003-4101-8656

###### Abstract

We present a construction of left braces of right nilpotency class at most two
based on suitable actions of an abelian group on itself with an invariance
condition. This construction allows us to recover the construction of a free
right nilpotent one-generated left brace of class two.

_Mathematics Subject Classification (2020):_ 16T25, 16N40, 81R50

_Keywords:_ left brace, right nilpotent left brace, one-generated left brace,
Smoktunowicz-nilpotent left brace.

##  1 Introduction

A _skew left brace_ $(A,{+},{\cdot})$ consists of a set $A$ with two binary
operations such that $(A,{+})$ and $(A,{\cdot})$ are groups related by the
modified distributive law $a(b+c)=ab-a+ac$ for all $a$, $b$, $c\in A$. If, in
addition, the group $(A,{+})$ is abelian, we speak of a _skew left brace of
abelian type_ or, simply, a _left brace_. When the operations $+$ and $\cdot$
are clear from the context, we will refer to the left brace $(A,{+},{\cdot})$
simply as $A$. The structure of left brace was introduced by Rump as a
generalisation of Jacobson radical rings in [11]. Guarnieri and Vendramin [8]
introduced skew left braces by removing the commutativity of the addition in
Rump’s left braces. An interesting survey on left braces can be found in the
paper of Cedó [5], where most of the definitions and results presented in this
section can be found.

The structure of a left brace depends critically on the lambda action of the
multiplicative group on the additive group. Let $a$, $b$ be elements of a left
brace $A$. We define $\lambda_{a}(b)=-a+ab$. It is well known that
$\lambda\colon(A,{\cdot})\longrightarrow\operatorname{Aut}(A,{+})$ given by
$\lambda(a)=\lambda_{a}$ for $a\in A$ is a group homomorphism.

Another interesting operation that gives structural information on a left
brace is the star operation. We define it as $a*b=-a+ab-b=\lambda_{a}(b)-b$
for $a$, $b\in B$. Given two subsets $L$ and $M$ of $A$, we define $L*M$ as
the subgroup of the additive group generated by $\\{l*m\mid l\in L,m\in M\\}$.
We set $A^{(1)}=A$, $A^{(r+1)}=A^{(r)}*A$ for a natural $r$, and $A^{1}=A$,
$A^{r+1}=A*A^{r}$ for a natural $r$.

These constructions allow us to define two different nilpotencies for left
braces: we say that a left brace $A$ is _left nilpotent_ if there exists a
natural $m$ such that $A^{m+1}=\\{0\\}$, and that $A$ is _right nilpotent_ if
there exists a natural $m$ such that $A^{(m+1)}=\\{0\\}$. If $A$ is right
nilpotent (respectively, left nilpotent), the smallest $m$ satisfying
$A^{(m+1)}=\\{0\\}$, but $A^{(m)}\neq\\{0\\}$ (respectively,
$A^{m+1}=\\{0\\}$, but $A^{m}\neq\\{0\\}$) is called the _right_
(respectively, _left_) _nilpotency class_ of $A$. A left brace that is both
left nilpotent and right nilpotent is said to be _nilpotent in the sense of
Smoktunowicz_ or, simply, _Smoktunowicz-nilpotent_. These left braces were
studied by Smoktunowicz in [13]. For left braces, this concept coincides with
the central nilpotency introduced in [2] for skew left braces (see also [9]).

We say that a subset $S$ of a left brace $(A,{+},{\cdot})$ is a _subbrace_ of
$(A,{+},{\cdot})$ if $S$ is a subgroup of $(A,{+})$ and of $(A,{\cdot})$. The
intersection of a family of subbraces of a given left brace $(A,{+},{\cdot})$
is again a subbrace of $(A,{+},{\cdot})$. If $X$ is a subset of
$(A,{+},{\cdot})$, the intersection of all subbraces of $(A,{+},{\cdot})$ is a
subbrace of $(A,{+},{\cdot})$ called the _subbrace generated by_ $X$. When
$X=\\{a\\}$ is a singleton, we simply speak of the _subbrace generated by
$a$_. The description of the subbrace generated by a set or the free object on
a set $X$ in the category of left braces seems to be a difficult problem in
general. In fact, Agata and Alicja Smoktunowicz ask about such description in
the class of left braces of right nilpotency class at most two [14, Question
5.5].

Stefanello and Trappeniers [15, Theorem 3.13] have given some interesting
characterisations of left braces of right nilpotency class at most two. We
present here their results in the form we need in the sequel.

######  Proposition 1.1.

Let $A$ be a left brace. The following statements are equivalent in pairs:

  1. 1.

$A^{(3)}=\\{0\\}$.

  2. 2.

If $z\in A^{(2)}$ and $a\in A$, then $z*a=0$.

  3. 3.

$A^{(2)}\subseteq\operatorname{Ker}\lambda$.

  4. 4.

$\lambda$ is a group homomorphism between $(A,{+})$ and
$\operatorname{Aut}(A,{+})$.

  5. 5.

$\lambda_{\lambda_{a}(b)}=\lambda_{b}$ for every $a$, $b\in A$.

The left braces satisfying Condition 4 of Proposition 1.1 have been studied by
Bardakov, Neshchadim, and Yadav in [1], who called them _$\lambda$
-homomorphic_. They turn out to be examples of _bi-skew left braces_ or
_symmetric skew left braces_ , that is, skew left braces $(A,{+},{\cdot})$ for
which $(A,{\cdot},{+})$ is again a skew left brace, studied for the first time
by Childs in [6] and later by Caranti [3], Koch [10], and Caranti and
Stefanello [4]. We will prove Proposition 1.1 in Section 2 for the sake of
completeness.

A subbrace $S$ of a left brace $A$ is called a _left ideal_ of $A$ if
$A*S\subseteq S$. This is equivalent to saying that $\lambda_{a}(S)\subseteq
S$ for each $a\in S$. A subbrace $S$ of a left brace $A$ is called an _ideal_
of $A$ if $A*S\subseteq S$ and $S*A\subseteq A$. This is equivalent to saying
that $S$ is a left ideal of $A$ and $S$ is a normal subgroup of the
multiplicative group of $S$. We obtain that $A^{(r)}$ is an ideal of $A$ for
all naturals $r$ and that $A^{r}$ is a left ideal of $A$ for all naturals $r$.
The _socle_ of $A$, defined as the kernel of $\lambda$ in the case of left
braces, is another example of an ideal of $A$.

The first result of this paper is a general construction of a left brace from
an abelian group $(A,{+})$ and an action of $(A,{+})$ on itself that satisfies
an invariance condition from which the universal constructions related to one-
generated subbraces of right nilpotent left braces of class at most two
directly follow.

######  Theorem A.

Suppose that $(B,{+})$ is an abelian group acting on itself by means of an
action $\lambda\colon(B,{+})\longrightarrow\operatorname{Aut}(B,{+})$ given by
$\lambda(x)=\lambda_{x}$ for $x\in B$. Suppose also that if $x$, $y\in B$,
then $\lambda_{\lambda_{y}(x)}=\lambda_{x}$. We can define a product $\cdot$
on $B$ by means of $xy=x+\lambda_{x}(y)$ for $x$, $y\in B$. Then
$(B,{+},{\cdot})$ is a left brace with $B^{(3)}=\\{0\\}$.

As a consequence of Proposition 1.1, all left braces of right nilpotency class
two can be described in this way.

Rump [12], answering the above-mentioned question of Smoktunowicz and
Smoktunowicz, has given a description of the free one-generated right
nilpotent left brace of class two and a description of subbrace generated by
an element of a right nilpotent left brace of class at most two.

######  Theorem B (Rump [12]).

Let $(A,{+},{\cdot})$ be a left brace with $A^{(3)}=\\{0\\}$. Let $a\in A$.
For $i\in\mathbb{Z}$, we define $a_{i}=\lambda_{a^{i}}(a)=-a^{i}+a^{i+1}$. In
particular, $a_{0}=-a^{0}+a^{1}=-0+a=a$. We observe that
$\lambda_{a^{j}}(a_{i})=\lambda_{a^{j}}(\lambda_{a^{i}}(a))=\lambda_{a^{j}a^{i}}(a)=\lambda_{a^{i+j}}(a)=a_{i+j}$.
The set

$B=\Bigl{\\{}\sum_{i\in\mathbb{Z}}x_{i}a_{i}\mid
x_{i}\in\mathbb{Z},i\in\mathbb{Z},\text{ and $\\{i\in\mathbb{Z}\mid x_{i}\neq
0\\}$ is finite}\Bigr{\\}}.$

coincides with the subbrace of $A$ generated by $a\in A$. Furthermore, if
$D=\\{\sum_{i\in\mathbb{Z}}x_{i}a_{i}\in
B\mid\sum_{i\in\mathbb{Z}}x_{i}=0\\}$, then $B*B=D$.

######  Theorem C (see Rump [12]).

Let $C$ be a free abelian group with basis $\\{c_{i}\mid i\in\mathbb{Z}\\}$.
Given $x\in C$, then $x=\sum_{i\in\mathbb{Z}}x_{i}c_{i}$ with
$x_{i}\in\mathbb{Z}$ for all $i\in\mathbb{Z}$ and $x_{i}\neq 0$ for a finite
number of $i\in\mathbb{Z}$. We define an action of $x\in C$ on $C$ by means of
$\lambda_{x}(y)=\sum_{i\in\mathbb{Z}}y_{i}c_{i+\sum_{j\in\mathbb{Z}}x_{j}}=\sum_{i\in\mathbb{Z}}y_{i-\sum_{j\in\mathbb{Z}}x_{j}}c_{i}$
for $y=\sum_{i\in\mathbb{Z}}y_{i}c_{i}$. Then $(C,{+},{\cdot})$ with the
multiplication given by $xy=x+\lambda_{x}(y)$ for $x$, $y\in C$ is a left
brace generated by $a=c_{0}$ with $C^{(3)}=\\{0\\}$ and $C^{j}\neq\\{0\\}$ for
all $j\in\mathbb{N}$. Moreover, if $A$ is a left brace with $A^{(3)}=\\{0\\}$
and $b\in A$, then there exists a unique left brace epimorphism $\alpha$ from
$C$ and the subbrace of $A$ generated by $b$ such that $\alpha(c_{0})=b$.

Rump’s constructions rely on other algebraic structures, like cycle sets or
q-braces. We recover Rump’s results with a different approach that strongly
depend on Proposition 1.1 and Theorem A (Sections 4 and 5).

The special case of Smoktunowicz-nilpotent left braces $A$ with
$A^{(3)}=\\{0\\}$ and $A^{m+1}=\\{0\\}$ for a natural number $m\geq 2$ becomes
particularly relevant. Of course, we can apply the descriptions of the
subbrace $S$ generated by one element corresponding to $A^{(3)}=\\{0\\}$ and
the free one-generated left brace (Theorems B and C) to this case, but in this
case it seems desirable to have a description that takes into account the left
nilpotency class and the terms $S^{r}$ for all natural numbers $r$ become
apparent from the description. An adaptation of the previous description to
the subbrace generated by an element in these left braces appears in Section
6. We prove:

######  Theorem D.

Let $A$ be a left brace such that $A^{(3)}=A^{m+1}=\\{0\\}$ for a natural
number $m\geq 2$ and let $a\in A$. Define $a_{1}=a$, $a_{j+1}=a*a_{j}$ for
$j\geq 1$. Then the set

$S=\Bigl{\\{}\sum_{k=1}^{m}t_{k}a_{k}\mid t_{k}\in\mathbb{Z},1\leq k\leq
m\Bigr{\\}}$

is the subbrace of $A$ generated by $a$.

The construction of the free one-generated Smoktunowicz-nilpotent left braces
with right nilpotency class two and a given left nilpotency class is given in
Section 7. We prove:

######  Theorem E.

Let $m\geq 2$ be a natural number. Consider in
$B_{m}=\overbrace{\mathbb{Z}\times\dots\times\mathbb{Z}}^{(m)}$ the usual
addition given by

$(n_{1},\dots,n_{m})+(t_{1},\dots,t_{m})=(n_{1}+t_{1},\dots,n_{m}+t_{m})$

and the multiplication

$(n_{1},\dots,n_{m})(t_{1},\dots,t_{m})=(w_{1},\dots,w_{m})$

where, for $1\leq i\leq m$,

$w_{i}=n_{i}+\sum_{k=0}^{i-1}\binom{n_{1}}{k}t_{i-k}.$

Then $(B,{+},{\cdot})$ is a left brace with $B_{m}^{(3)}=B_{m}^{m+1}=\\{0\\}$,
$B_{m}^{m}\neq\\{0\\}$; $B$ is generated by $\mathbf{b}=(1,0,\dots,0)$, and
given a left brace $(A,{+},{\cdot})$ with $A^{(3)}=A^{m+1}=0$ and $a\in A$,
there exists a unique left brace homomorphism from $B_{m}$ to $A$ mapping
$\mathbf{b}$ to $a$ and whose image is the subbrace of $A$ generated by $a$.

The construction of this free left brace also appears as a particular case of
Theorem A.

##  2 Preliminary results

For the sake of completeness, we present here a proof of Proposition 1.1.

###### Proof of Proposition 1.1.

_1 implies 2_. Suppose that $z\in A^{(2)}$ and $a\in A$. Then
$z*a=A^{(3)}=\\{0\\}$.

_2 implies 3_. Let $z\in A^{(2)}$ and $a\in A$. Since
$\lambda_{z}(a)-a=z*a=0$, we have that $\lambda_{z}(a)=a$. This implies that
$\lambda_{z}=\operatorname{id}$ and so $z\in\operatorname{Ker}\lambda$.

_3 implies 4_. Since $ab=a*b+a+b=(a*b)\lambda_{a*b}^{-1}(a+b)$, we have that
$\lambda_{a}\circ\lambda_{b}=\lambda_{ab}=\lambda_{(a*b)\lambda_{a*b}^{-1}(a+b)}=\lambda_{a*b}\circ\lambda_{\lambda_{a*b}^{-1}(a+b)}=\lambda_{a+b}$
and so $\lambda\colon(A,{+})\longrightarrow\operatorname{Aut}(A,{+})$ is a
group homomorphism.

_4 implies 5_. We have that
$\lambda_{a}\circ\lambda_{b}=\lambda_{ab}=\lambda_{a+\lambda_{a}(b)}=\lambda_{a}\circ\lambda_{\lambda_{a}(b)}$.
Therefore, $\lambda_{b}=\lambda_{a}(b)$.

_5 implies 3_. Let $a$, $b\in A$. We have that

$\displaystyle\lambda_{a*b}$
$\displaystyle=\lambda_{\lambda_{a}(b)-b}=\lambda_{\lambda_{a}(b)\lambda^{-1}_{\lambda_{a}(b)}(-b)}=\lambda_{\lambda_{a}(b)\lambda_{b}^{-1}(-b)}=\lambda_{\lambda_{a}(b)}\circ\lambda_{\lambda_{b}^{-1}(-b)}$
$\displaystyle=\lambda_{b}\circ\lambda_{\lambda_{b}^{-1}}=\lambda_{b\lambda_{b}^{-1}(-b)}=\lambda_{b-b}=\lambda_{0}=\operatorname{id}.$

Therefore, $a*b\in\operatorname{Ker}\lambda$. Since
$\operatorname{Ker}\lambda=\operatorname{Soc}A$ is an ideal of $A$, we have
that $A^{(2)}\subseteq\operatorname{Ker}\lambda$.

_3 implies 1_. Suppose that $A^{(2)}\subseteq\operatorname{Ker}\lambda$. It is
enough to show that all generators of $A^{(3)}$, that have the form $d*e$ for
$d\in A^{(2)}$ and $e\in A$, are zero. Since
$A^{(2)}\subseteq\operatorname{Ker}\lambda$, we have that
$d*e=\lambda_{d}(e)-e=e-e=0$ if $d\in A^{(2)}$ and $e\in A$. Therefore,
$A^{(3)}=\\{0\\}$. ∎

The lambda map of a skew left brace is always an action from the
multiplicative group of the left brace on the additive group. Our construction
of left braces in Theorem A is based on an action from the additive group on
itself. The following result shows the relation between both actions.

######  Proposition 2.1.

Let $(A,{+})$ be an abelian group, $\lambda\colon
A\longrightarrow\operatorname{Aut}(A,{+})$ and define a binary operation
$\cdot$ on $A$ by means of $ab=a+\lambda_{a}(b)$ (this happens when $\lambda$
is the map defined on a left brace $(A,{+},{\cdot})$ by $\lambda_{a}(b)=-a+ab$
for $a$, $b\in A$). If two of the following three statements hold, then so
does the other one:

  1. 1.

$\lambda_{a+b}=\lambda_{a}\circ\lambda_{b}$ for every $a$, $b\in A$ (that is,
$\lambda\colon(A,{+})\longrightarrow\operatorname{Aut}(A,{+})$ is a group
homomorphism).

  2. 2.

$\lambda_{ab}=\lambda_{a}\circ\lambda_{b}$ for every $a$, $b\in A$.

  3. 3.

$\lambda_{\lambda_{a}(b)}=\lambda_{b}$ for every $a$, $b\in A$.

###### Proof.

_1 and 2 imply 3_. Since
$\lambda_{a}\circ\lambda_{b}=\lambda_{ab}=\lambda_{a+\lambda_{a}(b)}=\lambda_{a}\circ\lambda_{\lambda_{a}(b)}$,
we conclude that $\lambda_{b}=\lambda_{\lambda_{a}(b)}$.

_2 and 3 imply 1_. We have that
$\lambda_{a+b}=\lambda_{a\lambda_{a}^{-1}(b)}=\lambda_{a}\circ\lambda_{\lambda_{a}^{-1}(b)}=\lambda_{a}\circ\lambda_{\lambda_{a}(\lambda_{a}^{-1}(b))}=\lambda_{a}\circ\lambda_{b}$.

_3 and 1 imply 2_. We have that
$\lambda_{ab}=\lambda_{a+\lambda_{a}(b)}=\lambda_{a}\circ\lambda_{\lambda_{a}(b)}=\lambda_{a}\circ\lambda_{b}$.
∎

Some of the results of Section 6 will be given in terms of generalised
binomial coefficients of the form $\binom{n}{k}$, where $n$ is an integer,
that could be eventually negative, and $k$ is a non-negative integer. They are
defined by means of

$\binom{n}{0}=1,\qquad\binom{n}{k}=\frac{n(n-1)\dotsm(n-k+1)}{k!}\quad\text{if
$k>0$}.$

Some of the well-known identities of the usual binomial coefficients are also
satisfied by the generalised binomial coefficients. The next lemma summarises
some of these identities (see, for example, [7]).

######  Lemma 2.2.

  1. 1.

Let $n\in\mathbb{Z}$ and $k$ a non-negative integer. Then:

$\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}.$

  2. 2.

(Chu-Vandermonde’s Theorem) Let $n$ and $t$ be integer numbers and $r$ a non-
negative integer. Then

$\binom{n+t}{r}=\sum_{k=0}^{r}\binom{n}{k}\binom{t}{r-k}.$

The following properties of the star operation can be found, for instance, in
[9, Lemma 2.1]. In the absence of parentheses, the star operation will be
performed before the additions.

######  Lemma 2.3.

Let $(A,{+},{\cdot})$ be a left brace. Then, the following identities hold for
$a$, $b$, $c\in A$:

  1. 1.

$a*(b+c)=a*b+a*c$.

  2. 2.

$(ab)*c=a*(b*c)+b*c+a*c$.

For left braces $A$ with $A^{(3)}=\\{0\\}$, we have the following result.

######  Proposition 2.4.

Let $A$ be a left brace such that $A^{(3)}=\\{0\\}$. Then, for every $x$, $y$,
$z\in A$, we have that

$x*(z*y)+x*y+z*y=(x+z)*y=(xz)*y.$

###### Proof.

The first equality is just Lemma 2.3 (2). The second equality follows from the
fact that $\lambda_{x+y}=\lambda_{x}\lambda_{y}$ by the equivalence between 1
and 4 in Proposition 1.1 and so
$(x+z)*y=\lambda_{x+z}(y)-y=\lambda_{x}(\lambda_{z}(y))-y=\lambda_{xz}(y)-y=(xz)*y$.
∎

######  Corollary 2.5.

Let $A$ be a left brace with $A^{(3)}=\\{0\\}$. Then, for every $x$, $y\in A$,
$b\in A^{(2)}$, we have that $(x+b)*y=x*y$.

###### Proof.

We have that $b*y=0$ by Proposition 1.1. By Proposition 2.4, we have that

$(x+b)*y=x*(b*y)+x*y+b*y=x*0+x*y+0=x*y.\qed$

Now we present some general results about left braces $A$ satisfying the
condition $A^{(3)}=\\{0\\}$. They will be used in the description of the
subbrace generated by an element of $A$.

######  Proposition 2.6.

Let $A$ be a left brace and suppose that $A^{(3)}=\\{0\\}$. Let $a$ be an
element of $A$. Define $a_{1}=a$, $a_{j+1}=a*a_{j}$, $j\geq 1$. Then, for
every positive integer $n$,

$a^{n}=\sum_{k=1}^{n}\binom{n}{k}a_{k}=\binom{n}{1}a_{1}+\binom{n}{2}a_{2}+\binom{n}{3}a_{3}+\dots+\binom{n}{n-1}a_{n-1}+\binom{n}{n}a_{n}.$

###### Proof.

We prove this proposition by induction on $n$. For $n=1$, the result is
clearly true. Suppose that

$a_{1}^{n}=\sum_{k=1}^{n}\binom{n}{k}a_{k}.$

Then, by using Lemma 2.3, we have:

$\displaystyle a^{n+1}$ $\displaystyle=a+a*a^{n}+a^{n}$
$\displaystyle=a+a*\left(\sum_{k=1}^{n}\binom{n}{k}a_{k}\right)+\sum_{k=1}^{n}\binom{n}{k}a_{k}$
$\displaystyle=a+\sum_{k=1}^{n}\binom{n}{k}(a*a_{k})+\sum_{k=1}^{n}\binom{n}{k}a_{k}$
$\displaystyle=a+\sum_{k=1}^{n}\binom{n}{k}a_{k+1}+\sum_{k=1}^{n}\binom{n}{k}a_{k}$
$\displaystyle=a+\sum_{k=2}^{n+1}\binom{n}{k-1}a_{k}+\sum_{k=1}^{n}\binom{n}{k}a_{k}$
$\displaystyle=\sum_{k=1}^{n+1}\binom{n}{k-1}a_{k}+\sum_{k=1}^{n}\binom{n}{k}a_{k}$
$\displaystyle=\sum_{k=1}^{n+1}\binom{n+1}{k}a_{k}.$

We conclude that the result is true for all positive integers $n$. ∎

######  Proposition 2.7.

Let $A$ be a left brace such that $A^{(3)}=\\{0\\}$. Let $a$ be an element of
$A$. Write $a_{1}=a$, $a_{j+1}=a_{1}*a_{j}$ for $j\geq 1$. Then, for $j\geq
1$, we have that

$(na)*a_{j}=\sum_{k=1}^{n}\binom{n}{k}a_{k+j}=\binom{n}{1}a_{j+1}+\binom{n}{2}a_{j+2}+\dots+\binom{n}{n-1}a_{j+n-1}+a_{j+n}.$

###### Proof.

By Proposition 2.6, we have that $na=a^{n}+c$, where

$c=-\left(\binom{n}{2}a_{2}+\binom{n}{3}a_{3}+\dots+\binom{n}{n-1}a_{n-1}+a_{n}\right)\in
A^{2}.$

By Corollary 2.5, we have that $(na)*x=a^{n}*x$ for every element $x\in A$. We
argue by induction on $n$. For $n=1$, the result is clear. Suppose that

$(a^{n})*a_{j}=\sum_{k=1}^{n}\binom{n}{k}a_{k+j}.$

Then, by Lemma 2.3,

$\displaystyle(a^{n+1})*a_{j}$ $\displaystyle=(aa^{n})*a_{j}$
$\displaystyle=a*(a^{n}*a_{j})+a^{n}*a_{j}+a*a_{j}$
$\displaystyle=a*\left(\sum_{k=1}^{n}\binom{n}{k}a_{k+j}\right)+\sum_{k=1}^{n}\binom{n}{k}a_{k+j}+a_{j+1}$
$\displaystyle=a_{j+1}+\sum_{k=1}^{n}\binom{n}{k}(a*a_{k+j})+\sum_{k=1}^{n}\binom{n}{k}a_{k+j}$
$\displaystyle=a_{j+1}+\sum_{k=1}^{n}\binom{n}{k}a_{k+j+1}+\sum_{k=1}^{n}\binom{n}{k}a_{k+j}$
$\displaystyle=a_{j+1}+\sum_{k=2}^{n+1}\binom{n}{k-1}a_{k+j}+\sum_{k=1}^{n}\binom{n}{k}a_{k+j}$
$\displaystyle=\sum_{k=1}^{n+1}\binom{n+1}{k}a_{k+j}.$

The result follows by induction. ∎

##  3 Proof of Theorem A

In this section we prove Theorem A.

###### Proof of Theorem A.

We show first that $(B,{\cdot})$ is a group. Clearly, $\cdot$ is an internal
operation on $B$. In addition, given $x$, $y$, $z\in B$, we have that

$\displaystyle(xy)z$
$\displaystyle=xy+\lambda_{xy}(z)=(x+\lambda_{x}(y))+\lambda_{x+\lambda_{x}(y)}(z)=x+\lambda_{x}(y)+\lambda_{x}(\lambda_{\lambda_{x}(y)}(z))$
$\displaystyle=x+\lambda_{x}(y)+\lambda_{x}(\lambda_{y}(z))=x+\lambda_{x}(y+\lambda_{y}(z))=x+\lambda_{x}(yz)=x(yz).$

It follows that $\cdot$ is associative. Furthermore,
$0x=0+\lambda_{0}(x)=0+x=x$ and $x0=x+\lambda_{x}(0)=x+0=x$ for each $x\in B$.
Consequently, $0$ is the neutral element for $\cdot$. Finally, if $x\in B$, we
have that
$x(-\lambda_{x}^{-1}(x))=x+\lambda_{x}(-\lambda_{x}^{-1}(x))=x-\lambda_{x}(\lambda^{-1}_{x}(x))=x-x=0$,
while
$(-\lambda_{x}^{-1}(x))x=-\lambda_{x}^{-1}(x)+\lambda_{-\lambda_{x}^{-1}(x)}(x)=-\lambda_{x}^{-1}(x)+\lambda_{\lambda_{x}^{-1}(x)}^{-1}(x)=-\lambda_{x}^{-1}(x)+\lambda_{x}^{-1}(x)=0$,
and so $x^{-1}=-\lambda_{x}^{-1}(x)$ is the inverse of $x$ under $\cdot$.
Therefore $(B,{\cdot})$ is a group. Furthermore, if $x$, $y$, $z\in B$, we
have that
$xy-x+xz=x+\lambda_{x}(y)-x+x+\lambda_{x}(z)=x+\lambda_{x}(y)+\lambda_{x}(z)=x+\lambda_{x}(y+z)=x(y+z)$.
Therefore, $(B,{+},{\cdot})$ is a left brace.

Let $a$, $b\in B$. We have that $a*b=\lambda_{a}(b)-b$ and so
$\lambda_{a*b}=\lambda_{\lambda_{a}(b)-b}=\lambda_{\lambda_{a}(b)}\circ\lambda_{b}^{-1}=\lambda_{b}\circ\lambda_{b}^{-1}=\operatorname{id_{B}}$.
It follows that $a*b\in\operatorname{Ker}\lambda$. Therefore
$B^{(2)}\subseteq\operatorname{Ker}\lambda$, as
$\operatorname{Ker}\lambda=\operatorname{Soc}(B)$ and $B^{(2)}$ are ideals of
$B$. Let $u\in B^{(2)}$ and $c\in B$, then $u*c=\lambda_{u}(c)-c=c-c=0$, which
implies that $B^{(3)}=\\{0\\}$. ∎

##  4 The subbrace generated by an element of a right nilpotent left brace
with nilpotency class at most two

Theorem B describes the subbrace generated by an element of a right nilpotent
left brace of right nilpotency class at most $2$. We present in this section
an alternative proof of this result as a rather easy consequence of
Proposition 1.1.

###### Proof of Theorem B.

By Proposition 1.1, the map
$\lambda\colon(A,{+})\longrightarrow\operatorname{Aut}(A,{+})$ is a group
homomorphism and $A^{2}\subseteq\operatorname{Ker}(\lambda)$. In particular,
for $i\in\mathbb{Z}$,

$\lambda_{a_{i}}=\lambda_{-a^{i}+a^{i+1}}=\lambda_{a^{i}}^{-1}\lambda_{a^{i+1}}=\lambda_{(a^{i})^{-1}a^{i+1}}=\lambda_{a}.$

Clearly, $B$ is contained in the left subbrace generated by $a$. Furthermore,
$B$ is closed under the addition and by taking additive opposites. If
$x=\sum_{i\in\mathbb{Z}}x_{i}a_{i}\in B$, then

$\lambda_{\sum_{i\in\mathbb{Z}}x_{i}a_{i}}=\prod_{i\in\mathbb{Z}}\lambda_{a_{i}}^{x_{i}}=\prod_{i\in\mathbb{Z}}\lambda_{a}^{x_{i}}=\lambda_{a}^{t},\qquad\text{where
$t={\sum_{i\in\mathbb{Z}}x_{i}}$.}$

Here the product symbol corresponds to the iterated composition of group
homomorphisms. It follows that, for $y=\sum_{i\in\mathbb{Z}}y_{i}a_{i}\in B$,

$\lambda_{\sum_{i\in\mathbb{Z}}x_{i}a_{i}}\Bigl{(}\sum_{j\in\mathbb{Z}}y_{j}a_{j}\Bigr{)}=\sum_{j\in\mathbb{Z}}y_{j}a_{j+t}=\sum_{j\in\mathbb{Z}}y_{j-t}a_{j},\qquad
t=\sum_{i\in\mathbb{Z}}x_{i}.$

Therefore, $xy=x+\lambda_{x}(y)\in B$.

Furthermore, if $x=\sum_{i\in\mathbb{Z}}x_{i}a_{i}$, then
$y=\sum_{i\in\mathbb{Z}}(-x_{i-\sum_{j\in\mathbb{Z}}x_{j}})a_{i}\in B$
satisfies that $xy=yx=0$. Consequently $y=x^{-1}$ is the multiplicative
inverse of $x$. We conclude that $B$ is a subbrace, and so $B$ is the subbrace
generated by $a$.

We note that if $x=\sum_{i\in\mathbb{Z}}x_{i}a_{i}\in B$ and $y=\sum_{i\in
I}y_{i}a_{i}$, then
$x*y=\lambda_{x}(y)-y=\sum_{i\in\mathbb{Z}}(y_{i-\sum_{j\in\mathbb{Z}}x_{i}}-y_{i})a_{i}$
satisfies that
$\sum_{i\in\mathbb{Z}}(y_{i-\sum_{j\in\mathbb{Z}}x_{i}}-y_{i})=\sum_{i\in\mathbb{Z}}y_{i}-\sum_{i\in\mathbb{Z}}y_{i}=0$.
It follows that $B*B\subseteq D$. Furthermore, if
$x=\sum_{i\in\mathbb{Z}}x_{i}a_{i}\in D$, then
$x=\sum_{i\in\mathbb{Z}}x_{i}a_{i}-\sum_{i\in\mathbb{Z}}(-x_{i}a)=\sum_{i\in\mathbb{Z}}x_{i}(a_{i}-a)=\sum_{i\in\mathbb{Z}}x_{i}(\lambda_{a^{i}}(a)-a)=\sum_{i\in\mathbb{Z}}x_{i}(a^{i}*a)\in
C*C$. It follows that $D=C*C$. ∎

##  5 A free one-generated right nilpotent left brace of class two

We prove Theorem C as an application of Theorem A.

###### Proof of Theorem C.

As $\sum_{i\in\mathbb{Z}}y_{i-t}=\sum_{i\in\mathbb{Z}}y_{i}$ for each
$t\in\mathbb{Z}$, we obtain that $\lambda_{\lambda_{x}(y)}=\lambda_{y}$ for
all $x$, $y\in C$ and so we conclude that $(C,{+},{\cdot})$ is a left brace by
Theorem A. In particular, for each $a$, $b\in C$, $-a+ab=-a+a+\lambda_{a}(b)$
and the map $\lambda\colon(C,{\cdot})\longrightarrow\operatorname{Aut}(C,{+})$
is also a group homomorphism.

If $i=0$, $a_{0}=\lambda_{a^{0}}(a)=\lambda_{0}(a)=a=c_{0}$. Assume that
$i>0$. Then
$a_{i}=\lambda_{a^{i}}(a)=\lambda_{a}\circ\overset{(i)}{\cdots}\circ\lambda_{a}(a)=\lambda_{ia}(c_{0})=c_{i}$
and
$a_{-i}=\lambda_{a^{-i}}(a)=\lambda_{a}^{-1}\circ\overset{(i)}{\cdots}\circ\lambda_{a}^{-1}=\lambda_{(-i)a}(c_{0})=c_{-i}$.
Consequently, for every $i\in\mathbb{Z}$, $\lambda_{a^{i}}(a)=c_{i}$ and so
$C$ coincides with the subbrace generated by $a=c_{0}$ by Theorem B.

Let $s_{1}=a$, $s_{j}=a*s_{j-1}$, $j\geq 2$. Let us show by induction that
$s_{j}=\sum_{k=0}^{j-1}(-1)^{k}\binom{j-1}{k}c_{j-k-1}$ for $j\geq 2$. For
$j=2$, we have that $s_{2}=a*a=\lambda_{a}(a)-a=c_{1}-c_{0}=-c_{0}+c_{1}$.
Suppose that $s_{j}=\sum_{k=0}^{j-1}(-1)^{k}\binom{j-1}{k}c_{j-k-1}$. Then

$\displaystyle s_{j+1}$
$\displaystyle=a*s_{j}-s_{j}=a*\sum_{k=0}^{j-1}(-1)^{k}\binom{j-1}{k}c_{j-k-1}-\sum_{k=0}^{j-1}(-1)^{k}\binom{j-1}{k}c_{j-k-1}$
$\displaystyle=\sum_{k=0}^{j-1}(-1)^{k}\binom{j-1}{k}(a*c_{j-k-1})-\sum_{k=0}^{j-1}(-1)^{k}\binom{j-1}{k}c_{j-k-1}$
$\displaystyle=\sum_{k=0}^{j-1}(-1)^{k}\binom{j-1}{k}c_{j+1-k-1}-\sum_{k=0}^{j-1}(-1)^{k}\binom{j-1}{k}c_{j-k-1}$
$\displaystyle=\sum_{k=0}^{j-1}(-1)^{k}\binom{j-1}{k}c_{j+1-k-1}-\sum_{k=1}^{j}(-1)^{k}\binom{j-1}{k-1}c_{j+1-k-1}$
$\displaystyle=\sum_{k=0}^{j}(-1)^{k}\binom{j}{k}c_{j+1-k-1}.$

We conclude that $0\neq s_{j}\in C^{j}$ for all $j\in\mathbb{N}$ and so
$C^{j}\neq\\{0\\}$ for all $j\in\mathbb{N}$.

The description of the subbrace generated by one element given in Theorem B
shows that, with the notation of that theorem, the assignment

$\sum_{i\in\mathbb{Z}}x_{i}c_{i}\longmapsto\sum_{i\in\mathbb{Z}}x_{i}a_{i}$

is a left brace epimorphism between $C$ and the subbrace of $A$ generated by
$a$ and maps $c_{0}$ to $a$. Furthermore it is clear that it is the unique
left brace epimorphism satisfying this condition. ∎

##  6 The subbrace generated by one element of a Smoktunowicz-nilpotent left
brace

In this section, we will consider a left brace $(A,{+},{\cdot})$ satisfying
$A^{m+1}=A^{(3)}=\\{0\\}$ for a natural number $m\geq 2$. These left braces
have been studied for the first time by Smoktunowicz in [13]. Our aim is to
give a description of the subbrace $S$ generated by one element
$(A,{+},{\cdot})$ that gives information about the left nilpotent series
$S^{k}$ of $S$.

Given $a\in A$, if we define $a_{1}=a$, $a_{j+1}=a*a_{j}$ for $j\geq 1$, we
see that $a_{j}=0$ for $j\geq m+1$. Since $\binom{n}{k}=0$ when $0<n<k$, by
Proposition 2.7 we have:

$(na)*a_{j}=\sum_{k=1}^{m-j}\binom{n}{k}a_{k+j}.$ (1)

Our next aim is to prove that this expression is also valid for all integers
$n$, even if $n<0$. Let us start with $n=-1$.

######  Proposition 6.1.

Let $A$ be a left brace such that $A^{(3)}=A^{m+1}=\\{0\\}$ for a natural
number $m\geq 2$. Define $a_{1}=a$, $a_{j+1}=a*a_{j}$ for $1\leq j\leq m-1$.
If $1\leq k\leq m-1$, then

$(-a)*a_{m-k}=\sum_{j=1}^{k}(-1)^{j}a_{m-k+j}.$

###### Proof.

We argue by induction on $k$. Note that, for all $k$, we have that

$0=((-a)+a)*a_{m-k}=(-a)*a_{m-k+1}+(-a)*a_{m-k}+a_{m-k+1}$ (2)

by Proposition 2.4.

We start with $k=1$. In this case, we have that $(-a)*a_{m}\in
A^{m+1}=\\{0\\}$ and so $(-a)*a_{m-1}=-a_{m}$, as desired.

Suppose now that the result is valid for $k-1$, that is,

$(-a)*a_{m-k+1}=\sum_{j=1}^{k-1}(-1)^{j}a_{m-k+1+j}.$

By Equation (2), we have that

$\displaystyle(-a)*a_{m-k}$ $\displaystyle=-a_{m-k+1}-(-a)*a_{m-k+1}$
$\displaystyle=-a_{m-k+1}+\sum_{j=1}^{k-1}(-1)^{j+1}a_{m-k+1+j}$
$\displaystyle=-a_{m-k+1}+\sum_{j=2}^{k}(-1)^{j}a_{m-k+j}$
$\displaystyle=\sum_{j=1}^{k}(-1)^{j}a_{m-k+j}.$

The result follows by induction. ∎

Now we can prove that Equation (1) is valid for all integers $n$.

######  Proposition 6.2.

Let $A$ be a left brace such that $A^{(3)}=A^{m+1}=\\{0\\}$. Define $a_{1}=a$,
$a_{j+1}=a*a_{j}$ for $1\leq j\leq m-1$. Let $n$ be an integer. Then

$(na)*a_{j}=\sum_{k=1}^{m-j}\binom{n}{k}a_{k+j}.$

###### Proof.

The result is obvious for $n=0$ and known for positive integers $n$ by
Equation (1). Hence it is enough to prove it for negative integers. We argue
by induction on $-n$. When $n=-1$, we observe that
$\binom{-1}{j}=(-1)^{j}\binom{j}{j}=(-1)^{j}$. The result follows by
Proposition 6.1.

Assume that the result is true for $-n=k$, that is,

$(na)*a_{j}=\sum_{k=1}^{m-j}\binom{n}{k}a_{k+j}.$

We want to prove that it is also true for $-n=k+1$. By Proposition 2.4, Lemma
2.3, Corollary 2.5, and Proposition 6.2, we have that

$\displaystyle((-1$
$\displaystyle+n)a)*a_{j}=(-a)*((na)*a_{j})+(-a)*a_{j}+(na)*a_{j}$
$\displaystyle=(-a)*\left(\sum_{k=1}^{m-j}\binom{n}{k}a_{k+j}\right)+\sum_{k=1}^{m-j}\binom{-1}{k}a_{k+j}+\sum_{k=1}^{m-j}\binom{n}{k}a_{k+j}$
$\displaystyle=\sum_{k=1}^{m-j}\binom{n}{k}((-a)*a_{k+j})+\sum_{k=1}^{m-j}\left(\binom{-1}{k}+\binom{n}{k}\right)a_{j+k}$
$\displaystyle=\sum_{k=1}^{m-j}\binom{n}{k}\sum_{t=1}^{m-k-j}\binom{-1}{t}a_{t+k+j}+\sum_{k=1}^{m-j}\left(\binom{-1}{k}+\binom{n}{k}\right)a_{k+j}$
$\displaystyle=\sum_{r=2}^{m-j}\biggl{(}\sum_{\begin{subarray}{c}1\leq k\leq
m-j\\\ 1\leq t\leq m-k-j\\\
t+k=r\end{subarray}}\binom{n}{k}\binom{-1}{t}\biggr{)}a_{r+j}+\sum_{k=1}^{m-j}\left(\binom{-1}{k}+\binom{n}{k}\right)a_{k+j}$
$\displaystyle=\sum_{r=2}^{m-j}\biggl{(}\sum_{t=1}^{r-1}\binom{n}{r-t}\binom{-1}{t}\biggr{)}a_{r+j}+\sum_{k=1}^{m-j}\left(\binom{-1}{k}+\binom{n}{k}\right)a_{k+j}$
$\displaystyle=\sum_{r=2}^{m-j}\biggl{(}\sum_{t=0}^{r}\binom{n}{r-t}\binom{-1}{t}\biggr{)}a_{r+j}+\left(\binom{-1}{1}+\binom{n}{1}\right)a_{1+j}$
$\displaystyle=\sum_{r=2}^{m-j}\binom{n-1}{r}a_{r+j}+(-1+n)a_{1+j}$
$\displaystyle=\sum_{r=1}^{m-j}\binom{n-1}{r}a_{r+j}=\sum_{k=1}^{m-j}\binom{n-1}{k}a_{k-j}.$

By induction, the result is valid for all negative integers $n$. ∎

We are now in a position to give a description of the subbrace generated by an
element $a$ in a left brace with $A^{(3)}=A^{m+1}=\\{0\\}$.

###### Proof of Theorem D.

It is clear that if $T$ is a subbrace of $A$ containing $a$, then $S\subseteq
T$. Observe also that $a_{k}=0$ if $k>m$. It is enough to check that $S$ is a
subbrace of $A$, that is, $S$ is closed by taking sums, products, additive
opposites, and multiplicative inverses. It is clear that $S$ is closed by
taking sums and additive opposites.

Since $xy=x+x*y+y$ for every $x$, $y\in A$ and $S$ is closed by taking sums,
it will be closed under taking products if it is closed under taking the star
operation $*$. Consider

$t=\sum_{k=1}^{m}t_{k}a_{k}\in S,\quad u=\sum_{k=1}^{m}u_{k}a_{k}\in S.$

Since $\sum_{k=2}^{m}t_{k}a_{k}\in A^{2}$, by Corollary 2.5 and Lemma 2.3 we
see that

$t*u=(t_{1}a_{1})*u=\sum_{k=1}^{m}u_{k}(t_{1}a)*a_{k}.$

By Proposition 6.2, this is a linear combination with integer coefficients of
elements of $S$, and so $t*u\in S$.

To conclude the proof, we must show that $S$ is closed under taking inverses.
Take $t=\sum_{i=1}^{m}t_{i}a_{i}$, we will show that

$t^{-1}=\sum_{j=1}^{m}\left(-\sum_{k=0}^{j-1}\binom{-t_{1}}{k}t_{j-k}\right)a_{j}.$

Call $n$ the right hand side of the previous equality. Since $nt=n+n*t+t$, we
compute

$\displaystyle n*t$
$\displaystyle=\left(-\binom{-t_{1}}{0}t_{1}a_{1}\right)*\sum_{i=1}^{m}t_{i}a_{i}$
$\displaystyle=(-t_{1}a_{1})*\sum_{i=1}^{m}t_{i}a_{i}$
$\displaystyle=\sum_{i=1}^{m}t_{i}((-t_{1}a_{1})*a_{i})$
$\displaystyle=\sum_{i=1}^{m}t_{i}\sum_{k=1}^{m-i}\binom{-t_{1}}{k}a_{i+k}$
$\displaystyle=\sum_{r=1}^{m}\biggl{(}\sum_{\begin{subarray}{c}1\leq i\leq
m\\\ 1\leq k\leq m-i\\\
i+k=r\end{subarray}}t_{i}\binom{-t_{1}}{k}\biggr{)}a_{r}$
$\displaystyle=\sum_{r=1}^{m}\biggl{(}\sum_{k=1}^{r-1}t_{r-k}\binom{-t_{1}}{k}\biggr{)}a_{r}.$

by Lemma 2.3, Corollary 2.5 and Proposition 6.2. It follows that
$nt=n+n*t+t=0$, the neutral element of $\cdot$, and so $n$ is the
multiplicative inverse of $t$ in $A$. As $n\in S$ and the inverse is unique,
$S$ is closed under taking inverses. This completes the proof. ∎

######  Remark 6.3.

We note that in Theorem D, if for $1\leq r\leq m$ we call

$S_{r}=\Bigl{\\{}\sum_{k=r}^{m}t_{k}a_{k}\mid t_{k}\in\mathbb{Z},r\leq k\leq
m\Bigr{\\}},$

we have that $S_{r}\subseteq S^{r}$ as $a_{k}\in S^{r}$ for $r\leq k\leq m$.

We prove now by induction on $r$ that $S^{r}\subseteq S_{r}$. For $r=1$, the
result is clear since $S=S_{1}$. Suppose that $S^{r}\subseteq S_{r}$. Then
$S^{r+1}=S*S^{r}\subseteq S*S_{r}$ and, since

$\displaystyle\left(\sum_{k=1}^{m}v_{k}a_{k}\right)*\left(\sum_{k=r}^{m}t_{k}a_{k}\right)$
$\displaystyle=v_{1}a*\left(\sum_{k=r}^{m}t_{k}a_{k}\right)$
$\displaystyle=\sum_{k=r}^{m}v_{1}t_{k}a_{k+1}=\sum_{k=r+1}^{m}v_{1}t_{k-1}a_{k}\in
S_{r+1},$

we conclude that $S^{r+1}\subseteq S_{r+1}$. By induction, we have that
$S_{r}=S^{r}$ for $1\leq r\leq m$. For $r\geq m+1$, we have that
$S^{r}=\\{0\\}$.

##  7 Free one-generated Smoktunowicz-nilpotent left braces with right
nilpotency class two

Our next step is to prove Theorem E, that gives the construction of a free
one-generated left brace $B_{m}$ with $B_{m}^{(3)}=B_{m}^{m+1}=\\{0\\}$ for a
natural number $m\geq 2$. As we have mentioned in the introduction, we will do
it by means of the construction in Theorem A.

###### Proof of Theorem E.

Consider an integer $m\geq 2$. We define in

$B_{m}=\overbrace{\mathbb{Z}\times\dots\times\mathbb{Z}}^{(m)}$

the addition in the usual form

$(n_{1},\dots,n_{m})+(t_{1},\dots,t_{m})=(n_{1}+t_{1},\dots,n_{m}+t_{m}).$

Given $\mathbf{n}=(n_{1},\dots,n_{m})\in B_{m}$ and
$\mathbf{t}=(t_{1},\dots,t_{m})\in B_{m}$, we define
$\lambda_{\mathbf{n}}(\mathbf{t})=(z_{1},\dots,z_{m})$ where, for $1\leq i\leq
m$,

$z_{i}=\sum_{k=0}^{i-1}\binom{n_{1}}{k}t_{i-k}.$

Let us denote $z_{i}$ by $(\lambda_{\mathbf{n}}(\mathbf{t}))_{i}$, $1\leq
i\leq m$. It is clear that $\lambda_{\mathbf{n}}$ is an endomorphism of
$(B,{+})$ for all $\mathbf{n}\in B_{m}$. We prove that $\lambda$ is a
homomorphism from $(B,{+})$ to $\operatorname{End}(B,{+})$. Consider the
elements of $B_{m}$ $\mathbf{n}=(n_{1},\dots,n_{m})$,
$\mathbf{t}=(t_{1},\dots,t_{m})$, and $\mathbf{w}=(w_{1},\dots,w_{m})$. Then

$(\lambda_{\mathbf{w}}(\mathbf{t}))_{1}=\sum_{k=0}^{0}\binom{w_{1}}{0}t_{1-k}=t_{1}$

and so, for $1\leq i\leq m$,

$\displaystyle(\lambda_{\mathbf{n}}(\lambda_{\mathbf{w}}(\mathbf{t})))_{i}$
$\displaystyle=\sum_{k=0}^{i-1}\binom{(\lambda_{\mathbf{w}}(\mathbf{t}))_{1}}{k}(\lambda_{\mathbf{w}}(\mathbf{t}))_{i-k}$
$\displaystyle=\sum_{k=0}^{i-1}\binom{t_{1}}{k}\sum_{l=0}^{i-k-1}\binom{w_{1}}{l}t_{i-k-l}$
$\displaystyle=\sum_{k=0}^{i-1}\sum_{l=0}^{i-k-1}\binom{t_{1}}{k}\binom{w_{1}}{l}t_{i-k-l}$
$\displaystyle=\sum_{k=0}^{i-1}\sum_{r=k}^{i-1}\binom{t_{1}}{k}\binom{w_{1}}{r-k}t_{i-r}$
$\displaystyle=\sum_{k=0}^{r-1}\binom{t_{1}+w_{1}}{r}t_{i-r}=(\lambda_{\mathbf{n}+\mathbf{w}}(\mathbf{t}))_{i}.$

We also note that if $\mathbf{0}=(0,\dots,0)\in B_{m}$, then

$(\lambda_{\mathbf{0}}(\mathbf{t}))_{i}=\sum_{k=0}^{i-1}\binom{0}{k}t_{i-k}=t_{i},$

therefore $\lambda_{\mathbf{0}}=\operatorname{id}$. Consequently, if
$\mathbf{n}\in B_{m}$, we have that $\lambda_{\mathbf{n}}$ has an inverse
$\lambda_{-\mathbf{n}}$ and so the image of $\lambda$ is contained in the
automorphism group of $(B,{+})$. Furthermore, for $\mathbf{n}$, $\mathbf{t}$,
$\mathbf{w}\in B_{m}$ and $1\leq i\leq m$,

$(\lambda_{\lambda_{\mathbf{w}}(\mathbf{n})}(\mathbf{t}))_{i}=\sum_{k=0}^{i-1}\binom{(\lambda_{\mathbf{w}}(\mathbf{n}))_{1}}{k}t_{i-k}=\sum_{k=0}^{i-1}\binom{n_{1}}{k}t_{i-k}=(\lambda_{\mathbf{n}}(\mathbf{t}))_{i}.$

We conclude that
$\lambda_{\lambda_{\mathbf{w}}(\mathbf{n})}=\lambda_{\mathbf{n}}$.

By Theorem A, we have that $B_{m}$ is a left brace with the product $\cdot$
given by $\mathbf{n}\mathbf{t}=\mathbf{n}+\lambda_{\mathbf{n}}(\mathbf{t})$
for $\mathbf{n}$, $\mathbf{t}\in B_{m}$ and $B^{(3)}=\\{0\\}$.

Let $\mathbf{b}=(1,0,\dots,0)$. If $\mathbf{b}_{1}=b$ and
$\mathbf{b}_{k+1}=\mathbf{b}*\mathbf{b}_{k}$ for $k\geq 1$, we prove that
${(\mathbf{b}_{i})}_{j}=\delta_{i,j}$ for $1\leq j\leq m$ by induction on $i$.
The result is clear for $i=1$. If ${(\mathbf{b}_{i})}_{j}=\delta_{i,j}$ for
$1\leq j\leq m$ and some $i$, then

$\displaystyle{(\mathbf{b}_{i+1})}_{j}$
$\displaystyle={(\mathbf{b}*\mathbf{b}_{i})}_{j}={(\lambda_{\mathbf{b}}(\mathbf{b}_{i})-\mathbf{b}_{i})}_{j}$
$\displaystyle=\sum_{k=0}^{j-1}\binom{1}{k}{(\mathbf{b}_{i})}_{j-k}-\delta_{i,j}=\binom{1}{0}\delta_{i,j}+\binom{1}{1}\delta_{i,j-1}-\delta_{i,j}$
$\displaystyle=\delta_{i,j-1}=\delta_{i+1,j}.$

By Theorem D, we have that the subbrace of $B_{m}$ generated by $\mathbf{b}$
coincides with $B_{m}$. By Remark 6.3, we obtain that

$B_{m}^{r}=\\{(t_{1},\dots,t_{m})\in B_{m}\mid\text{$t_{i}=0$ for $1\leq
i<m$}\\}$

if $r\leq m$, and $B_{m}^{r}=\\{\mathbf{0}\\}$ if $r>m$.

It is clear that, given a left brace $A$ with $A^{(3)}=A^{m+1}=\\{0\\}$ for a
natural number $m\geq 2$, with the same notation as in Theorem D, the map
$(n_{1},\dots,n_{m})\longmapsto\sum_{k=1}^{m}n_{k}a_{k}$ defines a left brace
epimorphism $\varphi$ from $B_{m}$ to the subbrace of $A$ generated by $a$
such that $\varphi(\mathbf{b})=a$, and this is the unique possible
homomorphism with this condition. ∎

## Acknowledgements

The third author is very grateful to the Conselleria d’Innovació,
Universitats, Ciència i Societat Digital of the Generalitat (Valencian
Community, Spain) and the Universitat de València for their financial support
and grant to host researchers affected by the war in Ukraine in research
centres of the Valencian Community. He is sincerely grateful to the first and
second authors for their hospitality, support and care. The third author is
also grateful for the support of the Isaac Newton Institute for Mathematical
Sciences and the University of Edinburgh provided in the frame of LMS
Solidarity Supplementary Grant Programme. He is sincerely grateful to Agata
Smoktunowicz.

We also thank Lorenzo Stefanello for some interesting conversations that have
helped us to improve the presentation of this paper.

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# Learnings from Technological Interventions in a Low Resource Language:
A Case-Study on Gondi

###### Abstract

The primary obstacle to developing technologies for low-resource languages is
the lack of usable data. In this paper, we report the adoption and deployment
of 4 technology-driven methods of data collection for Gondi, a low-resource
vulnerable language spoken by around 2.3 million tribal people in south and
central India. In the process of data collection, we also help in its revival
by expanding access to information in Gondi through the creation of linguistic
resources that can be used by the community, such as a dictionary, children’s
stories, an app with Gondi content from multiple sources and an Interactive
Voice Response (IVR) based mass awareness platform. At the end of these
interventions, we collected a little less than 12,000 translated words and/or
sentences and identified more than 650 community members whose help can be
solicited for future translation efforts. The larger goal of the project is
collecting enough data in Gondi to build and deploy viable language
technologies like machine translation and speech to text systems that can help
take the language onto the internet.

Keywords: Low-Resource Languages, Deployment, Applications

Learnings from Technological Interventions in a Low Resource Language:
A Case-Study on Gondi

Devansh Mehta††thanks: * Equal Contribution ††thanks: Correspondence:
<EMAIL_ADDRESS>Sebastin Santy2∗11footnotemark: 1, Ramaravind K.
Mothilal2, Brij M.L. Srivastava3,
---
Alok Sharma4, Anurag Shukla5, Vishnu Prasad1, Venkanna U.5, Amit Sharma2,
Kalika Bali2
1Voicedeck Technologies, 2Microsoft Research India, 3INRIA France, 4DN
Developers, 5IIIT Raipur

Abstract content

##  1\. Introduction

Around 40% of all the languages in the world face the danger of extinction in
the near future. Languages are not only a means of communication but also a
carrier of tradition and cultures like verbal art, songs, narratives, rituals
etc. When a language spoken in a particular community dies out, future
generations lose a vital part of the culture that is necessary to completely
understand it. This makes language a vulnerable aspect of cultural heritage
and hence calls for their preservation. When it comes to saving such
endangered languages, there are two aspects: Preservation and Revitalization
(also referred to as ‘revival linguistics’) [Zuckermann, 2013]. The former is
concerned with how languages can be archived using different linguistic
techniques so that it can serve as a lookup for future generations, while the
latter focuses on ensuring that the language is resurrected into the daily
fabric of people’s lives. The biggest success story of a language getting
revitalized is Hebrew [Fellman, 1973], which went from few native speakers to
several million.

Initiatives like SOAS’ Endangered Languages Documentation Programme (ELDP)
111SOAS ELDP: https://www.eldp.net/ and The Language Conservancy (TLC) project
222TLC: https://www.languageconservancy.org/ contribute mostly towards the
documentation of endangered languages. However, language evolves with culture,
and focusing solely on archival efforts misses out on how societies might have
evolved differently had their language continued to be in use. In the present
day and age of globalization and the integration of technology into almost
every aspect of life, native speakers are turning to dominant languages at a
faster rate than ever before to provide greater economic and social
opportunities to future generations. Any revitalization effort undertaken
today needs to include technological interventions that can comprehensively
reverse this degradation. At a bare minimum, these languages need to be
integrated with the Internet, which is becoming an ever dominant part of our
lives, to ensure their survival and continued usage.

Figure 1: UNESCO 2017 World Atlas

We focus our efforts on Gondi, a South-Central Dravidian tribal language
spoken by the Gond tribe in Central India. Gondi provides a unique outlook of
how a language can be in danger even after having all the ingredients of a
sustainable language like (1) long historical continuity (2) a population of 3
million people speaking it and, (3) is widely spoken in around 6 states of
India with various dialects and forms. The complexities arise as Gondi is a
predominantly spoken language with no single standard variety but a number of
dialects, some mutually unintelligible. [Beine, 1994].

Deploying technology is a non-trivial task and there are legitimate concerns
surrounding how language technologies should be implemented for low-resource
languages [Joshi et al., 2019]. It is difficult to simply transfer
technologies prevalent in high resource language communities to minority
communities for many reasons, the chief among them being the lack of data in
low-resource languages. Our focus when working with the Gond community is thus
centered more around devising novel approaches for data collection, unlike
well-resourced languages where the focus is more on engineering. There also
needs to be groundwork and identification of the real problems that can be
solved by deployment of language technologies in minority communities.
[Dearden and Tucker, 2015]. Technical systems working in isolation from social
contexts can be very dangerous to the ecosystem of these minority language
communities and hence aren’t ethically neutral exchanges of information. As
far as possible, we have ensured that the technology interventions described
in this study were led by the Gond community and non-profit organizations with
a long history of working with the community. We have a deep appreciation of
the fact that the problem we are trying to tackle is different from solving an
engineering problem, since the expected outcomes have social rather than
technical consequences.

In this paper, we deploy 4 technological interventions to help revitalize
Gondi. The interventions are designed to achieve two objectives: (1) create a
repository of linguistic resources in Gondi, with the intention of eventually
using them to build language technologies like machine translation or speech
to text systems that are essential for taking Gondi onto the internet; and (2)
expand the information available to the Gond community in their language.

The first linguistic resource created is a Gondi dictionary that is accessible
to the community as an Android app. The second is 230 children’s books that
were translated by the Gond community in a 10 day workshop. The third is an
Android app that serves as a one-stop shop for Gondi content on the internet
and also crowdsources translations from the community. The final intervention
is a phone number that community members could call to gain awareness about a
local election in their area, upon completion of which they earn mobile
credits. These interventions respectively resulted in identifying 80-100
community members that participated in creating the 3,500 word dictionary; 20
community members that translated about 8000 sentences from Hindi to Gondi; 7
community members that translated 601 words; and 557 native speakers of Gondi
that can be called for future workshops. The goal of the project is working
with the Gond community members to collect enough data to build machine
translation and speech to text systems that can help take Gondi onto the
internet.

##  2\. Context

According to the 2011 census [Chandramouli and General, 2011], the total
population of the Gond tribe is approximately 11.3 million. However, the total
Gondi speaking population is only around 2.7 million. That is, only about 25
percent of the entire tribe now speaks it as a first language. UNESCO’s Atlas
of the Worlds Languages in Danger [Moseley, 2010] lists Gondi as belonging to
the vulnerable category. There is an added difficulty of creating resources
for Gondi due to the linguistic heterogeneity within the Gonds. Spread over 6
states in India, Gondi is heavily influenced by the dominant language of each
state to the point where a Gond Adivasi from Telangana (a Telugu speaking
Southern state), finds it difficult to understand a Gond Adivasi from
Chhattisgarh (a Central state with Hindi as the dominant language).

Figure 2: Gondi speaking areas in India

As a predominantly oral language, the proportion of Gondi speakers is expected
to go down further as opportunities are shrinking to hear the language spoken
outside of their everyday surroundings. For example, All India Radio, the only
radio station in India allowed to broadcast news, does not have a single Gondi
news bulletin in their broadcasts. There is no TV station or channel catering
to Gondi speakers. There is also a severe dearth of online content in Gondi,
resulting in members of the tribe having to learn a mainstream language in
order to enjoy the benefits of internet connectivity. One of the few
exceptions to this is “The Wire”333Gondi Bulletin:
https://youtu.be/M3q2ycJ_U7g, an Indian news outlet that has a news bulletin
in Gondi that is published on Youtube. Views of their Gondi news broadcast
range from 2,500 to 7,500, although it should be kept in mind that the Gondi
spoken in the broadcast caters to the Gond Adivasis from the Central states of
India and it is difficult for Gonds from the Southern states to understand the
content.

Gondi is also not included in the 8th Schedule of the Indian Constitution,
with the result that education and exams for government jobs cannot be
administered in the language. The deleterious effects of this on their society
are manifold. Gondi is considered the lingua franca of the local insurgents,
who use the knowledge of the language and the perceived neglect by the
government to recruit candidates from the tribe to join them [Kumar, 2019].
Further, there are high dropout rates among children that speak Gondi as a
first language. A 2008 UNESCO study found that children whose mother tongue is
not the medium of instruction at primary school are more likely to fail in
early grades or drop out, which in turn increases the chances of them joining
the insurgency [Bühmann and Trudell, 2008]. Working with the Gond tribes on
reviving their language is thus important not just for cultural reasons, but
may also serve as an instrument for bringing peace to their society. Reports
from workshop participants attest to the younger generation in the Gond tribe
attaching less importance to developing their mother tongue. However, we have
also found committed community members determined to revive their language and
it proved easy to recruit participants for the 8 workshops we held with the
community.

##  3\. Technological Intervention

There are relatively few cases that provide a holistic view of technology
being used to help revive usage of an endangered or vulnerable language. The
larger framework our interventions fall under is that language resource
creation feeds into building language technologies, which in turn enhances
access to information in that language. To give an example of how this
framework might operate, ensuring that Gondi is usable and accessible online
greatly enhances the access to information. However, building the necessary
language technology to take it online requires copious amounts of language
data - by one estimate, a minimum of 100,000 translated sentences are required
to build a machine translation tool [Koehn and Knowles, 2017], while at least
500 hours of transcribed and translated Gondi audio content is needed for a
speech to text interface.[Huang et al., 2014]

Our efforts at collecting Gondi language resource creation and enhancing
access to information in Gondi can be placed under broadly four buckets:

###  3.1. Gondi Dictionary Development

The Gondi speaking community is spread over 6 states in India. The dominant
language in each state has crept into the dialect spoken by the Gond tribe
residing there, with the result that it has become difficult for speakers from
different parts to understand one another easily. In order to facilitate a
democratized process of mutual intelligibility between the Gond groups, CGNet
Swara organized seven workshops with Gondi speaking representatives from 6
states to develop a Gondi thesaurus containing all the different words used by
speakers from various Gond regions. Some words, such as water, had as many as
8 different words for it. At the 8th workshop in 2018, which saw more than 80
people in attendance, the thesaurus was developed into a dictionary containing
3,500 words that were understandable by all groups, in essence ‘purifying’ the
language of words that had entered their dialect due to the influence of the
dominant state language. This dictionary not only enables translation of key
words between Hindi and Gondi, but also understanding between the Gond tribes
themselves. 444aka.ms/indian-express-gondi-dictionary

To enhance reach, the dictionary was made into an Android app, Gondi Manak
Shabdkosh555aka.ms/gondi-dictionary, that allows users to enter a Hindi or
Gondi word and immediately hear or read its equivalent translation. This app
is in many ways similar to the Ma! Iwaidja dictionary app666ma-iwaidja-
dictionary.soft112.com/, except that there is no wheel based interface for
conjugation and sentence formation.

Figure 3: Gondi Manak Shabdkosh

One of the use cases being explored with this app is in primary education,
where tribal students and teachers are sometimes unable to communicate as the
teacher does not know Gondi while the student does not know any other
language. The hope is that the dictionary app will allow some basic
communication and learning to take place.

###  3.2. Creating children’s books in Gondi

One of the most effective ways of reviving a language is introducing it as a
medium of instruction in schools, or at least as a subject. However, creating
textbooks in that language is an important first step to achieving this goal.

Pratham Books is a non-profit publisher with a motto “Book in every child’s
hand”. Storyweaver777storyweaver.org.in/ is an initative of Pratham Books that
hosts more than 15,000 children’s stories in various languages and dialects. A
10-day workshop was organized where 20 bilingual Gond tribals from three
states of India came together and translated children’s books from Hindi to
Gondi on the Storyweaver platform. These books were published on Storyweaver’s
website888aka.ms/storyweaver-gondi, resulting in the first ever online
repository of children’s stories in Gondi.999Article: aka.ms/hindu-gondi-
nextgen

At the end of the workshop, 230 books and about 8000 sentences were translated
from Hindi to Gondi. These stories, many of which introduced Gondi children to
climate change for the first time, were printed out and distributed in primary
schools in the Central Gondi speaking belt of Chhattisgarh. Efforts are now
ongoing to convince the state government to include them as part of the school
curriculum across the tribal belts of the state. 101010Article:
aka.ms/news18-climatechange

###  3.3. Crowdsourcing Gondi translations

At the Pratham Books translation workshop, it was found that many participants
wanted to continue the translation work from home, but there was no avenue for
them do so. Taking inspiration from Aikuma[Bird et al., 2014], we developed
Adivasi Radio, an Android application that presents users with Hindi words or
sentences for which they need to provide the Gondi translation.

In the one month since its launch, there have been a total of 601 appearances
translated through the app from 7 unique users. However, 5 of these users are
either paid staff or volunteers hoping to become paid staff, indicating that
soliciting translations from the community without monetary compensation may
be a challenge. Moreover, the bulk of the translations, 493, have come from
one superuser. Future translation workshops will include a component on
continuing the translation work through the app, with the hope that attendees
continue translating even after they return to their village.

In addition to the translation role, Adivasi Radio also serves as a one-stop
shop for Gondi content on the internet. The translated books from Pratham are
accessible through the app, as are the Gondi stories from CGnet Swara. The
information on The Wire’s Gondi news bulletin are also available here.

Figure 4: Adivasi Radio

To ensure these are comprehensible to populations that cannot read or write, a
bootstrapped Devanagari text-to-speech system is integrated within the app to
read out the written reports in Gondi. As more Gondi content proliferates
online, we envisage Adivasi Radio as not only becoming the go-to place for
native speakers to find outlets and sites publishing Gondi content, but also
the primary medium for collecting the 100,000 sentences needed to build a
machine translation tool between Gondi and other mainstream languages.

###  3.4. Disseminating Gondi content

Many recent studies on information dissemination in low-resource settings have
relied on technologies, such as Interactive Voice Response (IVR), that could
be used with internet-less mobile phones to connect to people and deliver
information [Swaminathan et al., 2019, Chakraborty et al., 2019, Marathe et
al., 2015, Moitra et al., 2016, Patel et al., 2010, Raza et al., 2018, Raza et
al., 2019, Raza et al., 2013, Vashistha et al., 2015]. As an example,
Learn2Earn [Swaminathan et al., 2019] is an IVR based system that was used as
an awareness campaign tool to spread farmers’ land rights in rural India.
Learn2Earn awards mobile talktime to users who call a toll-free number and
listen to an awareness message, and answer all multiple-choice questions on
the message correctly. Further, it has a peer-to-peer referral component where
an additional recharge is provided for every new user successfully referred to
the system. Learn2Earn was successful in spreading land rights awareness to
17,000 farmers in 45 days, starting from an initial set of just 17 farmers
[Swaminathan et al., 2019].

In this intervention, we adapted the Learn2Earn technology to spread voter
awareness among Gondi speakers in Dantewada (a rural district in the state of
Chhattisgarh in India), during the time that a bypoll election was held. We
chose to disseminate voter rights messages in Gondi, as the bypoll was crucial
for Gondi speakers to establish representative and effective governance in
their area. Further, prior work has shown that larger the contexts, identities
and communicative functions associated with the language use, the more likely
the language is expected to survive. [Walsh, 2005] We believe voter rights
content in local language during elections has the potential to encourage
conversations on topics of wider contexts and functions.

Figure 5: Number of calls to the system (blue line) and the number of users
who answered all questions correctly (bar plot). The blue and orange bars
represent those users who came to know about the system through word of mouth
and through referrals, respectively.
---
Metrics | Value
Unique callers (total, during and after seeding) |  (557, 480, 77)
Unique callers per day - during seeding (min, mean, median, max) |  (13, 48, 49, 80)
Unique callers per day - after seeding (min, mean, median, max) |  (0, 4, 3, 20)
Callers answering all questions correctly |  313
Callers answering all questions correctly in their first call |  104
Calls made by callers (min, mean, median, max) |  (1, 3, 2, 64)
Table 1: Summary statistics for our IVR-based deployment.

Learn2Earn was previously deployed in Hindi to spread awareness for voting
awareness [Kommiya Mothilal et al., 2019]. We extrapolated these experiments
to Gondi to do that same in preparation for an upcoming bypoll in a Gondi
speaking district. Similar to the original Learn2Earn implementation, the
experiment started with a few ‘seed’ users and is intended to spread
voluntarily through word of mouth or through peer-referrals thereon, both
increasing the chances of language and content spread. Figure 5 shows the
number of unique calls that were made to our system and the number of users
who passed the awareness quiz over time. Further, the figure shows that a
majority of quiz passers came to know about our system organically through
word of mouth, though people knew that additional credits can be earned
through referrals. Only a very few users continued to use the system even
after the elections in September and after the seeding was stopped, probably
to earn additional mobile talktime.

Table 1 describes important descriptive statistics of our experiment. We were
able to spread voter awareness in Gondi to 557 speakers of which a majority
(86%) were reached when we actively seeded. Nearly 60% of the users correctly
answered all questions, either in their first attempt or in successive
attempts, indicating the effectiveness of the system for content
dissemination. Further, many users called the system more than once, with one
user calling 64 times (to refer someone, users had to place another call to
the system). A follow-up survey revealed that 22 of 113 users surveyed could
not vote as they did not have a voter ID card, potentially useful information
for authorities to increase voter turnout. One disappointing finding was that
only about 8.3 percent of users were women, indicating that a focus on
outreach towards this community is needed.

We see three clear benefits from conducting Learn2Earn pilots in endangered or
vulnerable languages. First, since it is entirely in the spoken form, only
speakers of the endangered or vulnerable language can comprehend the content
and earn the reward. Second, the phone numbers of the 557 users collected are
an important dataset of speakers of that language and can be used for future
translation workshops and related programs that help their economic and
linguistic development. Finally, an oral language has a tendency to die out
unless there is an opportunity for that language to be used outside of
everyday surroundings, which periodic Learn2Earn campaigns on important issues
help achieve.

##  4\. Prior Work

###  4.1. Preservation Programs

The Endangered Languages Documentation Programme (ELDP) engages in
documentation of endangered languages by creating a repository of resources
for linguistics, the social sciences, and the language communities themselves.
The Language Conservancy is a non-profit organization that strives towards
revitalization of the world’s endangered languages, restoring them to health
and stability, and safeguarding them for future generations. They intend to
implement this by introducing it as medium of instruction in schools by
developing appropriate teaching methodologies and promoting and supporting the
use of such languages beyond this setting - in homes and communities.

###  4.2. Language Revitalization Efforts and Successes

Hebrew is one of the languages which has shown the most promise in
revitalization. Not being used since the second century, the usage of language
in common conversations was kick-started by some Jewish communities in the
19th century [Fellman, 1973]. It has now risen to become the official language
for the state of Israel and is spoken fluently by 7 million people.

However, such organic growth is almost impossible when the usage of language
at this time is reinforced by many more factors. Ojibwe is an interesting
example, where computer-based language learning technology was used and
studies were conducted on how a particular multimedia tool might jumpstart
communication in the Ojibwe language at home.[Hermes and King, 2013]
Additionally, there were efforts where twitter was seeded with many Ojibwe
tweets and hashtags to motivate individuals to converse in Ojibwe on twitter.

###  4.3. Deployed Applications and their Effects

Documentation and easy access to it is the first step in reviving a language.
The Ma! Iwaidja is a mobile phone app which runs a dictionary of 1500 word and
450 English-Iwaidja translations along with audio for each. In addition, it
provides a functionality to add words to the dictionary by the lay user. Due
to their ubiquitous nature, crowd-sourcing translations through apps are ideal
due to their convenience and ease of use. Steven Bird’s Aikuma app has
leveraged its intuitive design to crowdsource 10 hours of audio or 100,000
words from indigenous communities in Brazil, Nepal and Papua New Guinea. [Bird
et al., 2014]

##  5\. Conclusion

While working with well-resourced languages, the main problem in designing
language technologies is engineering. For low-resource languages, however, the
main problem is one of designing methods for data collection upon which the
language technology can be built.

To keep community members motivated through the process of collecting the
large amounts of translations needed, our team strived to achieve 2
simultaneous goals in each of the four technology interventions: the
collection of data upon which language technologies such as speech to text or
machine translation can be built, and expanding the access to information in
that language, which the community members could point to as a demonstrable
success. For example, the Learn2Earn pilot in Gondi not only provided Gond
tribals with an opportunity to earn money for answering a quiz in their
language and referring others to it, but it also provided a dataset of native
Gondi speakers that can be called for future translation workshops. Similarly,
translating children’s stories and creating a standardized dictionary resulted
in both data upon which machine translation tools can be built, and also
tangible language resources that can be used by their community.

The Chhattisgarh government recently allowed primary education in 10 tribal
languages, including Gondi
111111https://indianexpress.com/article/governance/chhattisgarh-education-
reforms-tribal-languages-to-be-a-medium-of-education-in-pre-school-6271547/.
Our plan is to work with the government in creating new resources in Gondi
that can be used in schools and distributing the translated childrens stories
through the government apparatus. The larger goal of our research project is
integrating Gondi with the internet, which requires at least 100,000
translated sentences; and building a speech to text platform, requiring at
least 100 hours of transcribed and translated transcripts. Our most immediate
goal is getting more users on the Adivasi Radio app and crowdsourcing
translations through it. We plan to make all the data collected openly
available and invite other researchers to participate in building language
technologies that can benefit the Gond community.

##  6\. Acknowledgements

We would like to thank Shubhranshu Choudhury from CGnet Swara, William Thies
from Microsoft Research New England, Amna Singh from Pratham Books, and the
director of IIIT Naya Raipur, Pradeep Kumar Sinha, for their invaluable
assistance and advice. A word of thanks to all participants and workshop
attendees from the Gond community, without whom this would not have been
possible or meaningful. We are also grateful for the financial contributions
made by the National Mineral Development Corporation (NMDC), South Eastern
Coalfields Limited (SECL), Indira Gandhi National Center for Arts (IGNCA) and
Microsoft Research India (MSR) towards supporting this research.

##  7\. Bibliographical References

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