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\title[Multivalued groups and  Newton polyhedron]{\large Multivalued groups and  Newton polyhedron}


\author[V.G. BARDAKOV]{{\bf V.G. Bardakov}\protect\orcidicon{}}

\author[T.A. KOZLOVSKAYA]{{\bf T.A. Kozlovskaya}\protect\orcidicon{0000-0002-9614-7570}}




\address{Valeriy Georgievich Bardakov   % Контактные данные всех авторов и место работы указываются только на английском языке
\newline\hphantom{iii} Sobolev Institute of Mathematics,
\newline\hphantom{iii} pr. Koptyuga, 4,
\newline\hphantom{iii} 630090, Novosibirsk, Russia%
\newline\hphantom{iii} Novosibirsk State Agrarian University
\newline\hphantom{iii} Dobrolyubova Street, 160 Novosibirsk 630039, Russia}
\email{\textcolor{blue}{bardakov@math.nsc.ru}}%


\thanks{\sc Bardakov, V.G.,
Multivalued groups and  Newton polyhedron}
\thanks{\copyright \ 2023 Bardakov V.G}
\thanks{\rm  This work was  supported by  the state contract of the Sobolev Institute of Mathematics, SB RAS (no. I.1.5,
project FWNF-2022-0009).}

\thanks{\sc Kozlovskaya, T.A.,
Multivalued groups and  Newton polyhedron}
\thanks{\copyright \ 2023 Kozlovskaya T.A}
\thanks{\rm This study was supported by the Tomsk State University Development Programme (Priority-2030)
and the article was prepared within the framework of the project “Mirror Laboratories” HSE University, RF}
\thanks{\it Received January, 1, 2023, Published December, 31, 2023}%

\address{Tatyana Anatolevna Kozlovskaya  % Контактные данные всех авторов и место работы указываются только на английском языке
\newline\hphantom{iii} Regional Scientific and Educational Mathematical Center of Tomsk State University,
\newline\hphantom{iii} 36 Lenin Ave.,
\newline\hphantom{iii} 634050, Tomsk, Russia}%
\email{\textcolor{blue}{t.kozlovskaya@math.tsu.ru}}%

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\maketitle {\small

\vspace{-12pt}
%\centerline{{\it Communicated by} {\sc P.P. Petrov}
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\bigskip
\bigskip

\begin{quote}
\noindent{\bf Abstract:} On the set of complex number $\mathbb{C}$ it is possible to define $n$-valued group for any positive integer $n$. The $n$-multiplication defines a symmetric polynomial $p_n = p_n (x, y, z)$ with integer  coef\-fici\-ents. By the theorem on symmetric polynomials, one can present $p_n$ as polynomial in elementary symmetric polynomials $e_1$, $e_2$, $e_3$. V.~M.~Buchstaber formulated a question on description coefficients of this polynomial. Also, he formulated the next question:  How to describe the Newton polyhedron of $p_n$?  In the present paper we find all coefficients of $p_n$ under monomials of the form $e_1^i e_2^j$ and prove that  the Newton polyhedron of $p_n$ is a right  triangle.
\medskip

\noindent{\bf Keywords:} 
multi-set, multivalued group, symmetric polynomial, Newton polyhedron.


 \end{quote}
}

\bigskip

One  branch  of Abstract Algebra is studying algebraic systems with multi\-valued operations. Solutions of the Yang-Baxter equation (2-simplex equation) and its generalization, $n$-simplex equations, $n \geq 3$, are examples of multivalued operations. 
In 1971, V.~M.~Buchstaber and S.~P.~Novikov \cite{2} introduced a construction, suggested by
the theory of characteristic classes of vector bundles, in which the product of each pair of elements is an $n$-multi-set, the set of $n$ points with multiplicities. This construction leads to the notion of $n$-valued group.

A good survey on $n$-valued groups and its applications  can be found in \cite{1}. In Section 5 of this paper,  $n$-valued groups were constructed on the set of complex numbers $\mathbb{C}$ for any natural $n$. The $n$-valued multiplication is described by the polynomials $p_n = p_n(z;x,y)$ which are 
 $x, y, z$--symmetric polynomials with integer coefficients. If we introduce elementary symmetric polynomials
 $$
e_1 = x + y + z, ~~e_2 = xy + yz + zx,~~ e_3 = xyz,
$$
then $p_n = P_n(e_1, e_2, e_3)$ is a polynomial with integer coefficients in variables $e_1$, $e_2$, $e_3$. In \cite{1} two questions were  formulated on the description the coef\-fici\-ents of $P_n$ as well as a question on the Newton polyhedron of $p_n$.

In the present paper we find the coefficients for monomials of the form $e_1^{i} e_2^{j}$ in $P_n$. This gives particular answer to the first two questions. 
Also, we  prove that if  
$$
f = f(x_1, x_2, \ldots, x_n) \in \mathbb{Z}[x_1, x_2, \ldots, x_n]
$$
 is a symmetric homogeneous polynomial of degree $k$, which  contains a monomial $a x_1^k$ for some non-zero $a$, then  its Newton polyhedron  is the $k \Delta^{n-1}$-simplex.
From this theorem follows 
 that  the Newton polyhedron of $p_n$ is the right triangle with side which depend on $n$. This is the complete answer to the third question.
 
 At the end of the paper we formulate some open questions.





\section{Multivalued groups and Buchstaber's questions}\label{sec2}
\subsection{Multivalued groups}
Recall definitions and some facts from the theory of multivalued groups (see, for example, \cite{1}).

Let $X$ be a non-empty set. An $n$-{\it valued multiplication} on $X$ is a map
$$
\mu \colon X \times X \to (X)^n = Sym^nX,~~\mu(x, y) = x * y = [z_1, z_2, \ldots, z_n],~~z_k = (x*y)_k,
$$
where $(X)^n = Sym^nX$ is the $n$-th symmetric power of $X$, that is the quotient $X^n / S_n$ of the Cartesian power $X^n$ under the action of $S_n$ by permutations of components. 
The next axioms are  natural generalizations of the classical axioms of group multiplication.

{\it Associativity}. The $n^2$-multi-sets:
$$
[x * (y * z)_1, x * (y * z)_2,  \ldots, x * (y * z)_n], ~~[ (x * y)_1 * z, (x * y)_2 * z,  \ldots, (x * y)_n * z]
$$
are equal for all $x, y, z \in X$.

{\it Unit}. An element $e \in X$ such that 
$$
e * x = x * e = [x, x, \ldots, x]
$$
 for all $x \in X$.


{\it Inverse}. A map $inv \colon X \to X$ such that 
$$
e \in inv(x) * x~\mbox{and}~e \in x * inv(x)
$$
 for all $x \in X$.
 


The map $\mu$ defines $n$-{\it valued group structure} $\mathcal{X} = (X, \mu, e, inv)$ on $X$ if it is associative, has a unit and an inverse. 


\medskip

 Let $\mu$ be the multiplication
$$
\mu \colon \mathbb{C} \times \mathbb{C} \to (\mathbb{C})^n
$$
that is defined by the formula
$$
\mu(x, y) =  x * y = [ (\sqrt[n]{x} + \epsilon^r \sqrt[n]{y})^n,~~1 \leq r \leq n],
$$
where $\epsilon$ is a primitive $n$-th root of unity.
This multiplication endows $\mathbb{C}$ with the structure of an $n$-valued group with the unit $e = 0$. The inverse element is given by the map $inv(x) = (-1)^n x$.

The $n$-valued multiplication is described by the polynomials
$$
p_n = p_n(z; x, y) = \prod^n_{k=1} (z - (inv(x) * inv(y))_k),
$$
whence the product $x * y$ is defined by $z$-roots of the equation $p_n = 0$. The polynomials 
$p_n(z; x, y)$ are $x, y, z$--symmetric polynomials with integral coefficients, e.g.,
$$
p_1 = x + y + z, ~~p_2 = (x + y + z)^2 - 4(xy + yz + zx).
$$
Set
$$
e_1 = x + y + z, ~~e_2 = xy + yz + zx,~~ e_3 = xyz.
$$
Then

$p_1 = e_1,$\\

$p_2 = e_1^2 - 2^2 \,  e_2,$\\

$p_3 = e_1^3 - 3^3  \,  e_3,$\\

$p_4 = e_1^4 - 2^3  \,  e_1^2 e_2 + 2^4  \,  e_2^2 - 2^7  \,  e_1 e_3,$\\

$p_5 = e_1^5 - 5^4  \,  e_1^2 e_3 + 5^5  \,  e_2 e_3,$\\

$p_6 = e_1^6 - 2^2 \cdot 3  \,  e_1^4 e_2 + 2^4 \cdot 3  \,  e_1^2 e_2^2 - 2^6  \,  e_2^3 -   2 \cdot 3^4 \cdot 17  \,  e_1^3 e_3  - 2^3  \cdot 3^4 \cdot 19   \,  e_1 e_2 e_3 + 3^3 \cdot 19^3  \,  e_3^2 ,$\\

$p_7 = e_1^7 -  5 \cdot 7^4  \,  e_1^4 e_3 +    2 \cdot 7^6  \,   e_1^2 e_2 e_3  - 7^7    \,  e_2^2 e_3 + 7^8  \,  e_1 e_3^2.$\\


The following questions were formulated in \cite{1}.



\emph{(1) What is the relationship between prime factors of $n$ and prime factors of the
coefficients of the polynomials }$p_n$?


\emph{(2) How to distinguish the monomials that have zero coefficient? }


\emph{(3) How to describe the Newton polyhedron of} $p_n$?


\bigskip


\section{Coefficients  and the Newton polyhedron  of $p_n$} 

Since $p_n$ is a symmetric homogeneous  polynomial of degree $n$, by the theorem on symmetric polynomials we can present $p_n$ as a polynomial on the elementary symmetric polynomials $e_1$, $e_2$, and $e_3$,
$$
p_n = \sum_{\substack{k_1 \geq k_2 \geq k_3 \geq 0\\ k_1 + k_2 + k_3 = n}} A_{k_1,k_2,k_3} e_1^{k_1-k_2} e_2^{k_2-k_3} e_3^{k_3} \in \mathbb{Z}[e_1, e_2, e_3].
$$
The main problem is to find the coefficients $A_{k_1,k_2,k_3}$. 

We can write $p_n$ in the form
$$
p_n =  \prod^n_{k=1} \left(z - \left((inv(x) * inv(y)\right)_k\right) = \prod^n_{k=1} \left(z - ((-1)^n x * (-1)^n y)_k\right) =
$$
$$
= \prod^n_{k=1} \left(z - \left(\sqrt[n]{(-1)^n x} +  \epsilon^k \sqrt[n]{(-1)^n y}\right)^n\right).
$$
If $y = 0$, then
$$
\bar{p}_n = p_n(z; x, 0) = \prod^n_{k=1} \left(z - (\sqrt[n]{(-1)^n x})^n\right) = \prod^n_{k=1} \left(z - (-1)^n x\right) =   \left(z - (-1)^n x\right)^n.
$$

Denote by
$$
\bar{e}_1 = e_1(z;x,0) = x + z,~~\bar{e}_2 = e_2(z;x,0) = zx.
$$
We see that $e_3(z;x,0) = 0$.

The next proposition gives particular answers to the first two questions.


\begin{proposition}
1) If $n$ is odd, then all $A_{k_1,k_2,0}$, $k_2 \not= 0$, are zero, i.e. in this case $p_n$ does not contains monomials $e_1^i e_2^j$, $j > 0$.

2) If $n = 2k$ is even, then the coefficient $A_{2k-i,i,0}$ at $e_1^{2(k-i)} e_2^i$, is equal to
$$
A_{2k-i,i,0} = (-4)^i C_k^i =  (-4)^i \frac{k!}{i! (k-i)!},~~~i = 1, 2, \ldots, k.
$$
%$A_{k_1, k_2, 0}$ for $e_1^{k_1 - k_2} e_2^{k_2}$, where $k_1 \geq k_2 >0$ and $k_1 + k_2 = n$, is equal to
\end{proposition}

\begin{proof}
1) If $n$ is odd, then 
$$
\bar{p}_n =  (z + x)^n = \bar{e}_1^n.
$$
It means that in $p_n$ all coefficients  $A_{k_1,k_2,0}$, where $k_1 \geq k_2 >0$ and $k_1 + k_2 = n$ are zero.

2) If $n=2k$ is even, then
$$
\bar{p}_n =  (z - x)^n = (\bar{e}_1^2 - 4 \bar{e}_2)^k = \sum_{i=0}^k (-4)^i C_k^i (\bar{e}_1^2)^{k-i} (\bar{e}_2)^i.
$$
Hence, we have found the following coefficients in $p_n$.
$$
A_{2k-i,i,0} = (-4)^i C_k^i,~~~i = 1, 2, \ldots, k.
$$
\end{proof}

\textbf{Example.} From this proposition follows that for even $n$ hold\\

$\bar{p}_2 =   \bar{e}_1^2 - 2^2   \,  \bar{e}_2,$\\

$\bar{p}_4 =   \bar{e}_1^4 - 2^3   \,  \bar{e}_1^2 \bar{e}_2 + 2^4   \,  \bar{e}_2^2,$\\

$\bar{p}_6 =   \bar{e}_1^6 - 2^2 \cdot 3   \,  \bar{e}_1^4 \bar{e}_2 +   2^4 \cdot 3   \,  \bar{e}_1^2 \bar{e}_2^2 - 2^6   \,  \bar{e}_2^3,$\\

$\bar{p}_8 =   \bar{e}_1^8 - 2^4   \,  \bar{e}_1^6 \bar{e}_2 +   2^5 \cdot 3   \,  \bar{e}_1^4 \bar{e}_2^2 - 2^8   \,  \bar{e}_1^2 \bar{e}_2^3  + 2^8   \,  \bar{e}_2^4.$


It is easy to see that for even $n$ all coefficients of $\bar{p}_n$ except the coefficient at $\bar{e}_1^n$ are even. This is not true for polynomials $p_n$, as example $p_6$ shows. We can formulate

\textbf{Conjecture.} \emph{1) If} $n = p^m$ \emph{is a power of a prime} $p$, \emph{then all coefficients, except the coefficient at} $e_1^n$ \emph{are divided into} $p$. 
\emph{2) If }$n$\emph{ is even, then all coefficients }$A_{k_1,k_2,k_3}$ \emph{are non-zero.} 





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Newton polyhedron} In this subsection we give a complete answer to the third question in \cite{1}.
Recall the necessary definition. Let 
$$
f = f(x_1, x_2, \ldots, x_n) = \sum a_{i_1 \ldots i_n}  x_1^{i_1} \ldots x_n^{i_n} \in \mathbb{Z}[x_1, x_2, \ldots, x_n]
$$
 be a polynomial with integer coefficients. Denote by $I_f$ the set of multi indexes $(i_1, \ldots, i_n)$ such that $a_{i_1 \ldots i_n} \not= 0$. The convex hull 
$$
N_f = Conv(I_f) \subset \mathbb{R}^n
$$
is said to be a {\it Newton polyhedron} of $f$.

To find Newton polyhedra for polynomials $p_n$, consider them for small $n$,

$p_1 = x + y + z,$\\

$p_2 = x^2 + y^2 + z^2 - 2 x y - 2 y z - 2 z x,$\\

$p_3 =(z + x + y)^3 - 27 x y z,$\\

$p_4 = ((x + y + z)^2 - 4(x y + y z + z x))^2 - 2^7(x + y + z)x y z = p_2^2 - 2^7 p_1 x y z.$\\

Denote by $N_i \subset \mathbb{R}^3$ the Newton polyhedron for $p_i$. Then 

-- $N_1$ is the right triangle $A_1 B_1 C_1$ with the vertices $A_1 = (1, 0, 0)$, $B_1 = (0, 1, 0)$,
$C_1 = (0, 0, 1)$;

-- $N_2$ is the right triangle $A_2 B_2 C_2$ with the vertices $A_2 = (2, 0, 0)$, $B_2 = (0, 2, 0)$,
$C_2 = (0, 0, 2)$;

-- $N_3$ is the right triangle $A_3 B_3 C_3$ with the vertices $A_3 = (3, 0, 0)$, $B_3 = (0, 3, 0)$,
$C_3 = (0, 0, 3)$;

-- $N_4$ is the right triangle $A_4 B_4 C_4$ with the vertices $A_4 = (4, 0, 0)$, $B_4 = (0, 4, 0)$,
$C_4 = (0, 0, 4)$.

To describe $N_k$ for $k >2$ we introduce the next definition.

\begin{definition}
Let $k$ be a positive integer. The {\it standard $n$-simplex of size} $k$ is the subset of $\mathbb{R}^{n+1}$ given by
$$
k \Delta^n = \left\{ (t_0, t_1, \ldots, t_n) \in \mathbb{R}^{n+1}~|~\sum_{i=0}^n t_i = k ~\mbox{and}~t_i \geq 0~ \mbox{for}~i = 0, 1, \ldots, n \right\}.
$$
For simplicity we shall call the standard $n$-simplex of size $k$ by $k \Delta^n$-simplex.
\end{definition}
For $k=1$ we get the definition of the standard $n$-simplex (or unit simplex).

The $k \Delta^n$-simplex has $n+1$ vertices,
$$
E_0 =  (k, 0, 0, \ldots, 0, 0), E_1 =  (0, k, 0,  \ldots, 0, 0), \ldots E_n =  (0, 0, 0, \ldots, 0, k).
$$

Now we are ready to prove the main result of the present subsection.

\begin{theorem} 
 Let $f = f(x_1, x_2, \ldots, x_n) \in \mathbb{Z}[x_1, x_2, \ldots, x_n]$ be a symmetric homogeneous polynomial of degree $k$, which  contains a monomial $a x_1^k$ for some non-zero $a$. Then  its Newton polyhedron $N_f$ is the $k \Delta^{n-1}$-simplex.
\end{theorem} 


\begin{proof}
Since $a x_1^k$ is a monomial of $f$ and $f$ is symmetric, it contains monomials $a x_i^k$ for all $i = 1, 2, \ldots, n$. Hence,   $N_f$ contains the vertices 
$$
E_0 =  (k, 0, 0, \ldots, 0, 0), E_1 =  (0, k, 0,  \ldots, 0, 0), \ldots E_{n-1} =  (0, 0, 0, \ldots, 0, k) \in \mathbb{R}^n
$$
and hence it contains  $k \Delta^{n-1}$-simplex. Let us show that any other vertex of $N_f$, which corresponds a monomial in $f$ lies in this simplex. Indeed, any such monomial has the form 
$$
b x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n},~~b \in \mathbb{R},~~ b \not= 0.
$$
Since 
$$
k_1 + k_2 + \ldots + k_n = k,~~k_i \geq 0~ \mbox{for}~i = 1, 2, \ldots, n,
$$
the corresponding vertex lies in  $k \Delta^{n-1}$-simplex.
\end{proof}

We seen that the polynomial $p_k$ is homogeneous and has the form $p_k = e_1^k + \ldots$. Hence, the answer to the third question of V. M. Buchstaber follows from Theorem 1.


\textbf{Corollary.} \emph{The Newton polyhedron that corresponds to the polynomial }$p_k(x, y, z)$, $k \geq 1$\emph{, is the} $k \Delta^{2}$\emph{-simplex that is a right triangle  with sides of length} $\sqrt[]{2} \, k$.





\bigskip

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Some open questions}

The following questions seem interesting:
\begin{enumerate}
\item Let $f = f(x_1, x_2, \ldots, x_n) \in \mathbb{Z}[x_1, x_2, \ldots, x_n]$ be a symmetric polynomial, $N_f$ is its Newton polyhedron. Let us present $f$ as a polynomial in elementary symmetric polynomials, $f =F[e_1, e_2, \ldots, e_n] \in \mathbb{Z}[e_1, e_2, \ldots, e_n]$ and construct  its Newton polyhedron $N_F$. What is the relation between $N_f$ and $N_F$?

\item Let $f = f(x_1, x_2, \ldots, x_n) \in \mathbb{Z}[x_1, x_2, \ldots, x_n]$ be a symmetric polynomial of degree $k$, which does not contain $x_1^k$. What can we say about its Newton polyhedron $N_f$?


\item Since there is a homomorphism of one multivalued group to another multivalued group and the kernel of this homomorphism is defined, we can talk about extensions of multivalued groups. Construct a theory of extensions of multivalued groups. 

\item Is it possible to define (co)homology for  multivalued groups?

\end{enumerate}






\textbf{Acknowledgments.} The authors thank  V.~M.~Buchstaber and D.~V.~Talalaev for interesting discussions and useful suggestions.

\medskip





\bigskip
\begin{thebibliography}{1} % Ссылки на литературу приводятся только на английском языке (даже если у источника нет англоязычной версии).


\bibitem{1} Buchstaber, V. M.
\href{}{\it $n$-valued groups: theory and applications}, Moscow Math. J., 6, no. 1 (2006), 57--84.

\bibitem{2} V. M. Buchstaber, S. P. Novikov,  
\href{}{\it Formal groups, power systems and Adams operators}, Russian Mat. Sb. (N.S.), {\bf 84 (126)}, 1971, 81--118.


\end{thebibliography}
\end{document}
