# New Constructions of Exceptional Simple Lie Superalgebras in Low
Characteristic Via Tensor Categories

Arun S. Kannan  Department of Mathematics, Massachusetts Institute of
Technology, Cambridge, MA 02139<EMAIL_ADDRESS>
###### Abstract.

We present new constructions of several of the exceptional simple Lie
superalgebras in characteristic $p=3$ and $p=5$ (cf. [BGL+09]) by considering
the images of exceptional Lie algebras with a nilpotent derivation under the
semisimplification functor from
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{p}$ to
$\operatorname{Ver}_{p}$.

###### Contents

  1. 1 Introduction
  2. 2 Contragredient Lie Superalgebras
  3. 3 Constructing Lie Superalgebras from Lie Algebras Using Semisimplification
  4. 4 Exceptional Simple Lie Superalgebras in Characteristic $3$
  5. 5 An Exceptional Lie Superalgebra in Characteristic $5$

##  1\. Introduction

In [BGL+09], the finite-dimensional modular Lie superalgebras over
$\mathbb{K}$ with indecomposable Cartan matrix are classified, where
$\mathbb{K}$ is an algebraically closed field. In particular, if the Cartan
matrix is invertible, the Lie superalgebra is simple; otherwise, either the
derived algebra or second derived algebra of the Lie superalgebra modulo its
center is simple. In characteristic $5$, the classification includes Lie
superalgebras arising from reducing the classical simple Lie superalgebras
over $\mathbb{C}$ modulo $5$, the Brown superalgebras, and the Elduque
superalgebra $\mathfrak{el}(5;5)$. In characteristic $3$, there are the mod
$3$ reductions of the classical simple Lie superalgebras. The remaining are
the Brown superalgebras, the ten Elduque and Cunha superalgebras, and a
characteristic $3$ analog $\mathfrak{el}(5;3)$ of $\mathfrak{el}(5;5)$.

In this paper, we use the semisimplification of tensor categories to produce
new constructions of the ten Elduque and Cunha Lie superalgebras in
characteristic $3$, the Elduque Lie superalgebra in characteristic $5$, and
its characteristic $3$ analog discovered in [BGL+09]. In particular, we can
consider a Lie algebra $\mathfrak{g}$ over $\mathbb{K}$ with a nilpotent
derivation $d$ of order at most $p$, where $p>2$ is the characteristic of
$\mathbb{K}$; this can then be realized as a Lie algebra in the category
$\operatorname*{Rep}\mathbb{K}[t]/(t^{p})$ of $\mathbb{K}[t]/(t^{p})$ modules
by specializing $t$ to $d$. The semisimplification of this category is the
Verlinde category $\operatorname{Ver}_{p}$, which contains as a full
subcategory the category of supervector spaces
$\operatorname*{sVec}_{\mathbb{K}}$. Therefore, the image
$\overline{\mathfrak{g}}$ of $\mathfrak{g}$ under the semisimplification
functor projected onto this full subcategory is a Lie algebra in
$\operatorname*{sVec}_{\mathbb{K}}$, which is precisely a Lie superalgebra.
For more details, one can refer to [Eti18, EO19, Ost15].

In every case, we start with one of the exceptional simple Lie algebras
$\mathfrak{f}_{4},\mathfrak{e}_{6},\mathfrak{e}_{7}$, and $\mathfrak{e}_{8}$
and then choose an appropriate nilpotent element of order $3$ or $5$, and then
semisimplify. The software SuperLie (cf. [Gro13]) greatly simplified verifying
this process.

The original construction of the Elduque and Cunha Lie superalgebras are based
on symmetric composition algebras and the Elduque Supermagic Square, an analog
of Freudenthal’s Magic Square where division algebras are used to construct
the exceptional Lie superalgebras (cf. [Eld06, CE06, CE07]). In [BGL06,
Bou+20], explicit descriptions are given in terms of Cartan matrices and
generators and relations. In either case, it can be difficult to work with
descriptions of the Lie superalgebras, and we hope that the theoretical
framework of symmetric tensor categories makes approaching these new
constructions more tractable. Lastly, one can also use this approach to
construct new representations of these Lie superalgebras by suitably
semisimplifying a representation of the Lie algebra they came from; very
little is known about the representation theory of these Lie superalgebras, so
this may be particularly interesting.

Acknowledgements. The author would like to deeply thank his advisor, Pavel
Etingof, for pointing out that these constructions were possible and for his
patience and guidance. This paper is based upon work supported by The National
Science Foundation Graduate Research Fellowship Program under Grant No.
1842490 awarded to the author.

##  2\. Contragredient Lie Superalgebras

The Lie superalgebras listed above can all be realized as contragredient Lie
superalgebras, which means they arise from a Cartan matrix. We will use this
formulation to prove that each Lie superalgebra we construct from
semisimplification is isomorphic to the corresponding exceptional simple Lie
superalgebra with Cartan matrix. For the reader’s convenience, we recall the
basics on contragredient Lie superalgebras here.

Let $n$ be a positive integer, and let $A=(a_{ij})$ be an arbitrary $n\times
n$ matrix of rank $s$ with entries in $\mathbb{Z}$, called the Cartan matrix.
Let $\mathfrak{h}$ be a vector space over $\mathbb{K}$ of dimension $2n-s$,
where $\mathbb{K}$ is an algebraically closed field of characteristic $p\geq
0$. Let $\\{h_{i}\\}_{i=1}^{n}$ be a collection of $n$ linearly independent
vectors in $\mathfrak{h}$, and let $\\{\alpha_{i}\\}_{i=1}^{n}$ be vectors in
$\mathfrak{h}^{*}$ such that $\alpha_{j}(h_{i})=\overline{a_{ij}}$ for all
$1\leq i,j\leq n$, where the bar denotes the projection onto
$\mathbb{Z}/p\mathbb{Z}$ if $p>0$ and is the identity if $p=0$. Let
$I=\\{i_{1},\dots,i_{n}\\}\subset(\mathbb{Z}/2\mathbb{Z})^{n}$ be a collection
of parities, and let $\mathrm{par}(i)$ denote the parity of $i\in I$.

Then, we can construct the Lie superalgebra $\widetilde{\mathfrak{g}}(A,I)$
with generators $e_{1},\dots,e_{n},f_{1},\dots,f_{n}$, called Chevalley
generators, and $\mathfrak{h}$ such that the parity of $e_{j}$ and $f_{j}$ is
$i_{j}\in I$ for all $1\leq j\leq n$ and such that they are subject to the
following relations:

(2.1) $[e_{i},f_{j}]=\delta_{ij}h_{i};\ \ [h,e_{j}]=\alpha_{j}(h)e_{j};\ \
[h,f_{j}]=-\alpha_{j}(h)f_{j};\ \ [\mathfrak{h},\mathfrak{h}]=0,$

for all $1\leq i,j\leq n$ and $h\in\mathfrak{h}$. If $A$ has full rank (i.e.
$\det A\neq 0$), then the $\\{e_{i},f_{i},h_{i}\\}$ generate
$\widetilde{\mathfrak{g}}(A,I)$; otherwise, they generate the derived
subalgebra of $\widetilde{\mathfrak{g}}(A,I)$. It can be shown that
$\widetilde{\mathfrak{g}}(A,I)$ admits a triangular decomposition of Lie
subsuperalgebras
$\widetilde{\mathfrak{g}}(A,I)=\widetilde{\mathfrak{n}}^{-}\oplus\mathfrak{h}\oplus\widetilde{\mathfrak{n}}^{+}$,
where $\widetilde{\mathfrak{n}}^{-}$ is the Lie subsuperalgebra generated by
the $f_{j}$’s and $\widetilde{\mathfrak{n}}^{+}$ is the Lie subsuperalgebra
generated by the $e_{i}$’s. When we work with multiple different Cartan
matrices, we shall denote $\mathfrak{h}$ by $\mathfrak{h}(A,I)$ (and similarly
for other subalgebras). We shall refer to both $A$ and $(A,I)$ as the Cartan
matrix of $\widetilde{\mathfrak{g}}(A,I)$ (or of any quotient of
$\widetilde{\mathfrak{g}}(A,I)$). As usual, there is a root space
decomposition, which gives a $Q=\bigoplus_{i=1}^{n}\mathbb{Z}\alpha_{i}$
grading where $\mathrm{deg}(e_{i})=\alpha_{i}$,
$\mathrm{deg}(f_{i})=-\alpha_{i}$, and $\mathrm{deg}(h)=0$.

We define the contragredient Lie superalgebra $\mathfrak{g}(A,I)$ to be the
quotient $\mathfrak{g}(A,I)=\widetilde{\mathfrak{g}}(A,I)/\mathfrak{r}$, where
$\mathfrak{r}$ is the maximal graded ideal that trivially intersects
$\mathfrak{h}$ (indeed, such an ideal is unique because the sum of any two
such ideals also trivially intersects $\mathfrak{h}$). The quotienting
preserves the triangular decomposition in the sense that
$\mathfrak{r}=\mathfrak{r}\cap\widetilde{\mathfrak{n}}^{-}\oplus\mathfrak{r}\cap\widetilde{\mathfrak{n}}^{+}$.
Hence, the triangular decomposition descends to a triangular decomposition on
$\mathfrak{g}(A,I)=\mathfrak{n}^{-}\oplus\mathfrak{h}\oplus\mathfrak{n}^{+}$.
Similarly, we have a root space decomposition. The images of
$\\{e_{i},f_{i},h_{i}\\}$ for $1\leq i\leq n$ under the projection map are
still linearly independent, so we shall by abuse of notation use
$\\{e_{i},f_{i},h_{i}\\}$ to refer to these images as well. These clearly
generate $\mathfrak{g}(A,I)$.

If one takes $I$ to consist of solely zero parities, then the Lie superalgebra
is purely even and we recover the definition of a contragredient Lie algebra.
We point out that unlike the classical case, inequivalent choices of Cartan
matrices $A$ and parity set $I$ can produce isomorphic Lie superalgebras. We
can also do a similar construction where the entries of $A$ are arbitrary
elements of $\mathbb{K}$ (cf. [CE07a]), but the definitions provided shall
suffice for our purposes and are more convenient.

A Cartan matrix $A$ is indecomposable if there is no suitable permutation of
rows and columns of $A$ that can make $A$ block diagonal. A Cartan matrix $A$
with entries in $\mathbb{Z}$ is called a generalized Cartan matrix (with
respect to a parity set $I$) if it satisfies the following properties:

  1. (1)

$a_{ii}=2$ if $\mathrm{par}(i)=0$;

  2. (2)

$a_{ii}\in{0,1}$ if $\mathrm{par}(i)=1$;

  3. (3)

if $a_{ii}=2$ then $a_{ij}\in\mathbb{Z}_{\leq 0}$ for all $j$;

  4. (4)

$a_{ij}=0\Leftrightarrow a_{ji}=0$ for all $i\neq j$.

This is slightly different from the usual definition of a generalized Cartan
matrix (cf. [HS07]) but it shall be convenient for our purposes; for instance,
we can deduce the parity set from the Cartan matrix and shall henceforth
suppress it in the notation. Lastly, we shall say $A$ is symmetrizable if it
factors as the product of a diagonal matrix and a symmetric matrix.

Recall that a Kac-Moody Lie algebra over $\mathbb{K}$ is a contragredient Lie
algebra with symmetrizable, generalized Cartan matrix (with respect to the
purely even parity set). For Kac-Moody Lie algebras over $\mathbb{C}$ with
Cartan matrix $A$, which include the simple Lie algebras, the defining
relations of the ideal $\mathfrak{r}$ are well known; they are the Serre
relations (cf. [Kac90]):

$\displaystyle(\mathrm{ad}\ {e_{i}})^{1-a_{ij}}(e_{j})=(\mathrm{ad}\
{f_{i}})^{1-a_{ij}}(f_{j})=0$

for all $1\leq i\neq j\leq n$. Over $\mathbb{K}$ these relations also hold,
but there may be others. In general, the defining relations are not easily
determined. For the exceptional simple Lie superalgebras we consider, however,
there exists an analog of the Serre relations (cf. Proposition 2.5.1 in
[BGL06]).

For more details on the theory of contragredient Lie (super)algebras, one can
refer to [Kac90, Ser11].

##  3\. Constructing Lie Superalgebras from Lie Algebras Using
Semisimplification

In this section, we describe how one can construct a Lie superalgebra from a
Lie algebra using semisimplification, which will be the procedure used to
construct the exceptional simple Lie superalgebras. From now on, we shall
assume that $\mathbb{K}$ is an algebraically closed field of characteristic
$p>2$.

###  3.1. Operadic Lie Algebras

Recall that a Lie algebra over $\mathbb{K}$ is a vector space $\mathfrak{g}$
endowed with a $\mathbb{K}$-bilinear map
$[,]:\mathfrak{g}\times\mathfrak{g}\rightarrow\mathfrak{g}$ which is skew-
symmetric (assuming $\mathrm{char}\ \mathbb{K}\neq 2$) and satisfies the
Jacobi identity. Categorically speaking, the category
$\operatorname*{Vec}_{\mathbb{K}}$ of vector spaces over $\mathbb{K}$ is a
symmetric tensor category endowed with the usual braiding $c_{X,Y}:X\otimes
Y\rightarrow Y\otimes X$ given by interchanging $X$ and $Y$, a natural
isomorphism in objects $X$ and $Y$. Then, a Lie algebra (in the category
$\operatorname*{Vec}_{\mathbb{K}}$) is an object in $\mathfrak{g}$ equipped
with a morphism $B:\mathfrak{g}\otimes\mathfrak{g}\rightarrow\mathfrak{g}$
such that the following relations of morphisms hold:

$\displaystyle
B\circ(1_{\mathfrak{g}\otimes\mathfrak{g}}+c_{\mathfrak{g},\mathfrak{g}})=0,$
$\displaystyle B\circ(B\otimes 1_{\mathfrak{g}})\circ(1_{\mathfrak{g}^{\otimes
3}}+(123)_{\mathfrak{g}^{\otimes 3}}+(132)_{\mathfrak{g}^{\otimes 3}})=0,$

where the permutation $(123)_{\mathfrak{g}^{\otimes 3}}:\mathfrak{g}^{\otimes
3}\rightarrow\mathfrak{g}^{\otimes 3}$ is given by $(1_{\mathfrak{g}}\otimes
c_{\mathfrak{g}\otimes\mathfrak{g}})\circ(c_{\mathfrak{g}\otimes\mathfrak{g}}\otimes
1_{\mathfrak{g}})$, and the permutation $(132)_{\mathfrak{g}^{\otimes
3}}:\mathfrak{g}^{\otimes 3}\rightarrow\mathfrak{g}^{\otimes 3}$ is given by
$(c_{\mathfrak{g}\otimes\mathfrak{g}}\otimes
1_{\mathfrak{g}})\circ(1_{\mathfrak{g}}\otimes
c_{\mathfrak{g}\otimes\mathfrak{g}})$ (let us not concern ourselves with the
associativity morphisms). The first relation corresponds to the skew-symmetry
condition, and the second is the Jacobi identity. Using these as defining
axioms, we can extend the definition to any symmetric tensor category
$\mathcal{C}$ with braiding $c$, and call the pair $(\mathfrak{g},B)$ an
operadic Lie algebra in $\mathcal{C}$. We can also allow $\mathfrak{g}$ to be
an ind-object in the category.

Recall that the category of supervector spaces
$\operatorname*{sVec}_{\mathbb{K}}$ consists of
$\mathbb{Z}/2\mathbb{Z}$-graded vector spaces and morphisms, with braiding
$c_{X,Y}$ given by the Koszul sign rule:

(3.1) $c_{X,Y}(x\otimes y)=(-1)^{|x||y|}(y\otimes x),$

where $x,y$ are homogeneous (i.e. purely even or purely odd). We call a Lie
algebra in the category of supervector spaces
$\operatorname*{sVec}_{\mathbb{K}}$ a Lie superalgebra.

######  Remark 3.1.

One should note that in characteristic $3$, the usual definition of a Lie
superalgebra and the definition given above do not coincide, as for an odd
element $x$, the relation $[x,[x,x]]=0$ required for a Lie superalgebra does
not follow from the Jacobi identity. However, this will not be a major concern
for our considerations.

In a general symmetric tensor category we do not have the notion of elements
or vectors in an object. However, for the purposes of this paper, we will be
working with Lie algebras in the representation category of a certain finite
group scheme, so we can refer to vectors in objects of this category by
applying the forgetful functor into $\operatorname*{Vec}_{\mathbb{K}}$.
Furthermore, using vectors will make it easier to talk about and describe the
bracket.

For more details on operadic Lie algebras, one can consult [Eti18].

###  3.2. The Verlinde Category

We want to use the language of Lie algebras in a symmetric tensor category. To
do so, we need to pick suitable symmetric tensor categories. Let
$\mathbb{K}{\mbox{\boldmath$\alpha$}}_{p}=\mathbb{K}[t]/(t^{p})$, where
$\mathbb{K}$ is an algebraically closed field of characteristic $p$, and let
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{p}$ denote the category of
finite-dimensional representations of
$\mathbb{K}{\mbox{\boldmath$\alpha$}}_{p}$. As
$\mathbb{K}{\mbox{\boldmath$\alpha$}}_{p}$ is a Hopf algebra with
comultiplication defined by letting $t$ be primitive,
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{p}$ is a symmetric tensor
category with braiding given by the usual braiding of vector spaces (there is
a forgetful functor from $\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{p}$
to $\operatorname*{Vec}_{\mathbb{K}})$. Hence, an example of a Lie algebra
$(\mathfrak{g},B)$ in $\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{p}$ is a
Lie algebra in $\operatorname*{Vec}_{\mathbb{K}}$ equipped with a nilpotent
element $x\in\mathfrak{g}$ of order at most $p$; then $\mathfrak{g}$ is a
$\mathbb{K}{\mbox{\boldmath$\alpha$}}_{p}$-representation by letting $t$ act
as $\mathrm{ad}\ x$, and $B$ is naturally a morphism in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{p}$ by the Jacobi identity (as
a Lie algebra in $\operatorname*{Vec}_{\mathbb{K}}$). More generally, we can
take $t$ to be any nilpotent derivation of order at most $p$ (not necessarily
inner).

The category $\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{p}$ is not
semisimple; indeed, it contains non-simple indecomposable objects. The
pairwise non-isomorphic indecomposable objects are given by the
representations $J_{n}=\mathbb{K}^{n}$ where $t$ acts as the nilpotent Jordan
block of size $n$ ($1\leq n\leq p$). If $v_{1},v_{2},\dots,v_{n}$ is a basis
of $J_{n}$ such that $t.v_{i}=v_{i+1}$, we shall use the notation
$v_{1}\mapsto v_{2}\mapsto\cdots\mapsto v_{n}$ to refer to that particular
object $J_{n}$.

The semisimplification of this category is the Verlinde category
$\operatorname{Ver}_{p}$. Recall that in essence, the semisimplification of a
symmetric tensor category is the symmetric tensor category obtained by
declaring all indecomposable objects to be simple, except those with zero
dimension, which are removed (for more details, see [EO19]). More importantly,
there is a semisimplification functor from a symmetric tensor category
$\mathcal{C}$ to its a semisimplification $\overline{\mathcal{C}}$, and it is
symmetric and monoidal. We will denote the images of objects under this
functor with an overline over the original object.

Therefore, the simple objects in $\operatorname{Ver}_{p}$ are
$L_{1},\dots,L_{p-1}$, which are the images of $J_{1},\dots,J_{p-1}$ under the
semisimplification functor, respectively, i.e. $L_{i}=\overline{J_{i}}$. If
$v_{1}\mapsto v_{2}\mapsto\cdots\mapsto v_{i}$ denotes a $J_{i}$, we shall
refer to the corresponding copy of $L_{i}$ by $\overline{v_{1}\mapsto
v_{2}\mapsto\cdots\mapsto v_{i}}$ (for $i<p$). Note that $L_{p}$ is sent to
the zero object as it is $p$-dimensional, so its categorical dimension is $0$.
It is well known that the tensor product is given by the truncated Clebsch-
Gordan rule, which is similar to the usual Clebsch-Gordan rule of
$\mathfrak{sl}_{2}(\mathbb{C})$-modules (the truncation comes from the terms
in red):

$L_{m}\otimes
L_{n}=\bigoplus_{i=1}^{\min(m,n,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}p-m,p-n})}L_{|m-n|+2i-1}.$

In particular, we notice that $\mathbbm{1}\coloneqq L_{1}$ is the unit object
with respect to tensor product. More importantly, we have the following
proposition:

######  Proposition 3.2.

The category $\operatorname*{sVec}_{\mathbb{K}}$ is symmetric tensor
equivalent to the subcategory generated by the objects $L_{1}$ and $L_{p-1}$
in $\operatorname{Ver}_{p}$.

###### Proof.

This follows from results in [Ost15], but we offer a direct, linear algebraic
proof. We have $L_{p-1}\otimes L_{p-1}=L_{1}$ by the truncated Clebsch-Gordan
rule. Therefore, one needs to check that the induced braiding on
$\operatorname{Ver}_{p}$ from
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{p}$ under the
semisimplification functor (let us use $c$ to denote both the braiding and its
image) restricts to this subcategory appropriately.

In $\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{p}$, let $J_{p-1}$ be given
by the basis $v_{1}\mapsto v_{2}\mapsto\dots\mapsto v_{p-1}$. It is known that
in $\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{p}$, $J_{p-1}\otimes
J_{p-1}=J_{1}\oplus(p-2)J_{p}$ (cf. [Gre62]). One such decomposition is as
follows. For each $1\leq i\leq p-2$, a copy of $J_{p}$ arises from the
submodule generated by the vector $v_{i}\otimes v_{1}$. It is clear that
$t^{p}.(v_{i}\otimes v_{1})=0$ but $t^{p-1}.(v_{i}\otimes v_{1})$ is nonzero,
so these do indeed give a copy of $J_{p}$. Since $t^{p}=0$, these copies of
$J_{p}$ are automatically split. The copy of $J_{1}$ arises as the span of the
vector $\sum_{i=1}^{p-1}(-1)^{i}v_{i}\otimes v_{p-i}$. To see that this copy
of $J_{1}$ is split, let us define the degree of $v_{i}\otimes v_{j}$ to be
$i+j$. This gives a grading on $J_{p-1}\otimes J_{p-1}$, where the action of
$t$ increases the grading by $1$, and the copies of $J_{p}$ and $J_{1}$ above
are graded submodules. Suppose the copy of $J_{1}$ intersected the span of the
copies of $J_{p}$, meaning it lies in the graded subspace spanned by
$t^{p-1}.(v_{i}\otimes v_{1})$ for $1\leq i\leq p-2$. The vector
$\sum_{i=1}^{p-1}(-1)^{i}v_{i}\otimes v_{p-i}$ has degree $p$, but homogeneous
vectors in this graded subspace all have degree greater than or equal to
$p+1$. By way of contradiction, we deduce that
$\sum_{i=1}^{p-1}(-1)^{i}v_{i}\otimes v_{p-i}$ spans a split summand.

The copies of $J_{p}$ vanish in the semisimplification, so let’s just look at
the copy of $J_{1}$ spanned by the vector
$\sum_{i=1}^{p-1}(-1)^{i}v_{i}\otimes v_{p-i}$. Since $p$ is odd, this sum has
an even number of terms, and the braiding map
$c_{J_{p-1},J_{p-1}}:J_{p-1}\otimes J_{p-1}\rightarrow J_{p-1}\otimes J_{p-1}$
will negate this element. Therefore, in the semisimplification, we have
$L_{p-1}\otimes L_{p-1}=L_{1}$, and the image of the braiding
$c_{J_{p-1},J_{p-1}}$ under the semisimplification functor is
$c_{L_{p-1},L_{p-1}}:L_{p-1}\otimes L_{p-1}\rightarrow L_{p-1}\otimes L_{p-1}$
is multiplication by $-1$.

Therefore, the image of $c$ under the semisimplification functor restricted to
the subcategory generated by $L_{1}$ and $L_{p-1}$ behaves as described in
equation (3.1). ∎

Hence, from now on, we can safely refer to the full subcategory generated by
$L_{1}$ and $L_{p-1}$ as $\operatorname*{sVec}_{\mathbb{K}}$. By
functoriality, if we are given a Lie algebra $(\mathfrak{g},B)$ in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{p}$, the projection
$(\overline{\mathfrak{g}},\overline{B})$ of its semisimplification in
$\operatorname{Ver}_{p}$ is a Lie algebra in
$\operatorname*{sVec}_{\mathbb{K}}$, which is a Lie superalgebra. Therefore,
we can produce Lie superalgebras over fields of characteristic $p$ from Lie
algebras over fields of characteristic $p$ by specifying a nilpotent element
of order at most $p$.

######  Remark 3.3.

To get a Lie superalgebra in characteristic $3$, one must quotient out by the
further relation that $[x,[x,x]]=0$ for odd $x$. For instance, consider the
free Lie algebra $\mathfrak{g}$ on $x,y$ modulo degree $4$; this is a Lie
algebra with basis $\\{x,y,[x,y],[x,[x,y]],[y,[y,x]]\\}$. Define the
derivation $d$ by $d(x)=y$ and $d(y)=0$. Then, $\mathfrak{g}$ can be realized
as an object $\mathfrak{g}$ in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ with respect to $d$, where
we have a copy of $J_{1}$ for $[x,y]$ and copies of $J_{2}$ given by $x\mapsto
y$ and $[x,[x,y]]\mapsto-[y,[y,x]]$; semisimplifying gives a three-dimensional
operadic Lie superalgebra with odd generator $z$ and basis
$\\{z,[z,z],[z,[z,z]]\\}$.

###  3.3. Explicit Description of Semisimplification in Characteristic $3$

The language of symmetric tensor categories is naturally suited for talking
about semisimplification. However, since it is relatively new, we offer an
explicit description of what happens to the Lie bracket under
semisimplification using linear algebra. We will use this language in our
proofs below.

Let $\mathfrak{g}$ be a Lie algebra in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ with respect to some
derivation $d$ such that $d^{3}=0$. We can pick a non-canonical decomposition
$\mathfrak{g}=n_{1}J_{1}\oplus n_{2}J_{2}\oplus n_{3}J_{3}$. For each copy of
$J_{1}$, we pick a basis $x_{i}$, where $1\leq i\leq n_{1}$. For each copy of
$J_{2}$, we pick a basis $x_{i}\mapsto x_{i}^{\prime}$, where $1+n_{1}\leq
i\leq n_{1}+n_{2}$. Finally, for each copy of $J_{3}$, we pick a basis
$x_{i}\mapsto x_{i}^{\prime}\mapsto x_{i}^{\prime\prime}$ for
$1+n_{1}+n_{2}\leq i\leq n_{1}+n_{2}+n_{3}$. The collection
$\\{x_{i},x_{j},x_{j}^{\prime},x_{k},x_{k}^{\prime},x_{k}^{\prime\prime}\\}$
for $1\leq i\leq n_{1}$, $1+n_{1}\leq j\leq n_{1}+n_{2}$ , $1+n_{2}+n_{3}\leq
k\leq n_{1}+n_{2}+n_{3}$ is a basis of $\mathfrak{g}$.

After semisimplification, the $J_{3}$ terms vanish, and the $J_{1}$ terms and
$J_{2}$ terms give $L_{1}$ terms and $L_{2}$ terms, respectively, which
collectively give rise to a basis of the Lie superalgebra
$\overline{\mathfrak{g}}$. In particular, for $1\leq i\leq n_{1}$, the copy of
$J_{1}$ corresponding to $x_{i}$ gives an even basis vector $y_{i}$ in
$\overline{\mathfrak{g}}$ which spans the subspace $\overline{x_{i}}$. Here,
by abuse of notation, $x_{i}$ refers to the basis vector we picked above and
the $J_{1}$ it spans, so that the notation $\overline{x_{1}}$ makes sense. For
$n_{1}+1\leq n_{1}+n_{2}$, the copy of $J_{2}$ corresponding to $x_{i}\mapsto
x_{i}^{\prime}$ gives an odd basis vector $y_{i}$ in $\overline{\mathfrak{g}}$
which spans the subspace $\overline{x_{i}\mapsto x_{i}^{\prime}}$. Later in
the text, we will simply write $\overline{x_{i}}$ or $\overline{x_{i}\mapsto
x_{i}^{\prime}}$ in place of $y_{i}$, even though these are subspaces of
$\overline{\mathfrak{g}}$ (this is an another abuse of notation). For
instance, if $n_{1}=2$, then the bracket $[y_{1},y_{2}]$ will be written as
$[\overline{x_{1}},\overline{x_{2}}]$ and the bracket $[y_{2},y_{3}]$ will be
written as $[\overline{x_{2}},\overline{x_{3}\mapsto x_{3}^{\prime}}]$.

Now, we will describe the structure constants of this basis of
$\overline{\mathfrak{g}}$.

######  Proposition 3.4.

Let $C(i,j,k)$ denote the structure constants of the basis $\\{y_{i}\\}$ of
$\overline{\mathfrak{g}}$ above, i.e. $[y_{i},y_{j}]=\sum_{k}C(i,j,k)y_{k}$.
Then,

  1. (1)

if $i,j,k\leq n_{1}$, or $1\leq i\leq n_{1}$ and $j,k>n_{1}$, or $j<n_{1}$ and
$i,k>n_{1}$, then $C(i,j,k)$ is the coefficient of $x_{k}$ in $[x_{i},x_{j}]$;

  2. (2)

if $i,j>n_{1}$ and $k\leq n_{1}$, then $C(i,j,k)$ is the coefficient of
$x_{k}$ in $-[x_{i},x_{j}^{\prime}]+[x_{i}^{\prime},x_{j}]$;

  3. (3)

in all other cases, $C(i,j,k)=0$.

###### Proof.

The proofs of $1)$ and $3)$ are easy. The proof of $2)$ is a consequence of
the proof of Proposition 3.2. ∎

Let’s do this calculation explicitly for $\mathfrak{g}=\mathfrak{gl}_{3}$. Let
$e_{ij}$ refer to the elementary matrix with a $1$ in the $(i,j)$ entry and
zero elsewhere. Then, with respect to the adjoint action of $e_{23}$,
$\mathfrak{g}$ is an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$. We choose the following
decomposition: $e_{11}$, $e_{11}+e_{22}+e_{33}$ as copies of $J_{1}$;
$e_{12}\mapsto-e_{13}$ and $e_{31}\mapsto e_{21}$ as copies of $J_{2}$; and
$e_{32}\mapsto e_{22}-e_{33}\mapsto e_{23}$ as a copy of $J_{3}$. Then, the
basis vectors of $\overline{\mathfrak{g}}$ are $y_{1}=\overline{e_{11}}$,
$y_{2}=\overline{e_{11}+e_{22}+e_{33}}$, $y_{3}=\overline{e_{12}\mapsto-
e_{13}}$, and $y_{4}=\overline{e_{31}\mapsto e_{21}}$.

Applying the formula, we have: $[y_{1},y_{2}]=0$, $[y_{1},y_{3}]=y_{3}$, and
$[y_{1},y_{4}]=-y_{4}$. We can also compute $[y_{3},y_{4}]$. Both of these
vectors are odd, so we look at
$-[e_{12},e_{21}]+[-e_{13},e_{31}]=(e_{22}-e_{11})-(e_{11}-e_{33})=e_{11}+e_{22}+e_{33}$,
so $[y_{3},y_{4}]=y_{2}$. Therefore, the resulting Lie superalgebra is
$\mathfrak{gl}_{1|1}$.

###  3.4. An Example

Let us consider a more complicated example. Consider $\mathfrak{g}_{2}$, the
$14$-dimensional exceptional simple Lie algebra of rank $2$. It has Cartan
matrix $\begin{pmatrix}2&-1\\\ -3&1\end{pmatrix}$ and Dynkin diagram:

$\dynkin[labels={\beta,\alpha},scale=4,arrowshape/.style={-{jibladze[length=7pt]}},labeldistance=5pt]G{oo},$

where $\alpha$ corresponds to the index $1$ and $\beta$ to the index $2$. Its
root system can be visualized geometrically as the following:

$\alpha$$\beta$$\beta+\alpha$$\beta+2\alpha$$\beta+3\alpha$$2\beta+3\alpha$

In characteristic $3$, $\mathfrak{g}_{2}$ has non-equivalent reductions from
$\mathfrak{g}_{2}$ over $\mathbb{C}$. For instance, it can be viewed via the
Cartan matrix definition, which gives a $10$-dimensional simple Lie algebra
(note that this Cartan matrix is not symmetrizable in characteristic $3$). It
can also be viewed as the reduction of $\mathfrak{g}_{2}$ in characteristic
$0$ using the Kostant $\mathbb{Z}$-form (cf. [Hum94]). For our example, we
will take $\mathfrak{g}_{2}$ to be the reduction of the following basis of
$\mathfrak{g}_{2}$ over $\mathbb{C}$ to $\mathbb{K}$:
$\\{e_{\alpha},e_{\beta},[e_{\alpha},e_{\beta}],(\operatorname*{ad}e_{\alpha})^{2}(e_{\beta}),(\operatorname*{ad}e_{\alpha})^{3}(e_{\beta}),[[e_{\alpha},e_{\beta}],(\operatorname*{ad}e_{\alpha})^{2}(e_{\beta})]\\}$
for the upper triangular subalgebra; the corresponding basis for the lower
triangular subalgebra; and the Cartan subalgebra with basis
$\\{h_{\alpha},h_{\beta}\\}$.

With this definition, $\mathfrak{g}$ is not simple but rather has a four-
dimensional ideal $I$ spanned by the root spaces corresponding to the roots
$2\beta+3\alpha$ and $\beta+3\alpha$ and their negatives. Indeed, the Kac-
Moody Lie algebra definition will give a $10$-dimensional simple Lie algebra.
This is also known as the Brown algebra $\mathfrak{br}(2;0)$.

Now, let’s produce an example of semisimplification. For each positive root
$\gamma$, let $\\{e_{\gamma},h_{\gamma},f_{\gamma}\\}$ denote the
corresponding $\mathfrak{sl}_{2}$-triple. By the Serre relations, $e_{\beta}$
is nilpotent of degree $2$, so we can realize $\mathfrak{g}_{2}$ as an object
in $\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ w.r.t $e_{\beta}$.
Under this realization, a direct sum decomposition into indecomposables,
compatible with the root grading, is as follows: a copy of $J_{3}$
corresponding to $f_{\beta}\mapsto h_{\beta}\mapsto-2e_{\beta}$; four copies
of $J_{2}$ corresponding to $e_{\alpha}\mapsto e_{\beta+\alpha}$,
$e_{\beta+3\alpha}\mapsto e_{2\beta+3\alpha}$, $f_{\beta+\alpha}\mapsto
f_{\alpha}$, $f_{2\beta+3\alpha}\mapsto f_{\beta+3\alpha}$; and three copies
of $J_{1}$, spanned by $e_{\beta+2\alpha},f_{\beta+2\alpha}$, and
$h_{\beta+2\alpha}$. Therefore, $\mathfrak{g}_{2}=3J_{1}\oplus 4J_{2}\oplus
J_{3}$. Notice that the ideal $I$ is also an object in the category and
moreover it is a split summand of $\mathfrak{g}_{2}$ in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$.

In the semisimplification, the $J_{3}$ term vanishes and we are left with a
Lie superalgebra $\overline{\mathfrak{g}_{2}}$ of superdimension $(3|4)$.
Because $I$ is an ideal and a split summand, $\overline{I}$ is a subobject of
$\overline{\mathfrak{g}_{2}}$, so by the tensoriality of the
semisimplification functor, $\overline{I}$ is an ideal of
$\overline{\mathfrak{g}_{2}}$. It is clear that the even part of
$\overline{\mathfrak{g}_{2}}$ is the Lie algebra $\mathfrak{sl}_{2}$, as it is
the image of the $\mathfrak{sl}_{2}$-triple corresponding to the root
$\beta+2\alpha$. The odd part is then an $\mathfrak{sl}_{2}$-module visualized
below. Specifically, before semisimplification, the action of
$e_{\beta+2\alpha}$ on the $J_{2}$’s is labeled by the horizontal arrows (up
to a multiple, which may be zero). and reversing the arrows gives the action
of $f_{\beta+2\alpha}$ (again, up to a multiple); the vertical arrows
themselves are the $e_{\beta}$ action as usual:

$\begin{smallmatrix}e_{-2\beta-3\alpha}\\\
\rotatebox[origin={c}]{-90.0}{$\scriptstyle\mapsto$}\mkern 2.0mu\\\
e_{-\beta-3\alpha}\end{smallmatrix}\longmapsto\begin{smallmatrix}e_{-\beta-\alpha}\\\
\rotatebox[origin={c}]{-90.0}{$\scriptstyle\mapsto$}\mkern 2.0mu\\\
e_{-\alpha}\end{smallmatrix}\longmapsto\begin{smallmatrix}e_{\alpha}\\\
\rotatebox[origin={c}]{-90.0}{$\scriptstyle\mapsto$}\mkern 2.0mu\\\
e_{\beta+\alpha}\end{smallmatrix}\longmapsto\begin{smallmatrix}e_{\beta+3\alpha}\\\
\rotatebox[origin={c}]{-90.0}{$\scriptstyle\mapsto$}\mkern 2.0mu\\\
e_{2\beta+3\alpha}\end{smallmatrix}.$

Here, we write $e_{-\gamma}$ to denote $f_{\gamma}$ to make it clear what
happens to root spaces. This descends to the corresponding $L_{2}$’s in the
semisimplification. It turns out that this is a nonsplit extension of the
natural module, spanned by the middle two copies of $L_{2}$, with two copies
of the trivial representation, spanned by the outer two copies of $L_{2}$,
which also are a basis of the ideal $\overline{I}$.

Let us compute some brackets in the odd part. The bracket
$[\overline{e_{\alpha}\mapsto e_{\beta+\alpha}},\overline{e_{\alpha}\mapsto
e_{\beta+\alpha}}]$ is given by:

$[\overline{e_{\alpha}\mapsto e_{\beta+\alpha}},\overline{e_{\alpha}\mapsto
e_{\beta+\alpha}}]=\overline{-[e_{\alpha},e_{\beta+\alpha}]+[e_{\beta+\alpha},e_{\alpha}]}=-2\overline{e_{2\beta+\alpha}}$

A similar argument goes through for the $f$ counterpart. Finally, we have

$[\overline{e_{\alpha}\mapsto
e_{\beta+\alpha}},\overline{f_{\beta+\alpha}\mapsto
f_{\alpha}}]=\overline{-[e_{\alpha},f_{\alpha}]+[e_{\beta+\alpha},f_{\beta+\alpha}]}=\overline{-h_{\alpha}+h_{\beta+\alpha}}=\overline{-h_{\beta+2\alpha}}.$

These computations show that $\overline{\mathfrak{g}_{2}}/\overline{I}$ is
$\mathfrak{osp}_{1|2}$ (alternatively, semisimplifying $\mathfrak{br}(2;0)$
gives $\mathfrak{osp}_{1|2}$).

If one chooses the definition of $\mathfrak{g}_{2}$ over $\mathbb{K}$ to be
the one arising from reducing the Chevalley basis (cf. [Hum94]), then there is
a $7$-dimensional ideal $S$ isomorphic to $\mathfrak{psl}_{3}$ generated by
the root vectors associated to the short roots
$\alpha,\beta+\alpha,\beta+2\alpha$. Specifically, in $S$ we have copies of
$J_{1}$ corresponding to $h_{\alpha}$, $e_{\beta+2\alpha}$, and
$f_{\beta+2\alpha}$, and copies of $J_{2}$ corresponding to $e_{\alpha}\mapsto
e_{\alpha+\beta}$ and $f_{\alpha+\beta}\mapsto-f_{\alpha}$. With respect to
the adjoint action of $e_{\beta}$, this is a split ideal in $\mathfrak{g}_{2}$
in $\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$. Complementary to $S$
there is also a copy of $J_{3}$ associated to the $\mathfrak{sl}_{2}$-triple
$\\{f_{\beta},h_{\beta},e_{\beta}\\}$. Finally, $e_{\beta+3\alpha}\mapsto-
e_{2\beta+3\alpha}$ and $f_{2\beta+3\alpha}\mapsto f_{\beta+3\alpha}$ give
copies of $J_{2}$. Hence, we can write $\mathfrak{g}_{2}=S\oplus J_{3}\oplus
2J_{2}$. When we semisimplify $\mathfrak{g}_{2}$, we again get a Lie
superalgebra of superdimension $(3|4)$, which is isomorphic to the one
obtained by using the first reduction. This time, the two-dimensional ideal
$\overline{I}$ is spanned by $\overline{e_{\alpha}\mapsto e_{\alpha+\beta}}$
and $\overline{f_{\alpha+\beta}\mapsto-f_{\alpha}}$.

###  3.5. Main Theorem

We now state the main theorem of this paper. Let us first introduce some
notation. Let $\mathfrak{g}(A)$ be an exceptional Lie algebra with Cartan
matrix $A$ of rank $n>2$ over $\mathbb{Q}$ (i.e. $\mathfrak{f}_{4}$,
$\mathfrak{e}_{6}$, $\mathfrak{e}_{7}$, $\mathfrak{e}_{8}$). Let
$\\{i_{1},i_{2},\dots,i_{l}\\}$ be a subset of boundary nodes of the Dynkin
diagram of $\mathfrak{g}(A)$, and let $\\{j_{1},\dots,j_{l}\\}$ be the indices
such that $j_{r}$ is the node connected to $i_{r}$ for each $r$. The following
picture of the Dynkin diagram of $\mathfrak{e}_{7}$ illustrates an example:

[mark=o,labels=1,2,3,4,5,6,7,scale=4, label distance = 5pt]E7 [blue,
draw=black] (root 2) circle (0.05cm); [blue, draw=black] (root 7) circle
(0.05cm); [red, draw=black] (root 4) circle (0.05cm); [red, draw=black] (root
6) circle (0.05cm);

The boundary nodes are $1,2$ and $7$. The nodes $2$ and $7$, colored in blue,
are the chosen subset of boundary nodes, and the nodes $4$ and $6$, colored in
red, are the nodes attached to the boundary nodes $2$ and $7$, respectively.
Hence, $l=2$ and $\\{i_{1},i_{2}\\}=\\{2,7\\}$ and
$\\{j_{1},j_{2}\\}=\\{4,6\\}$.

Notice that the determinant of $A$ is always nonzero, so $\mathfrak{g}(A)$ is
generated by $\\{e_{i},f_{i}\\}$, and $\mathfrak{h}(A)$ has a basis
$\\{h_{1},\dots,h_{n}\\}$. Let $\widetilde{A}=(\widetilde{a}_{ij})$ be the
$(n-l)\times(n-l)$ matrix obtained from $A$ by setting $a_{j_{r},j_{r}}=0$ for
all $r$ and deleting the row and column attached to $i_{r}$ for all $1\leq
r\leq l$. Note that the $e_{i_{r}}$’s pairwise commute, as do the
$f_{i_{r}}$’s.

Before proceeding, it is useful to review the constructions in §4 and look
closely at the Cartan matrices of the exceptional Lie algebra and the
exceptional Lie superalgebra it semisimplifies to. The key idea is that we
have a copy of $J_{2}$ given by $e_{j_{r}}\mapsto[e_{i_{r}},e_{j_{r}}]$ for
each $1\leq r\leq l$, and in the semisimplification these merge to form an odd
generator. Now, let’s state some supporting lemmas.

######  Lemma 3.5.

The determinant of $\widetilde{A}$ is nonzero.

###### Proof.

Since there are a finite number of exceptional Lie algebras and a finite
number of boundary nodes, there are only a finite number of possibilities of
$\widetilde{A}$. Using a computer program, we compute the determinant of all
of these, and it is never zero. However, this can tediously also be done by
hand, as there are not too many cases. ∎

As a consequence, the Chevalley generators generate
$\widetilde{\mathfrak{g}}(\widetilde{A})$, and the dimension of its Cartan
subalgebra is $n-l$.

######  Lemma 3.6.

The element $e=\sum_{r=1}^{l}e_{i_{r}}$ is nilpotent of degree at most $3$ and
$\mathfrak{g}(A)$ can be realized as a Lie algebra in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ w.r.t.
$\operatorname*{ad}e$.

###### Proof.

Without loss of generality, by suitably reordering the indices, we may assume
that $i_{1}=1,i_{2}=2,\dots,i_{l}=l$ and $j_{1}=1+l,j_{2}=2+l,\dots,j_{l}=2l$.
Then, by the Serre relations, we have $[e_{i},e_{j}]=0$ for all $1\leq i\leq
l$ and $1\leq j\neq i+l\leq n$ and $[e_{i},e_{i+l}]\neq 0$ but
$[e_{i},[e_{i},e_{i+l}]]=0$. It follows that in characteristic $3$,
$\operatorname*{ad}e=\operatorname*{ad}e_{1}+\cdots+\operatorname*{ad}e_{l}$
is nilpotent of degree $3$. This shows that $\mathfrak{g}(A)$ can be realized
as an object in $\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$. ∎

For the remainder of this section, we shall assume that the indices are
reordered as in the proof of Lemma 3.6.

######  Lemma 3.7.

There exists a basis $B$ of $\mathfrak{g}(A)$ in which $\operatorname*{ad}e$
acts by the direct sum of Jordan blocks, and the generators $e_{i},f_{i}$
($1\leq i\leq n$) of $\mathfrak{g}(A)$ together with a suitable basis of the
Cartan subalgebra $\mathfrak{h}(A)$ of $\mathfrak{g}(A)$ collectively form a
subset of $B$.

###### Proof.

We will construct such a basis. It is useful to extend the action of $e$ to an
$\mathfrak{sl}_{2}$-triple $\\{e,f,h\\}$, where $f=f_{1}+f_{2}+\cdots+f_{l}$
and $h=[e,f]=h_{1}+h_{2}+\cdots+h_{l}$. We have the following indecomposables
in $\mathfrak{g}(A)$:

$J_{1}:\ \ \\{e_{1+2l},\dots,e_{n}\\},$ $J_{1}:\ \
\\{f_{1+2l},\dots,f_{n}\\},$ $J_{2}:\ \
\\{e_{1+l}\mapsto[e,e_{1+l}],\dots,e_{2l}\mapsto[e,e_{2l}]\\}.$

Since $a_{i_{r},j_{r}}=-1$ for all $r$, we also have:

$J_{2}:\ \ \\{[f,f_{1+l}]\mapsto f_{1+l},\dots,[f,f_{2l}]\mapsto f_{2l}\\}.$

We then have the following $\mathfrak{sl}_{2}$-triples giving $J_{3}$’s:

$J_{3}:\ \ \\{f_{1}\mapsto h_{1}\mapsto e_{1},\dots,f_{l}\mapsto h_{l}\mapsto
e_{l}\\}.$

Finally, we can consider the remaining part of the Cartan subalgebra. We have

$J_{1}:\ \ \\{h_{1+2l},\dots,h_{n}\\},$ $J_{1}:\ \
\\{h_{1+l}-h_{1},\dots,h_{2l}-h_{l}\\}.$

We claim that the sum $W$ of these copies of $J_{i}$ is a direct sum. First,
note that each $J_{i}$ is $Q$-homogeneous even though $e$ is not because
$[e,e_{i+l}]=[e_{i},e_{i+l}]$, $[f,f_{i+l}]=[f_{i},f_{i+l}]$,
$[e_{i},e_{j}]=[f_{i},f_{j}]=0$ for $1\leq i,j\leq l$ by the Serre relations.
Then, because each root space here is $1$-dimensional and because
$\\{h_{1},\dots,h_{l},h_{1+l}-h_{1},\dots,h_{l+l}-h_{l},h_{2l+1},\dots,h_{n}\\}$
is a basis of the Cartan subalgebra, this tells us that their sum is direct.

Now, we argue that $W$ is a direct summand of $\mathfrak{g}(A)$ in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$. First, note that $W$ is
$Q$-homogeneous because each $J_{i}$ is. This means that the height grading on
$\mathfrak{g}(A)$ also restricts to $W$. Let $\mathfrak{g}_{i}$ denote the
subspace of height $i$ in $\mathfrak{g}(A)$. Then, $W$ contains
$\mathfrak{g}_{-1},\mathfrak{g}_{0},\mathfrak{g}_{1}$ and has trivial
intersection with $\mathfrak{g}_{i}$ for $|i|>2$. Let $W^{\prime}$ be the span
of the root spaces that do not intersect $W$. $W^{\prime}$ is also
$Q$-homogeneous and graded by height, has trivial intersection with each of
$\mathfrak{g}_{-1},\mathfrak{g}_{0},\mathfrak{g}_{1}$, and contains
$\mathfrak{g}_{i}$ for $|i|>2$. Clearly, $\mathfrak{g}(A)=W\oplus W^{\prime}$
as vector spaces; we need to show that $W^{\prime}$ is
$\operatorname*{ad}e$-invariant.

To do so, consider any $x\in W^{\prime}$, and let $x=\sum_{|i|\geq 2}x_{i}$ be
a height decomposition with $x_{i}\in\mathfrak{g}_{i}$. Now, note that $e$
raises height by $1$ on homogeneous vectors, so for $i\leq-4$ and $i\geq 2$,
$[x,x_{i}]\in W^{\prime}$. Therefore, it suffices to assume that $x$ is of the
form $x=x_{-3}+x_{-2}$.

The terms $[e,x_{-2}]$ and $[e,x_{-3}]$ have different heights, so we can look
at them individually. Let’s start with $[e,x_{-2}]$. This has height $-1$, so
it lies in $W$. Hence, we can write $[e,x_{-2}]=\sum_{i=1}^{n}c_{i}f_{i}$ for
suitable $c_{i}$. Suppose $[e,x_{-2}]\neq 0$. By the $Q$-grading and because
$e=\sum_{k=1}^{l}e_{k}$, this means we can write
$x_{-2}=\sum_{k=1}^{l}\sum_{i=1}^{n}c_{ki}[f_{k},f_{i}]$. However, by the
Serre relations, $[f_{k},f_{i}]=0$ for all $1\leq k\leq l$ and $1\leq i\neq
k\leq n$. Hence, $x_{-2}=\sum_{k=1}^{l}c_{k,k+l}[f_{k},f_{k+l}]$ which lies in
$W$, a contradiction if $x_{-2}\neq 0$. Therefore, $[e,x_{-2}]=0$.

So we can actually assume that $x$ is homogeneous of height $-3$. Then, we can
write $[e,x]=w+w^{\prime}$, where $w\in W,w^{\prime}\in W^{\prime}$ both have
height $-2$. In particular, $w$ is of the form
$w=\sum_{i=1}^{l}d_{i}[f_{i},f_{i+l}]$ for suitable $d_{i}$. On the other
hand, the root decomposition of $w^{\prime}$ can only involve the root spaces
of height $-2$ with root not equal to $-\alpha_{i}-\alpha_{i+l}$ for any $i$
such that $1\leq i\leq l$. By the Serre relations, it follows that
$w^{\prime}=\sum_{i,j=l+1}^{n}y_{ij}$ where $y_{ij}$ lies in the root space
with root $-(\alpha_{i}+\alpha_{j})$. Now note that if
$-(\alpha_{k}+\alpha_{i}+\alpha_{i+l})$ is a root for $1\leq i,k\leq l$, then
the associated root space is one-dimensional. This root space is spanned by
$[f_{k},[f_{i},f_{i+l}]]$ because $[f_{i},f_{i+l}]$ is nonzero and $f_{i}$ and
$f_{k}$ commute by the Serre relations (alternatively, one can appeal to the
simplicity of $\mathfrak{g}(A)$). Hence, appealing to the $Q$-grading again,
we deduce $x$ is of the form:

$\displaystyle
x=\sum_{k=1}^{l}\sum_{i,j=l+1}^{n}y_{kij}+\sum_{k=1}^{l}\sum_{i=1}^{l}d_{ki}[f_{k},[f_{i},f_{i+l}]].$

where $y_{kij}$ lies is in the root space associated to the root
$-(\alpha_{k}+\alpha_{i}+\alpha_{j})$ and $d_{ki}$ are suitable constants. The
bracket of the first sum with $e$ is $w^{\prime}$ and the bracket of the
second sum with $e$ is $w$. However, $[f_{k},[f_{i},f_{i+l}]]=0$ for $1\leq
k,i\leq l$ by the Serre relations, so the second sum is zero, which means that
$w=0$. And therefore $[e,x]=w^{\prime}\in W^{\prime}$. This shows the claim.
Now, the basis $B$ is the basis of $W$ prescribed above together with any
Jordan basis of $W^{\prime}$. ∎

######  Remark 3.8.

We want to emphasize that at no point in the proof of Lemma 3.7 did we appeal
to the simplicity of $\mathfrak{g}(A)$. This will be important for Conjecture
4.13.

Let’s work out an explicit example. Consider the Dynkin diagram of
$\mathfrak{e}_{6}$, labeled as follows:

$\dynkin[labels={1,2,3,4,5,6},scale=4,labeldistance=5pt]E{oooooo}$

Suppose we semisimplify $\mathfrak{e}_{6}$ with respect to $e_{1}+e_{2}$.
Then, $l=2$, $\\{i_{1},i_{2}\\}=\\{1,2\\}$, and $\\{j_{1},j_{2}\\}={3,4}$. As
described in the lemma, we have the following split indecomposables, whose
basis vectors commute with $e_{1}$ and $e_{2}$, and therefore $e=e_{1}+e_{2}$:

$J_{1}:\ \ \\{e_{5},e_{6},f_{5},f_{6},h_{5},h_{6}\\}$

We also have these indecomposables, as their basis vectors commute with
$e=e_{1}+e_{2}$:

$J_{1}:\ \ \\{h_{3}-h_{1},h_{4}-h_{2}\\}.$

These indecomposables arise because 3 is connected to 1 by a single arrow and
4 is connected to 2 by a single arrow in the Dynkin diagram:

$J_{2}:\ \ \\{e_{3}\mapsto[e_{1},e_{3}],e_{4}\mapsto[e_{2},e_{4}]\\},$
$J_{2}:\ \ \\{[f_{1},f_{3}]\mapsto f_{3},[f_{2},f_{4}]\mapsto f_{4}\\}.$

Finally, we have two $\mathfrak{sl}_{2}$-triples associated to $e_{1}$ and
$e_{2}$.

$J_{3}:\ \ \\{f_{1}\mapsto h_{1}\mapsto e_{1},f_{2}\mapsto h_{2}\mapsto
e_{2}\\}.$

Notice how all the Chevalley generators are present in this decomposition and
that each subspace is $Q$-homogeneous. This concludes the example.

Now, we shall fix such a decomposition of $\mathfrak{g}(A)$ into
indecomposables as in Lemma 3.7. Each Chevalley generator $e_{i},f_{i}$ lies
in a copy of $J_{1}$, $J_{2}$, or $J_{3}$; in the semisimplification, the
image of each of these indecomposables will be an $L_{1}$ or $L_{2}$ or $0$.
Notice that those that do not lie in $J_{3}$ contain precisely one generator;
therefore, we shall refer to the basis vector associated to each of these (in
the language of §3.3 and Proposition 3.4) as the image of the corresponding
generator.

Let $\overline{\mathfrak{g}(A)}^{gen}$ denote the subquotient of
$\overline{\mathfrak{g}(A)}$ generated by the images of the generators, modulo
the additional relation that $[x,[x,x]]=0$ for all odd $x$ (recall that this
is not automatic in characteristic $3$). We note that when $[x,[x,x]]$ is
nonzero, it is never in the Cartan subalgebra because it is odd.

Now, recall the Lie algebra $\widetilde{\mathfrak{g}}(A)$ defined in §2 and
its generators and relations in (2.1). In particular, its upper triangular and
lower triangular subalgebras are freely generated. We claim the following:

######  Lemma 3.9.

There exists a surjective homomorphism from
$\widetilde{\mathfrak{g}}(\widetilde{A})$ to
$\overline{\mathfrak{g}(A)}^{gen}$.

###### Proof.

Referencing the proof of Lemma 3.7, let us label the images of generators in
the semisimplification. For $1+l\leq i\leq n-l$, let $\widetilde{e}_{i}$
denote the basis vector of $\overline{\mathfrak{g}(A)}^{gen}$ associated to
the copy of $L_{1}$ for $\overline{e_{i+l}}$ (resp. for the $f$’s and $h$’s);
for $1\leq i\leq l$, let $\widetilde{e}_{i}$ denote the basis vector of
$\overline{\mathfrak{g}(A)}^{gen}$ associated to the copy of $L_{2}$ for
$\overline{e_{i+l}\mapsto[e,e_{i+1}]}$ (resp. for the $f$’s), and let
$\widetilde{h}_{i}$ denote the basis vector of
$\overline{\mathfrak{g}(A)}^{gen}$ associated to the copy of $L_{1}$ for
$\overline{h_{i+l}-h_{i}}$. Then, we have generators
$\\{\widetilde{e}_{i},\widetilde{f}_{i},\widetilde{h}_{i}\\}_{1\leq i\leq
n-l}$ in $\overline{\mathfrak{g}(A)}^{gen}$. The first $l$ indices are odd,
and the last $n-2l$ are even.

For $\widetilde{\mathfrak{g}}(\widetilde{A})$, let us use the capital letters
$E,F,H$ instead of $e,f,h$ to avoid conflict of notation with the generators
of $\mathfrak{g}(A)$. By Lemma 3.5, $\\{E_{i},F_{i},H_{i}\\}$ generate
$\widetilde{\mathfrak{g}}(\widetilde{A})$. Again, by definition of
$\widetilde{A}$, $\widetilde{A}$ is an $(n-l)\times(n-l)$ matrix, where the
first $l$ indices are of odd parity and the last $n-2l$ are of even parity. We
claim that the surjection is given by the map $E_{i}\mapsto\widetilde{e}_{i}$,
$F_{i}\mapsto\widetilde{f}_{i}$ and $H_{i}\mapsto\widetilde{h}_{i}$.

To prove this, we need to check the relations in 2.1. We will check these
using the language of §3.3 and Proposition 3.4. Since these involve the
bracket of two generators, let us split this into four cases based on the
parity of each generator:

  1. (1)

Let $1\leq i,j\leq l$. The indices $i,j$ both have odd parity. Then, the
bracket $[\widetilde{e}_{i},\widetilde{f}_{j}]$ is given by the bracket
$\overline{-[e_{i+l},f_{j+l}]+[[e,e_{i+l}],[f,f_{j+l}]]}=\overline{-\delta_{ij}h_{i+l}+[[e_{i},e_{i+1}],[f_{j},f_{j+l}]]}$.
To compute the second bracket, we have by repeated applications of the Jacobi
identity and the relations:

$\displaystyle[[e_{i},e_{i+1}],[f_{j},f_{j+l}]]=[[e_{i},[f_{j},f_{j+l}]],e_{i+l}]+[e_{i},[e_{i+l},[f_{j},f_{j+l}]]]$
$\displaystyle=[[[e_{i},f_{j}],f_{j+l}]+[f_{j},[e_{i},f_{j+l}]],e_{i+l}]+[e_{i},[[e_{i+l},f_{j}],f_{j+l}]+[f_{j},[e_{i+l},f_{j+l}]]]$
$\displaystyle=[[\delta_{ij}h_{i},f_{j+l}],e_{i+l}]+[e_{i},[f_{j},\delta_{ij}h_{i+l}]]=\delta_{ij}a_{i,i+l}h_{i+l}+\delta_{ij}a_{i+l,i}h_{i}$
$\displaystyle=-\delta_{ij}(h_{i+l}+h_{i}),$

since $a_{i+l,i}=a_{i,i+l}=-1$. Therefore,
$-\delta_{ij}h_{i+l}+(-\delta_{ij}(h_{i+l}+h_{i}))$ =
$\delta_{ij}(h_{i+l}-h_{i})$ in characteristic $3$, and we deduce
$[\widetilde{e}_{i},\widetilde{f}_{j}]=\overline{\delta_{ij}(h_{i+l}-h_{i})}=\delta_{ij}\widetilde{h}_{i}$.
Now, let’s check that
$[\widetilde{h}_{i},\widetilde{e}_{j}]=\widetilde{a}_{ij}\widetilde{e}_{j}$.
Since $\widetilde{h}_{i}=\overline{h_{i+l}-h_{i}}$ and
$\widetilde{e}_{j}=\overline{e_{j+l}\mapsto[e,e_{j+l}]}$, we need to compute
the action of $h_{i+l}-h_{i}$ on both $e_{j+l}$ and $[e,e_{j+l}]$:

$\displaystyle[h_{i+l}-h_{i},e_{j+l}]=(a_{i+l,j+l}-a_{i,j+l})e_{j+l};$
$\displaystyle[h_{i+l}-h_{i},[e,e_{j+l}]]=[h_{i+l}-h_{i},[e_{j},e_{j+l}]]$
$\displaystyle=[[h_{i+l}-h_{i},e_{j}],e_{j+l}]+[e_{j},[h_{i+l}-h_{i},e_{j+l}]]$
$\displaystyle=(a_{i+l,j}-a_{ij}+a_{i+l,j+l}-a_{i,j+l})[e_{j},e_{j+l}]=(a_{i+l,j}-a_{ij}+a_{i+l,j+l}-a_{i,j+l})[e,e_{j+l}].$

If $i\neq j$, $a_{i+l,j+l}=\widetilde{a}_{ij}$ and
$a_{ij}=a_{i+l,j}=a_{i,j+l}=0$, and the coefficient of the RHS simplifies to
$\widetilde{a}_{ij}$ in both equations. If $i=j$, then $a_{i+l,i+l}=a_{ii}=2$
and $a_{i+l,i}=a_{i,i+l}=-1$, and again the coefficient of the RHS simplifies
to $0=\widetilde{a}_{ii}$ in both equations (remember we are in characteristic
$3$). This shows that
$[\widetilde{h}_{i},\widetilde{e}_{j}]=\widetilde{a}_{ij}\widetilde{e}_{j}$. A
similar argument goes through for the other relation
$[\widetilde{h}_{i},\widetilde{f}_{j}]=-\widetilde{a}_{ij}\widetilde{f}_{j}$.

Finally, the Cartan subalgebra will continue to be commutative, so the last
relation holds as well. Therefore, we deduce that the relations in 2.1 hold
between generators that correspond to indices of odd parity.

  2. (2)

Let $1\leq i\leq l$ and $1+l\leq j\leq n-l$. The index $i$ has odd parity and
the index $j$ has even parity. We proceed similarly to the first case. First,
we notice $[\widetilde{e}_{i},\widetilde{f}_{j}]=0$ as this must lie in the
Cartan subalgebra, which is purely even, but the first vector is odd and the
second vector is even. And indeed, the condition on $i$ and $j$ ensures $i\neq
j$, so we do have
$[\widetilde{e}_{i},\widetilde{f}_{j}]=\delta_{ij}\widetilde{h}_{i}$ in this
case.

Next, let’s check
$[\widetilde{h}_{i},\widetilde{e}_{j}]=\widetilde{a}_{ij}\widetilde{e}_{j}$.
Since $\widetilde{h}_{i}=\overline{h_{i+l}-h_{i}}$ and
$\widetilde{e}_{j}=\overline{e_{j+l}}$, we compute
$[h_{i+l}-h_{i},e_{j+l}]=(a_{i+l,j+l}-a_{i,j+l})e_{j+l}=a_{i+l,j+l}e_{j+l}=\widetilde{a}_{ij}e_{j+l}$,
from which the desired result follows. A similar argument shows that
$[\widetilde{h}_{i},\widetilde{f}_{j}]=-\widetilde{a}_{ij}{f}_{j}$. Finally,
the last relation holds again as the commutativity of the Cartan subalgebra is
preserved. This shows relations 2.1 in this case.

  3. (3)

Let $1+l\leq i\leq n-l$ and $1\leq j\leq l$. The index $i$ has even parity and
the index $j$ has odd parity. A similar argument to the previous case shows
$[\widetilde{e}_{i},\widetilde{f}_{j}]=\delta_{ij}\widetilde{h}_{i}=0$. Now,
let’s compute $[\widetilde{h}_{i},\widetilde{e}_{j}]$. Since
$\widetilde{h}_{i}=\overline{h_{i+l}}$ and
$\widetilde{e}_{j}=\overline{e_{j+l}\mapsto[e,e_{j+l}]}$, we check the action
of $h_{i+l}$. Therefore, we have

$\displaystyle[h_{i+l},e_{j+l}]=a_{i+l,j+l}e_{j+l};$
$\displaystyle[h_{i+l},[e,e_{j+l}]]=[h_{i+l},[e_{j},e_{j+l}]]$
$\displaystyle=[h_{i+l},e_{j}],e_{j+l}]+[e_{j},[h_{i+l},e_{j+l}]]$
$\displaystyle=(a_{i+l,j}+a_{i+l,j+l})[e_{j},e_{j+l}]=a_{i+l,j+l}[e,e_{j+l}]$
$\displaystyle=\widetilde{a}_{ij}e_{j+l},$

from which we deduce
$[\widetilde{h}_{i},\widetilde{e}_{j}]=\widetilde{a}_{ij}\widetilde{e}_{j}$. A
similar argument shows
$[\widetilde{h}_{i},\widetilde{f}_{j}]=-\widetilde{a}_{ij}\widetilde{f}_{j}$.
Finally, the last relation holds again as the commutativity of the Cartan
subalgebra is preserved. This shows relations 2.1 in this case.

  4. (4)

$1+l\leq i,j\leq n-l$. The indices $i,j$ both have even parity. This is the
easiest case, and the relations in 2.1 follow immediately.

We deduce that the generators
$\\{\widetilde{e}_{i},\widetilde{f}_{i},\widetilde{h}_{i}\\}$ satisfy the same
relations (and actually, they satisfy more relations) as the generators
$\\{E_{i},F_{i},H_{i}\\}$ of $\widetilde{\mathfrak{g}}(\widetilde{A})$. This
gives the desired surjection. ∎

We can now state the main theorem. If $V=V_{\overline{0}}\oplus
V_{\overline{1}}$ is a supervector space with even part $V_{\overline{0}}$ and
odd part $V_{\overline{1}}$, let $\operatorname*{sdim}V=(\dim
V_{\overline{0}}|\ \dim V_{\overline{1}})$ denote its superdimension. Two
supervector spaces have the same superdimension if the dimension of the even
parts agree and the dimensions of the odd parts agree.

######  Theorem 3.10.

If
$\operatorname*{sdim}\overline{\mathfrak{g}(A)}=\operatorname*{sdim}\mathfrak{g}(\widetilde{A})$,
then $\overline{\mathfrak{g}(A)}^{gen}=\overline{\mathfrak{g}(A)}$ and
$\overline{\mathfrak{g}(A)}$ is isomorphic to $\mathfrak{g}(\widetilde{A})$.

###### Proof.

By Lemma 3.9, we know there is a surjective, $Q$-grading-preserving
homomorphism from $\widetilde{\mathfrak{g}}(\widetilde{A})$ to
$\overline{\mathfrak{g}(A)}^{gen}$, and both have a Cartan subalgebra of the
same rank. Therefore, the kernel of this homomorphism is graded and trivially
intersects the Cartan subalgebra, and so this gives a well-defined surjection
from $\overline{\mathfrak{g}(A)}^{gen}$ to $\mathfrak{g}(\widetilde{A})$. On
the other hand, if the dimension of $\overline{\mathfrak{g}(A)}^{gen}$ is less
than the dimension of $\overline{\mathfrak{g}(A)}$, the surjection cannot
exist. Therefore, we deduce
$\overline{\mathfrak{g}(A)}^{gen}=\overline{\mathfrak{g}(A)}$, and by
comparing superdimensions, we deduce that this map must be an isomorphism. ∎

######  Remark 3.11.

It would be interesting to see how this theorem can be generalized (see
Conjecture 4.13), both for higher characteristic and more general Cartan
matrices. For instance, it also appears to be true for type $A_{n}$ (and for
$B_{n},C_{n},D_{n}$ with suitable restrictions on the boundary nodes).

###  3.6. Other Examples of Semisimplification

In this section, we offer some other examples of semisimplification, both for
completeness and to highlight some potential pitfalls. For instance, in the
proof of Theorem 3.10, we carefully showed that certain relations were
satisfied, although at first glance these were the “obvious” relations to be
satisfied; in general, this cannot be expected.

####  3.6.1. Example 1.

Consider the additive group scheme $\mathbb{G}_{a}$ over $\mathbb{C}$. The
representation category $\operatorname*{Rep}\mathbb{G}_{a}$ is a symmetric
tensor category, and its objects are described by nilpotent representations of
the algebra $\mathbb{C}[t]$. Any such representation is the direct sum of
Jordan blocks, and the tensor product of Jordan blocks is described by the
usual Clebsch-Gordan rule. It follows that the semisimplification of this
category is $\operatorname*{Rep}SL_{2}(\mathbb{C})$, where the Jordan block of
dimension $n$ semisimplifies to the $n$-dimensional representation of
$SL_{2}(\mathbb{C})$ (we are effectively deleting the maps between Jordan
blocks of different sizes and nilpotent endomorphisms, cf. [EO19]).

Any Lie algebra $\mathfrak{g}$ in $\operatorname*{Rep}\mathbb{G}_{a}$ is
therefore a Lie algebra equipped with a nilpotent derivation $d$, and
semisimplifying gives a Lie algebra $\overline{\mathfrak{g}}$ with an action
of $\mathfrak{sl}_{2}(\mathbb{C})$. If $\mathfrak{g}$ is semisimple, then $d$
is inner and we can write $d(x)=[e,x]$ for some $e\in\mathfrak{g}$. By the
Jacobson-Morozov lemma, $e$ can be included in an $\mathfrak{sl}_{2}$-triple
$\\{e,f,h\\}$, and the semisimplification of $\mathfrak{g}$ is isomorphic to
$\mathfrak{g}$ with the action of $\mathfrak{sl}_{2}$ prescribed by this
triple.

On the other hand, if $\mathfrak{g}$ is not semisimple, the action of $d$ may
not extend to an action of $\mathfrak{sl}_{2}$ by derivations (even if $d$ is
inner). This can result in $\overline{\mathfrak{g}}$ not being isomorphic to
$\mathfrak{g}$ unlike the previous case, and in fact,
$\overline{\mathfrak{g}}$ may even be abelian. In general,
$\overline{\mathfrak{g}}$ is the associated graded algebra of $\mathfrak{g}$
under the Deligne filtration by $d$, which on a vector space $V$ is defined by
$F_{k}V=\sum_{j-i=k}\ker d^{i}\cap\operatorname*{im}d^{j}$. If
$\overline{\mathfrak{g}}$ is semisimple, this filtration extends to a grading
by eigenvalues of $h$, so $\mathrm{gr}(\mathfrak{g})=\mathfrak{g}$. But when
$\overline{\mathfrak{g}}$ is not semisimple, it may not extend to such a
grading, and hence we may not have such an isomorphism.

Here is an explicit example. Consider the three-dimensional Heisenberg Lie
algebra spanned by $x,y,z$ with $z=[x,y]$ the central element. Let $d$ be the
derivation given by $\operatorname*{ad}x$. This does not extend to an
$\mathfrak{sl}_{2}$ action, and the semisimplification is abelian, which we’ll
explicitly check: using the notation of Jordan blocks from §3.2 in the obvious
way for characteristic $0$, we have a copy of $J_{1}$ spanned by $x$ and a
copy of $J_{2}$ given by $y\mapsto z$. The bracket $[x,y]=z$ and the bracket
$[x,z]=0$ are encapsulated by a morphism $J_{1}\otimes J_{2}\rightarrow J_{2}$
which is negligible, so it becomes zero in the semisimplification.

####  3.6.2. Example 2 (Duflo-Serganova, cf. [DS08]).

Consider the representation category
$\mathcal{C}=\operatorname*{Rep}\mathbb{G}_{a}^{0|1}$. A Lie algebra in
$\mathcal{C}$ is a Lie superalgebra with an odd derivation $d$ such that
$d^{2}=0$. The semisimplification of $\mathcal{C}$ is
$\operatorname*{sVec}_{\mathbb{C}}$ and the semisimplification of
$\mathfrak{g}$ is the cohomology of $d$, which is a new Lie superalgebra.

####  3.6.3. Example 3 (Entova-Aizenbud and Serganova, cf. [EAS20]).

Consider the representation category $\mathcal{C}$ of the supergroup scheme
$\mathbb{G}_{a}^{1|1}$. A Lie algebra $\mathfrak{g}$ in $\mathcal{C}$ is a Lie
superalgebra with an odd, nilpotent derivation $d$. The semisimplification of
$\mathcal{C}$ is $\operatorname*{Rep}\mathfrak{osp}_{1|2}$, and
$\overline{\mathfrak{g}}$ is a Lie superalgebra with an action of
$\mathfrak{osp}_{1|2}$. If $\mathfrak{g}$ is quasireductive and
$d=\operatorname*{ad}e$ is inner, where $e$ is a neat element (i.e. $[e,e]$
acts as the sum of odd dimensional Jordan blocks in every finite-dimensional
representation of $\mathfrak{g}$), then it is shown that
$\overline{\mathfrak{g}}=\mathfrak{g}$ and $e$ extends to action of
$\mathfrak{osp}_{1|2}$. However, this is nontrivial and only true under these
conditions. This is in a sense a super analog of Example 1 above.

##  4\. Exceptional Simple Lie Superalgebras in Characteristic $3$

In this section, we use Theorem 3.10 above to construct exceptional simple Lie
superalgebras via semisimplification. More accurately, we construct the
corresponding Kac-Moody Lie superalgebra that yields the simple Lie
superalgebra after potentially quotienting by the center. We are able to
construct all ten simple Elduque and Cunha Lie superalgebras. We shall use the
notation $\mathfrak{g}(a,b)$ to denote them, where $\mathfrak{g}$ denotes the
Lie superalgebra occupying the $(a,b)$-th slot in the Elduque-Freudenthal
magic super square (cf. [CE07, Eld06, CE06]). We are also able to construct
the Elduque Lie superalgebra.

For completeness, we include a description of the even part and odd part of
each Lie superalgebra. We shall use the notation $L(\omega_{i})$ to denote the
irreducible representation of a simple Lie algebra of highest weight a
fundamental weight $\omega_{i}$, whose labels shall follow that of Bourbaki
(cf. [Bou08]). The even parts of the Lie superalgebras we construct are
usually simple Lie algebras, and the odd parts are usually some fundamental
representation.

###  4.1. Construction $\mathfrak{g}(1,6)$ from $\mathfrak{f}_{4}$

In this section, we will construct the Elduque and Cunha Lie superalgebra
$\mathfrak{g}(1,6)$. This is a contragredient Lie superalgebra of
superdimension $(21|14)$ with a Cartan matrix and parity set:

$\displaystyle A=\begin{pmatrix}2&-1&0\\\ -1&2&-2\\\ 0&-1&0\end{pmatrix};\ \ \
I=\\{0,0,1\\}.$

The even part of $\mathfrak{g}(1,6)$ is $\mathfrak{sp}_{6}$ and the odd part
is the fundamental representation $L(\omega_{3})$. Applying our main theorem,
we consider the Lie algebra with Cartan matrix

$\displaystyle B=\begin{pmatrix}2&-1&0&0\\\ -1&2&-2&0\\\ 0&-1&2&-1\\\
0&0&-1&2\end{pmatrix}.$

This is the Cartan matrix of the $52$-dimensional simple Lie algebra
$\mathfrak{f}_{4}$. The Lie algebra $\mathfrak{f}_{4}$ has the following
Dynkin diagram, labeled in accordance with the Cartan matrix:

$\dynkin[labels={1,2,3,4},scale=4,arrowshape/.style={-{jibladze[length=7pt]}},labeldistance=5pt]F{oooo}$

$\mathfrak{f}_{4}$ can be realized as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ with respect to the
adjoint action of $e_{4}$ and decomposes as $21J_{1}\oplus 14J_{2}\oplus
J_{3}$, which can be checked using the software $SuperLie$. Therefore,
comparing dimensions, by Theorem 3.10, we have:

######  Theorem 4.1.

The semisimplification of $\mathfrak{f}_{4}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ under the adjoint action
of $e_{4}$ is $\mathfrak{g}(1,6)$.

###  4.2. Lie Superalgebras Arising from $\mathfrak{e}_{6}$

Recall that $\mathfrak{e}_{6}$ is the $78$ dimensional Lie algebra with Cartan
matrix:

$\displaystyle B=\begin{pmatrix}2&0&-1&0&0&0\\\ 0&2&0&-1&0&0\\\
-1&0&2&-1&0&0\\\ 0&-1&-1&2&-1&0\\\ 0&0&0&-1&2&-1\\\
0&0&0&0&-1&2\end{pmatrix}.$

In characteristic $3$, $\mathfrak{e}_{6}$ is not simple but rather has a one-
dimensional center in the Cartan subalgebra. When we quotient out by the
center, we get the simple Lie algebra $\mathfrak{p}\mathfrak{e}_{6}$. It has
the following Dynkin diagram, labeled in accordance with the Cartan matrix:

$\dynkin[labels={1,2,3,4,5,6},scale=4,labeldistance=5pt]E{oooooo}$

####  4.2.1. Constructing $\mathfrak{g}(2,3)$

In this section, we will construct the Elduque and Cunha Lie superalgebra
$\mathfrak{g}(2,3)$. This is a contragredient Lie superalgebra with a Cartan
matrix and parity set:

$\displaystyle A=\begin{pmatrix}0&-1&0\\\ -1&0&-1\\\ 0&-1&0\end{pmatrix};\ \ \
I=\\{1,1,1\\}.$

This is a Lie superalgebra of superdimension $(11|14)$ and contains a one-
dimensional center. Its even part is
$\mathfrak{gl}_{3}\oplus\mathfrak{sl}_{2}$ and its odd part is
$\mathfrak{psl}(3)\otimes\mathbb{K}^{2}$; here $\mathfrak{gl}_{3}$ acts on
$\mathfrak{psl}(3)$ by the adjoint representation and $\mathbb{K}^{2}$ is the
natural representation of $\mathfrak{sl}_{2}$. By comparing Cartan matrices,
we realize $\mathfrak{e}_{6}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ with respect to
$e_{1}+e_{2}+e_{6}$, where it decomposes as $\mathfrak{e}_{6}=11J_{1}\oplus
14J_{2}\oplus 13J_{3}$. Then, the hypothesis of Theorem 3.10 is satisfied, and
we have:

######  Theorem 4.2.

The semisimplification of $\mathfrak{e}_{6}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ under the adjoint action
of $e_{1}+e_{2}+e_{6}$ is $\mathfrak{g}(2,3)$.

It follows that when we semisimplify $\mathfrak{p}\mathfrak{e}_{6}$, we get
$\mathfrak{p}\mathfrak{g}(2,3)$, which is simple.

####  4.2.2. Constructing $\mathfrak{g}(3,3)$

In this section, we will construct the Elduque and Cunha Lie superalgebra
$\mathfrak{g}(3,3)$. This is a contragredient Lie superalgebra with a Cartan
matrix and parity set:

$\displaystyle A=\begin{pmatrix}0&-1&0&0\\\ -1&0&-1&0\\\ 0&-1&2&-1\\\
0&0&-1&2\end{pmatrix};\ \ \ I=\\{0,0,1,1\\}.$

This is a Lie superalgebra of superdimension $(22|16)$ and contains a one-
dimensional center. The even subalgebra is
$\mathfrak{so}_{7}\oplus\mathbb{K}d$, and the odd part, as a module of the
even subalgebra, is $L(\omega_{3})^{+}\oplus L(\omega_{3})^{-}$, where in
particular $L(\omega_{3})$ is the spinor representation of
$\mathfrak{so}_{7}$, and $d$ acts by $+1$ on the first copy and $-1$ on the
second copy.

By comparing Cartan matrices, we realize $\mathfrak{e}_{6}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ with respect to
$e_{1}+e_{2}$, where it decomposes as $\mathfrak{e}_{6}=22J_{1}\oplus
16J_{2}\oplus 8J_{3}$. Then, the hypothesis of Theorem 3.10 is satisfied, and
we have:

######  Theorem 4.3.

The semisimplification of $\mathfrak{e}_{6}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ under the adjoint action
of $e_{1}+e_{2}$ is $\mathfrak{g}(3,3)$.

It follows that when we semisimplify $\mathfrak{p}\mathfrak{e}_{6}$, we get
$\mathfrak{p}\mathfrak{g}(3,3)$, which is simple.

####  4.2.3. Constructing $\mathfrak{g}(2,6)$

In this section, we will construct the Elduque and Cunha Lie superalgebra
$\mathfrak{g}(2,6)$. This is a contragredient Lie superalgebra with a Cartan
matrix and parity set:

$\displaystyle A=\begin{pmatrix}2&-1&0&0&0\\\ -1&2&-1&0&0\\\ 0&-1&0&-1&0\\\
0&0&-1&2&-1\\\ 0&0&0&-1&2\end{pmatrix};\ \ \ I=\\{0,0,1,0,0\\}.$

This is a Lie superalgebra of superdimension $(35|20)$ and contains a one-
dimensional center. The even subalgebra is $\mathfrak{gl}_{6}$ and the odd
part is the module $L(\omega_{3})$. By comparing Cartan matrices, we realize
$\mathfrak{e}_{6}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ with respect to $e_{2}$,
where it decomposes as $\mathfrak{e}_{6}=35J_{1}\oplus 20J_{2}\oplus J_{3}$.
Then, the hypothesis of Theorem 3.10 is satisfied, and we have:

######  Theorem 4.4.

The semisimplification of $\mathfrak{e}_{6}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ under the adjoint action
of $e_{2}$ is $\mathfrak{g}(2,6)$.

It follows that when we semisimplify $\mathfrak{p}\mathfrak{e}_{6}$, we get
$\mathfrak{p}\mathfrak{g}(2,6)$, which is simple.

###  4.3. Lie Superalgebras Arising from $\mathfrak{e}_{7}$

Recall that $\mathfrak{e}_{7}$ is the $133$ dimensional simple Lie algebra
with Cartan matrix:

$\displaystyle B=\begin{pmatrix}2&0&-1&0&0&0&0\\\ 0&2&0&-1&0&0&0\\\
-1&0&2&-1&0&0&0\\\ 0&-1&-1&2&-1&0&0\\\ 0&0&0&-1&2&-1&0\\\ 0&0&0&0&-1&2&-1\\\
0&0&0&0&0&-1&2\end{pmatrix}.$

It has the following Dynkin diagram, labeled in accordance with the Cartan
matrix:

$\dynkin[labels={1,2,3,4,5,6,7},scale=4,labeldistance=5pt]E{ooooooo}$

####  4.3.1. Constructing $\mathfrak{g}(4,3)$

In this section, we will construct the Elduque and Cunha Lie superalgebra
$\mathfrak{g}(4,3)$. This is a contragredient Lie superalgebra with a Cartan
matrix and parity set:

$\displaystyle A=\begin{pmatrix}0&-1&0&0\\\ -1&0&-1&0\\\ 0&-1&2&-1\\\
0&0&-1&0\end{pmatrix};\ \ \ I=\\{1,1,0,1\\}.$

This is a Lie superalgebra of superdimension $(24|26)$. It has an even
subalgebra $\mathfrak{sp}_{6}\oplus\mathfrak{sl}_{2}$ and the module
$L(\omega_{2})\otimes\mathbb{K}^{2}$ of the even subalgebra is its odd part,
where $\mathbb{K}^{2}$ is the natural representation of $\mathfrak{sl}_{2}$.
By comparing Cartan matrices, we realize $\mathfrak{e}_{7}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ with respect to
$e_{1}+e_{2}+e_{7}$, where it decomposes as $\mathfrak{e}_{7}=24J_{1}\oplus
26J_{2}\oplus 19J_{3}$. Then, the hypothesis of Theorem 3.10 is satisfied, and
we have:

######  Theorem 4.5.

The semisimplification of $\mathfrak{e}_{7}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ under the adjoint action
of $e_{1}+e_{2}+e_{7}$ is $\mathfrak{g}(4,3)$.

####  4.3.2. Constructing $\mathfrak{g}(4,6)$

In this section, we will construct the Elduque and Cunha Lie superalgebra
$\mathfrak{g}(4,6)$. This is a contragredient Lie superalgebra with a Cartan
matrix and parity set:

$\displaystyle A=\begin{pmatrix}2&0&-1&0&0&0\\\ 0&0&-1&0&0&0\\\
-1&-1&2&-1&0&0\\\ 0&0&-1&2&-1&0\\\ 0&0&0&-1&2&-1\\\
0&0&0&0&-1&2\end{pmatrix};\ \ \ I=\\{0,1,0,0,0,0\\}.$

This is a Lie superalgebra of superdimension $(66|32)$. Its even subalgebra is
$\mathfrak{so}_{12}$ and its odd part is the module $L(\omega_{5})$. By
comparing Cartan matrices, we realize $\mathfrak{e}_{7}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ with respect to $e_{1}$,
where it decomposes as $\mathfrak{e}_{7}=66J_{1}\oplus 32J_{2}\oplus J_{3}$.
Then, the hypothesis of Theorem 3.10 is satisfied, and we have:

######  Theorem 4.6.

The semisimplification of $\mathfrak{e}_{7}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ under the adjoint action
of $e_{1}$ is $\mathfrak{g}(4,6)$.

####  4.3.3. Constructing $\mathfrak{el}(5;3)$

In this section, we will construct the Elduque superalgebra
$\mathfrak{el}(5;3)$. This is a contragredient Lie superalgebra with a Cartan
matrix and parity set:

$\displaystyle A=\begin{pmatrix}2&0&-1&0&0\\\ 0&0&-1&0&0\\\ -1&-1&2&-1&0\\\
0&0&-1&2&-1\\\ 0&0&0&-1&0\\\ \end{pmatrix};\ \ \ I=\\{0,1,0,0,1\\}.$

This is a Lie superalgebra of superdimension $(39|32)$. Its even subalgebra is
$\mathfrak{so}_{12}\oplus\mathfrak{sl}_{2}$ and its odd part is
$L(\omega_{4})\otimes\mathbb{K}^{2}$, where $\mathbb{K}^{2}$ is the natural
representation of $\mathfrak{sl}_{2}$. By comparing Cartan matrices, we
realize $\mathfrak{e}_{7}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ with respect to
$e_{1}+e_{7}$, where it decomposes as $\mathfrak{e}_{7}=39J_{1}\oplus
32J_{2}\oplus 10J_{3}$. Then, the hypothesis of Theorem 3.10 is satisfied, and
we have:

######  Theorem 4.7.

The semisimplification of $\mathfrak{e}_{7}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ under the adjoint action
of $e_{1}+e_{7}$ is $\mathfrak{el}(5;3)$.

###  4.4. Lie Superalgebras Arising From $\mathfrak{e}_{8}$

Recall that $\mathfrak{e}_{8}$ is the $248$ dimensional simple Lie algebra
with Cartan matrix:

$\displaystyle B=\begin{pmatrix}2&0&-1&0&0&0&0&0\\\ 0&2&0&-1&0&0&0&0\\\
-1&0&2&-1&0&0&0&0\\\ 0&-1&-1&2&-1&0&0&0\\\ 0&0&0&-1&2&-1&0&0\\\
0&0&0&0&-1&2&-1&0\\\ 0&0&0&0&0&-1&2&-1\\\ 0&0&0&0&0&0&-1&2\end{pmatrix}.$

It has the following Dynkin diagram, labeled in accordance with the Cartan
matrix:

$\dynkin[labels={1,2,3,4,5,6,7,8},scale=4,labeldistance=5pt]E{oooooooo}$

####  4.4.1. Constructing $\mathfrak{g}(8,3)$

$\mathfrak{g}(8,3)$ is a contragredient Lie superalgebra with a Cartan matrix
and parity set:

$\displaystyle A=\begin{pmatrix}0&-1&0&0&0\\\ -1&0&-1&0&0\\\ 0&-1&2&-1&0\\\
0&0&-1&2&-1\\\ 0&0&0&-1&0\\\ \end{pmatrix};\ \ \ I=\\{1,1,0,0,1\\}.$

This is a Lie superalgebra of superdimension $(55|50)$. Its even subalgebra is
$\mathfrak{f}_{4}\oplus\mathfrak{sl}_{2}$ and its odd part is the module
$L(\omega_{4})\otimes\mathbb{K}^{2}$, where $\mathbb{K}^{2}$ is the natural
representation of $\mathfrak{sl}_{2}$. By comparing Cartan matrices, we
realize $\mathfrak{e}_{8}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ with respect to
$e_{1}+e_{2}+e_{8}$, where it decomposes as $\mathfrak{e}_{8}=55J_{1}\oplus
50J_{2}\oplus 31J_{3}$. Then, the hypothesis of Theorem 3.10 is satisfied, and
we have:

######  Theorem 4.8.

The semisimplification of $\mathfrak{e}_{8}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ under the adjoint action
of $e_{1}+e_{2}+e_{8}$ is $\mathfrak{g}(8,3)$.

####  4.4.2. Constructing $\mathfrak{g}(6,6)$

$\mathfrak{g}(6,6)$ is a contragredient Lie superalgebra with a Cartan matrix
and parity set:

$\displaystyle A=\begin{pmatrix}0&-1&0&0&0&0\\\ -1&0&-1&0&0&0\\\
0&-1&2&-1&0&0\\\ 0&0&-1&2&-1&0\\\ 0&0&0&-1&2&-1\\\ 0&0&0&0&-1&2\end{pmatrix};\
\ \ I=\\{1,1,0,0,0,0\\}.$

This is a Lie superalgebra of superdimension $(78|64)$. Its even subalgebra is
$\mathfrak{so}_{13}$ and its odd part is the spinor representation
$L(\omega_{6})$. By comparing Cartan matrices, we realize $\mathfrak{e}_{8}$
as an object in $\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ with
respect to $e_{1}+e_{2}$, where it decomposes as
$\mathfrak{e}_{8}=78J_{1}\oplus 64J_{2}\oplus 14J_{3}$. Then, the hypothesis
of Theorem 3.10 is satisfied, and we have:

######  Theorem 4.9.

The semisimplification of $\mathfrak{e}_{8}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ under the adjoint action
of $e_{1}+e_{2}$ is $\mathfrak{g}(6,6)$.

####  4.4.3. Constructing $\mathfrak{g}(8,6)$

$\mathfrak{g}(8,6)$ is a contragredient Lie superalgebra with a Cartan matrix
and parity set:

$\displaystyle A=\begin{pmatrix}2&0&-1&0&0&0&0\\\ 0&0&-1&0&0&0&0\\\
-1&-1&2&-1&0&0&0\\\ 0&0&-1&2&-1&0&0\\\ 0&0&0&-1&2&-1&0\\\ 0&0&0&0&-1&2&-1\\\
0&0&0&0&0&-1&2\end{pmatrix};\ \ \ I=\\{0,1,0,0,0,0,0\\}.$

This is a Lie superalgebra of superdimension $(133|56)$. Its even subalgebra
is $\mathfrak{e}_{6}$ and its odd part is the module $L(\omega_{1})$. By
comparing Cartan matrices, we realize $\mathfrak{e}_{8}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ with respect to $e_{1}$,
where it decomposes as $\mathfrak{e}_{8}=133J_{1}\oplus 56J_{2}\oplus J_{3}$.
Then, the hypothesis of Theorem 3.10 is satisfied, and we have:

######  Theorem 4.10.

The semisimplification of $\mathfrak{e}_{8}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ under the adjoint action
of $e_{1}$ is $\mathfrak{g}(8,6)$.

####  4.4.4. Constructing $\mathfrak{g}(3,6)$

$\mathfrak{g}(3,6)$ is a contragredient Lie superalgebra with a Cartan matrix
and parity set:

$\displaystyle A=\begin{pmatrix}0&-1&0&0\\\ -1&0&-1&0\\\ 0&-1&0&-2\\\
0&0&-1&2\end{pmatrix};\ \ \ I=\\{1,1,1,0\\}.$

This is a Lie superalgebra of superdimension $(36|40)$. Its even subalgebra is
$\mathfrak{sp}_{8}$ and its odd part is $L(\omega_{3})$. We can construct this
Lie superalgebra from $\mathfrak{e}_{8}$, but this will slightly differ from
the main approach above. Let $x=e_{1}+e_{2}+e_{6}+e_{8}$. Then, it is easily
checked that with respect to the adjoint action of $x$, $\mathfrak{e}_{8}$ is
an object in $\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$, where it
decomposes as $\mathfrak{e}_{8}=36J_{1}+40J_{2}+44J_{3}$. We shall show that:

######  Theorem 4.11.

The semisimplification of $\mathfrak{e}_{8}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ under the adjoint action
of $e_{1}+e_{2}+e_{6}+e_{8}$ is $\mathfrak{g}(3,6)$.

###### Proof.

The main theorem does not go through because the node corresponding to $e_{6}$
on the Dynkin diagram is not a boundary node, so we will proceed manually
using the language of §3.3 and Proposition 3.4. However, they key point that
the generators $e_{1},e_{2},e_{6},e_{8}$ pairwise commute still holds. Let us
first consider what happens to the positive generators. Because of the new
situation, we will also need to consider root vectors that are not attached to
simple roots. Let:

  1. (1)

$e_{9}\coloneqq[x,e_{3}]=[e_{1},e_{3}]$,

  2. (2)

$e_{10}\coloneqq[x,e_{4}]=[e_{2},e_{4}]$,

  3. (3)

$e_{13}\coloneqq-[x,e_{5}]=[e_{5},e_{6}]$,

  4. (4)

$e_{14}\coloneqq[e_{6},e_{7}]$,

  5. (5)

$e_{15}\coloneqq[-e_{8},e_{7}]$,

  6. (6)

$e_{22}\coloneqq[e_{8},[e_{6},e_{7}]]$, which also happens to equal
$-[x,e_{14}-e_{15}]$.

and similarly for $f$; these new indices are chosen in accordance with
labeling in the SuperLie software and do not have any other meaning. In
particular, note that $e_{14}-e_{15}=[x,e_{7}]$.

Let’s consider the upper triangular subalgebra, completed to a split subobject
(by symmetry, this will tell us what happens to the lower triangular
subalgebra). We have the following copies of $J_{3}$: $f_{1}\mapsto
h_{1}\mapsto e_{1}$, $f_{2}\mapsto h_{2}\mapsto e_{2}$, $f_{6}\mapsto
h_{6}\mapsto e_{6}$, $f_{8}\mapsto h_{8}\mapsto e_{8}$, $e_{7}\mapsto
e_{14}-e_{15}\mapsto-e_{22}$; these will vanish in the semisimplification, and
in particular the information attached to the generator $e_{7}$ is
annihilated. The remaining generators yield odd generators, as we have the
following copies of $J_{2}$: $e_{3}\mapsto e_{9}$, $e_{4}\mapsto e_{10}$, and
$e_{5}\mapsto-e_{13}$.

In the semisimplification, the images of these odd generators do not generate
the image (in fact, it is easily seen that, along with their $f$ counter
parts, they generate the rank $3$ Lie superalgebra $\mathfrak{g}(2,3)$ above).
However, $\mathfrak{g}(3,6)$ is of rank $4$ and contains $\mathfrak{g}(2,3)$
as a subalgebra, which we know by just looking at their Cartan matrices, so we
should be able to find another indecomposable whose image will serve as our
final generator (it should span a split copy of $J_{1}$). In fact, the vector
that will serve as our generator is $e_{14}+e_{15}$ and its lower triangular
counterpart $-f_{14}-f_{15}$, which together form an
$\mathfrak{sl}_{2}$-triple (in characteristic $3$) with $h_{6}-h_{7}+h_{8}$.
$e_{14}+e_{15}$ should be treated as the positive generator. Hence, in the
semisimplification, we define the following vectors, where the LHS is the
vector and the RHS is the subspace it spans:

$\displaystyle\widetilde{e}_{1}\coloneqq\overline{e_{3}\mapsto e_{9}};\ \ \
\widetilde{e}_{2}\coloneqq\overline{e_{4}\mapsto e_{10}};\ \ \
\widetilde{e}_{3}\coloneqq\overline{e_{5}\mapsto-e_{13}};\ \ \
\widetilde{e}_{4}\coloneqq\overline{e_{14}+e_{15}}$
$\displaystyle\widetilde{f}_{1}\coloneqq\overline{f_{9}\mapsto f_{3}};\ \ \
\widetilde{f}_{2}\coloneqq\overline{f_{10}\mapsto f_{4}};\ \ \
\widetilde{f}_{3}\coloneqq\overline{-f_{13}\mapsto f_{5}};\ \ \
\widetilde{f}_{4}\coloneqq\overline{-f_{14}-e_{15}}$
$\displaystyle\widetilde{h}_{1}\coloneqq\overline{h_{3}};\ \ \
\widetilde{h}_{2}\coloneqq\overline{h_{4}};\ \ \
\widetilde{h}_{3}\coloneqq\overline{h_{5}};\ \ \
\widetilde{h}_{4}\coloneqq\overline{h_{6}-h_{7}+h_{8}}.$

Now, using SuperLie, we can check that the collection of vectors
$\\{\widetilde{e}_{i},\widetilde{f}_{i},\widetilde{h}_{i}\\}$ do indeed
generate the semisimplification. Furthermore, direct computations show that
they satisfy the same relations as their counterparts in $\mathfrak{g}(3,6)$,
so this means we have a morphism from $\mathfrak{g}(3,6)$ to the
semisimplification. Then, by comparing dimensions, we deduce that they are
isomorphic. As a remark, we do not need to even know that the generators
generate the image, as the Lie superalgebra $\mathfrak{g}(3,6)$ is simple. ∎

For completeness, we briefly describe another construction of
$\mathfrak{g}(3,6)$. Recall that as a Lie algebra-module pair,
$\mathfrak{e}_{8}$ splits as the $120$-dimensional simple Lie algebra
$\mathfrak{so}_{16}$ and its $128$-dimensional spinor representation
$L(\omega_{8})$. In our notation, the subalgebra $\mathfrak{so}_{16}$ is
generated by $e_{2},e_{3},e_{4},e_{5}$, $e_{6},e_{7},e_{8}$ and

$e_{100}\coloneqq[[[[e_{1},e_{3}],[e_{4},e_{5}]],[[e_{2},e_{4}],[e_{5},e_{6}]]],[[[e_{1},e_{3}],[e_{2},e_{4}]],[[e_{6},e_{7}],[e_{5},[e_{3},e_{4}]]]]],$

and their $f$ counter parts (again, the choice of index here is just based on
the program SuperLie and otherwise has no meaning).

Let $e_{27}\coloneqq[[e_{6},e_{5}],[e_{4},e_{3}]]$ and similarly for $f_{27}$.
Then, $\mathfrak{sp}_{8}$ is a subalgebra of $\mathfrak{so}_{16}$ generated by
$e_{100}-f_{5}$, $e_{8}-f_{4}$, $e_{7}-f_{3}$, $e_{27}$, and $f_{100}-e_{5}$,
$f_{8}-e_{4}$, $f_{7}-e_{3}$, $f_{27}$, where the first $4$ correspond to the
simple roots in the usual order and the last $4$ the negatives of the simple
roots. Then, one can take an $\mathfrak{sl}_{2}$-triple in
$\mathfrak{so}_{16}\subseteq\mathfrak{e}_{8}$ that centralizes
$\mathfrak{sp}_{8}$; semisimplifying with respect to say the positive root
vector in this triple gives the desired $\mathfrak{g}(3,6)$.

####  4.4.5. Summary

In this section, we summarize and extend the results above. Here we have the
appropriate Dynkin diagrams for easy reference, where
$\mathfrak{e}_{6}\subseteq\mathfrak{e}_{7}\subseteq\mathfrak{e}_{8}$ in the
obvious way:

$\displaystyle\mathfrak{f}_{4}:\hskip 14.22636pt$
$\displaystyle\dynkin[labels={1,2,3,4},scale=4,arrowshape/.style={-{jibladze[length=7pt]}},labeldistance=5pt]F{oooo}$
$\displaystyle\mathfrak{e}_{8}:\hskip 14.22636pt$
$\displaystyle\dynkin[labels={1,2,3,4,5,6,7,8},scale=4,labeldistance=5pt]E{oooooooo}$

In table below, we state results by specifying the starting Lie algebra, the
nilpotent element used to semisimplify, and the resulting Lie superalgebra. We
include all possible combinations of boundary nodes and some examples that do
not follow from the main theorem as well. Nilpotent elements that give the
same semisimplification are listed in the same row.

Lie algebra | Nilpotent element | Lie superalgebra
---|---|---
$\mathfrak{f}_{4}$ | $e_{1}$ | see ($\star$) below
| $e_{4}$ | $\mathfrak{g}(1,6)$
| $e_{1}+e_{4}$ | see ($\star$) below
$\mathfrak{e}_{6}$ | $e_{1},e_{2},e_{6}$ | $\mathfrak{g}(2,6)$
|  $e_{1}+e_{2}$, $e_{2}+e_{6}$, $e_{1}+e_{6}$ | $\mathfrak{g}(3,3)$
| $e_{1}+e_{2}+e_{6}$ | $\mathfrak{g}(2,3)$
$\mathfrak{e}_{7}$ | $e_{1},e_{2},e_{7}$ | $\mathfrak{g}(4,6)$
|  $e_{1}+e_{2}$, $e_{2}+e_{7}$, $e_{1}+e_{7}$ | $\mathfrak{el}(5;3)$
| $e_{1}+e_{2}+e_{7}$ | $\mathfrak{g}(4,3)$
| $e_{2}+e_{5}+e_{7}$ |  $\mathfrak{f}_{4}$; see $(\ast)$ below
| $e_{1}+e_{2}+e_{5}+e_{7}$ | $\mathfrak{g}(1,6)$
$\mathfrak{e}_{8}$ | $e_{1},e_{2},e_{8}$ | $\mathfrak{g}(8,6)$
|  $e_{1}+e_{2}$, $e_{2}+e_{8}$, $e_{1}+e_{8}$ | $\mathfrak{g}(6,6)$
| $e_{1}+e_{2}+e_{8}$ | $\mathfrak{g}(8,3)$
| $e_{1}+e_{2}+e_{6}+e_{8}$ | $\mathfrak{g}(3,6)$

  1. ($\ast$)

This is an interesting case not discussed in the paper, as we get a purely
even Lie algebra, and why this is the case is not clear from the Cartan
matrix. It suggests an idea of iterating semisimplification, as we can either
semisimplify $\mathfrak{e}_{7}$ to get $\mathfrak{f}_{4}$ and then
semisimplify again to get $\mathfrak{g}(1,6)$, or we can semisimplify
$\mathfrak{e}_{7}$ to directly get $\mathfrak{g}(1,6)$.

  2. ($\star$)

Semisimplifying $\mathfrak{f}_{4}$, which say has Cartan matrix $A$, with
respect to $e_{1}$ gives a Lie superalgebra $\overline{\mathfrak{f}_{4}}$ of
superdimension of $(15|8)$, whereas the Lie superalgebra
$\mathfrak{g}(\widetilde{A})=\mathfrak{sl}_{3|1}$ as described in the setup
for Theorem 3.10 is of superdimension $(9|6)$. Hence, Theorem 3.10 does not
apply. However, it can be computed by hand that
$\overline{\mathfrak{f}_{4}}^{gen}$ is of superdimension $(9|6)$ and is
isomorphic to $\mathfrak{sl}_{3|1}$. A similar statement applies when we
semisimplify $\mathfrak{f}_{4}$ with $e_{1}+e_{4}$. Although these may seem
like edge cases, we believe there is a more general conjecture which captures
these cases and the main theorem simultaneously, which we discuss in §4.5 in
Conjecture 4.13.

One may notice that for $\mathfrak{e}_{6},\mathfrak{e}_{7},\mathfrak{e}_{8}$,
when we picked fewer than three generators to sum, it did not matter which
generators we picked to semisimplify, but rather only how many we picked.
These are actually instances of a more general phenomenon. Let $\mathfrak{g}$
be an exceptional Lie algebra
$\mathfrak{f}_{4},\mathfrak{e}_{6},\mathfrak{e}_{7},\mathfrak{e}_{8}$, and let
$I$ be any subset of the nodes of the associated Dynkin diagram such that no
two nodes in $I$ are adjacent in the Dynkin diagram. Color the nodes in $I$
black, and all other nodes white. Finally, let $e_{I}\coloneqq\sum_{i\in
I}e_{i}$.

A legal swap of the colored Dynkin diagram is defined as any recoloring of the
diagram of the following form: if $i$ and $j$ are two adjacent nodes connected
by a single edge, with $i$ colored black, and $j$ colored white, and no other
node adjacent to $j$ is black, swap the colors of $i$ and $j$. It is clear
that the legal swaps generate a groupoid under composition; we will call its
elements legal recolorings. Let $\sigma(I)$ denote the set of nodes colored
black after applying the legal recoloring $\sigma$.

Here are some examples of legal swaps:

[mark=o,labels=1,2,3,4,5,6,7,scale=2, label distance = 5pt]E7 [black,
draw=black] (root 2) circle (0.05cm); [black, draw=black] (root 7) circle
(0.05cm); $\longrightarrow$ [mark=o,labels=1,2,3,4,5,6,7,scale=2, label
distance = 5pt]E7 [black, draw=black] (root 2) circle (0.05cm); [black,
draw=black] (root 6) circle (0.05cm); $\longrightarrow$
[mark=o,labels=1,2,3,4,5,6,7,scale=2, label distance = 5pt]E7 [black,
draw=black] (root 4) circle (0.05cm); [black, draw=black] (root 6) circle
(0.05cm);

[mark=o,labels=1,2,3,4,scale=2, arrow shape/.style=-jibladze[scale=0.58,
length=7pt], label distance = 5pt]F4 [black, draw=black] (root 4) circle
(0.05cm); $\longrightarrow$ [mark=o,labels=1,2,3,4,scale=2, arrow
shape/.style=-jibladze[scale=0.58, length=7pt], label distance = 5pt]F4
[black, draw=black] (root 3) circle (0.05cm);

Here are some examples of illegal swaps:

[mark=o,labels=1,2,3,4,5,6,7,scale=2, label distance = 5pt]E7 [black,
draw=black] (root 4) circle (0.05cm); [black, draw=black] (root 6) circle
(0.05cm); $\longrightarrow$ [mark=o,labels=1,2,3,4,5,6,7,scale=2, label
distance = 5pt]E7 [black, draw=black] (root 4) circle (0.05cm); [black,
draw=black] (root 5) circle (0.05cm);

[mark=o,labels=1,2,3,4,scale=2, arrow shape/.style=-jibladze[scale=0.58,
length=7pt], label distance = 5pt]F4 [black, draw=black] (root 3) circle
(0.05cm);F $\longrightarrow$ [mark=o,labels=1,2,3,4,scale=2, arrow
shape/.style=-jibladze[scale=0.58, length=7pt], label distance = 5pt]F4
[black, draw=black] (root 2) circle (0.05cm);

######  Theorem 4.12.

Let $I$ be a configuration of black nodes as above, and let $\sigma$ be a
legal recoloring. Then, $\mathfrak{g}$ can be realized as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ in two ways: with respect
to $e_{I}$, denoted $\mathfrak{g}_{I}$, or with respect to $e_{\sigma(I)}$,
denoted $\mathfrak{g}_{\sigma(I)}$. Furthermore, the semisimplifications
$\overline{\mathfrak{g}_{I}}$ and $\overline{\mathfrak{g}_{\sigma(I)}}$ are
isomorphic as Lie superalgebras.

###### Proof.

It suffices to assume that $\sigma$ is a legal swap that swaps the colors of
the adjacent nodes $i$ and $j$, where $i$ is black and $j$ is white. By the
initial configuration $I$ and the definition of a legal swap, the proof that
$e_{I}$ is nilpotent of degree at most three goes through just like the proof
of Lemma 3.6.

Now, let $G$ be a split simple linear algebraic group with Lie algebra
$\mathfrak{g}$. $G$ acts on $\mathfrak{g}$ by the adjoint representation as
Lie algebra automorphisms, and nilpotent elements are partitioned into orbits.
Hence, it suffices to show that $e_{I}$ and $e_{\sigma(I)}$ lie in the same
nilpotent orbit.

The Dynkin subdiagram formed by nodes $i$ and $j$ corresponds to a subgroup
$H$ of type $A_{2}$ (i.e. $SL_{3}$ or $PGL_{3}$) in $G$, and there is a Weyl
group element $w\in N(T)/T$, where $T$ is a maximal torus in $H$, that
permutes them. We can find a coset reprsentative $g\in H$ that corresponds to
$w$ such that its conjugation action on $\mathfrak{g}$ will send $e_{i}$ to
$e_{j}$. The element $g$ then lifts to an element $\widetilde{g}$ in $G$, and
because no other black nodes are connected to $i$ or $j$, the conjugation
action of $\widetilde{g}$ will not change any other $e_{k}$ for $k\neq i\in
I$. This shows the claim. ∎

Combining this theorem with Theorem 3.10 tells us the semisimplification with
respect to a large class of elements. For instance, we deduce that that
semisimplifying $\mathfrak{e}_{8}$ with respect to any $e_{i}$, regardless of
whether $i$ corresponds to a boundary node or not, gives $\mathfrak{g}(8,6)$.

###  4.5. Some Remarks

####  4.5.1. Kac-Moody Lie Algebras

Although we were able to construct these Lie superalgebras by semisimplifying
the exceptional Lie algebras, it is natural to see what Lie superalgebras we
get when we enlargen the class of Cartan matrices we consider.

Let $A$ be a purely even, symmetrizable, indecomposable, generalized Cartan
matrix of size $n$ with nonzero determinant whose off-diagonal entries are
between $1-p=-2$ and $0$, inclusive. Let $D$ be the corresponding Dynkin
diagram, and call a node in the diagram a boundary node if it is connected to
exactly one other node in the diagram via a single edge. Let
$I=\\{i_{1},\dots,i_{l}\\}$ be a subset of the boundary nodes of $D$ such that
no two nodes in $I$ share an adjacent node, and let
$J=\\{j_{1},\dots,j_{l}\\}$ be the corresponding adjacent nodes, respectively
(the elements of $J$ are necessarily distinct). Let $\widetilde{A}$ be the
$(n-l)\times(n-l)$ matrix obtained from $A$ by setting $a_{jj}=0$ for $j\in J$
and deleting the $i$-th row and column from $A$ for $i\in I$. Finally, let
$e_{I}=\sum_{i\in I}e_{i}$. Then, we have the following conjecture for
characteristic $3$:

######  Conjecture 4.13.

Under the preceding hypotheses, $\mathfrak{g}(A)$ can be realized as an object
in $\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ with respect to
$e_{I}$. Furthermore, if $\widetilde{A}$ has nonzero determinant, then
$\overline{\mathfrak{g}(A)}^{gen}$ is isomorphic to
$\mathfrak{g}(\widetilde{A})$. Furthermore, if $\sigma$ is any legal
recoloring of $I$, then the previous statement also holds for $e_{\sigma(I)}$
in place of $e_{I}$.

Most of the work for the first statement of the conjecture has already been
done when proving Theorem 3.10; in fact, if we assume that if for each
$n\in\mathbb{Z}$ that each subspace of height $n$ in
$\mathfrak{g}(\widetilde{A})$ and in $\overline{\mathfrak{g}(A)}^{gen}$ have
the same dimension, then the proof actually goes through directly. However,
loosening this dimension requirement would be interesting because it does not
appear to be necessary in our computations. In any case, there should be a
surjection from $\overline{\mathfrak{g}(A)}^{gen}$ to
$\mathfrak{g}(\widetilde{A})$. The difficulty lies in showing that the former
has no nontrivial ideal that intersects the Cartan subalgebra. The second
statement of the conjecture should follow from the proof of Theorem 4.12.

The reason for considering nonzero determinants is because then the Chevalley
generators generate the Lie superalgebras, and the proof of 3.10 can then be
used. However, it should be a straightforward modification for zero
determinant as well; one may have to consider the derived algebra instead.

It would be interesting to develop an $\mathfrak{sl}_{2}$-equivariant theory
of Kac-Moody Lie algebras, or more generally, a Kac-Moody Lie algebra in the
category of $\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{p}$; then one can
see what happens in the semisimplification in $\operatorname{Ver}_{p}$ and in
$\operatorname*{sVec}_{\mathbb{K}}$. Lastly, it would be interesting to see
what happens when we study this procedure for affine Lie algebras.

####  4.5.2. Other Exceptional Lie Superalgebras

There are some other exceptional Lie superalgebras in characteristic $3$ that
we did not attempt to construct, like the Brown Lie superalgebras, the
exceptional Lie superalgebras $G(3)$ and $F(4)$, which are also known as
$\mathfrak{ag}(2)$ and $\mathfrak{ab}(3)$ in the literature. One may attempt
to construct these in this fashion as well (however, unlike the Elduque and
Cunha Lie superalgebras, there are straightforward constructions for these).

####  4.5.3. Applications to Representation Theory

Lastly, one can use the representation theory of exceptional Lie algebras to
study representation theory of these exceptional Lie superalgebras. In
particular, one can start with a representation of an exceptional Lie algebra
in $\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{3}$ and semisimplify it to
construct representations of the corresponding Lie superalgebra. Although
these can be probably classified by a highest-weight argument, this approach
will give a construction to determine the size of these representations, which
to our knowledge is virtually unknown.

##  5\. An Exceptional Lie Superalgebra in Characteristic $5$

In this section, we construct the Elduque Lie superalgebra
$\mathfrak{el}(5;5)$ of superdimension $(55|32)$ by semisimplifying
$\mathfrak{e}_{8}$. The even subalgebra is the orthogonal Lie algebra
$\mathfrak{so}_{11}$, and the odd part is the $32$-dimensional spinor
representation $L(\omega_{5})$. The Lie superalgebra $\mathfrak{el}(5;3)$
constructed earlier is a characteristic $3$ analog of this discovered in
[BGL+09], in the sense that there is a suitable choice of Cartan matrix that
creates both, depending on the characteristic. Our main theorem has not been
extended to characteristic $5$, so we will construct this in a fashion similar
to the alternate construction of $\mathfrak{g}(3,6)$ described above.

As remarked earlier, $\mathfrak{e}_{8}$ splits as the $120$-dimensional simple
Lie algebra $\mathfrak{so}_{16}$ and its $128$-dimensional spinor
representation $L(\omega_{8})$. First, let’s consider the subalgebra
$\mathfrak{so}_{16}$. Recall from §4.4.4 that the root vectors
$e_{2},e_{3},e_{4},e_{5}$, $e_{6},e_{7},e_{8}$ and $e_{100}$, which along with
their $f$ counterparts generate $\mathfrak{so}_{16}$. Visually, the Dynkin
diagram looks like:

$\dynkin[labels={100,8,7,6,5,4,3,2},scale=4,labeldistance=5pt]D{oooooooo}$

Let $V=\mathbb{K}^{16}$, and let $x=e_{2}+e_{3}+e_{4}$. By definition,
$\mathfrak{so}_{16}$ is the subalgebra of $\mathfrak{gl}_{16}=V\otimes V^{*}$
that preserves the nondegenerate symmetric bilinear form $B$ (for instance,
consider the form given by $B=\begin{pmatrix}0&I_{8}\\\
I_{8}&0\end{pmatrix}$). This gives a matrix realization of
$x\in\mathfrak{so}_{16}$, which acts on $V$ by matrix multiplication. In
particular, as a matrix $x$ is nilpotent of degree $5$, so $V$ can be realized
as an object in $\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{5}$ with
respect to matrix multiplication action by $x$. It is easily checked that
$V=11J_{1}\oplus J_{5}$, such that the $11J_{1}$ and the $J_{5}$ are
orthogonal to each other, and the restriction of the form to each piece is
nondegenerate. One way to see this is that
$e_{2}+e_{3},e_{4},f_{2}+f_{3},f_{4}$ generate
$\mathfrak{so}_{5}=\mathfrak{sp}_{4}$ as a subalgebra, and $e_{2}+e_{3}+e_{4}$
acts as a Jordan block of size $4$ on the four-dimensional representation of
$\mathfrak{sp}_{4}$ and hence as a Jordan block of size $5$ on the five-
dimensional representation of $\mathfrak{so}_{5}$ (and once embedded in
$\mathfrak{so}_{16}$, it acts trivially on its orthogonal complement, which is
$11$-dimensional).

This in turn means that $\mathfrak{gl}_{16}$ and $\mathfrak{so}_{16}$ can be
realized as objects in $\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{5}$
with respect the adjoint action of $x$. Then, when we semisimplify,
$\mathfrak{so}_{16}$ becomes $\mathfrak{so}_{11}$. To see this (in more
generality), let use view our bilinear form $B$ as an isomorphism from
$B:V\rightarrow V^{*}$. This is a morphism in the category because $B$ is
stabilized by $x$ as $x$ lies in $\mathfrak{so}_{16}$. This gives us the
following composite map $\phi:\mathfrak{gl}(V)\rightarrow\mathfrak{gl}(V)$
given by $\phi=(1_{V}\otimes B^{-1})\circ(1_{V\otimes V}+c_{V\otimes
V})\circ(1_{V}\otimes B)$. Since the outside morphisms are isomorphisms and
the middle morphism is twice a projector, the kernel of $\phi$ is a split
summand of $\mathfrak{gl}(V)$. The kernel of the middle map is
$\wedge^{2}(V)$, and we can identify it with the kernel of $\phi$. On the
other hand, the kernel of $\phi$ is the definition of $\mathfrak{so}(B,V)$, so
via the form $B$, we can say $\mathfrak{so}(B,V)=\wedge^{2}(V)$.

When we semisimplify, $\overline{B}$ is a nondegenerate symmetric bilinear
form on $\overline{V}$, and
$\overline{\wedge^{2}(V)}=\wedge^{2}{\overline{V}}$ (because it is the degree
$2$ piece of the exterior algebra; for degrees greater than or equal to the
characteristic, this may not necessarily be true). We deduce
$\wedge^{2}(\overline{V})=\mathfrak{so}(\overline{B},\overline{V})$. So in our
case, $\overline{\mathfrak{so}_{16}}=\mathfrak{so}_{11}$.

Now, let’s check what happens to $L(\omega_{8})$, which we view as a subspace
of $\mathfrak{e}_{8}$. The adjoint action of $x$ on $\mathfrak{e}_{8}$ is
still nilpotent of degree $5$, so $\mathfrak{e}_{8}=\mathfrak{so}_{16}\oplus
L(\omega_{8})$ is a splitting in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{5}$. It can be checked using
the software SuperLie that $L(\omega_{8})=32J_{4}$ in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{5}$, so its semisimplification
is purely odd of dimension $32$. We claim that the semisimplification is
actually the spinor representation $L(\omega_{5})$ of $\mathfrak{so}_{11}$.

By the description above, the action of $\mathfrak{so}_{16}$ restricts to an
action of $\mathfrak{so}_{11}\times\mathfrak{so}_{5}$ on the spinor
representation $L(\omega_{8})$. It is well-known that as an
$\mathfrak{so}_{11}\times\mathfrak{so}_{5}$-module
$L(\omega_{8})=L(\omega_{5})\otimes L(\omega_{2})$ is the tensor product of
the corresponding spinor representations. The module $L(\omega_{5})$ in a
suitable basis is $32J_{1}$, and $L(\omega_{2})$ in a suitable basis is
$J_{4}$. After semisimplification, the claim follows.

Therefore, we deduce that the semisimplification of $\mathfrak{e}_{8}$ with
respect to the adjoint action of $x$ decomposes as $55L_{1}\oplus 32L_{4}$ in
$\operatorname{Ver}_{5}$, so it can be identified with a Lie superalgebra. The
even part is the simple Lie algebra $\mathfrak{so}_{11}$, and the odd part is
its simple module $L(\omega_{5})$.. Therefore, by the classification in
[BGL+09], this must be the Lie superalgebra $\mathfrak{el}(5;5)$. We have
proved the following theorem:

######  Theorem 5.1.

The semisimplification of $\mathfrak{e}_{8}$ as an object in
$\operatorname*{Rep}{\mbox{\boldmath$\alpha$}}_{5}$ under the adjoint action
of $e_{2}+e_{3}+e_{4}$ is a Lie algebra in $\operatorname{Ver}_{5}$ of the
form $55L_{1}\oplus 32L_{4}$; in particular, this means it is a Lie
superalgebra, and that Lie superalgebra is the Elduque Lie superalgebra
$\mathfrak{el}(5;5)$.

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