On Graded Lie Algebras of Characteristic Three With Classical Reductive Null Component

TL;DR

This paper investigates finite-dimensional irreducible transitive graded Lie algebras over algebraically closed fields of characteristic three, focusing on the structure of the null component and the restrictions on the representation of the commutator subalgebra.

Contribution

It establishes that for such Lie algebras with depth greater than one, the induced representation of the commutator subalgebra on the minus-one component must be restricted.

Findings

If the depth q > 1, the representation is restricted.

The null component L_0 is classical and reductive.

Structural constraints are derived for graded Lie algebras in characteristic three.

Abstract

We consider finite-dimensional irreducible transitive graded Lie algebras $L = \sum_{i=-q}^rL_i$ over algebraically closed fields of characteristic three. We assume that the null component $L_0$ is classical and reductive. The adjoint representation of $L$ on itself induces a representation of the commutator subalgebra $L_0'$ of the null component on the minus-one component $L_{-1}.$ We show that if the depth $q$ of $L$ is greater than one, then this representation must be restricted.

Equations

$$\frac{R\text{buildrel}def}{=\mathcal{O}(2:(1,1))\oplus H(2:(1,1))={⨁}_{i=-2-2(p-1)}^{2p-5}{R}_i,}$$

$$L=\underset{i=-q}{\overset{1}{⨁}}{L}_i$$

$$L_1=⟨{\partial }_j,j=1,\dots n⟩$$

$$L_0=S+T$$

$$L_{-i}=S\otimes ⟨x_1^{k_1}\dots x_n^{k_n},k_1+\dots +k_n=i⟩,$$

$$\mathfrak{g}={\mathfrak{g}}_1+\cdots +{\mathfrak{g}}_k+\mathfrak{z}(\mathfrak{g})$$

$$\frac{L_{<0}\text{buildrel}def}{={⨁}_{i=-q}^{-1}{L}_i,}$$

$$\frac{L_{>0}\text{buildrel}def}{={⨁}_{i=1}^r{L}_i.}$$

$$L_i=L_1^i=S_1^i\subseteq S_i\subseteq L_i,$$

$$[L_{-1},L_1]=[[L_{-1},A_0],L_1]=[[L_{-1},L_1],A_0]\subset A_0,$$

$$0\ne [S_{-1},L_1]\subseteq A_0\cap S_0\subseteq {\mathrm{Ann}}_{S_0}{S}_1.$$

$$[L_{-q},L_1]=[L_{-q},{\mathrm{Ann}}_{L_1}{L}_{-q}]=0,$$

$$0\ne (\mathrm{ad}{L}_{-1}{)}^{i-1}{Q}_i\subset {\mathrm{Ann}}_{L_1}{L}_{-q},$$

$$[(\mathrm{ad}{L}_1{)}^i{L}_{-q},(\mathrm{ad}{L}_1{)}^{q-i}{L}_{-q}]$$

$$L_{-q}$$

$$L_{-q}=[(\mathrm{ad}{L}_1{)}^i{L}_{-q},(\mathrm{ad}{L}_1{)}^{q-i}{L}_{-q}]\subseteq [L_{-q+i},L_{-i}]\subseteq L_{-q},1\leqq i\leqq q.$$

$$[V,L_{q-i}]$$

$$[V,[V,L_{q-i}]]$$

$$L_{-q}=[L_{-q+i},L_{-i}]=[[L_{-q},L_i],L_{-i}]=[L_{-q},[L_{-i},L_i]],$$

$$[L_{-q+i},L_{q-i}]$$

$$[L_{-q+i},[L_{-q+i},L_{q-i}]]$$

$$2(\mathrm{ad}[v,u]{)}^2{L}_{-q+i+1}$$

$$0=[[L_{-q+i},L_{q-i}],L_{-q+i+1}]=[L_{-q+i},[L_{q-i},L_{-q+i+1}]].$$

$$0=[[L_{-q+i},L_1],L_{q-i}]=[[L_{-q+i},L_{q-i}],L_1],$$

$$0\ne [V_{-j},L_{-1}]=[V_{-j},[L_{-q+1},L_{q-2}]]=[L_{-q+1},[V_{-j},L_{q-2}]].$$

$$0\ne [L_{-q},[V_{-j},L_{q-2}]]$$

$$[L_{-q+1},[V_{-2}^j,[V_{-2},L_q]]]=[V_{-2}^j,[V_{-2},[L_{-q+1},L_q]]=[V_{-2}^j,L_{-1}]\ne 0,$$

$$0$$

$$L_{-2}$$

$$L_{-q+1}$$

Full text

Keywords Lie algebras, graded Lie algebras, prime-characteristic Lie algebras

Subject Classification 180: Lie algebras and Lie superalgebras

0 Introduction

Over algebraically closed fields ${\mathbf{F}}$ of characteristic $p>0,$ the classification of the finite-dimensional simple Lie algebras relies on the classification of the finite-dimensional irreducible transitive graded Lie algebras $L=\bigoplus_{i=-q}^{r}L_{i}$ of depth $q\geqq 1$ with classical reductive null component $L_{0}.$ We recall some of the progress that has been made in the classification of such Lie algebras $L.$ In the case in which $L_{-1}$ is not only irreducible but also restricted as an $L_{0}$-module, such Lie algebras are described by the Recognition Theorem of Kac [10] for $p>5.$ (See also [3].) In [1] it is shown that for $p>5,$ $L_{-1}$ is necessarily a restricted $L_{0}^{\prime}$-module. (The assertion is also true for $p=5.$ [3]) When $p=3,$ the situation is more complicated. In characteristic three, there are series of simple graded Lie algebras which satisfy the conditions of Kac’s Recognition Theorem, but which are neither classical Lie algebras nor Lie algebras of Cartan type. (See [6], [12], [14].) Moreover, for $q=1$, examples exist in which $L_{-1}$ is not a restricted $L_{0}^{\prime}-$module. All simple depth-one graded Lie algebras over algebraically closed fields of characteristic three with non-restricted $L_{0}^{\prime}$-module $L_{-1}$ were determined in [4]. (In [11], the authors classified all simple depth-one graded Lie algebras over algebraically closed fields of characteristic three in which $L_{-1}$ is a restricted $L_{0}^{\prime}-$module.) In [2], two-graded (i.e., depth-two, graded) Lie algebras were examined, and it was proved that when $p=3$ and $q=2,$ the $L_{0}^{\prime}-$module $L_{-1}$ must be restricted. For $q=3,$ the corresponding statement was proved in [7]. It was conjectured in [2] that a non-restricted $L_{0}^{\prime}-$module $L_{-1}$ can exist only in Lie algebras of depth one. The present paper completes the proof of that conjecture for $r>1.$ We require that $r$ be greater than one in order to exclude, for example, $H(2:\mathbf{n},\omega)$ with the reverse gradation; see also Example 0.3 below. We note, as in [2], that because there are only finitely many irreducible restricted modules for the derived algebra of a classical reductive Lie algebra, what needs to be considered in classifying graded Lie algebras over algebraically closed fields of characteristic three is reduced. In this paper, we prove the following theorem, which we will henceforth refer to as the “Main Theorem.”

Theorem 0.1

(Main Theorem)* Let $L=L_{-q}\oplus L_{-q+1}\oplus\cdots\oplus L_{-1}\oplus L_{0}\oplus L_{1}\oplus\cdots\oplus L_{r},\,q>1,\,r>1,$ be a finite-dimensional graded Lie algebra over an algebraically closed field ${\mathbf{F}}$ of characteristic $p=3$ such that*

1. (A)

$L_{0}$* is classical reductive;* 2. (B)

$L_{-1}$* is an irreducible $L_{0}$-module (i.e., $L$ is irreducible);* 3. (C)

for all $j\geqq 0$, if $x\in L_{j}$ and $[x,L_{-1}]=(0)$, then $x=0$ (i.e., $L$ is transitive); 4. (D)

$L_{-i}=[L_{-i+1},L_{-1}]$* for all $i>1;$ and* 5. (E)

$L_{-2}\nsubseteq M(L),$* where $M(L)$ is the largest ideal of $L$ contained in the sum of the negative gradation spaces. (See Theorem 1.3 below.)*

Then $L_{-1}$ is a restricted module for $L_{0}^{\prime}$ under the adjoint action of $L$ on itself.

To help to motivate hypothesis (E) above, we offer the following

Example 0.2

For characteristic $p\geqq 3,$ consider the irreducible transitive graded Lie algebra

[TABLE]

where $R_{i}=H(2:(1,\,1))_{i}$ for $i\geqq-1,$ and $R_{i}={\mathcal{O}}(2:(1,\,1))_{i+2p}$ for $-2p=-2-2(p-1)\leqq i\leqq-2.$ Here, the divided-power algebra ${\mathcal{O}}(2:(1,\,1))$ is an abelian ideal of $R,$ and $H(2:(1,\,1))$ has its usual Lie algebra bracket operation and its usual action on ${\mathcal{O}}(2:(1,\,1)),$ except that $[D_{x_{1}},\,D_{x_{2}}]=x_{1}^{p-1}x_{2}^{p-1}\in R_{-2}.$ (See Theorem 1.3 below.) Then $R/M(R)\cong R/{\mathcal{O}}(2:(1,\,1))$ has depth one. In general, if we consider the free Lie algebra generated by the local part of any depth-one graded Lie algebra $L,$ and take a co-finite-dimensional subideal $C$ of the maximal ideal $D$ in the negative part (See [5].), then $M(L\oplus D/C)=D/C,$ and $(L\oplus D/C)/M(L\oplus D/C)\cong L$ has depth one.

To further illustrate the necessity of the requirement in the Main Theorem that $r$ be greater than one, we offer the following

Example 0.3

Consider a graded Lie algebra

[TABLE]

where

[TABLE]

[TABLE]

where $S$ and $T$ are classical, and

[TABLE]

and where $T$ is a subalgebra of $\mathfrak{gl}(n)=\langle x_{i}\partial_{j},\,i,j=1,\dots,n\rangle=W(n:1)_{0}$ with a non-restricted action on $\langle x_{1},\dots,x_{n}\rangle;$ i.e., the representation of $T$ in $\langle x_{1},\dots,x_{n}\rangle$ is a non-restricted representation of $T.$ (See (ii) of Theorem 1.3.) More succinctly, $L=S\otimes A(n:1)+1\otimes T+1\otimes\langle\partial_{i},\,i=1,\dots,n\rangle.$ The minimal example corresponds to $T=\mathfrak{sl}(2)$ and $n=3,$ so that, therefore, $q=6.$

We have noted that the Main Theorem has been proved for $q=2$ in [2] and for $q=3$ in [7]. When we refer to the Main Theorem to substantiate certain claims below, it will be for the cases already proved.

We conclude this section with a sketch of the plan of the rest of the work. In Section 1, we establish terminology and notation, and gather some previously known results. We continue to gather previous results at the beginning of Section 2; here, however, the results are of a more technical nature, and we immediately apply them to showing that under quite natural assumptions, the $L_{0}^{\prime}-$module $L_{-2}$ is irreducible. We conclude Section 2 by establishing other technical lemmas that we’ll use later on. In Sections 3, 4, and 5, we prove the Main Theorem under the “natural” assumptions just referred to, for the cases $q\geqq 6,$ $q=4,$ and $q=5,$ respectively. In Section 6, we complete the proof of the Main Theorem in its full generality.

1 Preliminaries

In this section, we recall definitions and introduce notation, after which we gather results from the literature that we will use later in the work.

Recall that if one takes a $\mathbb{Z}$-form (Chevalley basis) of a complex simple Lie algebra and reduces the scalars modulo $p,$ one obtains a Lie algebra over $I/(p).$ If ${\mathbf{F}}$ is any field of characteristic $p$ , then, by tensoring the Lie algebra we obtained over $I/(p)$ by ${\mathbf{F}},$ we obtain a Lie algebra over ${\mathbf{F}}.$ In characteristic $p,$ any such Lie algebra so obtained is referred to as classical, even those with root systems E6, E7, E8, F4, and G${}_{2}.$ This process may result in a Lie algebra with a non-zero center; such a Lie algebra is still referred to as classical, as is the quotient of such a Lie algebra by its center. For example, the Lie algebras $\mathfrak{gl}(pk)$ and $\mathfrak{pgl}(pk)$ are both considered to be classical Lie algebras. Thus, a classical Lie algebra ${\mathfrak{g}}$ may have a nontrivial center $\mathfrak{z}(\mathfrak{g}),$ as do the Lie algebras $\mathfrak{gl}(pk),\mathfrak{sl}(pk),$ and, if $p=3,$ E${}_{6}.$ It could also happen that a classical Lie algebra has a noncentral ideal, as do the Lie algebras $\mathfrak{gl}(pk),$ $\mathfrak{pgl}(pk),$ and, if $p=3,$ G${}_{2}.$ In characteristic three, G2 contains an ideal $I$ isomorphic to $\mathfrak{psl}(3),$ and $G_{2}/I\cong\mathfrak{psl}(3),$ as well. A classical reductive Lie algebra $\mathfrak{g}$ is the sum of commuting ideals $\mathfrak{g}_{j}$ which are classical Lie algebras, and an at-most-one-dimensional center $\mathfrak{z}(\mathfrak{g})$:

[TABLE]

For any classical Lie algebra $\mathfrak{g}_{j},$ the derived algebra $\mathfrak{g}_{j}^{\prime}$ is a restricted Lie algebra with a natural $p$-structure such that $e_{\alpha}^{[p]}$ $=0,$ and $h_{i}^{[p]}$ $=h_{i}$ for any Chevalley basis $\{e_{\alpha},\;h_{i}\;|\;\alpha$ $\in R,\;i=1,\dots,\text{rank}(\mathfrak{g}_{j}^{\prime})\}$ of $\mathfrak{g}_{j}^{\prime}$, where $R$ is the root system of the corresponding complex simple Lie algebra. For a classical reductive Lie algebra ${\mathfrak{g}},$ we will consider only that $p-$structure on ${\mathfrak{g}}^{\prime}$ $=$ ${\mathfrak{g}}_{1}^{\prime}$ $+$ $\cdots$ $+$ ${\mathfrak{g}}_{k}^{\prime}$ whose restriction to each classical summand is the natural $p$-structure on that summand.

Let $\pi:L\longrightarrow{\mathfrak{g}l}(n)$ be a finite-dimensional irreducible representation of a restricted Lie algebra $L$. The character of $\pi$ is the linear functional $\chi$ on $L$ such that $\chi(y)^{p}I=\pi(y)^{p}-\pi(y^{[p]})$ for all $y\in L$. The representation $\pi$ is restricted when the character $\chi$ equals [math].

Lemma 1.2

(See [1, Lemma 1].) Assume that $L$ is a graded Lie algebra satisfying conditions (A)-(D) of the Main Theorem. If $\chi$ is the character of $L_{0}^{\prime}$ on $L_{-1}$, then $L_{0}^{\prime}$ has character $-j\chi$ on $L_{j}$ for all $j$.

The following theorem of Weisfeiler [15] plays a fundamental rôle in the study of graded Lie algebras. In what follows, we will sometimes refer to Theorem 1.3 as “Weisfeiler’s Theorem.”

Theorem 1.3

(Weisfeiler’s Theorem)* Let $L=L_{-q}\oplus\cdots\oplus L_{-1}\oplus L_{0}\oplus L_{1}\oplus\cdots\oplus L_{r}$ be a graded Lie algebra such that conditions (B)-(D) of the Main Theorem hold. Let $M(L)$ denote the largest ideal of $L$ contained in $L_{-q}\oplus\cdots\oplus L_{-1}.$ Then*

1. (i)

$L/M(L)$* is semisimple and contains a unique minimal ideal $I=S\otimes{\mathcal{O}}(n:{\mathbf{1}})$, where $S$ is a simple Lie algebra, $n$ is a non-negative integer, and ${\mathcal{O}}(n:{\mathbf{1}})={\mathbf{F}}[x_{1},\ldots,x_{n}]/(x_{1}^{p},\ldots,x_{n}^{p}).$ The ideal $I$ is graded and $I_{i}=(L/M(L))_{i}$ for all $i<0.$* 2. (ii)

Degenerate Case* If $I_{1}=(0)$, then for some $\mathfrak{k}$, $1\leqq\mathfrak{k}\leqq n$, the algebra ${\mathcal{O}}(n:{\mathbf{1}})$ is graded by setting $\deg(x_{i})=-1$ for $1\leqq i\leqq\mathfrak{k}$ and $\deg(x_{i})=0$ for $\mathfrak{k}<i\leqq n$. Then $I_{i}=S\otimes{\mathcal{O}}(n:{\mathbf{1}})_{i}$ for all $i$,   $L_{2}=(0),$   $I_{0}=[L_{-1},L_{1}],$ and $L_{1}\subseteq\{D\in 1\otimes\hbox{\rm Der}{\mathcal{O}}(n:{\mathbf{1}})|\deg(D)=1\}.$* 3. (iii)

Non-Degenerate Case* If $I_{1}\neq(0)$, then $S$ is graded and $I_{i}=S_{i}\otimes{\mathcal{O}}(n:{\mathbf{1}})$ for all $i$. Moreover, $(0)\neq[L_{-1},L_{1}]\subseteq I_{0}.$*

Set

[TABLE]

and

[TABLE]

Set $L_{\leqq}\buildrel{\hbox{\small def}}\over{=}L_{<}+L_{0},$ and $L_{\geqq}\buildrel{\hbox{\small def}}\over{=}L_{0}+L_{>}.$

Theorem 1.4

(See Theorem 0.1 of [8].) Let $L=\oplus_{i\in{\mathbb{Z}}}L_{i}$ be a non-degenerate graded Lie algebra over an algebraically closed field of characteristic $p>2$ satisfying conditions (A)-(D) of the Main Theorem. If $[[L_{-1},V],V]=0$ for some proper $L_{0}$-submodule $V\subset L_{1}$, $\dim V>1,$ then $\dim L=\infty$.

By Lemma 1.2, the representation of $L_{0}^{\prime}$ on $L_{1}$ is restricted when and only when the representation of $L_{0}^{\prime}$ on $L_{-1}$ is restricted. Since no non-restricted representation of $L_{0}^{\prime}$ can have dimension one, we have the following corollary.

Corollary 1.5

Let $L$ be as in the above theorem, and suppose that the representation of $L_{0}^{\prime}$ on $L_{-1}$ is not restricted. If $[[L_{-1},V],V]=0$ for some proper $L_{0}$-submodule $V\subset L_{1}$, then $L$ is an infinite-dimensional Lie algebra.

We will make use of the following results from [2]. For definitions of the Lie algebras $L(\epsilon),$ $M,$ $H(2:{\mathbf{n}},\omega),$ and $CH(2:{\mathbf{n}},\omega)$ mentioned in the conclusion of Proposition 1.6 below, see, for example, Section 2 of [2]. When we make use of certain properties of these Lie algebras in later sections, we will explicitly state the properties we need.

Proposition 1.6

(See Lemma 2.12 of [2].) Let $L=L_{-1}\oplus L_{0}\oplus L_{1}\oplus\cdots\oplus L_{r}$ be a graded Lie algebra satisfying conditions (A), (B), and (C) of the Main Theorem, and suppose that $L_{1}\neq 0.$ If $L_{-1}$ is a nonrestricted $L_{0}^{\prime}-$module, then either $L$ is isomorphic to one of the Lie algebras L$(\epsilon)$ or $M$, or $L$ is a Hamiltonian Lie algebra such that $H(2:{\mathbf{n}},\omega)\subseteq L\subseteq CH(2:{\mathbf{n}},\omega),$ where ${\mathbf{n}}=(1,n_{2})$, $\omega=(\exp x^{(3)})dx\wedge dy,$ and the grading is of type (0,1).

Corollary 1.7

(See Corollary 2.13 of [2].) Under the assumptions of Proposition 1.6, $L_{0}^{\prime}$ $\cong\mathfrak{sl}(2),$ $L_{1}$ is an irreducible three-dimensional $L_{0}^{\prime}-$module, and $[L_{1},$ $L_{1}]$ $=0.$ In addition, $[L_{-1},$ $L_{1}]\cong\mathfrak{sl}(2)$ if and only if $L$ is a Hamiltonian Lie algebra; otherwise, $[L_{-1},\,L_{1}]\cong{\mathfrak{gl}}(2).$

Lemma 1.8

(See Lemma 2.14 of [2].) Let $L=L_{-1}\oplus L_{0}\oplus L_{1}\oplus\cdots\oplus L_{q}$ be one of the Lie algebras $L(\epsilon)$, $M,$ or $H(2:{\mathbf{n}},\omega)$ with ${\mathbf{n}}=(1,n_{2})$, let $\chi$ be the nonzero character of the $L_{0}$-module $L_{-1}$, and let $V$ be an $L$-module such that $l^{3}\cdot V=(0)$ for any $l\in L_{-1}\cup L_{1}.$ Suppose that $W$ is an irreducible $L_{0}$-submodule of $V$ with character $\chi_{W}=\zeta\chi,\,\zeta\in{\mathbf{F}}^{\times}$, and suppose that $L_{1}\cdot W=(0).$ Then $L_{-1}^{2}\cdot W\neq(0).$ Similarly, if $L_{-1}\cdot W=(0),$ then $L_{1}^{2}\cdot W\neq(0).$

In what follows, all Lie algebras will be finite-dimensional over an algebraically closed field ${\mathbf{F}}$ of characteristic $p=3$. The commutator ideal $[L,L]$ of a Lie algebra $L$ will be denoted by $L^{\prime},$ and the $i^{th}$ commutator $(\hbox{\rm ad\,}X)^{i-1}X$ of any set $X$ will be written as $X^{i}.$ The annihilator of an $L_{0}-$module $M\subseteq L$ in an $L_{0}-$module $N\subseteq L$ will be denoted by $\hbox{\rm Ann\,}_{N}M.$

2 Properties of irreducible transitive graded Lie algebras

This section contains technical lemmas and a proof that under hypotheses which we list here, the representation of the null component on the minus-two component must be irreducible.

We begin this section by recalling a few results from [1]. Let $L,$ $M(L),$ $I,$ and $S=\sum_{i=-q}^{s}S_{i}$ be as in Weisfeiler”s Theorem (Theorem 1.3). Throughout this section, we make the following two blanket assumptions:

1. (i)

$M(L)=0$ 2. (ii)

$I=S.$

In this regard, please see [13, (2.4.6)], and note that the Lie algebra $S+L_{0}$ satisfies hypothesis (B) of the Main Theorem.

Lemma 2.1

(See [1, Lemma 6].) For any $x$ in $L\backslash L_{-q},\,[L_{-1},\,x]\neq 0.$

Lemma 2.2

(See [1, Lemma 7].) $S_{j}=(\hbox{\rm ad\,}S_{-1})^{s-j}S_{s}$ for all $j,\,-q\leqq j\leqq s.$ If $q(t-1)\leqq s,$ then $(\hbox{\rm ad\,}S_{-q})^{t}S=0$ if and only if $(\hbox{\rm ad\,}S_{-q})^{t}S_{i}=0$ for some $i,\,q(t-1)\leqq i\leqq s.$

Lemma 2.3

(See [1, Lemma 8].) $[L_{-q},\,L_{i}]\neq 0$ for all $i=0,\dots r.$ In addition, $S_{j}=(\hbox{\rm ad\,}L_{-1})^{r-j}L_{r}$ for all $j,\,-q\leqq j\leqq r-1,$ so that $s=r-1$ or $r.$ $S_{s}$ is an irreducible $S_{0}$-module.

Lemma 2.4

(See [1, Lemma 9].) $S_{-q}$ is an irreducible $S_{0}$-module. In particular, $L_{-q}$ is an irreducible $L_{0}$-module.

Lemma 2.5

(See [1, Lemma 10].) $\hbox{\rm Ann\,}_{L_{0}}L_{i}\cap\hbox{\rm Ann\,}_{L_{0}}V_{i+1}=0$ for all $i=-q,\dots,r-1,$ where $V_{i+1}$ is any non-zero $L_{0}-$submodule of $L_{i+1}.$

Lemma 2.6

(See [1, Lemma 11].) $\hbox{\rm Ann\,}_{L_{i}}L_{-q}\cap\hbox{\rm Ann\,}_{L_{i}}V_{-q+1}=0$ for all $i=0,\dots,r,$ where $V_{-q+1}$ is any non-zero $L_{0}-$submodule of $L_{-q+1}.$

Lemma 2.7

(See [1, Lemma 12].) $\hbox{\rm Ann\,}_{L_{q-1}}L_{-q+1}=0.$

Lemma 2.8

(See [1, Lemma 13].) If $r\geqq q,$ then $L_{-q+i}=[L_{-q},\,L_{i}]$ for $i=0,1,\dots q-1.$

Lemma 2.9

(See [1, Lemma 14].) Let $U$ and $V$ be $L_{0}-$submodules of $L$ such that $[U,\,V]\subseteq L_{0}$ and $[U,\,[U,\,V]]=0.$ Then $\{\hbox{\rm ad\,}[u,\,v]|u\in U,\,v\in V\}$ is weakly closed (in the sense of [9, p.31]); consequently, if $(\hbox{\rm ad\,}[u,\,v])^{i}M=0,\,u\in U,\,v\in V,\,$ for some $i>1,$ and $L_{0}-$module $M,$ then $\hbox{\rm ad\,}_{M}[U,\,V]$ is “associative nilpotent.” (See Theorem II.2.1 of [9].)

Lemma 2.10

(See [1, Lemma 15].) If $r\geqq q,$ then $(\hbox{\rm ad\,}L_{-q})^{2}L\neq(0).$

Corollary 2.11

If $s\geqq q,$ then $[S_{-q},\,[S_{-q},\,S_{q}]]\neq(0).$ In particular, if $s\geqq q,$ then $[L_{-q},\,[L_{-q},\,L_{q}]]\neq(0).$

Proof By Lemma 2.10 applied to the Lie algebra $S+L_{0},$ $(\hbox{\rm ad\,}S_{-q})^{2}S\neq 0.$ Thus, in view of Lemma 2.2, it cannot be that $(\hbox{\rm ad\,}S_{-q})^{2}S_{q}=0.$ $~{}~{}\square$

Lemma 2.12

(See [1, Lemma 16].) Let $V$ be an $L_{0}-$submodule of $L_{-q+i}$ for some $i,$ where [math] $<$ $i$ $\leqq$ $\frac{q}{2},$ and suppose $[V,\,L_{q-i-1}]=0=[V,\,[V,L_{q-i}]].$ Suppose further that $L_{-q+i-1}$ is an irreducible $L_{0}-$submodule of $L,$ and that $[L_{-q+i-1},\,L_{q-i}]$ $\neq$ [math] (so that it equals $L_{-1}).$ Then $V$ $=$ $0.$

Lemma 2.13

(See [1, Lemma 17].) Suppose that $V$ is an irreducible $L_{0}-$submodule of $L_{-q+i}$ for some $i,$ where [math] $<i$ $<\frac{q-1}{2},$ such that $[V,$ $L_{q-i-1}]$ $\neq 0$ (so that it equals $L_{-1}$). Then $L_{-q+i}$ is an irreducible $L_{0}-$module; i.e., it equals $V.$

From our observations at the beginning of this section, we have that $S\subseteq L\subseteq\hbox{\rm Der}S$ where $S=S_{-q}\oplus S_{-q+1}\oplus\cdots\oplus S_{-1}\oplus S_{0}\oplus S_{1}\oplus\cdots\oplus S_{s}$ is a simple Lie algebra with $S_{i}=L_{i}$ for $i<0.$ Since $S_{1}$ is an $L_{0}$-submodule of $L_{1},$ it follows that if $L_{1}$ is an irreducible $L_{0}-$module, then $S_{1}=L_{1}.$ If, in addition, $L$ is generated by its local part $L_{-1}\oplus L_{0}\oplus L_{1},$ then for $i\geqq 1,$ we have

[TABLE]

so that $S_{i}=L_{i}$ for $i>0,$ and $L$ could differ from $S$ only in the null component. In particular, $s$ would equal $r.$ (See Lemma 2.3.)

In the lemmas that follow, we will consider graded Lie algebras $L=L_{-q}\oplus L_{-q+1}\oplus\cdots\oplus L_{-1}\oplus L_{0}\oplus L_{1}\oplus\cdots\oplus L_{r}$ satisfying assumptions (i) and (ii) below. Other assumptions will be noted in the statements of the results for which we use them. Note, for example, that, as noted in the paragraph above, assumption (iii) follows from assumptions (iv) and (v) (and assumptions (i) and (ii), of course). Also, assumption (viii) can be assumed whenever the previous assumptions are true, since if they hold, we can reverse the gradation, and have that all of the hypotheses of the Main Theorem continue to be true for the reversed gradation. Indeed, (iv) is the “reverse” of hypothesis (B) of the Main Theorem, (vi) is the “reverse” of (C), and (v) is the “reverse” of (D). Of course, by the transitivity (C) of $L$, there can be no ideals of $L$ in the positive part of $L$, so (E) holds in the “reverse” direction, also In addition, by Lemma 1.2, the representation of $L_{0}^{\prime}$ on L*-1* is restricted if and only if the representation of $L_{0}^{\prime}$ on $L_{1}$ is restricted, so (vii) holds in the “reverse” direction, as well.

1. (i)

$L$ satisfies conditions (A) - (E) of the Main Theorem. 2. (ii)

$L\subseteq\hbox{\rm Der}S$ where $S=S_{-q}\oplus S_{-q+1}\oplus S_{-1}\oplus S_{0}\oplus S_{1}\oplus\cdots\oplus S_{s}$ is a simple graded Lie algebra. 3. (iii)

$L_{i}=S_{i},\ i\neq 0.$ 4. (iv)

$L_{1}=S_{1}$ is an irreducible $L_{0}$-module. (See the discussion before (6.14).) 5. (v)

$L_{i+1}=[L_{i},L_{1}]$ for $i>0$. 6. (vi)

If $x$ is a non-zero element in $L_{-i}$ for some $i\geqq 0$, then $[L_{1},x]\neq(0).$ 7. (vii)

The character $\chi$ of $L_{0}^{\prime}$ on $L_{-1}$ is non-zero. 8. (viii)

$r\geqq q.$

Lemma 2.14

If assumptions (iv) and (v) hold, and $S_{1}$ $\neq 0,$ then $\hbox{\rm Ann\,}_{L_{0}}L_{1}$ $=0.$

Proof Suppose, on the contrary, that $A_{0}\buildrel{\hbox{\small def}}\over{=}\hbox{\rm Ann\,}_{L_{0}}L_{1}\neq 0.$ Then (as in [1, Lemma 18]) we have by transitivity (C) and irreducibility (B) that

[TABLE]

so that by transitivity (C),

[TABLE]

Since for $i>0,$ we have by assumptions (iv) and (v) that $S_{i}=S_{1}^{i},$ and since $J\buildrel{\hbox{\small def}}\over{=}S_{-q}\oplus S_{-q+1}\oplus\cdots\oplus S_{-1}+A_{0}\cap S_{0}$ is invariant under $\hbox{\rm ad\,}{S_{i}},\,-q\leqq i\leqq s,$ it follows, from our assumption that $S_{1}\neq 0,$ that $J$ is a proper ideal of the simple Lie algebra $S;$ i.e., we have obtained a contradiction. Thus, we must conclude that $\hbox{\rm Ann\,}_{L_{0}}L_{1}=0.$ $~{}~{}\square$

Lemma 2.15

If assumption (vi) holds, then $[V_{-2},$ $L_{1}]$ $=L_{-1}$ for any non-zero $L_{0}-$submodule $V_{-2}$ of $L_{-2}.$

Proof This lemma follows from assumptions (vi) and (B). $~{}~{}\square$

Lemma 2.16

If assumption (iv) holds, then $\hbox{\rm Ann\,}_{L_{i}}L_{-q}=0$ for all $i>0.$

Proof Consider first the case in which $i=1.$ If $\hbox{\rm Ann\,}_{L_{1}}L_{-q}\neq 0,$ then, since we are assuming (iv) that $L_{1}$ is an irreducible $L_{0}$-module, we would have $\hbox{\rm Ann\,}_{L_{1}}L_{-q}=L_{1}.$ But then

[TABLE]

to contradict Lemma 2.3. Consequently, $\hbox{\rm Ann\,}_{L_{1}}L_{-q}$ $=0.$ Now, if $Q_{i}$ $\buildrel{\hbox{\small def}}\over{=}\hbox{\rm Ann\,}_{L_{i}}L_{-q}$ $\neq 0$ for some $i$ $>1,$ then by transitivity (C), we would have

[TABLE]

to contradict what we just showed. Thus, $\hbox{\rm Ann\,}_{L_{i}}L_{-q}=0$ for all $i>0,$ which is what we wanted to show. $~{}~{}\square$

Lemma 2.17

If assumptions (vi) and (viii) hold, then $L_{-q}$ $=[L_{-q+i},$ $L_{-i}],$ [math] $\leqq i$ $\leqq q.$

Proof By Lemmas 2.3 and 2.4, $[L_{0},\,L_{-q}]=L_{-q},$ so the lemma is true for $i=0$ and $i=q.$ For $i=1,$ we use Lemmas 2.1 and 2.4. Now note that for $1\leqq i\leqq q-1,$ we have

[TABLE]

so that in view of (vi), (B), and Lemmas 2.1 and 2.4,

[TABLE]

Then

[TABLE]

$\square$

Lemma 2.18

Let $V$ be any (non-zero) irreducible $L_{0}-$submodule of $L_{-q+i},$ where [math] $\leqq i$ $<\frac{q-1}{2}.$ If assumptions (v) and (viii) hold, then $[V,$ $L_{q-(i+1)}]$ $\neq 0,$ so that $[V,$ $L_{q-(i+1)}]$ $=L_{-1}.$ Moreover, $L_{-q+i}$ is an irreducible $L_{0}-$module.

Proof If $[V,\,L_{q-(i+1)}]=0,$ then, since the positive gradation spaces are assumed (v) to be generated by $L_{1},$

[TABLE]

Consequently, since $i$ is assumed to be less than $\frac{q-1}{2},$ so that $2i+1<q,$

[TABLE]

Now, $L_{-q}=S_{-q}$ is an irreducible $L_{0}-$module by Lemma 2.4, so we can assume by induction on $i$ that $L_{-q+i-1}$ is an irreducible $L_{0}-$module. Also, since $[L_{-q},\,L_{q-1}]=L_{-1}$ by Lemma 2.3 and (B), we can assume by induction that $[L_{-q+i-1},\,L_{q-i}]$ $=$ $L_{-1}.$ But then Lemma 2.12 would imply that $V$ $=0,$ contrary to assumption. Thus, $[V,$ $L_{q-(i+1)}]$ is a non-zero $L_{0}-$submodule of $L_{-1},$ so that by irreducibility (B), $[V,$ $L_{q-(i+1)}]$ $=L_{-1}.$ The last assertion follows from Lemma 2.13. $~{}~{}\square$

Lemma 2.19

Suppose that assumptions (vi) and (viii) hold. Then for any $i,1\leqq i\leqq q,$ we have and $[L_{-q},\,[L_{-i},\,L_{i}]]=L_{-q};$ in particular, $[L_{-i},\,L_{i}]\neq 0.$

Proof The lemma will follow from Lemma 2.4 once we show that $[L_{-q},$ $[L_{-i},\,L_{i}]]\neq 0.$ For $i=q,$ the lemma follows from Corollary 2.11. Let $1\leqq i<q.$ Then by Lemma 2.17, we have $L_{-q}=[L_{-q+i},\,L_{-i}],$ and by Lemma 2.8, we have $L_{-q+i}=[L_{-q},\,L_{i}].$ Then we have

[TABLE]

so that $[L_{-q},[L_{-i},\,L_{i}]]\neq 0,$ as required. $~{}~{}\square$

Lemma 2.20

Suppose that assumptions (iv) , (v), (and therefore, (iii)), (vi), and (viii) hold and that $0<i<\frac{q-3}{2}.$ Then $[L_{-q+i},\ L_{q-i+1}]=L_{1}.$

Proof Suppose that $[L_{-q+i},\ L_{q-i+1}]=0.$ Then, since by (iii) and Lemma 2.3, $L_{j}=S_{j}=[S_{j+1},\,S_{-1}]=[L_{j+1},\,L_{-1}]\subseteq L_{j}$ for all $j,\,0<j<r,$ we have

[TABLE]

so that (since $i<\frac{q-3}{2}$ implies that $2i+3<q,$ so that a fortiori $2i-1<q)$

[TABLE]

Let $v\in L_{-q+i}$ and $u\in L_{q-i}.$ Then (since $2i+1<2i+3<q,$ so that $[v,\,L_{-q+i+1}]=0$),

[TABLE]

Consequently, $\hbox{\rm ad\,}_{L_{-q+i+1}}[L_{-q+i},\,L_{q-i}]$ is a nilpotent set of linear transformations by Lemma 2.9. Since we are assuming that $i<\frac{q-3}{2},$ we have $i+1<\frac{q-1}{2},$ so we can apply Lemma 2.18 to conclude that $L_{-q+i+1}$ is an irreducible $L_{0}-$module. It follows that $\hbox{\rm ad\,}_{L_{-q+i+1}}[L_{-q+i},\,L_{q-i}]$ annihilates $L_{-q+i+1}.$ Thus (since, again, $2i+1<2i+3<q)$

[TABLE]

If $[L_{q-i},\,L_{-q+i+1}]$ $\neq 0,$ then, since $L_{1}$ is assumed (iv) to be irreducible, $[L_{q-i},$ $L_{-q+i+1}]$ would have to equal $L_{1},$ and the above-displayed formula would imply a lack of $\{1\}-$transitivity (vi) of $L$ in its negative part. It follows that $[L_{q-i},\,L_{-q+i+1}]$ $=0.$ Then, in view of our initial assumption that $[L_{-q+i},\ L_{q-i+1}]$ $=$ $0,$ we would have

[TABLE]

to contradict Lemma 2.14, in view of Lemma 2.19. Thus, it must be that $[L_{-q+i},$ $L_{q-i+1}]$ $\neq 0,$ so that by the assumed irreducibility (iv) of $L_{1},$ $[L_{-q+i},\ L_{q-i+1}]$ $=L_{1},$ as required. $~{}~{}\square$

Lemma 2.21

Let $q>5,$ and suppose that assumptions (iv) , (v) (and therefore (iii)), (vi), and (viii) hold. If $q$ is even, then $L_{-2}^{\frac{q}{2}}=L_{-q},$ while if $q$ is odd, then $L_{-2}^{\frac{q-1}{2}}=L_{-q+1}.$

Proof We have by Lemma 2.8 that $L_{-2}=[L_{-q},\,L_{q-2}]$ and by Lemma 2.18 (since $q>3)$ that $L_{-1}=[L_{-q+1},\,L_{q-2}].$ Thus, for any $j,\,1<j<q-1,$ and any non-zero $L_{0}-$submodule $V_{-j}$ of $L_{-j},$ we have by $\{-1\}-$transitivity (Lemma 2.1) that

[TABLE]

Consequently, $[V_{-j},$ $L_{q-2}]$ $\neq 0.$ Then by Lemma 2.16 when $j$ $<q-2,$ or, when $j$ $=q-2,$ by Lemma 2.19,

[TABLE]

by Lemma 2.8. If, for $j=1,\,2,\,\dots,\,[\frac{q}{2}]-1,$ we successively let $V_{-2j}\buildrel{\hbox{\small def}}\over{=}L_{-2}^{j},$ we can conclude that $L_{-2}^{\frac{q}{2}}\neq 0$ if $q$ is even, and $L_{-2}^{\frac{q-1}{2}}\neq 0$ if $q$ is odd. Then, by Lemmas 2.4 and 2.18, respectively, $L_{-2}^{\frac{q}{2}}=L_{-q}$ if $q$ is even, or $L_{-2}^{\frac{q-1}{2}}=L_{-q+1}$ if $q$ is odd. $~{}~{}\square$

Lemma 2.22

Let $q>5,$ and suppose that assumptions (iv) , (v) (and therefore (iii)), (vi) and (viii) hold. Then $L_{-2}$ is an irreducible $L_{0}-$module.

Proof Let $V_{-2}$ be any irreducible $L_{0}-$submodule of $L_{-2}.$ Since $[L_{-q+1},$ $[V_{-2},$ $L_{q}]]$ $=[V_{-2},$ $[L_{-q+1},$ $L_{q}]]$ $=[V_{-2},$ $L_{1}]$ $=L_{-1}$ by Lemmas 2.20 and 2.15, it follows that for any $j,$ $0<j<\frac{q}{2}$ (i.e., $0<j$ $\leqq\frac{q-1}{2}$) for which $V_{-2}^{j}$ $\neq 0,$ we have by transitivity (Lemma 2.1) that

[TABLE]

so we conclude that $[V_{-2}^{j},\,[V_{-2},\,L_{q}]]$ $\neq 0$ and therefore that $(\hbox{\rm ad\,}V_{-2})^{j+1}L_{q}\neq 0.$ Thus, so long as $2(j+1)<q$ (i.e., $j<\frac{q}{2}-1),$ we have by Lemma 2.16 that

[TABLE]

Thus, $V_{-2}^{j}\neq 0$ for all $j,$ $0<j\leqq\frac{q-1}{2},$ and $(\hbox{\rm ad\,}V_{-2})^{j}L_{q}\neq 0$ for all $j,$ $0<j\leqq\frac{q+1}{2}.$ If $q$ is odd, then, since $q>5,$ we have by Lemma 2.18 that $V_{-2}^{\frac{q-1}{2}}=L_{-q+1},$ while if $q$ is even, we have $V_{-2}^{\frac{q}{2}-1}=L_{-q+2}.$

In the case of odd $q,$ we have, by the irreducibility (B) of $L$, that $L_{-1}$ $=(\hbox{\rm ad\,}V_{-2})^{\frac{q+1}{2}}L_{q},$ so that

[TABLE]

Thus, when $q$ is odd, we see that $L_{-2}$ is irreducible. In the case of even $q,$ we have by Lemma 2.18 (since $q>5)$ that $L_{-1}$ $=[L_{-q+2},\,L_{q-3}]$ $=[V_{-2}^{\frac{q}{2}-1},\,L_{q-3}]$ $\subseteq(\hbox{\rm ad\,}V_{-2})^{\frac{q}{2}-1}L_{q-3}$ $\subseteq L_{-1}.$ By (D) and transitivity (Lemma 2.1),

[TABLE]

so that (See also Lemma 2.18.) $(\hbox{\rm ad\,}V_{-2})^{q-2}L_{q-3}$ $=L_{-q+1}.$ Then, by Lemma 2.17,

[TABLE]

Now, by Lemma 2.16 and irreducibility (B), we have

[TABLE]

Consequently, we have

[TABLE]

as required. $~{}~{}\square$

Now suppose that (vii) holds. Note that by (D) of the Main Theorem, $L_{-2}=[L_{-1},\,L_{-1}];$ that is, $L_{-2}$ is spanned by brackets of elements of the three-dimensional $L_{0}-$module $L_{-1}.$ Consequently, $L_{-2}$ is at most three-dimensional. On the other hand, by Lemma 1.2, the character $\chi$ is non-zero on $L_{-2},$ so $L_{-2}$ is not a restricted $L_{0}-$module, so its dimension is, in fact, three, as are all irreducible non-restricted ${\mathfrak{sl}}(2)-$modules in characteristic three. Thus (Compare Lemma 2.22.), we have

Lemma 2.23

If assumption (vii) holds, then $L_{-2}$ is an irreducible three-dimensional non-restricted $L_{0}^{\prime}-$module.

Lemma 2.24

If $q>2$ and assumptions (vi) and (vii) hold, then $\hbox{\rm Ann\,}_{L_{1}}L_{-2}$ $=0.$

Proof Set $A_{1}=\hbox{\rm Ann\,}_{L_{1}}L_{-2},$ and suppose that $A_{1}\neq 0.$ Since

[TABLE]

we have

[TABLE]

Now, if $[L_{-q+1},\,A_{1}]\neq 0,$ then by (vi) and irreducibility (B), we would have $(\hbox{\rm ad\,}L_{1})^{q-3}$ $[L_{-q+1},\,A_{1}]$ $=L_{-1},$ so that $[L_{-q+1},\,L_{-1}]=0,$ to contradict transitivity (Lemma 2.1). Thus, we must have

[TABLE]

Now, since $[L_{-q},$ $[A_{1},$ $A_{1}]]$ $\subseteq[L_{-q+1},$ $A_{1}]$ $=0,$ and, clearly, $[L_{-q+1},$ $[A_{1},$ $A_{1}]]$ $=0,$ it follows from Lemma 2.6 that $[A_{1},$ $A_{1}]$ $=0.$ Then ${L}^{\dagger}$ $\buildrel{\hbox{\small def}}\over{=}$ $(L_{-q}$ $\oplus$ $\cdots$ $\oplus$ $L_{-1}$ $\oplus$ $L_{0}$ $\oplus$ $A_{1})/(L_{-q}$ $\oplus$ $\cdots$ $\oplus$ $L_{-2})$ is a depth-one Lie algebra which satisfies conditions (A) through (C) of the Main Theorem. Consequently, by Proposition 1.6, $({L}^{\dagger})^{\prime}$ is one of the Lie algebras enumerated in the hypothesis of Lemma 1.8. If we set $V$ $=$ $L_{-q}$ $\oplus$ $L_{-q+1},$ then the $({L}^{\dagger})^{\prime}-$module $V$ satisfies the hypotheses of Lemma 1.8. Set $W$ $=L_{-q+1},$ and note that by (2.25), $[A_{1},$ $W]$ $=$ $0.$

If $(\zeta=)\,q-1\not\equiv 0$ (mod 3), then Lemma 1.8 would imply that $[L_{-1},$ $[L_{-1},$ $L_{-q+1}]]$ $\neq$ $0.$ Since this is false, we must have $\zeta=q-1$ $\equiv 0$ mod 3. On the other hand, if we then set $W$ $=$ $L_{-q},$ we have $[L_{-1},\,L_{-q}]=0$ and, by (2.25),

[TABLE]

so we must, by similar reasoning, conclude that $(\zeta=)\,q\equiv 0$ mod 3. Since both $q-1$ and $q$ cannot be equivalent to zero modulo three, we have arrived at a contradiction. We therefore conclude that $\hbox{\rm Ann\,}_{L_{1}}L_{-2}$ $=A_{1}$ $=0,$ as required. $~{}~{}\square$

Lemma 2.26

If conditions (iv) , (vi) and (vii) hold, then

[TABLE]

Proof If $q=2$, this lemma follows from Lemma 2.16. If $q>2,$ it follows from Lemma 2.24.  $~{}~{}\square$

Lemma 2.27

If $M_{1}$ is a non-zero $L_{0}-$submodule of $L_{1}$ such that $\hbox{\rm Ann\,}_{L_{0}}M_{1}$ $\neq 0,$ then $[[L_{-1},M_{1}],\,M_{1}]$ $=0,$ and $[M_{1},\,M_{1}]$ $=0.$

Proof Set $X\buildrel{\hbox{\small def}}\over{=}\hbox{\rm Ann\,}_{L_{0}}M_{1}\neq 0.$ Then by transitivity (C) and irreducibility (B), $[L_{-1},\,X]=L_{-1},$ so

[TABLE]

Thus, $[L_{-1},[M_{1},\,M_{1}]]\subseteq[[L_{-1},\,M_{1}],\,M_{1}]\subseteq[X,\,M_{1}]=0,$ so, by transitivity (C), $[M_{1},\,M_{1}]=0.$  $~{}~{}\square$

Lemma 2.28

We may assume that $(S_{s}=)\,[[L_{-1},\,L_{1}],\,L_{r}]=L_{r}.$

Proof Suppose $[[L_{-1},\,L_{1}],\,L_{r}]\neq L_{r}.$ We distinguish two cases:

1. (i)

$[[L_{-1},\,L_{1}],\,L_{r}]=0.$ 2. (ii)

$[[L_{-1},\,L_{1}],\,L_{r}]\neq 0.$

(i) Suppose first that $[[L_{-1},\,L_{1}],\,L_{r}]$ were equal to zero. Then we would have

[TABLE]

so $\sum_{i\geqq 0}(\hbox{\rm ad\,}L_{-1})^{i}[L_{-1},\,L_{r}]\subseteq S$ would be an ideal in $S$ entailing equality by the definition of $S;$ in particular, we would have $[L_{-1},\,L_{r}]=S_{s}.$

If, in addition, $[[L_{-1},\,L_{1}],\,[L_{-1},\,L_{r}]]$ were also equal to zero, we could repeat the argument and get that $\sum_{i\geqq 0}(\hbox{\rm ad\,}L_{-1})^{i}[L_{-1},\,[L_{-1},\,L_{r}]$ would be a proper ideal of $S$, to contradict the simplicity of $S$. We conclude that $[[L_{-1},\,L_{1}],$ $[L_{-1},\,L_{r}]]\neq 0,$ and, in the case that $[[L_{-1},\,L_{1}],\,L_{r}]=0$, that $[L_{-1},\,L_{r}]=S_{s},$ which is an irreducible $L_{0}-$module by Lemma 2.3.

Then,

[TABLE]

Hence, if in the case that $[[L_{-1},\,L_{1}],\,L_{r}]=0$, we replace $L_{r}$ with $[L_{-1},L_{r}]$, the lemma follows.

(ii) Now suppose that $[[L_{-1},\,L_{1}],\,L_{r}]\neq 0.$ Since

[TABLE]

it would follow that $\sum_{i\geqq 0}(\hbox{\rm ad\,}L_{-1})^{i}[[L_{-1},\,L_{1}],\,L_{r}]$ would be an ideal of $S$ and hence all of $S.$ In particular, $s$ would equal $r,$ and $[[L_{-1},\,L_{1}],\,L_{r}]$ would equal $S_{s}.$ Thus, $[[L_{-1},\,L_{1}],\,L_{r}]=S_{s}$ is an irreducible $L_{0}-$module by Lemma 2.3. Consequently, if we replace $L$ by the Lie algebra generated by $L_{-1},$ $L_{0},$ $L_{1},$ and $[[L_{-1},\,L_{1}],\,L_{r}]$ $(=[S_{s-1},L_{1}]=S_{s}),$ then the highest gradation space will be of the form $[[L_{-1},\,L_{1}],\,L_{r}],$ as required.  $~{}~{}\square$

Lemma 2.29

Let $L$ be as in the statement of the Main Theorem, and suppose that $L_{2}$ $\neq 0,$ that $[L_{-2},\,L_{1}]$ $=0$ $=[L_{-2},$ $\,L_{2}],$ and that assumption (vii) holds. Let $\tilde{L}$ be the Lie subalgebra of $L$ generated by $L_{-1},$ $L_{0},$ and $L_{1}.$ If $M(\tilde{L})$ is as in Weisfeiler’s Theorem (Theorem 1.3), then $\tilde{L}/M(\tilde{L})$ is Hamiltonian, and we have $[L_{-1},\,L_{1}]\cong\mathfrak{sl}(2).$

Proof Let $\tilde{\tilde{L}}$ be the Lie subalgebra of $L$ generated by $L_{-1},$ $L_{0},$ $L_{1},$ and $L_{2}.$ Since $[L_{-2},$ $\,L_{1}]$ $=0$ $=[L_{-2},$ $\,L_{2}],$ we have $M(\tilde{\tilde{L}})$ $=L_{-q}$ $\oplus$ $\cdots$ $\oplus$ $L_{-2}$ $=M(\tilde{L}).$ But the depth $\tilde{\tilde{L}}/M(\tilde{\tilde{L}})$ is then one, so by Proposition 1.6, $\tilde{\tilde{L}}/M(\tilde{\tilde{L}})$ is either Hamiltonian (i.e., between $H(2:\underline{n},\,\omega)$ and $CH(2:\underline{n},\,\omega))$ or is isomorphic to a Lie algebra of type $L(\epsilon)$ or $M.$ However, the height of the latter two Lie algebras is one, and, since $L_{2}\neq 0,$ the height of $\tilde{\tilde{L}}/M(\tilde{\tilde{L}})$ is at least two. Thus, $\tilde{\tilde{L}}/M(\tilde{\tilde{L}})$ must be Hamiltonian, so $[L_{-1},\,L_{1}]\cong\mathfrak{sl}(2).$ It now follows from Corollary 1.7 that $\tilde{L}/M(\tilde{L})$ is Hamiltonian, as well. $~{}~{}\square$

Lemma 2.30

Let $L$ be as in the statement of the Main Theorem, and suppose that $\tilde{L}/M(\tilde{L})$ (as above) is isomorphic to $L(\epsilon)$ or $M$. Then $\hbox{\rm Ann\,}_{L_{2}}L_{-2}=0$ and in the proof of the Main Theorem, where we assume that $[L_{-2},\,L_{1}]=0,$ we may assume that $L_{2}$ is an irreducible $L_{0}-$module.

Proof Suppose first that $\hbox{\rm Ann\,}_{L_{2}}L_{-2}\neq 0.$ Then, as in the proof of the previous lemma, we may consider the Lie algebra $\tilde{\tilde{\tilde{L}}}$ generated by $L_{-1},$ $L_{0},$ $L_{1},$ and $\hbox{\rm Ann\,}_{L_{2}}L_{-2}.$ Then $\tilde{\tilde{\tilde{L}}}/M(\tilde{\tilde{\tilde{L}}})$ has depth one and height greater than one, but null component isomorphic to ${\mathfrak{gl}}(2)$, to contradict Corollary 1.7. Thus, $\hbox{\rm Ann\,}_{L_{2}}L_{-2}=0$, so that if $M_{2}$ is any irreducible $L_{0}-$module of $L_{2},$ then $[L_{-2},\,M_{2}]\neq 0.$ Replacing $L$ by the Lie algebra generated by $L_{-1},$ $L_{0},$ $L_{1},$ and $M_{2},$ we complete the proof of the lemma. $~{}~{}\square$

3 The Main Theorem under additional assumptions

In this and the following two sections, we assume that assumptions ((i) and (ii), of course), (iv) , (v) (and therefore (iii)), (vi), and (viii) of the previous section hold, so that, in particular, by Lemma 2.22, $L_{-2}$ is an irreducible $L_{0}-$module. We begin this section by forming the irreducible, transitive Lie algebra $B(L_{-2}).$ (See, for example, Section 3 of [1].) Indeed, consider the subalgebra

[TABLE]

of $L$ consisting of the gradation spaces $E_{i}=L_{-2}^{-i}$ for $i<0,$ and $E_{i}=L_{2i}$ for $i\geqq 0.$ Set $T_{0}=$ $\hbox{\rm Ann\,}_{E_{0}}E_{-1}=$ $\hbox{\rm Ann\,}_{L_{0}}L_{-2},$ and for $i=1,2,\dots$, let

[TABLE]

Then

[TABLE]

is an ideal of $E$, and the factor algebra

[TABLE]

is a transitive graded Lie algebra (See [1, Lemma 3].) Thus, the Lie algebra $B(L_{-2})\buildrel{\hbox{\small def}}\over{=}G$ satisfies conditions (A) - (D) of the Main Theorem. (It is shown in, for example, [3] that the process of forming $B(L_{-2})$ preserves condition (A).)

Lemma 3.1

Let $L$ be as in the statement of the Main Theorem, and suppose that assumptions (i) - (viii) hold. If $q\geqq 6,$ then $B(L_{-2})$ is an irreducible, transitive graded Lie algebra of height greater than or equal to two and depth greater than two, and $B(L_{-2})_{-2}\not\subseteq M(B(L_{-2})).$ Consequently, since the depth of $B(L_{-2})$ is no greater than half of the depth of $L,$ we can, using induction, apply the Main Theorem to conclude that the character of the representation of $B(L_{-2})_{0}^{\prime}$ on $B(L_{-2})_{-1}$ is equal to zero. Then the character of $L_{0}^{\prime}=B(L_{-2})_{0}^{\prime}$ on $L_{-1}$ is also zero.

Proof By Lemma 2.21, when $q$ is even, $L_{-2}^{\frac{q}{2}}=L_{-q},$ while when $q$ is odd, $L_{-2}^{\frac{q-1}{2}}=L_{-q+1},$ so, in either case, the depth of $B(L_{-2})$ is greater than or equal to three. In the even case, since by Corollary 2.11 $[L_{-q},\,[L_{-q},\,L_{q}]]\neq(0),$ we have

[TABLE]

so the height of $B(L_{-2})$ is greater than or equal to three, since $r\geq q\geq 6.$

In the odd case, since $q\geqq 7,$ we have by Lemma 2.18 that $[L_{-q+2},$ $L_{q-3}]$ $=L_{-1}.$ If $[L_{-q+1},$ $L_{q-3}]$ were equal to zero, then we would have

[TABLE]

to contradict transitivity (Lemma 2.1). We therefore conclude that

[TABLE]

Then $0\neq[L_{-q+1},\,L_{q-3}]=[L_{-2}^{\frac{q-1}{2}},\,L_{q-3}]\subseteq(\hbox{\rm ad\,}L_{-2})^{\frac{q-1}{2}}L_{q-3}$ so the height of $B(L_{-2})$ is at least two in the odd case.

We must now verify hypothesis (E) of the Main Theorem for the Lie algebra $B(L_{-2});$ that is, we must show that $[L_{-2},$ $L_{-2}]$ is not contained in $M(B(L_{-2})).$ Thus, suppose that $[L_{-2},$ $L_{-2}]$ is contained in $M(B(L_{-2}));$ i.e., that

[TABLE]

We will arrive at a contradiction by successively considering the two cases:

even $q$, and 2. 2)

odd $q.$

Suppose first that $q$ is even. Then by Lemma 2.21, $L_{-q}$ $=L_{-2}^{\frac{q}{2}}$ $=$ $(\hbox{\rm ad\,}L_{-2})^{\frac{q}{2}-2}[L_{-2},$ $L_{-2}],$ so that $L_{-q}$ $\subseteq M(B(L_{-2})).$ Then by Lemma 2.8, ($B(L_{-2})_{-1}$ $=)L_{-2}$ $=[L_{-q},$ $L_{q-2}]$ $\subseteq M(B(L_{-2})),$ so that we would have by the definition of $M(B(L_{-2}))$ that

[TABLE]

to contradict, for example, (3.2) above. Thus, $q$ cannot be even. 2. 2)

Next suppose that $q$ is odd. Then we have by Lemma 2.21 again,

[TABLE]

so that $L_{-q+1}$ $\subseteq M(B(L_{-2})).$ Since by Lemma 2.22, $L_{-2}$ is an irreducible $L_{0}-$module, it follows from (3.3) that $L_{-2}$ $=[L_{-q+1},$ $L_{q-3}]$ $\subseteq M(B(L_{-2})).$ But we saw at the conclusion of 1) above that $L_{-2}$ cannot be contained in $M(B(L_{-2})).$ This second contradiction shows that $B(L_{-2})_{-2}$ $=[L_{-2},$ $L_{-2}]$ is in fact not contained in $M(B(L_{-2})),$ no matter what the parity of $q$ is.

Consequently, we can conclude that $B(L_{-2})$ satisfies hypothesis (E), and therefore all of the hypotheses of the Main Theorem. Since the depth of $B(L_{-2})$ is greater than one, but less than or equal to $\frac{q}{2},$ we can now apply to conclude that the character $\chi$ of $B(L_{-2})_{0}^{\prime}$ on $B(L_{-2})_{-1}$ is zero. Then $\frac{1}{2}\chi,$ which is the character of $L_{0}^{\prime}$ on $L_{-1},$ must be zero as well, and Lemma 3.1 is proved. $~{}~{}\square$

We now address the depth-four and depth-five cases individually.

4 The depth-four case

Suppose

[TABLE]

satisfies conditions (i) through (viii) of Section 2. By Lemma 2.15,

[TABLE]

By Lemma 2.17, $[L_{-2},L_{-2}]$ $=L_{-4}.$ Furthermore, from Lemma 2.3 we have $[L_{-4},$ $L_{2}]$ $\neq 0.$ Then

[TABLE]

Now let $V_{-2}$ be any irreducible $L_{0}-$submodule of $L_{-2}.$ If $[V_{-2},$ $L_{3}]=0,$ then by Lemma 2.3 and irreducibility (B), [math] $=[L_{-4},$ $[V_{-2},$ $L_{3}]]$ $=[V_{-2},$ $[L_{-4},$ $L_{3}]]$ $=[V_{-2},$ $L_{-1}]$ to contradict transitivity (Lemma 2.1). Thus, we can assume that $[V_{-2},$ $L_{3}]$ $=L_{1},$ since we are assuming that $L_{1}$ is $L_{0}-$irreducible (iv). Then by Lemma 2.15, $L_{-1}$ $=[V_{-2},$ $[V_{-2},$ $L_{3}]].$ Then we have by condition (D) of the Main Theorem that

[TABLE]

so that $L_{-2}$ is an irreducible $L_{0}-$module.

Thus, in the depth-two irreducible, transitive graded Lie algebra $B(L_{-2}),$ we have by (4.1) that $B(L_{-2})_{1}$ $\neq 0.$ Furthermore, it follows again from (4.1) above that

[TABLE]

so that hypothesis (E) of the Main Theorem is satisfied for $B(L_{-2}),$ as are the other hypotheses of the Main Theorem. Then the Main Theorem (proved for the case $q=2$ in [2]) applies to show that the representation of $B(L_{-2})_{0}^{\prime}$ on $B(L_{-2})_{-1}$ is restricted. Consequently, the character $\chi$ of $L_{0}^{\prime}$ on $L_{-2}$ is zero, as must be the character $\frac{1}{2}\chi$ of $L_{0}^{\prime}$ on $L_{-1}.$

5 The depth-five case

Suppose

[TABLE]

satisfies conditions (i) through (viii) of Section 2. Since $L$ is transitive (C), $[L_{-1},\,L_{5}]\neq 0.$ Hence, by Lemma 2.16, $[L_{-5},\,[L_{-1},\,L_{5}]]\neq 0,$ so that, by irreducibility (B), $[L_{-5},\,[L_{-1},\,L_{5}]]=L_{-1}.$

Now suppose that $[L_{-4},\,L_{5}]=0.$ If also $[L_{-3},\,[L_{-3},\,L_{5}]]$ $=0,$ then we would have

[TABLE]

However, since $[L_{-5},\,[L_{-1},\,L_{5}]]$ $=L_{-1},$ we have by transitivity (Lemma 2.1) that

[TABLE]

so that $[L_{-3},\,[L_{-1},\,L_{5}]]$ $\neq 0.$ Since $L_{1}$ is assumed (iv) to be irreducible, we must have $[L_{-3},$ $[L_{-1},$ $L_{5}]]$ $=L_{1}.$ Then by $\{1\}-$transitivity (vi),

[TABLE]

contrary to what was derived above. Thus, we can assume that $[L_{-3},$ $[L_{-3},$ $L_{5}]]$ $\neq 0.$ But then, by the irreducibility (B) of $L$, we must have $[L_{-3},$ $[L_{-3},$ $L_{5}]]$ $=L_{-1}.$ Then, by $\{-1\}-$transitivity (Lemma 2.1),

[TABLE]

so that $[L_{-4},\,L_{5}]$ $\neq 0.$ Since we are assuming (iv) that $L_{1}$ is irreducible, it follows that

[TABLE]

Now let $V_{-2}$ be any non-zero $L_{0}-$submodule of $L_{-2}.$ Then we have by (5.1) and $\{1\}-$transitivity (vi) that

[TABLE]

so

[TABLE]

by the irreducibility (B) of $L.$ Then, by transitivity (Lemma 2.1),

[TABLE]

so that $[V_{-2},\,[V_{-2},\,L_{5}]]$ $\neq 0.$ Thus, by the assumed irreducibility (iv) of $L_{1},$ we must have $[V_{-2},\,[V_{-2},\,L_{5}]]$ $=L_{1}.$ Then, as above, by the $\{1\}-$transitivity (vi) of $L,$ we have

[TABLE]

so that

[TABLE]

by the irreducibility (B) of $L.$ Then, since the negative gradation spaces are generated (D) by $L_{-1},$ we have

[TABLE]

so that $L_{-2}$ is an irreducible $L_{0}-$module.

Now, by Lemma 2.17, $L_{-5}=[L_{-2},\,L_{-3}].$ Consequently, in view of (D) and (5.2) above

[TABLE]

Then by $\{1\}-$transitivity (vi), we have

[TABLE]

so that $[L_{-2},\,L_{-2}]\neq 0.$ Furthermore, since the negative gradation spaces are generated (D) by $L_{-1},$ we have by (5.2) above that

[TABLE]

so

[TABLE]

Now suppose that $[L_{-4},\,L_{2}]$ $=0,$ and suppose further that $[L_{-3},\,L_{2}]$ $=0.$ Then we would have by (C) and (D) that

[TABLE]

by the assumed irreducibility (iv) of $L_{1},$ to contradict $\{1\}-$transitivity (vi). Thus, $[L_{-3},\,L_{2}]$ $\neq 0,$ so that by the irreducibility (B) of $L,$ $[L_{-3},\,L_{2}]$ $=L_{-1}.$ Then

[TABLE]

by $\{-1\}-$transitivity (Lemma 2.1). Thus, it must be true that

[TABLE]

so that

[TABLE]

By (5.3) and construction, $B(L_{-2})$ is a transitive, irreducible depth-two Lie algebra. By (5.6) above, $B(L_{-2})_{1}$ $\neq 0.$ By (5.4) and (5.5) above, $B(L_{-2})$ satisfies hypothesis (E) of the Main Theorem. Therefore, as in the depth-four case above, it follows from the Main Theorem (proved for the case $q=2$ in [2]) that the character of $B(L_{-2})_{0}^{\prime}$ on $B(L_{-2})_{-1}$ $=L_{-2}$ is zero. Consequently, the character of $L_{0}^{\prime}$ on $L_{-1}$ is zero, as well, and $L_{-1}$ is a restricted $L_{0}^{\prime}-$module.

6 Conclusion of the proof of the Main Theorem

Let $L$ be as in the statement of the Main Theorem, and let $S=\sum_{i=-q}^{s}S_{i}$ be as in Weisfeiler’s Theorem (Theorem 1.3). If $[S_{>0},L_{-2}]$ were equal to zero, then $L_{-q}\oplus\cdots\oplus L_{-2}=\sum_{i\geqq 0}(\hbox{\rm ad\,}(L_{-1}))^{i}L_{-2}$ would be an ideal of the simple Lie algebra $S$. Consequently, it must be that $[S_{>0},L_{-2}]\neq 0.$ Let $j>0$ be minimal such that

[TABLE]

We wish to show that $j=1.$ Suppose not.

We begin by establishing a few basic properties. By the Jacobi Identity, if $1\leq k\leq j-1$ and $2\leq i\leq k+1,$ then (since in the sum that follows, $0\leqq\kappa\leqq i-2\leqq k-1\leqq j-2,$ so that $j-1\geqq k\geqq k-\kappa\geqq 1;$ here $\kappa$ is the number of $(\hbox{\rm ad\,}L_{-1})$s that act on $S_{k}$ before it brackets with $L_{-2}$ and annihilates it)

[TABLE]

i.e.,

[TABLE]

In particular, we have (letting $i=k$, $i=k+1$, and $k=j-1$, respectively)

[TABLE]

[TABLE]

[TABLE]

We will now show that $[L_{-i},\,S_{j}]\neq 0,\,1\leq i\leq j+1.$ Since $L$ is irreducible (B), it will follow that

[TABLE]

Note that (since we are assuming that $[L_{-2},\,S_{j}]\neq 0$ and $[L_{-2},\,S_{j^{\prime}}]$ = [math] for $1\leq j^{\prime}\leq j-1),$ we have by (D), by (6.5), by transitivity (C), and by induction, that

[TABLE]

Thus, we have (in view of transitivity (C) and (6.1)) that

[TABLE]

We will now show that

[TABLE]

Thus, suppose first that $T_{j}$ is an $L_{0}-$submodule of $L_{j}$ such that $[L_{-j},\,T_{j}]=0.$ Then, in view of $D$ and (6.5), we have

[TABLE]

so that by transitivity (C), we have $[L_{-j+1},\,T_{j}]=0.$ Then we may replace $L_{-j}$ in (6.9) by, successively, $L_{-j+1},$ $L_{-j+2},$ etc., to arrive at $[L_{-2},\,T_{j}]=0.$ On the other hand, if we rather define $T_{j}=\hbox{\rm Ann\,}_{L_{j}}L_{-2},$ then, again in view of (6.5), we may bracket the equation $[L_{-2},\,T_{j}]=0$ with $L_{-1}$ again and again to conclude that

[TABLE]

as required to establish (6.8).

Lemma 6.10

If $j$ is as above, then $[L_{-j-1},\,S_{j+1}]\neq 0.$

Proof Suppose that $[L_{-j-1},\,S_{j+1}]=0$. Then, since by (6.6) $[L_{-j-1},\,S_{j}]$ $=$ $L_{-1},$ we have

[TABLE]

Consequently, $[S_{j},\,S_{j+1}]\neq 0.$ Then we have

[TABLE]

Continuing in this manner, we get homogeneous non-zero $L_{0}-$submodules of arbitrarily high gradation degree. This of course cannot happen in a finite-dimensional Lie algebra, so it must be that $[L_{-j-1},\,S_{j+1}]\neq 0$.  $~{}~{}\square$

Lemma 6.12

If $j\geqq 2$, and $k$ is the smallest integer greater than one for which $[L_{-k},\,S_{r}]\neq 0,$ then $k=j.$

Proof Since by (D), $L_{<0}$ is generated by $L_{-1}$, we have by (6.7) and (6.3) and (iv) that

[TABLE]

so $[L_{-j},\,S_{r}]\neq 0,$ since by Lemma 2.28 we have

[TABLE]

Consequently,

[TABLE]

On the other hand, by definition of $k$, $[L_{-k},\,S_{r}]\neq 0$, and $[L_{-i},\,S_{r}]=0$ for $k-1\geqq i\geqq 2.$ It follows that if $[L_{-k},\,S_{k-1}]=0$ and $[L_{-k},\,S_{k}]=0$, then

[TABLE]

would be a proper ideal of $S$, to contradict the simplicity of $S$. Thus, it must be that either $[L_{-k},\,S_{k}]\neq 0$ or $[L_{-k},\,S_{k-1}]\neq 0$.

Now, by (6.3) and (6.7), $j$ is the smallest integer greater than or equal to two such that $[L_{-j},\,S_{j}]\neq 0.$ Consequently, if $[L_{-k},\,S_{k}]\neq 0,$ then $j\leqq k,$ so by (6.13), $j=k$. If, on the other hand, $[L_{-k},\,S_{k}]=0$, then, as we noted above, we must have $[L_{-k},\,S_{k-1}]\neq 0.$ But by (6.4) and (6.7), $j+1$ is the smallest $i\geqq 2$ such that $[L_{-i},\,S_{i-1}]\neq 0,$ so we must have $j+1\leqq k,$ so $j\leqq k-1$, to contradict (6.13). Thus,

[TABLE]

as required.  $~{}~{}\square$

Define $H$ to be the Lie algebra generated by $L_{-1}$ $\oplus$ $L_{0}$ $\oplus$ $S_{1}$ $\oplus$ $\dots$ $\oplus$ $S_{j-1}.$ Since $[L_{-2},\,S_{1}\oplus$ $\dots\oplus S_{j-1}]=0,$ it follows from (6.2) that if $M(H)$ is as in the statement of Weisfeiler’s Theorem (Theorem 1.3), then $M(H)$ $=L_{-q}\oplus\dots\oplus L_{-2},$ and $H/M(H)$ is a depth-one graded Lie algebra which inherits transitivity (C) and irreducibility (B) from $L$. From Proposition 1.6, we conclude that $H/M(H)$ is either between $H(2:\,(1,\,1),\,\omega)$ and $CH(2:\,(1,\,1),\,\omega),$ or is equal to $L(\epsilon)$ or $M.$ In each of these cases, $S_{1}$ is an irreducible abelian $L_{0}-$module. Thus, we can from now on assume (See Corollary 1.7.) that assumption (iv) of Section 2 holds and

[TABLE]

and, as a Lie algebra, $[L_{-1},\,L_{1}]$ lies between ${\mathfrak{sl}}(2)$ and ${\mathfrak{gl}}(2)$:

[TABLE]

Lemma 6.16

If $[L_{-2},\,L_{1}]=0$ and $[L_{-2},\,L_{2}]\neq 0$ and (vii) and (6.15) hold, then we may assume in what follows that $[L_{-2},\,L_{2}]=[L_{-1},\,L_{1}].$

Proof Let $i$ be any integer not equivalent to zero modulo three, and let $z$ be any element of the center of $L_{0}.$ By transitivity (C), $[L_{-1},\,z]\neq 0,$ and by irreducibility (B) and Schur’s Lemma, $\hbox{\rm ad\,}_{L_{-1}}z$ acts as multiplication by a non-zero scalar $a$, so that by (D), $\hbox{\rm ad\,}_{L_{-i}}z$ acts as multiplication by the non-zero scalar $ai$. Consequently, in view of (6.15), Lemma 1.2, and transitivity (C) (to deal with $\hbox{\rm ad\,}\,z$ when $i$ is positive),

[TABLE]

for any non-zero $L_{0}-$submodule $M_{i}$ of $L_{i},i\not\equiv 0\,\hbox{(mod 3)}.$ Since by (D), $[L_{-2},\,L_{2}]$ is an ideal of $[L_{-1},\,L_{1}]$, we have that $[L_{-2},$ $[L_{-2},\,L_{2}]]$ $\neq$ [math]. Similarly, (See Corollary 1.5.) $[L_{2},\,[L_{-2},L_{2}]]$ $\neq 0.$ It follows that

[TABLE]

Let $M(B(L_{-2}))$ be as in Weisfeiler’s Theorem (Theorem 1.3). We focus on

[TABLE]

whose depth is no greater than half of that of $L$. If the depth and height of $X$ are both greater than one, then we can apply the Main Theorem to conclude that the representation of $L_{0}^{\prime}$ on the minus-one component (namely, $L_{-2}$) of $X$ is restricted, so that (See Lemma 1.2.) the representation of $L_{0}^{\prime}$ on $L_{-1}$ is also restricted, to contradict assumption (vii).

If the depth of $X$ is one, then by Proposition 1.6, $X$ is isomorphic either to a Hamiltonian Lie algebra, or to $M,$ or to $L(\epsilon)$ for some $\epsilon.$ Then, by (6.15) and (D),

[TABLE]

Consequently, we are done unless $[L_{-2},\,L_{2}]\cong{\mathfrak{sl}}(2),$ and $[L_{-1},\,L_{1}]\cong{\mathfrak{gl}}(2).$ In that case, we have by Corollary 1.7 that $X$ is Hamiltonian. Let $e,$ $f,$ and $h$ be the usual basis of $L_{0}^{\prime}.$ Then, according to (1.4) of [4], $(\hbox{\rm ad\,}_{L_{-2}}f)^{3}$ acts as the identity on the minus-one component of $X,$ namely, $L_{-2}.$ On the other hand, if $\tilde{L}$ is as in Lemma 2.29, then $L_{-2}\subseteq M(\tilde{L}),$ and, again by Corollary 1.7, either $\tilde{L}/M(\tilde{L})\cong L(\epsilon),$ or $\tilde{L}/M(\tilde{L})\cong M.$ If $\tilde{L}/M(\tilde{L})\cong L(\epsilon)$, then $(\hbox{\rm ad\,}_{L_{-1}}f)^{3}$ acts as zero on $L_{-1}.$ On the other hand, if $\tilde{L}/M(\tilde{L})\cong M$, then $(\hbox{\rm ad\,}_{L_{-1}}f)^{3}$ acts as the identity on $L_{-1}.$ However, by (D) and the fact that $(\hbox{\rm ad\,}_{L_{-1}}f)^{3}$ is a derivation, we should have that $(\hbox{\rm ad\,}_{L_{-2}}f)^{3}$=$2(\hbox{\rm ad\,}_{L_{-1}}f)^{3}.$ These contradictions enable us to conclude that the lemma is true when the depth of $X$ is one.

Suppose, finally, that the height of $X$ is one. Since, as we saw above, $[L_{2},\,[L_{-2},L_{2}]]$ $\neq 0,$ it follows that $X$ is a non-degenerate Lie algebra. We may therefore apply Corollary 1.5 to $X$ to conclude that for any irreducible $L_{0}-$submodule $M_{2}$ of $L_{2}$ we have $[M_{2},\,[M_{2},\,L_{-2}]]\neq 0.$ We apply Corollary 1.7 to $B(M_{2})/M(B(L_{2})).$ Arguing as in the above paragraph, we obtain a contradiction also in this case. $~{}~{}\square$

Lemma 6.19

Suppose $j>2.$ Then $j$ cannot equal $r,$ so $L_{j+1}\neq 0.$ Similarly, $L_{j+2}\neq 0,$

Proof Suppose $j=r.$ Then by (6.7) $0\neq[L_{-j},\,S_{j}]=[L_{-r},\,S_{r}].$ Then we would have $[[L_{-r},\,S_{r}],S_{1}]=[[L_{-r},\,S_{1}],S_{r}]\subseteq[L_{-(r-1)},\,S_{r}]=0$ by the definition of $k.$ Since, again, $j=r,$ however, we would have $[L_{-j+1},\,L_{j}]=0,$ to contradict (6.7). Thus, $j\neq r.$

Now suppose that $j=r-1.$ Since $j>2,$ it follows that $r>3.$ By Lemma 6.10, $[L_{-r},\,L_{r}]\neq 0.$ Then by ((D) and) (6.17), $[L_{1},\,[L_{-r},\,L_{r}]]\neq 0.$ By (iv) (See the discussion before (6.14).), we have $L_{1}=[L_{-r+1},\,L_{r}],$ so by definition of $k$, $[L_{1},\,L_{i}]=0,\,1\leqq i\leqq k-2=j-2=r-3.$ If $r>4,$ then we would have, for example, $0=[L_{-1},\,[L_{1},\,L_{2}]]=[[L_{-1},\,L_{1}],\,L_{2}]],$ to contradict (6.17). If $r=4,$ then $j=3,$ so by (6.7), $[L_{-3},\,L_{3}]\neq 0,$ and by (6.5), $[L_{-3},\,L_{2}]=0.$ Thus, we would have $0=[L_{-3},\,0]=[L_{-3},\,[L_{2},\,L_{3}]]=[L_{2},\,[L_{-3},\,L_{3}]],$ to again contradict (6.17) by (D). $~{}~{}\square$

Now, by Lemma 2.29, $j\leq 2$ in the non-Hamiltonian cases. Thus, suppose that we are in the Hamiltonian case, and suppose, for a contradiction, that $j>2.$ For this (Hamiltonian, $j>2$) case, we will assume without loss of generality, that $L$ is generated by $L_{<0}\oplus L_{0}\oplus S_{1}\oplus\cdots\oplus S_{j-1}\oplus S_{j}.$ By definition of $j$ and the Jacobi Identity, $[S_{i},\,[L_{-2},\,L_{-2}]]=0,\,1\leqq i\leqq j-1.$ Furthermore, since $j$ is assumed to be greater than two, we have $[S_{j},\,[L_{-2},\,L_{-2}]]\subseteq[L_{-2},\,S_{j-2}]=0.$ Thus, again by the Jacobi Identity, $[L_{>0},\,[L_{-2},\,L_{-2}]]=0,$ so, if $[L_{-2},\,L_{-2}]$ were not equal to zero then $\sum_{i\geqq 0}(\hbox{\rm ad\,}L_{-1})^{i}[L_{-2},\,L_{-2}]$ would be a proper ideal of $S,$ to contradict the simplicity of $S.$ Thus, we may assume that

[TABLE]

In addition, $[L_{-2},\,S_{j-2}]=0$ also entails that

[TABLE]

Lemma 6.22

If $j>2,$ then

[TABLE]

Proof. Since we are assuming that $j>2,$ it follows that $2j>j+2,$ so the lemma follows from (6.21).  $~{}~{}\square$

We adopt the notation of [4].

Lemma 6.23

If $j>2,$ and $j\not\equiv 0$ (mod 3), then $[L_{-2},\,[L_{2},\,L_{j}]]\neq 0.$

Proof: Suppose not. Then

[TABLE]

since we are assuming that $j>2.$ But $0\neq[L_{-2},\,L_{j}]\subseteq H_{j-2}$ (See the discussion preceding (6.14).), and $H_{j-2}$ is $L_{0}-$irreducible. Consequently, we have $[H_{2},\,H_{j-2}]=0.$ But $\{y^{(3)},\,xy^{(j-1)}\}=y^{(2)}y^{(j-1)}\equiv jy^{(j+1)}\not\equiv 0$ (since $j\not\equiv 0$ (mod 3)).  $~{}~{}\square$

Lemma 6.24

If $M$ is an $L_{0}-$submodule of $L$ such that $[L_{-2},\,M]=0$ and $[L_{-2},\,[1,\,M]]$ = $0,$ then $(\hbox{\rm ad\,}1)^{3}$ is an $L_{0}-$homomorphism of $M$ into $(\hbox{\rm ad\,}1)^{3}M.$

Proof. Since $\{1,\,xy\}=0=\{1,\,x^{(2)}y\},$ we have $(\hbox{\rm ad\,}1)^{3}[L_{0},$ $M]$ $=$ $[(\hbox{\rm ad\,}1)^{3}L_{0},$ $M]$ + $[L_{0},\,(\hbox{\rm ad\,}1)^{3}M]$ $=$ $[(\hbox{\rm ad\,}1)^{2}x^{(2)},\,M]$ $+$ $[L_{0},\,(\hbox{\rm ad\,}1)^{3}M]$. Now, $[(\hbox{\rm ad\,}1)^{2}x^{(2)},$ $M]$ $\subseteq$ $[1,\,[L_{-2},\,M]]+[[1,\,M],\,L_{-2}]=0.$ $~{}~{}\square$

By transitivity (C), ($S_{j}\subseteq$) $L_{j}$ must be contained in the nine-dimensional $L_{0}-$module $L_{-1}^{*}\otimes S_{j-1}.$ (Here, and in what follows, we understand “is contained in” in this context to mean “is $L_{0}-$isomorphic to an $L_{0}-$submodule of” and “is acted on by $\hbox{\rm ad\,}L_{-1}$ as $\hbox{\rm ad\,}L_{-1}$ acts on” and we will often make use of this implication of (C) without comment.) Note that if the gradation degree of any of the following three-dimensional $L_{0}-$modules

[TABLE]

is not equivalent to zero modulo three, then the $L_{0}-$module is $L_{0}-$irreducible.

To begin our proof that $j=1,$ we will assume first that

[TABLE]

We consider separately the cases where $j$ is equivalent to one, two, or zero modulo three. We will show that in each case, if $j>2,$ then $L$ can be replaced with a subalgebra (containing $L_{-1}$ and $L_{0}$) in which the smallest integer $i>2$ such that $L_{-2}$ has nonzero bracket with the $i^{th}$ gradation space of the replacement is greater than $j.$

Note that

[TABLE]

Since $L$ contains $S_{j-1}\cong H(2:\mathbf{n},\omega)_{j-1},$ it follows that whenever $j\not\equiv 2$ (mod 3), we also have that $S_{j}$ contains the non-zero $L_{0}-$submodule (annihilated by $\hbox{\rm ad\,}L_{-2}$) $[S_{1},S_{j-1}]\cong H(2:\mathbf{n},\omega)_{j}.$ Furthermore, if we consider (6.26) with $j$ replaced by $j+1,$ we see that if $j\equiv 0$ (mod 3), then $S_{j+1}$ contains the non-zero $L_{0}-$submodule (also annihilated by $\hbox{\rm ad\,}L_{-2}$) $[S_{1},\,[S_{1},\,S_{j-1}]]\cong H(2:\mathbf{n},\omega)_{j+1}.$

When $j\equiv 1\,(\hbox{mod 3}),$ L_{-1}^{*}\otimes S_{j-1}$$(\supseteq L_{j}) contains the two $L_{0}-$submodules

[TABLE]

and

[TABLE]

(Acting here like $\frac{\partial}{\partial x}$, $\hbox{\rm ad\,}y$ maps $xy^{(j)}$, $x^{(2)}y^{(j)}$, and $y^{(j)}$, to $y^{(j)}$, $xy^{(j)}$, and zero, respectively. Also, adding one to the divided power modulo three and multiplying by two, $\hbox{\rm ad\,}y$ maps $1^{*}$, $x^{*}$, and $(x^{(2)})^{*}$, to $2\cdot x^{*}$, $2\cdot(x^{(2)})^{*}$ and $2\cdot 1^{*}$, respectively.)

Now,

[TABLE]

where modulo its submodule $X_{1,\,j,\,2},$ the indecomposable $L_{0}-$module $X_{1,\,j,\,3}$ is spanned by $(x^{(2)})^{*}\otimes x^{(2)}y^{(j)},\,2\cdot 1^{*}\otimes x^{(2)}y^{(j)}+(x^{(2)})^{*}\otimes xy^{(j)},\hbox{ and }1^{*}\otimes xy^{(j)}+x^{*}\otimes x^{(2)}y^{(j)}+(x^{(2)})^{*}\otimes y^{(j)}.$

Now $[L_{-2},\,X_{1,\,j,\,2}]=0,$ and, as an $L_{0}-$submodule of $L_{-1}^{*}\otimes S_{j-1},$ $X_{1,\,j,\,2}$ is unique in this regard. It follows from the discussion following (6.26) that $S_{j}\subseteq L_{j}$ must contain an $L_{0}-$submodule

[TABLE]

Since $S_{j}=[L_{-1},\,S_{j+1}]$ by Lemma 2.3, it follows from transitivity (C) that $S_{j+1}$ must be contained in

[TABLE]

Since $Q_{j}\subseteq S_{j},$ $S_{j+1}$ must have non-zero intersection with $L_{-1}^{*}\otimes Q_{j}\cong L_{-1}^{*}\otimes X_{1,\,j,\,2}.$ (Note that if $X_{1,\,j,\,2}\cap S_{j}=0,$ then $X_{1,\,j,\,3}\cap S_{j}=0;$ similarly, if $(L_{-1}^{*}\otimes X_{1,\,j,\,2})\cap S_{j+1}=0,$ then $(L_{-1}^{*}\otimes X_{1,\,j,\,3})\cap S_{j+1}=0$.)

If $S_{j+1}\cap(L_{-1}^{*}\otimes X_{1,\,j,\,2})$ were to contain an $L_{0}-$submodule $R_{j+1}$ such that $[L_{-2},\,R_{j+1}]\neq 0,$ we could replace $L$ with the Lie algebra generated by $L_{\leqq 0}\oplus S_{1}\oplus\cdots\oplus S_{j-1}\oplus Q_{j}\oplus R_{j+1},$ and we would have a Lie subalgebra of $L$ such that the minimal $i>2,$ such that $L_{-2}$ has non-zero bracket with the $i^{th}$ gradation space of the subalgebra, is $j+1.$

Define

[TABLE]

Then $y\cdot f=2d,$ and $L_{-1}^{*}\otimes X_{1,\,j,\,2}$ is the sum of the following $L_{0}-$submodules:

[TABLE]

[TABLE]

[TABLE]

Note that

[TABLE]

Now, since $(\hbox{\rm ad\,}[1,\,x])\cdot(1^{*}\otimes e)=[e,\,x]=2\cdot[x,\,e]=2xy^{(j)}\neq 0,$ and $(\hbox{\rm ad\,}[1,\,x])\cdot((x^{(2)})^{*}\otimes e+x^{*}\otimes f+2\cdot 1^{*}\otimes d)=[1,\,f]+2\cdot[d,\,x]=2x^{(2)}y^{(j)}\neq 0,$ if $L_{j+1}\cap X_{1,\,j+1,\,1}\neq 0,$ then $[L_{-2},\,L_{j+1}\cap X_{1,\,j+1,\,1}]\neq 0.$ Thus, if we set $R_{j+1}=L_{j+1}\cap X_{1,\,j+1,\,1}$ in the above argument, we see that we can assume in what follows that

[TABLE]

In addition, since $(\hbox{\rm ad\,}[x,\,x^{(2)}])\cdot(x^{*}\otimes e)=[e,\,x^{(2)}]=2\cdot[x^{(2)},\,e]=x^{(2)}y^{(j)}\neq 0,$ and $(\hbox{\rm ad\,}[1,\,x])\cdot(2\cdot(x^{(2)})^{*}\otimes e+x^{*}\otimes f)=[1,\,f]=x^{(2)}y^{(j)}\neq 0,$ if $L_{j+1}\cap X_{1,\,j+1,\,3}\neq 0,$ then $[L_{-2},\,L_{j+1}\cap X_{1,\,j+1,\,3}]\neq 0.$ Thus, if we set $R_{j+1}=L_{j+1}\cap X_{1,\,j+1,\,3}$ in the above argument, we see that we can assume in what follows that

[TABLE]

Now, $X_{1,\,j+1,\,1}$ and $X_{1,\,j+1,\,2}$ are isomorphic as $L_{0}-$modules, so suppose that, for scalars $A,$ and $B,$ $L_{j+1}\cap(X_{1,\,j+1,\,1}+X_{1,\,j+1,\,2})$ $\supseteq$ $AX_{1,\,j+1,\,1}+BX_{1,\,j+1,\,2},$ If this latter module had non-zero bracket with $L_{-2},$ then we could set $R_{j+1}$ equal to it and argue as in the previous two paragraphs. If, on the other hand, the bracket with $L_{-2}$ were zero, then, for example, we would have that the bracket of $[1,\,x]$ with

[TABLE]

was zero. Consequently, we would have

[TABLE]

from which we can conclude that $A=0,$ so that the only linear combination of $X_{1,\,j+1,\,1}$ and $X_{1,\,j+1,\,2}$ to have zero bracket with $L_{-2}$ is $X_{1,\,j+1,\,2}.$ It follows that we may assume (See (6.26)ff.) that $L_{j+1}$ has non-zero intersection with $X_{1,\,j+1,\,2}.$

Define

[TABLE]

Then $y\cdot\mu=\kappa,$ and $L_{-1}^{*}\otimes X_{1,\,j+1,\,2},$ which by transitivity (C) has non-zero intersection with $L_{j+2},$ is the sum of the following (non-isomorphic) $L_{0}-$modules:

[TABLE]

[TABLE]

[TABLE]

Now, since $(\hbox{\rm ad\,}[1,\,x])\cdot(x^{*}\otimes\mu)=[1,\,\mu]=d,$ and $(\hbox{\rm ad\,}[1,\,x])\cdot((x^{(2)})^{*}\otimes\kappa+x^{*}\otimes\lambda+1^{*}\otimes\mu)=[\mu,\,x]+[1,\,\lambda]=2f\neq 0,$ if $L_{j+2}\cap X_{1,\,j+2,\,1}\neq 0,$ then $[L_{-2},\,L_{j+2}\cap X_{1,\,j+2,\,1}]\neq 0,$ and we can replace $L$ with the Lie algebra generated by $L_{\leqq 0}\oplus L_{1}\oplus\cdots\oplus L_{j-1}\oplus(L_{j}\cap X_{1,\,j,\,2})\oplus(L_{j+1}\cap X_{1,\,j+1,\,2})\oplus(L_{j+2}\cap X_{1,\,j+2,\,1}),$ and we will have a Lie subalgebra of $L$ such that the minimal $i>2,$ such that $L_{-2}$ has non-zero bracket with the $i^{th}$ gradation space of the subalgebra, is $j+2.$ Thus, we may assume in what follows that

[TABLE]

Also, since $(\hbox{\rm ad\,}[1,\,x])\cdot(1^{*}\otimes\mu)=[\mu,\,x]=2f,$ and $(\hbox{\rm ad\,}[1,\,x])\cdot(2\cdot x^{*}\otimes\mu+1^{*}\otimes\kappa)=2\cdot[1,\,\mu]+[\kappa,\,x]=2d+2d\not\equiv 0,$ if $L_{j+2}\cap X_{1,\,j+2,\,3}\neq 0,$ then $[L_{-2},\,L_{j+2}\cap X_{1,\,j+2,\,3}]\neq 0,$ and we can replace $L$ with the Lie algebra generated by $L_{\leqq 0}\oplus L_{1}\oplus\cdots\oplus L_{j-1}\oplus(L_{j}\cap X_{1,\,j,\,2})\oplus(L_{j+1}\cap X_{1,\,j+1,\,2})\oplus(L_{j+2}\cap X_{1,\,j+2,\,3}),$ and we will have a Lie subalgebra of $L$ such that the minimal $i>2,$ such that $L_{-2}$ has non-zero bracket with the $i^{th}$ gradation space of the subalgebra, is $j+2.$ Thus, we may assume in what follows that

[TABLE]

Note that

[TABLE]

from which we can conclude that if $L_{j+2}$ has non-zero intersection with $X_{1,\,j+2,\,2}$, then by (6.34) and Lemma 6.24, we have that $L_{j+2}\cap X_{1,\,j+2,\,2}\cong S_{j-1}.$ We could then repeat the above argument (with $S_{j-1}$ replaced by $L_{j+2}\cap X_{1,\,j+2,\,2}$) to either obtain a Lie algebra in which the minimal gradation degree $i>2$ such that $[L_{-2},\,L_{i}]\neq 0$ is each time even greater, or, eventually, to find that the highest gradation space $S_{s}$ of $S$ contains an $L_{0}-$submodule $Q_{s}$ which has zero bracket with $L_{-2}.$ However, by Lemma 2.3, $S_{s}$ is irreducible as an $L_{0}-$module, so we would have that $[L_{-2},\,S_{s}]=0.$ Since $S$ is generated by $S_{s}$ and $L_{-1},$ it would follow from the construction of $Q_{s}$ $(=S_{s})$ (See also Lemma 2.3.) that $[L_{-2},\,S_{>0}]=0,$ to contradict, for example, (6.1).

In view of (6.39), (6.40), and (6.41)ff, if $j\equiv 1$ (mod 3), we have shown (since, from above, $L_{-1}^{*}\otimes X_{1,\,j+1,\,2}$ must have non-zero intersection with $L_{j+2}$) that we can assume that $L$ contains a Lie subalgebra such that if $i>2$ is minimal such that $L_{-2}$ has non-zero bracket with the $i^{th}$ gradation space of that subalgebra, then $i$ is greater than $j.$

If $j\equiv 2$ (mod 3), then $\{y,\,y^{(j)}\}=2x^{(2)}y^{(j)},$ and $L_{j}\subseteq(L_{-1})^{*}\otimes S_{j-1},$ which equals the sum of the $L_{0}-$submodules

[TABLE]

and

[TABLE]

and

[TABLE]

Since $[L_{-2},\,X_{2,\,j,\,2}]=0,$ it follows from the definition of $j$ (6.1) that $S_{j}\not\subset X_{2,\,j,\,2},$ so either $X_{2,\,j,\,1}\cap S_{j}\neq 0$ or $X_{2,\,j,\,3}\cap S_{j}\neq 0.$

Let us first suppose that $S_{j}\cap X_{2,\,j,\,1}\neq 0$. Define

[TABLE]

Then $y\cdot c=a,$ and by transitivity (C), $L_{j+1}$ has non-zero intersection with $L_{-1}^{*}\otimes X_{2,\,j,\,1},$ which is the sum of the following (non-isomorphic) $L_{0}-$submodules:

[TABLE]

[TABLE]

[TABLE]

We first suppose that $S_{j+1}$ has non-zero intersection with $X_{2,\,j+1,\,1},$ and we set

[TABLE]

Then $y\cdot\gamma=\alpha,$ and by transitivity (C), $S_{j+2}$ would have non-zero intersection with $L_{-1}^{*}\otimes X_{2,\,j+1,\,1},$ which is the sum of the following $L_{0}-$submodules:

[TABLE]

[TABLE]

[TABLE]

Focusing on $X_{2,\,j+2,\,1},$ we have $[1,\,x]\cdot(1^{*}\otimes\gamma)=[\gamma,\,x]=2\cdot[x,\,\gamma]=2b,$ and $(\hbox{\rm ad\,}[1,\,x])\cdot b=(\hbox{\rm ad\,}[1,\,x])\cdot(2\cdot x^{*}\otimes xy^{(j)}+1^{*}\otimes y^{(j)})=2\cdot[1,\,xy^{(j)}]+[y^{(j)},\,x]=y^{(j-1)}\neq 0,$ so by Lemma 6.22,

[TABLE]

Focusing on $X_{2,\,j+2,\,3},$ we have $(\hbox{\rm ad\,}[1,\,x])\cdot(x^{*}\otimes\alpha)=[1,\,\alpha]=a,$ and $(\hbox{\rm ad\,}[1,\,x])\cdot a=(\hbox{\rm ad\,}[1,\,x])\cdot(1^{*}\otimes xy^{(j)})=[xy^{(j)},\,x]=xy^{(j-1)}\neq 0,$ so by Lemma 6.22 again,

[TABLE]

It follows that if $X_{2,\,j+1,\,1}\subset S_{j+1},$ then $X_{2,\,j+2,\,2}\subset L_{j+2}.$ Note that

[TABLE]

Since also $[L_{-2},\,X_{2,\,j+1,\,1}]=0,$ it follows that if $L_{j+2}\cap X_{2,\,j+2,\,2}\neq 0,$ then we have by Lemma 6.24 that $(\hbox{\rm ad\,}1)^{3}$ is an isomorphism from $L_{j+2}\cap X_{2,\,j+2,\,2}$ to $S_{j-1}:$

[TABLE]

Since $X_{2,\,j+2,\,2}\cong S_{j-1}$ as an $L_{0}-$module, and since the construction of the “$X$”modules depends only on the $L_{0}-$module properties of $S_{j-1}$ and $L_{-1}^{*},$ and $L_{-1}$ (and its bracket with itself, $L_{-2}$) only interact with the first factor of the tensors, we can repeat the process with $X_{2,\,j+2,\,2}$ in place of $S_{j-1}$ and continue to repeat it until we arrive at the highest gradation space. Since $(\hbox{\rm ad\,}[1,\,x])\cdot(1^{*}\otimes xy^{(j)})=xy^{(j-1)}\neq 0,$ it follows that $[L_{-2},\,X_{2,j,1}]\neq 0.$ Consequently, when we arrive at the highest gradation space, $L_{-2}$ will have non-zero bracket with either $L_{r-2}$, or $L_{r-1}$, or $L_{r}$. Then by Lemma 6.12, we would have

[TABLE]

Since (6.54) is what we are presently trying to establish, we may assume in what follows that $L_{j+2}\cap X_{2,\,j+2,\,2}=0,$ which equation, together with (6.49) and (6.50), shows that $L_{-1}^{*}\otimes X_{2,\,j+1,\,1}\cap L_{j+2}=0,$ so that we may assume in what follows that $L_{j+1}$ has zero intersection with $X_{2,\,j+1,\,1}.$

Next, suppose that $X_{2,\,j+1,\,2}\cap L_{j+1}\neq 0.$ Then $\tilde{X}_{2,\,j+1,\,2}\cap L_{j+1}\neq 0,$ where

[TABLE]

since $\tilde{X}_{2,\,j+1,\,2}$ is the irreducible $L_{0}-$submodule of the indecomposable $L_{0}-$module $X_{2,\,j+1,\,2}.$ Set

[TABLE]

Then $y\cdot\delta=0,$ and $L_{-1}^{*}\otimes\tilde{X}_{2,\,j+1,\,2}$ is the sum of the following irreducible (non-isomorphic) $L_{0}-$submodules

[TABLE]

and

[TABLE]

If, in addition, we set

[TABLE]

then $y\cdot\zeta=\epsilon,$ and

[TABLE]

where

[TABLE]

Note that

[TABLE]

Consider $X_{2,\,j+2,\,4}.$ We have

[TABLE]

and

[TABLE]

so by Lemma 6.22,

[TABLE]

Next consider $X_{2,\,j+2,\,5}.$ Here we have

[TABLE]

and (from above) $(\hbox{\rm ad\,}[1,\,x])\cdot a\neq 0,$ so by Lemma 6.22 again,

[TABLE]

It follows that

[TABLE]

This, from above, implies that

[TABLE]

We next suppose that $X_{2,\,j+1,\,3}\cap L_{j+1}\neq 0$. We set

[TABLE]

Then $y\cdot\iota=0,$ and $L_{-1}^{*}\otimes X_{2,\,j+1,\,3}$ contains the following two (non-isomorphic) $L_{0}-$submodules:

[TABLE]

and

[TABLE]

Furthermore, $L_{-1}^{*}\otimes X_{2,\,j+1,\,3}=X_{2,\,j+2,\,7}+X_{2,\,j+2,\,9},$ where, modulo its irreducible submodule $X_{2,\,j+2,\,8},$ $X_{2,\,j+2,\,9}$ is spanned by $(x^{(2)})^{*}\otimes\eta,$ $(x^{(2)})^{*}\otimes\theta+2\cdot 1^{*}\otimes\eta,$ and $(x^{(2)})^{*}\otimes\iota+x^{*}\otimes\eta+1^{*}\otimes\theta.$ Now,

[TABLE]

so, since from above we know that $(\hbox{\rm ad\,}[1,\,x])\cdot a$ and $(\hbox{\rm ad\,}[1,\,x])\cdot b$ are both non-zero, it follows from Lemma 6.22 that

[TABLE]

Note that

[TABLE]

Focusing on $X_{2,\,j+2,\,7}$ we have

[TABLE]

so, since, again, from above we know that $(\hbox{\rm ad\,}[1,\,x])\cdot a$ is not zero, it follows from Lemma 6.22 that

[TABLE]

It follows from (6.65), (6.66), and (6.67) that $L_{-1}^{*}\otimes X_{2,\,j+1,\,3}$ has zero bracket with $L_{-2}.$

Now suppose that $X_{2,\,j,\,3}\cap L_{j}\neq 0.$ Set

[TABLE]

Then $y\cdot f=d,$ and $L_{-1}^{*}\otimes X_{2,\,j,\,3}$ contains the following submodules:

[TABLE]

[TABLE]

Furthermore, $L_{-1}^{*}\otimes X_{2,\,j,\,3}=X_{2,\,j+1,\,6}+X_{2,\,j+1,\,5},$ where, modulo $X_{2,\,j+1,\,4},$ $X_{2,\,j+1,\,6}$ is spanned by $\langle 1^{*}\otimes e+x^{*}\otimes d+(x^{(2)})^{*}\otimes f\rangle$ and $\langle 2\cdot(x^{(2)})^{*}\otimes e+x^{*}\otimes f,\,x^{*}\otimes e\rangle.$

Focusing first on $X_{2,\,j+1,\,4},$ we set

[TABLE]

Then $y\cdot\mu=0,$ and $L_{-1}^{*}\otimes X_{2,\,j+1,\,4}$ is the sum of the following three $L_{0}-$submodules:

[TABLE]

and

[TABLE]

and

[TABLE]

The irreducible $L_{0}-$submodules are the one-, two-, and three-dimensional subspaces noted respectively above. In particular, none of $X_{2,\,j+2,\,10}$, $X_{2,\,j+2,\,11}$, or $X_{2,\,j+2,\,12}$ is $L_{0}-$isomorphic to either of the others.

In the case of $X_{2,\,j+2,\,10},$ we have $(\hbox{\rm ad\,}[1,\,x])\cdot((x^{(2)})^{*}\otimes\lambda+x^{*}\otimes\mu)=[1,\,\mu]=d,$ and $(\hbox{\rm ad\,}[x,\,x^{(2)}])\cdot d=[xy^{(j)},\,x^{(2)}]=2x^{(2)}y^{(j-1)}\neq 0,$ so by Lemma 6.22,

[TABLE]

In the case of $X_{2,\,j+2,\,11},$ we have $(\hbox{\rm ad\,}[1,\,x])\cdot(2\cdot x^{*}\otimes\kappa+1^{*}\otimes\lambda)=2\cdot[1,\,\kappa]+[\lambda,\,x]=2e,$ and $2\cdot(\hbox{\rm ad\,}[1,\,x])\cdot e=(\hbox{\rm ad\,}[1,\,x])\cdot((x^{(2)})^{*}\otimes xy^{(j)}+2\cdot x^{*}\otimes y^{(j)})=2\cdot[1,\,y^{(j)}]=x^{(2)}y^{(j-1)}\neq 0,$ so by Lemma 6.22,

[TABLE]

In the case of $X_{2,\,j+2,\,12},$ we have $(\hbox{\rm ad\,}[x,\,x^{(2)}])\cdot(x^{*}\otimes\lambda)=[\lambda,\,x^{(2)}]\equiv d,$ and, as above, $(\hbox{\rm ad\,}[x,\,x^{(2)}])\cdot d=2x^{(2)}y^{(j-1)}\neq 0,$ so by Lemma 6.22,

[TABLE]

Thus, by transitivity (C),

[TABLE]

It follows that since $X_{2,\,j+1,\,4}$ is the irreducible $L_{0}-$submodule of the indecomposable $L_{0}-$module $X_{2,\,j+1,\,6},$ we must also have

[TABLE]

Focusing next on $X_{2,\,j+1,\,5},$ we set

[TABLE]

Then $y\cdot\pi=0,$ and $L_{-1}^{*}\otimes X_{2,\,j+1,\,5}$ contains the following (non-isomorphic) $L_{0}-$submodules:

[TABLE]

and

[TABLE]

Furthermore, $L_{-1}^{*}\otimes X_{2,\,j+1,\,5}=X_{2,\,j+2,\,13}+X_{2,\,j+2,\,15},$ where modulo $X_{2,\,j+2,\,14}$, $X_{2,\,j+2,\,15}$ is spanned by $(x^{(2)})^{*}\otimes\nu$, $2\cdot 1^{*}\otimes\nu+(x^{(2)})^{*}\otimes\xi,$ and $x^{*}\otimes\nu+1^{*}\otimes\xi+(x^{(2)})^{*}\otimes\pi.$ We have

[TABLE]

so, since from above $(\hbox{\rm ad\,}[x,\,x^{(2)}])\cdot d=2x^{(2)}y^{(j-1)}\neq 0,$ it follows from Lemma 6.22 that

[TABLE]

Focusing on $X_{2,\,j+2,\,14},$ we have $(\hbox{\rm ad\,}[1,\,x])\cdot(2\cdot(x^{(2)})^{*}\otimes\nu+x^{*}\otimes\xi)=[1,\,\xi]=e,$ and $(\hbox{\rm ad\,}[1,\,x])\cdot e=[1,\,y^{(j)}]=2x^{(2)}y^{(j-1)}\neq 0,$ so by Lemma 6.22,

[TABLE]

It follows from (6.81) and (6.82) that

[TABLE]

Note that

[TABLE]

We have observed before that $L_{j+2}$ is contained in $L_{-1}^{*}\otimes L_{-1}^{*}\otimes L_{-1}^{*}\otimes S_{j-1},$ whose intersection with $L_{j+2}$ has, in view of (6.51), (6.56), (6.66), and (6.84), zero bracket with $L_{-2}.$ In other words, $[L_{-2},\,L_{j+2}]=0.$ But this contradicts Lemma 6.23, and shows that $j\not\equiv 2$ (mod 3).

When $j\equiv 0\,(\hbox{mod 3)},$ there is only one irreducible three-dimensional $L_{0}$-submodule $Q_{j}$ of $(L_{-1})^{*}\otimes S_{j-1}$. It is spanned by

[TABLE]

and is the only $L_{0}-$submodule of $(L_{-1})^{*}\otimes S_{j-1}$ which has zero bracket with $L_{-2}.$ By what we observed above after (6.26), $Q_{j}\subseteq S_{j},$ Then $Q_{j}$ must equal $H(2:\mathbf{n},\omega)_{j}$ and the analysis of $L_{-1}^{*}\otimes Q_{j}$ would be the same as that of $L_{-1}^{*}\otimes S_{j-1}$ when $j\equiv 1$ (mod 3). In particular, $S_{j+1}\cap L_{-1}^{*}\otimes Q_{j}\neq 0.$ If $S_{j+1}$ were to contain an $L_{0}-$submodule $\Gamma_{j+1}$ corresponding to $X_{1,\,j,\,1}$ (which has non-zero bracket with $L_{-2}$), we could consider the Lie algebra generated by $L_{\leq 0}\oplus S_{1}\oplus\cdots\oplus S_{j-1}\oplus Q_{j}\oplus\Gamma_{j+1}$ and obtain a Lie algebra in which the minimal gradation degree $i>2$ such that $[L_{-2},\,S_{i}]\neq 0$ is greater than $j,$ as required.

We can therefore assume that $S_{j+1}$ contains an $L_{0}-$submodule $Q_{j+1}$ corresponding to $X_{1,j,2}$ (which has zero bracket with $L_{-2}$), so that $S_{j+2}\cap L_{-1}^{*}\otimes Q_{j+1}\neq 0.$ If $S_{j+2}$ were to contain an $L_{0}-$submodule $\Delta_{j+2}$ corresponding to $X_{1,\,j+1,\,1}$ or $X_{1,\,j+1,\,3}$ (both of which have non-zero bracket with $L_{-2}$), we can consider the Lie algebra generated by $L_{\leq 0}\oplus S_{1}\oplus\cdots\oplus S_{j-1}\oplus Q_{j}\oplus Q_{j+1}\oplus\Delta_{j+2}$ and obtain a Lie algebra in which the minimal gradation degree $i>2$ such that $[L_{-2},\,S_{i}]\neq 0$ is greater than $j,$ as required. We can therefore assume that $S_{j+2}$ contains an $L_{0}-$submodule $Q_{j+2}$ corresponding to $X_{1,\,j+1,\,2}$ (which has zero bracket with $L_{-2}$), so that $S_{j+3}\cap L_{-1}^{*}\otimes Q_{j+2}\neq 0,$ etc.

In view of Lemma 6.24 (which shows that $Q_{j+2}\cong Q_{j-1}$), we can repeat this process to either obtain a Lie in which the minimal gradation degree $i>2$ such that $[L_{-2},\,S_{i}]\neq 0$ is successively greater, or to eventually find that the highest gradation space $S_{s}$ of $S$ contains an $L_{0}-$submodule $Q_{s}$ which has zero bracket with $L_{-2}.$ However, by Lemma 2.3, $S_{s}$ is irreducible as an $L_{0}-$module, so we would have that $[L_{-2},\,S_{s}]=0.$ Since $S$ is generated by $S_{s}$ and $L_{-1},$ it would follow from the construction of $Q_{s}(=S_{s})$ (See also Lemma 2.3.) that $[L_{-2},\,S_{>0}]=0,$ to contradict, for example, (6.1).

Whatever the residue of $j$ modulo three, then, we have shown that $L$ can be assumed to contain a Lie subalgebra such that if $i\geqq 2$ is minimal such that $L_{-2}$ has non-zero bracket with the $i^{th}$ gradation space of that subalgebra, then $i$ is greater than $j.$

We can therefore conclude the following: Let $r^{\prime}=r$ if there have been no replacements of $L$, or let $r^{\prime}$ be the height of the most recent replacement of $L$ otherwise. Then, in either case, if $j\geqq 2$ is minimal such that $[L_{-2},\,S_{j}]\neq 0,$ and $j<r^{\prime}-2,$ we can find a subalgebra of $L$ such that the smallest $i\geqq 2,$ such that $L_{-2}$ has non-zero bracket with the $i^{th}$ gradation space of that subalgbra, is greater than $j.$ Thus, by induction, $L$ contains a subalgebra such that the smallest $i\geqq 2$ such that $L_{-2}$ has non-zero bracket with the $i^{th}$ gradation space of the subalgbra, is at least $r^{\prime}-2.$ Thus, by Lemma 6.12, $(j=)\,k\leqq 2,$ and we can now conclude that in any case

[TABLE]

To show that $j=1,$ we will, for a contradiction, assume that $j=2.$ We begin by using an inductive argument from [13, Lemma 2.14] to show that the centralizer of $S_{s}$ in $L_{<0}$ is zero. Denote by $Z$ the centralizer of $S_{s}$ in $L$. Then $Z=\oplus Z_{i}$ is a homogeneous subspace of $L.$ Since $S_{s}$ is stable under $\hbox{\rm ad\,}L_{\geqq 0}$, $Z$ is, as well. The component $Z_{-1}$ is an $L_{0}-$submodule of $L_{-1},$ and $Z_{-1}$ by definition has zero bracket with $S_{s}.$ Thus, by (B) and (C),

[TABLE]

We will show that (See [13, Lemma 2.14].)

[TABLE]

for all $i<0$, proceeding by (downward) induction on $i.$ Assume that $i<-1,$ and that

[TABLE]

By analogy with the previous argument, we here (where $j=2$) assume that $L$ is generated by $L_{<0}+L_{0}+S_{1}+S_{2}.$ If we show that $[S_{1},\,Z_{i}]=[S_{2},\,Z_{i}]=0,$ then $[{L}_{>0},Z_{i}]=0,$ so that

[TABLE]

would be an ideal of ${L}$ properly contained in ${S}$, the simple ideal of ${L}$, to contradict the simplicity of ${S}.$ Consequently, to verify that $Z_{i}=0$, we need only show that $[S_{1},\,Z_{i}]=[S_{2},\,Z_{i}]=0.$

We first show that $[S_{1},\,Z_{i}]=0.$ Indeed, when $i=-2,$ we have by (6.5) that $[Z_{-2},\,S_{1}]=0,$ and when $i<-2,$ we have by (6.89) that $[S_{1},\,Z_{i}]\subseteq Z_{i+1}=0.$ Similarly, $[S_{2},\,Z_{i}]\subseteq Z_{2+i}=0$ when $i<-2.$ If $i=-2$ and $[S_{2},\,Z_{-2}]\neq 0,$ then since by Lemma 2.23 $L_{-2}$ is an irreducible $L_{0}-$module, we would have by Lemma 6.16 that $0\neq[S_{2},\,Z_{-2}]=[S_{2},\,L_{-2}]=[S_{1},\,L_{-1}]=L_{0}.$ (See Corollary 1.7.) But then we would have $[L_{0},\,S_{s}]=[[S_{2},\,Z_{-2}],\,S_{s}]=[S_{2},\,[Z_{-2},\,S_{s}]]=0,$ to contradict [13, Lemma 2.13]. (See also Lemma 2.28.) This contradiction shows that $[Z_{-2},\,S_{2}]=0,$ and completes the verification of (6.88).

Since we are assuming that $j=2$, we have

[TABLE]

and

[TABLE]

We will first address the case in which $q>r,$ and then deal with the case in which $r\geqq q.$ Thus, assume first that

[TABLE]

Since (as we observed at the beginning of this section) $S_{1}$ is an irreducible $L_{0}-$module, it follows from (6.88) that

[TABLE]

Then by (6.91),

[TABLE]

If both sides of the equation were non-zero, then by irreducibility (B) they must both equal $L_{-1}.$ Then we would have, in view of Lemma 2.28 and (6.88) and (6.91), that, if $r>2,$

[TABLE]

which would imply that $[S_{1},\,S_{1}]\neq 0,$ contrary to (6.14). If, on the other hand, $r=2,$ then the third line above becomes “$\subseteq[[L_{-r+1},\,S_{r}],\,S_{1}],\,[L_{-2},\,S_{r}]]+[[L_{-r+1},\,S_{r}],\,S_{1}],$” which leads to a similar contradiction. Thus, we can conclude that

[TABLE]

Now note that under the present assumptions, $r$ cannot equal four. Indeed, by Lemma 2.28, $[[L_{-1},\,S_{1}],\,S_{r}]=[[L_{-1},\,S_{r}],\,S_{1}]$ can be assumed to be non-zero and

[TABLE]

If $r$ were four, we would have by (6.97), (6.91), and (6.14) that

[TABLE]

to contradict (6.88). Similarly, (6.14) and (6.97) contradict one another when $r=2$. Lastly, $r$ cannot equal three, either; indeed, by (6.91), $[[L_{-2},\,L_{-2}],\,L_{3}]\subseteq[L_{-2},\,L_{1}]=0,$ so if $r$ were three, then by (6.89), $[L_{-2},\,L_{-2}]=0.$ Now suppose that $M_{2}$ is a non-zero irreducible $L_{0}-$submodule of $L_{2}.$ Then by (6.14), $[L_{-1},\,[L_{1},\,M_{2}]]=[[L_{-1},\,L_{1}],\,M_{2}],$ which is non-zero by (6.17). It follows that $[L_{1},\,M_{2}]$ is a non-zero $L_{0}-$submodule of $S_{3},$ which is $L_{0}-$irreducible by Lemma 2.3. Then by (6.91) and (6.92), we have

[TABLE]

so that $[L_{-2},\,M_{2}]\neq 0.$ By (6.17) again, $[L_{-2},\,[L_{-2},\,M_{2}]]\neq 0.$ Consequently, in the Lie algebra $L_{-2}\oplus L_{0}\oplus M_{2},$ we have $B(L_{-2})_{1}\neq 0.$ Then (See Proposition 1.6.) setting $L=B(L_{-2})$ and $W=L_{1}$ in the latter option of Lemma 1.8, we arrive at a contradiction.

Since $r\geqq j=2,$ it follows that

[TABLE]

If we now bracket (6.96) by $S_{r-2}$, we obtain

[TABLE]

which (since in view of our assumption that $j=k=2$, we have $[L_{-2},\,S_{r}]\neq 0$) would imply that

[TABLE]

since otherwise it would equal $L_{-1}$ by irreducibility (B), and transitivity (C) would be violated.

If we next bracket (6.96) by $S_{r-1},$ we get

[TABLE]

Now suppose that

[TABLE]

Then we would have

[TABLE]

which by (B) and (C) would imply that

[TABLE]

which in turn would (in view of (D)) imply that

[TABLE]

the right-hand side of which we have assumed to be zero. But then we would have $[L_{-r-1},\,S_{r-1}]=0,$ so

[TABLE]

by (6.88) and irreducibility (B), to contradict transitivity (C). We conclude that

[TABLE]

to imply that (See (6.102).)

[TABLE]

so that by irreducibility (B),

[TABLE]

and to further imply that (See (6.101).)

[TABLE]

Then, since by Lemma 2.23 $L_{-2}$ is an irreducible $L_{0}-$module, it follows that

[TABLE]

and, in view of (6.103), that also

[TABLE]

If $[L_{-r},\,[L_{-2},\,S_{r}]]=0$, we would have by (6.104) that

[TABLE]

to contradict transitivity (C), since we are assuming that $(k=)j=2.$ Consequently,

[TABLE]

so that, as above

[TABLE]

By (6.107),

[TABLE]

Now, if $q>r+1,$ by (6.88) and Lemma 2.23, we have $[L_{-r-2},\,S_{r}]=L_{-2},$ so

[TABLE]

If, on the other hand, $q=r+1,$ then, since by (6.88) and irreducibility (B) $[L_{-r-1},\,S_{r}]=L_{-1},$ we have

[TABLE]

since by (D), $[L_{-i},\,S_{i}]\subseteq[L_{-1},\,S_{1}]$ for all $i$, $1\leqq i\leqq\min\{q,\,r\},$ so that in particular, $[L_{-2},\,S_{2}]\subseteq[L_{-1},\,S_{1}]=[L_{-r},\,S_{r}],$ so that (6.113) holds when $q=r+1,$ also. Thus, we have by (6.112) and (6.113), respectively, (since we are assuming that $j=2$)

[TABLE]

By Lemma 6.16, we may assume that $[L_{-2},\,S_{2}]=[L_{-1},\,S_{1}]$ and thus conclude that

[TABLE]

Then we have from (6.96), (6.111) and (6.116) that (See also (6.100).)

[TABLE]

to contradict (6.88). This final contradiction enables us to conclude that if $q>r$,

[TABLE]

Let us now assume that $q\leqq r.$ For $i\geqq 1,$ set

[TABLE]

Then we would have, by (6.91), that $[L_{-2},\,M_{-q+i}]=0$ for all $i.$ Furthermore, we would then have by induction that

[TABLE]

Now, if $[[L_{-i},\,S_{q-1}],\,M_{-q+i}]$ were not equal to zero, then it would have to equal $L_{-1}$ by the irreducibility (B) of $L$, and we would have, by Lemmas 2.1 and 2.4 and (6.125) that for $1<i<q-1,$

[TABLE]

This contradiction shows that

[TABLE]

so that for $i<\frac{q}{2},$ we have by (6.125) that

[TABLE]

Since for $i=\frac{q}{2}$, we have $[L_{-q+i},\,M_{-q+i}]=[L_{-\frac{q}{2}},\,M_{-q+\frac{q}{2}}]=0,$ also by (6.125), the above-displayed calculation is valid for $i\leqq\frac{q}{2}.$ Consequently, if $[L_{-q+i},$ $[L_{-i},\,S_{q-1}]]$ were not zero, then by the irreducibility (B) of $L$, it would equal $L_{-1}$, so that by Lemma 2.1, $M_{-q+i}$ would equal zero for $i\leqq\frac{q}{2}$. In [2] and [7], we proved the Main Theorem for all cases less than or equal to three. Conseqently, we may assume that $q\geqq 4,$ so that $M_{-q+i}$ would equal zero for all $i\leqq 2.$ However, when $i=1,$ it follows from Lemma 2.8 that $L_{-q+1}=[L_{-q},\,S_{1}]=M_{-q+1}$, so that $M_{-q+2}$ $=$ $[[L_{-q},\,S_{1}],\,S_{1}]$ $=$ $[L_{-q+1},\,S_{1}].$ Moreover, it follows from Lemma 1.8 (with $W=M_{-q+1}$ or $W=L_{-q}$) that $M_{-q+2}\neq 0$. We conclude that $[L_{-q+i},$ $[L_{-i},\,S_{q-1}]]=0$ for $i\leqq 2.$ In particular,

[TABLE]

Now, if $[L_{-q+2},\,L_{-2}]$ were not equal to zero, it would, by Lemma 2.4, equal $L_{-q},$ and we would have $[L_{-q},\,S_{q-1}]=0,$ to contradict Lemma 2.3. Thus, by Lemma 2.8, we must have

[TABLE]

By Lemma 6.16, $[L_{-2},\,S_{2}]=[L_{-1},\,S_{1}]$, so we have, in view of Lemmas 2.8,

[TABLE]

to contradict Lemma 2.1. This contradiction shows that here, too, (6.123) must be true.

Let $\tilde{L}$ be as in the statement of Lemma 2.29, and let $V_{1}$ be any irreducible $L_{0}-$submodule of $S_{1}.$ Because ${\tilde{L}}_{>0}$ is generated by $S_{1}$, we have $\{1\}-$transitivity (vi) in the negative part of $\tilde{L}/M(\tilde{L}),$ and since by (6.123) $[L_{-2},\,S_{1}]$ $\neq 0,$ we can apply Lemma 2.24 to $\tilde{L}/M(\tilde{L})$ to conclude that

[TABLE]

Consequently, if $\tilde{\tilde{L}}$ is the Lie algebra generated by $L_{-1}\oplus L_{0}\oplus V_{1},$ then the depth $\tilde{\tilde{q}}$ of $\tilde{\tilde{L}}/M(\tilde{\tilde{L}})$ (Again, see Theorem 1.3.) is (also) greater than one. Let $\tilde{\tilde{r}}$ denote the height of $\tilde{\tilde{L}}/M(\tilde{\tilde{L}}).$

Case I: $\tilde{\tilde{q}}<\tilde{\tilde{r}}.$ Suppose first that the depth $\tilde{\tilde{q}}$ is less than $\tilde{\tilde{r}}.$ If $\tilde{\tilde{q}}$ is less than $q,$ then we can apply the Main Theorem to $\tilde{\tilde{L}}/M(\tilde{\tilde{L}})$ to (inductively) conclude that the representation of $\tilde{\tilde{L}}_{0}=L_{0}$ on $\tilde{\tilde{L}}_{-1}=L_{-1}$ is restricted, and see that the Main Theorem is true in this case. If $\tilde{\tilde{q}}=q,$ the Main Theorem follows from [2], [7], Sections 4 and 5, and Lemma 3.1.

Case II: $\tilde{\tilde{q}}\geqq\tilde{\tilde{r}}.$ From now on, then, we will assume that $\tilde{\tilde{r}}$ is less than or equal to $\tilde{\tilde{q}}.$

Case IIA: $\tilde{\tilde{q}}\geqq\tilde{\tilde{r}}>1.$ Since $\tilde{\tilde{r}}>1,$ it follows from the definition of $\tilde{\tilde{L}}$ that $[V_{1},\,V_{1}]\neq 0,$ so that by Lemma 2.27, $\hbox{\rm Ann\,}_{L_{0}}V_{1}=0.$ Clearly, $B(V_{1})$ (See Section 3.) satisfies the conditions of the Main Theorem. (Condition (E), for example, follows from the transitivity (C) of $L,$ which shows that actually $M(B(V_{1}))=0.$) Now, $q$ is assumed to be greater than one, so we have

[TABLE]

so $B(V_{1})$ is not degenerate. Consequently, (since the depth $\tilde{\tilde{r}}$ of $B(V_{1})$ is less than or equal to $\tilde{\tilde{q}}$ which is less than or equal to $q$) we can, as in Case I, apply the Main Theorem to conclude that the representation of $L_{0}^{\prime}$ on $V_{1}$ is restricted, and that the representation of $L_{0}^{\prime}$ on $B(V_{1})_{1}=L_{-1}$ is restricted, as well (since $L_{-1}=B(V_{1})_{1}\subseteq\hbox{\rm Hom\,}(V_{1},\,L_{0});$ see also Lemma 1.2).

Case IIB: $\tilde{\tilde{q}}\geqq\tilde{\tilde{r}}=1.$ Since $\tilde{\tilde{r}}=1,$

[TABLE]

By Corollary 1.5, either $L$ is degenerate and $r=1,$ contrary to hypothesis, or $L$ and $\tilde{\tilde{L}}/M(\tilde{\tilde{L}})$ are not degenerate, in which case we can apply Proposition 1.6 to $B((\tilde{\tilde{L}}/M(\tilde{\tilde{L}}))_{1})=B(V_{1})$ to conclude that $B(V_{1})$ is isomorphic to $L(\epsilon)$ or $M,$ or is Hamiltonian (i.e., between $H(2:\mathbf{n},\omega)$ and $CH(2:\mathbf{n},\omega)$). But in those cases, the one-component $(B(V_{1}))_{1}=L_{-1}$ is abelian; i.e., $[L_{-1},\,L_{-1}]\subseteq M(\tilde{\tilde{L}}),$ so that by (D), $L_{-2}=[L_{-1},\,L_{-1}]\subseteq M(\tilde{\tilde{L}}),$ so $[L_{-2},\,V_{1}]=0,$ to contradict (6.134).

The proof of the Main Theorem is now complete.