Existence of Richardson elements in seaweed Lie algebras of type $\mathbb{B}$, $\mathbb{C}$ and $\mathbb{D}$

Seaweed Lie algebras are a natural generalisation of parabolic subalgebras of reductive Lie algebras. The well-known Richardson Theorem says that the adjoint action of a parabolic group has a dense open orbit in the nilpotent radical of its Lie algebra \cite{richardson}. We call elements in the open orbit Richardson elements. In \cite{JSY} together with Yu, we generalized Richardson's Theorem and showed that Richardson elements exist for seaweed Lie algebras of type $\mathbb{A}$. Using GAP, we checked that Richardson elements exist for all exceptional simple Lie algebras except $\mathbb{E}_8$, where we found a counterexample. In this paper, we complete the story on Richardson elements for seaweeds of finite type, by showing that they exist for any seaweed Lie algebra of type $\mathbb{B}$, $\mathbb{C}$ and $\mathbb{D}$. By decomposing a seaweed into a sum of subalgebras and analysing their stabilisers, we obtain a sufficient condition for the existence of Richarson elements. The sufficient condition is then verified using quiver representation theory. More precisely, using the categorical construction of Richardson elements in type $\mathbb{A}$, we prove that the sufficient condition is satisfied for all seaweeds of type $\mathbb{B}$, $\mathbb{C}$ and $\mathbb{D}$, except in two special cases, where we give a directproof.


Introduction
Throughout we work over k = C. Let g be a reductive Lie algebra and let G be a connected reductive algebraic group with Lie algebra g. Definition 1.1. [5,13] A Lie subalgebra q of g is called a seaweed subalgebra if there exists a pair (p, p ′ ) of parabolic subalgebras of g such that q = p ∩ p ′ and p + p ′ = g. Two such parabolic subalgebras are said to be opposite.
We are interested in the adjoint action of a seaweed Lie algebra on its nilpotent radical and the density of the action, continuing the work in [8] and [9]. Note that seaweed Lie algebras are also called biparabolic subalgebras by Joseph [10] and that the coadjoint action and the index of seaweed Lie algebra were considered by . A number of subsequent papers have studied the coadjoint action for these algebras, for instance [6] by Duflo and Yu,[10], [11], [12] by Joseph, [13] by Panyushev and [16] by Tauvel and Yu. Definition 1.2. An element x in the nilpotent radical n of a seaweed Lie algebra q is called a Richardson element if [q, x] = n.
This work was supported by EPSRC 1st grant EP/1022317/1. The results were written up during the authors' visit to the University of New South Wales. Both authors would like to thank J. Du and the School of Mathematics and Statistics for their hospitality. The first author would also like to thank R. W. Z. Yu for the invitation of presenting this work at their seminar, University of Reims Champagne-Ardenne.
In the case where q is parabolic, a well known theorem of Richardson [14] (see also [15,Chapter 33]) says that Richardson elements exist. The following natural question was raised independently by Duflo and Panyushev. Question 1.3. [9] Does a seaweed Lie algebra have a Richardson element?
Inspired by earlier work of Brüstle-Hille-Ringel-Röhrle [4] for parabolics in type A, Jensen-Su-Yu [9] give a positive answer to this question for all seaweeds of type A. In this paper we prove the existence of Richardson elements for seaweed Lie algebras of type B, C and D. Theorem 1.4. Let g be a Lie algebra of type B, C or D. Then any seaweed Lie algebra in g has a Richardson element.
We should mention that the method of proving the theorem is different from Richardson's [14]. Although the fact that the parabolic group is the normaliser of the nilpotent radical plays a crucial role in Richardson's proof, it is not a necessary condition. Evidently, abelian seaweeds have Richardson elements, but the corresponding seaweed groups are often not normalisers of the nilpotent radicals. Also, Brüstle-Hille-Ringel-Röhrle's [4] constructive proof of Richardson Theorem does not use this fact. For seaweed Lie algebras of type A, Jensen-Su-Yu [9] adapted the quiver approach in [4] and proved constructively the existence of Richardson elements. See Example 3.5 for an explicit construction in type A. The key ingredient of the quiver approach is the interplay between Lie algebras and quiver representation theory. For instance, endomorphism rings of representations that give us Richardson elements in seaweeds of type A correspond to stabilisers of the Richardson elements. In this paper we exclusively analyse properties of the endomorphisms at vertices in the quiver and apply these properties to prove the main theorem.
As a consequence of Theorem 1.4, Richardson elements exist for any seaweed Lie algebra of classical type. It is possible to show (using GAP) that Richardson elements exist in all exceptional types except E 8 (see [9]). We provide more detail on the counterexample at the end of this paper, including the GAP source code. At the moment, there is no conceptual explanation as to why E 8 is different. Also, no categorical realisation of Richardson elements similar to [4] and [9] in type A is known in the other classical types. See however [2] for partial results in the parabolic case.
The remainder of this paper is organised as follows. In Section 2, we recall the notion of a standard seaweed Lie algebras and prove some technical lemmas. We decompose standard seaweeds, including standard parabolics, as sums of two Lie subalgebra and consider how one subalgebra interacts with elements in the nilpotent radical of the others. Further, we show that a local property of stabilisers of Richardson elements of seaweeds of type A is sufficient for the existence of Richardson elements in seaweeds of other types. This condition can be verified using the categorical construction of Richardson elements [8,9] in type A. This is the key to the proof of the existence of Richardson elements in this paper. As such, we recall the categorical construction in type A in Section 3, and prove essential results on stabilisers in Section 4. In Section 5, we prove the main result. The proof is split into two cases. The first case deals with algebras with a decomposition as in Section 2. A completely different argument is needed for the remaining case, which is of type D. In the last section, we discuss the counterexample in E 8 .

2.
Richardson elements and decomposition of seaweeds 2.1. Standard seaweeds and parabolics. We fix a Borel subalgebra b of g and a Cartan subalgebra h contained in b. Denote by Φ, Φ + , Φ − and Π, respectively, the root system, the set of positive roots, the set of negative roots and the set of positive simple roots, determined by h, b and g. For α ∈ Φ, denote by g α the root space corresponding to α. Write α = We say that α is supported at a positive simple root α i if x i = 0 and call the set of all such simple roots the support of α. For S, T ⊂ Π, let Φ S be the set of roots with support in S, Note that p ± S are parabolic subalgebras and q S,T is a seaweed Lie algebra. Parabolics and seaweeds constructed in this way are said to be standard with respect to the choice of h and b.
Proposition 2.1. [13,15] Any seaweed Lie algebra in g is G-conjugate to a standard seaweed Lie algebra.
As a consequence, it suffices to consider standard seaweeds when proving the existence of Richardson elements. Let where n ± S,T = α∈Φ ± S,T g α and l S,T = h ⊕ α∈Φ S∩T g α . Then l S,T is the Levi-subalgebra of q S,T and n S,T = n + S,T ⊕ n − S,T is the nilpotent radical of q S,T . In the sequel, we will assume that neither S or T is equal to ∅ or Π. Note that two algebras of the same type may have different rank and we view By the type of a seaweed q ⊆ g we mean the type of g and thus the type of q is not well-defined without an embedding q ⊆ g.
We note the following symmetry with respect to the choice of S and T .
Lemma 2.2. The seaweed q S,T has a Richardson element if and only if so does the seaweed q T,S .
Proof. The lemma follows from the involution g → g mapping g α onto g −α .
2.2. A decomposition of seaweed subalgebras and Richardson elements. Let q S,T be a standard seaweed in a simple Lie algebra g of type B, C or D. Note that . Denote the positive simple roots of g by α 1 , . . . , α n , with the corresponding Dynkin graph numbered as follows.
be the Cartan matrix of g and e i , f i , h i be the Chevalley generators of g with g α i = span{e i }. That is, g is generated by the generators subject to the relations. ( We choose a new basis h 1 , · · · , h n of the Cartan subalgebra of g as follows. When g is of type B, h 1 = h 1 ; when g is of type C, h 1 = 1 2 h 1 ; when g is of type D, For all other types and for all other i when g is of type D, Proof. Direct computation gives the following, otherwise, except when i = 2 and g is of type D, where we have So the lemma follows.
Let h i be the subspace spanned by h i . Let ǫ = min{i | α i ∈ S} and η = min{i | α i ∈ T }.
By Lemma 2.2, we may assume that ǫ ≥ η ≥ 1. Let We define two subspaces of g, (1) If ǫ > 2 when g is of type D or ǫ > 1 for other types, then g 1 is a Lie subalgebra of the same type as g.
Proof. By the definition of h i , we have So the lemma follows.
Lemma 2.5. If ǫ = ω, then q S,T has a Richardson element.
Proof. If ω = ǫ = 1, then the root spaces in q S,T are supported on a subgraph of type A, and so q S,T is isomorphic to a seaweed of type A. Similarly, if ω = ǫ = 2 and g is of type D, then q S,T is also isomorphic to a seaweed of type A. If ω = ǫ = n, then q S,T is isomorphic to a parabolic of the same type as g. In all three cases, by Theorem 1.2 in [9] or Richardson's Theorem [14], q S,T has a Richardson element.
Otherwise, q S,T = q 2 ⊕ q 1 , where q 1 ⊆ g 1 is parabolic of the same type as g, and q 2 ⊆ g 2 is a seaweed of type A. Further, by ( * ), [g 1 , g 2 ] = 0 and so [q 1 , q 2 ] = 0. Since both q 1 and q 2 have Richardson elements, we can conclude that q S,T has a Richardson element.
Consequently we assume that ǫ > ω for the remainder of the paper. We also assume that (ǫ, ω) = (2, 1), when g is of type D. Let Note that S ′ = {α i |i < ǫ}. These subsets determine two subalgebras of q S,T , the positive parabolic subalgebra c S,T of g 1 determined by T ′ and the seaweed Lie subalgebra a S,T of g 2 determined by S ′′ and T ′′ . Example 2.6. Let g be a Lie algebra of type D 6 , S = {5, 3, 2, 1} and T = {6, 4, 2, 1}. Then ǫ = 4, ω = 3. The subalgebras a S,T and c S,T can for instance be described using matrices as follows, where a S,T is marked by * and †, and c S,T is marked by − and †. The one dimensional intersection is marked by †, and there is an anti-symmetry to the anti-diagonal.
Let n a and n c be the nilpotent radicals of a S,T and c S,T , respectively.
(1) Let α be a positive root such that g α ⊆ q S,T . Then α is not supported at α ǫ , since ǫ ∈ S. By the assumption on the root system, α must be supported on simple roots α i with all i < ǫ or all i > ǫ. So g α ⊆ a S,T or g α ⊆ c S,T . Similarly, a negative root β with g β ⊆ q S,T is supported on simple roots −α i for all i < ω or all i > ω, and so g β ⊆ a S,T or g β ⊆ c S,T . By construction, [g α i , g −α i ] ⊆ a S,T + c S,T for all simple roots α i , and so q S,T = a S,T + c S,T .
By Theorem 1.2 in [9], Richardson elements exist in a S,T . Let r 2 ∈ a S,T be a Richardson element and denote by stab a S,T (r 2 ) = {x ∈ a S,T |[x, r 2 ] = 0}, the stabiliser of r 2 in a S,T . For a subalgebra u ⊆ g given as a direct sum of root spaces and subspaces [g α i , g −α i ], let x |u be the canonical projection of x ∈ g onto u. Let Proof. Assume [c r 2 , r 1 ] = n c . Take any (x a , x c ) ∈ n a ⊕ n c . There exists y a ∈ a S,T such that [y a , r 2 ] = x a . Write y a = y ′ a + (y a ) |l . Note that r 1 | gα = 0 can occur only for positive roots α with support contained in {α ǫ−1 , . . . , α 1 } and y ′ a | g β = 0 can occur only for positive roots β with support contained in {α n , . . . , α ǫ+1 }, or negative roots with support contained in {α n , . . . , α ω+1 } and containing at least one α j for some j ≥ ǫ. So by the property ( * ) in Section 2.2 and the fact from Lemma 2.3 that [h i , g α j ] = 0 for i ≥ ǫ and j < ǫ, we have (1) [y ′ a , r 1 ] = 0 and so [y a , This completes the proof of the lemma.

A decomposition of parabolic subalgebras and Richardson elements.
Let S, T, ǫ, ω, g 1 , g 2 and l be defined as in Section 2.2 with ω < ǫ. We also continue the assumption that (ǫ, ω) = (2, 1) if g is of type D. Let g ′ be a Lie algebra of the same type as g, with rank at least ǫ and root system denoted by Φ ′ . We may assume that both g and g ′ are subalgebras of a Lie algebra of the same type as g such that g ⊆ g ′ or g ′ ⊆ g. Here all inclusions are induced by inclusions of Dynkin diagrams.
Let p + U ⊆ g ′ be the standard parabolic subalgebra determined by U with ǫ ∈ U and We choose a basis {h ′ i } i in the same manner as we did for the basis {h i } i of the Cartan subalgebra of g. Let g ′ 1 = g 1 and let g ′ 2 ⊆ g ′ be defined similarly to g 2 ⊆ g, i.e. ( These two sets determine the following standard parabolic subalgebras of g ′ 2 and g ′ 1 , Note that c U = c S,T . Let d U ⊆ p + U be the direct sum of all root spaces g α with α a positive root such that g α is neither contained in a U nor in c U . Let n ′ a be the nilpotent radical of a U . Recall that n c is the nilpotent radical of c U = c S,T . Lemma 2.9. The following are true. Proof. By the construction, d U is the direct sum of the root spaces g α with α positive and supported at both simple roots α ǫ and α ω . For any g −α ⊆ p + U with −α a negative root, the root −α not supported at α ǫ and α ω . So (1) follows. Similar, (2) holds.
By Richardson's theorem, there exists (1) (2), we may assume that r 1 is the Richardson element for c S,T from Section 2.2. Again by Lemma 2.9, we can identify where y ′′ + y |l ∈ a U and y |l + y ′ ∈ c U . Then similar to (1) in the proof of Lemma 2.8, Recall that r 2 is a Richardson element for a S,T . We have the following observation.
Then c r ′ 2 = c r 2 and so [c r 2 , r 1 ] = n c by Lemma 2.10. Then q S,T has a Richardson element, by Lemma 2.8.
The condition in the statement is actually true and can be verified using properties deduced from the categorical construction of Richardson elements [9]. Thus verifying the condition is a key step in the proof of the existence of Richardson elements in the seaweed q S,T and we recall the construction in next section.

Construction of Richardson elements in type A
Let Q be a quiver of type A m with vertices Q 0 = {1, · · · , m} and arrows Let A = kQ, the path algebra of Q. We denote the projective indecomposable Amodule associated to vertex i by P i . Let is a seaweed in a Lie algebra of type A, and a Richardson elements in radEnd A P (d) can be constructed from a good rigid representation X(d) [8,9] of a double quiver of Q with relations. We recall the double quiver with relations from [3,7] and the construction of X(d).
LetQ be the double quiver of for any arrow α ∈ Q 1 , and α * β for pairs of arrows α = β in Q 1 terminating at the same vertex. Let D = kQ/I. Any D-module is an A-module via the inclusion A ⊆ D and any A-module is a Dmodule via the surjection D ։ A mapping all arrows in Q * 1 to zero. We use the notation A X to indicate the A-module structure of a D-module X and note that for two A-modules M and N.
The algebra D is quasi-hereditary with Verma modules P 1 , . . . , P m (see [4]). The modules filtered by the Verma modules are called good modules. So for any good D-module M, we have We call d the ∆-dimension vector of M, denoted by dim ∆ M, and the set supp ∆ (M) = {i | d i = 0} the ∆-support of M. This definition is similar to the support of a module, which is defined using the usual dimension vector.
We identify modules with the corresponding quiver representations. So a D-module M is a collection of vector spaces M i , i ∈ Q 0 and linear maps M β , β ∈ Q 1 ∪ Q * 1 , satisfying the relations I, and a homomorphism f : M → N of D-modules is a collection of linear maps (f i ) i∈Q 0 commuting with the module structure on M and N.
3.1. The linear case [4]. Let Q be a linear quiver with m the unique sink vertex. Then I is generated by commutative relations at 2, . . . , m − 1, and a zero relation at 1. In this case the indecomposable projective D-module Q m , at vertex m, is injective.
Thus there is a natural bijection between subsets I ⊆ Q 0 and submodules of Q m . More precisely, under this bijection a subset I corresponds to the unique submodule X(I) ⊆ Q m with ∆-support I. For any vector d ∈ Z m ≥0 , define with dim ∆ X(d) = d and I 1 ⊆ I 2 ⊆ · · · ⊆ I t . We give an example to illustrate the construction. See [4] for more details.
Recall that a vertex is admissible if it is a source or a sink. Let  [8] Let u be a vertex with i v < u ≤ i v+1 . Suppose that two indecomposable rigid D-modules M and N, glued from X(I r )s and X(J r )s, respectively, are supported (but not necessarily ∆-supported) at u. We define M ≤ u N if for any s with both I s and J s nonempty, I s ⊆ J s when s − v is even and I s ⊇ J s when s − v is odd.
In general, using the order ≤ u we can construct rigid good D-modules as follows. Let M and N be two good rigid D-modules with dim ∆ (M) i = 0 for i > i j , dim ∆ (N) i = 0 for i < i j , and dim ∆ (N) i j = dim ∆ (M) i j . With respect to ≤ i j , we glue the ith biggest summand of M to the ith biggest summand of N. In this way, we obtain a rigid good D-module X(d) for any ∆-dimension vector d. We illustrate the construction by an example. See [8,9] for more details.
We have M 1 ≤ 3 M 2 and N 1 ≤ 3 N 2 . So X(d) is the direct sum of the gluings of M 1 , M 2 with N 1 and N 2 , respectively, i.e., By the construction of rigid good modules, we have the following lemma.
Lemma 3.4. [8,9] Let u be a source or sink vertex, the indecomposable summands of a rigid good D-modules X(d) which are supported at u are totally ordered by ≤ u .

Construction of Richardson elements.
We briefly decribe how to construct a Richardson element r(d) from X(d). Let be a decomposition of X(d) into indecomposable summands and let n = i d i . For each summand X i that is ∆-supported at j, choose a number x ij , where such that x ij = x lj for two different summands X i and X l . If X i is ∆-supported at both s and t with s < t, but not at s + 1, · · · , t − 1, then as a matrix r(d) ∈ gl n has a 1 at either (x is , x it ) or (x it , x is ), depending on which root-space belongs to q S,T = End A (P (d)). All other entries in r(d) are equal to 0. In this way, we construct a Richardson element r(d) in q S,T . Note that End D (X(d)) ∼ = stab q S.T (r(d)).
The matrix r(d) is also the adjacency matrix of an oriented graph with components corresponding to indecomposable summands of X(d). See the example below for an illustration and [1,9,4] for more detail.
The two subalgebras are direct sum of the Cartan subalgebras and root spaces in the bold faces, respectively. The corresponding oriented graphs are.

Stabilisers of Richardson elements in type A
Consider the quiver Q of type A m of arbitrary orientation, defined in Section 3. For D-modules M and N, let be the space of homomorphisms from N to M restricted to vertex i. Let End D (M) i and Aut D (M) i be defined similarly. We study the structure of the endomorphism algebra of a good rigid D-module and its restriction to a vertex. Three special cases are needed in the proof of the main result in the last section. Let We order the summands such that X i < m X i+1 for all indecomposable summands X i .  Proof. By the construction of X(d), Hom D (X i , X j ) m is one dimensional for i ≤ j and zero otherwise. Further we may choose basis elements r ji ∈ Hom D (X i , X j ) m for all i and j, such that r ji r kl = r jl if i = k and zero otherwise. The lemma follows.

4.2.
Restriction to m, when m is a sink. Recall that α m−1 : m − 1 → m is the arrow that ends at m. For D-modules M and N, let The proof is similar to the proof of Lemma 4.1.

Stabilisers of indecomposable rigid modules.
There is an obvious embedding of endomorphism rings As before let 1 = i 0 < i 1 < · · · < i t+1 = m be the admissible vertices of Q.
Let v be a sink in Q and let α be an arrow ending at v. Then αα * induces a nilpotent endomorphism x of X i . So we have such an endomorphism x s : X i → X i for each interval [i s−1 , i s ] with s = 1, · · · , t+1. Note that x s is zero on vertices that are not in the interval [i s−1 , i s ]. Let m s ≥ 1 be the smallest number such that x s ms = 0. In fact, Lemma 4.3. The map y s → x s induces an isomorphism k[y 1 , · · · , y t+1 ] < y s y l = 0 for s = l, y ms Proof. Let x s and x l be two arbitrary endomorphisms, induced by αα * and ββ * , respectively, where α and β are two different arrows. Clearly x s x l = 0 if α and β end at different sinks. Otherwise, x s x l = 0 follows from the relation α * β = 0. By definition, x ms s = 0. So the map is well-defined. By the construction of X i , End D (X i ) is generated by the x s and thus the map is surjective. The intersection of the images of x s and x l is zero if s = l, and so the injectivity follows.

By the embedding
End D (X(d)) ⊆ End A (P (d)) ⊆ gl n and the construction of Richardson elements discussed before Example 3.5, each element of i End D (X i ) n i can be explicitly described in terms of matrices. They can also be described in terms of the oriented graphs constructed from X i . Example 4.4. We use the two Richardson elements from Example 3.5. In both cases, there are two indecomposable summands X 1 and X 2 in X(d). We use two different colours to describe End D (X 1 ) and End D (X 2 ) in each case, stab(r 1 ) : where for instance in the first matrix, the a-entries are a · 1 X 1 the b-entries are b · x 1 with x 1 : X 1 → X 1 and the c-entries are c · x 2 1 . The non-zero off-diagonal entries in the matrices correspond to non-trivial paths in the oriented graphs as follows.

The main result
Theorem 5.1. Let g be a simple Lie algebra of type B, C or D. Then any seaweed in g has a Richardson element.
We continue to use the notation from Section 2. The proof of the theorem is split into two lemmas. The first lemma deals with type B, C, and type D when (ǫ, ω) = (2, 1). We will use quiver representations and results from Section 4 to verify that the condition in Lemma 2.11 holds, and so Richardson elements exist. The second lemma deals with the situation, where g is of type D and (ǫ, ω) = (2, 1). In this case, unlike the first one, there can be root spaces that are not contained in a S,T + c S,T , that is, q S,T is not necessarily equal to a S,T + c S,T . Lemma 5.2. Let q S,T ⊆ g be a seaweed, where g is of type B or C, or of type D with (ǫ, ω) = (2, 1). Then q S,T has a Richardson element.
Proof. Note that we only need to consider the situation where ǫ > ω and neither S nor T is not equal to Π or ∅.
Let P = P (d) be a projective A-module such that End A (P (d)) ≃ a S,T . Let r ∈ n a be a Richardson element constructed from the good rigid D-module X(d) as in Section 3.3. Fix an embedding End A (P (d)) ⊆ g such that End A (P (d) = a S,T and End A (P (d)) m = l, where l = a S,T ∩ c S,T is as in Section 2.2. By the condition ǫ > ω on a S,T , the vertex m is a source. So stab a S,T (r) |l = End D (X(d)) m is a parabolic in l, by Lemma 4.1.
We order the summands X(d) from big to small with respect to the order ≤ m , so that End D (X(d)) m is standard upper triangular. Suppose that the sizes of the blocks in the Levi subalgebra of End D (X(d)) m are c 1 , . . . , c l and let c = (c l , c l−1 + c l , . . . , j≥i c j , . . . , j≥1 c j ).
Let B be the path algebra of the linearly oriented quiver A l with the unique sink l and let P (ĉ) = Pĉ 1 1 ⊕ · · · ⊕ Pĉ l l , a projective representation of this quiver. Denote by F the algebra of the associated double quiver with relations, defined as D in Section 3, and let X(ĉ) be the rigid good F -module.
Choose g ′ and U (see Section 2.3), and an embedding End B (P (ĉ)) ⊆ g ′ such that End B (P (ĉ)) = a U and End B (P (ĉ)) 0 l = l. Let r ′ ∈ a U be a Richardson element corresponding to X(ĉ), where the summands of X(ĉ) are ordered from small to big with respect to the order ≤ l , so that End F (X(ĉ)) 0 l = stab a U (r ′ ) |l is standard upper triangular. Both End F (X(ĉ)) 0 l ⊆ l and End D (X(d)) 0 m ⊆ l are standard upper triangular with Levi blocks of equal sizes, and so stab a S,T (r) |l = stab a U (r ′ ) |l .
Then q S,T has a Richardson element by Lemma 2.11. Denote by E ij the elementary matrix with 1 at (i, j)-entry and 0 elsewhere. Lemma 5.3. If g has type D and (ǫ, ω) = (2, 1), then the seaweed q S,T ⊆ g has a Richardson element.
Proof. Let W = S\{1} and Then q S,T = q W,T ⊕ q 1 S,T and n S,T = n W,T ⊕ q 1 S,T . Note that q W,T is a seaweed of type A and so it has a Richardson element r ∈ n W,T . Since S,T . Indeed, let x + x ′ ∈ n W,T ⊕ q 1 S,T and let y ∈ q W,T and y ′ ∈ stab q W,T (r) such that
Fix an embedding gl n ⊆ g where E ij → E ij − E 2n−j+1,2n−i+1 . We may assume a Richardson element r(d) is constructed from a rigid good module X(d) = ⊕ i X i , as in Section 3.3. Choose an embedding End A (P (d)) ⊆ g so that End A (P (d)) = q W,T and End D (X(d)) = stab q W,T (r). Then where the x ij are constructed from the summand X i as in Section 3.3. Let By the construction of X(d) and r, As ǫ > ω, the vertex m is a source. There are now two cases to consider. First, when m − 1 is non-admissible, then each indecomposable summand P m−1 in P (d) is contained in a different indecomposable summand of X(d). Therefore each V i is at most one-dimensional and [k · 1 X i , r i ] = V i for any non-zero r i ∈ V i . Second, when m − 1 is admissible, then it is a sink, and we let β i be the smallest root with [1 X i , g β i ] = 0. Then for any non-zero r i ∈ g β i , where x is the endomorphism induced by αα * with α the arrow from vertex m − 2 to vertex m − 1 (see Lemma 4.3).
In both cases, [stab q W,T (r), This proves that q S,T has a Richardson element.

Remark 5.4.
a) The Richardson elements and their stabilisers can be explicitly constructed using results from [9], the proofs of Lemma 5.2 and Lemma 5.3. The work of Baur [1] on parabolic Lie algebras is also needed in the case Lemma 5.2. b) The method of Lemma 5.3 can be generalised to Lie algebras of exceptional types and therefore provide an explanation why Richardson elements do not exist for some seaweed Lie algebras of type E 8 .
We end this paper with an example of constructing Richardson elements, using the method discussed in Lemma 5.3.
(3) The Richardson elements of q S\{1},T and q K\{1},L are as below. Entries of the same colour come from the same indecomposable summand. Note that for X(d), one of the indecomposable summands is a Verma module, so both the 4th coloumn and row are zero.
(4) The stabilizers of the two Richardson elements are as follows.
There are two indecomposable direct summands in each of X(c) and X(d). The different colours indicate homomorphisms between different pairs of summands. to the natural action of a 0 0 e on k 2 , and r ′ = 1 1 ∈ q 1 K,L . So we have the following Richardson elements for the two seaweeds q S,T and q K,L ,

A counterexample
In our joint work with Yu [9], we gave a counterexample to the existence of dense orbits for a seaweed of type E 8 , but the detail of the computation did not appear in the published version [9]. In this section, we add explanations, including a GAP source file.
Recall that if r is a Richardson element for a seaweed q S,T , then r = r + + r − with r + ∈ n + S,T , r − ∈ n − S,T , [q S,T , r + ] = n + S,T and [q S,T , r − ] = n − S,T . Lemma 6.1. Let r − ∈ n − S,T with [q S,T , r − ] = n − S,T . Then r − + r + for r + ∈ n + S,T is a Richardson element if and only if [stab q S,T (r − ), r + ] = n + S,T . Proof. Assume that r − + r + is a Richardson element. Then for any y + ∈ n + S,T , there exists a q ∈ q S,T such that [q, r − + r + ] = y + . As [q, r − ] ∈ n − S,T and [q, r + ] ∈ n + S,T , we have [q, r − ] = 0 and [q, r + ] = y + .

Then
[q + q ′ , r − + r + ] = y − + 0 + [q, r + ] − [q, r + ] + y + = y. Therefore r − + r + is a Richardson element. Lemma 6.2. Suppose that r − ∈ n S,T with [q S,T , r − ] = n − S,T and q S,T has a Richardson element. Then there exists r + ∈ n + S,T such that r − + r + is a Richardson element. Proof. Assume that x = x − + x + is a Richardson element. Then there exists group element g ∈ Q S,T , where Q S,T is a Lie group corresponding to q S,T such that Ad g (x − ) = r − .
Therefore r − + r − is a Richardson element.
We need an element r − with a dense open orbit in n − S,T , i.e. [q S,T , r − ] = n − S,T . We choose where p i is the i'th prime and v i is a basis element of the i'th root space of n − S,T . The numbering of root spaces follows that of GAP. Much of the calculation below is done, using GAP [17]. We have dim stab q S,T (r − ) = 25 and so dim[q S,T , r − ] = 81 − 25 = dim n − S,T . Hence [q S,T , r − ] = n − S,T . Next we produce a basis of the centralizer of r − and compute [x, n + S,T ] for all the basis elements x. We have (i) [x, −] |n + S,T = Id n S,T for exactly one basis element x. (ii) [y, n S,T ] = [z, n S,T ] = span{v} is 1-dimensional for exactly two basis elements y, z and v = 89 · w 1 + 67 · w 2 + 3 · w 3 , where w i denotes the basis of the i'th root space in n + S,T produced by GAP. (iii) [−, n S,T ] = 0 for the other basis elements. This shows that for any r + ∈ n + S,T , we will have [stab q S,T (r − ), r + ] ⊆ span{r + , v}, which is at most 2-dimensional. So dim[stab q S,T (r − ), r + ] < dim n + S,T for any r + ∈ n + S,T . Thus r − cannot be completed to a Richardson element. Now by Lemma 6.2, we conclude that q S,T does not have a Richardson element.
Below we give the full source code in GAP [17] for the computation in the example.