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\begin{document}

\title{On simple $15$-dimensional Lie algebras in characteristic $2$}

\author{Alexander Grishkov}
\address[Alexander Grishkov]{University of S\~ao Paulo, Brazil and 
Omsk F.M. Dostoevsky State University, Russia}
\email{shuragri@gmail.com}

\author{Henrique Guzzo Jr.}
\address[Henrique Guzzo Jr.]{University of S\~ao Paulo, Brazil}
\email{guzzo@ime.usp.br}

\author{Marina Rasskazova}
\address[Marina Rasskazova]{Omsk State Technical University, Russia}
\email{marinarasskazova@yandex.ru}

\author{Pasha Zusmanovich}
\address[Pasha Zusmanovich]{University of Ostrava, Czech Republic}
\email{pasha.zusmanovich@gmail.com}

\date{First written March 26, 2021; last revised November 26, 2021}
\thanks{J. Algebra, to appear}

\keywords{Simple Lie algebra; characteristic $2$; tori; automorphisms; 
thin decomposition; GAP}
\subjclass[2020]{17B20; 17B40; 17B50; 17-04}

\begin{abstract}
Motivated by the recent progress towards classification of simple 
finite-dimensional Lie algebras over an algebraically closed field of 
characteristic $2$, we investigate such $15$-dimensional Skryabin algebras.
\end{abstract}

\maketitle

\section*{Introduction}

The classification of finite-dimensional simple Lie algebras over an 
algebraically closed field has a long and interesting history. The 
characteristic zero case, due to Killing, Cartan, and Dynkin, is nowadays a 
classic. The case of characteristic $p>3$ was accomplished relatively 
recently, due to the efforts of dozens of people, spread over more than 50 years
and hundreds of papers, and culminated in the three volumes by Strade, 
\cite{strade}. The cases of characteristics $2$ and $3$ remain open, although a
lot of efforts were done recently to augment the classification program by 
small characteristics specifics, in particular, to put Lie superalgebras into 
play (see, for example, the monumental treatises \cite{leites-and-co} and 
\cite{blls}, and references therein).

In \cite{GZ}, we did a small step towards the classification of 
finite-dimensional simple Lie algebras over an algebraically closed field of 
characteristic $2$: it was proved that any such algebra of absolute toral rank 
$2$, and having a Cartan subalgebra of toral rank one, is $3$-dimensional. In 
this, we relied on the paper by Skryabin, \cite{skryabin}, where it was proved, 
among other things, that in characteristic $2$ there are no simple Lie algebras
of absolute toral rank $1$, and simple Lie algebras having a Cartan subalgebra 
of toral rank $1$ were characterized as certain filtered deformations of 
semisimple Lie algebras with the socle of the form $S \otimes \mathcal O$, where
$S$ is a simple Lie algebra either of Zassenhaus or Hamiltonian type, and 
$\mathcal O$ is the algebra of truncated polynomials.

In the process of the proof of the main result of \cite{GZ}, we have constructed
a $2$-parameter family $\mathscr L(\beta,\delta)$ of $15$-dimensional simple 
Lie algebras (\cite[\S 6]{GZ}). The algebra $\mathscr L(0,0)$ in this family 
coincides with the smallest algebra in the series constructed by Skryabin 
(\cite[Example at pp. 691--692]{skryabin}).

The purpose of this paper is to study the family $\mathscr L(\beta,\delta)$.
Among other things, we prove that all algebras within the family are isomorphic
to the same algebra $\mathscr L$ (\S \ref{sec-2param}), and we determine the 
absolute toral rank (\S \ref{sec-subalg}), and the automorphism group of 
$\mathscr L$ (\S \ref{sec-aut}). In passing, we introduce the notion of a thin 
decomposition of a simple Lie algebra with respect to a torus (\S \ref{ss-thin})
which, we suggest, should play a role in classification efforts (see open 
questions in \S \ref{s-quest}).

Throughout the paper, the ground field $K$ is assumed to be perfect of 
characteristic $2$, unless stated otherwise. Although the initial problem 
assumes the ground field is algebraically closed, for our purposes it will be 
enough to assume that square roots exist in $K$. This allows us to include in 
our consideration the case $K = \GF(2)$; this can be useful in some 
circumstances (cf. \S \ref{sec-subalg} and \S \ref{ss-eick}).

Our terminology and notation is mostly standard: $C_L(X)$ denotes the 
centralizer of a set $X$ in a Lie algebra $L$; the linear span of a set $X$ is 
denoted either by $KX$ or by $\langle X \rangle$; the $2$-envelope of a Lie 
algebra $L$ is denoted by $L_2$; $\id_L$ denotes the identity map on $L$. Other
notation is explained as soon as it is introduced in the text.





\section{The $2$-parameter family $\mathscr L(\beta,\delta)$}\label{sec-2param}

\subsection{Definition of the $2$-parameter family}

Recall the definition of the family $\mathscr L(\beta,\delta)$, depending of two
parameters $\beta, \delta \in K$, of $15$-di\-men\-si\-o\-nal simple Lie 
algebras constructed in \cite[\S 6]{GZ}. These algebras are filtered 
deformations of the semisimple Lie algebra of the form
\begin{equation}\label{eq-ss}
\ess \otimes \mathcal O_1(2) + g \otimes \langle 1,x \rangle + \partial , 
\end{equation}
where $\ess$ is the $3$-dimensional simple Lie algebra with the basis 
$\{e,f,h\}$ and multiplication table
\begin{equation}\label{eq-mult}
[e,h] = e, \quad [f,h] = f, \quad [e,f] = h ,
\end{equation}
$g = (\ad f)^2$ is an outer derivation of $\ess$, $\mathcal O_1(2)$ is the 
$4$-dimensional divided power algebra with the basis $\{1,x,x^{(2)},x^{(3)}\}$,
with the multiplication given by
$$
x^{(i)} x^{(j)} = \binom{i+j}{i} x^{(i+j)} ,
$$
and $\partial: x^{(i)} \mapsto x^{(i-1)}$ is the special derivation of 
$\mathcal O_1(2)$.

For our purpose, it will be convenient to relabel the basis elements of the
algebras $\mathscr L(\beta,\delta)$ as follows:
\begin{alignat}{7}\label{eq-b}
b_1 &= e \otimes 1 ,\qquad  &c_1 &= e \otimes x^{(3)} , \notag \\
b_2 &= f \otimes 1 ,\qquad  &c_2 &= f \otimes x^{(3)} , \notag \\
b_3 &= h \otimes 1 ,\qquad  &c_3 &= h \otimes x^{(3)} , \notag \\
b_4 &= e \otimes x ,\qquad  &c_4 &= g \otimes 1       , \notag \\
b_5 &= f \otimes x ,\qquad  &c_5 &= g \otimes x       ,        \\
b_6 &= h \otimes x ,\qquad  &d &= \partial            , \notag \\
b_7 &= e \otimes x^{(2)}                              , \notag \\
b_8 &= f \otimes x^{(2)}                              , \notag \\
b_9 &= h \otimes x^{(2)}                              . \notag
\end{alignat}

In terms of this basis, the multiplication table of $\mathscr L(\beta,\delta)$
(see \cite[(5.4)]{GZ}) reads:

\begin{equation}\label{eq-m}
\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline
&b_2&b_3&b_4&b_5&b_6&b_7&b_8&b_9&c_1&c_2&c_3&c_4&c_5&d \\ \hline\hline
b_1&b_3&b_1&\beta c_5&b_6&b_4&\delta c_4&b_9&b_7&\delta c_5 + d&c_3&c_1&b_2&b_5&\beta c_2 
\\ \hline
b_2&&b_2&b_6&0&b_5&b_9&0&b_8&c_3&0&c_2&0&0&0            \\ \hline
b_3&&&b_4&b_5&0&b_7&b_8&0&c_1&c_2&0&0&0&0               \\ \hline
b_4&&&&0&0&\delta c_5 + d&c_3&c_1&b_3&c_4&b_2&b_5&0&b_1 \\ \hline
b_5&&&&&0&c_3&0&c_2&c_4&0&0&0&0&b_2                     \\ \hline
b_6&&&&&&c_1&c_2&0&0&0&c_4&0&0&b_3                      \\ \hline
b_7&&&&&&&0&0&b_6&c_5&b_5&b_8&c_2&b_4                   \\ \hline
b_8&&&&&&&&0&c_5&0&0&0&0&b_5                            \\ \hline
b_9&&&&&&&&&0&0&c_5&0&0&b_6                             \\ \hline
c_1&&&&&&&&&&0&b_8&c_2&0&b_7                            \\ \hline
c_2&&&&&&&&&&&0&0&0&b_8                                 \\ \hline
c_3&&&&&&&&&&&&0&0&b_9                                  \\ \hline
c_4&&&&&&&&&&&&&0&0                                     \\ \hline
c_5&&&&&&&&&&&&&&c_4                                    \\ \hline
\end{array}
\end{equation}

\medskip


\subsection{Isomorphisms within the family}

\begin{lemma}\label{lemma-delta}
For any $\beta,\delta \in K$, we have 
$\mathscr L(\beta,\delta) \simeq \mathscr L(\beta,0)$.
\end{lemma}

\begin{proof}
Let us consider a new basis 
$$
\{
b_1^\prime, b_2, b_3^\prime, b_4^\prime, b_5, b_6, b_7^\prime, b_8, b_9^\prime,
c_1^\prime, c_2, c_3, c_4, c_5, d^\prime
\}
$$
of the algebra $\mathscr L(\beta,\delta)$, where
\begin{align*}
b_1^\prime &= b_1 + \sqrt{\delta} b_6 + \delta b_8 ,\\
b_3^\prime &= b_3 + \sqrt{\delta} b_5              ,\\
b_4^\prime &= b_4 + \delta c_2                     ,\\
b_7^\prime &= b_7 + \sqrt{\delta} c_3              ,\\
b_9^\prime &= b_9 + \sqrt{\delta} c_2              ,\\
c_1^\prime &= c_1 + \sqrt{\delta} c_4              ,\\
d^\prime   &= d + \sqrt{\delta} b_2 .
\end{align*}

It is straightforward to check that in this basis the multiplication table 
coincides with the multiplication table (\ref{eq-m}) with $\delta = 0$. 
\end{proof}


\begin{lemma}\label{lemma-beta}
For any $\beta \in K$, we have $\mathscr L(\beta,0) \simeq \mathscr L(0,0)$.
\end{lemma}

\begin{proof}
Consider a new basis
$$
\{
b_1^\prime, b_2, b_3^\prime, b_4^\prime, b_5, b_6^\prime, b_7, b_8, b_9, 
c_1^\prime, c_2, c_3, c_4, c_5, d^\prime
\}
$$
of the algebra $\mathscr L(\beta,0)$, where
\begin{align*}
b_1^\prime &= b_1 + \beta^2 b_9  ,\\
b_3^\prime &= b_3 + \beta^2 b_8  ,\\
b_4^\prime &= b_4 + \beta^2 c_3  ,\\
b_6^\prime &= b_6 + \beta^2 c_2  ,\\
c_1^\prime &= b_1 + \beta^2 c_5  ,\\
d^\prime   &= d + \beta^2 b_5 .
\end{align*}

It is straightforward to check that in this basis the multiplication table 
coincides with the multiplication table (\ref{eq-m}) with $\beta = \delta = 0$.
\end{proof}

An immediate corollary of these two lemmas is

\begin{theorem}\label{t1}
For any $\beta,\delta \in K$, we have 
$\mathscr L(\beta,\delta) \simeq \mathscr L(0,0)$.
\end{theorem}

Since the algebra $\mathscr L = \mathscr L(0,0)$ is isomorphic to the smallest 
$15$-dimensional algebra in the family of simple Lie algebras constructed by 
Skryabin in \cite[pp.691--692]{skryabin}, we will refer to it in the rest of the
paper as the \emph{Skryabin algebra}. The rest of the paper is devoted to
elucidation of some properties, and computation of some invariants of the 
Skryabin algebra.


\subsection{Note on cohomology and deformations}

As explained in \cite[\S 5]{GZ}, the algebras $\mathscr L(\beta,\delta)$ are
obtained as a linear deformation of the algebra (\ref{eq-ss}) by a linear 
combination of three $2$-cocycles; one of them has to enter the linear 
combination with nonzero coefficient (to ensure simplicity of the deformed 
algebra) which can be normalized to be equal to $1$, and the other two enter 
with coefficients $\beta$ and $\delta$. In terms of the basis (\ref{eq-b}), 
these $2$-cocycles are:
\begin{equation}\label{eq-2coc}
\begin{aligned}
&b_1 \wedge b_4 &\mapsto c_5 ,\\
&b_1 \wedge d   &\mapsto c_2 \phantom{,}
\end{aligned}
\end{equation}
and
\begin{equation}\label{eq-2coc1}
\begin{aligned}
&b_1 \wedge b_7 &\mapsto c_4  ,\\
&b_4 \wedge b_7 &\mapsto c_5  ,\\
&b_1 \wedge c_1 &\mapsto c_5  \phantom{,}
\end{aligned}
\end{equation}
(only nonzero values on all possible pairs of the basis elements are given).

Either direct calculations, or simple reasonings based on the relationship 
between cohomology of the graded algebra (\ref{eq-ss}) and its filtered 
deformation $\mathscr L$  (the Massey bracket between all the involved 
$2$-cocycles vanishes) show that the cocycles (\ref{eq-2coc}) and 
(\ref{eq-2coc1}) are also cohomologically independent $2$-cocycles of the 
algebra $\mathscr L$ (note in passing that the computer calculation in GAP shows
that the dimension of the whole cohomology $\Homol^2(\mathscr L, \mathscr L)$ is
equal to $13$). Thus, each $\mathscr L(\beta,\delta)$ is a deformation of the 
algebra $\mathscr L$ by a nontrivial $2$-cocycle, which turns out to be isomorphic to the original
algebra $\mathscr L$ via ad-hoc isomorphism. Such deformations are dubbed in
\cite{blls} (see also \cite[\S 3.1.2]{leites-and-co}, and references therein)
as \emph{semitrivial}. According to Lemmas \ref{lemma-delta} and 
\ref{lemma-beta}, the isomorphism between $\mathscr L(\beta,\delta)$ and 
$\mathscr L$ is polynomial (actually, quadratic) in $\beta$ and $\sqrt{\delta}$;
hence by \cite[Lemma 2.2]{blls}, each of the cocycles (\ref{eq-2coc}) and 
(\ref{eq-2coc1}) is cohomologous to a cocycle of the form 
$x \wedge y \mapsto [D(x),D(y)]$ for some derivation $D$ of $\mathscr L$.





\section{$2$-envelope, derivations, sandwich subalgebra}\label{sec-misc}

We continue to employ the basis $\{b_1,\dots,b_9,c_1,\dots,c_5,d\}$ of the 
Skryabin algebra $\mathscr L$ as given by (\ref{eq-b}) (referred as the 
\emph{standard basis} in what follows). The multiplication table is given by 
(\ref{eq-m}), where $\beta = \delta = 0$.

The Skryabin algebra is not a $2$-algebra; its $2$-envelope $\mathscr L_2$ has 
dimension $19$, with additional basis elements 
$b_1^{[2]}, b_4^{[2]}, b_7^{[2]}, c_3^{[2]}$, the $2$-map:
\begin{gather*}
b_2^{[2]} = c_4 ,\quad
b_3^{[2]} = b_3 ,\quad
c_1^{[2]} = b_9 ,\quad
d^{[2]} = b_4^{[2]} ,
\\
b_5^{[2]} = b_6^{[2]} = b_8^{[2]} = b_9^{[2]} = c_2^{[2]} = c_4^{[2]} = 
c_5^{[2]} = 0 ,
\\
b_7^{[4]} = b_7^{[2]} ,\quad b_1^{[4]} = b_4^{[4]} = c_3^{[4]} = 0 ,
\end{gather*}
and the multiplication: 
\begin{equation*}
\begin{array}[t]{|c||c|c|c|c|c|c|}\hline
    & b_1^{[2]} & b_4^{[2]} & b_7^{[2]} & c_3^{[2]} \\ \hline\hline
b_1 & 0         & 0         & 0         & b_8  \\ \hline
b_2 & b_1       & 0         & 0         & 0    \\ \hline
b_3 & 0         & 0         & 0         & 0    \\ \hline
b_4 & 0         & 0         & b_4       & c_2  \\ \hline
b_5 & b_4       & 0         & b_5       & 0    \\ \hline
b_6 & 0         & 0         & b_6       & 0    \\ \hline
b_7 & 0         & b_1       & 0         & 0    \\ \hline
b_8 & b_7       & b_2       & 0         & 0    \\ \hline
b_9 & 0         & b_3       & 0         & 0    \\ \hline
c_1 & 0         & b_4       & c_1       & 0    \\ \hline
c_2 & c_1       & b_5       & c_2       & 0    \\ \hline
c_3 & d         & b_6       & c_3       & 0    \\ \hline
c_4 & b_3       & 0         & 0         & 0    \\ \hline
c_5 & b_6       & 0         & c_5       & 0    \\ \hline
d   & 0         & 0         & d         & c_5  \\ \hline
\end{array}
\qquad
\begin{array}[t]{|c||c|c|c|} \hline
          & b_4^{[2]} & b_7^{[2]} & c_3^{[2]} \\ \hline\hline
b_1^{[2]} & 0         & 0         & b_9       \\ \hline
b_4^{[2]} &           & 0         & c_4       \\ \hline
b_7^{[2]} &           &           & 0         \\ \hline
\end{array}
\end{equation*}

\bigskip

In what follows, we will employ the following standard notation for arbitrary
elements of $\mathscr L$:
\begin{align}
&\lambda_1 b_1 + \dots + \lambda_9 b_9 + \mu_1 c_1 + \dots + \mu_5 c_5 + \eta d
,
\notag
\intertext{and of $\mathscr L_2$:}
&\lambda_1 b_1 + \dots + \lambda_9 b_9 + \mu_1 c_1 + \dots + \mu_5 c_5 + \eta d
+ \xi_1 b_1^{[2]} + \xi_4 b_4^{[2]} + \xi_7 b_7^{[2]} + \xi_3 c_3^{[2]} ,
\label{eq-genel}
\end{align}
where 
$\lambda_1,\dots,\lambda_9,\mu_1,\dots,\mu_5,\eta,\xi_1,\xi_4,\xi_7,\xi_3 
\in K$.


\begin{proposition}\label{prop-der}
The derivation algebra of the Skryabin algebra coincides with its $2$-envelope.
\end{proposition}

\begin{proof}
This is amenable to straightforward, though tedious, calculations. A short 
computer-assisted proof can be utilized instead: we can check in GAP that the 
derivation algebra of the Skryabin algebra over $\GF(2)$ is $19$-dimensional. 
Since derivation algebra does not change under field extensions, the same is 
true for the Skryabin algebra over an arbitrary field. For any Lie algebra, its 
$2$-envelope is contained in the derivation algebra. But the $2$-envelope of
the Skryabin algebra is $19$-dimensional, as specified above. Hence, the 
derivation algebra coincides with the $2$-envelope.

Yet another, more conceptual, proof will be given below in \S \ref{ss-thin}; it
utilizes a different realization of the Skryabin algebra.
\end{proof}


Recall that an element $x$ of a Lie algebra $L$ is called a \emph{sandwich} if 
$(\ad x)^2 = 0$ and $[[L,x],[L,x]] = 0$. (It is well-known -- and easy to see --
that if the characteristic of the ground field is different from $2$, then the 
second condition follows from the first one, but in characteristic $2$ this is 
not true). The set of all sandwiches is multiplicatively closed; this implies 
that the \emph{sandwich subalgebra}, i.e., the subalgebra of $L$ generated by 
all sandwiches, is just the linear span of sandwiches.

It follows from the result first proved by Kostrikin and Zelmanov, that a Lie 
algebra generated by sandwiches is nilpotent. See \cite[\S 3.2]{vaughan-lee} for
details and further references.

More generally, we will call a derivation $D$ of a Lie algebra $L$ a 
\emph{sandwich derivation}, if $D^2 = 0$ and $[D(L),D(L)]$ = 0.

\begin{lemma}\label{prop-sand}
The set of elements $x \in \mathscr L_2$ such that
\begin{equation}\label{eq-lx}
[[\mathscr L,x],[\mathscr L,x]] = 0
\end{equation}
coincides with the linear span of elements $c_2$, $c_4$, $c_5$, $c_3^{[2]}$.
\end{lemma}

\begin{proof}
It is a matter of straightforward verification that any linear span of elements
$c_2$, $c_4$, $c_5$, $c_3^{[2]}$ satisfies the condition (\ref{eq-lx}).

Conversely, let $x$ be an arbitrary element (\ref{eq-genel}) of $\mathscr L_2$ 
satisfying this condition. We perform the following calculations:
\begin{itemize}

\item
Collecting in the equality $[[x,b_8],[x,c_5]] = 0$ the terms containing $c_1$ 
and $c_2$, we get, respectively, $\xi_1^2 = 0$ and $\lambda_1^2 + \xi_1\xi_7 = 0$,
whence $\xi_1 = 0$ and $\lambda_1 = 0$. 

\item
Collecting in the equality $[[x,b_6],[x,c_4]] = 0$ the terms containing $c_5$, 
we get $\lambda_7^2 = 0$, whence $\lambda_7 = 0$.

\item
Collecting in the equality $[[x,b_1],[x,c_4]] = 0$ the terms containing $b_8$, 
we get $\mu_1^2 = 0$, whence $\mu_1 = 0$. 

\item
Collecting in the equality $[[x,b_6],[x,c_2]] = 0$ the terms containing $b_8$, 
we get $\eta^2 = 0$, whence $\eta = 0$.

\item
Collecting in the equality $[[x,b_8],[x,c_3]] = 0$ the terms containing $b_5$ 
and $c_2$, we get, respectively, $\xi_4^2 = 0$ and 
$\lambda_4^2 + \lambda_3\xi_4 = 0$, whence $\xi_4 = 0$ and $\lambda_4 = 0$.

\item
Collecting in the equality $[[x,b_1],[x,b_8]] = 0$ the terms containing $b_9$ 
and $c_4$, we get, respectively, $\lambda_3^2 = 0$ and $\lambda_6^2 = 0$, whence
$\lambda_3 = 0$ and $\lambda_6 = 0$.

\item
Collecting in the equality $[[x,b_4],[x,c_2]] = 0$ the terms containing $c_4$, 
we get $\xi_7^2 = 0$, whence $\xi_7 = 0$.

\item
Collecting in the equality $[[x,b_1],[x,b_6]] = 0$ the terms containing $b_5$ 
and $c_2$, we get, respectively, $\lambda_2^2 = 0$ and $\mu_3^2 = 0$, whence 
$\lambda_2 = 0$ and $\mu_3 = 0$.

\item
Collecting in the equality $[[x,b_1],[x,c_3]] = 0$ the terms containing $c_2$, 
we get $\lambda_9^2 = 0$, whence $\lambda_9 = 0$.

\item
Collecting in the equality $[[x,b_1],[x,b_4]] = 0$ the terms containing $c_5$, 
we get $\lambda_8^2 = 0$, whence $\lambda_8 = 0$.

\item
Collecting in the equality $[[x,b_1],[x,b_7]] = 0$ the terms containing $c_4$, 
we get $\lambda_5^2 = 0$, whence $\lambda_5 = 0$.

\end{itemize}

We are left with a linear combination of $c_2$, $c_4$, $c_5$, $c_3^{[2]}$.
\end{proof}

As an immediate corollary of this lemma, we have

\begin{proposition}\label{prop-sandr}\hfill
\begin{enumerate}[\upshape(i)]
\item\label{it-sk}
The sandwich subalgebra of $\mathscr L$ is $3$-dimensional abelian, linearly
spanned by elements $c_2$, $c_4$, $c_5$.

\item\label{it-der}
The set of sandwich derivations of $\mathscr L$ forms a $4$-dimensional abelian
subalgebra of $\mathscr L_2$ spanned by the inner derivations $\ad c_2$, 
$\ad c_4$, $\ad c_5$, and by the outer derivation $\ad c_3^{[2]}$.
\end{enumerate}
\end{proposition}

\begin{proof}
(i) By Lemma \ref{prop-sand}, the sandwich subalgebra of $\mathscr L$ lies in
the linear span of $c_2$, $c_4$, $c_5$. Since these elements pairwise commute,
they span the $3$-dimensional abelian subalgebra of $\mathscr L$, and 
$$
(\mu_1 c_2 + \mu_4 c_4 + \mu_5 c_5)^{[2]} = 
\mu_1^2 c_2^{[2]} + \mu_4^2 c_4^{[2]} + \mu_5^2 c_5^{[2]} = 0 ,
$$
so any element $x$ in this $3$-dimensional subalgebra satisfies the condition
$(\ad x)^2 = 0$.

(ii) By Proposition \ref{prop-der}, any derivation is an element of 
$\mathscr L_2$. Apply Lemma \ref{prop-sand} and reason as above.
\end{proof}


Note that from Lemma \ref{prop-sand} and Proposition \ref{prop-sandr} it follows
that in the Skryabin algebra $\mathscr L$, the condition 
$[[\mathscr L,x],[\mathscr L,x]] = 0$ implies $[[\mathscr L,x],x] = 0$. In 
general, this is, of course, not true: take, for example, any metabelian 
non-nilpotent Lie algebra.





\section{Computations over $\GF(2)$, the absolute toral rank, 
thin decomposition}\label{sec-subalg}

\subsection{Some numerology}\label{ss-num}

In this section we report on computations over $\GF(2)$, performed on the 
computer in GAP, \cite{gap}\footnote{
The GAP code is available at 
\texttt{https://web.osu.cz/$\sim$Zusmanovich/papers/15dim/} .
}.
A brute-force search on the computer shows that in the $2$-envelope of the 
Skryabin algebra $\mathscr L$ over $\GF(2)$ there are 384 toral elements, $6144$
$2$-dimensional tori, 21504 $3$-dimensional tori, 26880 $4$-dimensional tori, 
and no $5$-dimensional tori (no attempt was made to determine their conjugacy 
classes with respect to the automorphism group). 

The centralizer in $\mathscr L$ of each of the $384$ toral elements is 
$7$-dimensional, and for $240$ toral elements the centralizer is (central) 
simple. Among these $240$ simple algebras, $48$ have absolute toral rank $2$,
and $192$ have absolute toral rank $3$. As proved in \cite[Theorem 1]{vl-comp}
(and confirmed by computations in \cite{eick}), over $\GF(2)$ there exist two 
simple $7$-dimensional Lie algebras. These algebras are identified as (forms) of
the Zassenhaus algebra $W_1^{\prime}(3)$, and the Hamiltonian algebra 
$H_2^{\prime\prime}\big((2,1),(1 + x_1^{(3)}x_2)\, dx_1 \wedge dx_2\big)$, 
denoted by us here simply as $W$ and $H$, respectively (see \cite{gg-2010} and 
\cite{gga} for further info, including explicit multiplication tables of these 
algebras, and their identification with some other simple $7$-dimensional Lie 
algebras from the literature). Both algebras have absolute toral rank $3$ over 
an algebraically closed field, but over $\GF(2)$, the algebra $W$ has absolute 
toral rank $2$, while the algebra $H$ has absolute toral rank $3$; thus the 
absolute toral rank can be used to distinguish them as subalgebras of the Skryabin algebra in our 
computations.


\subsection{Some examples}

Let us exhibit explicitly one of the maximal tori, and one of the 
$7$-di\-men\-si\-o\-nal simple subalgebras mentioned in the previous subsection.

Here is just one of the $4$-dimensional tori, linearly spanned by the toral 
elements
\begin{equation}\label{eq-tor}
b_1 + b_3 + b_1^{[2]} , \quad
b_2 + b_3 + c_4 + b_7^{[2]} , \quad 
b_4 + b_6 + b_4^{[2]} + b_7^{[2]} , \quad
b_8 + b_9 + c_1 + c_3 + b_7^{[2]} + c_3^{[2]} .
\end{equation}

Now take the first toral element in this torus, $h = b_1 + b_3 + b_1^{[2]}$. Its
centralizer $C_{\mathscr L}(h)$ has the basis
$$
b_1 + b_3       ,\quad 
b_2 + c_3 + c_4 ,\quad 
b_4 + b_6       ,\quad
b_5 + b_6 + c_5 ,\quad
b_7 + b_9       ,\quad 
c_1 + c_3       ,\quad 
d .
$$

It is straightforward to check that in this basis the multiplication table of
$C_{\mathscr L}(h)$ coincides with the multiplication table of the simple
$7$-dimensional Hamiltonian algebra $H$ (see \cite[\S 3]{gg-2010} or 
\cite[\S 1]{gga}) via the identification
\begin{gather*}
V_0 \leftrightarrow b_5 + b_6 + c_5 ,\quad
V_1 \leftrightarrow c_1 + c_3       ,\quad 
E_1 \leftrightarrow b_7 + b_9       ,\quad
E_0 \leftrightarrow b_2 + c_3 + c_4 ,\quad
\\
F_1 \leftrightarrow b_4 + b_6       ,\quad
F_0 \leftrightarrow d               ,\quad
G   \leftrightarrow b_1 + b_3 .
\end{gather*}



\subsection{The absolute toral rank}

The computer calculation in \S \ref{ss-num} shows that the absolute toral rank 
of the Skryabin algebra over $\GF(2)$ is equal to $4$. However, a bit of 
additional work allows to establish this result over an arbitrary field.

\begin{theorem}
The absolute toral rank of the Skryabin algebra is equal to $4$.
\end{theorem}

\begin{proof}
As we want to establish this result over an arbitrary field $K$, we will 
distinguish the Skryabin algebra 
$\overline{\mathscr L} = \mathscr L \otimes_{\GF(2)} K$ over $K$, and its 
$\GF(2)$-form $\mathscr L$.

A direct computer verification shows that each of the 26880 $4$-dimensional tori
$T$ in $\mathscr L_2$ coincides with its normalizer, i.e., is a Cartan 
subalgebra of $\mathscr L_2$. Consequently, $\overline T = T \otimes_{\GF(2)} K$
is a Cartan subalgebra of $\overline{\mathscr L_2} = (\overline{\mathscr L})_2$.
By \cite[Theorem 2(ii)]{premet-cartan}, $\overline T$ is a torus of the maximal
possible dimension in $\overline{\mathscr L_2}$, and hence the absolute toral 
rank of $\overline{\mathscr L}$ is equal to $4$.

Alternatively, a direct simple proof free from reference to the computer can be
provided by just looking at one of the $4$-dimensional tori, for example, at 
(\ref{eq-tor}). Indeed, a direct calculation shows that (\ref{eq-tor}) coincides
with its normalizer in the $2$-envelope of the Skryabin algebra 
$\overline{\mathscr L}$ over an arbitrary ground field $K$, and hence by the 
same reference to \cite{premet-cartan} is a torus of the maximal possible 
dimension.
\end{proof}




\subsection{Thin decomposition}\label{ss-thin}

Let us consider the following situation. Assume that a Lie algebra $L$ has
dimension $2^n-1$, the absolute toral rank of $L$ is $n$, $T$ is a torus in the
$2$-envelope of $L$ of (the maximal) dimension $n$, and $T \cap L = 0$. Assume 
further that the roots of the action of $T$ on $L$ are exactly nonzero tuples in
$\GF(2)^n$ (in particular, the centralizer of $T$ in $L$ is zero), and each root
space is one-dimensional. Thus, the root space decomposition is of the form
\begin{equation}\label{eq-thin}
L = \bigoplus_{\substack{\alpha \in \GF(2)^n \\ \alpha \ne (0,\dots,0)}} 
K e_\alpha
\end{equation}
for some elements $e_\alpha \in L$. In particular, the multiplication table of 
$L$ in the basis $\set{e_{\alpha}}{\alpha \in \GF(2)^n \backslash (0,\dots,0)}$
has the following form: either $[e_\alpha,e_\beta] = e_{\alpha + \beta}$, or 
$[e_\alpha,e_\beta] = 0$.

In such a situation, we will call the decomposition (\ref{eq-thin}) \emph{thin}.
We suggest that this is an 
important property of simple Lie algebras which should be taken into account in
the classification efforts. 

A direct computer verification shows that the Skryabin algebra admits a thin 
decomposition with respect to \emph{each} of the 26880 $4$-dimensional tori in 
its $2$-envelope. We will explicitly provide one of them, corresponding to the 
torus (\ref{eq-tor}).

The corresponding generators of the one-dimensional root spaces are:
\begin{align*}
&e_{0001} = b_2 + b_3 + b_4 + b_6 + c_4                       \\
&e_{0010} = b_2 + b_3 + c_1 + c_3 + c_4                       \\
&e_{0011} = b_2 + b_3 + b_4 + b_6 + c_1 + c_3 + c_4           \\
&e_{0100} = b_1 + b_3 + b_7 + b_9 + d                         \\
&e_{0101} = b_7 + b_9 + d                                     \\
&e_{0110} = b_1 + b_3 + b_5 + b_6 + c_5 + d                   \\
&e_{0111} = b_5 + b_6 + c_5                                   \\
&e_{1000} = b_2 + b_3 + b_8 + b_9 + c_1 + c_2 + c_3           \\
&e_{1001} = b_2 + b_3 + b_4 + b_5 + b_6                       \\
&e_{1010} = b_2 + b_3 + c_1 + c_2 + c_3                       \\
&e_{1011} = b_2 + b_3 + b_4 + b_5 + b_6 + c_1 + c_2 + c_3     \\
&e_{1100} = b_1 + b_2 + b_3 + b_7 + b_8 + b_9 + c_2 + c_3 + d \\
&e_{1101} = b_5 + b_6 + b_7 + b_8 + b_9 + c_2 + c_3 + d       \\
&e_{1110} = b_1 + b_2 + b_3 + b_5 + b_6 + c_2 + c_3 + d       \\
&e_{1111} = b_5 + b_6
\end{align*}
and the corresponding multiplication table of $\mathscr L$ reads:

{\small
\begin{equation*}
\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline
& e_{0010} & e_{0011} & e_{0100} & e_{0101} & e_{0110} & e_{0111} & e_{1000} & 
e_{1001} & e_{1010} & e_{1011} & e_{1100} & e_{1101} & e_{1110} & e_{1111} 
\\ \hline\hline
e_{0001} & e_{0011} & e_{0010} & e_{0101} & e_{0100} & 0 & 0 & e_{1001} & 0 &
e_{1011} & e_{1010} & e_{1101} & e_{1100} &e_{1111} & 0
\\ \hline
e_{0010} && e_{0001} & e_{0110} & e_{0111} & e_{0100} & 0 & 0 & e_{1011} & 0 & 
e_{1001} & e_{1110} & e_{1111} & e_{1100} & 0            
\\ \hline
e_{0011} &&& e_{0111} & e_{0110} & e_{0101} & 0 & e_{1011} & e_{1010} & e_{1001}
& 0 & 0 & e_{1110} & e_{1101} & 0
\\ \hline
e_{0100} &&&& 0 & e_{0010} & e_{0011} & e_{1100} & e_{1101} & e_{1110} & 0 & 
e_{1000} & e_{1001} & 0 & e_{1011} 
\\ \hline
e_{0101} &&&&& e_{0011} & e_{0010} & 0 & e_{1100} & e_{1111} & e_{1110} & 0 & 
e_{1000} & e_{1011} & e_{1010}
\\ \hline
e_{0110} &&&&&& e_{0001} & e_{1110} & e_{1111} & e_{1100} & e_{1101} & 0 & 0 & 
e_{1000} & e_{1001}
\\ \hline
e_{0111} &&&&&&& 0 & 0 & 0 & 0 & e_{1011} & e_{1010} & e_{1001} & 0
\\ \hline
e_{1000} &&&&&&&& e_{0001} & 0 & e_{0011} & e_{0100} & 0 & e_{0110} & 0
\\ \hline
e_{1001} &&&&&&&&& e_{0011} & e_{0010} & e_{0101} & e_{0100} & 0 & 0
\\ \hline
e_{1010} &&&&&&&&&& e_{0001} & e_{0110} & e_{0111} & e_{0100} & 0
\\ \hline
e_{1011} &&&&&&&&&&& e_{0111} & e_{0110} & e_{0101} & 0
\\ \hline
e_{1100} &&&&&&&&&&&& e_{0001} & e_{0010} & e_{0011}
\\ \hline
e_{1101} &&&&&&&&&&&&& 0 & e_{0010}
\\ \hline
e_{1110} &&&&&&&&&&&&&& e_{0001}
\\ \hline
\end{array}
\end{equation*}
}

Using this realization of the Skryabin algebra, it is possible to provide an 
alternative proof of the fact that its derivation algebra coincides with its 
$2$-envelope, not utilizing computer or tedious calculations.

\begin{proof}[Another proof of Proposition \ref{prop-der}]
Consider the torus $T$, and the corresponding  thin decomposition of 
$\mathscr L$ as above. We shall prove that the derivation algebra of 
$\mathscr L$ is isomorphic to the semidirect sum $T \ltimes \mathscr L$.

As follows, for example, from \cite[Proposition 1.2]{farnsteiner}, any 
derivation of $\mathscr L$ can be represented as a sum of a derivation $D$ 
preserving the thin decomposition, and an inner derivation. We have 
$D(e_\alpha) = \lambda_\alpha e_\alpha$ for any nonzero $\alpha \in \GF(2)^4$, 
and some $\lambda_\alpha \in K$. Writing the derivation condition for each pair
of basic elements $e_\alpha, e_\beta$, we have
\begin{equation}\label{eq-lambda}
\lambda_{\alpha + \beta} = \lambda_\alpha + \lambda_\beta \quad\text{if}\quad
[e_\alpha, e_\beta] \ne 0 .
\end{equation}

Let us denote by $h_1, h_2, h_3, h_4$ the respective basis elements of $T$ 
(listed in (\ref{eq-tor})). Modifying $D$ by the action of the toral element
$\ad \big(\lambda_{0001} h_1 + \lambda_{0010} h_2 + \lambda_{0100} h_3 + 
          \lambda_{1000} h_4\big)$,
we may assume that
\begin{equation}\label{eq-zero}
\lambda_{0001} = \lambda_{0010} = \lambda_{0100} = \lambda_{1000} = 0 .
\end{equation}

Now, looking at the multiplication table above, and using the conditions 
(\ref{eq-lambda}) and (\ref{eq-zero}), it is easy to see that 
$\lambda_\alpha = 0$ for any nonzero $\alpha \in \GF(2)^4$, i.e., $D=0$.
\end{proof}


Other examples of simple Lie algebras admitting a thin decomposition include
the $7$-di\-men\-si\-o\-nal algebra $H$ (as shown in \cite{gg-2010}; but not the
other $7$-dimensional algebra $W = W_1(3)^\prime$, see below), and certain 
$15$-dimensional Hamiltonian algebras and algebras constructed by Eick, see 
\S\ref{ss-eick}. 

Recall that the well-known Zassenhaus algebra $W_1(n)^\prime$ of dimension 
$2^n-1$ has a realization in the basis 
$\set{e_\alpha}{\alpha \in \GF(2^n)^\times}$ with multiplication 
$[e_\alpha,e_\beta] = (\alpha + \beta)e_{\alpha + \beta}$. The grading 
$W_1(n)^\prime = \bigoplus_{\alpha \in GF(2^n)^\times} Ke_\alpha$ has most of 
the properties of a thin decomposition: it is the root space decomposition with
respect to the maximal $n$-dimensional torus $\langle e_0, e_0^{[2]}, \dots, e_0^{[2^{n-1}]} \rangle$, and the root spaces 
are one-dimensional. However, it fails to be a thin decomposition: the roots lie
in $\GF(2^n)$, and not in $\GF(2)^n$.






\section{The automorphism group}\label{sec-aut}

The goal of this section is to determine the automorphism group of the Skryabin
algebra. First, we define three types of automorphisms -- exponential 
automorphisms, certain explicitly defined three one-parameter families, and 
diagonal automorphisms, determine the group generated by them, and then prove 
that they exhaust the whole automorphism group.

\subsection{Exponential automorphisms}

If $D$ is a sandwich derivation of a Lie algebra $L$, then $\exp(D) = 1 + D$ is
an automorphism of $L$, called an \emph{exponential automorphism}. Since $
\exp(D)^2 = \exp(2D) = 1$, exponential automorphisms are of order $2$, and 
generate a unipotent subgroup of exponent $2$, denoted by $\Exp(L)$, of the automorphism group $\Aut(L)$. Note also that if $D_1,D_2 \in L$ are two 
commuting sandwich derivations, then the corresponding automorphisms also 
commute: $\exp(D_1) \circ \exp(D_2) = \exp(D_2) \circ \exp(D_1)$.

According to Proposition \ref{prop-sandr}(\ref{it-der}), the group 
$\Exp(\mathscr L)$ is $4$-dimensional abelian, isomorphic to the additive group
$K^4 = K \oplus K \oplus K \oplus K$. Let us write down its one-parameter 
generators explicitly (here and below we indicate only those basis elements on
which the automorphism acts non-identically):
$$
\exp (\ad \alpha c_2): 
\begin{array}{l}
b_1 \mapsto b_1 + \alpha c_3  \\ 
b_3 \mapsto b_3 + \alpha c_2  \\
b_4 \mapsto b_4 + \alpha c_4  \\ 
b_7 \mapsto b_7 + \alpha c_5  \\ 
d\>\>\> \mapsto d + \alpha b_8 
\end{array}
$$
$$
\exp (\ad \alpha c_4): 
\begin{array}{l}
b_1 \mapsto b_1 + \alpha b_2  \\ 
b_4 \mapsto b_4 + \alpha b_5  \\ 
b_7 \mapsto b_7 + \alpha b_8  \\ 
c_1 \mapsto c_1 + \alpha c_2  \\
\end{array}
$$
$$
\exp (\ad \alpha c_5): 
\begin{array}{l}
b_1 \mapsto b_1 + \alpha b_5   \\
b_7 \mapsto b_7 + \alpha c_2   \\
d\>\>\> \mapsto d + \alpha c_4
\end{array}
$$
$$
\exp (\ad \alpha c_3^{[2]}): 
\begin{array}{l}
b_1 \mapsto b_1 + \alpha b_8   \\
b_4 \mapsto b_4 + \alpha c_2   \\
d\>\>\> \mapsto d + \alpha c_5 .
\end{array}
$$
Here $\alpha \in K$ is a parameter.


\subsection{On $\exp(\ad \alpha c_4)$}

For a moment, let us return to the realization of $\mathscr L$ as a filtered 
deformation of the semisimple Lie algebra (\ref{eq-ss}). The exponential 
automorphisms $\exp(\ad \alpha c_4)$ are the only automorphisms of $\mathscr L$ 
which are ``lifted'' from automorphisms of the algebra (\ref{eq-ss}). Using the
results about automorphisms of the tensor product of a simple Lie algebra and a
divided power algebra (see, for example, \cite[\S 2.2]{weisf}), it is not 
difficult to describe the automorphism group of (\ref{eq-ss}) (roughly, those 
are automorphisms of $\ess \otimes \mathcal O_1(2)$ invariant under the action 
of $g$ and of $\partial$). Automorphisms which are preserved by the cocycles 
defining the deformation, are also automorphisms of $\mathscr L$. Among the 
automorphisms of (\ref{eq-ss}), the only automorphisms satisfying this 
condition, are automorphisms acting on $\ess \otimes \mathcal O_1(2)$ as 
$\phi(\alpha) \otimes \id_{\mathcal O_1(2)}$, where $\phi(\alpha)$ is an 
automorphism of $\ess$ of the form
$$
e \mapsto e + \alpha f, \quad f \mapsto f, \quad h \mapsto h ,
$$
and leaving $g \otimes \langle 1,x \rangle$ and $\partial$ invariant. In terms
of the standard basis of $\mathscr L$, this is exactly $\exp(\ad \alpha c_4)$. 
In general, the automorphism group of the algebra (\ref{eq-ss}) is much smaller 
than the automorphism group of its deformation $\mathscr L$, which shows that, 
generally, there is no strong relationship between automorphisms of a Lie 
algebra and of its deformation.


\subsection{Automorphisms $\Phi$, $\Psi$, and $\Theta$}

Consider the following three one-parameter families of linear maps on 
$\mathscr L$, depending on the parameter $\alpha \in K$:
\begin{equation*}
\Phi(\alpha):
\begin{array}{l}
b_1 \mapsto b_1 + \alpha b_4                                \\
b_2 \mapsto b_2 + \alpha b_5                                \\
b_7 \mapsto b_7 + \alpha^2 b_2 + \alpha^3 b_5 + \alpha c_1  \\
b_8 \mapsto b_8 + \alpha c_2                                \\
b_9 \mapsto b_9 + \alpha^2 c_4                              \\
c_1 \mapsto c_1 + \alpha^2 b_5                              \\
c_3 \mapsto c_3 + \alpha c_4                                \\
d\>\>\> \mapsto d + \alpha b_3 + \alpha^2 b_6 
\end{array}
\end{equation*}
$$
\hspace{25pt}
\Psi(\alpha):
\begin{array}{l}
b_1 \mapsto	b_1 + \alpha b_6 + \alpha c_1 + \alpha^2 c_4 \\
b_2 \mapsto b_2 + \alpha c_2                             \\
b_3 \mapsto b_3 + \alpha b_5                             \\
b_4 \mapsto b_4 + \alpha b_2 + \alpha^2 c_2              \\
b_6	\mapsto b_6 + \alpha c_4                             \\
b_7 \mapsto b_7 + \alpha b_5 + \alpha c_3                \\
b_9 \mapsto b_9 + \alpha c_2 + \alpha c_5                \\
c_1 \mapsto c_1 + \alpha b_8 + \alpha c_4                \\
d\>\>\> \mapsto	d + \alpha b_2 + \alpha b_9 + \alpha^2 c_2 + \alpha^2 c_5
\end{array}
$$

\smallskip

\begin{equation*}
\hspace{34pt}
\Theta(\alpha):
\begin{array}{l}
b_1 \mapsto b_1 + \alpha b_7 + \alpha^3 b_8 + \alpha^2 b_9                 \\ 
b_2 \mapsto b_2 + \alpha b_8                                               \\
b_3 \mapsto b_3 + \alpha^2 b_8                                             \\
b_4 \mapsto b_4 + \alpha^2 b_5 + \alpha c_1 + \alpha^2 c_3 + \alpha^3c_5   \\
b_5 \mapsto b_5 + \alpha c_2                                               \\
b_6 \mapsto b_6 + \alpha^2 c_2 + \alpha^2 c_5                              \\
c_1 \mapsto c_1 + \alpha^2 c_2 + \alpha^2 c_5                              \\
c_3 \mapsto c_3 + \alpha c_5                                               \\
d\>\>\> \mapsto d + \alpha^2 b_5 + \alpha b_6 + \alpha^3 c_2 + \alpha^2 c_3 
\end{array}
\end{equation*}

\bigskip

Direct calculations show that all of them are automorphisms of $\mathscr L$, and
\begin{alignat}{3}
&\Phi(\alpha) \>\circ\> \Phi(\alpha^\prime) &\>=\>& \Phi(\alpha + \alpha^\prime)
\notag \\
&\Psi(\alpha) \>\circ\> \Psi(\alpha^\prime) &\>=\>& 
\Psi(\alpha + \alpha^\prime) \>\circ\>
\exp\big(\ad \alpha\alpha^\prime c_3^{[2]}\big)
\label{eq-psi}
\\
&\Theta(\alpha) \>\circ\> \Theta(\alpha^\prime) &\>=\>&
\Theta(\alpha+\alpha^\prime) \>\circ\>
\exp\big(\ad (\alpha^2\alpha^\prime + \alpha{\alpha^\prime}^2) c_3^{[2]}\big)
\notag
\end{alignat}
for any $\alpha, \alpha^\prime \in K$. In particular, the automorphisms
$\Phi(\alpha)$ and $\Theta(\alpha)$ are of order $2$, and the automorphisms 
$\Psi(\alpha)$ are of order $4$.



\subsection{Diagonal automorphisms}\label{ss-diag}

Let $L$ be a Lie algebra with a basis $B$. We will call an automorphism of $L$ 
\emph{diagonal with respect to $B$} (or just \emph{diagonal} if it is clear 
which basis $B$ is meant), if it leaves invariant each one-dimensional subspace
$Kx$, where $x \in B$.

\begin{lemma}\label{lemma-diag}
Each diagonal automorphism of $\mathscr L$ with respect to the standard basis 
is of the form
\begin{alignat}{7}\label{eq-diag}
b_1&\mapsto \lambda^{-2} b_1 \qquad &c_1 &\mapsto \lambda c_1   \notag \\
b_2&\mapsto \lambda^2 b_2    \qquad &c_2 &\mapsto \lambda^5 c_2 \notag \\
b_3&\mapsto b_3              \qquad &c_3 &\mapsto \lambda^3 c_3 \notag \\
b_4&\mapsto \lambda^{-1} b_4 \qquad &c_4 &\mapsto \lambda^4 c_4 \notag \\
\Delta(\lambda): \quad 
b_5&\mapsto \lambda^3 b_5    \qquad &c_5 &\mapsto \lambda^5 c_5        \\
b_6&\mapsto \lambda b_6      \qquad &d &\mapsto \lambda^{-1} d  \notag \\
b_7&\mapsto b_7              \notag \\
b_8&\mapsto \lambda^4 b_8    \notag \\
b_9&\mapsto \lambda^2 b_9    \notag
\end{alignat}
where $\lambda \in K^\times$.
\end{lemma}

\begin{proof}
Let $x \mapsto \alpha(x)x$, where $x$ is an element in the standard basis, be a 
diagonal automorphism of $\mathscr L$. Denote $\alpha(b_4) = \lambda^{-1}$.

We perform the following calculations:
\begin{itemize}
\item 
$\alpha(b_3) = \alpha(b_1)\alpha(b_2)$, 
$\alpha(b_1) = \alpha(b_1)\alpha(b_3)$, and
$\alpha(b_2) = \alpha(b_2)\alpha(b_3)$
imply $\alpha(b_2) = \alpha(b_1)^{-1}$ and \linebreak $\alpha(b_3) = 1$.

\item
$\alpha(b_1) = \alpha(b_4)\alpha(d)$ and $\alpha(d) = \alpha(b_1)\alpha(c_1)$
imply $\alpha(c_1) = \lambda$.

\item
$\alpha(b_7) = \alpha(c_1)\alpha(d) = \alpha(b_1)\lambda^2$.

\item
$b_7^{[4]} = b_7^{[2]}$ implies $\alpha(b_7)^4 = \alpha(b_7)^2$, thus 
$\alpha(b_7)^2 = 1$ and $\alpha(b_7) = 1$, $\alpha(b_1) = \lambda^{-2}$, 
$\alpha(b_2) = \lambda^2$, and $\alpha(d) = \lambda^{-1}$.

\item
$\alpha(c_3) = \alpha(b_1)\alpha(c_2) = \alpha(c_2)\lambda^{-2}$. 

\item
$\alpha(c_1) = \alpha(b_1)\alpha(c_3)$ implies 
$\lambda = \alpha(c_2)\lambda^{-4}$, thus $\alpha(c_2) = \lambda^5$, and 
$\alpha(c_3) = \lambda^3$.

\item
$\alpha(b_2) = \alpha(b_1)\alpha(c_4)$ implies $\alpha(c_4) = \lambda^4$.
 
\item
$\alpha(b_9) = \alpha(b_2)\alpha(b_7) = \lambda^2$.

\item
$\alpha(b_8) = \alpha(b_2)\alpha(b_9) = \lambda^4$.

\item
$\alpha(c_2) = \alpha(b_7)\alpha(c_5)$ implies $\alpha(c_5) = \lambda^5$.

\item
$\alpha(b_5) = \alpha(b_4)\alpha(c_4) = \lambda^3$.

\item
$\alpha(b_6) = \alpha(b_1)\alpha(b_5) = \lambda$.
\end{itemize}

\end{proof}

Conversely, it is straightforward to verify that each map of the form 
(\ref{eq-diag}) is an automorphism of $\mathscr L$. Therefore, the group of 
diagonal automorphisms $\Diag(\mathscr L)$ is isomorphic to $K^\times$, the 
multiplicative group of $K$.


\subsection{Putting all these automorphisms together}

Denote by $\Aut_0(\mathscr L)$ the group generated by all automorphisms of 
$\mathscr L$ we have defined so far: exponential, $\Phi$, $\Psi$, $\Theta$, and
diagonal.

Direct calculations show that for each $\alpha, \gamma \in K$, any of 
$\Phi(\alpha)$, $\Psi(\alpha)$, and $\Theta(\alpha)$ commutes with any of 
$\exp(\ad \gamma c_2)$, $\exp(\ad \gamma c_4)$, $\exp(\ad \gamma c_5)$, and 
$\exp(\ad \gamma c_3^{[2]})$. Additionally,
\begin{alignat*}{4}
&\Psi(\gamma) &\>\circ\>& \Phi(\alpha) = &\Phi(\alpha) &\>\circ\> \Psi(\gamma) 
\>\circ\> \exp\big(\ad \alpha\gamma c_4\big)
\\
&\Theta(\gamma) &\>\circ\>& \Phi(\alpha) = 
&\Phi(\alpha) &\>\circ\> \Theta(\gamma)
\>\circ\> 
\exp\big(\ad \alpha\gamma^2 c_2\big) \>\circ\> 
\exp\big(\ad \alpha^2\gamma c_4\big) \>\circ\>
\exp\big(\ad \alpha\gamma^2 c_5\big) \>\circ\>
\exp\big(\ad \alpha^2\gamma^2 c_3^{[2]}\big)
\\
&\Theta(\gamma) &\>\circ\>& \Psi(\alpha) = 
&\Psi(\alpha) &\>\circ\> \Theta(\gamma)
\>\circ\> 
\exp\big(\ad \alpha\gamma c_2\big) \>\circ\> 
\exp\big(\ad \alpha\gamma c_5\big) .  
\end{alignat*}

Together with (\ref{eq-psi}) this implies that the group $\mathcal N$ generated
by the exponential automorphisms, and automorphisms $\Phi$, $\Psi$ and $\Theta$,
is a $7$-dimensional unipotent algebraic group. Further, taking into account that 
$\Delta(\lambda)^{-1} = \Delta(\lambda^{-1})$, we have:
\begin{alignat}{5}\label{eq-act}
&\exp(\ad \alpha c_2)^{\Delta(\lambda)} &\>=\>& \exp(\ad \lambda^{-5}\alpha c_2)
\notag \\
&\exp(\ad \alpha c_4)^{\Delta(\lambda)} &\>=\>& \exp(\ad \lambda^{-4}\alpha c_4)
\notag \\
&\exp(\ad \alpha c_5)^{\Delta(\lambda)} &\>=\>& \exp(\ad \lambda^{-5}\alpha c_5)
\notag \\
&\exp(\ad \alpha c_3^{[2]})^{\Delta(\lambda)} &\>=\>& 
\exp(\ad \lambda^{-6}\alpha c_3^{[2]})
\\
&\Phi(\alpha)^{\Delta(\lambda)} &\>=\>& \Phi(\lambda^{-1}\alpha)
\notag \\
&\Psi(\alpha)^{\Delta(\lambda)} &\>=\>& \Psi(\lambda^{-3}\alpha)
\notag \\
&\Theta(\alpha)^{\Delta(\lambda)} &\>=\>& \Theta(\lambda^{-2}\alpha) .
\notag
\end{alignat}
(here the superscript denotes the group conjugation: 
$\varphi^\psi = \psi^{-1} \circ \varphi \circ \psi$).

Therefore, $\mathcal N$ is a normal subgroup in $\Aut_0(\mathscr L)$, and 
$\Aut_0(\mathscr L)$ is isomorphic to the semidirect product 
$K^\times \ltimes \mathcal N$, with the action of $K^\times$ on $\mathcal N$ 
defined by (\ref{eq-act}).



\subsection{Invariant subspaces}

Now we are going to prove that the automorphisms constructed in the previous 
subsections exhaust all automorphisms of $\mathscr L$. To this aim, we determine
certain invariant subspaces in the Skryabin algebra.

\begin{proposition}\label{flag}
The Skryabin algebra $\mathscr L$ has the following 
$\Aut(\mathscr L)$-invariant subspaces:
{\setlength{\jot}{4pt}
\begin{gather*}
\langle c_2 \rangle \quad 
\langle c_4 \rangle \quad 
\langle c_5 \rangle 
\\
\langle b_5,c_2 \rangle \quad \langle b_8,c_2 \rangle
\\
V_4 = \langle b_2,b_5,b_8,c_2 \rangle  
\\
V_5 = \langle b_8,c_2,c_3,c_4,c_5 \rangle
\\
V_6 = \langle b_2,b_3,b_5,b_8,c_2,c_4 \rangle
\quad
V_6^\prime = \langle b_5,b_6,b_8,c_2,c_4,c_5 \rangle
\quad
V_6^{\prime\prime} = \langle b_5,b_8,b_9,c_2,c_4,c_5 \rangle
\\
V_7 = \langle b_5,b_6,b_8,b_9,c_2,c_4,c_5 \rangle
\quad
V_7^\prime = \langle b_5,b_8,b_9,c_2,c_3,c_4,c_5 \rangle
\\
V_8 = \langle b_2,b_3,b_5,b_6,b_8,c_2,c_4,c_5 \rangle
\quad
V_8^\prime = \langle b_2,b_5,b_6,b_8,b_9,c_2,c_4,c_5 \rangle 
\quad
V_8^{\prime\prime} = \langle b_2,b_5,b_6,b_8,c_2,c_3,c_4,c_5 \rangle
\\
V_9 = \langle b_2,b_3,b_5,b_6,b_8,b_9,c_2,c_4,c_5 \rangle
\quad
V_9^\prime = \langle b_2,b_5,b_6,b_8,b_9,c_2,c_3,c_4,c_5 \rangle 
\\
V_{11} = \langle b_2,b_3,b_4,b_5,b_6,b_8,c_1,\dots,c_5 \rangle
\quad
V_{11}^\prime = \langle b_2,b_3,b_5,b_6,b_8,b_9,c_1,\dots,c_5 \rangle
\\
V_{11}^{\prime\prime} = 
\langle b_2,b_3,b_5,b_6,b_8,b_9,c_2,c_3,c_4,c_5,d \rangle
\\
V_{12} = \langle b_2,b_3,b_4,b_5,b_6,b_8,b_9,c_1,\dots,c_5 \rangle 
\end{gather*}
}
\end{proposition}

\begin{proof}
By Proposition \ref{prop-sandr}(\ref{it-sk}), the sandwich subalgebra $S$ of 
$\mathscr L$ coincides with $\langle c_2,c_4,c_5 \rangle$. Starting from this, 
rewrite the specified subspaces in invariant terms:

\begin{itemize}

\item $\langle c_5 \rangle = \set{x\in S}{\dim [\mathscr L,x] \leq 3}$;

\item 
$\langle c_4,c_5 \rangle$ is the subspace (actually, the abelian subalgebra) 
linearly spanned by elements $x\in S$ such that $\dim [\mathscr L,x] \leq 4$;

\item $\langle c_4 \rangle = [\mathscr L,c_5] \cap \langle c_4,c_5 \rangle$;

\item $V_4 = [\mathscr L,c_4]$;

\item $\langle c_2 \rangle = S \cap V_4$;

\item $V_5 = [\mathscr L,c_2]$;

\item $\langle b_8,c_2 \rangle = V_4 \cap V_5$;

\item $V_7^\prime = [\mathscr L,\langle b_8,c_2 \rangle]$

\item $V_9^\prime = \set{x\in \mathscr L}{[x,S] = 0}$;

\item 
$\langle b_5,c_2 \rangle = 
[\mathscr L,c_5] \cap V_4 \cap [V_9^\prime,V_9^\prime]$;

\item $V_8^\prime = \set{x \in V_9^\prime}{x^{[2]} \in \mathscr L}$;

\item $V_8^{\prime\prime} = [\mathscr L, \langle b_5,c_2 \rangle]$;

\item 
Note that $V_8^\prime$ is a subalgebra of $\mathscr L$ with the center $S$. The
subspace $V_7$ (actually, a subalgebra) coincides with the set of elements 
$x \in V_8^\prime$ such that the induced adjoint map 
$\ad x: V_8^\prime/S \to V_8^\prime/S$ has rank $\le 1$;

\item $V_6^\prime = V_7 \cap V_8^{\prime\prime}$;

\item $V_6^{\prime\prime} = V_7 \cap V_7^\prime$;

\item $V_{11}^\prime = \set{x \in \mathscr L}{[x,S] \subseteq S}$;

\item $V_{11}^{\prime\prime} = C_{\mathscr L}(c_4)$;

\item $V_9 = [V_{11}^{\prime\prime},V_{11}^{\prime\prime}]$;

\item 
$V_{12} = 
\set{x \in \mathscr L}{[x,V_{11}^\prime] \subseteq V_{11}^\prime + Kx}$;

\item $V_{11} = [V_{12},V_{12}]$;

\item $V_8 = V_9 \cap V_{11}$;

\item 
$V_6$ is a linear span of elements $x^{[2]}$, where $x \in V_8$;

\end{itemize}

All these items are verified by straightforward computations.
\end{proof}

Many more $\Aut(\mathscr L)$-invariant subspaces of $\mathscr L$ can be produced
in a similar fashion, here we confine ourselves only to those which will be 
needed in the sequel.



\subsection{No other automorphisms}

\begin{theorem}\label{l-b3}
Assume that any quadratic equation with coefficients in the ground field $K$ has
a solution in $K$. Then $\Aut(\mathscr L) = \Aut_0(\mathscr L)$.
\end{theorem}

\begin{proof}
Let $\varphi$ be an automorphism of $\mathscr L$. Our strategy is to 
consecutively ``twist'' $\varphi$ by taking compositions with various 
automorphisms from $\Aut_0(\mathscr L)$, and eventually arrive at the conclusion
$\varphi = \id_{\mathscr L}$.

By Proposition \ref{flag}, $\varphi(b_3)$ lies in $V_6$ and does not lie in 
$V_4 + \langle c_4 \rangle$, i.e., is of the form
$$
\varphi(b_3) = 
\lambda_2 b_2 + \lambda_3 b_3 + \lambda_5 b_5 + \lambda_8 b_8 + 
\mu_2 c_2 + \mu_4 c_4 ,
$$
where $\lambda_3 \ne 0$. As $b_3$ is toral, $\varphi(b_3)$ is toral, and the 
equality $\varphi(b_3) = \varphi(b_3)^{[2]}$ is equivalent to the following 
quadratic system:
\begin{align*}
&\lambda_2 = \lambda_2\lambda_3     \\
&\lambda_3 = \lambda_3^2            \\
&\lambda_5 = \lambda_3\lambda_5     \\
&\lambda_8 = \lambda_3\lambda_8     \\
&\mu_2 = \lambda_3\mu_2             \\
&\mu_4 = \lambda_2^2 .
\end{align*}

Consequently,
\begin{equation}\label{eq-x1}
\varphi(b_3) = \lambda_2 b_2 + b_3 + \lambda_5 b_5 + \lambda_8 b_8 + \mu_2 c_2 
+ \lambda_2^2 c_4 .
\end{equation}

Assume that $\lambda_2 \ne 0$. Applying to both sides of the equality 
(\ref{eq-x1}) the automorphism
$$
\exp\Big(
\ad \big(\lambda_2^{-\frac 32}\lambda_5 + \lambda_2^{-\frac 52}\mu_2\big) c_2
\Big) \>\circ\> 
\Theta(\alpha) \>\circ\> 
\Phi\big(\lambda_2^{-\frac 32}\lambda_5\big) \>\circ\> 
\Delta\big(\lambda_2^{\frac 12}\big) ,
$$
where $\alpha$ satisfies the quadratic equation 
$\alpha^2 + \alpha + \lambda_2^{-2}\lambda_8 = 0$ (this is the only place
where we need the assumption that any quadratic equation with coefficients in 
$K$ has a solution), we may assume that
$$
\varphi(b_3) = b_2 + b_3 + c_4 .
$$

The automorphism $\varphi$ maps the subalgebra $C_{\mathscr L}(b_3)$ to the 
subalgebra $C_{\mathscr L}(b_2 + b_3 + c_4)$. We have
$$
C_{\mathscr L}(b_3) = \langle b_3, b_6, b_9, c_3, c_4, c_5, d \rangle
$$
and
$$
C_{\mathscr L}(b_2 + b_3 + c_4) = 
\langle b_2 + b_3 + c_4, b_5 + b_6, b_8 + b_9, c_2 + c_3, c_4, c_5, d \rangle .
$$

By Proposition \ref{flag}, $\varphi(b_6)$ lies in $V_6^\prime$, so does not 
contain terms with $b_2 + b_3 + c_4$, $b_8 + b_9$, $c_2 + c_3$, and $d$. 
Similarly, $\varphi(b_9)$ lies in $V_6^{\prime\prime}$, so does not contain 
terms with $b_2 + b_3 + c_4$, $b_5 + b_6$, $c_2 + c_3$, and $d$. Therefore, we 
can write
\begin{alignat*}{4}
&\varphi(b_6) &\>=\>& \alpha_6(b_5 + b_6) + \alpha_4 c_4 + \alpha_5 c_5  
\\
&\varphi(b_9) &\>=\>& \beta_9(b_8 + b_9) + \beta_4 c_4 + \beta_5 c_5     
\\
&\varphi(d) &\>=\>& \gamma_3(b_2 + b_3 + c_4) + \gamma_6(b_5 + b_6) 
+ \gamma_9(b_8 + b_9) + \delta_3(c_2 + c_3) + \delta_4 c_4 + \delta_5 c_5 
+ \gamma d 
\end{alignat*}
for certain $\alpha_i, \beta_i, \gamma_i, \delta_i, \gamma \in K$. Then, the 
equalities $[\varphi(b_6),\varphi(d)] = \varphi(b_3)$ and 
$[\varphi(b_9),\varphi(d)] = \varphi(b_6)$ are equivalent to
\begin{gather}
\alpha_6\gamma = 1                    \label{eq-1} \\
\alpha_6\delta_3 + \alpha_5\gamma = 1 \label{eq-2}
\end{gather}
and
\begin{gather}
\beta_9\gamma = \alpha_6   \label{eq-bga}   \\
\beta_5\gamma = \alpha_4      \\
\beta_9\delta_3 = \alpha_5 \label{eq-last}
\end{gather}
respectively. It is easy to see that the system (\ref{eq-1})--(\ref{eq-last}) is
contradictory: for example, multiplying (\ref{eq-last}) by $\alpha_6$, and 
taking into account (\ref{eq-2}), we get 
$\beta_9 + \alpha_5\beta_9\gamma = \alpha_5\alpha_6$, which, in its turn, 
together with (\ref{eq-bga}) gives $\beta_9 = 0$, hence 
$\alpha_5 = \alpha_6 = 0$, contradicting condition (\ref{eq-2}).

Therefore, $\lambda_2 = 0$. Applying to both sides of (\ref{eq-x1}) the 
automorphism
$$
\Psi(\lambda_5) \>\circ\> 
\exp\Big(\ad \big(\lambda_5\sqrt{\lambda_8} + \mu_2\big) c_2\Big) \>\circ\> 
\Theta\big(\sqrt{\lambda_8}\big) ,
$$
we may assume that $\varphi(b_3) = b_3$. Consequently, the eigenspace 
$\langle b_1, b_2, b_4, b_5, b_7, b_8, c_1, c_2 \rangle$ corresponding to the 
eigenvalue $1$ of $\ad b_3$, is invariant under $\varphi$, and we may write
$$
\varphi(b_1) = \lambda_1 b_1 + \lambda_2 b_2 + \lambda_4 b_4 + \lambda_5 b_5 + 
               \lambda_7 b_7 + \lambda_8 b_8 + \mu_1 c_1 + \mu_2 c_2 .
$$

Applying to both sides of this equality the automorphism
\begin{align*}
&\exp\Big(\ad \big(
  \lambda_1^{\frac 12}\lambda_2\lambda_4 
+ \lambda_1^{-\frac 32}\lambda_4^3\lambda_7
+ \lambda_1^{-\frac 12}\lambda_4^2\mu_1 
+ \lambda_1^{\frac 32}\lambda_5
\big) c_5\Big) 
\\ \circ\> 
&\exp\Big(\ad \big(
  \lambda_1\lambda_2\lambda_7 
+ \lambda_1^{-1}\lambda_4\lambda_7^2
+ \lambda_1^2\lambda_8 
\big) c_3^{[2]}\Big) 
\\ \circ\> 
&\exp\Big(
\ad \big(\lambda_1\lambda_2 + \lambda_1^{-1}\lambda_4\lambda_7\big) c_4
\Big) 
\\ \circ\> 
&\Phi\Big(\lambda_1^{-\frac 12}\lambda_4\Big)
\\ \circ\>
&\Delta\big(\lambda_1^{-\frac 12}\big) ,
\end{align*}
we may assume that $\lambda_1 = 1$ and 
$\lambda_2 = \lambda_4 = \lambda_5 = \lambda_8 = 0$.

Then we have
$$
0 = \varphi(b_1^{[4]}) = \varphi(b_1)^{[4]} = 
\mu_1^2 b_4^{[2]} + \lambda_7^4 b_7^{[2]} + \mu_2^2 c_3^{[2]} + 
\text{ (terms lying in $\mathscr L$)} ,
$$
which implies $\lambda_7 = \mu_1 = \mu_2 = 0$, and $\varphi(b_1) = b_1$. 
Further:
\begin{itemize}

\item $\varphi(c_4) \in \langle c_4 \rangle$ by Proposition \ref{flag};

\item 
$\varphi(b_2) = [\varphi(b_1),\varphi(c_4)] \in [b_1,\langle c_4 \rangle] = 
 \langle b_2 \rangle$;

\item 
$\varphi(c_2) \in \langle c_2 \rangle$ by Proposition \ref{flag};

\item 
$\varphi(c_3) = [\varphi(b_1),\varphi(c_2)] \in [b_1, \langle c_2 \rangle] =
 \langle c_3 \rangle$;

\item
$\varphi(c_1) = [\varphi(b_1),\varphi(c_3)] \in [b_1, \langle c_3 \rangle] =
 \langle c_1 \rangle$;

\item 
$\varphi(c_5) \in \langle c_5 \rangle$ by Proposition \ref{flag};

\item 
$\varphi(b_5) = [\varphi(b_1),\varphi(c_5)] \in [b_1, \langle c_5 \rangle] =
 \langle b_5 \rangle$;

\item
$\varphi(b_6) = [\varphi(b_1),\varphi(b_5)] \in [b_1, \langle b_5 \rangle] =
 \langle b_6 \rangle$;

\item 
$\varphi(d) = [\varphi(b_1),\varphi(c_1)] \in [b_1, \langle c_1 \rangle] =
 \langle d \rangle$;

\item
$\varphi(b_7) = [\varphi(c_1),\varphi(d)] \in 
[\langle c_1 \rangle, \langle d \rangle] = \langle b_7 \rangle$;

\item
$\varphi(b_4) = [\varphi(b_7),\varphi(d)] \in 
[\langle b_7 \rangle, \langle d \rangle] = \langle b_4 \rangle$;

\item
$\varphi(b_8) = [\varphi(c_2),\varphi(d)] \in 
[\langle c_2 \rangle, \langle d \rangle] = \langle b_8 \rangle$;

\item
$\varphi(b_9) = [\varphi(c_3),\varphi(d)] \in 
[\langle c_3 \rangle, \langle d \rangle] = \langle b_9 \rangle$.

\end{itemize}

Hence, $\varphi$ is a diagonal automorphism, and by Lemma \ref{lemma-diag} we 
have $\varphi = \Delta(\lambda)$, where $\lambda^{-2} = 1$. But then 
$\lambda = 1$ and $\varphi = \id_{\mathscr L}$.
\end{proof}




\section{Gradings}

Having a supply of automorphisms of $\mathscr L$ at hand, and using the known 
correspondence between automorphisms and group gradings (formulated in full 
generality in the language of affine group schemes -- see, for example, 
\cite[Proposition 1.36]{elduque-kochetov}), we may try to construct group 
gradings of $\mathscr L$.

In practice, this is achieved by extending the ground field $K$ to a suitable 
commutative ring $R$ (not necessary a field -- we are dealing with the group 
scheme $R \to \Aut_R(\mathscr L \otimes_K R)$), extending an automorphism 
$\varphi$ of $\mathscr L$ to the automorphism $\overline\varphi$ of 
$\overline{\mathscr L} = \mathscr L \otimes_K R$ via 
$\overline\varphi(x \otimes r) = \varphi(x) \otimes r$, and considering 
eigenspaces 
$
\overline{\mathscr L}_\lambda = 
\set{x \in \overline{\mathscr L}}{\overline\varphi(x) = \lambda x} .
$
For a suitable choice of $\varphi$ and $R$, and a suitable homomorphism $\chi$ 
from the group generated by all eigenvalues $\lambda \in R$ to a group $G$, the
eigenspaces are ``rational'': 
$\overline{\mathscr L}_\lambda = \mathscr L_{\chi(\lambda)} \otimes_K R$. The 
ensued grading 
$\mathscr L = \bigoplus_{\chi(\lambda)} \mathscr L_{\chi(\lambda)}$ is a 
grading of $\mathscr L$ by $G$, even in the case where the eigenvalues $\lambda$
do not necessarily belong to the ground field.

For example, the diagonal automorphism $\Delta(\lambda)$ for the ``generic'' 
value of $\lambda$ (or, which is the same, for $\lambda \in K$ such that the 
order of $\lambda$ in the multiplicative group $K^\times$ is $>7$) produces a 
$\mathbb Z$-grading
\begin{alignat}{10}\label{eq-z}
&\hspace{2pt}\scriptstyle{-2}& 
&\hspace{7pt}\scriptstyle{-1}& 
&\hspace{10pt}\scriptstyle{0}&
&\hspace{11pt}\scriptstyle{1}&
&\hspace{10pt}\scriptstyle{2}&
&\hspace{11pt}\scriptstyle{3}&
&\hspace{10pt}\scriptstyle{4}&
&\hspace{9pt}\scriptstyle{5}
\notag \\
\mathscr L =\>
\langle &b_1&     \rangle \oplus 
\langle &b_4,d&   \rangle \oplus 
\langle &b_3,b_7& \rangle \oplus 
\langle &b_6,c_1& \rangle \oplus 
\langle &b_2,b_9& \rangle \oplus 
\langle &b_5,c_3& \rangle \oplus 
\langle &b_8,c_4& \rangle \oplus 
\langle &c_2,c_5& \rangle .
\end{alignat}

Specializing the automorphism $\Delta(\lambda)$ to the cases $\lambda^n = 1$, 
$n \le 7$, we obtain a $\mathbb Z / n\mathbb Z$-grading of $\mathscr L$. 

As the product of any two elements from the standard basis is either zero, or
again an element from the standard basis, the decomposition of $\mathscr L$ into
one-dimensional subspaces spanned by the basis element is a grading. This is a 
group grading, and its universal group (\cite[\S 1.2]{elduque-kochetov}) is 
isomorphic to the (additive) abelian group with generators $x,y$ (corresponding
to elements $b_1$, $b_4$ respectively), and the relation $2x = 4y$, which is 
clearly isomorphic to the direct sum $\mathbb Z \oplus \mathbb Z / 2\mathbb Z$.
 On the other hand, the thin decomposition exhibited in \S \ref{ss-thin} is a 
$(\mathbb Z / 2\mathbb Z)^4$-grading.

In general, it seems to be a difficult task to classify all group gradings of 
the Skryabin algebra. Even to determine whether the automorphisms of order $2$
(exponential, $\Phi$, and $\Theta$) lead to $\mathbb Z / 2\mathbb Z$-gradings 
seems to be far from trivial. Perhaps, it could be approached with the method
of \cite{krutov-leb}.






\section{Comparison of the Skryabin algebra with algebras from the Eick list. 
A bit more numerology}\label{ss-eick}

In \cite{eick}, a computer-generated list of simple Lie algebras over $\GF(2)$ 
of dimension $\le 20$ is presented. The Skryabin algebra (defined over $\GF(2)$)
is not isomorphic to any of the $15$-dimensional algebras in the list.

One way to see this, using data from \cite{eick}, is to look at automorphisms, 
either at the group of exponential automorphisms $\Exp(\mathscr L)$, or at the 
whole group $\Aut(\mathscr L)$. As the ground field is $\GF(2)$, 
$\Exp(\mathscr L)$ is isomorphic to the additive group 
$\GF(2) \oplus \GF(2) \oplus \GF(2) \oplus \GF(2)$, thus having order $16$, 
which is different from all the $15$-dimensional algebras in the list 
(including the new ones, number 7 and 8, dubbed by us here as $Eick_7$ and 
$Eick_8$) except for the non-alternating Hamiltonian algebra $P(2,1,1)$ (number
4). The order of $\Aut(\mathscr L)$ is $2^7 = 128$ (over $\GF(2)$, there are no nontrivial 
diagonal automorphisms), which is, again, different from all the 
$15$-dimensional algebras in the list.

Another way to distinguish between all these algebras, is to repeat for them the
same pedestrian, but informative computations concerning tori and the sandwich
subalgebra, as in \S \ref{ss-num} and \S \ref{sec-misc}. The following table 
accumulates some information about the $15$-dimensional central simple algebras
from the Eick list, whose derivation algebra is $19$-dimensional (and coincides
in all the cases with the $2$-envelope). Here $TR$ denotes the absolute toral
rank of the algebra, $N_1$ and $N_m$ denote, respectively, the number of toral 
elements, and of tori of the maximal dimension $TR$ in the $2$-envelope, and $S$
is the dimension of the sandwich subalgebra. We keep Eick's notation for 
algebras; in particular, $W(4)$ is, in a more common notation, the Zassenhaus 
algebra $W_1(4)^\prime$.

$$
\begin{tabular}{|l|r|r|r|r|}
\hline
\multicolumn{1}{|c|}{\small{algebra}} & \multicolumn{1}{|c|}{\small{$TR$}} & 
\multicolumn{1}{|c|}{\small{$N_1$}} & \multicolumn{1}{|c|}{\small{$N_m$}} & 
\small{$S$}
\\ \hline
$W(4)$     & 2 & 256 &  1,536 & 10
\\ \hline
$P(2,1,1)$ & 4 & 448 & 43,680 & 1
\\ \hline
$P(3,1)$   & 3 & 384 & 10,752 & 5
\\ \hline
$P(2,2)$   & 4 & 384 & 13,440 & 3
\\ \hline
$Eick_7$   & 4 & 464 & 87,360 & 1
\\ \hline
$Eick_8$   & 4 & 464 & 67,200 & 1
\\ \hline
\end{tabular}
$$

\medskip

Interestingly enough, all of these algebras which are of absolute toral rank 
$4$, i.e., $P(2,1,1)$, $P(2,2)$, $Eick_7$, and $Eick_8$, also admit thin 
decompositions with respect to a lot (conjecturally with respect to all) of the
$4$-dimensional tori.

Non-alternating Hamiltonian algebras in characteristic $2$ were investigated in 
depth in the recent interesting preprints \cite{kkc} and \cite{kondrat}. We 
believe that over $\GF(2)$ all $15$-dimensional algebras from \cite{kondrat} are 
isomorphic to $P(3,1)$, but have not ventured into proving it.




\section{Open questions}\label{s-quest}

{\bf 1.} Is it true that two toral elements $h$ and $h^\prime$ in $\mathscr L$ 
are conjugate with respect to the automorphism group, if and only if 
$C_{\mathscr L}(h) \simeq C_{\mathscr L}(h^\prime)$ ?

\medskip

{\bf 2.} Describe all gradings of the Skryabin algebra.

\medskip

{\bf 3.} \emph{Conjecture}.
Let $L$ be a simple Lie algebra admitting a thin decomposition with respect to 
a torus $T$. Then the derivation algebra of $L$ is isomorphic to the semidirect 
sum $T \ltimes L$.

The conjecture is true for the Skryabin algebra, as shown in \S \ref{ss-thin}.
It is easy to see that the condition of simplicity is essential here: for 
example, the $3$-dimensional nilpotent Lie algebra admits a thin decomposition,
but not satisfies the conclusion of the conjecture.

\medskip

{\bf 4.} \emph{Conjecture}. 
Any simple Lie algebra of dimension $>3$ over a field of characteristic $2$ 
admitting a thin decomposition has:
\\
a) a proper simple graded subalgebra (with respect to this decomposition);
\\
b) a graded subalgebra isomorphic either to $W$ or to $H$.

\medskip

{\bf 5.} Classify simple finite-dimensional Lie algebras over an algebraically 
closed field of characteristic $2$, admitting a thin decomposition.

\medskip

{\bf 6.} 
Classify simple finite-dimensional $\mathbb Z$-graded Lie algebras over an 
algebraically closed field of characteristic $2$, such that all homogeneous 
components are of dimension $<3$. (Note that the Skryabin algebra belongs to 
this class, due to 
(\ref{eq-z})).






\section*{Acknowledgements}

Thanks are due to Bettina Eick, Dimitry Leites, Alexander Premet, and the 
anonymous referee for useful remarks. 
Alexander Grishkov was supported by FAPESP, grant number 2018/23690-6 and 
CNPq, grant number 307824/2016-0;
Marina Rasskazova during the work on this paper was visiting the University of
S\~ao Paulo, and was supported by FAPESP, grant number 2018/11292-6; 
Pasha Zusmanovich was supported by the Ministry of Education and Science of the
Republic of Kazakhstan, grant number AP09259551.




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