LIE ALGEBRAS GENERATED BY EXTREMAL ELEMENTS

ARJEH M. COHEN, ANJA STEINBACH, ROSANE USHIROBIRA, DAVID WALES

Abstract. We study Lie algebras generated by extremal elements (i.e., elements spanning inner ideals of L) over a field of characteristic distinct from 2. We prove that any Lie algebra generated by a finite number of extremal elements is finite dimensional. The minimal number of extremal generators for the Lie algebras of type A_{n} (n ≥ 1), B_{n} (n ≥ 3), C_{n} (n ≥ 2), D_{n} (n ≥ 4), E_{n} (n = 6, 7, 8), F_{4} and G_{2} are shown to be n + 1, n + 1, 2n, n, 5, 5, and 4 in the respective cases. These results are related to group theoretic ones for the corresponding Chevalley groups.

1. Introduction

Let k be a field of characteristic not 2, and L a Lie algebra over k. We study the rôle of extremal elements in L, that is, those x ∈ L with [x, [x, L]] ⊆ kx. Since an inner ideal of L is by definition ([1]) a linear subspace I of L such that [I, [I, L]] ⊆ I, this amounts to kx being an inner ideal. By E or, if necessary to express dependence on L, by E(L), we shall denote the set of all nonzero extremal elements of L.

We are mostly interested here in Lie algebras generated by extremal elements. The motivation stems from the fact that long root elements are extremal in Lie algebras of Chevalley type (i.e., those Lie algebras over k that are given by the multiplication table of a Chevalley basis, coming from a simple Lie algebra in characteristic 0). They were used by Chernousov [4] in his proof of the Hasse principle for E_{8}. The associated root groups were studied in a more abstract group theoretic setting in [14]. Sandwiches, that is, elements x ∈ L with [x, [x, L]] = 0 are extremal elements of a special kind; they are prominent in the classification of finite dimensional simple modular Lie algebras over algebraically closed fields of characteristics 5 and 7 and could well be useful in a similar way for other positive characteristics, cf. [10].

We shall be particularly concerned with Lie algebras which are generated by a finite set of extremal elements. In §4, we show that in this case the Lie algebra has finite dimension. For this we use the work of Zel’manov and Kostrikin [15]. It follows from their results that there is a universal Lie algebra L_{r} generated by a finite number of sandwich elements x_{1}, …, x_{r} ∈ L_{r}; it is nilpotent and of finite dimension. This leads to the question of a description of these universal Lie algebras. Some information about these for small values of r is given.

Using functions arising from the definition of extremal elements, we are able to define an associative bilinear form f on L. In particular, if x ∈ E and y ∈ L, then [x, [x, y]] is a multiple of x and so we can set [x, [x, y]] = f(x, y)x. In §9, we discuss the connection of f with the Killing form, and give some information about the radical of this form. In §8, we determine the minimal number of generators by extremal elements for the Lie algebras of Chevalley type. Exponentiation by ad_{x} for x an extremal element gives an automorphism of L. The group generated by these automorphisms satisfies some conditions defined by Timmesfeld [14], which we discuss in §10.

Date: March 29, 1999.

We gratefully acknowledge some extremely useful discussions with Gabor Ivanyos as well as computer verifications of our results on L_{5} by Willem de Graaf.

2. General properties of extremal elements

Throughout the remainder of this paper, k is a field of characteristic distinct from 2 and L is a Lie algebra over k. By linearity of [x, [x, y]] in y, an element x of L is extremal if and only if there is a linear functional f_{x} : y ↦ f_{x}(y) on L such that, for all y ∈ L, we have

[x, [x, y]] = f_{x}(y)x. (1)

Note that f_{x}(y) = 0 if x and y commute. We also write f(x, y) instead of f_{x}(y).

Lemma 2.1. If x, y ∈ E then f_{x}(y) = f_{y}(x).

Proof. By (1), [y, [x, [x, y]]] = f_{x}(y)[y, x], but by Jacobi, antisymmetry, and (1),

[y, [x, [x, y]]] = −[x, [[x, y], y]] − [[x, y], [y, x]] = −[x, [y, [y, x]]] = −f_{y}(x)[x, y].

This shows that the lemma holds if x and y do not commute. Otherwise, f_{x}(y) = 0 = f_{y}(x). □

We give some general identities and properties. At least the first two rewriting rules go back to Premet, see [4].

Lemma 2.2. Let x ∈ E. Then, for all y, z ∈ L,

(2)  2[[x, y], [x, z]] = f_{x}([y, z])x + f_{x}(z)[x, y] − f_{x}(y)[x, z],
(3)  2[x, [y, [x, z]]] = f_{x}([y, z])x − f_{x}(z)[x, y] − f_{x}(y)[x, z].

Proof. Let y, z ∈ L. Then

[[x, y], [x, z]] = −[[y, [x, z]], x] − [[[x, z], x], y] = [x, [y, [x, z]]] + f_{x}(z)[x, y].

Similarly, interchanging y and z,

[[x, y], [x, z]] = −[x, [z, [x, y]]] − f_{x}(y)[x, z].

Furthermore,

[x, [z, [x, y]]] = −[x, [x, [y, z]]] − [x, [y, [z, x]]] = −f_{x}([y, z])x + [x, [y, [x, z]]].

Hence

[[x, y], [x, z]] = [x, [y, [x, z]]] + f_{x}(z)[x, y],
[[x, y], [x, z]] = f_{x}([y, z])x − [x, [y, [x, z]]] − f_{x}(y)[x, z].

From this the two equations follow. □

Lemma 2.3. Let x, y ∈ E and z ∈ L. Then

2[[x, y], [x, [y, z]]] = f_{y}(z)f_{x}(y)x + f_{x}([y, z])[x, y] − f_{x}(y)[x, [y, z]], (4)

(5) 2[[x, y], [[x, y], z]] = (f_{x}([y, z]) − f_{y}([x, z]))[x, y]

+ f_{x}(y) (f_{x}(z)y + f_{y}(z)x − [y, [x, z]] − [x, [y, z]]) .

Proof. Expression (4) is the first identity of Lemma 2.2 with [y, z] replacing z. Identity (5) follows from two applications of (4) and Jacobi:

2[[x, y], [[x, y], z]] = 2[[y, x], [y, [x, z]]] + 2[[x, y], [x, [y, z]]]
= f_{x}(z)f_{x}(y)y + f_{y}([x, z])[y, x] − f_{x}(y)[y, [x, z]]
+f_{y}(z)f_{x}(y)x + f_{x}([y, z])[x, y] − f_{x}(y)[x, [y, z]]
= (f_{x}([y, z]) − f_{y}([x, z]))[x, y]
+f_{x}(y) (f_{x}(z)y + f_{y}(z)x − [y, [x, z]] − [x, [y, z]]) .
□

The lemmas above also make clear why we are assuming that the characteristic of k is not 2.

Corollary 2.4. If x, y ∈ E with f_{x}(y) = 0 but [x, y] ≠ 0, then [x, y] ∈ E, with f_{[x,y]}(z) = \frac{1}{2}(f_{x}([y, z]) − f_{y}([x, z])) for z ∈ L.

Proof. Immediate from Identity (5) of Lemma 2.3. □

Let x ∈ E. Because ad_{x}^{3} = 0, the exponential of the derivation s ad_{x} for s ∈ k is given by

exp(x, s) := 1 + s ad_{x} +\frac{s^{2}}{2} ad_{x}^{2} .

This is an automorphism of L in view of Identity (2) of Lemma 2.2. Denote by G (or G(L) if necessary) the group of automorphisms of L generated by all exp(x, s) for x ∈ E and s ∈ k.

We use these automorphisms to show that if a Lie algebra is generated (as a Lie algebra) by extremal elements, it is also linearly spanned by extremal elements.

Lemma 2.5. If L is generated as a Lie algebra by extremal elements, then it is linearly spanned by the set E of all extremal elements.

Proof. We use induction on the length of z ∈ L as a bracketing of elements from E (that is, obtained by forming successive brackets of elements from E). It readily reduces to the case where this length is 2, the case of length 1 being z ∈ E. Suppose z = [x, y] with x, y ∈ E. Now let m = exp(x, 1)y = y +[x, y]+ \frac{1}{2}f_{x}(y)x. As exp(x, 1) is an automorphism, m is extremal and so [x, y] is in the linear span of the extremal elements x, m, y. □

We show how the map f defined in the beginning of this section gives rise to a symmetric bilinear associative form.

Theorem 2.6. Suppose that L is generated by E. There is a unique bilinear symmetric form f : L × L → k such that, for each x ∈ E, the linear form f_{x} coincides with y ↦ f(x, y). This form is associative, in the sense that f(x, [y, z]) = f([x, y], z) for all x, y, z ∈ L.

Proof. Take {u_{i}}_{i} to be a basis of L consisting of elements from E. Its existence is guaranteed by Lemma 2.5. Notice that if x is extremal, then [λx, [λx, y]] = λf(x, y)λx and so λx is also extremal with f(λx, y) = λf(x, y). Now suppose x = ∑_{i} λ_{i}u_{i}. Define f_{x} by ∑_{i} λ_{i}f_{u_{i}}. Notice from above this is ∑_{i} f_{λ_{i}u_{i}}. Now suppose ∑_{i} u_{i} = ∑_{i} v_{i} are two ways of writing an element x of L as a sum with v_{1}, …, v_{m}, extremal elements. By Lemma 2.1, for z ∈ E, we have

f_{x}(z) = ∑_{i} f_{u_{i}}(z) = ∑_{i} f_{z}(u_{i}) = f_{z}(∑_{i} u_{i}) = f_{z}(∑_{i} v_{i}) = ∑_{i} f_{z}(v_{i}) = ∑_{i} f_{v_{i}}(z).

Since E spans L, we conclude ∑_{i} f_{u_{i}} = ∑_{i} f_{v_{i}}. Thus, f_{x} is a well defined linear functional. It readily follows that f(x, y) = f_{x}(y) defines a bilinear form. It is symmetric by Lemmas 2.1 and 2.5.

It remains to establish that f is associative. Take x, y, z ∈ E. Interchanging x and y in (5) of Lemma 2.3, we obtain

2[[x, y], [y, [x, z]]] = −f(x, z)f(x, y)y + f(y, [x, z])[x, y] + f(x, y)[y, [x, z]].

On the other hand, by Jacobi and Lemma 2.2,

2[[x, y], [y, [x, z]]] = −2[y, [[x, z], [x, y]]] − 2[[x, z], [[x, y], y]]
= [y, f(x, [y, z])x + f(x, z)[x, y] − f(x, y)[x, z]]
−2f(x, y)[[x, z], y]
= −f(x, [y, z])[x, y] − f(x, z)f(x, y)y + f(x, y)[y, [x, z]].

Suppose now [x, y] ≠ 0. Comparing the coefficients of [x, y] in the two expressions for 2[[x, y], [y, [x, z]]], we find −f(x, [y, z]) = f(y, [x, z]). But then, by symmetry of f, we have f(x, [z, y]) = −f(x, [y, z]) = f(y, [x, z]) = f([x, z], y), proving f(x, [z, y]) = f([x, z], y) whenever [x, y] ≠ 0. Similarly, f(x, [y, z]) = f([x, y], z), whenever [x, z] ≠ 0 and f(y, [x, z]) = f([y, x], z), whenever [y, z] ≠ 0.

Suppose that [x, y] = 0. We have to show f(x, [z, y]) = f([x, z], y). If [x, z] ≠ 0 and [y, z] ≠ 0, then f(x, [z, y]) = 0 = f([x, z], y) by the previous equations. Hence (up to symmetry) we are left with the case [x, y] = 0, [x, z] = 0. Then f([x, z], y) = 0, so we must show that f(x, [z, y]) = 0. By Jacobi, [x, [y, z]] = 0. Applying ad_{x} yields f(x, [y, z]) = 0 as needed.

We conclude that f(x, [z, y]) = f([x, z], y) for all x, y, z ∈ E. Since E linearly spans L, this proves that f is associative. □

Corollary 2.7. If x, y ∈ E with f_{x} = 0 and [x, y] ≠ 0, then [x, y] ∈ E with f_{[x,y]} = 0.

Proof. By Corollary 2.4, we have [x, y] ∈ E. Since f is associative by Theorem 2.6, we find f_{[x,y]}(z) = f([x, y], z) = f(x, [y, z]) = 0 for z ∈ L. □

3. Some examples

We first discuss Lie algebras generated by two extremal elements.

Lemma 3.1. Let L be generated by two extremal elements x, y ∈ E. Then one of the following three assertions holds.

(i) L = kx + ky is Abelian, E = L \ {0}, and f = 0. (ii) L = kx + ky + kz with z = [x, y] ≠ 0, its center Z(L) and commutator subal-

gebra [L, L] coincide with kz and L/Z(L) is Abelian. Thus, L is a Heisenberg algebra. Moreover E = L \ {0} and f = 0.

(iii) L ≅ 𝔰𝔩_{2} and E consists of all nilpotent elements of L. Thus,

E ∪ {0} = kx ∪ ky ∪ ⋃_{δ∈k\{0}} k (δx + δ^{−1}λy + [x, y]) ,

where λ = \frac{1}{2}f(x, y) ≠ 0.

Proof. Clearly, x, y and [x, y] linearly span L. The rest is straightforward. □

Example 3.2. To see that there are real differences in the characteristic 2 case, let us take k = ℤ/2ℤ and look at the case where L is generated by two extremal elements x, y with [x, [x, y]] = x. Then L = kx + k[x, y] + ky is not isomorphic to 𝔰𝔩_{2}, as it is simple whereas 𝔰𝔩_{2} is not.

Next we deal with finite dimensional Lie algebras of Chevalley type. A long root element of such a Lie algebra is an element of the form x_{α} with α a long root, or the image of such an element under a Lie algebra automorphism.

Proposition 3.3. Let L be a Lie algebra of Chevalley type over k. Then E contains the long root elements of L. In particular, L is generated by E. Moreover, if k has characteristic distinct from 2 and 3, every element of E is a long root element.

Proof. Consider a Chevalley basis B of L with respect to a given Cartan subalgebra H of L, write Φ for the corresponding root system, and x_{β} for the element of B associated with a given root β ∈ Φ. Suppose that α is a long root in Φ. Then, for h ∈ H, we have [x_{α}, [x_{α}, h]] = α(h)[x_{α}, x_{α}] = 0. Moreover, for β ∈ Φ, the sum β + 2α is not a root in Φ unless β = −α, so [x_{α}, [x_{α}, x_{β}]] = 0 unless β = −α, in which case the result is a multiple of x_{α}.

To see that L is generated by the long root elements, observe that L is generated by the root elements of the basis B and (by analysis of the Lie rank 2 case) that any short root element can be written as a sum of three long root elements, as is clear from the following computations in the Lie algebras of type B_{2} and G_{2}, respectively.

for B_{2} :  \underbrace{exp(x_{ε_{1}}, 1)x_{−(ε_{1}−ε_{2})}}_{long} = \underbrace{x_{−(ε_{1}−ε_{2})}}_{long} − \underbrace{x_{ε_{2}}}_{short} + \underbrace{x_{ε_{1}+ε_{2}}}_{long}

for G_{2} :  \underbrace{exp(x_{α}, 1)x_{β}}_{long} + \underbrace{exp(x_{α}, −1)x_{β}}_{long} = 2 \underbrace{x_{β}}_{long} −2 \underbrace{x_{2α+β}}_{short}

As for the last assertion, let e ∈ E. Under the given characteristic restriction for k, the proof of Lemma 2.1 of [1] and the paragraph following Step 1 in the proof of Theorem 3.2 of [1] show that there exists a Cartan subalgebra of L with eigenspace ke. Thus e is a root element. If the diagram of L is not simply laced, short root elements of L are easily seen not to be extremal. Hence e is a long root element. □

We expect that, in characteristic 3, each extremal element is also a long root element. Such an assertion might follow from a classification of nilpotent elements in the Lie algebras of Chevalley type, but we have not been able to verify this completely.

4. L is finite dimensional in the finitely generated case

We use the work of Zel’manov and Kostrikin [15] to show the following main result.

Theorem 4.1. If L is generated as a Lie algebra by a finite number of extremal elements, then L is finite dimensional.

Proof. Suppose that L is generated by extremal elements x_{1}, …, x_{r}. Denote by f the associated bilinear form. We first consider the case in which f is identically zero.

Lemma 4.2. Suppose that L is generated by elements x_{1}, …, x_{r} with (ad_{x_{i}})^{2} = 0 for i = 1, …, r. Then L is finite dimensional and is nilpotent. Furthermore, there is a universal such Lie algebra, L_{r}.

Proof. Observe that each x_{i} is in E. By Corollary 2.7 (ad_{w})^{2} = 0 for any bracketing w in x_{1}, …, x_{r}. (We consider also x_{i} as a bracketing.) Since char k ≠ 2, this implies that ad_{w} ad_{u} ad_{w} = 0 for any two of these bracketings, see (3) of Lemma 2.2. In the language of Zel’manov and Kostrikin [15], each ad_{w} is a crust of a thin sandwich, and by [15, Theorem 2] the associative algebra generated by ad_{x_{1}}, …, ad_{x_{r}} is nilpotent. Hence all products of some fixed length, say l, of the ad_{x_{i}} are 0. This yields that L is a nilpotent Lie algebra with bracketings of length l + 1 being 0. In particular, L is finite dimensional.

To see that there is a universal such Lie algebra, let F be the free Lie algebra generated by r elements f_{1}, f_{2}, …, f_{r}. Now factor out by the ideal J generated by all [f_{i}, [f_{i}, u]] where u is a bracketing in the f_{i}. Now F/J is a universal Lie algebra spanned by r extremal elements with f identically 0. This completes the proof of the lemma. □

We now show how this implies Theorem 4.1.

Lemma 4.3. Suppose that L is generated by r extremal elements where the values of f need not all be 0. Then the dimension of L is at most dim L_{r}. In particular L is finite dimensional.

Proof. For a nonzero bracketing in F, its length is understood to be the total number of the f_{i} which appear, counting multiplicities (e.g., the length of [[f_{1}, f_{2}], [f_{1}, f_{3}]] is 4). Choose a basis {w_{1} + J, …, w_{t} + J} of F/J and let U be the linear span of w_{1}, …, w_{t}. Then F = U +J. Consider the surjective homomorphism : F → L determined by f_{i} ↦ x_{i}. We claim that {w̃_{1}, …, w̃_{t}} linearly spans L, i.e., Ũ = L.

Any element of J is a linear combination of terms

[u_{s}, […, [u_{1}, [f_{i}, [f_{i}, u]]] …]] (6)

with u, u_{1}, …, u_{s} bracketings in the f_{i}. Suppose that Ũ is properly contained in L. Then there exists a nonzero bracketing g ∈ F of minimal length with g̃ ∉ Ũ. We may write g as a linear combination of the w_{i} and terms T_{i} as in (6) with all w_{i}, T_{i} of the same length as g. (Note that in the free Lie algebra F, the linear span of the bracketings of a fixed length ℓ trivially intersects the linear span of all bracketings of lengths distinct from ℓ.) Among these terms T_{i} there is a term T = [u_{s}, […, [u_{1}, [f_{i}, [f_{i}, u]]] …]] as in (6) with T̃ ∉ Ũ. Set T_{0} := [u_{s}, […, [u_{1}, f_{i}] …]]. Then f(x_{i}, ũ)T̃_{0} = T̃ ∉ Ũ. Hence f(x_{i}, ũ) ≠ 0 and T̃_{0} ∉ Ũ with T_{0} a bracketing of shorter length than g, a contradiction with the choice of g. □

Combining these two lemmas proves Theorem 4.1. □

Remark 4.4. We have worked out the Lie algebra for up to five generators and summarize the dimensions below. Let L_{r} be the universal Lie algebra generated by r extremal generators subject to f being identically 0. Also, let R_{r} be the free associative algebra over k generated by r elements y_{1}, y_{2}, …, y_{r} for which y_{i}^{2} = 0 and y_{i}wy_{i} = 0 where w is any bracketing in the y_{i}s. Then L_{r}/Z(L_{r}), the image of L_{r} under ad, is a quotient of R_{r} in view of (3) of Lemma 2.2. Our computations have found the values of the corresponding dimensions in the table below.

Number of generators dim L_{r} dim R_{r}
1  1  2
2  3  5
3  8  19
4  28  193
5  537  ?

By way of example, consider the case r = 2. Put A_{1} = ad_{x_{1}} and A_{2} = ad_{x_{2}}. Then a basis for L_{2} is x_{1}, x_{2}, [x_{1}, x_{2}] of size 3. A basis for R_{2} is I, A_{1}, A_{2}, A_{1}A_{2}, A_{2}A_{1}, of size 5, as can be seen immediately. This explains the corresponding entries for r = 2 in the above table.

We will explain in later sections how some of the other entries have been determined. For the remainder we refer to the algorithmic methods described in [12] and [8] using Lyndon words, and to the computational algebra packages GAP [5] (including the FPSLA program by Gerdt & Kornyak), and LiE [9].

Let Rad(f) denote the radical of f, that is, the set of all y ∈ L for which f(y, z) = 0 for all z ∈ L.

Corollary 4.5. If L is simple, then f is non-degenerate and there is no x ∈ L, x ≠ 0, with ad_{x}^{2} = 0. (This means that L is non-degenerate in the sense of [1].)

Proof. We know that f is non-trivial by Lemma 4.2. (Otherwise f is identically 0 and L is nilpotent.) Thus, Rad(f) is a proper ideal of L and so, as L is simple, Rad(f) = 0. This means that f is non-degenerate.

Suppose x ∈ L with ad_{x}^{2} = 0. Then x is extremal and [x, [x, y]] = 0 for all y ∈ L. Hence x ∈ Rad(f). This means x = 0. □

Remark 4.6. By the above corollary and the proof of Theorem 3.2 in [1], for p > 5, the only modular finite dimensional simple Lie algebras over an algebraically closed field of characteristic p and generated by extremal elements, are quotients of those of Chevalley type.

5. The 3 generator case

Suppose that L is generated by three extremal elements x, y, z. We wish to determine the various possibilities for L. The identities of Lemmas 2.2 and 2.3, together with the identity

2[[x, [y, z]], [y, [x, z]]] = −\frac{1}{2} (f_{y}(z)f_{x}([y, z])x + f_{x}([y, z])f_{x}(z)y + f_{x}([y, z])f_{x}(y)z)

− f_{y}(z)f_{x}(z)[x, y] + f_{y}(z)f_{x}(y)[x, z] − f_{x}(z)f_{x}(y)[y, z], show that

x, y, z, [x, y], [x, z], [y, z], [x, [y, z]], [y, [x, z]]

linearly span L. In particular, L is at most 8-dimensional. Hence, L_{3} is also of dimension at most 8. It is readily checked though, that the above 8 bracketings provide a basis in the free case, so dim L_{3} = 8.

Example 5.1. The algebra 𝔰𝔩_{3} can be generated by 3 elements. It can be realized as follows:

x = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ) , y = ( \begin{matrix}0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ) , z = ( \begin{matrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ −2 & −2 & −2 \end{matrix} ) .

In this example, we have

f(x, y) = f(x, z) = f(y, z) = −2 and f(x, [y, z]) = 0.

The actions of ad_{x} and ad_{y} on the linear generators of L can be fully described by means of the identities in terms of the four parameters f(x, y), f(x, z), f(y, z), and f(x, [y, z]).

We can describe the four parameters pictorially by drawing a triangle with vertices x, y, z, and labeling the edge {x, y} by the ‘edge’ parameter f(x, y) and so on, and putting the ‘central’ parameter f(x, [y, z]) in the middle, with an indication of orientation (note that f(x, [y, z]) = f([x, y], z) = f(z, [x, y]), so the value is invariant under cyclic permutations of the nodes).

We shall reduce f(x, [y, z]) to zero by transforming the generators using elementary transformations. To begin reduction, consider the triple x, y, exp(x, s)z, where s ∈ k. It has parameters:

f(x, y) = f(x, y)
f(x, exp(x, s)z) = f(x, z)
f(y, exp(x, s)z) = f(y, z) − sf(x, [y, z]) + \frac{1}{2}s^{2}f(x, y)f(x, z)
f(x, [y, exp(x, s)z]) = f(x, [y, z]) − sf(x, z)f(x, y).

Clearly, this triple again consists of extremal elements, and generates the same algebra as x, y and z. If at least two of the three edges have nonzero labels (e.g., f(x, z) and f(x, y)), then we can transform the central parameter to 0 (by taking s = f(x, [y, z])/(f(x, z)f(x, y))).

On the other hand, if at most one edge is nonzero, say f(x, y), and the central parameter f(x, [y, z]) is also nonzero, then the above transformation shows that we can move to three extremal generators x, y, exp(x, s)z with one more edge (namely f(y, exp(x, s)z) and, if applicable, f(x, y)) nonzero. Hence, we can reduce to the previous case (if necessary in two steps), and so we may assume that the central parameter is zero: f(x, [y, z]) = 0.

Next, we scale x, y, z to αx, βy, γz for nonzero α, β, γ ∈ k. This leaves f(αx, βγ[y, z]) = 0 and changes the edge labels to

αβf(x, y), αγf(x, z), βγf(y, z).

We claim that, at the cost of a field extension of k, all nonzero edge labels may be transformed into −2. If at least one of them is zero, this is obvious. Otherwise, take

α = \sqrt{\frac{−2f(y, z)}{f(x, y)f(x, z)}}, β = \sqrt{\frac{−2f(x, z)}{f(x, y)f(y, z)}}, γ = \sqrt{\frac{−2f(x, y)}{f(x, z)f(y, z)}}.

Thus we are left with four essentially different cases, distinguished by the number of nonzero labeled edges in the triangle. Straightforward computation using GAP [5] leads to the following descriptions of the resulting four Lie algebras.

Theorem 5.2. Suppose that L is generated by three extremal elements. Then after extending the field if necessary, L is generated by three extremal elements whose central parameter is zero, and whose nonzero edge parameters are −2. In particular, L is a quotient of a Lie algebra M generated by extremal elements x, y, z with f(x, [y, z]) = 0 and dim M = 8. Moreover, according as the number of nonzero edge parameters is 0, 1, 2, 3, the Lie algebra M has the following form.

(0) f = 0 and M ≅ L_{3}. Thus, M is nilpotent, with [M, M] = k[x, y] + k[x, z] + k[y, z] + Z where Z = k[x, [y, z]] + k[y, [x, z]] = [[M, M], M] is the center of M.

(1) f(x, y) = −2, f(x, z) = f(y, z) = 0, and M = Z ⊕ [M, M] where Z = k(z − [x, [y, z]] − [y, [x, z]]) is the center of M. The solvable radical of M is R = kz + k[x, z] + k[y, z] + k[x, [y, z]] + k[y, [x, z]]. The subalgebra S = kx+k[x, y]+ky is isomorphic to 𝔰𝔩_{2} and M is the semi-direct product of S and R. The S-modules k[x, z] + k[y, [x, z]] and k[y, z] + k[x, [y, z]] are irreducible.

(2) f(x, y) = f(x, z) = −2, f(y, z) = 0, and M is the semi-direct product of S = kx + k[x, y] + ky, which is isomorphic to 𝔰𝔩_{2}, and the solvable radical R of M. Moreover, R = k(y − \frac{1}{2}[y, [x, z]])+k(z − \frac{1}{2}[y, [x, z]])+k([x, y]−[x, z])+ k[y, z] + k[x, [y, z]], [R, R] = k(y + z − [y, [x, z]]) + k[y, z] + k[x, [y, z]] and [R, [R, R]] = k[y, z] + k[x, [y, z]]. The center of M is trivial and [M, M] = M. The subspace k(y + z − [x, [y, z]] − [y, [x, z]]) of [R, R] is centralized by S.

(3) f(x, y) = f(x, z) = f(y, z) = −2 and M ≅ 𝔰𝔩_{3} as described in Example 5.1.

The algebra R_{3} can be determined easily. Again let A_{i} = ad_{x_{i}}.

Words  Conditions Number
I identity 1
A_{i}  3
A_{i}A_{j} i, j distinct 6
A_{i}A_{j}A_{k} i, j, k distinct  6
A_{i}A_{j}A_{k}A_{i} i, j, k distinct  3

Notice that in the last line as A_{i}[A_{j}, A_{k}]A_{i} = 0, we get A_{i}A_{j}A_{k}A_{i} = A_{i}A_{k}A_{j}A_{i} and so there are only 3 of these. Notice these are the only possibilities as multiplying one of the words of length 4 by any A_{l} obviously gives 0 on one side and gives 0 on the other side after using the commutation rule just given. There is such an algebra, as we found using the 4 generator Lie algebra L_{4} to be described in the next section.

6. The 4 generator case

Definition 6.1. A monomial of length s is a bracketing of the form

[x_{1}, [x_{2}, … [x_{s−1}, x_{s}] …]].

A monomial is reducible if it is a linear combination of monomials of strictly smaller length.

Lemma 6.2. Let L be generated by a subset D. Then L is the linear span of all monomials in elements of D.

Proof. By induction on the length of a bracketing (with respect to D) and Jacobi. □

Proposition 6.3. Let L be generated by the extremal elements x, y, z, u. Then L is linearly spanned by the following 28 monomials of length ≤ 5:

x, y, z, u,
[x, y], [x, z], [x, u], [y, z], [y, u], [z, u],

[x, [y, z]], [x, [y, u]], [x, [z, u]], [y, [x, z]], [y, [x, u]], [y, [z, u]], [z, [x, u]], [z, [y, u]],

[x, [y, [z, u]]], [x, [z, [y, u]]], [y, [x, [z, u]]], [y, [z, [x, u]]], [z, [x, [y, u]]], [z, [y, [x, u]]],

[x, [y, [z, [x, u]]]], [y, [x, [z, [y, u]]]], [z, [x, [y, [z, u]]]], [u, [x, [y, [z, u]]]].

Proof. Since L is linearly spanned by the monomials in x, y, z, u, we have to show that each monomial may be written as a linear combination of the given 28 elements.

All monomials of length 1 are on the list. There are 4 · 3 monomials of length 2 with different factors. Since [a, b] = −[b, a] for all a, b ∈ L, all of them may be expressed by the 6 monomials of length 2 on the list.

All monomials of length 3 which involve only 2 letters are reducible, since x, y, z, u are extremal. There are 4 · 3 · 2 monomials of length 3 with three different letters. With antisymmetry, all [x, [a, b]] and all [y, [a, b]] may be expressed. By Jacobi this holds also for the remaining ones.

All monomials of length 4 which involve only three letters are reducible, see the identity for 2[x, [y, [x, z]]] in Lemma 2.2. There are 4 · 3 · 2 monomials of length 4 with different factors.

With Jacobi we have the following equations

[x, [y, [z, u]]] − [x, [z, [y, u]]] + [x, [u, [y, z]]] = 0, (7)

[y, [x, [z, u]]] − [y, [z, [x, u]]] + [y, [u, [x, z]]] = 0, (8)

[z, [x, [y, u]]] − [z, [y, [x, u]]] + [z, [u, [x, y]]] = 0. (9)

The elements in the first two columns are on the list. Hence all monomials beginning with x, y, or z may be expressed as a linear combination of the given 28 basis elements. For the monomials beginning with u, we calculate:

[u, [x, [y, z]]] − [x, [u, [y, z]]] + [[y, z], [u, x]] = 0, (10)

[u, [y, [x, z]]] − [y, [u, [x, z]]] + [[x, z], [u, y]] = 0, (11)

[u, [z, [x, y]]] − [z, [u, [x, y]]] + [[x, y], [u, z]] = 0. (12)

The products of the form [[a, b], [c, d]] may be expressed as follows.

[x, [y, [z, u]]] − [y, [x, [z, u]]] + [[z, u], [x, y]] = 0, (13)

[x, [z, [y, u]]] − [z, [x, [y, u]]] + [[y, u], [x, z]] = 0, (14)

[y, [z, [x, u]]] − [z, [y, [x, u]]] + [[x, u], [y, z]] = 0. (15)

Hence also the monomials of length 4 beginning with u may be expressed as a linear combination of the monomials on the list.

We obtain monomials of length 5 by multiplying a letter from the left to a monomial [a, [b, [c, d]]] of length 4. This yields four possibilities: First [a, [a, [b, [c, d]]]], which obviously is reducible. Second [b, [a, [b, [c, d]]]], which is of the form [b, [a, [b, e]]] and hence it is reducible. Third [c, [a, [b, [c, d]]]]. There is no obvious way to rewrite this. Fourth [d, [a, [b, [c, d]]]], which yields the previous case by interchanging c and d. We are left with the monomials of the form [c, [a, [b, [c, d]]]] or [d, [a, [b, [c, d]]]], where [a, [b, [c, d]]] is one of the monomials of length 4 on the list. This yields the following 12 monomials m_{11}, m_{12}, …, m_{34} (m_{ij} is in row i and column j):

[z, [x, [y, [z, u]]]], [y, [x, [z, [y, u]]]], [z, [y, [x, [z, u]]]], [x, [y, [z, [x, u]]]],

[y, [z, [x, [y, u]]]], [x, [z, [y, [x, u]]]], [u, [x, [y, [z, u]]]], [u, [x, [z, [y, u]]]],

[u, [y, [x, [z, u]]]], [u, [y, [z, [x, u]]]], [u, [z, [x, [y, u]]]], [u, [z, [y, [x, u]]]].

Our intended basis vectors are m_{11}, m_{12}, m_{14}, m_{23}. We may express the other 8 elements as follows:

m_{24}:  with (7),
m_{13}, m_{31}: with (13) and (12),
m_{32}:  with (8),
m_{21}, m_{33}: with (14) and (11),
m_{22}, m_{34}: with (15) and (10).

Finally, we show that all monomials of length 6 are reducible. We multiply the monomials of length 5 on the list with a letter. Note that all these monomials of length 5 are of the form ±[c, [a, [b, [c, d]]]]. Multiplication from the left with c or a yields reducible monomials. Next, we deal with [b, [c, [a, [b, [c, d]]]]]. With Jacobi we may pass from [a, [b, [c, d]]] to [b, [[c, d], a]] and [[c, d], [a, b]]. From [[c, d], [a, b]], we pass to [d, [c, [a, b]]] and [c, [[a, b], d]]. Now we multiply first with c, then with b from the left. This yields products of the form [b, [c, [b, [[c, d], a]]]], [b, [c, [d, [c, [a, b]]]]] and [b, [c, [c, [[a, b], d]]]], which are all reducible. (Look for patterns of the type [u, [v, [u, w]]].) The last monomial we have to reduce is of the form [d, [c, [a, [b, [c, d]]]]]. We pass from [b, [c, d]] by Jacobi to v_{1} = [c, [d, b]] and v_{2} = [d, [c, b]]. The products [d, [c, [a, v_{1}]]] and [d, [c, [a, v_{2}]]] are reducible. (In the second one, the two letters d are at distance 3.)

As a consequence all monomials of length ≥ 6 are reducible and may be written as a linear combination of the 28 vectors on the list. □

Remark 6.4. Removing all elements involving a letter u from the list for the 4-generator case, yields a spanning set of size 8 for the 3-generator case. More generally, for r ∈ ℕ, r > 1, the Lie algebra L_{r−1} is a Lie subalgebra of L_{r}, generated by the first r − 1 elements of a set {x_{1}, …, x_{r}} of extremal generators for L_{r}. The module generated by x_{r} under the ad action of L_{r−1} gives the algebra R_{r−1} defined in Remark 4.4. (This can be proved by observing that the defining relations for L_{r} are homogeneous with respect to the multidegree counting the number of each x_{i}, so that a non-trivial relation amongst bracketings in R_{r−1}x_{r} would lead to a nontrivial homogeneous relation of x_{r}-degree 1. Thus, no relation in R_{r−1}x_{r} involves relations coming from ad_{x_{r}}^{2} = 0, and so any linear relation amongst bracketings in R_{r−1}x_{r} is a consequence of the defining relations of R_{r−1}.)

Counting the monomials with a single u in the list of Proposition 6.3, we find dim R_{3} = 19, distributed according to length as follows, where length of course stands for the length in R_{3}, which is one less than the length of these monomials of Proposition 6.3.

length 0 1 2 3 4
number 1 3 6 6 3

The algebra R_{4} has dimension 193. This can be seen by routine enumeration which we do not include. However we list the numbers of words of each length in the table below.

length 0 1 2 3 4 5 6 7 8 9
number 1 4 12 24 36 40 36 24 12 4

Notice that in all of these algebras R_{2}, R_{3}, R_{4} the number of words with a given length is symmetric about the highest number after deleting the one word of length 0 which is the identity. We wonder if this behavior persists for R_{m} with m ≥ 5.

7. Quadratic modules

Definition 7.1. Let L be a Lie algebra generated by extremal elements. We call an L-module U quadratic if there is a generating set D of extremal elements of L such that x · (x · U) = 0 for all x ∈ D.

Remark 7.2. Suppose that L is generated by n extremal elements. Consider the subalgebra A generated by the set D of the first n − 1 extremal generators of L. Since A is an A-submodule of L, the quotient L/A is also an A-module. Clearly, x · (x · L/A) = 0 for all x ∈ D, and D ⊆ E(L) ∩ A ⊆ E(A). Therefore, L/A is a quadratic A-module.

We give the irreducible quadratic modules for Lie algebras of Chevalley type.

Proposition 7.3. Let L be a Lie algebra of Chevalley type and k a field of characteristic distinct from 2 and 3. The highest weights of its non-trivial quadratic highest weight modules of finite dimension are given in the table below.

Type of L Highest weights

A_{n}  ω_{1}, …, ω_{n} B_{n}  ω_{1}, ω_{n} C_{n}  ω_{1}, …, ω_{n} D_{n}  ω_{1}, ω_{n−1}, ω_{n} E_{6}  ω_{1}, ω_{6} E_{7}  ω_{7} F_{4}  ω_{4} G_{2}  ω_{1}

Proof. (Sketch; a strongly related result for groups can be found in [11].) Let V be a non-trivial quadratic highest weight module for L of finite dimension. By Proposition 3.3, there is a single G-orbit of extremal elements in L. Moreover, all these (long root) elements have the same nilpotency index on V (because the highest weight representation is equivalent to a composition of itself with conjugation by an element of G). Consequently, each element x ∈ E(A) satisfies x · (x · V) = 0.

In particular, for every long root α, the corresponding Chevalley basis element x_{α} is nilpotent of index 2. It follows that if μ is a weight of V, then μ − 2α is not. Since the set of weights for V is a convex subset of the coset of the root lattice containing the highest weight, only small weights can occur. Thus, the result can be readily established by use of LiE, cf. [9], and the following argument for the case of a single root length. □

Lemma 7.4. Suppose that L is a Lie algebra of Chevalley type with only one root length. If U is an irreducible quadratic finite dimensional L-module, then U is a minuscule weight representation.

Proof. Let λ be the highest weight of U. Suppose μ is a weight for U. Then there is a path of weights from μ to λ, such that, for each adjacent pair (μ_{1}, μ_{2}) from the path, the difference μ_{2} − μ_{1} is a positive root. Since there is only one root length, each of these differences is a root whose root element is extremal. Now take the fundamental 𝔰𝔩_{2}-triplet containing this root element. Quadraticity means that the non-trivial irreducible sub-representations in U of the corresponding subalgebra isomorphic to 𝔰𝔩_{2} all have dimension 2. But μ_{1}, μ_{2} are in the same representation, so they belong to a 2-dimensional module. But then, they are conjugate by an element of the corresponding subgroup of type A_{1}. This establishes that U has a basis of weight spaces which are all conjugate, which can be taken as a definition of minuscule. □

Remark 7.5. For the Lie algebra of type E_{8}, there are no non-trivial finitedimensional quadratic modules. Thus, this Lie algebra can only occur as a direct component of a bigger Lie algebra generated by extremal elements.

8. The minimal number of extremal generators for Lie algebras of Chevalley type

Let 𝔤 be a Lie algebra of Chevalley type over k (of characteristic distinct from 2). As we have seen in Proposition 3.3, the long root elements of 𝔤 are extremal elements and 𝔤 is generated by these extremal elements. Write t(𝔤) for the minimal number of these extremal generators of 𝔤. In this section, we determine t(𝔤). We discuss implications for the group analog in §10.

Fix a Cartan subalgebra 𝔥 of 𝔤 and let Φ be the corresponding root system. Throughout this section x_{α} and h_{α} denote, respectively, the root element and the element in 𝔥 of 𝔤 corresponding to the root α in Φ. Also exp(x_{α}) denotes exp(x_{α}, 1) in the notation of Section 2. The signs appearing here in the multiplication rules for the x_{α} are the ones arising from the implementations in the relevant packages. In general, the signs depend on the choice of a Chevalley basis.

Lemma 8.1. Let 𝔤 be a Lie algebra of Chevalley type and let α_{1}, …, α_{n} be the simple roots in Φ. Denote by β the root of highest height. Then 𝔤 is generated by the root elements x_{α_{1}}, x_{α_{2}}, …, x_{α_{n}}, x_{−β}.

Proof. The simple root elements with respect to Φ generate the subalgebra 𝔫 = ∑_{α∈Φ^{+}} 𝔤_{α} where Φ^{+} is the set of positive roots and 𝔤_{α} is a root space. The lemma follows from the fact that the 𝔫-submodule of 𝔤 generated by the root element corresponding to the root of lowest height, coincides with 𝔤. □

Theorem 8.2. Let 𝔤 be a Lie algebra of Chevalley type over the field k of characteristic distinct from 2. Then the number t(𝔤) is as given in the table below.

Type of 𝔤 t(𝔤) condition
A_{n}  n + 1  n ≥ 1
B_{n}  n + 1  n ≥ 3
C_{n}  2n  n ≥ 2
D_{n}  n  n ≥ 4
E_{n}  5  n = 6, 7, 8
F_{4}  5
G_{2}  4

The proof of the theorem will be given in the rest of this section.

Lemma 8.3. If 𝔤 has an irreducible representation of dimension N and its extremal elements have rank m in this representation, then the number of extremal elements generating 𝔤 is at least N/m.

Proof. If 𝔤 is generated by the extremal elements x_{1}, …, x_{t}, then the image of 𝔤 in V, the underlying N-dimensional vector space, is generated by

⟨Im x_{1}, …, Im x_{t}⟩ = ⟨Im x_{1}⟩ + ⋯ + ⟨Im x_{t}⟩,

of dimension at most tm. Thus, irreducibility of the representation implies tm ≥ N, whence the lemma. □

Here is an upper bound for groups without multiple bonds.

Lemma 8.4. If the root system of 𝔤 has just one root length, then t(𝔤) ≤ n + 1, where n is the rank of 𝔤.

Proof. This follows immediately from Lemma 8.1. □

Lemma 8.5. If 𝔤 is generated by t extremal elements, then t(𝔤) ≥ 4 if dim 𝔤 ≥ 9, and t(𝔤) ≥ 5 if dim 𝔤 ≥ 29.

Proof. This is immediate from the dimensions given in Section 5 and in Proposition 6.3. □

The previous lemmas suffice for the proof of all exact lower bounds on t(𝔤). For A_{n}, consider the natural representation of dimension N = n + 1. Since

extremal elements have rank 1 in this representation, Lemma 8.3 gives t(A_{n}) ≥ n+1.

For B_{n}, Lemma 8.3 applied to the natural module for 𝔤 (of dimension 2n + 1, in which the extremal elements have rank 2), we find t(𝔤) ≥ (2n + 1)/2, whence t(B_{n}) ≥ n + 1. Observe that for n = 2, this bound is not sharp. Since the Lie algebra of type B_{2} has dimension 10, at least 4 extremal generators are needed in view of Lemma 8.5. This lower bound coincides with the result for the Lie algebra of type C_{2} (which is isomorphic to the one of type B_{2}).

For C_{n}, Lemma 8.3 applied to the natural module for 𝔤 (of dimension 2n, in which the extremal elements have rank 1), we find t(C_{n}) ≥ 2n.

For D_{n}, recall that the Lie algebra has a natural representation of dimension 2n, in which extremal elements have rank 2; hence, by Lemma 8.3, t(D_{n}) ≥ n.

For the exceptional Lie algebras, Lemma 8.5 shows that the Lie algebra of type G_{2} is generated by no fewer than 4 and the other four (types F_{4} and E_{6}, E_{7}, E_{8}) are generated by no fewer than 5 extremal elements. When k has characteristic 3, the Lie algebra 𝔤 of Chevalley type G_{2} has a 7-dimensional simple quotient. This quotient is isomorphic to a quotient of 𝔰𝔩_{3} by its center, and is generated by 3 extremal elements. Nevertheless, 𝔤 itself, being 14-dimensional, cannot be generated by fewer than four extremal elements.

To prove the theorem, it remains to show that there is a generating set of extremal elements of the size indicated in the table. To this end, we often argue along the following pattern. Let 𝔤 be the Lie algebra under consideration. We shall work with a fixed Chevalley basis B of 𝔤 and a fixed root system whose simple roots α_{1}, …, α_{n} are labeled as in [3]. When we talk about root elements corresponding to specific roots we mean elements from this basis.

We shall select a Lie subalgebra M of 𝔤 of Chevalley type generated by extremal elements, usually chosen from B. Let 𝔤 = M ⊕V_{1} ⊕⋯⊕V_{r} be the decomposition of the M-module 𝔤 into irreducible M-modules. The modules V_{i} are often spanned by certain elements of B and usually the index i corresponds to coefficients of simple roots not supported in M. We shall write down an element d which is the image of an extremal element under a composition of exponentials of root elements. It is an extremal element whose projections onto many of the modules V_{j} (j = 1, …, r) are nonzero. Next, we take C to be the Lie algebra generated by M and d and, by suitably bracketing d by elements of C, we find vectors from each V_{j}, usually again elements of B. Once those are found, we see that C coincides with 𝔤, so t(𝔤) ≤ t(M)+1. (In case 𝔤 is of type C_{n}, we need two additional extremal elements instead of 1 outside M and there we show t(𝔤) ≤ t(M) + 2.)

We start with the classical Lie algebras. In this case, with exception of A_{n}, we will write the simple roots using the unit orthonormal vectors in ℂ^{n}, ε_{1}, …, ε_{n} with α_{i}’s expressed in terms of ε_{i}’s as in Bourbaki.

All computations were made using the ELIAS routines [6] in GAP [5], and LiE [9]. The signs appearing here are the ones arising from the implementations in the relevant packages. In general, the signs depend on the choice of Chevalley basis.

8.1. Type A_{n}. By Lemma 8.4 we know that A_{n} can be generated by n+1 extremal elements, that is, t(A_{n}) ≤ n + 1. So there is nothing left to prove.

8.2. Type B_{n}. Let 𝔤 be the Lie algebra of type B_{n}.

First assume n = 3. Take M to be the Lie subalgebra of 𝔤 generated by the root elements corresponding to the first two simple roots, ε_{1} − ε_{2} and ε_{2} − ε_{3} and their negatives. Then M is of type A_{2} and contains the root elements x_{±(ε_{1}−ε_{2})}, x_{±(ε_{2}−ε_{3})} and x_{±(ε_{1}−ε_{3})}. Moreover, 𝔤 decomposes as

𝔤 = M ⊕ V_{−2} ⊕ V_{−1} ⊕ V_{1} ⊕ V_{2} ⊕ H_{0},

where V_{1} is the natural M-module linearly spanned by root elements x_{ε_{1}}, x_{ε_{2}} and x_{ε_{3}}; V_{2} is spanned by root elements x_{ε_{1}+ε_{2}}, x_{ε_{1}+ε_{3}} and x_{ε_{2}+ε_{3}}. Their dual Mmodules V_{−1} and V_{−2} are natural M-modules linearly spanned by x_{−ε_{1}}, x_{−ε_{2}}, x_{−ε_{3}} and x_{−(ε_{1}+ε_{2})}, x_{−(ε_{1}+ε_{3})}, x_{−(ε_{2}+ε_{3})}, respectively and H_{0} is the trivial M-submodule of L, spanned by the torus element h_{ε_{3}} centralizing M.

Now write:

d = exp(x_{−(ε_{1}+ε_{2})}) exp(x_{ε_{1}})x_{−(ε_{1}−ε_{2})} = x_{−(ε_{1}−ε_{2})} − x_{ε_{2}} + x_{ε_{1}+ε_{2}} + x_{−ε_{1}} + h_{−(ε_{1}+ε_{2})} − x_{−(ε_{1}+ε_{2})}.

Consider the Lie subalgebra C of 𝔤 generated by M and d. As x_{ε_{1}−ε_{2}} ∈ M, we have −x_{ε_{1}} − x_{−ε_{2}} + M = [x_{ε_{1}−ε_{2}}, d] + M ⊆ C, whence x_{ε_{1}} + x_{−ε_{2}} ∈ C. Bracketing by the element x_{ε_{2}−ε_{3}} of M, we find x_{−ε_{3}} ∈ C. In particular, the M-submodule V_{−1} generated by this element belongs to C, whence also x_{−ε_{1}}, and so e := d − x_{−(ε_{1}−ε_{2})} − x_{−ε_{1}} ∈ C.

Taking brackets of x_{−ε_{1}} with e gives 2x_{−(ε_{1}−ε_{2})} − x_{ε_{2}} + x_{−ε_{1}} ∈ C, whence x_{ε_{2}} ∈ C. In particular, x_{ε_{3}}, which lies in the same M-submodule, belongs to C, and hence so do all root elements corresponding to positive roots. But then e + C = x_{−(ε_{1}+ε_{2})} + C ⊆ C, and so also the lowest root element belongs to C. The root elements corresponding to the simple roots and to the lowest root generate 𝔤, so C = 𝔤. Thus t(𝔤) ≤ t(M) + 1 = 4.

Next, assume n ≥ 4. In the Lie algebra 𝔤 of type B_{n} we have the Lie subalgebra M of type D_{n} generated by all long root elements (extremal elements) of the form x_{ε_{i}±ε_{j}} (1 ≤ i < j ≤ n). As an M-representation space, 𝔤 decomposes into the adjoint module M and the natural representation V_{1} of degree 2n. Here, V_{1} is linearly spanned by the elements x_{±ε_{j}} (j = 1, …, n) from the Chevalley basis B.

Now

d := exp(x_{ε_{2}})x_{ε_{1}−ε_{2}} = x_{ε_{1}−ε_{2}} − x_{ε_{1}} − x_{ε_{1}+ε_{2}}

is an extremal element. Let C be the Lie subalgebra of 𝔤 generated by M and d.

We show it is 𝔤. For, d + M = x_{ε_{1}} + M, so x_{ε_{1}} ∈ C, whence V_{1} ∩ C ≠ {0}. By irreducibility of V_{1}, we find that x_{±ε_{j}} ∈ C for all j. Thus, C contains all root elements of the standard Chevalley basis B of 𝔤, and so coincides with 𝔤.

The conclusion is that t(B_{n}) ≤ t(D_{n})+1. In particular, the proof of the theorem for B_{n} is complete once the theorem is shown to hold for D_{n} (as then t(D_{n}) = n).

8.3. Type C_{n}. Suppose n ≥ 2 and let 𝔤 be the Lie algebra of type C_{n}. Take M to be the Lie subalgebra of type C_{n−1} generated by the root elements with roots in the linear span of ε_{2}, …, ε_{n}. Then 𝔤 decomposes as

𝔤 = M ⊕ V_{1} ⊕ V_{−1} ⊕ V_{2} ⊕ V_{−2} ⊕ V_{0},

where V_{1} and V_{−1} are natural M-modules, spanned by all root elements with first simple root coordinate 1, respectively −1, and V_{2}, V_{0}, V_{−2} are trivial modules, spanned by x_{2ε_{1}}, h_{2ε_{1}}, and x_{−2ε_{1}}, respectively.

Now take

d_{1} = exp(x_{−(ε_{1}+ε_{2})})x_{2ε_{1}} = x_{2ε_{1}} − x_{ε_{1}−ε_{2}} − x_{−2ε_{2}}

and

d_{2} = exp(x_{ε_{1}+ε_{2}})x_{−2ε_{1}} = x_{−2ε_{1}} + x_{−(ε_{1}−ε_{2})} − x_{2ε_{2}}.

Let C be the Lie subalgebra generated by M, d_{1}, and d_{2}. Then d_{1} + M = x_{2ε_{1}} − x_{ε_{1}−ε_{2}} + M and d_{2} + M = x_{−2ε_{1}} + x_{−(ε_{1}−ε_{2})} + M.

Now C = [x_{−2ε_{2}}, d_{2}] + C = x_{−(ε_{1}+ε_{2})} + C, and C = [x_{2ε_{2}}, d_{1}] + C = x_{ε_{1}+ε_{2}} + C, so generators for each natural M-submodule of 𝔤 are in C. But then C = d_{1} + C = x_{2ε_{1}} + C and C = d_{2} + C = x_{−2ε_{1}} + C, so V_{2} and V_{−2} are also contained in C. We have established that 𝔤 ⊆ C, whence C = 𝔤.

The conclusion is t(𝔤) ≤ t(M) + 2. If n = 2, then M is of type A_{1}, so we have t(M) = 2 and t(𝔤) ≤ 4. By induction, we find t(𝔤) ≤ 2n for arbitrary n.

8.4. Type D_{n}. First we consider the case n = 4. Let M be the subalgebra of 𝔤 generated by the root elements corresponding to the roots x_{±(ε_{1}−ε_{2})}, x_{±(ε_{2}−ε_{3})} and x_{±(ε_{1}−ε_{3})}. Then M is of type A_{2} and 𝔤 decomposes as

𝔤 = M ⊕ V_{ε_{3}−ε_{4}} ⊕ V_{ε_{3}+ε_{4}} ⊕ V_{1,0} ⊕ V_{0,1} ⊕ V_{1,1} ⊕ V_{−1,0} ⊕ V_{0,−1} ⊕ V_{−1,−1},

where V_{i,j} is a 3-dimensional module linearly spanned by the root elements corresponding to simple roots of the form λ(ε_{1} − ε_{2}) + μ(ε_{2} − ε_{3}) + i(ε_{3} − ε_{4}) + j(ε_{3} + ε_{4}) where λ, μ are scalars and V_{ε_{3}−ε_{4}} and V_{ε_{3}+ε_{4}} are one-dimensional M-modules.

Consider

d = exp(x_{ε_{1}+ε_{4}}) exp(x_{−(ε_{3}−ε_{4})}) exp(x_{−(ε_{1}+ε_{3})})x_{ε_{3}−ε_{4}} =
−x_{ε_{1}+ε_{3}} − 2x_{ε_{1}+ε_{4}} + x_{ε_{3}−ε_{4}} − 2x_{−(ε_{3}−ε_{4})}
−x_{−(ε_{1}+ε_{3})} + x_{−(ε_{1}+ε_{4})} + h_{−(ε_{3}−ε_{4})} + h_{ε_{1}+ε_{4}}.

This is an extremal element as it is the image of a root element under an automorphism of 𝔤.

Denote by C the Lie subalgebra of 𝔤 generated by M and d. A computation shows

[x_{−(ε_{1}−ε_{3})}, [x_{ε_{1}−ε_{2}}, d]] = −x_{−(ε_{2}−ε_{3})} + x_{−(ε_{1}+ε_{2})}.

Since x_{−(ε_{1}−ε_{3})}, x_{ε_{1}−ε_{2}}, x_{−(ε_{2}−ε_{3})} ∈ M, we derive x_{−(α_{1}+2α_{2}+α_{3}+α_{4})} = x_{−(ε_{1}+ε_{2})} ∈ C; that is, the lowest root element belongs to C. It follows that V_{−1,−1} ⊆ C.

Another computation yields

[x_{−(ε_{2}−ε_{3})}, [x_{−(ε_{1}−ε_{2})}, d]] = −2x_{ε_{3}+ε_{4}} + x_{−(ε_{1}−ε_{3})}.

Similar to the above, we derive x_{α_{4}} = x_{ε_{3}+ε_{4}} ∈ C. So V_{0,1} ⊆ C.

By computation, we have

[x_{ε_{1}−ε_{2}}, [h_{ε_{1}−ε_{3}}, [x_{−(ε_{1}+ε_{3})}, d]]] = x_{−(ε_{2}+ε_{4})}.

Since x_{ε_{1}−ε_{2}}, h_{ε_{1}−ε_{3}} ∈ M and x_{−(ε_{1}+ε_{3})} ∈ V_{−1,−1} ⊂ C, we see x_{−(α_{2}+α_{4})} = x_{−(ε_{2}+ε_{4})} ∈ C, whence V_{0,−1} ⊂ C.

Since x_{−(ε_{1}+ε_{4})} ∈ V_{0,−1} ⊆ C and h_{ε_{1}−ε_{3}} ∈ M, we have

C = [x_{−(ε_{1}−ε_{3})}, [x_{ε_{1}−ε_{3}}, d]] + C = x_{−(ε_{1}+ε_{4})} + 2h_{ε_{1}−ε_{3}} + x_{ε_{3}−ε_{4}} + C = x_{ε_{3}−ε_{4}} + C,

and so x_{α_{3}} = x_{ε_{3}−ε_{4}} ∈ C. So V_{1,0} ⊆ C.

We have seen that the root elements corresponding to all four simple roots and to the lowest root belong to C. Since they generate 𝔤, we conclude C = 𝔤, whence t(𝔤) ≤ t(M) + 1 = 4.

Assume, from now on, that n > 4. We show that the Lie algebra 𝔤 of type D_{n} satisfies t(𝔤) ≤ n.

Let M be the Lie subalgebra of 𝔤 generated by all root elements corresponding to the subrootsystem on ε_{i} ± ε_{j} for 2 ≤ i < j ≤ n. Then M has type D_{n−1}. As an M-module, 𝔤 decomposes as

𝔤 = M ⊕ V_{1} ⊕ V_{−1} ⊕ H_{0},

where H_{0} is a 1-dimensional toral subalgebra centralizing M and V_{1}, V_{−1} are natural modules spanned by all root elements whose root vectors have first coordinate (with respect to the basis of simple roots) equal to 1, −1, respectively.

By induction t(M) ≤ n − 1, so we need to find an extremal element of 𝔤 that, together with M, generates 𝔤. Take

d = exp(x_{−(ε_{1}−ε_{2})})x_{ε_{1}−ε_{2}} = x_{ε_{1}−ε_{2}} − h_{ε_{1}−ε_{2}} − x_{−(ε_{1}−ε_{2})}.

Now [x_{ε_{2}−ε_{3}}, d] + M = −x_{ε_{1}−ε_{3}} + M, so x_{ε_{1}−ε_{3}} ∈ C, whence every root element whose root has first root coordinate 1 belongs to C. That is, V_{1} ⊆ C and therefore x_{ε_{1}−ε_{2}} ∈ C. Similarly, C = [x_{−(ε_{2}−ε_{3})}, d] + C = −x_{−(ε_{1}−ε_{3})} + C, and so also every root element whose root has first coordinate −1 belongs to C. That is, V_{−1} ⊆ C and therefore x_{−(ε_{1}+ε_{2})} ∈ C. Thus, all root elements from the standard Chevalley basis belong to C, proving C = 𝔤, and t(𝔤) ≤ t(M) + 1 = n.

8.5. Type E_{6}. Let us denote by α_{1}, α_{2}, α_{3}, α_{4}, α_{5}, α_{6} the simple roots of 𝔢_{6}. Let M be a subalgebra of 𝔢_{6} generated by M = ⟨x_{α_{2}}, x_{α_{3}}, x_{α_{4}}, x_{α_{5}}, x_{−(α_{2}+α_{3}+2α_{4}+α_{5})}⟩. Then M is of type D_{4}. One has the following decomposition of 𝔢_{6}:

𝔢_{6} = M ⊕ W ⊕ V_{0,1} ⊕ V_{1,0} ⊕ V_{1,1} ⊕ V_{0,−1} ⊕ V_{−1,0} ⊕ V_{−1,−1},

where V_{a,b} is the module generated by all x_{α} with a coefficient of α_{1} and b coefficient of α_{6} in α. The action of M on W is trivial.

We need one more extremal element:

d = exp(x_{−(α_{1}+α_{3}+α_{4})}) exp(x_{α_{3}+α_{4}+α_{5}+α_{6}}) exp(x_{−(α_{1}+α_{2}+α_{3}+2α_{4}+2α_{5}+α_{6})})x_{α_{1}}
= x_{α_{1}} − x_{−(α_{2}+α_{3}+2α_{4}+2α_{5}+α_{6})} + x_{α_{1}+α_{3}+α_{4}+α_{5}+α_{6}} + x_{−(α_{2}+α_{4}+α_{5})}
−x_{−(α_{3}+α_{4})} − x_{−(α_{1}+α_{2}+2α_{3}+3α_{4}+2α_{5}+α_{6})}
+x_{α_{5}+α_{6}} − x_{−(α_{1}+α_{2}+α_{3}+2α_{4}+α_{5})}.

We claim that the subalgebra C generated by M and d is the whole algebra 𝔢_{6}. The root elements x_{−(α_{2}+α_{4}+α_{5})} and x_{−(α_{3}+α_{4})} belong to M, so e := d −

x_{−(α_{2}+α_{4}+α_{5})}+x_{−(α_{3}+α_{4})} is in C. Since x_{α_{3}}, x_{−α_{5}} and x_{−α_{2}} are in M, the following brackets are in C:

[x_{α_{3}}, e] = x_{α_{1}+α_{3}}, [x_{−α_{5}}, e] = x_{α_{6}} and [x_{−α_{2}}, e] = −x_{−(α_{1}+2α_{2}+2α_{3}+3α_{4}+2α_{5}+α_{6})}

But x_{α_{1}+α_{3}} ∈ C implies V_{1,0} ⊂ C, and so x_{α_{1}} ∈ C.

Now C contains all the simple root elements and the lowest root element. By Lemma 8.1, we are done.

8.6. Type E_{7}. Let us denote by α_{1}, α_{2}, α_{3}, α_{4}, α_{5}, α_{6}, α_{7} the simple roots of 𝔢_{7}. Take the subalgebra M of type D_{4} in 𝔢_{7} exactly as in 𝔢_{6}.

To generate the whole algebra 𝔢_{7}, consider the extremal element

d = exp(x_{α_{2}+α_{3}+α_{4}+α_{5}+α_{6}+α_{7}}) exp(x_{−(α_{1}+α_{3})})
exp(x_{−(α_{1}+α_{2}+α_{3}+2α_{4}+α_{5}+α_{6})}) exp(x_{α_{3}+α_{4}+α_{5}+α_{6}})
exp(x_{−(α_{1}+α_{2}+α_{3}+2α_{4}+2α_{5}+α_{6}+α_{7})})x_{α_{1}}
= x_{α_{1}} + x_{α_{4}+α_{5}+α_{6}} + x_{α_{1}+α_{3}+α_{4}+α_{5}+α_{6}} + x_{α_{2}+α_{4}+α_{5}+α_{6}+α_{7}}
+x_{α_{1}+α_{2}+α_{3}+α_{4}+α_{5}+α_{6}+α_{7}} − x_{α_{1}+α_{2}+2α_{3}+3α_{4}+2α_{5}+α_{6}}
+x_{α_{1}+2α_{2}+2α_{3}+3α_{4}+2α_{5}+α_{6}+α_{7}} − x_{−α_{3}} + x_{−(α_{2}+α_{4})} − x_{−(α_{4}+α_{5})}
−x_{−(α_{1}+α_{2}+α_{3}+α_{4})} + x_{−(α_{1}+α_{3}+α_{4}+α_{5})} − x_{−(α_{2}+α_{3}+2α_{4}+α_{5}+α_{6})}
−x_{−(α_{1}+α_{2}+2α_{3}+2α_{4}+α_{5}+α_{6})} − x_{−(α_{2}+α_{3}+2α_{4}+2α_{5}+α_{6}+α_{7})} −
x_{−(α_{1}+α_{2}+2α_{3}+2α_{4}+2α_{5}+α_{6}+α_{7})} − x_{−(α_{1}+2α_{2}+2α_{3}+4α_{4}+3α_{5}+2α_{6}+α_{7})} +
x_{−(2α_{1}+2α_{2}+3α_{3}+4α_{4}+3α_{5}+2α_{6}+α_{7})}.

We show that the subalgebra C generated by M and d is the whole algebra 𝔢_{7}. The root elements x_{−α_{3}}, x_{−(α_{2}+α_{4})} and x_{−(α_{4}α_{5})} belong to M, so e := d + x_{−α_{3}} − x_{−(α_{2}+α_{4})} + x_{−(α_{4}+α_{5})} is in C. The Lie algebra 𝔢_{7} decomposes as:

𝔢_{7} = M ⊕ W ⊕ V_{0,0,1} ⊕ V_{0,1,0} ⊕ V_{0,1,1} ⊕ V_{0,2,1} ⊕ V_{1,0,0} ⊕ V_{1,1,0}
⊕V_{1,1,1} ⊕ V_{1,2,1} ⊕ V_{2,2,1} ⊕ V_{0,0,−1}
⊕V_{0,−1,0} ⊕ V_{0,−1,−1} ⊕ V_{0,−2,−1}
⊕V_{−1,0,0} ⊕ V_{−1,−1,0} ⊕ V_{−1,−1,−1}
⊕V_{−1,−2,−1} ⊕ V_{−2,−2,−1},

where W is a 3-dimensional module with trivial M action. The modules V_{1,0,0}, V_{0,1,1}, V_{0,1,0} and V_{1,2,1} have dimension 8 and V_{a,b,c} is generated by all roots elements not in M such that a is the coefficient of α_{1}, b of α_{6} and c of α_{7}.

The bracket [x_{α_{2}+α_{3}+2α_{4}+α_{5}}, [x_{α_{3}}, e]] is easily computed to be a nonzero scalar multiple of x_{α_{1}+α_{2}+2α_{3}+2α_{4}+α_{5}}, and, since both arguments of the bracket belong to M, the element x_{α_{1}+α_{2}+2α_{3}+2α_{4}+α_{5}} belongs to C. Consequently, V_{1,0,0} is contained in C and so x_{α_{1}} ∈ C.

Since x_{−(α_{4}+α_{5})}, x_{α_{5}} ∈ C, we have [x_{α_{5}}, [x_{−(α_{4}+α_{5})}, e]] = −x_{α_{5}+α_{6}} ∈ C. So V_{0,1,0} ⊂ C and x_{α_{6}} ∈ C. Also, [x_{−α_{5}}, [x_{α_{3}}, e]] belongs to C. Expanding it, we find x_{−(α_{1}+α_{2}+α_{3}+2α_{4}+2α_{5}+α_{6})} ∈ C. Hence C contains V_{−1,−1,0} and taking brackets with x_{α_{1}} we obtain V_{0,−1,0} ⊂ C.

Now x_{α_{7}} ∈ C, as [x_{−(α_{3}+α_{4}+α_{5}+α_{6})}, [x_{−(α_{1}+α_{2}+α_{3}+α_{4}+α_{5}+α_{6})}, e]] is a nonzero element of kx_{α_{7}} + V_{0,−1,0} which also belongs to C.

Finally, as x_{α_{5}+α_{6}} ∈ C and [x_{−α_{4}}, e] = x_{α_{5}+α_{6}} − x_{−(α_{1}+α_{2}+2α_{3}+3α_{4}+2α_{5}+α_{6}+α_{7})}, we find V_{−1,−1,−1} ⊂ C. Taking brackets with a convenient element of V_{−1,−1,0}, we see that also the lowest root element is in C. Now C = 𝔢_{7} by Lemma 8.1.

8.7. Type E_{8}. Let α_{1}, α_{2}, α_{3}, α_{4}, α_{5}, α_{6}, α_{7}, α_{8} be the simple roots of 𝔢_{8}. Adopt here the notation (a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, a_{8}) for a root α = ∑_{i=1}^{8} a_{i}α_{i}. Consider the subalgebra M of type D_{4} in 𝔢_{8} as in 𝔢_{6}. We have the following decomposition of 𝔢_{8} into M-modules.

𝔢_{8} = M ⊕ W ⊕ ⨁_{a,k,l,m∈I} V_{a,k,l,m},

where W is a 28 dimensional trivial module, V_{a,k,l,m} is an irreducible 8 dimensional module generated by all roots elements x_{α} where a, k, l, m are respectively the coefficients of α_{1}, α_{6}, α_{7}, α_{8} in α, and

I = ±{(0, 1, 0, 0), (0, 1, 1, 0), (0, 1, 1, 1), (1, 0, 0, 0), (1, 1, 0, 0),
(1, 1, 1, 0), (1, 1, 1, 1), (1, 2, 1, 0), (1, 2, 1, 1), (1, 2, 2, 1),
(1, 3, 2, 1), (2, 3, 2, 1)}.

To generate the whole algebra 𝔢_{8}, consider the extremal element

d = exp(x_{01111110}) exp(x_{−11121110}) exp(x_{01122111}) exp(x_{−10111111})
exp(x_{12343321}) exp(x_{−23354321})x_{10000000}
= x_{10000000} − x_{01011000} + x_{01111100}
+x_{00111111} + x_{11111110} + x_{11122111} − x_{11222211} − x_{12232210}
−x_{12233221} − x_{22343321} − x_{13354321} − x_{23465432} + x_{−10000000}
−x_{−00010000} − x_{−01011000} − x_{−00111111} + x_{−11121100} − x_{−01121110}
−x_{−11122111} − x_{−12232210} − x_{−11232221} + x_{−12243211} + x_{−22343321}
+x_{−13354321} − x_{−23465432}.

Let C be the Lie subalgebra generated by M and d. Bracketing and using the decomposition of 𝔢_{8} as an M-module, one can again derive equality between C and 𝔢_{8}. We omit further details, as they are very similar to the other cases dealt with (but lengthier).

8.8. Type F_{4}. Write α_{1}, α_{2}, α_{3} and α_{4} for the simple roots of 𝔣_{4}. So, α_{1} = ε_{2} −ε_{3}, α_{2} = ε_{3} − ε_{4}, α_{3} = ε_{4}, α_{4} = \frac{1}{2} (ε_{1} − ε_{2} − ε_{3} − ε_{4}).

Let M be the subalgebra of 𝔣_{4} generated by x_{α_{1}}, x_{α_{2}}, x_{α_{2}+2α_{3}}, x_{α_{2}+2α_{3}+2α_{4}} and x_{−(2α_{1}+3α_{2}+4α_{3}+2α_{4})}. That is, M = ⟨x_{ε_{1}−ε_{2}}, x_{ε_{2}−ε_{3}}, x_{ε_{3}−ε_{4}}, x_{ε_{3}+ε_{4}}, x_{−(ε_{1}+ε_{2})}⟩. According to Lemma 8.1, M is of type D_{4}. By Section 8.4 we know then that M is generated by four extremal elements.

Define the following element in 𝔣_{4}:

d = exp(x_{−(α_{1}+2α_{2}+3α_{3}+α_{4})}) exp(x_{α_{4}})(x_{α_{2}+2α_{3}}) = x_{α_{2}+2α_{3}} − x_{α_{2}+2α_{3}+α_{4}} − x_{α_{2}+2α_{3}+2α_{4}} − x_{−(α_{1}+α_{2}+α_{3})} − x_{−(α_{1}+α_{2}+α_{3}+α_{4})} − x_{−(2α_{1}+3α_{2}+4α_{3}+2α_{4})}.

The element d is extremal since α_{2} + 2α_{3} is long. Consider the subalgebra C generated by M and d.

The root elements x_{α_{2}+2α_{3}}, x_{α_{2}+2α_{3}+2α_{4}} and x_{−(2α_{1}+3α_{2}+4α_{3}+2α_{4})} are in M. So f := d − x_{α_{2}+2α_{3}} + x_{α_{2}+2α_{3}+2α_{4}} + x_{−(2α_{1}+3α_{2}+4α_{3}+2α_{4})} = −x_{α_{2}+2α_{3}+α_{4}} − x_{−(α_{1}+α_{2}+α_{3})} − x_{−(α_{1}+α_{2}+α_{3}+α_{4})} is in C.

The Lie algebra 𝔣_{4} decomposes as:

𝔣_{4} = M ⊕ V_{0,1} ⊕ V_{1,0} ⊕ V_{0,0} ⊕ V_{1,1} ⊕ V_{1,2}
⊕V_{0,−1} ⊕ V_{−1,0} ⊕ W_{0,0} ⊕ V_{−1,−1} ⊕ V_{−1,−2},

where V_{0,1} is the module generated by x_{α_{4}}, x_{α_{3}+α_{4}}, x_{α_{2}+α_{3}+α_{4}} and x_{α_{2}+2α_{3}+α_{4}}, V_{1,0} = ⟨x_{α_{1}+α_{2}+α_{3}}⟩, V_{0,0} = ⟨x_{α_{3}}, x_{α_{2}+α_{3}}⟩, V_{1,1} = ⟨x_{α_{1}+α_{2}+α_{3}+α_{4}}, x_{α_{1}+α_{2}+2α_{3}+α_{4}}, x_{α_{1}+2α_{2}+2α_{3}+α_{4}}, x_{α_{1}+2α_{2}+3α_{3}+α_{4}}⟩ and V_{1,2} = ⟨x_{α_{1}+2α_{2}+3α_{3}+2α_{4}}⟩. The module W_{0,0} is generated by x_{−α_{3}}, x_{−(α_{2}+α_{3})} and obviously the remaining modules are generated by the negative root elements.

We have [x_{α_{1}+α_{2}+2α_{3}}, f] + C = x_{α_{3}} + C since x_{α_{1}+α_{2}+2α_{3}} ∈ M. Hence x_{α_{3}} is in C and then V_{0,0} ⊂ C.

Now the bracket of x_{α_{1}+α_{2}+2α_{3}+2α_{4}} ∈ M with f leads to the conclusion that x_{α_{3}+α_{4}} ∈ C, thus V_{0,1} ⊂ C. In particular, x_{α_{4}} ∈ C.

By Lemma 8.1, the algebra 𝔣_{4} is generated by x_{α_{1}}, x_{α_{2}}, x_{α_{3}}, x_{α_{4}} and x_{−(2α_{1}+3α_{2}+4α_{3}+2α_{4})}. But C contains all these root elements. Therefore C = 𝔣_{4}. The conclusion is t(𝔣_{4}) ≤ t(D_{4}) + 1 = 5.

8.9. Type G_{2}. Let α and β be the simple roots of 𝔤_{2}. Let M be the Lie subalgebra of 𝔤_{2} generated by the long root elements x_{β}, x_{3α+β}, x_{−(3α+2β)}. This subalgebra is of type A_{2}.

Consider the extremal element d = exp(x_{−(2α+β)})(x_{3α+2β}); it is extremal since 3α + 2β is a long root. We have

d = x_{3α+2β} + x_{α+β} − x_{−α}.

Let C be the subalgebra of 𝔤_{2} generated by M and d. We will prove that C is actually 𝔤_{2}. Since x_{α}, x_{β} and x_{−(3α+2β)} generate 𝔤_{2} and the latter two are obviously in C, it is enough to show that x_{α} is also in C.

Since d − x_{3α+2β} ∈ C, so does the bracket

[x_{−(3α+2β)}, d − x_{3α+2β}] = [x_{−(3α+2β)}, x_{α+β} − x_{−α}] = −x_{−(2α+β)}.

Therefore [x_{3α+β}, x_{−(2α+β)}] = x_{α} is also in C.

As before we conclude that C = 𝔤_{2} and so 𝔤_{2} is generated by four extremal elements.

9. The bilinear form and the Killing form

Throughout this section, L is a finite dimensional Lie algebra (over the field k of characteristic not 2) generated by extremal elements. Recall that the Killing form κ is defined by κ(x, y) := tr(ad_{x} ad_{y}) for x, y ∈ L. We consider connections between the associative bilinear form f defined in Theorem 2.6 and the Killing form κ.

Fix x ∈ E and y ∈ L. Set φ := ad_{x} ad_{y}.

Lemma 9.1. We have φ^{2} + \frac{1}{2}f(x, y)φ : L → kx + k[x, y].

Proof. For z ∈ L, using Lemma 2.2, we find

φ^{2}(z) = [x, [y, [x, [y, z]]]] = \frac{1}{2}(f(x, [y, [y, z]])x − f(x, [y, z])[x, y] − f(x, y)[x, [y, z]]).

This implies that φ^{2} + \frac{1}{2}f(x, y)φ maps L to kx + k[x, y]. □

Lemma 9.2. Let x ∈ E, y ∈ L. Then φ satisfies the following properties. a) If f(x, y) = 0, then all eigenvalues of φ are 0. In particular, κ(x, y) = 0.

b) If f(x, y) = −2, then φ has eigenvalue 2 with multiplicity 2, eigenvalue 1 with multiplicity s − 2 and the remaining eigenvalues are 0. Here s = dim ad_{x}(L). In particular, κ(x, y) = s + 2.

Proof. Note that φ(x) = −f(x, y)x and (by Lemma 2.2) φ([x, y]) = −f(x, y)[x, y]. Hence U = kx+k[x, y] is invariant under φ. Furthermore, φ^{2} + \frac{1}{2}f(x, y)φ is zero on L/U by Lemma 9.1. We write φ on L/U in Jordan normal form with eigenvalues 0 or −\frac{1}{2}f(x, y). Now a) is obvious. Next suppose f(x, y) = −2. Then x, [x, y] are linearly independent. Denote by E the eigenspace of φ (with eigenvalue 1). Then E + U/U = ad_{x}(L)/U and dim E = dim ad_{x}(L) − 2. This proves the lemma. □

We now consider connections between the radicals of the two forms. Recall that Rad(f) denotes the radical of f; similarly, we write Rad(κ) for the radical of κ.

Corollary 9.3. We have Rad(f) ⊆ Rad(κ).

Proof. Let y ∈ Rad(f). Then f(x, y) = 0 for all x ∈ E. Hence κ(x, y) = 0 for all x ∈ E by Lemma 9.2. Therefore y ∈ Rad(κ). □

Remark 9.4. Lemma 9.2 gives a way to tell if one of the Lie algebras of Chevalley type is the radical of its Killing form. (In this case Rad(f) is properly contained in Rad(κ).) For a root α, set γ_{α} = dim ad_{x_{α}}(L)−2. Then by Lemma 9.2, κ(x_{α}, x_{−α}) = γ_{α} + 4. When there is only one root length, γ_{α} is the number of roots which have inner product −1 with α. We look at the case of 𝔰𝔩_{3}. For α = e_{i} − e_{j}, there are exactly two such roots, e_{k} − e_{i}, and e_{j} − e_{k}, so κ(x_{α}, x_{−α}) = 2 + 4 = 6. Hence 𝔰𝔩_{3} has Killing form zero if k is of characteristic 3.

Lemma 9.5. If char(k) = 0, then Rad(f) = Rad(κ).

Proof. One inclusion comes from Corollary 9.3. Let y ∈ Rad(κ). Assume that y ∉ Rad(f). Then there exists x ∈ E such that f(x, y) = −2. Hence κ(x, y) = dim ad_{x}(L) + 2 ≠ 0 in characteristic 0. This is a contradiction. □

In what follows we determine some more properties about the radicals of the Killing form and of f.

Lemma 9.6. Let J be an ideal of L and N := span_{k}{x ∈ E | x ∉ J}. Then f(N, J) = 0.

Proof. It is enough to show for x ∈ E \ J and y ∈ J that f(x, y) = 0. We have f(x, y)x = [x, [x, y]] ∈ J. Hence f(x, y) = 0. □

Lemma 9.7. Suppose that K is a solvable ideal of L. Then E ∩ K ⊆ Rad(f).

Proof. Let x ∈ E ∩ K. Take y ∈ E. If y ∉ K, then f(x, y) = 0 by Lemma 9.6. If y ∈ K, suppose f(x, y) ≠ 0. Then ⟨x, y⟩ ≃ 𝔰𝔩_{2}. Hence ⟨x, y⟩ is a non-solvable Lie subalgebra of K which is a contradiction. So f(x, y) = 0.

This means that f(x, y) = 0 for all x ∈ E ∩ K and y ∈ E. The lemma follows as E linearly spans L, cf. Lemma 2.5. □

Let Rad(L) be the maximal solvable ideal in L.

Proposition 9.8. We have Rad(L) ⊆ Rad(f).

Proof. Suppose that K is a solvable ideal of L and let x ∈ K. For y ∈ E, either y ∈ E \ K, in which case f(x, y) = 0 by Lemma 9.6, or y ∈ E ∩ K and then f(x, y) = 0 by Lemma 9.7. □

Remark 9.9. An example where Rad(L) is properly contained in Rad(f) is the 14-dimensional Lie algebra L of type G_{2} over a field k of characteristic 3. Then, Rad(L) = 0 but Rad(f) is a 7-dimensional ideal, generated by the short root elements of a Chevalley basis. Proposition 9.12 shows that characteristic 3 is indeed exceptional.

In fact, Rad(f) and Rad(L) are just two ideals of a chain of five. Denote by SanRad(L) the linear span of all sandwiches of E, that is, those x ∈ E, for which ad_{x}^{2} = 0. We claim that SanRad(L) is an ideal of L. Indeed, if x ∈ SanRad(L) ∩ E and y ∈ E, then f_{x} = 0, so Corollary 2.7 shows that either [x, y] = 0 or [x, y] ∈ E with f_{[x,y]} = 0, that is, [x, y] ∈ SanRad(L). Since the restriction of f to SanRad(L) vanishes, the latter is a nilpotent Lie subalgebra of L by Lemma 4.2, so SanRad(L) is contained in NilRad(L), the nilpotent radical of L. We thus have the chain

SanRad(L) ⊆ NilRad(L) ⊆ Rad(L) ⊆ Rad(f) ⊆ Rad(κ).

We know that the last two inclusions may be proper, and, in view Case (2) of Theorem 5.2, we have examples where SanRad(L) is strictly contained in Rad(L).

Lemma 9.10. Suppose x ∈ E \ Rad(f) and y ∈ Rad(f). Then [x, y] satisfies ad_{[x,y]}^{4} = 0.

Proof. Put X = ad_{x} and Y = ad_{y}. Then, by (3) of Lemma 2.2, for z ∈ L,

2XY XY z = 2[x, [y, [x, [y, z]]]
= f(x, [y, [y, z]])x − f(x, [y, z])[x, y] − f(x, y)[x, [y, z]] = 0.

For the last equality, use that f is associative and y ∈ Rad(f). We obtain XY XY = 0. Also X^{2}Y = 0, as X^{2}Y (L) ⊆ Rad(f) ∩ kx = 0. Consequently,

ad_{[x,y]}^{4} = (XY − Y X)^{4} = (XY XY − XY^{2}X − Y X^{2}Y + Y XY X)^{2}
= (Y XY X − XY^{2}X)^{2}
= XY^{2}X^{2}Y^{2}X + (Y X)^{4} − XY^{2}(XY)^{2}X − Y XY X^{2}Y^{2}X = 0.
□

Lemma 9.11. Let K be an ideal of L which is contained in Rad(f). Then \overline{L} := L/K is linearly spanned by extremal elements with induced form \overline{f} defined by \overline{f}(\overline{x}, \overline{y}) := f(x, y) for x, y ∈ L.

Proof. Since K ⊆ Rad(f), the expression \overline{f}(\overline{x}, \overline{y}) := f(x, y) is well-defined for x, y ∈ L. For x ∈ E, y ∈ L, [\overline{x}, [\overline{x}, \overline{y}]] = \overline{f}(\overline{x}, \overline{y}) · \overline{x}, whence \overline{x} ∈ E(\overline{L}). □

We owe the proof of the following result to Gabor Ivanyos.

Proposition 9.12. If the characteristic of k is not 2 or 3, then Rad(f) = Rad(L).

Proof. Recall that Rad(L) ⊆ Rad(f), so if Rad(f) is contained in the center Z(L) of L, there is nothing to prove. Suppose, therefore, Rad(f) ⊈ Z(L). We show that SanRad(L) ≠ 0.

By Lemma 9.10, there is a nonzero element y ∈ Rad(f) with ad_{y}^{4} = 0. (If E ∩Rad(f) ≠ ∅, any element of the intersection will do, otherwise, take y ∈ Rad(f), and x ∈ E such that [x, y] ≠ 0, whose existence is guaranteed by the hypotheses that L = ⟨E⟩ and Rad(f) ⊈ Z(L), and apply Lemma 9.10.) By Proposition 2.1.5 of [7] (see also Proposition 1.5 of [1]), if the characteristic of k is not 2 or 3, the element z = ad_{y}^{3}(x) for any x ∈ L, satisfies ad_{z}^{3} = 0. In particular, there is a nonzero element y ∈ Rad(f) with ad_{y}^{3} = 0.

If ad_{y}^{2} = 0, then we are done. Otherwise, there is b ∈ E with x = ad_{y}^{2}(b) ≠ 0. If b ∈ Rad(f), then b ∈ Rad(f) ∩ E ⊆ SanRad(L), and we are done. So,

assume b ∉ Rad(f). By Lemma 1.7(iii) of [1], as k has characteristic not 2 or 3, we have ad_{x}^{2} = ad_{y}^{2} ad_{b}^{2} ad_{y}^{2}. Since y ∈ Rad(f), we have ad_{y}^{2}(L) ⊆ Rad(f) and since b ∈ E \ Rad(f), we have ad_{b}^{2}(Rad(f)) = 0, whence ad_{x}^{2} = 0, proving that E ∩ Rad(f) ≠ 0. This establishes SanRad(L) ≠ 0.

Thus, if Rad(f) ⊈ Z(L), it contains a nonzero sandwich, and so NilRad(L) ≠ 0. But then, in view of Lemma 9.11 and by induction on the dimension, Rad(f) is solvable.

□

We finish this section with a result identifying Rad(f) and Rad(L) in arbitrary characteristic distinct from 2 under additional hypotheses on the structure of L. It shows that f plays a role similar to κ in the theory of Lie algebras of characteristic 0.

Lemma 9.13. Let L = L_{1} ⊕ L_{2} be a direct sum of ideals. Then L_{1} and L_{2} are linearly spanned by extremal elements (with form f restricted to L_{1} and L_{2}, respectively). Furthermore, L_{1} and L_{2} are orthogonal with respect to f.

Proof. As [L_{1}, L_{2}] = 0, we have E(L_{i}) ⊆ E for i = 1, 2. Let x ∈ E(L) and y ∈ L. Write x = x_{1} + x_{2}, y = y_{1} + y_{2} with x_{1}, y_{1} ∈ L_{1}, x_{2}, y_{2} ∈ L_{2}. We calculate f(x, y)x_{1} + f(x, y)x_{2} = [x_{1}, [x_{1}, y]] + [x_{2}, [x_{2}, y]]. Hence [x_{i}, [x_{i}, y]] = f(x, y)x_{i} for all y ∈ L, and x_{i} ∈ E (i = 1, 2).

Since L is linearly spanned by E, each L_{i} is linearly spanned by the projections. Hence L_{i} is linearly spanned by extremal elements (with form f restricted to L_{i}).

Finally, for z ∈ E(L_{1}), l_{2} ∈ L_{2}, we have f(z, l_{2})z = [z, [z, l_{2}]] = 0. Since L_{1} is linearly spanned by extremal elements, this shows that the decomposition L = L_{1} ⊕ L_{2} is an orthogonal one with respect to f. □

Proposition 9.14. We have Rad(f) = 0 if and only if L is a direct sum of simple ideals.

Proof. Assume that L = L_{1} ⊕ L_{2} ⊕ ⋯ ⊕ L_{r} is a direct sum of simple ideals L_{i}. By Lemma 9.13, the decomposition is an orthogonal one with respect to f. Suppose L_{i} ⊆ Rad(f) for some i. Then the form f restricted to L_{i} is trivial. But then L_{i} is nilpotent by Lemma 4.2, a contradiction. This means that the form f restricted to each L_{i} has a trivial radical. Hence Rad(f) = 0.

As for the converse, assume Rad(f) = 0. Suppose that I is a minimal nonzero ideal of L. By Proposition 9.8, I is not Abelian. Let J be the orthoplement of I with respect to f. As f is associative, J is an ideal of L. We claim L = I ⊕ J. For, I ∩ J ≠ 0 would imply I ⊆ J by minimality of I and hence [I, J] = I as I is not Abelian, so f(I, L) = f([I, J], L) = f(I, [J, L]) ⊆ f(I, J) = 0, yielding the contradiction I ⊆ Rad(f).

Hence, if K is a nonzero ideal of I, it also is an ideal of L, and so coincides with I. This means that I is simple. Moreover, by Lemma 9.13, J is generated by extremal elements with form f|_{J}. Notice that Rad(f|_{J}) ⊆ Rad(f) = 0, so by induction on the dimension, we find that L is a direct sum of simple ideals. □

Corollary 9.15. We have that L/ Rad(L) is a direct sum of simple ideals if and only if Rad(L) = Rad(f).

Proof. In Lemma 9.8, we have already proved that Rad(L) ⊆ Rad(f). Set \overline{L} = L/ Rad(L) and let \overline{f} be as in Lemma 9.11.

Since \overline{Rad(f)} = Rad(\overline{f}), we have Rad(L) = Rad(f) if and only if Rad(\overline{f}) = 0, so the corollary is a direct consequence of Proposition 9.14. □

10. Analysis of root groups

Definition 10.1. For y ∈ E, we define U_{y} := {exp(y, t) | t ∈ k} to be the root group associated to y. Since exp(cy, t) = exp(y, ct) for all c ∈ k, the group only depends on the 1-dimensional subspace ky.

By Σ := {U_{y} | y ∈ E} we denote the set of root groups associated to extremal elements of L. Set G := ⟨Σ⟩ ≤ Aut(L). For calculations in G, we use (·, ·) for commutators in their group theoretic meaning. Thus for g, h ∈ G and A, B ≤ G, we write (g, h) := g^{−1}h^{−1}gh and (A, B) = ⟨(a, b) | a ∈ A, b ∈ B⟩. Furthermore, we denote the conjugate of A under g by A^{g}, so A^{g} = g^{−1}Ag. Observe that G preserves the bilinear form f.

Definition 10.2. Let A, B be two Abelian subgroups of G. Following Timmesfeld [14], we call ⟨A, B⟩ a rank 1 group, if the following holds: For each 1 ≠ a ∈ A there exists some 1 ≠ b ∈ B with A^{b} = B^{a}; and vice versa. If in addition a^{b} = (b^{−1})^{a}, then we call the rank 1 group special.

Theorem 10.3. For x, y ∈ E and s, t ∈ k, the group G has the following properties.

(1) exp(y, s) exp(y, t) = exp(y, s + t). In particular, U_{y} is isomorphic to the additive group of k.

(2) exp(y, s)x ∈ E with (U_{x})^{exp(y,s)} = U_{exp(y,−s)x}.

(3) If [x, y] = 0, then (U_{x}, U_{y}) = 1.

(4) If f(x, y) = 0, but [x, y] ≠ 0, then (exp(y, t), exp(x, s)) = exp([y, x], ts). In particular, the group ⟨U_{x}, U_{y}⟩ is nilpotent of class 2 and (U_{x}, U_{y}) = (u, U_{y}) = (U_{x}, v) = U_{[x,y]} for all 1 ≠ u ∈ U_{x}, 1 ≠ v ∈ U_{y}.

(5) If f(x, y) = −2, then, for s ∈ k, s ≠ 0,

exp(y, −s) exp(x, s^{−1}t) exp(y, s) = exp(x, −s^{−1}) exp(y, −ts) exp(x, s^{−1}).

In particular, the group ⟨U_{x}, U_{y}⟩ is a special rank 1 group in G.

Proof. All equations can be verified in a straightforward manner, by use of the definition of exp(y, t). In the calculations we use the Jacobi identity, associativity of f (e.g., f(y, [y, z]) = 0, f(x, [y, [x, z]]) = 0 when f(x, y) = 0, and f(x, [y, [x, z]]) = 2f(x, z) when f(x, y) = −2) and the rewriting rule for [x, [y, [x, z]]] of Lemma 2.2. □

Remark 10.4. Theorem 10.3 shows that the set Σ of root groups in G associated to extremal elements of L is a set of so-called abstract root subgroups in the sense of Timmesfeld [14]. If Σ is a conjugacy class and G is quasi-simple, then we may apply Timmesfeld’s classification [14] of groups generated by abstract root subgroups to determine the structure of L.

Whenever G is an algebraic group its long root subgroups correspond to projective points (that is, 1 dimensional linear subspaces) spanned by extremal elements in its Lie algebra L(G). The geometry of these long root subgroups, which is well studied, can thus also be studied in the Lie algebra L(G).

For instance, in a Chevalley group, consider the subgroup M generated by two different long root subgroups U_{α}, U_{β}, with α + β not a root. Then either M is partitioned by the long root subgroups contained in it or it contains no other long root subgroups than U_{α} and U_{β}, depending on whether α − β is a root or not. In the lemma below, we describe the corresponding behavior in Lie algebras for the case of a partitioning. For recognition of such pairs, we can take the general setup of a Lie algebra containing extremal elements and do not need to assume that it is the Lie algebra of an algebraic group.

Lemma 10.5. Let x, y ∈ E with [x, y] = 0. Then the following are equivalent:

(1) For s, t ∈ k, s, t ≠ 0, the element sx + ty is extremal. (1)’ There are s, t ∈ k, s, t ≠ 0, such that the element sx + ty is extremal.

(2) 2[y, [x, z]] = f(x, z)y + f(y, z)x for all z ∈ E. (2)’ 2[y, [x, z]] = f(x, z)y + f(y, z)x for all z ∈ L. Moreover, in these cases, we have exp(y, t) exp(x, s) = exp(sx + ty, 1) for s, t ∈ k.

Proof. Note that (2) and (2)’ are equivalent by Lemma 2.5. For z ∈ E, 0 ≠ s, t ∈ k, we have [sx + ty, [sx + ty, z]] = s^{2}f(x, z)x + t^{2}f(y, z)y + 2st[y, [x, z]] and f(sx + ty, z)(sx + ty) = s^{2}f(x, z)x + t^{2}f(y, z)y + st(f(x, z)y + f(y, z)x). Hence (1’) implies (2). If (2) holds, then sx + ty is extremal and a short calculation shows exp(y, t) exp(x, s)z = exp(sx+ty, 1)z. Notice [x, [y, z]] = [y, [x, z]] here. This yields the result. □

Corollary 10.6. If three points on a projective line of L represent commuting extremal elements, then the whole line consists of commuting extremal elements.

An example of the occurrence of projective lines consisting fully of extremal elements as described in the previous lemma is given in the next lemma. Note the correspondence with the group geometries of [14].

Lemma 10.7. Let x, y ∈ E with f(x, y) = 0, but [x, y] ≠ 0. Then the conditions of Lemma 10.5 hold for x and [x, y].

Proof. By Lemma 2.2 we have 2[[x, y], [x, z]] = f(x, [y, z])x + f(x, z)[x, y] + 0. By associativity of f (cf. Theorem 2.6), we see f(x, [y, z]) = f([x, y], z), which proves Condition (2)’ of Lemma 10.5, as required. □

Many more results in this direction can be derived, such as the non-existence of a chain x_{1}, x_{2}, x_{3} of extremal elements such that x_{1}, x_{2} are as x, y in Lemma 10.5, [x_{2}, x_{3}] = 0, and f(x_{1}, x_{3}) ≠ 0.

To finish, we relate the results on the generation of L by extremal elements and to corresponding properties for G. Let F(Σ) denote the graph whose vertices are the elements of Σ and whose edges are the unordered pairs {A, B} with ⟨A, B⟩ a rank 1 group.

Lemma 10.8. Assume that the graph F(Σ) is connected. If G = ⟨U_{x_{i}} | i ∈ I⟩, then L is generated as a Lie algebra by the x_{i} (i ∈ I).

Proof. Denote by W the Lie algebra generated by {x_{i} | i ∈ I}. Let z ∈ E with f(z, x_{i}) ≠ 0 for some i ∈ I. Since W is invariant under U_{x_{j}} for j ∈ I, it is also invariant under G. Hence [z, x_{i}] + \frac{1}{2}f(z, x_{i})z = exp(z, 1)x_{i} − x_{i} ∈ W. We apply exp(z, 1) to this vector and see that [z, x_{i}] + \frac{1}{2}f(z, x_{i})z + f(z, x_{i})z is in W. Hence also the difference, which is f(z, x_{i})z, is in W. Since f(z, x_{i}) ≠ 0, we conclude that z ∈ W.

The graph F(Σ) is connected, so we obtain that E ⊆ W, whence L = W. □

For the case where k is algebraically closed and L is of Chevalley type, we can prove the converse as well.

Theorem 10.9. Suppose that k is algebraically closed. Let L be a Lie algebra of Chevalley type over k of characteristic distinct from 2, 3, and let G be the corresponding group of automorphisms generated by long root groups. Then t(L), the minimal number of extremal elements generating L, determined in Theorem 8.2, is the minimal number of long root groups needed to generate G.

Proof. Every element of E is a long root element by Proposition 3.3. Suppose that L is generated by the extremal elements x_{1}, …, x_{t}. Write G = ⟨U_{x} | x ∈ E⟩ (as usual) and H = ⟨U_{x_{i}} | i = 1, …, t⟩. It suffices to show that H coincides with G. Put D = {x ∈ E | U_{x} ⊆ H}. Notice that, for x, y ∈ D, also exp(x, 1)y ∈ D. Hence, by the argument of the proof of Lemma 2.5 with D instead of E, we obtain that L is linearly spanned by D.

Subgroups of the algebraic group GL(L) generated by connected algebraic subgroups are connected algebraic subgroups (see [2], Proposition I.2.2), so H is a closed algebraic subgroup of the connected linear algebraic group G. Clearly, the derivative dι of the embedding ι : H → G at the identity of H is the embedding L(H) → L(G) of the Lie algebra of H in the Lie algebra of G. As D linearly spans L, we have L(H) = L = L(G) and so dι is surjective. By Theorem 3.2.21 of [13], this implies that ι is dominant, that is, H is dense in G. But H is closed as well, so H = G. This establishes that G is generated by at most t(L) long root subgroups.

The converse is handled by the previous lemma. □

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Addresses of the authors

(correspondence to:) Arjeh M. Cohen, Fac. Wisk. en Inf., TUE, POBox 513, 5600 MB Eindhoven, The Netherlands (email amc@win.tue.nl)

Anja Steinbach, Math. Inst., Justus-Liebig-Universität, Arndtstr. 2, D-35392, Giessen, Germany (email anja.steinbach@math.uni-giessen.de)

Rosane Ushirobira, Fac. Wisk. en Inf., TUE, POBox 513, 5600 MB Eindhoven, The Netherlands (email rosane@win.tue.nl)

David Wales, Sloan Lab, Caltech, Pasadena, CA 91125, USA (email dbw@caltech.edu)