# On the existence of sympathetic Lie algebras with dimension less than 25

A. L. García-Pulido and G. Salgado  Department of Computer Science; University
of Liverpool; Ashton Street, Liverpool, L69 3BX; United Kingdom  A.L.Garcia-
Pulido<EMAIL_ADDRESS>Facultad de Ciencias, UASLP;
Av. Parque Chapultepec 1570, Priv. del Pedregal, CP 78210, San Luis Potosí,
SLP, México<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this article we investigate the question of the lowest possible dimension
that a sympathetic Lie algebra $\mathfrak{g}$ can attain, when its Levi
subalgebra $\mathfrak{g}_{L}$ is simple. We establish the structure of the
nilradical of a perfect Lie algebra $\mathfrak{g}$, as a
$\mathfrak{g}_{L}$-module, and determine the possible Lie algebra structures
that one such $\mathfrak{g}$ admits. We prove that, as a
$\mathfrak{g}_{L}$-module, the nilradical must decompose into at least $4$
simple modules. We explicitly calculate the semisimple derivations of a
perfect Lie algebra $\mathfrak{g}$ with Levi subalgebra
$\mathfrak{g}_{L}=\mathfrak{sl}_{2}(\mathbb{C})$ and give necessary conditions
for $\mathfrak{g}$ to be a sympathetic Lie algebra in terms of these
semisimple derivations. We show that there is no sympathetic Lie algebra of
dimension lower than $15$, independently of the nilradical’s decomposition. If
the nilradical has $4$ simple modules, we show that a sympathetic Lie algebra
has dimension greater or equal than $25$.

###### Key words and phrases:

Sympathetic Lie algebras; Equivariant maps; Inner derivations

###### 1991 Mathematics Subject Classification:

Primary: 17B05, 17B30. Secondary: 17B20 17B60

##  1\. Introduction

A Lie algebra $\mathfrak{g}$ is sympathetic if its centre is trivial, if
$\mathfrak{g}$ is perfect and if $\mathfrak{g}$ has no outer derivations; that
is,

  1. (1)

$Z(\mathfrak{g})=0$,

  2. (2)

$\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$

  3. (3)

$\operatorname{Der}(\mathfrak{g})=\operatorname{ad}(\mathfrak{g})$.

It is well known that any semisimple Lie algebra satisfies properties (1), (2)
and (3). Sympathetic Lie algebras were introduced to determine if properties
(1), (2) and (3) completely characterise semisimple Lie algebras. In [1] the
first non semisimple Lie algebra satisfying the above properties appears; it
has dimension 35. This was followed by an example in dimension 48 [2], where
the term sympathetic was first used. The lowest known dimension of a
sympathetic Lie algebra is 25 and can be found in [3]. This algebra has much
smaller dimension that the previous examples due to the careful construction
with Levi subalgebra equal to $\mathfrak{sl}_{2}(\mathbb{C})$, which is the
lowest dimensional simple Lie algebra.

In [3] and [2], Benayadi develops theory describing structural properties of
sympathetic Lie algebras. However, the existing results do not lead to a
classification, or even showing existence, of these algebras in low
dimensions. Indeed, the non-existence of sympathetic Lie algebras of
dimensions less than 10 is attributed to J. Simon in [3] and it remains an
open problem to determine if 25 is the smallest dimension possible in a
sympathetic Lie algebra. The lack of progress could be due to two factors.
Little was known of the classification of Lie algebras at that time and the
available computational resources and power were very limited.

In this paper we give a solution to the problem of dimension minimality of
sympathetic Lie algebras which are not semisimple. To do this, we restrict our
attention to the class of algebras whose Levi factor is a simple Lie algebra.

By taking full advantage of the representation theory of simple Lie algebras,
we find very precise necessary conditions that a Lie algebra
$\mathfrak{g}=\mathfrak{g}_{L}\ltimes\operatorname{Rad}(\mathfrak{g})$ must
satisfy in order to be sympathetic. In 3.10, 3.12, we show that if
$\mathfrak{g}$ is perfect and $\operatorname{Rad}(\mathfrak{g})$ decomposes
into 1, 2 or 3 simple $\mathfrak{g}_{L}$-modules, then $\mathfrak{g}$ is not
sympathetic. Then $\operatorname{Rad}(\mathfrak{g})$ must decompose into at
least $4$ simple $\mathfrak{g}_{L}$-modules.

This condition completely characterises sympathetic Lie algebras where
$\operatorname{Rad}(\mathfrak{g})$ decomposes into 4 representations. In 3.14
we show that, up to isomorphism, there are only $6$ possible algebra
structures. We are therefore able to determine that the smallest possible
dimension such a sympathetic Lie algebra can have is 25, see 3.32. Moreover,
we show there is only one such algebra in dimension 25 and it is the example
given in [3]. The proof of the non-existence in dimensions smaller than 10
follows easily from our results. These results are independent of the choice
of simple Levi subalgebra.

We now describe the techniques used to prove our main theorems. We make
systematic use of each of the properties that define a sympathetic Lie algebra
$\mathfrak{g}=\mathfrak{g}_{L}\ltimes\operatorname{Rad}(\mathfrak{g})$ to
derive necessary conditions on its algebra structure. Recall that
$\operatorname{Rad}(\mathfrak{g})$ admits a $\mathfrak{g}_{L}$-module
structure induced by the Lie bracket and therefore can be decomposed into a
direct sum of simple $\mathfrak{g}_{L}$-modules.

A crucial point of our argument is to use the fact that the product between
any two simple $\mathfrak{g}_{L}$-modules $V,W$ is completely determined by
bilinear $\mathfrak{g}_{L}$-equivariant morphisms $V\otimes W\to U$, where $U$
is one of the simple $\mathfrak{g}_{L}$-modules appearing in the decomposition
of $\operatorname{Rad}(\mathfrak{g})$. In other words, the multiplication is
completely determined by the spaces
$\operatorname{Hom}_{\mathfrak{g}_{L}}(V\otimes W,U)$, where $V,W,U$ are
simple $\mathfrak{g}_{L}$-modules appearing in the decomposition of
$\operatorname{Rad}(\mathfrak{g})$. In addition, the nilpotency of
$\operatorname{Rad}(\mathfrak{g})$ enables us to derive a general form of the
multiplicative structure of $\operatorname{Rad}(\mathfrak{g})$. This process
further highlights how, when working under the assumption that $\mathfrak{g}$
is sympathetic, one would construct such an algebra.

Combining the above multiplicative structure and Clebsh-Gordan-type theorems
for simple $\mathfrak{g}_{L}$-modules, we obtain explicit forms of the
possible multiplication tables in terms of the number of simple
$\mathfrak{g}_{L}$-modules in the decomposition of
$\operatorname{Rad}(\mathfrak{g})$. The final step is to verify the (non)
existence of sympathetic Lie algebras with this explicit multiplication tables
in the case of at most 4 modules.

The outline of the paper is as follows. For completeness, we begin Section 2
with preliminary notions of classical Lie algebras. In Section 3, we prove our
main theorem 3.1: if $\mathfrak{g}$ is a sympathetic Lie algebra such that its
Levi subalgebra $\mathfrak{g}_{L}$ is simple and its nilradical decomposes
into four simple $\mathfrak{g}_{L}$-modules, then
$\operatorname{dim}_{\mathbb{C}}(\mathfrak{g})\geq 25$. From this point, we
work with Lie algebras which Levi subalgebra is simple. In Section 3.1, we
show that the nilradical of any sympathetic Lie algebra decomposes into at
least four simple modules and determine the possible multiplicative structures
for this case 3.14. In Section 3.2 and Section 3.3 we investigate whether any
of these possible algebras are sympathetic Lie algebras. For part of this
analysis, we designed algorithms that enable us to do explicit calculations to
this end. In Section 4, we include these algorithms and show in a concrete
example how to put to practice our theoretical results and use our algorithms.
An implementation of our algorithms can be found in
https://github.com/winsy17/Sympathetic_Lie_Algebras.

##  2\. Preliminaries

Throughout this article we only consider finite dimensional vector spaces over
$\mathbb{C}$.

######  Definition 2.1.

A Lie algebra $\mathfrak{g}$, is a vector space together with a bilinear map
$[,]:\mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}$ satisfying:

  1. (1)

$[x,y]=-[y,x]$ for every $x,y\in\mathfrak{g}$,

  2. (2)

$[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0$ for every $x,y,z\in\mathfrak{g}$.

Equation (2) is known as the Jacobi identity. Notice that (1) is equivalent to
$[x,x]=0$ for every $x\in\mathfrak{g}$.

One of the typical examples of a Lie algebra is the following. Consider a
vector space $V$ and the associative algebra of endomorphims of $V$ denoted by
$\operatorname{End}_{\mathbb{C}}(V)$. Define the commutator on
$\operatorname{End}_{\mathbb{C}}(V)$, by $[T,S]:=T\circ S-S\circ T$ for every
$T,S\in\operatorname{End}_{\mathbb{C}}(V)$. The resulting Lie algebra is
denoted by $\mathfrak{gl}(V)$ and called the general linear algebra associated
to $V$.

Let $\mathfrak{g}$ and $\mathfrak{h}$ be two Lie algebras. A Lie algebra
homomorphism, is a linear transformation
$\rho\colon\mathfrak{g}\to\mathfrak{h}$ such that
$\rho([x,y]_{\mathfrak{g}})=[\rho(x),\rho(y)]_{\mathfrak{h}}$. An isomorphism
of Lie algebras is a bijective homomorphism. A linear representation of a Lie
algebra $\mathfrak{g}$ on a vector space $V$ is a Lie algebra homomorphism
$\rho\colon\mathfrak{g}\to\mathfrak{gl}(V)$.

Let $\alpha\colon\mathfrak{g}\to\mathfrak{gl}(U)$ and
$\beta\colon\mathfrak{g}\to\mathfrak{gl}(W)$ be two linear representations of
$\mathfrak{g}$ on $U$ and $W$, respectively. The space of equivariant
homomorphisms from $U$ to $W$ is

$\operatorname{Hom}_{\mathfrak{g}}(U,W):=\\{T\in\operatorname{Hom}_{\mathbb{C}}(U,W)\
|\ T\circ\alpha(x)=\beta(x)\circ T,\ x\in\mathfrak{g}\\}.$

We can induce a new representation $\alpha\otimes\beta$ from $\mathfrak{g}$ on
$U\otimes W$ defined by $(\alpha\otimes\beta)(x)(u\otimes
w):=\alpha(x)(u)\otimes w+u\otimes\beta(x)(w)$ for every $x\in\mathfrak{g}$,
$u\in U$ and $w\in W$. This representation is the tensor product of $\alpha$
and $\beta$.

Given any Lie algebra $\mathfrak{g}$ the Lie bracket induces a natural linear
representation $\operatorname{ad}:\mathfrak{g}\to\mathfrak{gl}(\mathfrak{g})$
defined by $\operatorname{ad}_{x}(y):=[x,y]$ for any $x,y\in\mathfrak{g}$. We
call this representation the adjoint representation. Notice that
$\operatorname{Ker}(\operatorname{ad})=Z(\mathfrak{g})=\\{x\in\mathfrak{g}\mid[x,y]=0,\forall
y\in\mathfrak{g}\\}$.

A derivation on $\mathfrak{g}$ is a linear transformation
$D\colon\mathfrak{g}\to\mathfrak{g}$ which satisfies the Leibniz rule with
respect to the Lie bracket, $D([x,y])=[Dx,y]+[x,Dy],$ for every
$x,y\in\mathfrak{g}$.

######  Lemma 2.2.

Let $\mathfrak{g}$ be a Lie algebra. The following statements are equivalent.

  1. (1)

$D$ is a derivation in $\mathfrak{g}$,

  2. (2)

$D([e_{i},e_{j}])=[De_{i},e_{j}]+[e_{i},De_{j}]$ for every
$e_{i},e_{j}\in\beta$,

  3. (3)

$[D,\operatorname{ad}(e_{i})]-\operatorname{ad}(De_{i})=0$ for every
$e_{i}\in\beta$.

Where $\beta=\\{e_{1},\ldots,e_{n}\\}$ is a basis of $\mathfrak{g}$.

######  Remark 2.3.

Item 3 shows that obtaining a basis for the vector space
$\operatorname{Der}(\mathfrak{g})=\\{D\in\operatorname{End}(\mathfrak{g})\mid
D([x,y])=[Dx,y]+[x,Dy]\\}$ is equivalent to finding the solutions of a
homogeneous system of linear equations.

Given a linear representation of $\mathfrak{g}$ on $V$
$\rho:\mathfrak{g}\to\mathfrak{gl}(V)$, we say that a subspace $W\subset V$ is
invariant if $\rho(x)W\subset W$ for any $x\in\mathfrak{g}$. The trivial
subspaces $\\{0\\}$ and $V$ are invariant subspaces. We call a representation
$\rho\colon\mathfrak{g}\to\mathfrak{gl}(V)$ irreducible if the only invariant
subspaces of $V$ are the trivial ones. We say that the representation is
completely reducible if every invariant subspace $W$ admits an invariant
complement $U$ such that $V=W\oplus U$.

From now on, we say that $V$ is a $\mathfrak{g}$-module if there exists a
linear representation $\rho:\mathfrak{g}\to\mathfrak{gl}(V)$. We will say that
the $\mathfrak{g}$-module $V$ is simple if $\rho$ is an irreducible
representation and, semisimple, if $\rho$ is completely reducible.

The very well known classification of simple Lie algebras over $\mathbb{C}$ is
given as follows.

######  Theorem 2.4 (See [5], [6], [7]).

If $\mathfrak{g}$ is a finite dimensional simple Lie algebra over
$\mathbb{C}$, then $\mathfrak{g}$ is isomorphic to one of the following Lie
algebras.

  1. (1)

$A_{n}:=\mathfrak{sl}(n+1)$, $n\geq 1$ ,

  2. (2)

$B_{n}:=\mathfrak{so}(2n+1)$, $n\geq 2$ ,

  3. (3)

$C_{n}:=\mathfrak{sp}(2n)$, $n\geq 3$,

  4. (4)

$D_{n}:=\mathfrak{so}(2n)$, $n\geq 4$,

  5. (5)

$E_{6}$, $E_{7}$, $E_{8}$,

  6. (6)

$F_{4}$,

  7. (7)

$G_{2}$.

The corresponding dimensions in each case are given by

  1. (1)

$\operatorname{dim}(\mathfrak{sl}(n+1))=n^{2}+2n$,

  2. (2)

$\operatorname{dim}(\mathfrak{so}(2n+1))=2n^{2}+n$,

  3. (3)

$\operatorname{dim}(\mathfrak{sp}(2n))=2n^{2}+n$,

  4. (4)

$\operatorname{dim}(\mathfrak{so}(2n))=2n^{2}-n$,

  5. (5)

$\operatorname{dim}(E_{6})=78$,

  6. (6)

$\operatorname{dim}(E_{7})=133$,

  7. (7)

$\operatorname{dim}(E_{8})=248$,

  8. (8)

$\operatorname{dim}(F_{4})=52$,

  9. (9)

$\operatorname{dim}(G_{2})=14$.

The only existing isomorphisms between semisimple Lie algebras are the
following.

$A_{1}\simeq B_{1}\simeq C_{1}\simeq D_{1},$ $B_{2}\simeq C_{2},\qquad
D_{2}\simeq A_{1}\oplus A_{2},\qquad A_{3}\simeq D_{3}$

######  Theorem 2.5 (Weyl).

Let $\mathfrak{g}$ be a semisimple Lie algebra. Then, any finitely dimensional
linear representation of $\mathfrak{g}$ is completely reducible.

######  Theorem 2.6 (Schur’s Lemma).

Let $\mathfrak{g}$ be a Lie algebra and let

$\rho_{V}\colon\mathfrak{g}\to\mathfrak{gl}(V)\qquad\text{and}\qquad\rho_{W}\colon\mathfrak{g}\to\mathfrak{gl}(W)$

be two irreducible representations. Let
$T\in\operatorname{Hom}_{\mathbb{C}}(V,W)$ be such that
$T\circ\rho_{V}=\rho_{W}\circ T$.

  1. (1)

If $\rho_{V}\not\sim\rho_{W}$ then $T=0$,

  2. (2)

If $\rho_{V}=\rho_{W}$ then $T=\lambda\operatorname{Id}_{V}$.

##  3\. Main Theorem

We divide this section in two parts. In the first part we assume that
$\mathfrak{g}=\mathfrak{g}_{L}\ltimes\operatorname{Rad}(\mathfrak{g})$ is a
Lie algebra such that its Levi subalgebra $\mathfrak{g}_{L}$ is simple and its
radical $\operatorname{Rad}(\mathfrak{g})$ is nilpotent. We show that, if
$\mathfrak{g}$ is sympathetic, the decomposition of
$\operatorname{Rad}(\mathfrak{g})$ into simple $\mathfrak{g}_{L}$-modules must
contain at least four non-trivial modules.

In the second part, we fix a simple Lie algebra $\mathfrak{g}_{L}$ and
determine all the possible multiplicative structures that a sympathetic Lie
algebra $\mathfrak{g}$ can have assuming that
$\operatorname{Rad}(\mathfrak{g})$ decomposes into four simple
$\mathfrak{g}_{L}$-modules. These possible Lie brackets significantly reduce
the candidate sympathetic Lie algebras. We verify if any of the remaining
candidates is a sympathetic Lie algebra.

The main result of this work is the following.

######  Theorem 3.1 (Main Theorem).

Let $\mathfrak{g}$ be a sympathetic Lie algebra with simple Levi subalgebra
$\mathfrak{g}_{L}$, then $\operatorname{dim}_{\mathbb{C}}\mathfrak{g}\geq 25$.

From now on, we will only consider Lie algebras $\mathfrak{g}$ with a simple
Levi subalgebra $\mathfrak{g}_{L}$ and we will denote the radical
$\operatorname{Rad}(\mathfrak{g})$ by $\mathfrak{h}$. By 2.5, $\mathfrak{h}$
can be decomposed, with respect to the $\mathfrak{g}_{L}$-action on
$\mathfrak{h}$ induced by the Lie bracket, into a sum of simple
$\mathfrak{g}_{L}$-modules:

$\mathfrak{h}=V_{n_{1}}\oplus\cdots\oplus V_{n_{k}}.$

Denote by $s(\mathfrak{g}):=k$ to the number of simple
$\mathfrak{g}_{L}$-modules in the decomposition of $\mathfrak{h}$. Throughout
this work we denote by $V_{n_{i}}$ to the irreducible representation of
$\mathfrak{g}_{L}$ with highest weight $n_{i}$.

Now assume that $\mathfrak{g}$ is perfect. It is well known that if
$\mathfrak{g}$ is perfect then $\operatorname{Rad}(\mathfrak{g})$ is nilpotent
(see [6] (Chapter 3, Section 9 Corollary 2)) and therefore
$\operatorname{Rad}(\mathfrak{g})=\operatorname{NilRad}(\mathfrak{g})$. Since
$\mathfrak{h}$ is a nilpotent subalgebra, $0\neq
Z(\mathfrak{h})\subset\mathfrak{h}$ and $Z(\mathfrak{h})$ inherits a
decomposition as a $\mathfrak{g}_{L}$-submodule of $\mathfrak{h}$:

$Z(\mathfrak{h})=V_{z_{1}}\oplus\cdots\oplus V_{z_{r}},$

where $\\{z_{1},\ldots,z_{r}\\}\subseteq\\{n_{1},\ldots,n_{k}\\}$.

######  Proposition 3.2.

Let $\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$ be a Lie algebra with
simple Levi subalgebra $\mathfrak{g}_{L}$. Given the decomposition
$\mathfrak{h}=V_{n_{1}}\oplus\cdots\oplus V_{n_{k}}$ into simple
$\mathfrak{g}_{L}$-modules, then

  1. (1)

$\operatorname{dim}(V_{n_{i}})=1$ if, and only if,
$[\mathfrak{g}_{L},V_{n_{i}}]=0$,

  2. (2)

if $\operatorname{dim}(V_{n_{i}})>1$, then
$[\mathfrak{g}_{L},V_{n_{i}}]=V_{n_{i}}$,

  3. (3)

if $[V_{n_{i}},V_{n_{j}}]\neq 0$, then there exists a
$\mathfrak{g}_{L}$-module $V_{n_{p}}$, $1\leq p\leq k$, such that
$V_{n_{p}}\subset V_{n_{i}}\otimes V_{n_{j}}$; and a unique
$\mathfrak{g}_{L}$-equivariant morphism $\Gamma_{i,j}^{p}\colon
V_{n_{i}}\otimes V_{n_{j}}\to V_{n_{p}}$ such that
$\pi_{p}([V_{n_{i}},V_{n_{j}}])=\Gamma_{i,j}^{p}(V_{n_{i}},V_{n_{j}})$, where
$\pi_{p}\colon V_{n_{i}}\otimes V_{n_{j}}\to V_{n_{p}}$ is the natural
projection to $V_{n_{p}}$. If $i=j$, then
$V_{n_{p}}\subset\Lambda^{2}(V_{n_{i}})$ and $\Gamma_{i,i}^{p}$ is skew-
symmetric.

###### Proof.

The proof of Item 1 and Item 2 is straightforward.

The fact that any triple $x\in V_{n_{i}}$, $y\in V_{n_{j}}$,
$z\in\mathfrak{g}_{L}$ satisfies the Jacobi identity is equivalent to the
existence of a $\mathfrak{g}_{L}$-equivariant morphism $[,]\colon
V_{n_{i}}\otimes V_{n_{j}}\to\mathfrak{h}$. As a $\mathfrak{g}_{L}$-module,
there is a decomposition

$V_{n_{i}}\otimes V_{n_{j}}=V_{m_{1}}\oplus\cdots\oplus V_{m_{l}}.$

By Schur’s lemma,
$\\{m_{1},\ldots,m_{l}\\}\cap\\{n_{1},\ldots,n_{k}\\}=\emptyset$ implies
$[V_{n_{i}},V_{n_{j}}]=0$. ∎

######  Corollary 3.3.

Let $\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$ be a Lie algebra with
simple Levi subalgebra $\mathfrak{g}_{L}$. Given the decomposition
$\mathfrak{h}=V_{n_{1}}\oplus\cdots\oplus V_{n_{k}}$ into simple
$\mathfrak{g}_{L}$-modules, if $[V_{n_{i}},V_{n_{j}}]\neq 0$, then there exist
simple $\mathfrak{g}_{L}$-modules $V_{m_{1}},\ldots,V_{m_{r}}\subset
V_{n_{i}}\otimes V_{n_{j}}$ and unique $\mathfrak{g}_{L}$-equivariant
morphisms $\Gamma_{i,j}^{s}:V_{n_{i}}\otimes V_{n_{j}}\to V_{m_{s}}$ such that

$[V_{n_{i}},V_{n_{j}}]=V_{m_{1}}\oplus\cdots\oplus
V_{m_{r}},\quad\text{and}\quad[x,y]=\sum\limits_{s=1}^{r}\Gamma_{i,j}^{s}(x,y),$

for every $x\in V_{n_{i}}$, $y\in V_{n_{j}}$. In the case when $i=j$,
$V_{m_{1}},\ldots,V_{m_{r}}\subset\Lambda^{2}(V_{n_{i}})$ and we have a skew-
symmetric $\mathfrak{g}_{L}$-equivariant morphism
$\Gamma_{i}^{s}:V_{n_{i}}\otimes V_{n_{i}}\to V_{m_{s}}$.

######  Remark 3.4.

Given $\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$, the Lie algebra
structure of $\mathfrak{h}$ is uniquely determined by a collection of
$\mathfrak{g}_{L}$-equivariant morphisms, when considering $\mathfrak{h}$ as a
$\mathfrak{g}_{L}$-module. Conversely, in order to construct a Lie algebra
$\mathfrak{h}$ from a $\mathfrak{g}_{L}$-module, one first needs to determine
the possible equivariant morphisms which define an algebra structure on
$\mathfrak{h}$. In fact, any scalar multiple of such morphisms defines a
multiplicative structure on $\mathfrak{h}$. Therefore, $\mathfrak{h}$ is a Lie
algebra when there exist an appropriate choice of multiples for which the
Jacobi identity holds.

######  Proposition 3.5.

Let $\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$ be a finite dimensional
sympathetic Lie algebra with simple Levi subalgebra $\mathfrak{g}_{L}$.
Suppose $D\colon\mathfrak{g}\to\mathfrak{g}$ is a derivation such that
$D|_{\mathfrak{g}_{L}}=0$ and
$D|_{V_{n_{i}}}=\lambda_{n_{i}}\operatorname{Id}_{V_{n_{i}}}$ for any simple
$\mathfrak{g}_{L}$-module in the decomposition of $\mathfrak{h}$. Then $D$ is
an inner derivation of $\mathfrak{g}$ if, and only if, $D|_{\mathfrak{h}}=0$.

###### Proof.

Let $D$ be an inner derivation. Write $D$ as the sum of its semisimple and
nilpotent parts $D=\operatorname{ad}_{\mathfrak{g}}(s+h)$ for some
$s\in\mathfrak{g}_{L}$ and $h\in\mathfrak{h}$. By definition of $D$, for any
$y\in\mathfrak{g}_{L}$,

$0=D(y)=[s+h,y]=[s,y]+[h,y].$

Since $[s,y]\in\mathfrak{g}_{L}$ and $[h,y]\in\mathfrak{h}$, this implies
$[s,y]=0$ and $[h,y]=0$ for every $y\in\mathfrak{g}_{L}$. Thus $s\in
Z(\mathfrak{g}_{L})$ and hence $s=0$ because $\mathfrak{g}_{L}$ is a simple
Lie algebra. Consequently, $D$ is nilpotent and this implies
$\lambda_{n_{i}}=0$ for every $i$. The converse is straightforward.

∎

######  Proposition 3.6.

Let $\mathfrak{g}=\mathfrak{sl}_{2}(\mathbb{C})\ltimes\mathfrak{h}$ be a
finite dimensional Lie algebra and $V_{m},V_{p},V_{k}$ simple
$\mathfrak{sl}_{2}(\mathbb{C})$-modules in the decomposition of $\mathfrak{h}$
such that $[V_{m},V_{p}]_{\mathfrak{g}}=V_{k}$, or
$[V_{p},V_{p}]_{\mathfrak{g}}=V_{k}$. For any semisimple derivation
$D\in\operatorname{Der}(\mathfrak{g})$ there exist
$\lambda,\alpha,\beta,\gamma\in\mathbb{F}$ such that

$\displaystyle\alpha+\beta-\gamma$
$\displaystyle=\lambda(m+p-k),\quad\text{or}$ $\displaystyle 2\beta-\gamma$
$\displaystyle=\lambda(2p-k).$

###### Proof.

Since $\mathfrak{sl}_{2}(\mathbb{C})$ is a subalgebra of $\mathfrak{g}$, there
exists $\lambda\in\mathbb{F}$ such that

$D|_{\mathfrak{sl}_{2}(\mathbb{C})}=\lambda\operatorname{ad}_{\mathfrak{sl}_{2}(\mathbb{C})}(H).$

Suppose that $V_{m}=\langle e_{i}\rangle$, $V_{p}=\langle f_{i}\rangle$ and
$V_{k}=\langle g_{i}\rangle$, where the basis for $V_{m},V_{p}$ and $V_{k}$
are as in 3.19 Since $D$ is a semisimple derivation, we assume without loss of
generality that $D$ is of the form

$D(e_{i})=\alpha_{i}e_{i},\quad D(f_{i})=\beta_{i}f_{i},\quad
D(g_{i})=\gamma_{i}g_{i}.$

Given any element $e_{i}\in V_{m}$, we have

$D([E,e_{i}])=\alpha_{i-1}[E,e_{i}],$

and

$\displaystyle\hbox{}[DE,e_{i}]+[E,De_{i}]$
$\displaystyle=2\lambda[E,e_{i}]+\alpha_{i}[E,e_{i}]$
$\displaystyle=(2\lambda+\alpha_{i})[E,e_{i}].$

Using the fact that $D$ is a derivation we obtain

$\alpha_{i-1}=2\lambda+\alpha_{i}.$

Equivalently,

$[D|_{V_{m}}]=\operatorname{Diag}(\alpha_{0},\alpha_{0}-2\lambda,\cdots,\alpha_{0}-m(2\lambda)).$

Analogously,

$\displaystyle\hbox{}[D|_{V_{p}}]$
$\displaystyle=\operatorname{Diag}(\beta_{0},\beta_{0}-2\lambda,\cdots,\beta_{0}-p(2\lambda))$
$\displaystyle[D|_{V_{k}}]$
$\displaystyle=\operatorname{Diag}(\gamma_{0},\gamma_{0}-2\lambda,\cdots,\gamma_{0}-k(2\lambda)).$

Set $\alpha=\alpha_{0}$, $\beta=\beta_{0}$ and $\gamma=\gamma_{0}$. Notice
that the product $[V_{m},V_{p}]_{\mathfrak{g}}=V_{k}$ is completely determined
by an $\mathfrak{sl}_{2}(\mathbb{C})$-equivariant morphism which leaves
weights invariant. In consequence, if

(1) $[e_{i},f_{j}]=C_{i,j}g_{l(i,j)},\quad C_{i,j}\in\mathbb{F},$

then $e_{i}\otimes f_{j}\in V_{m}\otimes V_{p}$ and $g_{l(i,j)}\in V_{k}$ have
equal weights for every $i,j$. A direct calculation shows that
$l(i,j)=i+j-\frac{1}{2}(m+p-k)$. Applying $D$ to both sides of Equation 1 and
using the previous expression for $l(i,j)$, we can conclude that

$\alpha+\beta-\gamma=\lambda(m+p-k).$

Similarly, when $m=p$ one obtains

$2\beta-\gamma=\lambda(2p-k).$

∎

######  Remark 3.7.

Let $\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$ be a Lie algebra with
simple Levi subalgebra $\mathfrak{g}_{L}$. Recall that $\mathfrak{g}$ is a
sympathetic Lie algebra if $\mathfrak{g}$ does not admit any outer
derivations. We use 3.5 and 3.6 as criteria to show the existence of outer
derivations for certain Lie algebras $\mathfrak{g}$ as follows. When
$\mathfrak{g}_{L}=\mathfrak{sl}_{2}(\mathbb{C})$, 3.6 enable us to describe
semisimple derivations in terms of a system of homogenous linear equations,
which we use to construct outer derivations. When
$\mathfrak{g}_{L}\neq\mathfrak{sl}_{2}(\mathbb{C})$, we construct outer
derivations of the form described in 3.5.

######  Remark 3.8.

Solving the system from 3.6 requires solving a homogenous system of at most
$k$ equations with $k+1$ unknowns, where $s(\mathfrak{g})=k$. This has a huge
computational advantage to calculating directly
$\operatorname{Der}(\mathfrak{g})$, which requires solving a system of
$(\operatorname{dim}_{\mathbb{C}}\mathfrak{g})^{3}$ equations, for
$\operatorname{dim}_{\mathbb{C}}\mathfrak{g}=n_{1}+\cdots+n_{k}+k+3$.

######  Proposition 3.9.

Let $\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$ be a perfect Lie
algebra such that its Levi subalgebra $\mathfrak{g}_{L}$ is simple and
$[\mathfrak{h},\mathfrak{h}]\subset Z(\mathfrak{h})$. Then $\mathfrak{g}$ is
not sympathetic.

###### Proof.

As $\mathfrak{g}_{L}$-modules, $\mathfrak{h}$, $[\mathfrak{h},\mathfrak{h}]$
and $Z(\mathfrak{h})$ are completely reducible. This implies that there exists
a $\mathfrak{g}_{L}$-submodule $V$ such that $\mathfrak{h}=V\oplus
Z(\mathfrak{h})$. Define $D\colon\mathfrak{g}\to\mathfrak{g}$ by

$D|_{\mathfrak{g}_{L}}\equiv 0,\quad
D|_{Z(\mathfrak{h})}=\operatorname{Id}_{Z(\mathfrak{h})},\quad
D|_{V}=\frac{1}{2}\operatorname{Id}_{V}.$

A straightforward calculation shows that $D$ is a derivation and by 3.5, $D$
is an outer derivation. ∎

###  3.1. The radical of a sympathetic Lie algebra

####  3.1.1. Case $s(\mathfrak{g})=1,2$

######  Corollary 3.10.

Let $\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$ be a perfect Lie
algebra with simple Levi subalgebra $\mathfrak{g}_{L}$ is simple. If
$s(\mathfrak{g})=1,2$, then $\mathfrak{g}$ is not sympathetic.

###### Proof.

If $s(\mathfrak{g})=1,2$, then $[\mathfrak{h},\mathfrak{h}]\subset
Z(\mathfrak{h})$ and 3.9 implies that $\mathfrak{g}$ fails to be sympathetic.
∎

####  3.1.2. Case $s(\mathfrak{g})=3$

######  Convention 3.11.

From now on, we assume that if $[V_{n_{i}},V_{n_{j}}]\neq 0$, there exists a
unique $p$, $1\leq p\leq k$, such that $[V_{n_{i}},V_{n_{j}}]=V_{n_{p}}$.
Under this convention we will prove that if $s(\mathfrak{g})\leq 4$, then
$Z(\mathfrak{h})$ is a simple $\mathfrak{g}_{L}$-module. For simplicity, we
assume that $Z(\mathfrak{h})$ is a simple $\mathfrak{g}_{L}$-module whenever
$s(\mathfrak{g})\geq 5$.

In this subsection we show the following

######  Theorem 3.12.

Suppose that $\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$ is a Lie
algebra, with simple Levi subalgebra $\mathfrak{g}_{L}$, satisfying
$[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, $Z(\mathfrak{g})=0$ and
$s(\mathfrak{g})=3$. Then $\mathfrak{g}$ is not sympathetic.

###### Proof.

We determine the multiplication table for $\mathfrak{h}$ using that
$\mathfrak{h}$ is nilpotent. Suppose that $\mathfrak{h}=V_{m}\oplus
V_{p}\oplus V_{k}$. Since $\mathfrak{h}$ is a nilpotent subalgebra,
$Z(\mathfrak{h})=V_{m}\oplus V_{p}\oplus V_{k}$ or
$Z(\mathfrak{h})=V_{p}\oplus V_{k}$ imply $[\mathfrak{h},\mathfrak{h}]\subset
Z(\mathfrak{h})$ and, by 3.9, the conclusion of 3.12 holds.

Assume that $Z(\mathfrak{h})=V_{k}$ and notice that
$\mathfrak{h}^{2}=[\mathfrak{h},\mathfrak{h}]\neq\mathfrak{h}$. Now, by 3.9 if
$\mathfrak{h}^{2}=Z(\mathfrak{h})=V_{k}$ then $\mathfrak{g}$ is not
sympathetic. As a result, $\mathfrak{h}^{2}$ must be equal to exactly two
simple $\mathfrak{g}_{L}$-modules

$\mathfrak{h}^{2}=[\mathfrak{h},\mathfrak{h}]=V_{p}\oplus
V_{k},\quad\text{and}\quad\mathfrak{h}^{3}=V_{k}=Z(\mathfrak{h}).$

The first equation implies that $[V_{m},V_{m}]=V_{p}$ whereas the second
implies that $[V_{m},V_{p}]=V_{k}$. From this follows,

$\displaystyle 0$
$\displaystyle=[V_{m},[V_{m},V_{p}]]+[V_{m},[V_{p},V_{m}]]+[V_{p},[V_{m},V_{m}]]$
$\displaystyle=[V_{m},V_{k}]+[V_{m},V_{k}]+[V_{p},V_{p}]$
$\displaystyle=[V_{p},V_{p}].$

Therefore the only possible multiplication is given by

$[,]$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---
$V_{m}$ | $V_{p}$ | $V_{k}$ | 0
$V_{p}$ | $V_{k}$ | $0$ | 0
$V_{k}$ | $0$ | $0$ | 0

Notice that, by 3.5, the derivation $D|_{\mathfrak{g}_{L}}\equiv 0$,
$D|_{V_{m}}=\frac{1}{2}\operatorname{Id}|_{V_{m}}$,
$D|_{V_{p}}=\operatorname{Id}|_{V_{p}}$ and
$D|_{V_{k}}=\frac{3}{2}\operatorname{Id}|_{V_{k}}$ is an outer derivation. ∎

####  3.1.3. Case $s(\mathfrak{g})=4$

A consequence of 3.12 is the following.

######  Theorem 3.13.

If $\mathfrak{g}$ is a sympathetic Lie algebra, then $s(\mathfrak{g})\geq 4$.

Now we are interested in determining all the possible multiplicative
structures of nilpotent Lie algebras $\mathfrak{h}$ in the case when
$s(\mathfrak{g})=4$, i.e. $\mathfrak{h}=V_{n}\oplus V_{m}\oplus V_{p}\oplus
V_{k}$.

######  Theorem 3.14.

Suppose that $\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$ is a perfect
Lie algebra with simple Levi subalgebra $\mathfrak{g}_{L}$,
$\mathfrak{h}=V_{n}\oplus V_{m}\oplus V_{p}\oplus V_{k}$ and
$Z(\mathfrak{h})=V_{k}$. The algebra structure of $\mathfrak{h}$ is completely
determined by one of the following

  1. (1)

$[\mathfrak{h},\mathfrak{h}]=Z(\mathfrak{h})$ or,

  2. (2)

if $[\mathfrak{h},\mathfrak{h}]=V_{m}\oplus V_{p}\oplus V_{k}$, then

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{p}$ | $V_{k}/0$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

  3. (3)

or if $[\mathfrak{h},\mathfrak{h}]=V_{p}\oplus V_{k}$, then

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}/V_{k}/0$ | $V_{p}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{p}$ | $V_{p}/V_{k}/0$ | $V_{k}$ | $0$
$V_{p}$ | $V_{k}$ | $V_{k}$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$
$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}/V_{k}/0$ | $V_{p}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{p}$ | $V_{k}/0$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$
$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}$ | $0$ | $V_{k}$ | $0$
$V_{m}$ | $0$ | $V_{k}$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$
$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}$ | $V_{k}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{k}$ | $V_{k}/0$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

###### Proof.

Suppose that $[\mathfrak{h},\mathfrak{h}]\neq Z(\mathfrak{h})$. Since
$\mathfrak{g}$ is perfect, $\mathfrak{h}$ is nilpotent and therefore

$\mathfrak{h}\supsetneq[\mathfrak{h},\mathfrak{h}]=\mathfrak{h}^{2}\supsetneq\mathfrak{h}^{3}\supsetneq\cdots.$

As a consequence, $[\mathfrak{h},\mathfrak{h}]$ must be the sum of at most $3$
simple modules and, by hypothesis, $[\mathfrak{h},\mathfrak{h}]$ can only
contain $2$ or $3$ simple modules.

(1) First suppose that $\mathfrak{h}^{2}=V_{m}\oplus V_{p}\oplus V_{k}$ and
$\mathfrak{h}^{3}=V_{p}\oplus V_{k}$ y
$\mathfrak{h}^{4}=V_{k}=Z(\mathfrak{h})$.

From

$V_{m}\subset\mathfrak{h}^{2},\qquad
V_{m}\not\subset\mathfrak{h}^{3}=V_{p}\oplus V_{k}$

follows that $[V_{n},V_{n}]=V_{m}$. Using that

$[\mathfrak{h},\mathfrak{h}^{2}]=[V_{n}\oplus V_{m}\oplus V_{p},V_{m}\oplus
V_{p}]=V_{p}\oplus V_{k}$

together with

$[\mathfrak{h},\mathfrak{h}^{3}]=[V_{n}\oplus V_{m}\oplus
V_{p},V_{p}]=V_{k}=Z(\mathfrak{h})$

we conclude that

$[V_{p},\mathfrak{h}^{2}]=0\quad\text{and}\quad[V_{n},V_{p}]=V_{k}$

Hence the possible multiplicative structure of $\mathfrak{h}$ is described in
the following table.

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{m}$ | $V_{p}/V_{k}/0$ | $V_{k}$ | $0$
$V_{m}$ | $V_{p}/V_{k}/0$ | $V_{p}/V_{k}/0$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

(1.a) Assume that $[V_{n},V_{m}]=V_{p}$. Then

$0=[V_{m},[V_{n},V_{m}]]+[V_{n},[V_{m},V_{m}]]+[V_{m},[V_{m},V_{n}]]=[V_{n},[V_{m},V_{m}]]$

and thus $[V_{m},V_{m}]\subseteq V_{k}$. In this case, the possible
multiplication table is given by

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{p}$ | $V_{k}/0$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

(1.b) By a similar argument to the previous case, if $[V_{n},V_{m}]=V_{k}$,
then $[V_{m},V_{m}]=V_{k}/0$.

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{m}$ | $V_{k}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{k}$ | $V_{k}/0$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

However, this is impossible as we are assuming $\mathfrak{h}^{2}$ decomposes
into $3$ simple modules.

(1.c) Now suppose that $[V_{n},V_{m}]=0$. In this case the possible
multiplication table is

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{m}$ | $0$ | $V_{k}$ | $0$
$V_{m}$ | $0$ | $V_{p}/V_{k}/0$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

Since we are assuming that $\mathfrak{h}^{2}=V_{m}\oplus V_{p}\oplus V_{k}$,
we have $[V_{m},V_{m}]=V_{p}$. Therefore

$\displaystyle 0$
$\displaystyle=[V_{n},[V_{m},V_{m}]]+[V_{m},[V_{m},V_{n}]]+[V_{m},[V_{n},V_{m}]]$
$\displaystyle=[V_{n},[V_{m},V_{m}]]=V_{k}=Z(\mathfrak{h}),$

and, in consequence, this case is impossible.

(2) Now suppose that $\mathfrak{h}^{2}=V_{p}\oplus V_{k}$. Since
$\mathfrak{h}$ is nilpotent and $V_{p}\neq Z(\mathfrak{h})$, we have
$[\mathfrak{h},V_{p}]\neq V_{p},0$ and therefore

$\mathfrak{h}^{3}=[\mathfrak{h},\mathfrak{h}^{2}]=[\mathfrak{h},V_{p}]=V_{k}.$

Hence we may assume $[V_{n},V_{p}]=V_{k}$ without loss of generality.

Since $V_{p}\subset[\mathfrak{h},\mathfrak{h}]$, $V_{p}=[V,W]$ for some
$V,W=V_{n},V_{m}$.

$0=[V_{p},[V,W]]+[V,[W,V_{p}]]+[W,[V_{p},V]]=[V_{p},V_{p}].$

In summary, the possible multiplication table of $\mathfrak{h}$ is of the
form,

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}/V_{k}/0$ | $V_{p}/V_{k}/0$ | $V_{k}$ | $0$
$V_{m}$ | $V_{p}/V_{k}/0$ | $V_{p}/V_{k}/0$ | $V_{k}/0$ | $0$
$V_{p}$ | $V_{k}$ | $V_{k}/0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

(2.a) Suppose that $[V_{m},V_{p}]=V_{k}$. In this case we have that
$[V_{n},V_{m}]=V_{p}$. Otherwise, $[V_{n},V_{m}]=V_{k}/0$ implies

$0=[[V_{n},V_{m}],V_{n}]+[[V_{m},V_{n}],V_{n}]+[[V_{n},V_{n}],V_{m}]=[[V_{n},V_{n}],V_{m}],$

and

$0=[[V_{m},V_{n}],V_{m}]+[[V_{n},V_{m}],V_{m}]+[[V_{m},V_{m}],V_{n}]=[[V_{m},V_{m}],V_{n}].$

Hence $[V_{n}\oplus V_{m},V_{n}\oplus V_{m}]\subset V_{k}$, contradicting the
fact $V_{p}\subset[V_{n}\oplus V_{m},V_{n}\oplus V_{m}]$.

In this case, the multiplication table is

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}/V_{k}/0$ | $V_{p}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{p}$ | $V_{p}/V_{k}/0$ | $V_{k}$ | $0$
$V_{p}$ | $V_{k}$ | $V_{k}$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

(2.b) Now suppose that $[V_{m},V_{p}]=0$. Then

$\displaystyle 0$
$\displaystyle=[[V_{m},V_{n}],V_{m}]+[[V_{n},V_{m}],V_{m}]+[[V_{m},V_{m}],V_{n}]$
$\displaystyle=[[V_{m},V_{m}],V_{n}].$

This implies that $[V_{m},V_{m}]=V_{k}/0$. Since
$V_{p}\subset[\mathfrak{h},\mathfrak{h}]$, either $V_{p}=[V_{n},V_{m}]$ or
$V_{p}=[V_{n},V_{n}]$. In the first case, the following possible
multiplication table is given by

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}/V_{k}/0$ | $V_{p}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{p}$ | $V_{k}/0$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

If $V_{p}\neq[V_{m},V_{n}]$, then using the fact that $V_{m}\neq
Z(\mathfrak{h})$, we obtain two possible multiplicative structures

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}$ | $0$ | $V_{k}$ | $0$
$V_{m}$ | $0$ | $V_{k}$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$
$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}$ | $V_{k}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{k}$ | $V_{k}/0$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

∎

######  Corollary 3.15.

Let $\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$ be a perfect Lie with
simple Levi subalgebra $\mathfrak{g}_{L}$ and such that
$\mathfrak{h}=V_{n}\oplus V_{m}\oplus V_{p}\oplus V_{k}$,
$Z(\mathfrak{h})=V_{k}$. If $[\mathfrak{h},\mathfrak{h}]=V_{m}\oplus
V_{p}\oplus V_{k}$, then $\mathfrak{g}$ admits semisimple outer derivations.

###### Proof.

By 3.5, any derivation of the form

$\displaystyle D|_{\mathfrak{g}_{L}}\equiv 0,\quad
D|_{V_{n}}=\frac{\lambda}{4}\operatorname{Id}_{V_{n}},\quad
D|_{V_{m}}=\frac{\lambda}{2}\operatorname{Id}_{V_{m}},$ $\displaystyle
D|_{V_{p}}=\frac{3\lambda}{4}\operatorname{Id}_{V_{p}},\quad
D|_{V_{k}}=\lambda\operatorname{Id}_{V_{k}},$

with $\lambda\neq 0$ is an outer derivation. ∎

######  Corollary 3.16.

Let $\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$ be a perfect Lie such
that $\mathfrak{h}=V_{n}\oplus V_{m}\oplus V_{p}\oplus V_{k}$,
$Z(\mathfrak{h})=V_{k}$ and $[\mathfrak{h},\mathfrak{h}]=V_{p}\oplus V_{k}$.
Then $\mathfrak{g}$ has semisimple outer derivations, if the multiplicative
structure of $\mathfrak{h}$ is isomorphic to one of the following.

  1. (1)

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}/0$ | $V_{p}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{p}$ | $V_{p}/0$ | $V_{k}$ | $0$
$V_{p}$ | $V_{k}$ | $V_{k}$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

  2. (2)

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}/V_{k}/0$ | $V_{p}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{p}$ | $0$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

  3. (3)

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}$ | $0$ | $V_{k}$ | $0$
$V_{m}$ | $0$ | $V_{k}$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

  4. (4)

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}$ | $V_{k}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

###### Proof.

To show the existence of outer derivations, we now define derivations
satisfying 3.5 in each of the above cases. As in 3.5, we will take
$D|_{\mathfrak{g}_{L}}\equiv 0$ in each of the cases.

For the first case consider

$D|_{V_{n}\oplus V_{m}}=\frac{\lambda}{2}\operatorname{Id}_{V_{n}\oplus
V_{m}},D|_{V_{p}}=\lambda\operatorname{Id}_{V_{p}},\quad
D|_{V_{k}}=\frac{3\lambda}{2}\operatorname{Id}_{V_{k}}$

with $\lambda\neq 0$. In the second case, define $D$ as before if
$[V_{n},V_{n}]=V_{p}$. Otherwise, set

$D|_{V_{m}}\equiv 0,\quad D|_{V_{n}\oplus
V_{p}}=\frac{\lambda}{2}\operatorname{Id}_{V_{n}\oplus V_{p}},\quad
D|_{V_{k}}=\lambda\operatorname{Id}_{V_{k}}$

with $\lambda\neq 0$. For the third case define

$D|_{V_{n}}=\frac{\lambda}{2}\operatorname{Id}_{V_{n}},\quad
D|_{V_{m}}=\frac{3\lambda}{4}\operatorname{Id}_{V_{m}},\quad
D|_{V_{p}}=\lambda\operatorname{Id}_{V_{p}}\quad
D|_{V_{k}}=\frac{3\lambda}{2}\operatorname{Id}_{V_{k}}$

with $\lambda\neq 0$. Finally, set

$D|_{V_{n}}=\frac{\lambda}{2}\operatorname{Id}_{V_{n}},\quad D|_{V_{m}\oplus
V_{p}}=\lambda\operatorname{Id}_{V_{m}\oplus V_{p}}\quad
D|_{V_{k}}=\frac{3\lambda}{2}\operatorname{Id}_{V_{k}}$

with $\lambda\neq 0$ for the fourth case. ∎

######  Proposition 3.17.

Suppose that $\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$ is a Lie
algebra with simple Levi subalgebra $\mathfrak{g}_{L}$ such that
$\mathfrak{h}=V_{n}\oplus V_{m}\oplus V_{p}\oplus V_{k}$, as a
$\mathfrak{g}_{L}$-module. If $\mathfrak{g}$ is sympathetic, then there exist
two $\mathfrak{g}_{L}$-modules, say $V_{p}$ and $V_{k}$, such that
$[\mathfrak{h},\mathfrak{h}]=V_{p}\oplus V_{k}$ and $Z(\mathfrak{h})=V_{k}$.
Moreover, up to isomorphism, the multiplicative structure of $\mathfrak{h}$
must be given by one of the $6$ following tables.

  1. (1)

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}$ | $V_{p}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{p}$ | $V_{k}$ | $V_{k}$ | $0$
$V_{p}$ | $V_{k}$ | $V_{k}$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

  2. (2)

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{k}$ | $V_{p}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{p}$ | $0$ | $V_{k}$ | $0$
$V_{p}$ | $V_{k}$ | $V_{k}$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

  3. (3)

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{k}$ | $V_{p}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{p}$ | $V_{k}$ | $V_{k}$ | $0$
$V_{p}$ | $V_{k}$ | $V_{k}$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

  4. (4)

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}$ | $V_{p}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{p}$ | $V_{k}$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

  5. (5)

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{k}$ | $V_{p}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{p}$ | $V_{k}$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

  6. (6)

$[,]$ | $V_{n}$ | $V_{m}$ | $V_{p}$ | $V_{k}$
---|---|---|---|---
$V_{n}$ | $V_{p}$ | $V_{k}$ | $V_{k}$ | $0$
$V_{m}$ | $V_{k}$ | $V_{k}$ | $0$ | $0$
$V_{p}$ | $V_{k}$ | $0$ | $0$ | $0$
$V_{k}$ | $0$ | $0$ | $0$ | $0$

###### Proof.

First notice that if $Z(\mathfrak{h})=\mathfrak{h}$ or if
$Z(\mathfrak{h})=V_{m}\oplus V_{p}\oplus V_{k}$, then
$[\mathfrak{h},\mathfrak{h}]\subset Z(\mathfrak{h})$ and by 3.9 $\mathfrak{g}$
is not sympathetic. If $Z(\mathfrak{h})=V_{p}\oplus V_{k}$, then similar
arguments to those in 3.12 show that $\mathfrak{g}$ is not sympathetic.

Since $\mathfrak{g}$ is sympathetic, then $\mathfrak{g}$ is perfect and
therefore the Lie algebra structure of $\mathfrak{h}$ is one of the
multiplications given in 3.14. 3.9 implies that, either
$[\mathfrak{h},\mathfrak{h}]=V_{p}\oplus V_{k}$, or
$[\mathfrak{h},\mathfrak{h}]=V_{m}\oplus V_{p}\oplus V_{k}$. Given that
$\mathfrak{g}$ does not contain any outer derivations, 3.15 implies that
$[\mathfrak{h},\mathfrak{h}]=V_{p}\oplus V_{k}$. For the same reason, the
possible multiplicative structures of $\mathfrak{h}$ in 3.14 (Item 3) must
exclude those listed in 3.16. ∎

###  3.2. Sympathetic Lie algebras with simple Levi subalgebra
$\mathfrak{g}_{L}\neq\mathfrak{sl}_{2}(\mathbb{C})$

######  Theorem 3.18.

Suppose that $\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$ is a
sympathetic Lie algebra with simple Levi subalgebra $\mathfrak{g}_{L}$,
$\mathfrak{g}_{L}\neq\mathfrak{sl}_{2}(\mathbb{C})$ and $\mathfrak{h}$ does
not contain any trivial $\mathfrak{g}_{L}$-modules. Then
$\operatorname{dim}(\mathfrak{g})>25$.

###### Proof.

In this proof we use several standard results from representation theory,
which can be found for example in [4]. A simple calculation shows that, if
$\mathfrak{g}_{L}\neq\mathfrak{sl}_{3}(\mathbb{C}),\mathfrak{sl}_{4}(\mathbb{C}),\mathfrak{so}(5)$,
then $\operatorname{dim}(\mathfrak{g})\geq 21$. By 3.13, the decomposition of
$\mathfrak{h}$ into simple $\mathfrak{g}_{L}$-modules has at least $4$
modules. By 3.2, since $\mathfrak{h}$ does not contain any trivial modules, we
obtain a lower bound $\operatorname{dim}(\mathfrak{h})\geq 8$. Hence,
$\operatorname{dim}(\mathfrak{g})\geq 21+8=29$, as required.

Next we prove the statement for the $3$ remaining cases. First suppose that
$\mathfrak{g}_{L}=\mathfrak{sl}_{4}(\mathbb{C}),\mathfrak{so}(5)$. In this
case, $\operatorname{dim}(\mathfrak{g}_{L})=15$ and any non trivial simple
$\mathfrak{g}_{L}$-module has dimension at least $4$. Using 3.13 we get a
bound on the dimension of $\mathfrak{g}$,

$\operatorname{dim}(\mathfrak{g})\geq 15+4\cdot 4=31.$

Finally, suppose that $\mathfrak{g}_{L}=\mathfrak{sl}_{3}(\mathbb{C})$. Let
$V_{(a_{1},a_{2})}$ denote the simple $\mathfrak{sl}_{3}(\mathbb{C})$-module
of highest weight $(a_{1},a_{2})$. It is a standard result that

$\operatorname{dim}(V_{(a_{1},a_{2})})=\frac{1}{2}(a_{1}+1)(a_{2}+1)(a_{1}+a_{2}+2).$

Notice that $V_{(0,1)}$ and $V_{(1,0)}$ are the non-trivial simple
$\mathfrak{g}_{L}$-modules with lowest dimension and
$\operatorname{dim}(V_{(1,0)})=\operatorname{dim}(V_{(0,1)})=3$. We show that
if the $\mathfrak{g}_{L}$-module decomposition of $\mathfrak{h}$ is of the
form $\mathfrak{h}=V_{(1,0)}^{r}\oplus V_{(0,1)}^{s}$, then $\mathfrak{g}$
cannot be a sympathetic Lie algebra. Assume $\mathfrak{h}=V_{(1,0)}^{r}\oplus
V_{(0,1)}^{s}$ as $\mathfrak{g}_{L}$-modules, and set
$\mathfrak{h}_{(1,0)}=V_{(1,0)}^{r}$ and $\mathfrak{h}_{(0,1)}=V_{(0,1)}^{s}$.

From the tensor product decompositions

$\displaystyle V_{(1,0)}\otimes V_{(1,0)}$ $\displaystyle=V_{(2,0)}\oplus
V_{(0,1)}$ $\displaystyle V_{(0,1)}\otimes V_{(0,1)}$
$\displaystyle=V_{(0,2)}\oplus V_{(1,0)}$ $\displaystyle V_{(1,0)}\otimes
V_{(0,1)}$ $\displaystyle=V_{(1,1)}\oplus V_{(0,0)},$

we see that

$[\mathfrak{h}_{(1,0)},\mathfrak{h}_{(1,0)}]\subset\mathfrak{h}_{(0,1)},\quad[\mathfrak{h}_{(1,0)},\mathfrak{h}_{(0,1)}]=0,\quad[\mathfrak{h}_{(0,1)},\mathfrak{h}_{(0,1)}]\subset\mathfrak{h}_{(1,0)}.$

For any two simple modules $A,B\subset\mathfrak{h}$, $[A,B]=C\neq 0$ if, and
only if, $A,B\subset\mathfrak{h}_{(1,0)}$ or $A,B\subset\mathfrak{h}_{(0,1)}$.
Let $A,B\subset\mathfrak{h}$ be simple modules satisfying $[A,B]=C\neq 0$ and
suppose, without loss of generality, that $A,B\subset\mathfrak{h}_{(1,0)}$ and
thus $C\in\mathfrak{h}_{(0,1)}$. Since $[B,D]=[A,D]=0$ for any
$D\in\mathfrak{h}_{(0,1)}$, we have

$0=[[A,B],D]+[[B,D],A]+[[D,A],B]=[C,D].$

Therefore $C\subset Z(\mathfrak{h})$ and, more generally, we have
$[\mathfrak{h},\mathfrak{h}]\subset Z(\mathfrak{h})$. By 3.9, $\mathfrak{g}$
is not a sympathetic Lie algebra. As a consequence of this, $\mathfrak{h}$
must contain at least one simple $\mathfrak{g}_{L}$-module with dimension
strictly greater than $3$. The smallest possible such simple modules are
$V_{(1,1)}$, $V_{(2,0)}$ and $V_{(0,2)}$ with
$\operatorname{dim}(V_{(1,1)})=\operatorname{dim}(V_{(0,2)})=\operatorname{dim}(V_{(2,0)})=6$.
Using this and 3.13 we obtain a lower bound

$\operatorname{dim}(\mathfrak{g})=\operatorname{dim}(\mathfrak{g}_{L})+\operatorname{dim}(\mathfrak{h})\geq
8+3\cdot 3+6=23.$

In particular, if $s(\mathfrak{g})\geq 5$ and $\mathfrak{g}$ is a sympathetic
Lie algebra, then $\operatorname{dim}(\mathfrak{g})>25$. Hence a Lie algebra
$\mathfrak{g}$ with $\operatorname{dim}(\mathfrak{g})\leq 25$ is sympathetic
if, and only if, $\mathfrak{h}$ decomposes exactly into $4$ simple modules:
one of dimension 6, and three of dimension 3. We will show that $\mathfrak{g}$
cannot have this structure.

We will only prove the case where $V_{(1,1)}$ is contained in the
decomposition of $\mathfrak{h}$, as the other cases are similar. Recall that
the algebra structure of $\mathfrak{h}$ is one of the multiplications
determined in 3.17. We use the following decompositions of tensor products

$\displaystyle V_{(1,1)}\otimes V_{(1,1)}$ $\displaystyle=V_{(2,2)}\oplus
V_{(3,0)}\oplus V_{(0,3)}\oplus V_{(0,0)}$ $\displaystyle V_{(1,1)}\otimes
V_{(1,0)}$ $\displaystyle=V_{(2,1)}\oplus V_{(1,0)}\oplus V_{(0,2)},$
$\displaystyle V_{(1,1)}\otimes V_{(0,1)}$ $\displaystyle=V_{(1,2)}\oplus
V_{(0,1)}\oplus V_{(2,0)}.$

Notice that

$V_{(1,0)}\subsetneq V_{(1,1)}\otimes V_{(1,1)},\qquad V_{(0,1)}\subsetneq
V_{(1,1)}\otimes V_{(1,1)},$

implies $[V_{(1,1)},V_{(1,1)}]=0$. Therefore, if the multiplicative structure
of $\mathfrak{h}$ is isomorphic that of tables 1,3-6, then
$V_{(1,1)}=V_{p},V_{k}$; and $V_{(1,1)}=V_{m},V_{p},V_{k}$ otherwise. The fact
that

$V_{(1,1)}\subsetneq V_{(1,0)}\otimes V_{(1,0)},\qquad V_{(1,1)}\subsetneq
V_{(0,1)}\otimes V_{(0,1)},$

implies $V_{(1,1)}\neq[V_{(0,1)},V_{(0,1)}]$ and
$V_{(1,1)}\neq[V_{(1,0)},V_{(1,0)}]$. In consequence, $V_{(1,1)}\neq V_{k}$
and the algebra structure of $\mathfrak{h}$ cannot be isomorphic to tables
1,3,4,6. In tables 2 and 5, $V_{(1,1)}=V_{p}=[V_{n},V_{m}]$ implies that
$V_{n}\not\cong V_{m}$. If $V_{n}=V_{(1,0)}$ then
$[V_{n},V_{p}]=[V_{(1,0)},V_{(1,1)}]=V_{k}$ implies $V_{k}=V_{(1,0)}$. However
$V_{k}=[V_{n},V_{n}]$ but this contradicts $V_{(1,0)}\subsetneq
V_{(1,0)}\otimes V_{(1,0)}$. Similarly, $V_{(1,1)}\neq V_{m}$. ∎

###  3.3. Sympathetic Lie algebras with Levi subalgebra
$\mathfrak{g}_{L}=\mathfrak{sl}_{2}(\mathbb{C})$

For completeness, we begin this subsection with standard results of the
representation theory of $\mathfrak{sl}_{2}(\mathbb{C})$.

####  3.3.1. Irreducible representations of $\mathfrak{sl}_{2}(\mathbb{C})$

Let

$\mathfrak{sl}_{2}(\mathbb{C})=\\{A\in\operatorname{Mat}(2\times
2,\mathbb{C})\ |\ \operatorname{Tr}(A)=0\\}$

and let

$\\{H=\left(\begin{smallmatrix}1&0\\\
0&-1\end{smallmatrix}\right),E=\left(\begin{smallmatrix}0&1\\\
0&0\end{smallmatrix}\right),F=\left(\begin{smallmatrix}0&0\\\
1&0\end{smallmatrix}\right)\\}$

be a basis of $\mathfrak{sl}_{2}(\mathbb{C})$. Notice that
$\mathfrak{sl}_{2}(\mathbb{C})$ is a simple Lie algebra equipped with the
commutator $[,]$. Moreover, this choice of basis satisfies

$[H,E]=2E,\quad[H,F]=-2F,\quad[E,F]=H.$

Denote by $V_{n}$ to the $n+1$ dimensional vector space with basis
$\\{e_{0},e_{1},\ldots,e_{n}\\}$. Set $e_{-1}:=e_{n+1}:=0$. We now recall
classical results, which can be found in [4], [5], [6] or [7].

######  Proposition 3.19.

Let $n\in\mathbb{N}$ and
$\rho_{n}:\mathfrak{sl}_{2}(\mathbb{C})\to\mathfrak{gl}(V_{n})$ be the linear
representation given by

  * •

$\rho_{n}(H)(e_{i})=(n-2i)e_{i}$,

  * •

$\rho_{n}(F)(e_{i})=e_{i+1}$,

  * •

$\rho_{n}(E)(e_{i})=i(n+1-i)e_{i-1}$,

for every $0\leq i\leq n$. Then

  1. (1)

$\rho_{n}$ is an irreducible representation of
$\mathfrak{sl}_{2}(\mathbb{C})$,

  2. (2)

if $\rho:\mathfrak{sl}_{2}(\mathbb{C})\to\mathfrak{gl}(V)$ is an irreducible
representation of $\mathfrak{sl}_{2}(\mathbb{C})$ with
$\operatorname{dim}V\geq 1$, then there exists $n\in\mathbb{N}$ such that
$V=V_{n}$

We will write $V_{0}=\mathbb{C}$ and
$\rho_{0}:\mathfrak{sl}_{2}(\mathbb{C})\to\mathfrak{gl}(V_{0})$ for the
trivial representation $\rho_{0}(x)=0$ for every
$x\in\mathfrak{sl}_{2}(\mathbb{C})$.

Given any finite dimensional linear representation
$\rho:\mathfrak{sl}_{2}(\mathbb{C})\to\mathfrak{gl}(V)$ of
$\mathfrak{sl}_{2}(\mathbb{C})$ there is a unique decomposition

$V=V_{n_{1}}\oplus\cdots\oplus V_{n_{k}}.$

with $\operatorname{dim}_{\mathbb{C}}V=(n_{1}+1)+\cdots+(n_{k}+1)$ for some
$n_{1}\leq n_{2}\leq\ldots\leq n_{k}$.

######  Theorem 3.20 (Clebsh-Gordan).

For any $n\geq m$,

(2) $V_{n}\otimes V_{m}\simeq\bigoplus_{i=0}^{m}V_{n+m-2i}\simeq V_{n+m}\oplus
V_{n+m-2}\oplus V_{n+m-4}\oplus\cdots\oplus V_{n-m}.$

######  Corollary 3.21.

For $n=m$,

  1. (1)

$\Lambda^{2}V_{n}\simeq\bigoplus_{i=1}^{[(n+1)/2]}V_{(2n+2)-4i}$,

  2. (2)

$S^{2}V_{n}\simeq\bigoplus_{j=0}^{[n/2]}V_{2n-4j}$,

  3. (3)

$V_{n}\otimes V_{n}\simeq\Lambda^{2}V_{n}\oplus S^{2}V_{n}$.

Given a Lie algebra $\mathfrak{g}$ with Levi decomposition
$\mathfrak{g}=\mathfrak{sl}_{2}(\mathbb{C})\ltimes\mathfrak{h}$ and an
$\mathfrak{sl}_{2}(\mathbb{C})$-module decomposition
$\mathfrak{h}=V_{n_{1}}\oplus\cdots\oplus V_{n_{k}}$, we define

$\mathfrak{h}^{0}:=\bigoplus\limits_{\\{n_{i}\mid\ n_{i}\equiv
0\mod{2}\\}}V_{n_{i}}\qquad\mathfrak{h}^{1}:=\bigoplus\limits_{\\{n_{i}\mid\
n_{i}\equiv 1\mod{2}\\}}V_{n_{i}}.$

We show that the product in $\mathfrak{h}$ is $\mathbb{Z}_{2}$-graded:

######  Lemma 3.22.

Let $\mathfrak{g}$ be a Lie algebra, with Levi decomposition given by
$\mathfrak{g}=\mathfrak{sl}_{2}(\mathbb{C})\ltimes\mathfrak{h}$ and
$\mathfrak{h}=\mathfrak{h}^{0}\oplus\mathfrak{h}^{1}$. Then $\mathfrak{h}^{0}$
is a subalgebra of $\mathfrak{h}$ and
$[\mathfrak{h}^{1},\mathfrak{h}^{1}]\subset\mathfrak{h}^{0}$ and
$[\mathfrak{h}^{1},\mathfrak{h}^{0}]\subset\mathfrak{h}^{1}$. In particular,
if $[V_{n_{i}},V_{n_{j}}]=V_{n_{l}}$ then $n_{i}+n_{j}\equiv n_{l}\mod{2}$.

###### Proof.

By the 3.20, the decompositions

$V_{2n+1}\otimes V_{2m+1}=V_{2(n+m)+2}\oplus\cdots\oplus V_{2(m-n)}$
$V_{2n}\otimes V_{2m}=V_{2(n+m)}\oplus\cdots\oplus V_{2(m-n)}$

only contain simple $\mathfrak{sl}_{2}(\mathbb{C})$-modules with $p\equiv
0\mod{2}$.

∎

######  Lemma 3.23.

Let $\mathfrak{g}$ be a Lie algebra with Levi decomposition given by
$\mathfrak{g}=\mathfrak{sl}_{2}(\mathbb{C})\ltimes\mathfrak{h}$. If
$\mathfrak{h}=\mathfrak{h}^{1}$, then the radical $\mathfrak{h}$ is abelian.
In particular, $\mathfrak{g}$ is not sympathetic.

###### Proof.

Combining 3.22 with Schur’s Lemma 2.6 we conclude that any bilinear
$\mathfrak{sl}_{2}(\mathbb{C})$-equivariant morphism $\Gamma\colon
V_{2n+1}\otimes V_{2m+1}\to V_{2l+1}$ is trivial. In particular, the Lie
bracket $[\ ,\ ]\colon V_{n}\otimes V_{m}\to\mathfrak{h}$ is zero for every
$n,m\in\\{n_{1},\ldots,n_{k}\\}$. By 3.9, $\mathfrak{g}$ is not a sympathetic
Lie algebra.

∎

######  Corollary 3.24.

Let $\mathfrak{g}$ be a Lie algebra with Levi decomposition
$\mathfrak{g}=\mathfrak{sl}_{2}(\mathbb{C})\ltimes\mathfrak{h}$ and
$\mathfrak{h}=\mathfrak{h}^{0}\oplus\mathfrak{h}^{1}$. If
$\mathfrak{h}^{0}\subset Z(\mathfrak{h})$, then $\mathfrak{g}$ is not a
sympathetic Lie algebra.

###### Proof.

From Clebsh-Gordan’s 3.20 follows that $[\mathfrak{h},\mathfrak{h}]\subset
Z(\mathfrak{h})$. By 3.9, $\mathfrak{g}$ is not a sympathetic Lie algebra.

∎

######  Corollary 3.25.

Let $\mathfrak{g}$ be a sympathetic Lie algebra with Levi decomposition given
by $\mathfrak{g}=\mathfrak{sl}_{2}(\mathbb{C})\ltimes\mathfrak{h}$ and with
$\mathfrak{h}=V_{n}\oplus V_{m}\oplus V_{p}\oplus V_{k}$ and
$Z(\mathfrak{h})=V_{k}$. Then $n\equiv m\equiv p\equiv k\equiv 0\mod{2}$.
Furthermore, in each case of 3.17 the following additional conclusions hold.

  1. (1)

$p\equiv k\equiv 2\mod{4}$,

  2. (2)

$k\equiv 2\mod{4}$,

  3. (3)

$p\equiv k\equiv 2\mod{4}$,

  4. (4)

$k\equiv 2\mod{4}$,

  5. (5)

$p\equiv k\equiv 2\mod{4}$.

###### Proof.

The fact that $n\equiv m\equiv p\equiv k\equiv 0\mod{2}$ follows directly by
applying 3.22 on each of the possible multiplicative structures of
$\mathfrak{h}$ from 3.17.

To prove the remaining statements, it is sufficient to observe that if two
simple $\mathfrak{sl}_{2}(\mathbb{C})$-modules $V_{r},V_{s}$ satisfy
$V_{r}=[V_{s},V_{s}]$ then the skew-symmetry of the Lie bracket implies that
$V_{r}\subset\Lambda^{2}(V_{s})$. Using the decomposition from 3.20, we
conclude that $r\equiv 2\mod{4}$.

∎

######  Corollary 3.26.

Let $\mathfrak{g}$ be a sympathetic Lie algebra with Levi decomposition given
by $\mathfrak{g}=\mathfrak{sl}_{2}(\mathbb{C})\ltimes\mathfrak{h}$, such that
$\mathfrak{h}=V_{n}\oplus V_{m}\oplus V_{p}\oplus V_{k}$, and
$Z(\mathfrak{h})=V_{k}$. Then $\operatorname{dim}\mathfrak{g}=2N+1$.

######  Proposition 3.27.

Let $\mathfrak{g}$ be a sympathetic Lie algebra with Levi decomposition given
by $\mathfrak{g}=\mathfrak{sl}_{2}(\mathbb{C})\ltimes\mathfrak{h}$. Suppose
that $\mathfrak{h}=V_{n}\oplus V_{m}\oplus V_{p}\oplus V_{k}$, as an
$\mathfrak{sl}_{2}(\mathbb{C})$-module, and that $Z(\mathfrak{h})=V_{k}$,
where $n,m,p,k\in\mathbb{N}$. If $\operatorname{dim}(\mathfrak{g})\leq 25$
then one of the following cases must necessarily hold:

  * •

the multiplication is given by one of the tables 1-6 and $(n,m,p,k)$ equals
one of

$\displaystyle(4,4,2,6),(4,6,2,6),(6,4,2,6),(4,4,6,2),(6,4,6,2),(4,6,6,2),$
$\displaystyle(2,2,2,2),(2,4,2,2),(4,2,2,2),(4,4,2,2)$

  * •

the multiplication is given by one of the tables 2,3,5 and $(n,m,p,k)$ equals
one of

$\displaystyle(4,4,4,6),(2,2,4,2),(2,4,4,2),(2,6,4,2),(4,2,4,2),(4,4,4,2),$
$\displaystyle(4,6,4,2),(6,2,4,2),(6,4,4,2),(6,6,4,2)$

  * •

the multiplication is given by one of the tables 4,5,6 and $(n,m,p,k)$ equals
one of

$(4,2,6,2),(4,6,2,2)$

  * •

the multiplication is given by one of the tables 4,5 and $(n,m,p,k)$ equals
one of

$(6,2,6,2),(8,2,6,2)$

  * •

the multiplication is given by table 2 and $(n,m,p,k)$ equals one of

$(4,2,4,6),(6,2,4,6),(4,2,6,6)$

  * •

the multiplication is given by table 5 and $(n,m,p,k)$ equals one of

$(6,2,8,2),(4,8,4,2).$

###### Proof.

Using the parity of $n,m,k,p$ found in 3.25, we write $n=2a$, $m=2b$, $p=2c$
and $k=4d+2$. Recall that the possible multiplicative structure of
$\mathfrak{h}$ is one of the tables in 3.17. Notice that
$\operatorname{dim}(\mathfrak{g})=2(a+b+c+2d)+9\leq 25$ if and only if
$a+b+c+2d\leq 8$.

We only give the proof for $d\geq 1$ as the case $d=0$ is very similar. By
Clebsh-Gordan theorem, $V_{4d+2}\subsetneq V_{2}\otimes V_{2}$. In each of the
above tables we see that $V_{4d+2}=[V_{2a},V_{2c}]$ and hence $a\geq 2$ or
$c\geq 2$. Similarly, in tables 1,3,4,5,6,

$V_{4d+2}=[V_{2b},V_{2b}]$

and therefore $b\geq 2$. In particular, $a+b+c\geq 5$. Further, in table 2

$V_{4d+2}=[V_{2a},V_{2a}]\qquad\text{ and }\qquad V_{4d+2}=[V_{2b},V_{2c}]$

and so $a+b+c\geq 5$ in this case too. Since we require $a+b+c+2d\leq 8$ then
$d=1$ and hence

(3) $5\leq a+b+c\leq 6.$

In tables 1,4,6, $c\equiv 1\mod{2}$ since $[V_{2a},V_{2a}]=V_{2c}$. By similar
arguments to before, $c\geq 3$ and $[V_{2a},V_{2a}]=V_{2c}$ imply $a\geq 2$
and $a+b+c\geq 7$. Hence $c=1$ so that Equation 3 is satisfied. Since
$[V_{2a},V_{2c}]=V_{6}$, then $a\geq 2$. Then we have $a,b\geq 2$, $c=1$ and
$d=1$. In this case, the only possibilities for $\mathfrak{h}$ are

Multiplications 1,4,6, $d=1$
---
$a$ | $b$ | $c$ | $d$ | $\mathfrak{h}=V_{2a}\oplus V_{2b}\oplus V_{2c}\oplus V_{4d+2}$ | $\operatorname{dim}(\mathfrak{g})$
$2$ | $2$ | $1$ | $1$ | $V_{4}\oplus V_{4}\oplus V_{2}\oplus V_{6}$ | $23$
$2$ | $3$ | $1$ | $1$ | $V_{4}\oplus V_{6}\oplus V_{2}\oplus V_{6}$ | $25$
$3$ | $2$ | $1$ | $1$ | $V_{6}\oplus V_{4}\oplus V_{2}\oplus V_{6}$ | $25$

Using the same arguments than before in tables 3 and 5, we obtain $a,b\geq 2$.
Therefore, the $a,b,c$ that satisfy Equation 3 are

Multiplications 3,5, $d=1$
---
$a$ | $b$ | $c$ | $d$ | $\mathfrak{h}=V_{2a}\oplus V_{2b}\oplus V_{2c}\oplus V_{4d+2}$ | $\operatorname{dim}(\mathfrak{g})$
$2$ | $2$ | $1$ | $1$ | $V_{4}\oplus V_{4}\oplus V_{2}\oplus V_{6}$ | $23$
$2$ | $3$ | $1$ | $1$ | $V_{4}\oplus V_{6}\oplus V_{2}\oplus V_{6}$ | $25$
$3$ | $2$ | $1$ | $1$ | $V_{6}\oplus V_{4}\oplus V_{2}\oplus V_{6}$ | $25$
$2$ | $2$ | $2$ | $1$ | $V_{4}\oplus V_{4}\oplus V_{4}\oplus V_{6}$ | $25$

For Table 2, we previously showed that $a\geq 2$ and either $b\geq 2$ or
$c\geq 2$. In this case, the possible $a,b,c,d$ are

Multiplication 2, $d=1$
---
$a$ | $b$ | $c$ | $d$ | $\mathfrak{h}=V_{2a}\oplus V_{2b}\oplus V_{2c}\oplus V_{4d+2}$ | $\operatorname{dim}(\mathfrak{g})$
$2$ | $2$ | $1$ | $1$ | $V_{4}\oplus V_{4}\oplus V_{2}\oplus V_{6}$ | $23$
$2$ | $3$ | $1$ | $1$ | $V_{4}\oplus V_{6}\oplus V_{2}\oplus V_{6}$ | $25$
$3$ | $2$ | $1$ | $1$ | $V_{6}\oplus V_{4}\oplus V_{2}\oplus V_{6}$ | $25$
$2$ | $2$ | $2$ | $1$ | $V_{4}\oplus V_{4}\oplus V_{4}\oplus V_{6}$ | $25$
$2$ | $1$ | $2$ | $1$ | $V_{4}\oplus V_{2}\oplus V_{4}\oplus V_{6}$ | $23$
$3$ | $1$ | $2$ | $1$ | $V_{6}\oplus V_{2}\oplus V_{4}\oplus V_{6}$ | $25$
$2$ | $1$ | $3$ | $1$ | $V_{4}\oplus V_{2}\oplus V_{6}\oplus V_{6}$ | $25$

∎

######  Remark 3.28.

Notice that any $\mathfrak{sl}_{2}(\mathbb{C})$-module
$\mathfrak{h}=V_{n}\oplus V_{m}\oplus V_{p}\oplus V_{k}$, with any choice of
multiplicative structure in 3.17, is an algebra but not necessarily a Lie
algebra.

For the analysis of the remaining cases, we use 3.29 to show that most of the
cases from 3.27 are not Lie algebras. Indeed, given a fixed value of
$(n,m,p,k)$, we consider the associated vector space $\mathfrak{h}$ and make a
choice of $\mathfrak{g}_{L}$-equivariant morphisms between the simple modules
of $\mathfrak{h}$ which is compatible with the respective multiplication
prescribed in 3.27. Recall from 3.4 that in order to show that $\mathfrak{h}$
with a prescribed multiplicative structure is not a Lie algebra, it is
necessary to prove that after any non-zero rescaling of these morphisms,
$\mathfrak{h}$ does not satisfy the Jacobi identity. In other words, it is
necessary to show that any algebra in the family resulting from the non
trivial rescalings does not satisfy the Jacobi identity. The following Lemma
gives us a criterion to show no element of the family is a Lie algebra.

######  Lemma 3.29.

Let $\mathfrak{g}_{L}$ be a simple Lie algebra and
$\mathfrak{h}=V_{n_{1}}\oplus\cdots\oplus V_{n_{k}}$ be a
$\mathfrak{g}_{L}$-module with $V_{n_{j}}$ simple. For every $1\leq i\leq
j\leq k$, suppose that there is $1\leq p\leq k$ and a
$\mathfrak{g}_{L}$-equivariant morphism

$\Gamma_{i,j}^{p}:V_{n_{i}}\otimes V_{n_{j}}\to V_{n_{p}},$

such that $\Gamma_{i,i}^{p}$ is skew-symmetric. Consider the resulting algebra
$\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$, which product $[,]$ is
constructed from the $\Gamma_{i,j}^{p}$.

Suppose that

  1. (1)

there exist $a,b,c\in\mathfrak{g}$, such that

$[a,[b,c]]=[b,[c,a]]=0\qquad\text{and}\qquad[c,[a,b]]\neq 0,$

  2. (2)

or, there exist $1\leq i,j\leq k$ and $a,b\in V_{n_{i}}$ $c\in V_{n_{j}}$ such
that

$[a,[b,c]]+[b,[c,a]]\neq 0\qquad\text{and}\qquad[c,[a,b]]=0,$

  3. (3)

or there exist $1\leq i\leq k$ and $a,b,c\in V_{n_{i}}$ that do not satisfy
the Jacobi identity.

Then any algebra $\tilde{\mathfrak{g}}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$
obtained from non-zero rescaling $\Gamma_{i,j}^{p}$ does not satisfy the
Jacobi identity.

###### Proof.

The proof of Item 1 is straightforward. Let $i,j$ and $a,b,c$ as in Item 2.
The equation

$0=[c,[a,b]]=\Gamma^{q}_{p,m}(c\otimes\Gamma^{p}_{n,n}(a,b))$

is homogeneous and therefore satisfied by any rescaling of
$\Gamma^{q}_{p,m},\Gamma^{p}_{n,n}$. Similarly, by definition of $[\ ,\ ]$ we
have that

$\displaystyle\hbox{}[a,[b,c]]+[b,[c,a]]$
$\displaystyle=\Gamma^{s}_{n,r}(a\otimes\Gamma^{r}_{n,m}(b,c))-\Gamma^{s}_{n,r}(b\otimes\Gamma^{r}_{n,m}(a,c))$
$\displaystyle=\Gamma^{s}_{n,r}\left(a\otimes\Gamma^{r}_{n,m}(b,c)-b\otimes\Gamma^{r}_{n,m}(a,c)\right),$

is also homogeneous with respect to $\Gamma^{s}_{n,r},\Gamma^{r}_{n,m}$ and
therefore non-zero when rescaling.

The proof of Item 3 is very similar. ∎

Applying the previous lemma to each of the cases listed in 3.27 we obtain the
following.

######  Corollary 3.30.

Suppose $\mathfrak{g}=\mathfrak{sl}_{2}(\mathbb{C})\ltimes\mathfrak{h}$ is an
algebra where $\mathfrak{h}=V_{n}\oplus V_{m}\oplus V_{p}\oplus V_{k}$ is one
of the algebras listed in the conclusion of 3.27. If $\mathfrak{g}$ is a Lie
algebra then one of the following cases must necessarily hold

  1. (1)

the algebra structure of $\mathfrak{h}$ is given by table 5 and
$(n,m,p,k)=(2,6,4,2),$ or,

  2. (2)

the algebra structure of $\mathfrak{h}$ is given by table 6 and
$(n,m,k,p)=(6,4,6,2),(2,2,2,2)$ or $(2,4,2,2).$

######  Proposition 3.31.

Suppose $\mathfrak{g}=\mathfrak{sl}_{2}(\mathbb{C})\ltimes\mathfrak{h}$ is an
algebra where $\mathfrak{h}=V_{n}\oplus V_{m}\oplus V_{p}\oplus V_{k}$ is one
of the algebras listed in the conclusion of 3.27. The following algebras admit
outer derivations.

  1. (1)

the multiplication of $\mathfrak{h}$ is given by table 6 and
$(n,m,p,k)=(2,2,2,2),(2,4,2,2),$

  2. (2)

the multiplication of $\mathfrak{h}$ is given by table 5 and
$(n,m,p,k)=(2,6,4,2)$.

######  Theorem 3.32.

Let $\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$ be a sympathetic, non
semisimple Lie algebra, with simple Levi subalgebra $\mathfrak{g}_{L}$ and
$\mathfrak{h}=V_{n}\oplus V_{m}\oplus V_{p}\oplus V_{k}$, as a
$\mathfrak{g}_{L}$-module. Then $dim(\mathfrak{g})\geq 25$. Moreover,
$\operatorname{dim}(\mathfrak{g})=25$ if and only if $(n,m,p,k)=(6,4,6,2)$ and
$\mathfrak{g}_{L}=\mathfrak{sl}_{2}(\mathbb{C})$.

######  Theorem 3.33.

Let $\mathfrak{g}=\mathfrak{g}_{L}\ltimes\mathfrak{h}$ be a sympathetic, non
semisimple Lie algebra, with simple Levi subalgebra $\mathfrak{g}_{L}$. Then
$\operatorname{dim}(\mathfrak{g})\geq 15$.

###### Proof.

The case $\mathfrak{g}_{L}\neq\mathfrak{sl}_{2}(\mathbb{C})$ is an immediate
consequence of 3.18.

Suppose $\mathfrak{g}_{L}=\mathfrak{sl}_{2}(\mathbb{C})$ and set
$\mathfrak{h}=V_{n_{1}}\oplus\ldots\oplus V_{n_{k}}$. By 3.13,
$s(\mathfrak{g})=k\geq 4$. 3.32 gives us the bound for the case
$s(\mathfrak{g})=4$. Assume that $s(\mathfrak{g})=k\geq 5$. Since
$\mathfrak{g}$ is sympathetic, 3.23 implies that $n_{i}>1$ for at least one
$i=1,\ldots,k$. In fact, there exist at least two $i,j$ such that
$n_{i},n_{j}>1$. Otherwise, using the notation from 3.22, either
$\mathfrak{h}^{0}=Z(\mathfrak{h})$ or $\mathfrak{g}$ admits an outer
derivation similar to those we constructed before. In consequence, we get the
required bound

$\operatorname{dim}(\mathfrak{g})=8+\sum_{i=1}^{k}{n_{i}}\geq 15.$

∎

######  Theorem 3.34.

Let $\mathfrak{g}$ be a perfect Lie algebra with Levi decomposition
$\mathfrak{g}=\mathfrak{sl}_{2}(\mathbb{C})\ltimes\mathfrak{h}$ and
$Z(\mathfrak{g})=0$. Suppose that $\mathfrak{h}=V_{n}\oplus V_{m}\oplus
V_{p}\oplus V_{k}$, $Z(\mathfrak{h})=V_{k}$, as
$\mathfrak{sl}_{2}(\mathbb{C})$-modules, where $n,m,p,k\in\mathbb{N}$. If
$\operatorname{dim}(\mathfrak{g})<25$ then $\mathfrak{g}$ is not a sympathetic
Lie algebra.

##  4\. Appendix: Algorithmic Construction and Verification of Sympathetic Lie
Algebras

In this section we include the algorithms that we use to construct and verify
particular cases of Lie algebras with Levi subalgebra
$\mathfrak{g}_{L}=\mathfrak{sl}_{2}(\mathbb{C})$. Our Python implementation of
these algorithms can be found in
https://github.com/winsy17/Sympathetic_Lie_Algebras.

###  4.1. Algorithm to construct simple
$\mathfrak{sl}_{2}(\mathbb{C})$-submodules of $V_{n}\otimes V_{m}$

This algorithm takes integers $n\geq m\geq 0$ and $k$ as input, and returns a
basis of $V_{k}$ as a simple submodule of $V_{n}\otimes V_{m}$. We require
$k=n+m-2r$ for some $0\leq r\leq m$, in order to guarantee that $V_{k}$ is a
simple submodule of $V_{n}\otimes V_{m}$ (see 3.20).

We use the notation $e_{i}\odot f_{j}$ to indicate an element of the basis of
the respective vector space $V_{n}\otimes V_{m}$, $S^{2}(V_{n})$ or
$\Lambda^{2}(V_{n})$. In the case of $S^{2}(V_{n})$, when $i\neq j$, we use
$e_{i}\odot e_{j}=e_{i}\otimes e_{j}+e_{j}\otimes e_{i}$ and $e_{i}\odot
e_{i}=e_{i}\otimes e_{i}$. For $\Lambda^{2}(V_{n})$ we use the standard basis
$e_{i}\odot e_{j}=e_{i}\otimes e_{j}-e_{j}\otimes e_{i}$, where $1\leq i<j\leq
n$.

Algorithm 1 Finding a basis of $V_{k}$ as a submodule of $V_{n}\otimes V_{m}$,
$S^{2}(V_{n})$ or $\Lambda^{2}(V_{n})$.

1: Input: $n\geq m$, $k$, corresponding to $V_{n},V_{m},V_{k}$;

2: Output: Basis of $V_{k}$ from decompositions 3.20, 3.21

3: $W_{k}:=\\{e_{i}\odot f_{j}:(\rho_{n}\otimes\rho_{m})(H)(e_{i}\odot
f_{j})=k\cdot(e_{i}\odot f_{j})\\}$

4: $W_{k+2}:=\\{e_{i}\odot f_{j}:(\rho_{n}\otimes\rho_{m})(H)(e_{i}\odot
f_{j})=(k+2)\cdot(e_{i}\odot f_{j})\\}$

5: if $W_{k+2}=\emptyset$ then

6: $u_{k}\leftarrow W_{k}$

7: else

8: $u_{k}\leftarrow\operatorname{Ker}(\rho_{n}\otimes\rho_{m})(E)|_{W_{k}}$

9: end if

10: $\mathcal{B}:=\\{(\rho_{n}\otimes\rho_{m})^{j}(F)(u_{k}):j=0,\ldots,k\\}$

11: return $\mathcal{B}$

###  4.2. Algorithm to construct $\mathfrak{sl}_{2}(\mathbb{C})$-equivariant
morphisms

This algorithm receives integers $n\geq m$ and $k$ as input. If $V_{k}$ is a
simple submodule of $V_{n}\otimes V_{m}$, then the algorithm returns a matrix
corresponding to a non trivial equivariant morphism $\Gamma_{n,m}^{k}\colon
V_{n}\otimes V_{m}\to V_{k}$. Otherwise, it returns the zero matrix.

We use $[\mathcal{A}]$ to denote the matrix which rows are the elements of an
ordered basis $\mathcal{A}$; $\mathbb{O}_{r,s}$ the zero matrix of dimension
$r\times s$. In steps 13-21 we construct the matrix $I$ of dimension
$(n+1)(m+1)\times(k+1)$. With the exception of an identity submatrix of
dimension $(k+1)\times(k+1)$, the matrix $I$ only contains zeros. These zeros
correspond to the morphisms

$V_{r}\hookrightarrow V_{n}\otimes V_{m}\xrightarrow{0}V_{k}$

when $r\neq k$ and

$V_{k}\hookrightarrow V_{n}\otimes V_{m}\xrightarrow{\operatorname{Id}}V_{k}.$

Algorithm 2 Constructing an $\mathfrak{sl}_{2}(\mathbb{C})$-equivariant
morphism $\Gamma\colon V_{n}\otimes V_{m}\to V_{k}$.

1: Input: $n\geq m$, $k$, corresponding to $V_{n},V_{m},V_{k}$

2: Output: Matrix associated to $\Gamma$ in terms of the standard basis of
$V_{n}\otimes V_{m}$.

3: if $V_{k}$ is not a simple submodule of $V_{n}\otimes V_{m}$ then

4: return $\mathbb{O}_{(k+1),(n+1)(m+1)}$

5: end if

6: if $n=m$ and $2n-k\equiv 0\operatorname{mod}4$ then

7: $W=S^{2}(V_{n})$

8: else if $n=m$ and $2n+2-k\equiv 0\operatorname{mod}4$ then

9: $W=\Lambda^{2}(V_{n})$

10: else

11: $W=V_{n}\otimes V_{m}$

12: end if

13: $\mathcal{B}=\emptyset$, $\text{I}=[]$

14: for all $V_{r}$ simple submodules of $W$ do

15:
$\mathcal{B}=\mathcal{B}\cup\\{u_{r},\rho(F)(u_{r}),\ldots,\rho^{r}(F)(u_{r})\\}$

16: if $r\neq k$ then

17: $\text{I}=\text{I}+[\mathbb{O}_{r+1,k+1}]$

18: else

19: $\text{I}=\text{I}+[\operatorname{Id}_{k+1,k+1}]$

20: end if

21: end for

22: $\Gamma=[\mathcal{B}]^{-1}\cdot\text{I}$

23: if $W=V_{n}\otimes V_{m}$ then

24: return $\Gamma$

25: else if $W=S^{2}(V_{n})$ then

26: return Extension of $\Gamma$ to $V_{n}\otimes V_{m}$ as a symmetric
function.

27: else

28: return Extension of $\Gamma$ to $V_{n}\otimes V_{m}$ as an antisymmetric
function.

29: end if

To reduce the number of calculations when $n=m$, we use the fact that
$\Gamma\colon V_{n}\otimes V_{m}\to V_{k}$ is either symmetric or skew-
symmetric, depending only on $2n-k\mod{4}$. Therefore we restrict our
construction of $\Gamma$ to the corresponding subspace $S^{2}(V_{n})$ or
$\Lambda^{2}(V_{n})$ and extend to $V_{n}\otimes V_{n}$ by symmetry or
antisymmetry, see Steps 6 -11.

A further optimisation of our algorithm is our implementation of step 22. In
order to calculate $[\mathcal{B}]^{-1}$, we order the basis $\mathcal{B}$ by
decreasing weights to obtain a block matrix and then calculate the inverses of
each block. With this process we invert $n+m+1$ blocks of size at most
$(m+1)\times(m+1)$ instead of inverting the full matrix, which has size
$(n+1)(m+1)\times(n+1)(m+1)$. This gives a significant improvement in
complexity. Indeed, when using the Gauss-Jordan algorithm, which has
complexity $O(N^{3})$ on the size of the matrix, our optimisation leads to a
complexity of $O(m^{3}n)$ in contrast to the direct calculation that has
complexity $O(m^{3}n^{3})$.

###  4.3. Algorithm to calculate $\operatorname{Der}(\mathfrak{g})$

The input for this algorithm is the set of adjoint matrices of the basis of
$\mathfrak{g}$ and its output is a matrix containing a basis of
$\operatorname{Der}(\mathfrak{g})$.

Let $e_{1},\ldots,e_{n}$ be a basis for $\mathfrak{g}$ and
$\operatorname{ad}_{e_{i}}$ the associated adjoint to $e_{i}$. A linear map
$D\colon\mathfrak{g}\to\mathfrak{g}$ is a derivation if and only if
$D=(d_{r,s})_{1\leq r,s\leq n}$ satisfies

$\displaystyle D\circ\operatorname{ad}_{e_{i}}(e_{j})$
$\displaystyle=[D(e_{i}),e_{j}]+\operatorname{ad}_{e_{i}}\circ D(e_{j})$
$\displaystyle=\sum_{k=1}^{n}d_{k,i}[e_{k},e_{j}]+\operatorname{ad}_{e_{i}}\circ
D(e_{j})$
$\displaystyle=\sum_{k=1}^{n}d_{k,i}\operatorname{ad}_{e_{k}}(e_{j})+\operatorname{ad}_{e_{i}}\circ
D(e_{j})$

for every $1\leq i,j\leq n$. This identity is equivalent to equality of linear
operators

$D\circ\operatorname{ad}_{e_{i}}=\sum_{k=1}^{n}d_{k,i}\operatorname{ad}_{e_{k}}+\operatorname{ad}_{e_{i}}\circ
D,$

for all $1\leq i\leq n$. Using the adjoint matrices as an input, the above
identities allow us to define a system of linear equations, where $D$ is the
unknown. We then use the isomorphism
$\mathbb{C}^{n}\times\mathbb{C}^{n}\simeq\mathbb{C}^{n^{2}}$ to obtain a basis
for the space of solutions of $D$. That is, we obtain a set of generators of
the vector space $\operatorname{Der}(\mathfrak{g})$ and as a consequence, its
dimension.

###  4.4. Algorithm to verify if a vector space is a Lie algebra.

Given candidate structure constants as input, this algorithm verifies if these
are the structure constants of a Lie algebra. That is, it checks the Jacobi
identity.

Using candidate structure constants $[e_{i},e_{j}]=\sum_{k}{c^{i}_{jk}}e_{k}$
for $1\leq i<j\leq n$, we first construct associated adjoint matrices
$\operatorname{ad}_{e_{i}}$ using the antisymmetry of a Lie bracket. In our
code we verify if the identity

$\operatorname{ad}_{[e_{i},e_{j}]}=\operatorname{ad}_{e_{i}}\circ\operatorname{ad}_{e_{j}}-\operatorname{ad}_{e_{j}}\circ\operatorname{ad}_{e_{i}}$

holds for every $1\leq i<j\leq n$.

###  4.5. The sympathetic Lie algebra of dimension 25

In this section we present how our algorithms construct, in practice, the
isomorphisms given by the Weyl and Clebsh-Gordan theorems and Schur’s lemma.
We work with a concrete example: Benayadi’s $25$-dimensional sympathetic Lie
algebra.

Consider the vector space $V_{6}$, generated by
$\\{e_{0},e_{1},\ldots,e_{6}\\},$ and its corresponding representation
$\rho_{6}:\mathfrak{sl}_{2}(\mathbb{C})\to\mathfrak{gl}(V_{6})$, defined in
3.19. By 3.21 we have a decomposition

$None$ $\Lambda^{2}(V_{6})\simeq V_{10}\oplus V_{6}\oplus V_{2}.$

In order to define an $\mathfrak{sl}_{2}(\mathbb{C})$-equivariant morphism
between $\Lambda^{2}(V_{6})$ and $V_{6}$, we use that

$\displaystyle\operatorname{Hom}_{\mathfrak{sl}_{2}(\mathbb{C})}(\Lambda^{2}(V_{6}),V_{6})$
$\displaystyle\simeq\operatorname{Hom}_{\mathfrak{sl}_{2}(\mathbb{C})}(V_{10}\oplus
V_{6}\oplus V_{2},V_{6})$
$\displaystyle\simeq\operatorname{Hom}_{\mathfrak{sl}_{2}(\mathbb{C})}(V_{10},V_{6})\oplus\operatorname{Hom}_{\mathfrak{sl}_{2}(\mathbb{C})}(V_{6},V_{6})$
$\displaystyle\qquad\oplus\operatorname{Hom}_{\mathfrak{sl}_{2}(\mathbb{C})}(V_{2},V_{6})$

From Schur’s Lemma, it follows that

$\operatorname{Hom}_{\mathfrak{sl}_{2}(\mathbb{C})}(V_{10},V_{6})\equiv\operatorname{Hom}_{\mathfrak{sl}_{2}(\mathbb{C})}(V_{2},V_{6})\equiv
0,$

and hence

$None$
$\operatorname{Hom}_{\mathfrak{sl}_{2}(\mathbb{C})}(\Lambda^{2}(V_{6}),V_{6})\simeq\operatorname{Hom}_{\mathfrak{sl}_{2}(\mathbb{C})}(V_{6},V_{6})$

In order to construct this morphism, it is necessary to obtain the explicit
decomposition from eq. (1), which can be achieved using standard
representation theory: find a basis of highest weight vectors of
$\operatorname{Ker}(\hat{\rho}_{6}(E))$, where
$\hat{\rho}_{6}=\rho_{6}\otimes\rho_{6}$.

Define $e_{i}\wedge e_{j}:=e_{i}\otimes e_{j}-e_{j}\otimes e_{i}$. A
straightforward calculation shows

######  Lemma 4.1.

$\operatorname{dim}\operatorname{Ker}(\hat{\rho}_{6}(E))=3$, and,
$\operatorname{Ker}(\hat{\rho}_{6}(E))=\langle e_{0}\wedge e_{1},-e_{0}\wedge
e_{3}+2e_{1}\wedge e_{2},3e_{0}\wedge e_{5}-5e_{1}\wedge e_{4}+6e_{2}\wedge
e_{3}\rangle$.

######  Corollary 4.2.

Let $u_{10}:=e_{0}\wedge e_{1}$, $u_{6}:=-e_{0}\wedge e_{3}+2e_{1}\wedge
e_{2}$ y $u_{2}:=3e_{0}\wedge e_{5}-5e_{1}\wedge e_{4}+6e_{2}\wedge e_{3}$.
Then

$V_{i}\simeq\langle
u_{i},\hat{\rho}_{6}(F)(u_{i}),\ldots,\hat{\rho}_{6}(F)^{i}(u_{i})\rangle$

for $i=2,6,10$.

######  Corollary 4.3.

The morphism in Eq. (2) is given by:

  * •

$\hat{\rho}_{6}(F)^{j}(u_{10})\mapsto 0$, for $j=0,\ldots,10$,

  * •

$\hat{\rho}_{6}(F)^{j}(u_{6})\mapsto e_{j}$ for $j=0,\ldots,6$,

  * •

$\hat{\rho}_{6}(F)^{j}(u_{2})\mapsto 0$ for $j=0,1,2$.

4.3 allows us to define a skew-symmetric
$\mathfrak{sl}_{2}(\mathbb{C})$-equivariant morphism
$\Gamma_{6,6}^{6}:V_{6}\times V_{6}\to V_{6}$. To construct
$\Gamma_{6,6}^{6}:V_{6}\times V_{6}\to V_{6}$ we solve the linear system of
equations originating from 3.21. For example, it is easy to verify the
identities

$u_{6}=-e_{0}\wedge e_{3}+2e_{1}\wedge e_{2},\quad\text{ and
}\quad\hat{\rho}_{6}(F)^{2}(u_{10})=e_{0}\wedge e_{3}+e_{1}\wedge e_{2}.$

Thus

$-e_{0}\wedge e_{3}+2e_{1}\wedge e_{2}\mapsto e_{0},\quad e_{0}\wedge
e_{3}+e_{1}\wedge e_{2}\mapsto 0$

In terms of $\Gamma_{6,6}^{6}$ this is equivalent to:

$\displaystyle-\Gamma_{6,6}^{6}(e_{0},e_{3})+2\Gamma_{6,6}^{6}(e_{1},e_{2})$
$\displaystyle=e_{0}$
$\displaystyle\Gamma_{6,6}^{6}(e_{0},e_{3})+\Gamma_{6,6}^{6}(e_{1},e_{2})$
$\displaystyle=0.$

From these equations one can easily obtain $\Gamma_{6,6}^{6}(e_{0},e_{3})$ y
$\Gamma_{6,6}^{6}(e_{1},e_{2})$. Moreover,

######  Lemma 4.4.

Up to scalar multiples, the only skew-symmetric
$\mathfrak{sl}_{2}(\mathbb{C})$-equivariant morphism
$\Gamma_{6,6}^{6}:V_{6}\times V_{6}\to V_{6}$ is given by:

$\begin{matrix}\Gamma_{6,6}^{6}&e_{0}&e_{1}&e_{2}&e_{3}&e_{4}&e_{5}&e_{6}\\\
&&&&&&&\\\ e_{0}&0&0&0&-e_{0}&-2e_{1}&-2e_{2}&-e_{3}\\\ &&&&&&&\\\
e_{1}&&0&e_{0}&e_{1}&0&-e_{3}&-e_{4}\\\ &&&&&&&\\\
e_{2}&&&0&e_{2}&e_{3}&0&-e_{5}\\\ &&&&&&&\\\ e_{3}&&&&0&e_{4}&e_{5}&-e_{6}\\\
&&&&&&&\\\ e_{4}&&&&&0&2e_{6}&0\\\ &&&&&&&\\\ e_{5}&&&&&&0&0\\\ &&&&&&&\\\
e_{6}&&&&&&&0\\\ \end{matrix}$

Analogously, we conclude that
$\operatorname{Hom}_{\mathfrak{sl}_{2}(\mathbb{C})}(\Lambda^{2}V_{6},V_{2})\simeq\operatorname{Hom}_{\mathfrak{sl}_{2}(\mathbb{C})}(V_{2},V_{2})$
and obtain:

######  Lemma 4.5.

Up to scalar multiples, the only bilinear, skew-symmetric
$\mathfrak{sl}_{2}(\mathbb{C})$-equivariant morphism
$\Gamma_{6,6}^{2}:V_{6}\times V_{6}\to V_{2}:=\langle
f_{0},f_{1},f_{2}\rangle$ is given by:

$\begin{matrix}\Gamma_{6,6}^{2}&e_{0}&e_{1}&e_{2}&e_{3}&e_{4}&e_{5}&e_{6}\\\
&&&&&&&\\\ e_{0}&0&0&0&0&0&f_{0}&3f_{1}\\\ &&&&&&&\\\
e_{1}&&0&0&0&-f_{0}&-2f_{1}&3f_{2}\\\ &&&&&&&\\\
e_{2}&&&0&f_{0}&f_{1}&-5f_{2}&0\\\ &&&&&&&\\\ e_{3}&&&&0&6f_{2}&0&0\\\
&&&&&&&\\\ e_{4}&&&&&0&0&0\\\ &&&&&&&\\\ e_{5}&&&&&&0&0\\\ &&&&&&&\\\
e_{6}&&&&&&&0\\\ \end{matrix}$

######  Lemma 4.6.

Up to scalar multiples, there is a unique bilinear
$\mathfrak{sl}_{2}(\mathbb{C})$-equivariant morphism
$\Gamma_{6,4}^{2}:V_{6}\times V_{4}\to V_{2}=\langle g_{0},g_{1},g_{2}\rangle$
and it is given by:

$\begin{matrix}\Gamma_{6,4}^{2}&f_{0}&f_{1}&f_{2}&f_{3}&f_{4}\\\ &&&&&&&\\\
e_{0}&0&0&0&0&g_{0}\\\ &&&&&&&\\\ e_{1}&0&0&0&-g_{0}&g_{1}\\\ &&&&&&&\\\
e_{2}&0&0&g_{0}&-2g_{1}&g_{2}\\\ &&&&&&&\\\ e_{3}&0&-g_{0}&3g_{1}&-3g_{2}&0\\\
&&&&&&&\\\ e_{4}&g_{0}&-4g_{1}&6g_{2}&0&0\\\ &&&&&&&\\\
e_{5}&5g_{1}&-10g_{2}&0&0&0\\\ &&&&&&&\\\ e_{6}&15g_{2}&0&0&0&0\\\
\end{matrix}$

One can then extend $\Gamma_{6,4}^{2}$ to a unique bilinear skew-symmetric and
$\mathfrak{sl}_{2}(\mathbb{C})$-equivariant morphism

$\Gamma_{(6,4)}^{2}:(V_{6}\oplus V_{4})\times(V_{6}\oplus V_{4})\to V_{2},$

by defining

$\Gamma_{(6,4)}^{2}|_{V_{6}\times V_{6}}\equiv
0,\quad\Gamma_{(6,4)}^{2}|_{V_{4}\times V_{4}}\equiv 0,$

and

$\Gamma_{(6,4)}^{2}|_{V_{6}\times
V_{4}}=\Gamma_{6,4}^{2}\quad\Gamma_{(6,4)}^{2}|_{V_{4}\times
V_{6}}=-\Gamma_{6,4}^{2},$

where

$\Gamma_{(6,4)}^{2}(u,w):=-\Gamma_{6,4}^{2}(w,u),$

for $u\in V_{4}$, $w\in V_{6}$.

######  Lemma 4.7.

Up to scalar multiples, the only skew-symmetric
$\mathfrak{sl}_{2}(\mathbb{C})$-equivariant morphism
$\Gamma_{4,4}^{2}:V_{4}\times V_{4}\to V_{2}$ is given by:

$\begin{matrix}\Gamma_{4,4}^{2}&e_{0}&e_{1}&e_{2}&e_{3}&e_{4}\\\ &&&&&\\\
e_{0}&0&0&0&-f_{0}&-2f_{1}\\\ &&&&&\\\ e_{1}&&0&f_{0}&f_{1}&-2f_{2}\\\
&&&&&\\\ e_{2}&&&0&3f_{2}&0\\\ &&&&&\\\ e_{3}&&&&0&0\\\ &&&&&\\\
e_{4}&&&&&0\\\ \end{matrix}$

We modified the notation used in Benayadi’s original construction, [3]. Here
$V_{6}=D(3)$, $V_{4}=D(2)$ and $V_{2}=D(1)$.

As a vector space, $\mathfrak{g}=\mathfrak{sl}_{2}(\mathbb{C})\oplus
V_{6}\oplus V_{4}\oplus V_{6}\oplus V_{2}$ and so
$\operatorname{dim}(\mathfrak{g})=25$. The algebra structure of $\mathfrak{g}$
is then defined as:

$\begin{matrix}[,]_{\mathfrak{g}}&\mathfrak{sl}_{2}(\mathbb{C})&&V_{6_{1}}&&V_{4}&&V_{6_{2}}&V_{2}\\\
&&&&&&&&\\\
\mathfrak{sl}_{2}(\mathbb{C})&[,]_{\mathfrak{sl}_{2}(\mathbb{C})}&&\rho_{6}&&\rho_{4}&&\rho_{6}&\rho_{2}\\\
&&&&&&&&\\\
V_{6_{1}}&&&\Gamma_{6_{1},6_{1}}^{6_{2}}&&\Gamma_{6_{1},4}^{2}&&\Gamma_{6_{1},6_{2}}^{2}&0\\\
&&&&&&&&\\\ V_{4}&&&&&\Gamma_{4,4}^{2}&&0&0\\\ &&&&&&&&\\\
V_{6_{2}}&&&&&&&0&0\\\ &&&&&&&&\\\ V_{2}&&&&&&&&0\end{matrix}$

Here $\Gamma_{a,b}^{c}:V_{a}\times V_{b}\to V_{c}$ denotes the unique, up to
scalar multiples, bilinear $\mathfrak{sl}_{2}(\mathbb{C})$-equivariant
morphism obtained from Schur’s Lemma. For example, $\Gamma_{6,6}^{6}$ is
defined by 4.4 and we are using the notation $V_{6_{2}}$ to indicate that
projection is onto the second copy of $V_{6}$.

Using the algorithm Section 4.1 we are able to explicitly calculate basis for
$\Lambda^{2}V_{6}$ and $\Lambda^{2}V_{4}$ and therefore establish 4.4, 4.6,
4.6.

Given the structure constants that we obtained, the algorithm in Section 4.4
then verifies that $\mathfrak{g}$ is indeed a $25$-dimensional Lie algebra.

Finally, we use the algorithm 4.3 to explicitly calculate
$\operatorname{Der}(\mathfrak{g})$ and obtain its dimension
$\operatorname{dim}_{\mathbb{C}}\operatorname{Der}(\mathfrak{g})=25$. In this
case we know
$0=Z(\mathfrak{g})=\operatorname{Ker}(\operatorname{ad}_{\mathfrak{g}})$ and
so $\operatorname{dim}_{\mathbb{C}}\operatorname{ad}(\mathfrak{g})=25$. Since
$\operatorname{dim}_{\mathbb{C}}\operatorname{Der}(\mathfrak{g})=\operatorname{dim}_{\mathbb{C}}\operatorname{ad}(\mathfrak{g})$,
we can conclude:

######  Proposition 4.8.

Let $\mathfrak{g}$ be as in the previous construction. Then,

  1. (1)

$Z(\mathfrak{g})=\\{0\\}$,

  2. (2)

$\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$,

  3. (3)

$\operatorname{Der}(\mathfrak{g})=\operatorname{ad}(\mathfrak{g})$.

That is, $\mathfrak{g}$ is a sympathetic Lie algebra.

###### Proof.

Notice that by construction (2) $[\mathfrak{sl}_{2}(\mathbb{C}),V_{i}]=V_{i}$
and therefore $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$. ∎

### Acknowledgements

ALGP is a member of the Centre for Topological Data Analysis funded by the
EPSRC grant “New Approaches to Data Science: Application Driven Topological
Data Analysis” number EP/R018472/1. For the purpose of Open Access, the
authors have applied a CC BY public copyright licence to any Author Accepted
Manuscript (AAM) version arising from this submission. GS would like to
acknowledge the partial support received by CONACyT grant A-S1-45886 and by
PRODEP grant UASLP-CA-228.

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