Journal of Lie Theory Volume 14 (2004) 11–23 \begin{matrix}c \\ ◯ \end{matrix} 2004 Heldermann Verlag

On the Nilpotency of Certain Subalgebras of Kac-Moody Lie Algebras

Yeonok Kim, Kailash C. Misra, and Ernie Stitzinger^{∗}

Communicated by J. D. Lawson

Abstract. Let 𝔤 = 𝔫_{−} ⊕ 𝔥 ⊕ 𝔫_{+} be an indecomposable Kac-Moody Lie algebra associated with the generalized Cartan matrix A = (a_{ij}) and W be its Weyl group. For w ∈ W, we study the nilpotency index of the subalgebra S_{w} = 𝔫_{+} ∩ w(𝔫_{−}) and find that it is bounded by a constant k =k(A) which depends only on A but not on w for all A = (a_{ij}) finite, affine of type other than E or F and indefinite type with |a_{ij}| ≥ 2. In each case we find the best possible bound k. In the case when A = (a_{ij}) is hyperbolic of rank two we show that the nilpotency index is either 1 or 2.

Introduction

Let A = (a_{ij})_{i,j∈I} be an indecomposable generalized Cartan matrix and 𝔤 = 𝔤(A) denote the associated Kac-Moody Lie algebra over the field of complex numbers [2]. Following the usual convention we will take the index set I to be {0, 1, ⋯, n} when A is of affine type and I to be {1, 2, ⋯, n} otherwise. Let 𝔤 = 𝔫_{−} ⊕ 𝔥 ⊕ 𝔫_{+} be the triangular decomposition of 𝔤 with respect to the Cartan subalgebra 𝔥 and let Δ = Δ_{+} ∪ Δ_{−} denote the set of roots with Δ_{+} and Δ_{−} denoting the set of positive and negative roots respectively. Let Π = {α_{i}|i ∈ I} denote the set of simple roots and Π̌ = {h_{i}|i ∈ I} be the set of simple coroots. Note that α_{i}(h_{j}) = a_{ji} for i, j ∈ I. Let 𝔤_{α} denote the α-root space, and Q = ∑_{i∈I} ℤα_{i} denote the root

lattice. For α, β ∈ Q, we define α > β if α − β ∈ Q_{+} = ∑_{i∈I}ℤ_{≥0}α_{i} and α ≠ β.

For α = ∑_{i∈I}k_{i}α_{i} ∈ Q_{+} define ht(α) = ∑_{i∈I}k_{i} to be the height of α. Let W be the

Weyl group of 𝔤 generated by the simple reflections {r_{i}|i ∈ I}. For w ∈ W we

denote Δ_{+}(w) = \{α ∈ Δ_{+}|w^{−1}α < 0\} and w(𝔫_{±}) = \begin{matrix}⨁ \\ α ∈ Δ_{±} \end{matrix} 𝔤_{wα}. In this paper,

for w ∈ W we study the nilpotency index of the nilpotent subalgebras

ISSN 0949–5932 / $2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag

∗ The authors thank the referees for many valuable suggestions which improved the paper a great deal. Also Misra acknowledges the partial support by NSA grant MDA904-02-1-0072.

S_{w} = 𝔫_{+} ∩ w(𝔫_{−}) = ⨁_{α∈Δ_{+}(w)} 𝔤_{α}.

In [3], Billig and Pianzola conjectured that the nilpotency index of the subalgebra S_{w} is bounded by a constant k = k(A) which depends only on the Cartan matrix A and not on w. This information helps in determining the existence of certain subroot system within the root system of a Kac-Moody Lie algebra. In this paper we prove the following theorem which settles the conjecture in most cases. Although we believe the conjecture to be true in the remaining few cases we have not been able to prove it using our approach. Main Theorem. For w ∈ W the nilpotency index of the subalgebras S_{w} is bounded by k = k(A) where

1. k is the height of the highest long root when A is of finite type.

2. k = h − 1, h being the Coxeter number, when A is of affine type 1 other than E, F or A is of type A_{2n}^{(2)} and k = ȟ − 1, ȟ being the dual Coxeter number, when A is of affine type 2 or 3 other than A_{2n}^{(2)} or E_{6}^{(2)}.

3. k = 1 when A is of indefinite type with |a_{ij}| ≥ 2 and k = 2 when

A = ( \begin{matrix}2 & −a \\ −1 & 2 \end{matrix} ) , a > 4 .

Furthermore, in each case k is the best bound possible.

In the following three sections we prove the three parts of our main theorem in Theorem 1.2, Theorem 2.5, Theorem 3.5 and Theorem 3.6.

1. A of finite type

Let A be of finite type and 𝔤 = 𝔤(A) be the associated simple Lie algebra. Let 𝔥 be the Cartan subalgebra and Π = {α_{1}, ⋯, α_{n}} be the set of simple roots. Let θ be the highest long root. Since −Π = {−α_{1}, ⋯, −α_{n}} is also a root basis and since W acts transitively on the root bases [1], there exists w_{0} ∈ W such that w_{0}^{−1}(Π) = −Π. Note that in this case clearly the nilpotency index has a bound. Our interest is to find the best possible bound k.

Proposition 1.1. Each positive root α can be written as α_{i_{1}} + ⋯ + α_{i_{k}}, α_{i_{j}} ∈ Π, such that each partial sum α_{i_{1}} + ⋯ + α_{i_{j}} is also a root.

Theorem 1.2. Let 𝔤 be a finite dimensional simple Lie algebra. If k = ht(θ), then S_{w}^{(k)} = 0 for all w ∈ W. Moreover, k is the least such integer.

Proof. Let w_{0} ∈ W such that w_{0}(Π) = −Π. Then S_{w_{0}} = \begin{matrix}⊕ \\ α ∈ Δ_{+} \end{matrix} 𝔤_{α}.

By Proposition 1.1 there is a sequence of positive roots β_{1}, ⋯, β_{k} = θ with ht(β_{i}) = i. Then S_{w_{0}}^{(t)} = ∑_{ht(α)>t}𝔤_{α}. Hence S_{w_{0}}^{(k)} = 0. On the other hand, if

s < k, then [𝔤_{α_{i}}, 𝔤_{β_{s}}] = 𝔤_{β_{s}+α_{i}} ≠ 0 for some α_{i} ∈ Π with β_{s} + α_{i} = β_{s+1}. Hence S_{w_{0}}^{(k−1)} ≠ 0. Since w_{0}^{−1}(Π) = −Π and Δ_{+}(w_{0}) = Δ_{+}, it follows that S_{w_{0}}^{(t)} ⊇ S_{w}^{(t)} for t = 1, ⋯, k, and all w ∈ W. Hence S_{w}^{(k)} = 0 for all w ∈ W.

2. A of affine type

Let A = (a_{ij})_{i,j∈I} , with I = {0, 1, ⋯, n}, be of affine type, 𝔤 = 𝔤(A) be

the associated affine Lie algebra, A^{∘} = (a_{ij}), 1 ≤ i, j ≤ n, and 𝔤^{∘} = 𝔤(A^{∘})

be the corresponding simple Lie algebra with Cartan matrix A^{∘}. Let Δ and Δ^{∘} denote the sets of roots, W and W^{∘} the Weyl groups, and Π = {α_{0}, ⋯, α_{n}}

and Π^{∘} = {α_{1}, ⋯, α_{n}} the sets of simple roots of 𝔤 and 𝔤^{∘} respectively. Let the subscripts + and − stand for positive and negative roots and let the superscripts ’re’ and ’im’ stand for real and imaginary roots respectively. Thus Δ_{+}^{re} denotes the

set of real positive roots in Δ. Let δ = ∑_{i=0}^{n}a_{i}α_{i} where each a_{i} is a positive integer,

gcd(a_{0}, ⋯, a_{n}) = 1 and A(a_{0}, ⋯, a_{n})^{t} = 0. Then Δ_{+}^{im} = {jδ|j ∈ ℤ_{>0}}. Since Δ_{+}^{im} is W -invariant, Δ_{+}(w) = {α ∈ Δ_{+}|w^{−1}(α) < 0} ⊆ Δ_{+}^{re} for each w ∈ W.

It is known that a_{0} = 1 unless A = A_{2n}^{(2)} in which case a_{0} = 2. Furthermore,

if 𝔤 is of type 1 or of type A_{2n}^{(2)}, then δ − a_{0}α_{0} is the highest long root in Δ^{∘}_{+}. Otherwise, it is the highest short root.

The real roots for 𝔤 = 𝔤(A), where A is of affine type are described in [2].

They are of the form γ = β + jδ where β ∈ Δ^{∘} and j is an integer, and when A = A_{2n}^{(2)}, we also have roots of the form γ = \frac{1}{2}(β + (2j − 1)δ) where β is a long

root in Δ^{∘} and j is an integer.

Let θ denote the highest long root in Δ^{∘}_{+}, k = ht(θ) and w_{0} ∈ W^{∘} such

that w_{0}^{−1}(Π^{∘}) = −Π^{∘}. By the description of real roots in Δ and Theorem 1.2 it follows that S_{w_{0}}^{(k−1)} ≠ 0. It is known that k is related to the Coxeter number h and the dual Coxeter number ȟ as follows. If A is of type 1, then k = h − 1. If A is of type A_{2n}^{(2)}, then k = h − 2 and k = ȟ − 1 otherwise. For the algebras in the Main Theorem, we will show that S_{w}^{(k)} = 0 for all w ∈ W (S_{w}^{(k+1)} = 0 for 𝔤 = A_{2n}^{(2)}) and that S_{w}^{(k)} ≠ 0 for some w ∈ W when 𝔤 = A_{2n}^{(2)}.

For w ∈ W, define ℕ_{0} = {0, 1, 2, … } and

X = \{β ∈ Δ^{∘}_{+}|(∃j ∈ ℕ_{0}) β + jδ ∈ Δ_{+}(w) or \frac{1}{2}(β + jδ) ∈ Δ_{+}(w)\}
Y = \{β ∈ Δ^{∘}_{−}|(∃j ∈ ℕ_{0}) β + jδ ∈ Δ_{+}(w) or \frac{1}{2}(β + jδ) ∈ Δ_{+}(w)\}

and −X = {−α|α ∈ X}, −Y = {−α|α ∈ Y }. Let {β_{i}} be a sequence in X ∪ Y

such that s_{m} = ∑_{l=1}^{m}β_{i} ∈ X ∪ Y for each m. The β_{i} are not necessarily distinct.

Proposition 2.1. Let 𝔤 = 𝔤(A) be an affine Lie algebra with Weyl group W and let w ∈ W. Then

1. X ∩ (−Y) = Ø

2. If j ≠ r, then s_{j} ≠ ±s_{r}

3. No partial sum of the β_{i} is 0.

Proof. (1) Let β ∈ X ∩(−Y). Then β+n_{1}δ or \frac{1}{2}(β+n_{1}δ) is in Δ_{+}(w) for some non-negative integer n_{1} and −β+n_{2}δ or \frac{1}{2}(−β+n_{2}δ) is in Δ_{+}(w) for some positive integer n_{2}. Suppose γ_{1} = \frac{1}{2}(β + n_{1}δ) ∈ Δ_{+}(w) and γ_{2} = (−β + n_{2}δ) ∈ Δ_{+}(w) for n_{1} ≥ 0, n_{2} > 0. Then w^{−1}(−β + n_{2}δ) < 0 and w^{−1}(\frac{1}{2}(β + n_{1}δ)) < 0. Hence either 0 < (2n_{1} + n_{2})δ < 0 or 0 < (n_{1} + n_{2})δ < 0, a contradiction. The other cases are treated in the same manner.

(2) Suppose that s_{j} = s_{r} for j > r. Then s_{r} = s_{r} + β_{r+1} + · + β_{j} and 0 = β_{r+1}+⋯+β_{j}. Thus β_{r+1} = −(β_{r+2}+⋯+β_{j}). But w^{−1}(β_{r+i}+n_{i}δ) < 0, n_{i} ≥ 0 for i = 2, ⋯, j − r. Therefore w^{−1}((β_{r+2} + ⋯ + β_{j}) + (n_{2} + ⋯ + n_{j−r})δ) < 0. If β_{r+1} ∈ X, then β_{r+2} + ⋯ + β_{j} ∈ Y and β_{r+1} ∈ X ∩ (−Y), a contradiction. Similarly, if β_{r+1} ∈ Y , then −β_{r+1} ∈ X ∩ (−Y), a contradiction. Now assume that s_{j} = −s_{r} for some j > r. Then s_{r} + β_{r+1} + ⋯ + β_{j} = −s_{r} and β_{r+1} = −(2s_{r} + β_{r+2} + · + β_{j}). Thus either β_{r+1} ∈ X ∩ (−Y) or −β_{r+1} ∈ X ∩ (−Y). Either case is a contradiction.

(3) Suppose that s_{i_{p}} = ∑_{j=1}^{p}β_{i_{j}} = 0. Then β_{i_{p}} = −∑_{j=1}^{p−1} β_{i_{j}} and β_{i_{p}} ∈

X ∩ (−Y) or −β_{i_{p}} ∈ X ∩ (−Y). Both cases yield contradictions.

For each 𝔤^{∘} we associate a graph using the highest long root θ. For 𝔤 ≠ D_{n}^{(1)} the graph has vertices θ_{0} = 0, θ_{1}, ⋯, θ_{k} = θ where θ_{j} = α_{i_{1}} + · + α_{i_{j}}, α_{i_{l}} ∈ Π^{∘}, each θ_{j} ≠ 0 for all j > 0 and k =ht(θ). As we have seen, S_{w_{0}}^{(k−1)} ≠ 0 where

w_{0}(Π^{∘}) = −Π^{∘}. For D_{n}^{(1)}, there is an extra vertex since both α_{1} + ⋯ + α_{n−1} and α_{1} + ⋯ + α_{n−2} + α_{n} are roots. The graphs and associated θ_{j} are as follows: A = A_{n}^{(1)}

θ_{0} θ_{1} θ_{2} ⋯ θ_{n−1} θ_{n}
θ_{i} = α_{1} + ⋯ + α_{i}, 1 ≤ i ≤ n, θ = α_{1} + ⋯ + α_{n}

A = B_{n}^{(1)} or D_{n+1}^{(2)}

θ_{0} θ_{1} θ_{2} ⋯ θ_{2n−2} θ_{2n−1}
θ_{i} = α_{1} + ⋯ + α_{i}, 1 ≤ i ≤ n, θ = α_{1} + 2α_{2} + ⋯ + 2α_{n}
θ_{n+i+1} = α_{1} + ⋯ + α_{n} + α_{n} + ⋯ + α_{n−i} 0 ≤ i ≤ n − 2 .

A = C_{n}^{(1)} or A_{2n}^{(2)} or A_{2n−1}^{(2)}

θ_{0} θ_{1} ⋯ θ_{2n−2} θ_{2n−1}
θ_{i} = α_{1} + ⋯ + α_{i} , 1 ≤ i ≤ n, θ = 2α_{1} + 2α_{2} + ⋯ + 2α_{n−1} + α_{n}
θ_{n+i+1} = α_{1} + ⋯ + α_{n} + ⋯ + α_{n−(i+1)}, 0 ≤ i ≤ n − 2.

A = D_{n}^{(1)}

[figure: θ_{0}; θ_{1}; θ_{2}; θ_{n−1}; θ̂_{n−1}; θ_{2n−2}; θ_{2n−1}]

θ_{i} = α_{1} + ⋯ + α_{i}, 1 ≤ i ≤ n − 1, θ̂_{n−1} = α_{1} + ⋯ + α_{n−2} + α_{n}

θ_{n} = α_{1} +⋯+α_{n}, θ_{n+i} = α_{1} +⋯+α_{n−1} +α_{n} +α_{n−2} +⋯+α_{n−i−1} 1 ≤ i ≤ n − 3,

θ = α_{1} + 2α_{2} + ⋯ + 2α_{n−2} + α_{n−1} + α_{n}

A = G_{2}^{(1)}

θ_{0} θ_{1} θ_{2} θ_{3} θ_{4} θ_{5}
θ_{1} = α_{1}, θ_{2} = α_{1} + α_{2}, θ_{3} = α_{1} + α_{2} + α_{2}
θ_{4} = α_{1} + α_{2} + α_{2} + α_{2}, θ_{5} = α_{1} + α_{2} + α_{2} + α_{2} + α_{1} = θ.

A = D_{4}^{(3)}

θ_{0} θ_{1} θ_{2} θ_{3} θ_{4} θ_{5}

θ_{1} = α_{2}, θ_{2} = α_{2} + α_{1}, θ_{3} = α_{2} + α_{1} + α_{1}

θ_{4} = α_{2} + α_{1} + α_{1} + α_{1}, θ_{5} = α_{2} + α_{1} + α_{1} + α_{1} + α_{2} = θ.

Note that the Coxeter numbers and dual Coxeter numbers for these algebras are listed on page 80 of [2] and correspond to the respective heights as previously noted.

Each root in Δ^{∘} gives a connected directed subgraph as is seen by checking each case using the results on p. 64-65 of [1].

Example 2.2. Let A = A_{5}^{(1)} and let θ_{i} = α_{1} + ⋯ + α_{i} for i = 1, ⋯, 5. Let θ = θ_{5}. Then ht(θ) = 5 and the graph is

θ_{0} θ_{1} θ_{2} θ_{3} θ_{4} θ_{5}

A typical root is β = α_{3} + α_{4} = θ_{4} − θ_{2} and the subgraph is
θ_{2} θ_{3} θ_{4}

β uses two vertices. Generally if β_{1} + ⋯ + β_{s} is a root for s = 1, ⋯, t, then t + 1 vertices are used. Since cycles are not permitted, only 5 β ’s may occur in a string. These remarks hold in the general case.

Suppose that 𝔤 ≠ A_{2n}^{(2)} and that S_{w}^{(t−1)} ≠ 0. Then there exists t roots of

the form β_{i} + m_{i}δ in Δ_{+}(w), β_{i} ∈ Δ^{∘} and m_{i} ∈ Z, whose sum is a root and the corresponding ∑_{i=1}^{t} β_{i} ∈ X ∪ Y . In case 𝔤 ≠ D_{n}^{(1)}, t < k + 1 where k = ht(θ) which is seen as follows (see also the example). s_{1} = β_{1} contains two vertices on the graph. Then s_{2} = β_{1} + β_{2} forms a connection of one of the original vertices with a new vertex since the graph is connected with no cycles using Proposition 2.1. Continuing in the manner, s_{t} uses t+1 vertices. Hence t < k+1 and S_{w}^{(k)} = 0 for all w ∈ W.

Suppose that 𝔤 = D_{n}^{(1)}. The number of vertices is k + 2. As in the preceding paragraph, t ≤ k + 1. Suppose that t = k + 1. Then there are roots

β_{1}, ⋯, β_{t} ∈ X ∪ Y with s_{j} = ∑_{i=1}^{j}β_{i} ∈ X ∪ Y for j = 1, ⋯, t.

Let z_{0} be the initial vertex of s_{t}. Hence z_{0} is the initial vertex of some β_{i_{1}} , with terminal vertex z_{1}. Then z_{1} is the initial vertex of some β_{i_{2}} with terminal vertex z_{2}. Continue this process to obtain the sequence S = {z_{0}, z_{1}, ⋯, z_{t}} with β_{i_{j}} connecting z_{j−1} to z_{j} and all vertices on the graph are listed in S. To simplify the notation we will write that β_{j} connects z_{j−1} to z_{j}.

Two vertices will be called symmetric if a subgraph joining them is of the form

B : ±(2α_{i} + ⋯ + 2α_{n−2} + α_{n−1} + α_{n}) or C : ±(α_{n−1} − α_{n}) .

We denote symmetric vertices by x_{i} and y_{i}.

Lemma 2.3. Adjacent vertices in S are not symmetric.

Proof. Otherwise a root would be of the form B or C, a contradiction.

Define z_{i} < z_{j} in S if i < j. Thus β_{i+1} + ⋯ + β_{j} connects z_{i} to z_{j}.

Lemma 2.4. x_{i} < x_{j} if and only if y_{j} < y_{i}.

Proof. Suppose that x_{i} < x_{j} and y_{i} < y_{j}. Then some β_{k_{1}} + ⋯ + β_{k_{b}} connects y_{i} to y_{j}. Then β_{i+1} + ⋯ + β_{j} = −(β_{k_{1}} + ⋯ + β_{k_{b}}) which yields a partial sum of β equal to 0. This contradicts Proposition 2.1.

The following Corollary is an immediate consequence of Lemma 2.4.

Corollary 2.5. x_{i} < x_{j} < x_{l} if and only if y_{l} < y_{j} < y_{i}.

Now suppose that z_{0} = 0 and let z_{1} = x_{i}. By Lemma 2.3, y_{i} is the final vertex in S. Let z_{2} = x_{j}. Then y_{j} is the next to last vertex in S. Continue this process and finally we obtain adjacent symmetric vertices, a contradiction. If the final vertex in S is 0, this same process yields another contradiction.

Therefore, suppose that neither the initial nor final vertex in S is 0. Then the initial and final vertices are symmetric by Lemma 2.4. Thus s_{t} has the form B or C and is not a root. But s_{t} ∈ X ∪ Y and hence is a root. Therefore t ≠ k + 1. Consequently, t < k + 1 and only k roots are allowed in a non-zero sum. Hence S_{w}^{(k)} = 0 for each w ∈ W.

Now we consider the case 𝔤 = 𝔤(A_{2n}^{(2)}). The existence of half roots makes this case special. The real roots divide into the cases

1. Δ_{s} = Δ_{s}^{re} = \{\frac{1}{2}(α + (2j − 1)δ)|α ∈ Δ^{∘}_{l}, j ∈ ℤ\}

2. Δ_{m} = Δ_{m}^{re} = {α + jδ|α ∈ Δ^{∘}_{s}, j ∈ ℤ}

3. Δ_{l} = Δ_{l}^{re} = {α + 2jδ|α ∈ Δ^{∘}_{l}, j ∈ ℤ}

where Δ^{∘}_{s} = {±(α_{i} + ⋯ + α_{n} + ⋯ + α_{j})|1 ≤ i < j ≤ n} ∪ {±(α_{i} + ⋯ + α_{j})|1 ≤

i < j ≤ n − 1} and Δ^{∘}_{l} = {±(α_{i} + ⋯ + α_{n} + ⋯ + α_{i})|1 ≤ i ≤ n}.

Define Δ_{i} +Δ_{j} = Δ_{k} if there exist γ_{i} ∈ Δ_{i}, γ_{j} ∈ Δ_{j} such that γ_{i} +γ_{j} ∈ Δ_{k}. If γ_{i} + γ_{j} is never a root when γ_{i} ∈ Δ_{i} and γ_{j} ∈ Δ_{j}, then let Δ_{i} + Δ_{j} = 0. We obtain the following addition table.

Δ_{s}  Δ_{m}  Δ_{l}
Δ_{s} Δ_{m} ∪ Δ_{l}  Δ_{s}  0
Δ_{m}  Δ_{s}  Δ_{m} ∪ Δ_{l} Δ_{m}
Δ_{l}  0  Δ_{m}  0

This table is verified case by case. The material in [1, p. 64] is helpful. Let γ_{i} = \frac{1}{2}(β_{i} + (2n_{i} − 1)δ) ∈ Δ_{s} for i = 1, 2. Then γ_{1} + γ_{2} = \frac{1}{2}(β_{1} + β_{2} +

2(n_{1} + n_{2} − 1)δ). If γ_{i} + γ_{j} is a root, then either \frac{1}{2}(β_{i} + β_{j}) ∈ Δ^{∘}_{s} and γ_{i} + γ_{j} ∈ Δ_{m}

or \frac{1}{2}(β_{i} + β_{j}) ∈ Δ^{∘}_{l} and γ_{i} + γ_{j} ∈ Δ_{l}. Hence Δ_{s} + Δ_{s} = Δ_{m} ∪ Δ_{l}.

Let γ_{1} = \frac{1}{2}(β_{1} + (2n_{1} − 1)δ) ∈ Δ_{s} and γ_{2} = β_{2} + n_{2}δ ∈ Δ_{m}. Then

γ_{1} + γ_{2} = \frac{1}{2}[(β_{1} + 2β_{2}) + (2n_{1} + 2n_{2} − 1)δ]. If γ_{1} + γ_{2} is a root, then β_{1} + 2β_{2} ∈ Δ^{∘}_{l} and γ_{1} + γ_{2} ∈ Δ_{s}. Hence Δ_{s} + Δ_{m} = Δ_{s}.

Let γ_{1} = \frac{1}{2}(β_{1} + (2n_{1} − 1)δ) ∈ Δ_{s} and γ_{2} = β_{2} + 2n_{2}δ ∈ Δ_{l}. Then γ_{1} + γ_{2} = \frac{1}{2}[(β_{1} + 2β_{2}) + (2n_{1} − 1 + 4n_{2})δ]. But β_{1} + 2β_{2} is not a root since

β_{1}, β_{2} ∈ Δ^{∘}_{l}. Hence Δ_{s} + Δ_{l} = 0.

Let γ_{i} = β_{i} + n_{i}α ∈ Δ_{m} for i = 1, 2. Then γ_{1} + γ_{2} = (β_{1} + β_{2}) + (n_{1} + n_{2})δ.

It is possible that β_{1} + β_{2} ∈ Δ^{∘}_{l} or in Δ^{∘}_{s}. Hence γ_{1} + γ_{2} ∈ Δ_{m} ∪ Δ_{l}. Therefore, Δ_{m} + Δ_{m} = Δ_{m} ∪ Δ_{l}.

Let γ_{1} = β_{1} + 2n_{1}δ ∈ Δ_{l}, γ_{2} = β_{2} + n_{2}δ ∈ Δ_{m}. Then γ_{1} + γ_{2} = (β_{1}+β_{2})+(2n_{1}+n_{2})δ. If γ_{1}+γ_{2} is a root, then β_{1}+β_{2} ∈ Δ^{∘}_{s}. Hence γ_{1}+γ_{2} ∈ Δ_{m}. So Δ_{m} + Δ_{l} = Δ_{m}.

Let γ_{i} = β_{i} + 2n_{i}δ ∈ Δ_{l}, for i = 1, 2. Then γ_{1} + γ_{2} = β_{1} + β_{2} + 2(n_{1} + n_{2})δ. But β_{1} + β_{2} is not a root. Hence γ_{1} + γ_{2} is not a root and Δ_{l} + Δ_{l} = 0.

In the graph of A_{2n}^{(2)} there are 2n vertices and k = ht(θ) = 2n − 1. As in previous cases no more than k β_{i}’s can exist in a given sum that is a non-zero root. We proceed by considering how the number of summands in a sum of γ_{i} ∈ Δ^{+}(w) translates into the number of summands in the corresponding β_{i}. From the results in the table and their proofs, adding a γ adds one β unless

D : γ_{1} + ⋯ + γ_{j} ∈ Δ_{s} and γ_{j+1} ∈ Δ_{s}
or E : γ_{1} + ⋯ + γ_{j} ∈ Δ_{m} and γ_{j+1} ∈ Δ_{s}
or F : γ_{1} + ⋯ + γ_{j} ∈ Δ_{s} and γ_{j+1} ∈ Δ_{m}.

In the first case, the number of β ’s do not decrease. In the other cases, the number of β ’s increase by at least two.

Again consider γ_{1} + ⋯ + γ_{t}. Let n_{t} be the number of corresponding β ’s and set n_{0} = 0. Then n_{1} = 1. Checking cases it is seen that n_{2} < 2 implies n_{2} = 1 and γ_{1} + γ_{2} ≠ Δ_{s}.

We claim that n_{j} < j implies n_{j} = j − 1 and γ_{1} + ⋯ + γ_{j} ∉ Δ_{s}. Assume the result holds for j and consider γ_{1} + ⋯ + γ_{j+1}. Suppose that n_{j} = j − 1. By assumption γ_{1} +⋯+γ_{j} ∈ Δ_{m} ∪Δ_{l}. If γ_{j+1} ∈ Δ_{s}, then n_{j+1} ≥ j +1. If γ_{j+1} ∉ Δ_{s}, then n_{j+1} = j. Now suppose that n_{j} ≠ j − 1. Then n_{j} ≥ j implies n_{j+1} ≥ j. If n_{j+1} = j = n_{j}, then γ_{1} + ⋯ + γ_{j}, γ_{j+1} ∈ Δ_{s} and γ_{1} + ⋯ + γ_{j+1} ≠ Δ_{s}. Hence the result holds.

Now if S_{w}^{(t−1)} ≠ 0, then t−1 ≤ n_{t} < k +1 and t < k +2. Hence S_{w}^{(k+1)} = 0. It remains to show that S_{w}^{(k)} ≠ 0 for some w ∈ W. We will work with the

graph for A_{2n}^{(2)} and the discussion of the roots of C_{n} in [1, p. 64]. The roots are of the form ±2e_{i} and ±(e_{i} ± e_{j}), i ≠ j with base α_{j} = e_{j+1} − e_{j}, j = 1, ⋯, n − 1

and α_{n} = 2e_{n}. Let w_{0} ∈ W^{∘} such that w_{0}^{−1}(Π^{∘}) = −Π^{∘}. Let γ_{1} = \frac{1}{2}(θ + (2j − 1)δ), γ_{2} = \frac{1}{2}(θ + (2j + 1)δ), γ_{3} = \frac{1}{2}(−α_{1} + j_{1}δ), ⋯, γ_{k+2} = (−α_{k} + j_{k}δ) where k = ht(θ) = 2n − 1. The first two of these roots are in Δ_{s} and γ_{1} + γ_{2} = θ + 2jδ ∈ Δ_{l}. The remaining roots are in Δ_{m} and γ_{1} + ⋯ + γ_{j} ∈ Δ_{m} for all j > 2. On the graph, θ goes from θ_{0} to θ_{k} and the second θ does not move from vertex θ_{k}. Each remaining γ_{i} moves one vertex to the left. Hence γ_{1} + ⋯ + γ_{k+1} is a root and S_{w_{0}}^{(k)} ≠ 0 but γ_{1} + ⋯ + γ_{k+2} is a cycle and S_{w_{0}}^{(k+1)} = 0.

Due to the relation between the heights of the highest long roots with the Coxeter numbers and dual Coxeter numbers pointed out earlier in the paper, we have completed the proof of the following theorem.

Theorem 2.6. Let 𝔤 = 𝔤(A) be any affine Lie algebra other than type E or F. Then there is a smallest positive integer m such that S_{w}^{(m)} = 0 for all w ∈ W. Furthermore,

1. if A is of affine type 1 or A_{2n}^{(2)} then m = h−1 where h is the Coxeter number, and

2. if A is of affine type 2 or 3 and A ≠ A_{2n}^{(2)}, E_{6}^{(2)}, then m = ȟ−1 where ȟ is the dual Coxeter number.

3. A of indefinite type

Let A = (a_{ij}), 1 ≤ i, j ≤ n be an indecomposable generalized Cartan matrix of indefinite type and 𝔤 = 𝔤(A) be the associated Kac-Moody Lie algebra. Let Π = {α_{1}, α_{2}, ⋯, α_{n}} denote the set of simple roots and W = ⟨r_{1}, r_{2}, ⋯, r_{n}⟩ denote the Weyl group of 𝔤, where r_{i} = r_{α_{i}} are the simple reflections. The following lemma is well known.

Lemma 3.1. 2 . For w ∈ W, let w = r_{i_{1}} ⋯ r_{i_{t}} be a reduced expression. Then Δ_{+}(w) = {β_{1}, β_{2}, ⋯, β_{t}}, where β_{p} = r_{i_{1}} ⋯ r_{i_{p−1}}(α_{i_{p}}), 1 ≤ p ≤ t, are distinct real positive roots.

Proposition 3.2. Let w = r_{i_{1}} ⋯ r_{i_{t}} ∈ W be reduced and

β_{p} = r_{i_{1}} ⋯ r_{i_{p−1}}(α_{i_{p}}), 1 ≤ p ≤ t

as above. Then for 1 ≤ k < j ≤ t, 1 ≤ l ≤ t, we have β_{k} + β_{j} = β_{l} ⇒ k < l < j.

Proof. Clearly l ≠ k, j. Suppose l < k. Then

r_{i_{1}} ⋯ r_{i_{k−1}}α_{i_{k}} + r_{i_{1}} ⋯ r_{i_{j−1}}α_{i_{j}} = r_{i_{1}} ⋯ r_{i_{l−1}}α_{i_{l}}

implies

r_{i_{l}} ⋯ r_{i_{k−1}}α_{i_{k}} + r_{i_{l}} ⋯ r_{i_{j−1}}α_{i_{j}} = α_{i_{l}} .

Since r_{i_{l}} ⋯ r_{i_{k−1}}r_{i_{k}} and r_{i_{l}} ⋯ r_{i_{j−1}}r_{i_{j}} are reduced by Lemma 3.1 we have

r_{i_{l}} ⋯ r_{i_{k−1}}α_{i_{k}} > 0 and r_{i_{l}} ⋯ r_{i_{j−1}}α_{i_{j}} > 0 ,

which implies that

\begin{matrix}1 = ht(α_{i_{l}}) = ht(r_{i_{l}} ⋯ r_{i_{k−1}}α_{i_{k}} + r_{i_{l}} ⋯ r_{i_{j−1}}α_{i_{j}}) \\ = ht(r_{i_{l}} ⋯ r_{i_{k−1}}α_{i_{k}}) + ht(r_{i_{l}} ⋯ r_{i_{j−1}}α_{i_{j}}) ≥ 1 + 1 = 2 , \end{matrix}

a contradiction. Now suppose j < l. Then β_{k} + β_{j} = β_{l} implies

α_{i_{j}} = r_{i_{j}} ⋯ r_{i_{l−1}}α_{i_{l}} + r_{i_{j−1}} ⋯ r_{i_{k+1}}α_{i_{k}} .

As before, since r_{i_{j}} ⋯ r_{i_{l−1}}r_{i_{l}} and r_{i_{j−1}} ⋯ r_{i_{k+1}}r_{i_{k}} are reduced, by Lemma 3.1 we have

ht(r_{i_{j}} ⋯ r_{i_{l−1}}α_{i_{l}}) ≥ 1 and ht(r_{i_{j−1}} ⋯ r_{i_{k+1}}α_{i_{k}}) ≥ 1 ,

which gives a contradiction. Hence k < l < j.

Corollary 3.3. Let w = r_{i_{1}} ⋯ r_{i_{t}} ∈ W be reduced and β_{p} = r_{i_{1}} ⋯ r_{i_{p−1}}α_{i_{p}}, 1 ≤ p ≤ t as before. If β_{1} < β_{2} < ⋯ < β_{t}, then S_{w}^{(1)} = 0.

Proof. Since {β_{p}} is an increasing sequence, it follows from Proposition 3.2 that β_{i} + β_{j} ∉ Δ^{+}(w), 1 ≤ i, j ≤ t. Hence S_{w}^{(1)} = [S_{w}, S_{w}] = [𝔤_{β_{1}} ⊕ ⋯ ⊕ 𝔤_{β_{t}}, 𝔤_{β_{1}} ⊕ ⋯ ⊕ 𝔤_{β_{t}}] = 0.

Lemma 3.4. Let A = (a_{ij}) with |a_{ij}| ≥ 2, 1 ≤ i, j ≤ n. Then for β ∈ Q_{+}, r_{i}β ≥ β ⇒ r_{k}(r_{i}β) ≥ r_{i}β for all k ≠ i.

Proof. Let β = ∑_{j=1}^{n} x_{j}α_{j} ∈ Q^{+}. Then r_{i}β ≥ β implies β(h_{i}) ≤ 0, which implies 2x_{i} ≤ ∑_{j≠i}(−a_{ij})x_{j}. To show that r_{k}(r_{i}β) ≥ r_{i}β, we need to show that (β − β(h_{i})α_{i})(h_{k}) = β(h_{k}) − β(h_{i})a_{ki} ≤ 0. We have

β(h_{i})a_{ki} − β(h_{k}) = (2x_{i} + ∑_{j≠i}(a_{ij})x_{j})a_{ki} − (2x_{k} + ∑_{j≠k}(a_{kj})x_{j})
= x_{i}a_{ki} + (a_{ik}a_{ki} − 2)x_{k} + ∑_{j≠k,i}(a_{ij}a_{ki} − a_{kj})x_{j}
≥ \frac{1}{2} ∑_{j≠i}(−a_{ij})a_{ki}x_{j} + (a_{ik}a_{ki} − 2)x_{k} + ∑_{j≠k,i}(a_{ij}a_{ki} − a_{kj})x_{j}
= \frac{1}{2} [(a_{ik}a_{ki} − 4)x_{k} + ∑_{j≠k,i}(a_{ij}a_{ki} − 2a_{kj})x_{j}]
≥ 0 ,

since |a_{ij}| ≥ 2, hence (a_{ik}a_{ki} − 4) ≥ 0 and (a_{ij}a_{ki} − 2a_{kj}) ≥ 0. Therefore, the result follows.

Theorem 3.5. Let A = (a_{ij}) with |a_{ij}| ≥ 2, 1 ≤ i, j ≤ n. Then S_{w}^{(1)} = 0 for all w ∈ W.

Proof. Let w = r_{i_{1}}r_{i_{2}} ⋯ r_{i_{t}} ∈ W be reduced. Then Δ^{+}(w) = {β_{1}, β_{2}, ⋯, β_{t}} where β_{p} = r_{i_{1}} ⋯ r_{i_{p−1}}(α_{i_{p}}), 1 ≤ p ≤ t. Consider

\begin{aligned}β_{p+1} − β_{p} & = & r_{i_{1}} ⋯ r_{i_{p}}α_{i_{p+1}} − r_{i_{1}} ⋯ r_{i_{p−1}}α_{i_{p}} \\ & = & r_{i_{1}} ⋯ r_{i_{p−1}}β , \end{aligned}

where β = (−a_{i_{p}i_{p+1}} − 1)α_{i_{p}} + α_{i_{p+1}} ∈ Q_{+} since a_{i_{p}i_{p+1}} ≤ −2. Now

\begin{aligned}r_{i_{p−1}}β − β & = −β(h_{i_{p−1}})α_{i_{p−1}} \\ & = [(−a_{i_{p}i_{p+1}} − 1)(−a_{i_{p−1}i_{p}}) + (−a_{i_{p−1}i_{p+1}})]α_{i_{p−1}}. \end{aligned}

Since −a_{ij} ≥ 2 for i ≠ j, we have r_{i_{p−1}}β − β ≥ 0, hence r_{i_{p−1}}β ≥ β . Therefore, by Lemma 3.4, we have β_{p+1} − β_{p} = r_{i_{1}} ⋯ r_{i_{p−1}}β ≥ β > 0 for 1 ≤ p ≤ t . This implies that β_{1} < β_{2} < ⋯ < β_{t} and hence by Corollary 3.3 we have S_{w}^{(1)} = 0.

Now consider the case A = ( \begin{matrix}2 & −a \\ −b & 2 \end{matrix} ) , ab > 4, a, b ≥ 0, any rank two

hyperbolic GCM. If a ≥ 2 and b ≥ 2, then by Theorem 3.5 we have S_{w}^{(1)} = 0 for all w ∈ W. Hence without loss of generality assume b = 1 and a > 4. In this case the Weyl group W is an infinite dihedral group with presentation W = {(r_{1}r_{2})^{j}, r_{2}(r_{1}r_{2})^{j}|j ∈ ℤ}. We define the sequence of integers {A_{j}}_{j∈ℤ} by the recurrence relations

A_{0} = 0, A_{1} = 1, A_{2j+2} = A_{2j+1} − A_{2j} and A_{2j+1} = aA_{2j} − A_{2j−1} .

Note that, A_{−j} = −A_{j} for j ∈ ℤ and

r_{2}(α_{1}) = A_{1}α_{1} + A_{2}α_{2} .

Assume that

r_{2}(r_{1}r_{2})^{j}(α_{1}) = A_{2j+1}α_{1} + A_{2j+2}α_{2} .

Then

r_{2}(r_{1}r_{2})^{j+1}(α_{1}) = (r_{2}r_{1})r_{2}(r_{1}r_{2})^{j}(α_{1})
= (r_{2}r_{1})(A_{2j+1}α_{1} + A_{2j+2}α_{2})
= −A_{2j+1}(α_{1} + α_{2}) + A_{2j+2}(aα_{1} + aα_{2} − α_{2})
= (aA_{2j+2} − A_{2j+1})α_{1} + (aA_{2j+2} − A_{2j+1} − A_{2j+2})α_{2}
= A_{2j+3}α_{1} + (A_{2j+3} − A_{2j+2})α_{2} = A_{2j+3}α_{1} + A_{2j+4}α_{2} .

Hence by induction for j ≥ 0 we have

r_{2}(r_{1}r_{2})^{j}(α_{1}) = A_{2j+1}α_{1} + A_{2j+2}α_{2} .

Similarly,

(r_{1}r_{2})^{j}(α_{1}) = A_{2j+1}α_{1} + A_{2j}α_{2} .

Note that

\begin{aligned}r_{2}(r_{1}r_{2})^{−j}(α_{1}) & = & −(r_{1}r_{2})^{j−1}(α_{1}) = −A_{2j−1}α_{1} − A_{2j−2}α_{2} \\ & = & A_{2(−j)+1}α_{1} + A_{2(−j)+2}α_{2} , \end{aligned}

and

\begin{aligned}(r_{1}r_{2})^{−j}(α_{1}) & = & −r_{2}(r_{1}r_{2})^{j−1}(α_{1}) \\ & = & −A_{2j−1}α_{1} − A_{2j}α_{2} = A_{2(−j)+1}α_{1} + A_{2(−j)}α_{2} . \end{aligned}

Therefore, for all j ∈ ℤ we have

\begin{aligned}r_{2}(r_{1}r_{2})^{j}(α_{1}) & = & A_{2j+1}α_{1} + A_{2j+2}α_{2} , \\ (r_{1}r_{2})^{j}(α_{1}) & = & A_{2j+1}α_{1} + A_{2j}α_{2} . \end{aligned}

Similarly,

\begin{aligned}r_{2}(r_{1}r_{2})^{j}(α_{2}) & = & −aA_{2j}α_{1} − A_{2j+1}α_{2} , \\ (r_{1}r_{2})^{j}(α_{2}) & = & −aA_{2j}α_{1} − A_{2j−1}α_{2} . \end{aligned}

Theorem 3.6. Let A = ( \begin{matrix}2 & −a \\ −1 & 2 \end{matrix} ) , a > 4. Then S_{w}^{(1)} ≠ 0 and S_{w}^{(2)} = 0

for all w ∈ W.

Proof. In this case W = {r_{2}(r_{1}r_{2})^{j}, (r_{1}r_{2})^{j}|j ∈ ℤ}. We consider the case w = r_{2}(r_{1}r_{2})^{j}, j ≥ 0. The proof for the other cases are similar. For w = r_{2}(r_{1}r_{2})^{j}, l(w) = 2j + 1 and

Δ_{+}(w) = {β_{1}, β_{2}, ⋯, β_{2j+1}}

where

\begin{aligned}β_{2p−1} & = & (r_{2}r_{1})^{p−1}α_{2} = (r_{1}r_{2})^{−p+1}α_{2} = −aA_{−2p+2}α_{1} − A_{−2p+1}α_{2} \\ & = & aA_{2p−2}α_{1} + A_{2p−1}α_{2} , \end{aligned}

for p = 1, 2, ⋯, j + 1 and for p = 1, 2, ⋯, j,

β_{2p} = (r_{2}r_{1})^{p−1}r_{2}(α_{1}) = r_{2}(r_{1}r_{2})^{p−1}(α_{1}) = A_{2p−1}α_{1} + A_{2p}α_{2} .

Note that since β_{p} ∈ Δ_{+}, 1 ≤ p ≤ 2j + 1, we have A_{p} ≥ 0. Observe that since a > 4, we have

β_{4} = A_{3}α_{1} + A_{4}α_{2} = (a − 1)α_{1} + (a − 2)α_{2} > α_{1} + α_{2} = β_{2} .

Assume

β_{2p} = A_{2p−1}α_{1} + A_{2p}α_{2} > β_{2p−2} = A_{2p−3}α_{1} + A_{2p−2}α_{2} .

Hence

A_{2p−1} ≥ A_{2p−3} and A_{2p} ≥ A_{2p−2} .

Then

\begin{aligned}A_{2p+1} − A_{2p−1} & = & aA_{2p} − 2A_{2p−1} = aA_{2p} − 2(A_{2p} + A_{2p−2}) \\ & = & (a − 2)A_{2p} − 2A_{2p−2} > 2(A_{2p} − A_{2p−2}) ≥ 0 , \end{aligned}

and

A_{2p+2} − A_{2p} = A_{2p+1} − 2A_{2p} = A_{2p+1} − \frac{2}{a}(A_{2p+1} + A_{2p−1})
= \frac{1}{a} ((a − 2)A_{2p+1} − 2A_{2p−1})
> \frac{2}{a}(A_{2p+1} − A_{2p−1}) ≥ 0.

Hence,

β_{2p+2} − β_{2p} = (A_{2p+1} − A_{2p−1})α_{1} + (A_{2p+2} − A_{2p})α_{2} > 0

Therefore, by induction the subsequence {β_{2p}}_{p=1}^{j} is an increasing sequence. Similarly, the subsequence {β_{2p−1}}_{p=1}^{j+1} is also an increasing sequence. Also since a > 4

β_{4} − β_{1} = (a − 1)α_{1} + (a − 3)α_{2} > 0,

and (a − 1)(a − 3) > (a − 1). Note that β_{2p+2} − β_{2p−1} = (r_{2}r_{1})(β_{2p} − β_{2p−3}). Hence using induction we have β_{2p+2} − β_{2p−1} > 0.

\begin{aligned}β_{2p} + β_{2p+2} & = & (A_{2p−1} + A_{2p+1})α_{1} + (A_{2p} + A_{2p+2})α_{2} \\ & = & aA_{2p}α_{1} + A_{2p+1}α_{2} = β_{2p+1} \end{aligned}

Therefore, S_{w}^{(1)} ≠ 0 and also β_{2p−1} < β_{2p+2} < β_{2p+1} for p = 1, 2, ⋯, j − 1. Observe that

\begin{aligned}β_{2p−1} + β_{2p+1} & = & a(A_{2p−2} + A_{2p})α_{1} + (A_{2p−1} + A_{2p+1})α_{2} \\ & = & aA_{2p−1}α_{1} + aA_{2p}α_{2} = aβ_{2p} ∉ Δ_{+}(w). \end{aligned}

Furthermore, since {β_{2p}} and {β_{2p−1}} are increasing sequences and β_{2p−1} < β_{2p+2} < β_{2p+1} for p = 1, 2, ⋯, j − 1, by Proposition 3.2 we have β_{2k−1} + β_{2p+1} ∉ Δ_{+}(w), β_{2k} + β_{2p+1} ∉ Δ_{+}(w), β_{2k−1} + β_{2p} ∉ Δ_{+}(w) and β_{2k} + β_{2p} ≠ β_{2m} for k < m < p. For example, if β_{2k−1} + β_{2p+1} = β_{2m} then β_{2m} > β_{2p+1} > β_{2p+2} which is a contradiction since {β_{2p}} is an increasing sequence. If β_{2k−1} + β_{2p} = β_{2m−1} for some k < m ≤ p then since β_{2p} > β_{2p−3} the only possibility is m = p. However, β_{2k−1} + β_{2p} = β_{2p−1} = β_{2p−2} + β_{2p} will imply β_{2k−1} = β_{2p−2} which is not true. Therefore,

S_{w}^{(1)} ⊆ ⨁_{p=1}^{j+1} 𝔤_{β_{2p−1}}

and hence by Corollary 3.3, S_{w}^{(2)} = [S_{w}, S_{w}^{(1)}] = 0, since {β_{2p−1}} is an increasing sequence.

References

[1] Humphreys, J. E., “Introduction to Lie Algebras and Representation Theory,” Springer-Verlag, 1972.

[2] Kac, V. G., “Infinite-Dimensional Lie Algebras,” Third ed., Cambridge University Press, 1990.

[3] Billig, Yuly, and Arturo Pianzola, Root Strings with Two Consecutive Real roots, Tôhoku Math. J., 47 (1995), 391–403.

[4] Kim, Y., A Note on the Root System of an Affine Lie Algebra of Type A_{2l}^{(2)}, Commun. Korean Math. Soc. 16 (2001), 85–94.

Yeonok Kim Department of Mathematics Soong Sil University Seoul 151 Korea yokim@ssu.ac.kr Kailash C. Misra and Ernie Stitzinger Department of Mathematics North Carolina State University Raleigh, NC 27695-8205 USA misra@math.ncsu.edu stitz@math.ncsu.edu

Received January 21, 2002 and in final form September 19, 2003