Luca Francone

Nicolas Ressayre

On the multiplicity spaces for branching to a spherical subgroup of minimal rank

Let g be a complex semi-simple Lie algebra and g be a semisimple subalgebra of g. Consider the branching problem of decomposing the simple g-representations V as a sum of simple grepresentations V . When g = g × g, it is the tensor product decomposition. The multiplicity space Mult(V, V ) satisfies

where the sum runs over the isomorphism classes of simple g-representations. In the case when g is spherical of minimal rank, we describe Mult(V, V ) as the intersection of kernels of powers of root operators in some weight space of the dual space V * of V . When g = g × g, we recover by geometric methods a well known result.

Introduction

Let G be a connected reductive subgroup of a complex semisimple group G. The branching problem consists in decomposing irreducible representations of G as sum of irreducible G-representations.

Fix maximal tori T ⊂ T and Borel subgroups B ⊃ T and B ⊃ T of G and G respectively. Let X(T ) denote the group of characters of T and let X(T ) + denote the set of dominant characters. For ν ∈ X(T ) + , V ν denotes the irreducible representation of highest weight ν. Similarly, we use notation X( T ), X( T ) + , V ν relatively to G. For any G-representation V , the subspace of G-fixed vectors is denoted by V G . Given ν ∈ X(T ) + and ν ∈ X( T ) + , set

Mult(ν, ν) = Hom(V ν , V * ν ) G = (V * ν ⊗ V * ν ) G , (1)
where V * ν and V * ν denote the dual representations of V ν and V ν respectively. The branching problem is equivalent to the knowledge of these spaces. Indeed, there is a natural G-equivariant isomorphism:

ν∈X(T ) + Hom(V ν , V * ν ) G ⊗ V ν -→ V * ν f ⊗ v -→ f (v).
Let G/ B denote the complete flag variety of G. In this article, we are interested in the case when the pair ( G, G) is spherical of minimal rank. In other words, we assume that there exists x in G/ B such that the orbit G.x of x is open in G/ B and the stabilizer G x of x in G contains a maximal torus of G. An important example is when G = G × G in which G is diagonally embedded. Then the point x = (B, B -) ∈ G/ B works, where B -denotes the opposite Borel subgroup of B containing T . More generally, the spherical pairs of minimal rank have been classified by the second author in [START_REF] Ressayre | Spherical homogeneous spaces of minimal rank[END_REF]. The complete list, assuming in addition that G is semisimple simply connected, G is simple and G = G is:

1. G is simple, simply connected and diagonally embedded in G = G × G; 2. (SL 2n , Sp 2n ) with n ≥ 2;

3. (Spin 2n , Spin 2n-1 ) with n ≥ 4; 4. (Spin 7 , G 2 ); 5. (E 6 , F 4 ).

Our aim is to present a uniform description of the multiplicity spaces Mult(ν, ν) for given ( G, G) in this list. Let us first fix some notation. Recall that T ⊂ T and denote by ρ : X( T )-→X(T ) the restriction map. Let ∆ (resp. ∆) denote the set of simple roots of G (resp. G). For each positive root α of G, we fix an sl 2 -triple (X α , H α , X -α ) such that H α (resp. X α ) belongs to the Lie algebra of T (resp. B). Given µ ∈ X(T ), we denote by V ν (µ) = {v ∈ V ν : ∀t ∈ T tv = µ(t)v} the corresponding multiplicity space for the action of the maximal torus T .

Let W denote the Weyl group of G. Fix also y 0 ∈ W such that G y 0 B/ B is dense in G/ B and such that y 0 has minimal length with this property (see Section 2.3 for details).

For cases 2 to 5, we also denote by Φ 1 the set of long roots of G. In case 1, we define Φ 1 to be the empty set. Set

D = { α ∈ ∆ : ρ( y 0 α) ∈ Φ 1 }. Theorem 1. For ν ∈ X(T ) + and ν ∈ X( T ) + , there is a natural isomorphism from Mult(ν, ν) onto the subspace of V * ν (ρ( y 0 ν)) consisting in the vectors v such that 1. X α • v = 0, for any α ∈ Φ + ∩ Φ 1 ; 2. X m -ρ( y 0 α) • v = 0 ∀ m > ν, α ∨ , for any α ∈ D.
Actually, the conditions 2 are not pairwise independant. For example, in the case of the tensor product, the conditions associated to (α, 0) ∈ D and (0, -w 0 α) ∈ D are equivalent for any simple root α. Here w 0 denotes the longest element of the Weyl group. For each example, we describe in Section 4 an explicit subset of D giving an irredundant set of inequalities.

In the case of the tensor product, Theorem 1 is well known. See [PRRV67, Theorem 2.1, p. 392] or [Ž73, Theorem 5, p. 384]. The usual proofs are algebraic, based on properties of the enveloping algebra. Our geometric proof seems to be new even in this case.

The branching rule for (Spin 2n , Spin 2n-1 ) is multiplicity free and hence easy to determine (see e.g. [START_REF] Fulton | Representation theory[END_REF]). That of (SL 2n , Sp 2n ) has been the subject of much attention in the literature. The first positive rule in terms of dominos was obtained by Sundaram [START_REF] Sundaram | Tableaux in the representation theory of the classical Lie groups[END_REF]. Naito-Sagaki conjectured a rule in terms of Littelmann's patches [START_REF] Naito | An approach to the branching rule from sl 2n (C) to sp 2n (C) via Littelmann's path model[END_REF]. Later, B. Schumann and J. Torres proved this conjecture by obtaining a bijection with Sundaram's model. A nonpositive rule for (Spin 7 , G 2 ) were obtained by McGovern in [START_REF] William | A branching law for Spin(7, C) → G 2 and its applications to unipotent representations[END_REF]. Hopefuly, Theorem 1 could be the first step toward combinatorial rules for these branching problems. On the following diagrams the restriction of ρ to the set of simple roots is the vertical projection.

A 2n-1 C n D n B n-1 E 6 F4 B 3 G 2
By [Res10, Lemma 4.4], for any α ∈ Φ, ρ-1 (α) has cardinality one or two. Hence, Φ splits in the following two subsets

Φ 1 := {α ∈ φ h : ♯ρ -1 (α) = 1} and Φ 2 := {α ∈ φ h : ♯ρ -1 (α) = 2}.
Set also Φ 1 = ρ-1 (Φ 1 ) and Φ 2 = ρ-1 (Φ 2 ). It is easy to check using the pictures, that this set Φ 1 , coincides with that defined in the introduction.

Isotropies

Assume, in addition, that G is connected.

Lemma 2. Let x ∈ G/ B. Then the isotropy G x is connected and contains a maximal torus of G.

Proof. The fact that G x contains a maximal torus T of G follows from the monotonicity properties of the rank of orbit closures of G in G/ B (see [Kno95, Theorem 2.2]).

Let L be a Levi subgroup (maximal reductive subgroup) of G x . Then L is isomorphic to the quotient of G x by its unipotent radical. A reference for the existence of L is [OV90, Theorem 4. p286]. Up to conjugacy, one may assume that T ⊂ L.

But L is a reductive subgroup of G x = B ′ , that is a Borel subgroup of G. Hence L maps injectively into B ′ / U ′ ≃ T and L is abelian. The torus T being its own centralizer in G, we deduce that L = T . Now G x is connected as the product of T and its unipotent radical.

Orbits of G in G/ B

We are now interested in the set G\ G/ B of G-orbits in G/ B. Let W and W be the Weyl groups of G and G respectively. Since T is a regular torus in G (see [Bri99, Lemma 2.3]), W naturally embeds in W .

Lemma 3. The map

W \ W -→ G\ G/ B w -→ G w B/ B
is a well-defined bijection.

Proof. It is clear that the G-orbit G w B/ B does not depend on the representative w in its class W w. Hence, the map of the lemma is well-defined. Since T is a regular torus in G, its fixed points in G/ B are the points w B/ B for w ∈ W . But Lemma 2 implies that each G-orbit in G/ B contains a T -fixed point. The surjectivity follows.

Let now w and w ′ in W such that G w B/ B = G w ′ B/ B. To get the injectivity and finish the proof, it remains to prove that W w = W w ′ . Choose g ∈ G such that g w B/ B = w ′ B/ B. Let H and H ′ denote the isotropies in G of the points w B/ B and w ′ B/ B, respectively. Observe that T is a maximal torus of both H and H ′ . Then, T and gT g -1 are two maximal tori of H ′ , and there exists h ′ ∈ H ′ such that h ′ T h ′-1 = gT g -1 . Thus, n := h ′-1 g normalizes T and satisfies n w B/ B = h ′-1 w ′ B/ B = w ′ B/ B. It follows that w ′ ∈ W w.

Let ℓ : W -→N denote the length function. It is well known that ℓ( w) = ♯( Φ + ∩ w -1 Φ -), where

Φ -= -Φ + .
We fix, once for all, y 0 ∈ W such that G y 0 B/ B is the open G-orbit in G/ B and such that y 0 has minimal length with this property. Let H 0 denote the stabilizer of y 0 B/ B in G.

Given a root α (resp. α) of g (resp. g), denote by g α (resp. g α ) the corresponding root space.

Lemma 4. The Lie algebra of H 0 is

Lie(T ) ⊕ α∈Φ + ∩Φ 1 g α . Proof. It is clear that H 0 contains T ∩ G = T . Then Lie(H 0 ) = Lie(T ) ⊕ α∈S g α ,
for some subset S of Φ.

Let now α ∈ Φ 1 ∩ Φ + . We want to prove that α ∈ S. Let α ∈ Φ such that ρ( α) = α. We have g ± α = g ±α . But, exactly one between g α and g -α is contained in the Borel Lie algebra Lie( y 0 B y 0 -1 ). Assume, for a contradiction that g -α ⊂ Lie( y 0 B y 0 -1 ).

Set β = -y 0 -1 α and y ′ 0 = s α y 0 . Since β is positive and y 0 β is negative, ℓ( y ′ 0 ) < ℓ( y 0 ). But, g ± α = g ±α implies that s α = s α and that y ′ 0 ∈ W y 0 . This contradicts the minimality assumption on the length of y 0 .

At this point, we proved that the Lie algebra of the lemma is contained in Lie

(H 0 ). But dim(G/H 0 ) = dim( G/ B), hence dim(H 0 ) = dim(G) -dim( G/ B) = dim(T ) + 2♯Φ + -♯ Φ + = dim(T ) + ♯(Φ + ∩ Φ 1 )
and we can conclude.

Lemma 5. Fix β ∈ Φ 2 . Then there is exactly one root in ρ-1 (β) which is sent in Φ + by the action of y 0 -1 . Proof. Write ρ-1 (β) = { β 1 , β 2 }. Because of [Res10, Lemma 4.4], we have that g β ⊆ g β 1 ⊕ g β 2 and g β = g β i for i = 1 , 2. Lemma 4 implies that g β ⊆ Lie(H 0 ). But since H 0 = H ∩ y 0 B y 0 -1
, this means that y 0 -1 β 1 and y 0 -1 β 2 are not both positive roots, equivalently:

{ y 0 -1 β 1 , y 0 -1 β 2 } ⊆ Φ + . The same argument applied to -β implies that {-y 0 -1 β 1 , -y 0 -1 β 2 } ⊆ Φ + . The lemma follows.
Lemma 5 allows us to distinguish between the roots in the fiber of ρ by means of the action of y 0 -1 . In particular we introduce the following notation.

If β ∈ Φ 2 ∩ Φ + , then β + (resp. β -) is the unique element in ρ-1 (β) that satisfies: y 0 -1 β + ∈ Φ + (resp. y 0 -1 β -∈ Φ -).

The graph of G-orbits in G/ B

We recall the definition given in [START_REF] Ressayre | Sur les orbites d'un sous-groupe sphérique dans la variété des drapeaux[END_REF] of a graph Γ( G/G) whose vertices are the elements of G\ G/ B. The original construction of Γ( G/G) due to M. Brion is equivalent but slightly different (see [START_REF] Brion | On orbit closures of spherical subgroups in flag varieties[END_REF]).

If α belongs to ∆, we denote by P α the associated minimal standard parabolic subgroup of G. Consider the unique G-equivariant map π α : G/ B-→ G/P α which is a P 1 -bundle.

Let O ∈ G\ G/ B and α ∈ ∆. Consider π α -1 (π α (O)). Two cases occur.

• G acts transitively on π α -1 (π α (O)).

• π α -1 (π α (O) contains two G-orbits, one closed V and one open V ′ . Then we says that α raises V to V ′ . In this case, dim(V ′ ) = dim(V ) + 1, and for any x ∈ V , π α -1 (π α (x)) ∩ V = {x}.
Definition. Let Γ( G/G) be the oriented graph with vertices the elements of G\ G/ B and edges labeled by ∆, where V is joined to V ′ by an edge labeled by

α if α raises V to V ′ . Lemma 6. Let w ∈ W . We have dim G w B/ B -dim G/B ≤ ℓ( w).

Moreover, for any

G-orbit O in G/ B there exists w ∈ W such that O = G w and dim G w B/ B -dim G/B = ℓ( w).
Proof. We consider the B-orbits in G/G. The closed orbit is B/B. Choose a reduced expression of

w = s αs • • • s α 1 . Consider the quotient P α 1 × B • • • × B P αs × B B/B of P α 1 × • • • × P αs × B/B
P α 1 × B • • • × B P αs × B B/B -→ G/G (p 1 , . . . , p s , x) -→ p 1 • • • p s x.
The dimension of the left hand side is ℓ( w) + dim( B/B). The right hand side is a B-orbit closure containing B w -1 G/G. The first inequality follows. This equality is reached when the expression is obtained by reading the labels on some path from the closed orbit to O in the graph

Γ( G/G). Such a path exists by [Res10, Proposition 2.2]. Set ℓ m ( w) = min{ℓ(w w) : w ∈ W }. By Lemma 6, ℓ m ( w) = dim(G w B/ B) -dim(G/B). If an element in w ∈ W satisfies ℓ( w) = ℓ m ( w),
we say that w has minimal length.

Lemma 7.

1. There are ♯∆ 2 codimension one G-orbits in G/ B.

2. Let α ∈ ∆. Then G y 0 s α B/ B has codimension one if and only if y 0 α ∈ Φ 2 . 3. Let α ∈ ∆ such that y 0 α ∈ Φ 2 .
Then, the Lie algebra of the stabilizer of

y 0 s α B/ B is Lie(H 0 ) ⊕ g -ρ( y 0 α) .
Proof. By [Res10, Proposition 2.3], the number of codimension one G-orbits in G/ B is the number of G-orbits O of dimension dim(G/B) + 1. This is also

♯({W s α ∈ W \ W ; α ∈ ∆} -{W }). But s α ∈ W if and only if α ∈ ∆ 1 . Moreover, if α 1 = α 2 ∈ ∆ 2 then s α 1 s α 2 ∈ W if and only if ρ( α 1 ) = ρ( α 2 ). Since G y 0 B/ B is dense, either G y 0 s α B/ B = G y 0 B/ B or G y 0 s α B/ B has codimension one. But y 0 s α B/ B belongs to the open G-orbit if and only if y 0 s α ∈ W y 0 if and only if y 0 α ∈ Φ 1 .
Let α be like in the last assertion. Then

y 0 s α Φ + = ( y 0 Φ + ∪ {-y 0 α}) -{ y 0 α}.
Now, the fact that Lie(H 0 ) is contained in the Lie algebra of the stabilizer of y 0 s α B/ B follows from Lemma 4.

The root y 0 α belonging to Φ 2 , there exists a second root β in the fiber ρ-1 (ρ(-y 0 α)). But, Lemma 5 implies that β belongs to y 0 Φ + and hence to y 0 s α Φ + . We deduce that

g ρ(-y 0 α) ⊂ g β ⊕ g -y 0 α ⊂ Lie( y 0 s α B( y 0 s α ) -1 ).
In particular, g ρ(-y 0 α) is contained in the Lie algebra of the stabilizer of y 0 s α B/ B. Now, the last assertion follows by equality of the dimensions.

Compatible sl 2 -triples

Recall that we have fixed an sl 2 -triple (X β , H β , Y β ) associated to any positive root β of G.

Lemma 8. If β ∈ Φ 2 ∩ Φ + then there exist sl 2 -triples (X β + , H β + , X -β + ) and (X β -, H β -, X -β -)
for β + and β -such that:

• X β = X β + + X β -; H β = H β + + H β -; X -β = X -β+ + X -β -. • exp(tX β ) = exp(tX β + ) • exp(tX β -) for any t ∈ C
Proof. Since g β ⊆ g β + ⊕ g β -and is not equal to any of the 2 root spaces on the right hand side,

X β = X β + +X β -for nonzero X β ± ∈ g β ± .
Then there exist sl 2 -triples of the form

(X β + , H β + , X -β + )
and (X β -, H β -, X -β -) for β + and β -. Moreover, X -β = aX -β + + bX -β -with nonzero constants a and b, and

H β = [X β , X -β ] = [X β + + X β -, aX -β + + bX -β -] = aH β + + bH β -.
For the last equality we use that

[X β + , X -β -] = 0 and [X β -, X -β + ] = 0. Indeed, if β + -β -(reps. β --β + ) was a root in Φ, then ρ( β + -β -) = 0 (resp. ρ( β --β + ) = 0). But this is absurd since ρ sends Φ in Φ.
Using that ρ( β + ) = ρ( β -) = β, we get that:

2 = β(H β ) = β + (aH β + + bH β -) = a β + (aH β + ) = 2a.
Where for the second-last inqueality we used that β + and β -are ortogonal, hence β + (H β -) = 0. Then a = 1, and similarly we get b = 1. This proves the first point.

For the second one notice that [X β + , X β -] = 0. In fact if β + + β -was a root, then ρ( β

+ + β -) = 2β.
But this is absurd since 2β is not a root. Hence for any t ∈ C, exp(tX

β + + tX β -) = exp(tX β + ) • exp(tX β -)
, and the second point follows immediately.

Remark. For the proof we could also used [Res10, Lemma 4.1] and [Res10, Lemma4.3] to reduce the problem to the case of PSL 2 diagonally embedded in PSL 2 × PSL 2 .

3 Proof of Theorem 1

An embedding of the multiplicity space

Fix dominant weights ν ∈ X(T ) + and ν ∈ X( T ) + . Observe that H 0 is a subgroup of y 0 B y 0 -1 and y 0 ν is a character of this last group. In particular, y 0 ν restricts as a character of H 0 . Lemma 9. In the G-representation V * ν , the subspaces

(V * ν ) (H 0 ) y 0 ν = {v ∈ V * ν : ∀h ∈ H 0 h.v = ( y 0 ν)(h)v} and {v ∈ V * ν ( y 0 ν) : ∀α ∈ Φ + ∩ Φ 1 X α • v = 0} coincide.
Proof. Since T is a maximal torus of H 0 , the first subspace is contained in V * ν ( y 0 ν). Furthermore H 0 = R u (H 0 )T , where R u (H 0 ) denotes the unipotent radical. Hence

(V * ν ) (H 0 ) y 0 ν = V * ν ( y 0 ν) R u (H 0 ) = V * ν ( y 0 ν) Lie(R u (H)) .
Now, Lemma 4 allows to conclude.

Consider on G/B the G-linearized line bundle L ν such that B acts on the fiber over B/B with weight -ν. Similarly, define L ν . By the Borel-Weyl theorem, the space Lemma 5 allows to distinguish two situations that lead to two slightly different local descriptions of the corresponding divisor. 1) y 0 α is negative. Then β = -ρ( y 0 α) and y 0 α = -β -.

H 0 (G/B × G/ B, L ν ⊗ L ν ) G ≃ (V * ν ⊗ V * ν ) G of G-invariant sections identifies with Mult(ν, ν). The orbit G. y 0 B/ B being open, the restriction map H 0 (G/B × G/ B, L ν ⊗ L ν ) G -→H 0 (G/B × G. y 0 B/ B, L ν ⊗ L ν ) G is injective. Moreover H 0 (G/B × G. y 0 B/ B, L ν ⊗ L ν ) G ≃ H 0 (G/B, L ν ) (H 0 ) y 0 ν ≃ (V * ν ) (H 0 ) y 0 ν . For ϕ ∈ (V * ν ) (H 0 ) y 0 ν and denote by σ ∈ H 0 (G/B × G. y 0 B/ B, L ν ⊗ L ν ) G the associated section. To describe the image of H 0 (G/B × G/ B, L ν ⊗ L ν ) G in (V * ν ) (H 0 ) y 0 ν ,

Local description along the divisors

2) y 0 α is positive. Then β = ρ( y 0 α) and y 0 α = β + . Given a root α (resp. α) of G (resp. G), we denote by U α (resp. U α ) the corresponding subgroup isomorphic to (C, +). The aim of this section is to describe an open subset of G/ B intersecting the divisor D α . In particular we will prove that U α is a common transverse slice to D α at any point of this subset. Before doing so we need some preparatory work.

Lemma 10. Let α, β and D α be as above. Set S( α) := (Φ + ∩ Φ 2 ) \ {β}. Index the elements of S( α) = {γ 1 , . . . , γ s }. Then: 1) If y 0 α is negative then the map

i α : U -× γ∈S( α) U γ -→ G y 0 s α B/ B (u -, (u γ ) γ ) -→ (u - s i=1 u γ i ) y 0 s α B/ B
is an open immersion.

2) If y 0 α is positive then the map

i α : U -× γ∈S( α) U γ -→ G y 0 s α B/ B (u -, (u γ ) γ ) -→ s β (u - s i=1 u γ i )s β y 0 s α B/ B
is an open immersion.

Our proof of Lemma 10 depends on the following well-known lemma.

Lemma 11.

The product induces an open immersion

U -× U × T -→ G (u -, u, t) -→ u -ut.
2. Let H be a T -stable closed subgroup of U . Number γ 1 , . . . , γ s the positive roots of G that are not roots of H. Then the map

U γ 1 × • • • × U γs × H -→ U ((u i ), u) -→ u 1 . . . u
U γ × G y 0 s α B/ B -→G
is an open embedding. Taking the quotient by G y 0 s α B/ B we conclude.

The second case is obtained similarly by applying Lemma 11 to the maximal unipotent subgroups s β U ± s β in place of U ± . Proposition 12. Keep the setting as in Lemma 10. 1) If y 0 α is negative then the map:

f α : U -× γ∈S( α) U γ × U -α -→ G/ B (u -, (u γ ) γ , u -α ) -→ (u - s i=1 u γ i ) y 0 s α u -α B
is an open immersion. Furthermore the image of f α is contained in G y 0 B/ B ∪ D α , and

f α -1 (D α ) = U -× γ∈S( α) U γ × {1}.
2) If y 0 α is positive then the map:

f α : U -× γ∈S( α) U γ × U -α -→ G/ B (u -, (u γ ) γ , u -α ) -→ s β (u - s i=1 u γ i ) s β y 0 s α u -α B is an open immersion. Furthermore the image of f α is contained in G y 0 B/ B ∪ D α , and f α -1 (D α ) = U -× γ∈S( α) U γ × {1}.
Proof. We prove part 1). The proof of the positive case is obtained from the following one by replacing every appearance of y 0 with s β y 0 . Identify G/ B with the fibered product G × P α P α / B. Via this identification f α is the map:

f α : U -× γ∈S( α) U γ × U -α -→ G × P α P α / B (u -, (u γ ) γ , u -α ) -→ (u - s i=1 u γ i y 0 s α : u -α B/ B) .
Since y 0 s α B/ B and π α ( y 0 s α B/ B) have the same isotropy in G, we can identify their G-orbits. In particular we will think about i α , the open immersion of Lemma 10, as a map to G. y 0 s α P α /P α ⊆ G/P α . Call V α ⊆ G/P α the image of i α . By Lemma 10, V α is open in G/P α . Denote the two components of

i α -1 : V α -→ U -× γ∈S( α) U γ , by j 1 : V α -→ U -and j 2 : V α -→ γ∈S( α) U γ , respectively. Observe that U -α ≃ U -α B/ B = P α / B \ ( B/ B) is open in P α / B. The image of f α is Ω α := { [( g : x)] ∈ G × P α P α / B : [ g] ∈ V α and s α -1 y 0 -1 j 2 ([ g])) -1 j 1 ([ g])) -1 gx ∈ U -α B/ B}, where [ g] = gP α /P α ∈ G/P α . We deduce that Ω α is open in G × P α P α / B.
Finally we prove that f α is an isomorphism. If we call φ α : U -α B/ B-→U -α the inverse of the natural projection, then it's easy to see that the map

Ω α -→ U -× γ∈S( α) U γ × U -α
that sends ( g, x) to:

(i α -1 ) 1 ([ g]) , (i α -1 ) 2 ([ g]) , φ α s α -1 y 0 -1 j 2 ([ g]) -1 j 1 ([ g[) -1 gx is the inverse of f α .
To conclude, notice that for u -α = 1, f α restrict to i α , hence Ω α ∩ D α is an open dense subset of the divisor. For u -α = 1, y 0 s α u -α B/ B = y 0 s α B/ B in y 0 s α P α / B. Hence y 0 s α u -α B/ B has to be a point of the open G-orbit, that is the orbit of y 0 B/ B. We conclude that f α maps U -× γ∈S( α)

U γ × U -α \{1} in the open G-orbit and that f α -1 (D α ) = U -× γ∈S( α) U γ × {1}.

Conclusion

In the setting of the end of Section 3.1, we are now in position to characterize the image of the embedding:

H 0 (G/B × G/ B, L ν ⊗ L ν ) G -→H 0 (G/B, L ν ) (H 0 ) y 0 ν . ( 2
)
Fix ϕ ∈ (V * ν ) (H 0 ) y 0 ν and follow it throught the following isomorphisms:

(V * ν ) (H 0 ) y 0 ν ≃ H 0 (G/B, L ν ) (H 0 ) y 0 ν ≃ H 0 (G/B × G. y 0 B/ B, L ν ⊗ L ν ) G ϕ → σ → σ. Fix v 0 ∈ V (B) ν
and ỹ0 ∈ (L ν ) y 0 -{0}. Explicitly, σ and σ are given by the formulas

∀g ∈ G σ(gB/B) = [g : ϕ(gv 0 )] ∀g 1 , g 2 ∈ G σ(g 1 B/B, g 2 y 0 B/ B) = [g 1 : ϕ(g 2 -1 g 1 v 0 )] ⊗ g 2 ỹ0 .
(3)

We want to determine when σ extends to a global section. Take α ∈ ∆ that is a label of some edge descending from the open orbit. Then D α := G y 0 s α B/ B is a divisor of G/ B along which, we want to determine the vanishing order of σ. Consider the image V α of the map ι α defined by Proposition 12.

Proposition 13.

1. Assume that y 0 α is negative. Then, the section σ extends to a regular section on G/B × V α if and only if

X m β ϕ = 0 for m > -y 0 • ν, ( β -) ∨ .
2. Assume that y 0 α is positive. Then, the section σ extends to a regular section on

G/B × V α if and only if X m -β ϕ = 0 for m > y 0 • ν, ( β + ) ∨ .
Proof. Denote by ǫ ±β := C g ±β U ±β exp the additive one-parameter subgroups, and think about β ∨ : C * -→T ⊂ G as a one-parameter subgroup. These three morphisms glue to give a group homomorphism φ β : SL 2 (C)-→G. The same notation is used, in the obvious way, also for

G. Now fix an sl 2 -triple (X α , H α , X -α ) for α, with X α ∈ g α .
Suppose first that we are in the negative case, that is y 0 α = -β -, where β = -ρ( y 0 α). Because of Proposition 12, σ extends to a section on G/B × V α if and only if the map:

G/B × U -× γ∈S( α) U γ × C * -→ L ν ⊗ L ν (gB/B, u -, (u γ ) γ , t) -→ σ (gB/B, u -s i=1 u γ i y 0 s α ǫ -α (t) B/ B
) extends at t = 0. Notice also that U -and U γ are subgroups of G and σ is G-invariant, hence the function above extend at t = 0 if and only if the following map, that with a little abuse will still be called σ, extends at t = 0.

G/B × C * -→ L ν ⊗ L ν (gB/B, t) -→ σ (gB/B , y 0 s α ǫ -α (t) B/ B) . (4)
Now fix sl 2 -triples for β ± as in Lemma 8, so that for any

t ∈ C * , ǫ ±β (t) = ǫ ± β + (t)ǫ ± β -(t).
In SL 2 , we have

1 t 0 1 0 1 -1 0 = 1 0 t -1 1 -t 0 0 -t -1 1 -t -1 0 1 , for any t ∈ C * . Hence ǫ α (t)s α = ǫ -α (t -1 ) α ∨ (-t)ǫ α (-t -1 ). (5)
Where α can be replaced by any positive root of G or G for which a corresponding sl 2 -triple has been fixed. Now take (gB/B, t) ∈ G/B × C * , then

(gB/B , y 0 s α ǫ -α (t) B/ B) =(gB/B , y 0 ǫ α (t)s α B/ B) =(gB/B , y 0 ǫ -α (t -1 ) B/ B) =(gB/B , ǫ β -(c -1 t -1 ) y 0 B/ B)
where c is the nonzero constant satisfying y

0 X α = cX -β -. But since U β + ⊂ y 0 B y 0 -1 : ǫ β -(c -1 t -1 ) y 0 B/ B = ǫ β -(c -1 t -1 )ǫ β + (c -1 t -1 ) y 0 B/ B = ǫ β (c -1 t -1 ) y 0 B/ B. Now, Formula (3) gives σ (gB/B , y 0 s α ǫ -α (t) B/ B) = [g : ϕ(ǫ β (-c -1 t -1 )gv 0 )] ⊗ ǫ β (c -1 t -1 )ỹ 0 .
Moreover,

ǫ β (c -1 t -1 )ỹ 0 = ǫ β -(c -1 t -1 )ǫ β + (c -1 t -1 )ỹ 0 = ǫ -β -(ct)(s β -) -1 ( β -) ∨ (-ct)ǫ -β -(c -1 t -1 )ỹ 0 by formula (5) = ǫ -β -(ct)(s β -) -1 (-ct) -y 0 ν,( β -) ∨ since U -β -⊂ y 0 U y 0 -1 and
T acts with weight -y 0 ν.

Rewrite the term ϕ(ǫ β (-c -1 t -1 )gv 0 ) as

ϕ(ǫ β (-c -1 t -1 )gv 0 ) = ǫ β (c -1 t -1 )ϕ, gv 0 = n≥0 (c -1 t -1 ) n n!
X n β ϕ, gv 0 . Finally, we get

σ (gB/B , y 0 s α ǫ -α (t) B/ B) = [g : n≥0 ± (c -1 t -1 ) n+ y 0 ν,( β -) ∨ n! X n β ϕ, gv 0 ] ⊗ (ǫ -β -(ct)(s β -) -1 ỹ0 ).
The term ǫ -β -(ct)(s β -) -1 ỹ0 is regular on C. Hence σ has no pole along t = 0 if and only if

∀n > -y 0 ν, ( β -) ∨ =⇒ X n β ϕ, gv 0 = 0 ∈ C[G]. But Gv 0 spans V ν that is irreducible. As a consequence, X n β ϕ, gv 0 = 0 if and only if X n β ϕ = 0.
Suppose now that we are in the positive case, hence y 0 • α = β + with β = ρ( y 0 α). The outline of the proof doesn't change. By the same argument of the previous case, σ extends to a section on G/B × V α if and only if the following map, that will still be called σ, extends at t = 0.

G/B × C * -→ L ν ⊗ L ν (gB/B, t) -→ σ (gB/B , s β y 0 s α ǫ -α (t) B/ B) (6)
Again fix sl 2 -triples for β ± as in Lemma 8. Take where now c is the nonzero constant satisfying y 0 X α = cX β + . Hence:

σ (gB/B , s β y 0 s α ǫ -α (t) B/ B) = [g : ϕ(ǫ -β (-c -1 t -1 )(s β ) -1 gv 0 )] ⊗ s β ǫ -β (c -1 t -1 )ỹ 0 .
Similarly to the previous computations:

ǫ -β (c -1 t -1 )ỹ 0 = ǫ -β + (c -1 t -1 )ǫ -β -(c -1 t -1 )ỹ 0 = ǫ β + (ct)s β + ǫ β + (c -1 t -1 )( β + ) ∨ (-c -1 t -1 )ỹ 0 = ǫ β + (ct)s β + (-c -1 t -1 ) -y 0 ν,( β + ) ∨ .
While for the other term

ϕ(ǫ -β (-c -1 t -1 )s β -1 gv 0 ) = ǫ -β (c -1 t -1 )ϕ, s β -1 gv 0 = n≥0 (c -1 t -1 ) n n! X n -β ϕ, s β -1 gv 0 .
Finally we get σ (gB/B , s

β y 0 s α ǫ -α (t) B/ B) = [g : n≥0 ± (c -1 t -1 ) n-y 0 ν,( β + ) ∨ n! X n -β ϕ, s β -1 gv 0 ]⊗(s β ǫ β + (ct)s β + )ỹ 0 ).
By the same argument of the previous case we conclude that σ has no pole along t = 0 if and only if ∀ n > y 0 ν, ( β + ) ∨ =⇒ X n β ϕ = 0.

Remark. If y 0 α = -β -, then y 0 • α ∨ = -( β -) ∨ , hence:

-

y 0 • ν, ( β -) ∨ = y 0 • ν, y 0 • α ∨ = ν, α ∨ .
Similarly, if y 0 α = β + , then y 0 • α ∨ = ( β + ) ∨ , which implies:

y 0 • ν, ( β + ) ∨ = y 0 • ν, y 0 • α ∨ = ν, α ∨ .
Recall that in the introduction we set D = { α ∈ Φ : ρ( y 0 α) ∈ Φ 1 }, for α ∈ D, let V α be as in Proposition 13. We prove the theorem of the introduction.

Proof of Theorem 1. The first condition of the theorem is implied by Lemma 9. Then set

V := α∈D G/B × V α , which is open in G/B × G/ B.
The complement of V is of codimension strictly larger then 1 and G/B × G/ B is normal, hence the restriction map

H 0 (G/B × G/ B, L ν ⊗ L ν )-→H 0 (V, L ν ⊗ L ν )
is an isomorphism. Then, from Proposition 13 and from the previous remark we deduce that ϕ

∈ (V * ν ) (H 0 ) y 0 ν is in the image of H 0 (G/B × G/ B, L ν ⊗ L ν )-→(V * ν ) (H 0 ) y 0 ν
if and only if the second condition of the theorem holds.

As remarked in the introduction, the conditions of Theorem 1 are in general redundant. This is linked to the fact in Γ( G/G) we may have different edges between the same vertices. It seems not easy to determine a minimal set of conditions in a uniform way. The following lemma checks that, for the tensor product case, a minimal set of conditions is described by the set:

{(α, 0) ∈ Φ × Φ : α ∈ ∆}.
Lemma 14. For β ∈ Φ + , and σ ∈ V * ν ( y 0 • ν), the following are equivalent:

1) X m β • σ = 0 for m > -y 0 • ν, ( β -) ∨ . 2) X m -β • σ = 0 for m > y 0 • ν, ( β + ) ∨ .
Remark. If β ∈ Φ + and α 1 , α 2 ∈ ∆ satisfy:

y 0 • α 1 = -β -and y 0 • α 1 = β +
then y 0 s α 1 = s β y 0 s α 2 , hence their G-orbit is the same, and by the proof we have done we expect that α 1 and α 2 give the same condition in Theorem 1. This is directly checked in previous lemma.

true for any k, this stabilizer consists in block diagonal matrices with blocks of size 2. Moreover, in each block the matrix has to be triangular because of its action on e 1 , . . . , e n . Moreover, these blocks belong to Sp(2) = SL(2). We just proved that G y 0 B/ B is contained in a group of dimension 2n. For dimension reasons, we deduce that G y 0 B/ B is equal to this subgroup and that G.

y 0 B/ B in open in G/ B.
The length of y 0 is the number of inversions, that is the set of pairs (i < j) such that j occurs before i in the word. Fix i ∈ {1, . . . , n}. The number of j > i such that j occurs before i in the word y 0 is i -1. The contribution to these pairs to ℓ( y 0 ) is n(n-1)

The stabilizer of

H in G 0 is G 0 = SO 2n-1 (C). Its maximal torus T 0 is {diag(t 1 , . . . , t n-1 , 1, 1, t -1 n-1 , . . . , t -1 1 ) : t i ∈ C * }.
The groups G and G are the universal covers of G 0 and G 0 respectively. Here

Φ + = {ε i ± ε j : 1 ≤ i < j ≤ n -1} ∪ {ε i : 1 ≤ i ≤ n -1}, and ∆ = {α 1 = ε 1 -ε 2 , α 2 = ε 2 -ε 3 , . . . , α n-2 = ε n-2 -ε n-1 , α n-1 = ε n-1 }.
Note that ∆ 2 = {α n-1 } and there is only one G 0 -stable divisor in G/ B. We have

dim G/ B = n(n -1) dim(G/B) = (n -1) 2 . Set y 0 = s α n-1 . . . s α 2 s α 1 = (1 n n -1 . . . 2)( 1 n . . . 2).
We have y 0 ǫ i = ǫ i-1 for i = 2, . . . , n and y 0 ǫ 1 = ǫ n . In particular

y 0 Φ + = { ǫ n ± ǫ j : j = 1, . . . , n -1} ∪ { ǫ i ± ǫ j : 1 ≤ i < j = n -1}.
One easily deduce that the stabilizer of

y 0 in G 0 is H 0 . The only descent is α 1 . Then D 0 = D = { α 1 } and -ρ( y 0 α 1 ) = -ρ(ε n -ε1 ) = ε 1 . 4.4 G 2 in Spin 7
Here G is the group of type G 2 embedded in Spin 7 (C) = G using the first fundamental representation which is 7-dimensional. We label the simple roots as follows:

B 3 1 2 3 G 2 1 2
The map ρ is characterized by ρ( α 2 ) = α 2 ρ( α 1 ) = ρ( α 3 ) = α 1 , and satisfies ρ( α 1 + α 2 ) = ρ( α 2 + α 3 ) = α 1 + α 2 ρ( α 2 + 2 α 3 ) = ρ( α 1 + α 2 + α 3 ) = 2α 1 + α 2 In particular

Φ + ∩ Φ 1 = { α 2 , α 1 + α 2 + 2 α 3 , α 1 + 2 α 2 + 2 α 3 }.
The working y 0 are y 0 = s 1 s 2 s 3 and y 0 = s 3 s 2 s 3 Indeed, one can check that ℓ( y 0 ) = 3 and dim(G ∩ y 0 B y 0 -1 ) = 3. Moreover, the only simple roots α such that dim(G ∩ w B w -1 ) = 4 where w = y 0 s α is α 3 . The so obtained divisor is the only G-stable divisor accordingly to ♯∆ 2 = 1.

There are ♯ W ♯W = 4 G-orbits in G/ B. The closed orbit has dimension 6 and the open one 9: hence there is one orbit in each dimension from 6 to 9. Computing the dimensions of the stabilizers we get the graph Γ( G/G):

F 4 in E 6

Here G is the group of type F 4 embedded in G of type E 6 . We label the simple roots as follows:

E 6 1 3 4 5 6 2 F 4 1 2 3 4
The root system E 6 lies in R 8 with basis ( ǫ i ) 1≤i≤8 . More precisely it spans the space x 6 = x 7 = -x 8 . See [START_REF] Bourbaki | Lie groups and Lie algebras[END_REF]. The roots are ± ǫ i ± ǫ j 1 ≤ i < j ≤ 5 and ± 1 2 ( ǫ 8 -ǫ 7 -ǫ 6 + 5 i=1 (-1) ν(i) ǫ i ν(i) even.

Set ǫ6 = ǫ 8 -ǫ 6 -ǫ 7 . Then, the simple roots are α 1 = 1 2 ǫ 1 + ǫ 8 -( ǫ 2 + ǫ 3 + ǫ 4 + ǫ 5 + ǫ 6 + ǫ 7 ) = 1 2 ǫ 1 + ǫ8 -( ǫ 2 + ǫ 3 + ǫ 4 + ǫ 5 ) and α 2 = ǫ 1 + ǫ 2 α 3 = ǫ 2 -ǫ 1 α 4 = ǫ 3 -ǫ 2 α 5 = ǫ 4 -ǫ 3 α 6 = ǫ 5 -ǫ 4 .

Root system F 4 . The roots are 1 2 (±ǫ 1 ± ǫ 2 ± ǫ 3 ± ǫ 4 ) and ± ǫ i ± ǫ i ± ǫ j 1 ≤ i < j ≤ 4.

The simple roots are

α 1 = ǫ 2 -ǫ 3 α 2 = ǫ 3 -ǫ 4 α 3 = ǫ 4 α 4 = 1 2 (ǫ 1 -ǫ 2 -ǫ 3 -ǫ 4 ).
The map ρ is characterized by ρ( α 2 ) = α 1 ρ( α 4 ) = α 2 ρ( α 3 ) = ρ( α 5 ) = α 3 ρ( α 1 ) = ρ( α 6 ) = α 4 . We have 12 short positive roots in F 4 :

ǫ i 1 2 (ǫ 1 ± ǫ 2 ± ǫ 3 ± ǫ 4 )
A working y 0 is y 0 = s 1 s 5 s 3 s 4 s 2 s 3 s 4 s 5 s 4 s 3 s 1 s 6 .

Indeed, one can check that ℓ( y 0 ) = 12 and dim(G ∩ y 0 B y 0 -1 ) = 16. Moreover, the only simple roots α such that dim(G ∩ w B w -1 ) = 17 where w = y 0 s α are α 1 and α 6 . The so obtained divisors are distinct since ♯∆ 2 = 2. Set D 0 = { α 1 , α 6 }.

One checks that -y 0 α 1 = α 1 + α 2 + 2 α 3 + 2 α 4 + α 5 ρ -→ α 1 + α 2 + 3α 3 + α 4 ; -y 0 α 6 = α 1 + α 2 + α 3 + 2 α 4 + 2 α 5 + α 6 ρ -→ α 1 + 2α 2 + 3α 3 + 2α 4 .

2

  Reminder and complements on spherical homogeneous spaces of minimal rank 2.1 Roots of G and G Fix a spherical pair of minimal rank (G, G) with G and G reductive. Choose a maximal torus T in G and a Borel subgroup T ⊂ B ⊂ G. Denote by Φ (resp. Φ + ) the set of roots (resp. positive roots). Recall that ∆ denotes the set of simple roots. Fix also a maximal torus T of G containing T and let ρ : X( T )-→X(T ) denote the restriction map. The set Φ of roots of G maps onto Φ (see [Res10, Lemma 4.2]). Let ρ denote the restriction of ρ to Φ. By setting Φ + = ρ-1 (Φ + ), one gets a choice of positive roots for G. Let B denote the corresponding Borel subgroup. Then B contains B. By [Res10, Lemma 4.6], ρ( ∆) = ∆, where ∆ denote the set of simple roots of G.

  by the action of B s given by (b 1 , . . . , b s ).(p 1 , . . . , p s , B/B) = (p 1 b 1 -1 , . . . , b s-1 p s b -1 s , b s B/B) (with obvious notation). Consider the regular map

  we have to understand what sections σ extend, and hence the order of the poles of σ along the divisors of (G/B × G/ B)-(G/B × G. y 0 B/ B).

Fix

  α ∈ ∆ that is a label of some edge descending from the open orbit in Γ( G/G). Then D α := G y 0 s α B/ B is a divisor of G/ B. By Lemma 7 this happens if and only if y 0 s α ∈ Φ 2 . Set β = ±ρ( y 0 α) the sign being chosen to get β ∈ Φ + .

  (gB/B, t) ∈ G/B × C * , then (gB/B , s β y 0 s α ǫ -α (t) B/ B) =(gB/B , s β y 0 ǫ -α (t -1 ) B/ B) =(gB/B , s β ǫ -β (c -1 t -1 ) y 0 B/ B)

  D = { α 3 } and -ρ( y 0 α 3 ) = α 1 + α 2 .

It is the quotient by ǫ 2

 2 + ǫ 3 -ǫ 1 -ǫ 4 1 2 ǫ 1 + ǫ6 + ǫ 4 -(ǫ 2 + ǫ 3 + 3ǫ 5 ) Moreover ♯ Φ + = 36 ♯ W = 51 840 ♯∆ 2 = 2 ♯W = 1 152 ♯Φ + = 24 ♯ W /W = 45.

Proof. Call sl β the subalgebra of g spanned by X β , H β , X -β . Decompose V * ν into a direct sum of sl β irreducible representations: V * ν = V δ . Write σ = δ σ δ accordingly to this decomposition. Observe that σ δ ∈ V δ ( y 0 • ν |CH β ). Since V δ is an sl 2 irreducible representation, if σ δ = 0, then:

Hence condition 1) is equivalent to:

Similarly condition 2) is equivalent to:

then we easily conclude that 1) and 2) are equivalent.

Explicit description on the examples

For each example, we determine a working y 0 and a set of simple roots parametrizing the G-stable divisors of G/ B. With the notation of the introduction, we give a subset D 0 of D such that the map α -→ G y 0 s α B/ B is a bijection from D 0 onto the set of G-stable divisors.

Tensor product case

Note that y 0 = y 0 -1 and that, for α ∈ ∆, y 0 (0, α) = (0, w 0 α) ∈ Φ -, while y 0 (α, 0) = (α, 0) ∈ Φ + . So, according to our notations, (0, α) = α -and (α, 0) = α + . If ν ∈ X(T ) + and λ, µ ∈ X(T ) we define two sets:

In Theorem 2.1 of [START_REF] Parthasarathy | Representations of complex semi-simple Lie groups and Lie algebras[END_REF] the autors realized, by algebraic methods isomorphisms:

We explain how this can be recovered from Theorem 1, using D + 0 and D - 0 as parametrizations of the G-stable divisors. Notice that, in this case, the conditions 1 of the theorem are empty. Then ρ( y 0 ν) = (ν 1 -ν * 2 ) and for α + = (α, 0) ∈ D + 0 , ρ( y 0 α + ) = α and ν, ( α) ∨ = ν 1 , α ∨ . Hence, from Theorem 1, we deduce that

) is isomorphic to Mult(ν * , ν), we recover (7). Now, ρ( y 0 (0, -w 0 α)) = -α, and ν, (0, -w 0 α) ∨ = ν 2 , (-w 0 α) ∨ = ν * 2 , α ∨ . Then, if we use D - 0 to get a minimal number of conditions in the second point of the theorem, we deduce that:

And since Mult(ν, ν * ) ≃ Mult(ν * , ν) ≃ Mult(ν * , (ν 2 , ν 1 )), we recover (8).

Sp 2n in SL 2n

Fix n ≥ 2. Let V be a 2n-dimensional vector space with fixed basis B = (e 1 , . . . , e 2n ). Consider the following matrices

of size n × n and 2n × 2n respectively. View ω as a symplectic bilinear form of V .

Let G be the associated symplectic group. Set T = {diag(t 1 , . . . , t n , t -1 n , . . . , t -1 1 ) : t i ∈ C * }. Let B be the Borel subgroup of G consisting of upper triangular matrices of G. For i ∈ [1, n], let ε i denote the character of T that maps diag(t 1 , . . . , t n , t -1 n , . . . , t -1 1 ) to t i ; then X(T ) = ⊕ i Zε i . Here

This is consistent with the fact that ♯Φ

Since the ω-orthogonal of e 1 , e1, . . . , ek is e k+1 , . . . , e n the stabilizer of y 0 B/ B is diagonal by blocks with 2 blocks of size 2k and 2(n -k) (in the basis ordered according to y 0 ). Since this is . Similarly, the contribution of the pairs j > ī with ī ∈ {n + 1, . . . , 2n} is n(n-1)

and y 0 has minimal length.

Let T be the maximal torus of G consisting in diagonal matrices. Write X( T ) = ⊕ 2n i=1 Zε i /(ε 1 + • • • + ε2n ) with the usual notation. Let α i = εi -εi+1 be a simple root of G. If i = 2k is even then

Spin 2n-1 in Spin 2n

Let V be a 2n-dimensional vector space endowed with a basis B = (e 1 , . . . , e 2n ). Denote by (x 1 , . . . , x 2n ) the dual basis. For i ∈ [1; 2n], set i = 2n + 1 -i. Let G 0 be the orthogonal group associated to the quadratic form

x i x i .

Set T 0 = {diag(t 1 , . . . , t n , t -1 n , . . . , t -1 1 ) : t i ∈ C * } in G 0 . Let B 0 be the Borel subgroup of G 0 consisting in upper triangular matrices of G 0 . Let εi denote the character of T 0 that maps diag(t 1 , . . . , t n , t -1 n , . . . , t -1 1 ) to t i ; then X( T 0 ) = n i=1 Zε i . Here Φ + = {ε i ± εj : 1 ≤ i < j ≤ n}, and ∆ = { α 1 = ε1 -ε2 , α 2 = ε2 -ε3 , . . . , α n-1 = εn-1 -εn , α n = εn-1 + εn }.

The Weyl group W of G 0 is a subgroup of the Weyl group S 2n of SL(V ). More precisely

) is even }.

For i = 1, . . . , n -1, s α i = (i, i + 1)(i + 1 ī). Moreover s αn = (n -1 n)(n n -1).

Let H = e 1 , . . . , e n-1 , en+e n+1 √ 2

, e n+1 , . . . , e 2n with coordinates (x i ) 1≤i≤n-1 ∪ (y) ∪ (xī) 1≤i≤n-1 . The restriction of Q on H is