Sel. math., New ser. 12 (2006), 517–540 c 2007 Birkh¨

auser Verlag Basel/Switzerland 1022-1824/040517-24, published online March 13, 2007 DOI 10.1007/s00029-006-0027-z

Selecta Mathematica New Series

Adjacency of Young tableaux and the Springer fibers N. G. J. Pagnon and N. Ressayre Abstract. We connect different results about irreducible components of the Springer fibers of type A. Firstly, we show a relation between the Spaltenstein partition of the fibers and a total order ≺ on the set of standard Young tableaux. Next, using a result of Steinberg, we connect a work of the first author to the Robinson–Schensted map. We also perform the Spaltenstein study of the relative position of the Springer fibers and P1 -fibrations of the flag manifold. This leads us to consider the adjacency relation on the set of standard Young tableaux and to define oriented and labeled graphs with the standard Young tableaux as vertices. Using this adjacency relation, we describe some smooth irreducible components of the Springer fibers. Finally, we show that these graphs can be identified with some full subgraphs of the Bruhat graph. Mathematics Subject Classification (2000). 14M15, 05E10. Keywords. Springer fiber, Robinson–Schensted correspondence, Schubert cell, Spaltenstein correspondence, Bruhat graph.

1. Introduction Let V be an n-dimensional vector space over a field k, F = F(V ) be the (full) flag manifold of V , and x be a nilpotent endomorphism of V . In this article, we study the Springer fiber Fx := {({0} = V0 , V1 , . . . , Vn ) ∈ F : x(Vi ) ⊆ Vi−1 ∀i = 1, . . . , n}. N. Spaltenstein has defined in each irreducible component of Fx a non-singular open subvariety. Moreover, these subvarieties form a partition of Fx and are parametrized. To explain Spaltenstein’s results more precisely, let us introduce some notation. Let λ1 ≥ · · · ≥ λr be the sizes of the Jordan blocks of x in decreasing order. So, λ1 + · · · + λr = n; in other words, we have a partition of n denoted by λ ⊢ n. This partition is called the type of x. To the partition λ is associated its

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Young diagram Y (λ) = Y (x) whose r lines are composed respectively of λ1 , . . . , λr squares. A filling of the Young diagram Y (λ) with the integers 1, . . . , n such that the entries along any line or column are strictly increasing is called a standard Young tableau of shape λ. Let Stλ denote the set of standard Young tableaux of shape λ. For any flag ξ in Fx , by considering the relative position of the subspaces of ξ and the images of the powers of x, N. Spaltenstein associated to ξ a standard Young tableau σξ . He showed that for any σ ∈ Stλ , the set Fx,σ of ξ such that σξ = σ is a non-singular open irreducible subvariety of Fx . Moreover, each irreducible component of Fx contains exactly one Fx,σ ; that is, the irreducible components of Fx are the closures Fσ of the Fx,σ . Let us recall from [6] the definition of a total order ≺ on Stλ . For each σ ∈ Stλ and for i ∈ {1, . . . , n}, let cσ (i) (resp. lσ (i)) denote the number of the column (resp. line) in which i lies in σ. For σ1 , σ2 ∈ Stλ , we write σ1 ≺ σ2 if for some 1 ≤ i0 ≤ n we have lσ1 (i0 ) < lσ2 (i0 ), and lσ1 (j) = lσ2 (j) for i0 < j ≤ n. The notation σ1  σ2 means σ1 ≺ σ2 or σ1 = σ2 . We can now state our first result: Theorem 1. For any σ ∈ Stλ , we have Fσ ⊆

[

Fx,γ .

γσ

By Theorem 1, the knowledge S of the order ≺ determines the open subset Fx,σ of Fσ ; indeed, Fx,σ = Fσ − γ≺σ Fγ . Let σmax (resp. σmin ) denote the maximal (resp. minimal) element of Stλ for the order ≺. Theorem 1 shows that Fx,σmin = Fσmin ; thus, by a result of Spaltenstein, this irreducible component of Fx is smooth. In Section 4 we connect the result in [6] with the Robinson–Schensted map. Let us fix a base B of V such that the matrix of x in the base B has Jordan’s form with decreasing block sizes. Let B denote the subgroup of GL(V ) consisting of the endomorphisms with upper triangular matrices in the base B. The orbits of B in F are the Schubert cells; they are parametrized by the group Sn of permutations of {1, . . . , n}. For every σ ∈ Stλ , in [5] for the hook case and in [6] for the general case, we can find a description of the element wσ ∈ Sn corresponding to the Schubert cell Cwσ which intersects Fσ in an open dense subset of Fσ . In [1] and [9], one can find a combinatorial definition and a geometric interpreS tation of the Robinson–Schensted bijection RS : µ⊢n Stµ × Stµ → Sn , (σ, γ) 7→ RS(σ, γ). In the classical combinatorial theory of standard Young tableaux, the Sch¨ utzenberger involution σ 7→ σ ∨ of the set Stλ is known. Our main result in Section 4 is Theorem 2. For all σ ∈ Stλ , we have wσ = RS(σmax , σ ∨ ). In Section 5 we are interested in the open subset Cwσ ∩ Fx of Fx . Since the map RS is injective, Cwσ ∩ Fx is not dense in Fγ for all γ 6= σ. The following theorem makes this observation precise:

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Theorem 3. For every σ ∈ Stλ , we have Cwσ ∩ Fx ⊆ Fx,σ . In particular, Cwσ ∩ Fx is smooth. Let k ∈ [1; n − 1] be an integer. We denote by Fk the variety consisting of the partial flags (V1 ⊆ · · · ⊆ Vk−1 ⊆ Vk+1 ⊆ · · · ⊆ Vn = V ) such that dim(Vi ) = i for all i. We denote by φk : F → Fk the P1 -fibration which omits the subspace Vk . A subset X in F is said to be φk -stable if X = φ−1 k (φk (X)). We recover Spaltenstein’s result (see [8]) about the φk -stability of Fσ and give a more precise description of it (see Proposition 6.1). Combined with the geometric interpretation of the Sch¨ utzenberger involution due to van Leeuwen (see [10]), Spaltenstein’s result implies that cσ (k − 1) ≥ cσ (k) if and only if cσ∨ (n − k) ≥ cσ∨ (n + 1 − k), for any 2 ≤ k ≤ n. For any 3 ≤ k ≤ n, two standard Young tableaux of shape λ are said to be k-adjacent if they are obtained from each other by switching the places of k and k − 1. From our description of the φk -stability, we deduce that if σ and γ are k-adjacent then Fσ intersects Fγ in a set of codimension one. Let Γλ be the oriented graph with vertex set Stλ and the edges labeled by the integers 3, . . . , n, where σ is joined to γ by an edge labeled by k if σ and γ are k-adjacent and σ ≺ γ. The graph Γλ is called the adjacency graph of λ. We show that σmin (resp. σmax ) is the unique minimal (resp. maximal) vertex of Γλ . An irreducible component Fσ of Fx which is homogeneous under the action of a parabolic subgroup of GL(V ) is called a Richardson component. In Section 7.1, we recall the standard Young tableaux corresponding to the Richardson components. We determine other smooth components: Theorem 4. Let σ0 be a standard Young tableau such that Fσ0 is a Richardson component. Let σ be a standard Young tableau k-adjacent to σ0 such that σ0 ≺ σ. Then Fσ is smooth. For i = 1, . . . , n−1, let si denote the transposition in Sn which permutes i and i + 1. The si ’s generate Sn ; the length l(w) of the element w in Sn is the minimal number of si necessary to write w. A decomposition of w into transpositions is said to be reduced if it makes use of l(w) transpositions si . In Section 8, we show that the expression of wσ given in [6] is reduced. Generalizing this result, we give a (new?) reduced canonical decomposition of the elements of Sn . Let Γn be the oriented graph with vertex set Sn and the edges labeled by the integers 2, . . . , n, where w is joined to w′ by an edge labeled by k if w′ = wsn+1−k and l(w′ ) = l(w) + 1. The graph Γn is called the Bruhat graph. Our main result connects the Bruhat graph and the adjacency graphs: Theorem 5 (Main Theorem). The map Stλ → Sn , σ 7→ wσ , makes the adjacency graph a full subgraph of the Bruhat graph. In particular, for any σ and γ in Stλ , wσ and wγ are not joined by an edge labeled by 2 in the Bruhat graph. Let us recall that in [6], the first author gives an expression of wσ as a product of simple reflexions in terms of the standard Young tableau σ. The study of this

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decomposition is central to this article. Our last result gives an alternative reduced expression of wσ (see Theorem 8.9).

2. Spaltenstein map In this section, we recall some useful facts about the Spaltenstein map. For later use, we also include some proofs. Other proofs can be found in [2, 8, 11]. A base B is said to be adapted to x if the matrix of x in B is a Jordan matrix with decreasing block sizes. For any base B, the flag whose ith subspace is spanned by the first i vectors of B is called the canonical flag associated to B. A flag is said to be adapted to x if it is the canonical flag associated to an adapted base B. Let us fix a line V1 in Ker x. Consider V ′ = V /V1 , the canonical projection q : V → V ′ and x′ : V ′ → V ′ , v + V1 7→ x(v) + V1 . Let (F1 , . . . , Fn ) be a flag adapted to x. Set ( (Fi + V1 )/V1 if 1 ≤ i ≤ K − 1, ′ (2.1) Fi := Fi+1 /V1 if K ≤ i ≤ n − 1, where K denotes the minimum i such that V1 ⊆ Fi . Since V1 ⊆ Ker x, we have K = λ1 + · · · + λi0 + 1 for some i0 ≤ r − 1 (obviously, λ1 + · · · + λi0 = 0 if i0 = 0). ′ ) is adapted to x′ . Lemma 2.1. With the above notation, the flag (F1′ , . . . , Fn−1 Moreover, the partition λ′ associated to x′ is obtained from λ by replacing λi1 by λi1 − 1 (obviously if λi1 equals one we have to omit λi1 to obtain λ′ ), where i1 := max{j : λj = λi0 }.

Proof. Let B = (e1 , . . . , en ) be a base of V adapted to x such that (F1 , . . . , Fn ) is the canonical flag associated to B. Consider the following base B′ of V ′ : B′ = (q(e1 ), . . . , q(eK−1 ), q(eK+1 ), . . . , q(en )). The matrix of x′ in B′ has the following form:  ⋆ 0  Jλ .. 1  .   0  ..  .. .  .  ..  .   Jλi0 +1  



..

. By an easy change of base, one can check the lemma.

            



Let ξ = (V1 , . . . , Vn ) ∈ Fx . For i = 1, . . . , n, we set αi := max{k : Vi ⊆ Vi−1 + Im(xk−1 )}. Notice that by convention V0 = {0}.

(2.2)

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Proposition 2.2. There exists a unique standard Young tableau σ with the property that cσ (n + 1 − i) = αi for all i = 1, . . . , n. Proof. The unicity is obvious. Let us prove the existence by induction on n. We use the notation of Lemma 2.1. For i = 1, . . . , n − 1, set βi = max{k : q(Vi+1 ) ⊆ q(Vi ) + Im(x′k−1 )}. By induction, there exists a standard Young tableau σ ′ of shape λ′ such that cσ′ (n − i) = βi for all i = 1, . . . , n − 1. One easily checks that the standard Young tableau σ obtained from σ ′ by adding a box indexed by n in line i1 satisfies the lemma.  Corollary 2.3 (see [2, 11]). Let ξ = (V1 , . . . , Vn ) ∈ Fx and σ ∈ Stλ . The following are equivalent: (i) Vi ⊆ Vi−1 + Im(xcσ (n+1−i)−1 ) for all i = 1, . . . , n − 1.  Vi ⊆ Vi−1 + Im(xcσ (n+1−i)−1 ) (ii) for all i = 1, . . . , n − 1. Vi * Vi−1 + Im(xcσ (n+1−i) ) Proof. We assume that ξ satisfies (i). Consider the integers αi defined by (2.2). By Proposition 2.2, there exists a standard Young tableau γ such that ξ satisfies (ii) with γ in place of σ. Since ξ satisfies (i), we have cγ (n + 1 − i) = αi ≤ cσ (n + 1 − i) for all i = 1, . . . , n. One easily deduces that γ = σ.  We can now explain the Spaltenstein partition of Fx . For σ ∈ Stλ , we denote by Fx,σ the set of flags ξ which satisfy the conditions of Corollary 2.3. Proposition 2.2 shows the following result of Spaltenstein: (Fx,σ )σ∈Stλ is a partition of Fx . Moreover, Spaltenstein showed that the Fx,σ are smooth, open and irreducible subvarieties of Fx . For any σ ∈ Stλ , the closure Fσ of Fx,σ is an irreducible component of Fx . Let IFx denote the set of irreducible components of Fx . Spaltenstein showed in [7] that the map Spal : Stλ → IFx , σ 7→ Fσ , is a bijection. Corollary 2.4. Let (V1 , . . . , Vk ) be a partial flag stable under x and satisfying conditions (ii) of Corollary 2.3 for i = 1, . . . , k. Then there exists ξ ∈ Fx,σ starting with (V1 , . . . , Vk ). Proof. First we notice that the map Fx,σ → P(Ker x ∩ Im(xcσ (n)−1 )) − P(Ker x ∩ Im(xcσ (n) )),

(Vi ) 7→ V1 ,

is a surjective fibration with a typical fiber isomorphic to Fx′ ,σ′ . If (V1 , . . . , Vk ) is a partial flag stable under x and satisfies conditions (ii) of Corollary 2.3, then ′ := Vk /V1 ) is stable under x′ and satisfies the partial flag (V1′ := V2 /V1 , . . . , Vk−1 ′ (ii) for x and the standard Young tableau σ ′ obtained by deleting the square labeled by n. By induction on dim V , there exists ξ ′ = ξ/V1 ∈ Fx′ ,σ′ starting with ′ (V1′ , . . . , Vk−1 ) and the result follows for ξ. 

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3. Closure of Spaltenstein cells 3.1. Let us recall from [6] the definition of a total order ≺ on Stλ . For σ1 , σ2 ∈ Stλ , we write σ1 ≺ σ2 if for some 1 ≤ i0 ≤ n we have lσ1 (i0 ) < lσ2 (i0 ), and lσ1 (j) = lσ2 (j) for i0 < j ≤ n. Let σmax (resp. σmin ) denote the maximal (resp. minimal) element of Stλ for the order ≺. Notice that σmax is obtained by filling the first line of the Young diagram Y (λ) with the integers {1, . . . , λ1 }, the second with {λ1 + 1, . . . , λ1 + λ2 }, and so on. The tableau σmin is obtained by filling the first column of Y (λ) with {1, . . . , r}, the second with r + 1, r + 2, . . . , and so on. Lemma 3.1. Let σ be a standard Young tableau (different from σmin ) and γ its predecessor for the order ≺. Let i0 denote the integer such that lγ (i0 ) < lσ (i0 ) and lγ (j) = lσ (j) for i0 < j ≤ n. Let σ ′ (resp. γ ′ ) denote the standard Young tableau obtained from σ (resp. γ) by deleting the squares labeled by the integers {i0 , . . . , n}. (i) The boxes occupied by i = i0 + 1, . . . , n are the same in γ and σ. (ii) Consider the Young diagram Y obtained from σ ′ (or γ ′ ) by adding the square labeled by i0 in σ (or γ). In Y , there is no corner between the boxes labeled by i0 in σ and γ. (iii) σ ′ (resp. γ ′ ) is the maximal (resp. minimal) element for the corresponding shape. Proof. Assertion (i) is obvious. By omitting the boxes occupied by i0 + 1, . . . , n, we may assume that i0 = n. By contradiction, assume that there exists a corner between the two squares labeled by n in γ and σ. Let δ be a standard Young tableau for which this corner is occupied by n. We have γ ≺ δ ≺ σ; this is absurd and assertion (ii) follows. Any tableau δ of shape Y (λ) with n in the same box as in γ satisfies δ ≺ σ. Therefore, γ ′ is maximal. Any tableau δ of shape Y (λ) with n in the same box as in σ satisfies γ ≺ δ. Therefore, σ ′ is minimal.  If i0 = n, Lemma 3.1 can be summarized by Figure 1. σmin

σmax

n

n

σ

Predecessor of σ Fig. 1. Predecessor for ≺.

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3.2. Let σ ∈ Stλ be a standard Young tableau. Define the boundary of Fσ as ∂Fσ := Fσ − Fx,σ . The goal of this section is to prove the following Theorem 3.2. ∂Fσ ⊆

[

Fγ .

γ≺σ

First, we prove the following

Lemma 3.3. There exist integers d1 , . . . , dn−1 such that for all ξ = (V1 , . . . , Vn ) ∈ Fx,σ and for all i = 1, . . . , n − 1 we have dim(Vi−1 + Im(xcσ (n+1−i)−1 )) = di . Proof. We proceed by induction on i. For i = 1, there is nothing to prove. Consider V ′ , x′ , σ ′ ξ ′ as before. For 2 ≤ i ≤ n − 1 we have cσ (n + 1 − i) = cσ′ (n + 1 − i) = ′ ) ∈ Fx′ ,σ′ and for every cσ′ (n − (i − 1)). By induction for all ξ ′ = (V1′ , . . . , Vn−1 i = 1, . . . , n − 2, we have ′ + Im(x′cσ′ (n−(i−1))−1 )) d′i−1 = dim(Vi−2

= dim(Vi−1 /V1 + (Im(xcσ (n+1−i)−1 ) + V1 )/V1 ) = dim((Vi−1 + Im(xcσ (n+1−i)−1 ))/V1 ) = dim(Vi−1 + Im(xcσ (n+1−i)−1 )) + 1. These equalities show that di = d′i−1 − 1 and the proof is complete.



We also need the following well known Lemma 3.4. Let F be a vector subspace of V . Let Grk (V )i := {W ∈ Grk (V ) : dim(F + W ) = i}. Then Grk (V )i is a locally closed in Grk (V ). Moreover, the map φ : Grk (V )i → Gri (V ), W 7→ F + W , is a morphism. We can now prove Theorem 3.2: Proof of Theorem 3.2. Let ξ = (V1 , . . . , Vn ) ∈ Fσ and αi be the integers defined by (2.2). Set i0 := min{i : αi 6= cσ (n + 1 − i) − 1}. Consider the integers di defined in Lemma 3.3. Denote by Ω the set of (W1 , . . . , Wn ) ∈ F such that ∀i ≤ i0

dim(Wi−1 + Im(xcσ (n+1−i)−1 )) = di .

By Lemma 3.4, Ω is a locally closed subvariety of F. Moreover, for all i = 1, . . . , i0 , the map φi : Ω → Grdi (V ), (W1 , . . . , Wn ) 7→ Wi−1 + Im(xcσ (n+1−i)−1 ), is a morphism. We deduce that Ω0 := {ζ = (W1 , . . . , Wn ) ∈ Ω : Wi ⊆ φi (ζ) ∀i ≤ i0 } is closed in Ω. But, since i0 is minimal, Corollary 2.4 and Lemma 3.3 imply that ξ ∈ Ω. So, ξ belongs to the closure of Fx,σ in Ω. Therefore, it belongs to Ω0 . We deduce that αi0 > cσ (n + 1 − i0 ) − 1. The theorem follows. 

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Corollary 3.5. We have Fx,σmin = Fσmin . In particular, Fσmin is a smooth subvariety.

4. Robinson–Schensted map 4.1. We now introduce a dual Spaltenstein map. Let V ∗ be the dual of V . If F is a vector subspace of V , we denote by F ⊥ the orthogonal of F in V ∗ . Consider the isomorphism η : F(V ) → F(V ∗ ),

⊥ , . . . , V1⊥ , V ∗ ). (V1 , . . . , Vn ) 7→ (Vn−1

Consider the transposed map t x ∈ Hom(V ∗ , V ∗ ) of x. One easily checks that η induces by restriction an isomorphism from Fx onto Ft x . In particular, it induces a bijection η˜ : IFx → IFt x . By composing the Spaltenstein map of t x with η˜, one obtains a new bijection Spal∗ : Stλ → IFx ,

σ 7→ Fσ∗ .

In [10], one can find a combinatorial definition of the Sch¨ utzenberger involution Stλ → Stλ , σ 7→ σ ∨ . We now recall from [10] the following geometric interpretation of this involution: for all σ ∈ Stλ , we have Fσ∗ = Fσ∨ . 4.2. Let RS : Stλ × Stλ → Sn denote the Robinson–Schensted map. We now explain the Steinberg geometric interpretation of the Robinson–Schensted map. Let us start with a geometric interpretation of permutations. Let (ξ, ξ ′ ) ∈ F ×F. There exists a unique w ∈ Sn such that there exists a basis (e1 , . . . , en ) of V such that ξ (resp. ξ ′ ) is the canonical flag associated to (e1 , . . . , en ) (resp. (ew(1) , . . . , ew(n) )). The permutation w is called the relative position of ξ and ξ ′ , or of the pair (ξ, ξ ′ ). In fact, by Bruhat’s lemma w determines the GL(V )-orbit of (ξ, ξ ′ ). Let (C, C ′ ) ∈ IFx × IFx . Since GL(V ) has finitely many orbits in F × F, there exists a unique orbit O(C,C ′ ) such that O(C,C ′ ) ∩ (C × C ′ ) is open and dense in C × C ′ . Let w(C,C ′ ) ∈ Sn be the relative position of an element of O(C,C ′ ) . Consider the Robinson–Schensted–Steinberg map RSS : IFx × IFx → Sn ,

(C, C ′ ) 7→ w(C,C ′ ) .

We can now state the result of Steinberg: for any (σ, σ ′ ) ∈ Stλ × Stλ , we have RSS(Fσ∗ , Fσ∗′ ) = RS(σ, σ ′ ). In other words, RS is obtained by composing RSS and the parametrization Spal∗ of IFx .

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4.3. We now recall from [6] the definition of a map Pa : Stλ → Sn . Let us fix a flag ξ0 adapted to x. For any w ∈ Sn , we denote by Ow the set of flags ξ such that w is the relative position of (ξ0 , ξ). For any σ ∈ Stλ , there exists a unique wσ ∈ Sn such that Owσ ∩ Fσ is open and dense in Fσ . Obviously, wσ does not depend on the choice of the adapted flag ξ0 . We set Pa : Stλ → Sn , σ 7→ wσ . We now recall from [6] a combinatorial description of wσ . Set nσ (i) := card{j > i : lσ (j) > lσ (i)}. Let ci denote the increasing cycle of Sn of length nσ (i) ending with n + 1 − i. In other words, ci = sn−(i+nσ (i)−1) · · · sn−(i+1) sn−i . The main result of [6] is the following formula for wσ : wσ := c1 · · · cn−1 .

(4.1)

We now introduce a diagram useful to understand wσ . First, we number n vertical lines from 1 to n. Then the cycle ci is represented by a horizontal arrow ending on line n + 1 − i and of length nσ (i). We draw successively cn−1 , . . . , c1 . For example, if n = 6, nσ (5) = 1, nσ (4) = 2, nσ (3) = 0, nσ (2) = 2 and nσ (1) = 4 we obtain: 1 2 3 4 5 6 nσ (5) = 1 nσ (4) = 2 nσ (3) = 0 nσ (2) = 2 nσ (1) = 4

4.4. Let σ0 be the unique standard Young tableau such that an adapted flag ξ0 belongs to Fx,σ0 . One can describe σ0 as follows: First, n, . . . , n − λ1 + 1 are placed such that cσ (n) = λ1 , cσ (n − 1) = λ1 − 1, . . . , cσ (n − λ1 + 1) = 1. Next, n − λ1 , . . . , n − λ1 − λ2 + 2 are placed in the same way in the free boxes; and so on. As an example we give σ0 for the partition (6, 4, 3, 1): 1 3 4 8 13 14 2 6 7 12 5 10 11 9 We can now state a relation between RSS and Pa: Theorem 4.1. For all σ ∈ Stλ , we have RSS(Fσ , Fσ0 ) = wσ .

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Proof. Let Ωx denote the set of flags adapted to x. First, we notice that Ωx is one orbit of Z(x) := {g ∈ GL(V ) : gxg −1 = x}. Indeed, let ξ and ξ ′ be two points of Ωx . Let B and B′ be two bases adapted to x such that ξ and ξ ′ are respectively the flags associated to B and B′ . Let g ∈ GL(V ) be such that g(B) = B′ . Since the matrices of x in B and B′ are the same, g ∈ Z(x). But ξ ′ = gξ. Conversely, if g ∈ Z(x) and ξ ∈ Ωx , then gξ ∈ Ωx . Since Ωx is one orbit, it is a locally closed subvariety of Fx . We claim that dim Ωx =

r X

(i − 1)λi .

i=2

We argue by induction on r. Consider the following locally closed subvariety of Grλ1 (V ): Gx := {F ∈ Grλ1 (V ) : x(F ) ⊆ F and xλ1 −1 (F ) 6= {0}}, and the morphism Θ : Ωx → G x ,

(V1 , . . . , Vn ) 7→ Vλ1 .

′

Fix F in Gx . Set V = V /F and denote by q : V → V ′ the canonical projection. Let y denote the restriction of x to F , and x′ ∈ Hom(V ′ , V ′ ) the linear map induced by x. Let ξ = (V1 , . . . , Vn ) ∈ Θ−1 (F ). Then Vi = Ker(y i ) for i = 1, . . . , λ1 . On the other hand, (Vλ1 +1 /F, . . . , Vn−1 /F ) is a flag of V ′ adapted to x′ . Conversely, ′ ) is a flag of V ′ adapted to x′ then the flag if (Vλ′1 +1 , . . . , Vn−1 ′ )) (Ker(y), . . . , Ker(y λ1 −1 ), F, q −1 (Vλ′1 +1 ), . . . , q −1 (Vn−1

is adapted to x. One easily deduces that Θ−1 (F ) is isomorphic to Ωx′ . In particular, dim Ωx = dim Ωx′ + dim Gx . P But, by induction, dim Ωx′ = i≥3 (i − 2)λi . It remains to prove that dim Gx = λ2 + · · · + λr = n − λ1 . Consider the set V0 of vectors v in V such that (v, x(v), . . . , xλ1 −1 (v)) are linearly independent, and the map Γ : V0 → Gx which associates to each v ∈ V0 the subspace spanned by (v, x(v), . . . , xλ1 −1 (v)). The set V0 is non-empty and open in V . Moreover, Γ is surjective and for all F ∈ Gx , Γ−1 (F ) is an open subset of F . We deduce that dim Gx = dim V − dim F = n P− λ1 . The claim follows. By Proposition 2.2 of [10], dim Fx = (i − 1)λi . So, we just proved that dim Ωx = dim Fx . Moreover, we have already noticed that Ωx is contained in Fx,σ0 . Finally, Ωx is open and dense in Fσ0 . Let σ ∈ Stλ . By definition, for all ξ ∈ Fσ ∩ Cwσ and ξ0 ∈ Ωx , the relative position of ξ and ξ0 is wσ . Since Fσ ∩ Cwσ and Ωx are open and dense respectively in Fσ and Fσ0 , we have RSS(Fσ , Fσ0 ) = wσ .  Remark 4.2. We have proved that Fσ0 contains a dense orbit of the neutral component Z(x)◦ of Z(x). By duality, Fσmax has the same property. An interesting question is to determine all the irreducible components Fσ with this property.

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5. Robinson–Schensted and Spaltenstein maps For every irreducible component Fσ of Fx , Cwσ ∩ Fσ is an open subset of Fσ . On the other hand, Spaltenstein has defined another open subset Fx,σ in Fσ . Our next theorem compares these two open subsets of Fσ : Theorem 5.1. For every σ ∈ Stλ , we have Cwσ ∩Fx ⊆ Fx,σ . In particular, Cwσ ∩Fx is smooth. Before proving Theorem 5.1, we show some lemmas. 5.1. Let σ be a standard Young tableau of shape λ. In this subsection, we collect some useful results about wσ and Cwσ . Lemma 5.2. Let ξ = (V1 , . . . , Vn ) ∈ Cwσ . Set K := min{k : V1 ⊆ Fk }. Then K = λ1 + · · · + λlσ (n)−1 + 1

and

wσ (1) = K.

Proof. In this proof, we will say that a flag (W1 , . . . , Wn ) has Property (∗) if λ1 + · · · + λlσ (n)−1 + 1 = min{k : W1 ⊆ Fk }. By Corollary 2.3 and since cσ (n) = λlσ (n) , generically a point of Fx,σ has Property (∗). So, there exists a flag in Cwσ ∩ Fx,σ which has Property (∗). By definition of the Bruhat cells, any point of Cwσ has Property (∗). Let us fix a base B = (e1 , . . . , en ) adapted to x such that ξ0 is the canonical flag associated to B. Then the canonical flag associated to (ewσ (1) , . . . , ewσ (n) ) belongs to Cwσ and has Property (∗). One easily deduces that wσ (1) = K.  Lemma 5.3. Let φ : {1, . . . , n} − {wσ (1)} → {1, . . . , n − 1} be defined by  k if k < wσ (1), φ(k) = k − 1 if k > wσ (1).

Let σ ′ be the standard Young tableau obtained from σ by deleting the box occupied by n. Then wσ′ (i) = φ(wσ (i + 1)) for all i = 1, . . . , n − 1.

Proof. Set w′ (i) = φ(wσ (i + 1)). We have to prove that w′ = wσ′ . Let η = (V1 , . . . , Vn ) ∈ Cwσ ∩ Fx,σ . Consider the subvariety FV1 of F consisting of the flags with V1 as line. The map π : FV1 → F(V ′ ), η = (V1 , W2 , . . . , Wn ) 7→ η ′ = (W2 /V1 , . . . , Wn /V1 ), is an isomorphism. ′ We use the notation of Lemma 2.1 with V1 and set ξ0′ = (F1′ , . . . , Fn−1 ). ′ ′ We consider the Schubert decomposition of F(V ) associated to ξ0 . Since by Lemma 2.1, ξ0′ is adapted to x′ , Cwσ′ ∩ Fx′ ,σ′ is open and dense in Fσ′ . Since the stabilizer of V1 and ξ0 in GL(V ) acts on F(V ′ ) just as the stabilizer ′ of ξ0 in GL(V ′ ) does, π(Cwσ ∩ FV1 ) is a Schubert cell Cw′ of F(V ′ ). Fix a non-zero vector v1 in V1 . Let B = (e1 , . . . , en ) be a base of V such that ξ0 is the canonical flag associated to B. By Lemma 5.2, the canonical flag associated to (e1 , . . . , eK−1 , v1 , eK+1 , . . . , en ) is also ξ0 : from now on, we assume that eK = v1 .

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By Lemma 5.2 again, the canonical flag associated to (ewσ (1) , . . . , ewσ (n) ) belongs to FV1 . Moreover, its image under π is the canonical flag associated to (εw′ (1) , . . . , εw′ (n−1) ) where εi = q(ei ) if i ≤ K − 1 and εi = q(ei+1 ) if i ≥ K + 1. As the canonical flag associated to (ε1 , . . . , εn−1 ) is ξ0′ , we infer that π(Cwσ ∩ FV1 ) = Cw′ . On the other hand, by Corollary 2.3, π(Fx,σ ∩ FV1 ) = Fx′ ,σ′ . We deduce that Cw′ ∩ Fx′ ,σ′ is dense in Fx′ ,σ′ , and so w′ = wσ′ . Moreover, by Corollary 2.3, π(FV1 ∩ Fx,σ ) = Fx′ ,σ′ . So, π(FV1 ∩ Cwσ ∩ Fx,σ ) is open in Fx′ ,σ′ and contains η ′ . Therefore, π(FV1 ∩ Cwσ ) equals Cwσ′ . The lemma follows.  5.2. Proof of Theorem 5.1. The proof proceeds by induction on the dimension of V . Let σ ∈ Stλ . Let ξ = (V1 , . . . , Vn ) ∈ Cwσ ∩ Fx . It remains to prove that for all k = 1, . . . , n we have  Vk ⊆ Vk−1 + Im(xcσ (n+1−k)−1 ), (5.1) Vk 6⊆ Vk−1 + Im(xcσ (n+1−k) ). First, V1 ⊆ Fwσ (1) ∩ Ker x. But, by Lemma 5.2, Fwσ (1) ∩ Ker x ⊆ Im(xcσ (n)−1 ). It follows that V1 ⊆ Im(xcσ (n)−1 ). On the other hand, V1 6⊆ Fwσ (1)−1 ; so V1 6⊆ Im(xcσ (n) ). Hence, relations (5.1) are fulfilled for k = 1. Let V ′ = V /V1 and x′ be as in Lemma 2.1. Obviously, ξ ′ = (V2 /V1 , . . . , Vn /V1 ) belongs to Fx′ . Let σ ′ be the standard Young tableau obtained from σ by deleting the box occupied by n. By Lemma 5.3, ξ ′ ∈ Cwσ′ . So, by induction, ξ ′ ∈ Fx′ ,σ′ . One easily deduces that relations (5.1) hold for k ≥ 1. The theorem is proved. 

6. P1 -fibrations of F and Fx 6.1. Notation. Let k ∈ [1; n − 1] be an integer. We denote by Fk the variety consisting of the partial flags (V1 ⊆ · · · ⊆ Vk−1 ⊆ Vk+1 ⊆ · · · ⊆ Vn = V ) such that dim Vi = i for all i. The group GL(V ) acts naturally on Fk . We denote by φk : F → Fk the map which omits the subspace Vk . This map is a P1 -fibration and is GL(V )-equivariant. Let X be a subset of F. The subset φ−1 k (φk (X)) is called the φk -saturation of X, and X is said to be φk -stable if it is equal to its φk -saturation. 6.2. φk -stability of Fσ . Proposition 6.1. Let σ be a standard Young tableau of shape λ and 2 ≤ k ≤ n. Then: (i) Fx,σ is φn+1−k -stable if and only if cσ (k − 1) = cσ (k). (ii) Fσ is φn+1−k -stable if and only if cσ (k − 1) ≥ cσ (k). (iii) If cσ (k − 1) > cσ (k) then by switching the places of k and k − 1 in σ, we obtain a standard Young tableau γ. Moreover, for all ξ ∈ Fx,σ , the φn+1−k saturation of ξ minus one point which belongs to Fx,γ is contained in Fx,σ .

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Proof. Set V ′ := V /Vn−k and x′ : V ′ → V ′ , v + Vn−k 7→ x(v) + Vn−k . By replacing x by x′ , we may assume that k = n. First, we assume that cσ (n − 1) ≥ cσ (n). Let ξ = (V1 , . . . , Vn ) ∈ Fx,σ . We claim that V2 ⊆ Ker x. By Corollary 2.3 we have x(V2 ) ⊆ x(V1 ) + x(Im(xcσ (n−1)−1 )) = Im(xcσ (n−1) ) ⊆ Im(xcσ (n) ), since cσ (n − 1) ≥ cσ (n). But, by Corollary 2.3, V1 6⊆ Im(xcσ (n) ), and so V1 ∩ Im(xcσ (n) ) = {0}. Since x(V2 ) ⊆ V1 , the claim follows. By the claim, the φ1 -saturation of Fx,σ is contained in Fx . But it is irreducible, and so contained in Fσ . Since Fσ is the closure of Fx,σ , we deduce that Fσ is φ1 -stable. Notice that V2 ⊆ Im(xcσ (n)−1 ) since Im(xcσ (n−1)−1 ) ⊆ Im(xcσ (n)−1 ). Let W1 be a line in V2 . Then ζ := (W1 , V2 , . . . , Vn−1 ) ∈ Fx . Let γ be the unique standard Young tableau such that ζ ∈ Fx,γ . By Corollary 2.3, cσ (k) = cγ (k) for all k = 1, . . . , n−2. Therefore, either γ = σ or γ is obtained from σ by switching the places of n and n − 1. By Corollary 2.3, V1 6⊆ Im(xcσ (n) ), and so the dimension of V2 ∩ Im(xcσ (n) ) equals zero or one. We distinguish these two cases. Case 1: V2 ∩ Im(xcσ (n) ) = {0}. Since W1 ⊆ V2 , W1 6⊆ Im(xcσ (n) ), and W1 ⊆ Im(xcσ (n)−1 ). By Corollary 2.3, we deduce that cγ (n) = cσ (n), and so γ = σ. It follows that Fx,σ is φ1 -stable. By the assumption on the dimension of V2 ∩Im(xcσ (n) ), V2 6⊆ W1 +Im(xcσ (n) ). But V2 ⊆ Im(xcσ (n)−1 ) ⊆ V1 + Im(xcσ (n)−1 ). We deduce that cγ (n − 1) = cσ (n); but γ = σ, and so cσ (n − 1) = cσ (n). Case 2: V2 ∩ Im(xcσ (n) ) is a line denoted by l. If W1 = l, we have W1 ⊆ Im(xcσ (n) ), and so cγ (n) > cσ (n). In particular γ 6= σ, and γ is obtained from σ by switching the places of n and n − 1. Hence, cσ (n − 1) > cσ (n). If W1 6= l, then cγ (n) = cσ (n), and so γ = σ. To complete the proof of the proposition, it is sufficient to prove that if cσ (n − 1) < cσ (n) then Fσ is not φ1 -stable. Let v in Im(xcσ (n)−1 ) − Im(xcσ (n) ) be such that x(v) = 0. Since x(Im(xcσ (n−1)−1 )) = Im(xcσ (n−1) ) ⊇ Im(xcσ (n)−1 ), there exists w in cσ (n−1)−1 Im(x ) such that x(w) = v. Let V1 (resp. V2 ) be the vector spaces spanned by v (resp. v and w). By Corollary 2.4, we can complete (V1 , V2 ) to a flag ξ of Fx,σ . Since x(V2 ) = V1 , we have ζ 6∈ Fx for all ζ ∈ φ1−1 (φ1 (ξ)) − {ξ}. This completes the proof.  Proposition 6.1 has the following purely combinatorial corollary:

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Corollary 6.2. Let σ be a standard Young tableau of shape λ and 2 ≤ k ≤ n. Then (with the notation of Section 4.1), cσ (k − 1) ≥ cσ (k)

if and only if

cσ∨ (n − k) ≥ cσ∨ (n + 1 − k).

Proof. By Proposition 6.1, cσ (k − 1) ≥ cσ (k) if and only if Fσ is φn+1−k -stable, which is equivalent to η(Fσ ) being φk -stable, and so to Fσ∨ being φk -stable.  Remark 6.3. With the notation of Proposition 6.1, assertion (iii) shows that Fγ intersects Fσ in a set of codimension one. It would be interesting to find all the standard Young tableaux δ such that Fδ intersects Fσ in a set of codimension one. Assertion (ii) of Proposition 6.1 deals with the Spaltenstein parametrization of the irreducible components of Fx (and was already obtained in [7]), but not with the Spaltenstein partition into the Fx,σ . Assertion (iii) is more precise but concerns the Spaltenstein partition. It would be interesting to be able to read off from the standard Young tableaux γ and σ whether or not Fγ ∩ Fσ contains an irreducible component C whose φn+1−k -saturation is Fσ . If cσ (k − 1) 6= cσ (k) Proposition 6.1 answers this question. If cσ (k − 1) = cσ (k) the following two examples show that the situation is less clear. Example 6.4. Consider σ= 1 3. 2 4 Then Fx,σ = Fσ is φ1 -stable. Set γ= 1 2 3 4

and

C = Fγ ∩ Fσ .

One easily checks that the φ1 -saturation of C equals Fσ . Example 6.5. Consider the three standard Young tableaux 1 2 σmax := 3 , 4

1 3 σ := 2 , 4

1 4 σmin := 2 3

of shape

.

One easily checks that Fσmin − (Fσmax ∪ Fσ ) is φ1 -stable. 6.3. From now on, we focus our attention on the case cσ (k − 1) > cσ (k). Definition 6.6. Let 3 ≤ k ≤ n. Two standard Young tableaux of shape λ are said to be k-adjacent if they are obtained from each other by switching the places of k and k − 1. Definition 6.7. Let Γλ be the oriented graph with vertex set Stλ and the edges labeled by the integers 3, . . . , n, where σ is joined to γ by an edge labeled by k if σ and γ are k-adjacent and σ ≺ γ. The graph Γλ is called the adjacency graph of λ.

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Example 6.8. The adjacency graph of the partition (3, 2) is 1 2 3 4 5 4 1 2 4 3 5 3

5

1 3 4 2 5

1 2 5 3 4 5

3 1 3 5 2 4

6.4. We can state the first property of Γλ : Proposition 6.9. Each vertex in Γλ is joined to σmax (resp. σmin ) by an increasing (resp. decreasing) path. Before showing Proposition 6.9, we prove two lemmas: Lemma 6.10. Let σ ∈ Stλ and 3 ≤ k ≤ n. Consider the following partition of the boxes of σ: | | | | | | |

| | | | | | |

| | | | | | |

| | | | | | |

| | | | | | |

| | | | | | |

| | | | | | |

| | | | | | |

k ++++++++ ++++++++ ++++++++ ++++++++ ++++++++ ++++++++ ++++++++ ++++++++

Then: (i) The number k − 1 cannot belong to a box (ii) If k − 1 belongs to a box tableau.

.

, then σ is k-adjacent to no standard Young

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++

(iii) If k − 1 belongs to a box + + , then σ is k-adjacent to a standard Young tableau γ such that γ ≺ σ. | | (iv) If k − 1 belongs to a box | | , then σ is k-adjacent to a standard Young tableau γ such that σ ≺ γ. Proof. This is trivial.



Lemma 6.11. Let σ be a standard Young tableau and γ be its predecessor for ≺. Let i0 be such that lγ (i0 ) < lσ (i0 ) and lγ (j) = lσ (j) for all j > i0 . Then there exists a standard Young tableau δ + (resp. δ − ) which is i0 -adjacent to γ (resp. σ). Moreover, σ  δ + and δ −  γ. Proof. By omitting the boxes occupied by i0 +1, . . . , n in each tableau, we may assume that i0 = n. By Figure 1, n−1 belongs to the last line of γ. Then Lemma 6.10 shows that there exists δ + n-adjacent to γ such that γ ≺ δ + . Since γ is the predecessor of σ, one deduces that σ  δ + . The proof for δ − is similar.  Proof of Proposition 6.9. We give the proof for σmax . The reader can easily deduce the case of σmin . Let E denote the set of standard Young tableaux joined to σmax by an increasing path. We have σmax ∈ E; assume by contradiction that E = 6 Stλ . Consider the maximal element γ of Stλ − E and let σ be its successor. By Lemma 6.11, there exists δ + i0 -adjacent to γ such that γ ≺ δ + (with some 3 ≤ i0 ≤ n). Since γ ≺ δ + , δ + belongs to E, and so γ belongs to E, a contradiction. 

7. Some smooth components of Fx 7.1. Some homogeneous components of Fx . In this subsection, we will describe all the irreducible components of Fx which are homogeneous under a parabolic subgroup of GL(V ). These components are already classified by Kraft and Hesselink in [4] and [3]. Here, we recover their results and specify the standard Young tableaux corresponding to those components. Let us first introduce some notation. ˆ = (λ ˆ1, λ ˆ 2 , . . . ) the dual partition defined If λ is a partition of n, we denote by λ ˆ ˆ ˆ by λi := card{j : λj ≥ i}; notice that λ1 ≥ λ2 ≥ · · · . An ordered partition of n is a tuple (µ1 , . . . , µr ) of positive integers whose sum equals n. If we ignore the order on the set of µi ’s, we obtain the underlying partition of the form µi1 ≥ · · · ≥ µir . ˆ as underlying partition. We denote by O(λ) the set of ordered partitions with λ In other words, an element of O(λ) is a λ1 -uple of integers whose components are ˆ i ’s. the λ Fix a base B = (e1 , . . . , en ) of V adapted to x. We number the boxes of Y (λ) as in σmax . In box i, we put the vector ei . Let o = (µ1 , . . . , µλ1 ) ∈ O(λ). We define a partial flag ξo = ({0} = Wo0 ⊆ Wo1 ⊂ · · · ⊆ Woλ1 = V ) as follows: • Wo1 is the subspace spanned by the vectors in the first µ1 lines and in the first column of Y (λ);

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• Wo2 is the subspace spanned by Wo1 and the vectors in the first µ2 lines and on the left among the vectors not in Wo1 ; and so on. Pj For all j = 1, . . . , λ1 , we set dj := i=1 µi = dim Woj . Consider Fo := {(V1 , . . . , Vn ) ∈ F : Vdj = Woj ∀j = 1, . . . , λ1 }.

Notice that since x(Woj ) = Woj−1 , Fo is contained in Fx . Moreover, Fo is homogeneous under the action of the stabilizer in GL(V ) of (Wo1 ⊆ · · · ⊆ Woλ1 ); it is P ˆ ˆ isomorphic to a product of flag manifolds and its dimension equals i λ i (λi −1)/2, which is the dimension of each irreducible component of Fx (see [7]). We deduce that Fo is a Richardson component of Fx . We now define a standard Young tableau σo associated to o. In the first µr lines of the first column of σo , we put the numbers 1, . . . , µr . In the first µr−1 lines and on the left among the non-numbered boxes, we put the numbers µr + 1, . . . , µr + µr−1 , and, so on. Let us give an example of ξo and σo . Example 7.1. For the partition λ = (4 ≥ 4 ≥ 3 ≥ 1), its dual partition is (4 ≥ 3 ≥ 3 ≥ 2). Then there are exactly twelve elements in O(λ). Set o = (3, 2, 4, 3) ∈ O(λ). We have: e1 e2 e3 e4 e5 e6 e7 e8 e9 e10 e11 e12

1 4 8 10 2 5 9 11 σo = 3 6 12 7

and ξo = (Wo1 ⊆ Wo2 ⊆ Wo3 ⊆ Wo4 = W ) is described by: Wo1 = he1 , e5 , e9 i,

Wo2 = hWo1 , e2 , e6 i,

Wo3 = hWo2 , e3 , e7 , e10 , e12 i,

Wo4 = W.

We can now describe the Richardson components: Theorem 7.2. For any o ∈ O(λ), we have Fo = Fσo . Moreover, any irreducible component of Fx homogeneous under a parabolic subgroup equals Fσo for some o ∈ O(λ). Proof. Fix o ∈ O(λ). We have already noticed that Fo is a Richardson component. Conversely, let σ ∈ Stλ be such that Fσ is a Richardson component. There exists a partial flag ({0} = W 0 ⊆ W 1 ⊆ · · · ⊆ W s = V ) such that Fσ is the set of flags containing the W j ’s. Let d1 , . . . , ds denote the dimensions of the W j ’s. Consider the projection π : F(V ) → P(V ), (V1 , . . . , Vn ) 7→ V1 . By Corollary 2.4, W 1 = ˆ i for some i1 . π(Fσ ) = Im(xcσ (n)−1 ) ∩ Ker x. We deduce that d1 = lσ (n) equals λ 1 ′ 1 1 1 Consider the endomorphism x : V /W → V /W , v + W 7→ x(v) + W 1 . One easily checks that the dual partition of the partition associated to x′ is obtained ˆ by omitting λ ˆ i . We deduce by induction on the dimension of V that from λ 1

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o := (d1 , d2 − d1 , . . . , ds − ds−1 ) belongs to O(λ). Moreover, from the equality W 1 = Im(xcσ (n)−1 ) ∩ Ker x (and by an immediate induction) we deduce that Fσ = Fo . We now prove by induction on n that σo = σ. Since lσ (n) = d1 , we have cσo (n) = cσ (n). Set o′ = (d1 − 1, d2 − d1 , . . . , ds − ds−1 ). Notice that σo′ is obtained from σo by deleting the box occupied by n. Let σ ′ be the standard Young tableau obtained from σ by deleting the square occupied by n. Fix V1 ∈ (Im(xcσ (n)−1 ) − Im(xcσ (n) )) ∩ Ker x. By considering V /V1 , the induction shows that σo′ = σ ′ . The theorem follows.  Remark 7.3. By the above description we also see that Fσmin is the Richardson ˆ1, . . . , λ ˆ λ ). This is another component corresponding to the ordered partition (λ 1 way to show that Fσmin is smooth (cf. Corollary 3.5). 7.2. Some smooth components of Fx . The main result of this section is Theorem 7.4. Let o ∈ O(λ). Let σ be a standard Young tableau k-adjacent to σo such that σo ≺ σ. Then Fσ is smooth. Proof. We use the notation of Section 7.1. By Lemma 6.10, since σo ≺ σ, we have lσo (k − 1) > lσo (k). By the definition of σo , we deduce that lσo (k) = 1. Set j0 = cσo (k) and k′ = dj0 = n + 1 − k. By Proposition 6.1, Fσ is φk′ -stable. We will prove that Fσ is the φk′ saturation of the following smooth subvariety of Fo : C = {(V1 , . . . , Vn ) ∈ Fo : x(Vk′ +1 ) ⊆ Vk′ −1 }. We first prove that C is smooth. Consider the set X of partial flags obtained from those of Fo by omitting Vdj0 +1 , . . . , Vdj0 +1 −1 . Notice that X is isomorphic to a product of flag varieties. Since x(Woj0 +1 ) = Woj0 , Woj0 +1 ∩ x−1 (Vk′ −1 ) is a hyperplane of Woj0 +1 containing Woj0 . Moreover, the map Vk′ −1 7→ Woj0 +1 ∩x−1 (Vk′ −1 ) is a morphism from the projective space of hyperplanes of Woj0 containing Woj0 −1 to the projective space P((Woj0 +1 /Woj0 )∗ ) of hyperplanes of Woj0 +1 containing Woj0 . Set ϕ : X → P((Woj0 +1 /Woj0 )∗ ), Vk′ −1 7→ Woj0 +1 ∩ x−1 (Vk′ −1 ). Notice that C is isomorphic to the subvariety of (x, ζ) ∈ X × F(Woj0 +1 /Woj0 ) such that V1 ∈ φ(x), where V1 is the line of ζ. One easily deduces that C is a smooth irreducible subvariety of Fo . Moreover, since σ exists, we have µj0 +1 > 1. Therefore, the codimension of C in Fo equals one. Since for all (V1 , . . . , Vn ) ∈ C, we have x(Vk′ +1 ) ⊆ Vk′ −1 , the φk′ -saturation of C is an irreducible component Fγ of Fx . Moreover, the restriction of φk′ to C is an isomorphism. We deduce that Fγ is smooth. It remains to prove that γ = σ. Set D := Fx,σo ∩ Fσ ; then D is a closed subvariety of Fx,σo . By Proposition 6.1, for any ξ ∈ Fx,σ the line φ−1 k (φk (ξ)) intersects Fx,σo in exactly one point of D. In particular, Fσ is the closure of the φk -saturation of D.

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Notice that C = {ξ ∈ Fo : φ−1 k (φk (ξ)) ⊆ Fx }. We deduce that D is contained in C. Therefore, Fσ , which is contained in Fγ , is the φk -saturation of C. It follows that σ = γ. 

8. The Robinson–Schensted map and the adjacency graph 8.1. Let Γn be the oriented graph with vertex set Sn and the edges labeled by the integers 2, . . . , n, where w is joined to w′ by an edge labeled by k if w′ = wsn+1−k and l(w′ ) = l(w) + 1. The graph Γn is called the Bruhat graph of Sn . We can now state Theorem 8.1 (Main Theorem). The map Stλ → Sn , σ 7→ wσ , identifies the graph Γλ with a full subgraph of Γn . In particular, for all σ and γ in Stλ , wσ and wγ are not joined by an edge labeled by 2 in the Bruhat graph. Remark 8.2. Since the Robinson–Schensted map is injective, Theorem 4.1 shows that when λ varies among the partitions of n, the subgraphs of Γn obtained by Theorem 8.1 are pairwise disjoint. Before showing Theorem 8.1, we prove some useful combinatorial properties of Sn . 8.2. Consider the set ∆ := {(n1 , . . . , nn−1 ) ∈ Nn−1 : 0 ≤ ni ≤ n − i}. Lemma 8.3. The map π : ∆ → Sn , (ni ) 7→ c1 c2 · · · cn−1 , where  sn−(i+ni −1) · · · sn−(i+1) sn−i if ni ≥ 1, ci := id otherwise, is a bijection.

Proof. Set ∆′ := {(ni ) ∈ ∆ : n1 = 0}. Then for every w ∈ π(∆′ ) we have w(n) = n; by induction the restriction of π to ∆′ induces a bijection onto H = {w ∈ Sn : w(n) = n}. Since c1 (n) = n−(i+ni −1), every element w ∈ Sn such that w(n) = k can be uniquely written as w = c1 w′ with w′ ∈ H and c1 = sk sk+1 · · · sn−1 . The lemma follows.  We now introduce some structure in the set ∆. For k = 2, . . . , n − 1, we set ∆k := {(n1 , . . . , nn−1 ) ∈ ∆ : nk < n − k}, ∆k := {(n1 , . . . , nn−1 ) ∈ ∆ : nk−1 6= 0}. We define τk : ∆k → ∆, (n1 , . . . , nn−1 ) 7→ (n′1 , . . . , n′n−1 ), where n′i = ni for i 6∈ {k − 1, k}, n′k−1 = nk and n′k = nk−1 − 1. We also define ρk : ∆k → ∆, (n1 , . . . , nn−1 ) 7→ (n′1 , . . . , n′n−1 ), where n′i = ni for i 6∈ {k − 1, k}, n′k−1 = nk and n′k = nP k−1 + 1. For α = (n1 , . . . , nn−1 ) ∈ ∆, we define the length l(α) of α by l(α) = ni . Notice that l(τk (α)) = l(α) − 1 and l(ρk (α)) = l(α) + 1. Via π, we can read the Bruhat order on ∆. We obtain

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Lemma 8.4. Let α ∈ ∆ and k ∈ {2, . . . , n}. (i) If nk−1 > nk then α ∈ ∆k and π(τk (α)) = π(α)sn+1−k . (ii) If nk−1 ≤ nk then α ∈ ∆k and π(ρk (α)) = π(α)sn+1−k . Proof. We use the notation of Lemma 8.3 for α. Set w1 = c1 · · · ck−2 and w2 = ck+1 · · · cn−1 ; so we have π(α) = w1 ck−1 ck w2 . But w2 (n − k + 1) = n − k + 1 and w2 (n − k + 2) = n − k + 2, so w2 and sn−k+1 commute. Therefore, π(α)sn+1−k = w1 ck−1 ck sn+1−k w2 . We compute ck−1 ck sn+1−k with the help of the diagrams of Figure 2.

=

=

nk−1 > nk

nk−1 ≤ nk

Fig. 2. A product of cycles. Notice that these two computations can be deduced from each other. The lemma follows easily.  Proposition 8.5. With the above notation, for all α ∈ ∆, we have l(α) = l(π(α)). In other words, the expression of π(α) given in Lemma 8.3 is reduced. Proof. First, notice that since the definition of π gives a formula of length l(α) for π(α), we have l(π(α)) ≤ l(α). By contradiction, we assume that the proposition is false. Let us consider an element α = (n1 , . . . , nn−1 ) ∈ ∆ such that l(α) > l(π(α)) of maximal length. Two cases may occur. Case A: there exists k ∈ {2, . . . , n − 1} such that nk ≥ nk−1 . Since α ∈ ∆k , we can set β = ρk (α). By Lemma 8.4, we have π(β) = π(α)sn−k+1 . In particular, l(π(α)) = l(π(β)) ± 1. Since l(β) > l(α), it follows that l(π(β)) = l(β) = l(α) + 1. So, l(π(α)) equals l(α) or l(α) + 2, a contradiction. Case B: nk < nk−1 for all k = 2, . . . , n − 1. We have n − 1 ≥ n1 > · · · > nn−1 ≥ 0. So, there exists k ∈ {0, . . . , n − 1} such that ni = n − i for all i ≤ k − 1 and ni = n − i − 1 for all i ≥ k. We can draw the cycles ci in Figure 3. One easily reads off from the picture the following values of π(α)(i): i 1 ··· k−1 k k+1 ··· n π(α)(i) n · · · n + 2 − k 1 n + 1 − k · · · 2

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2

k−1

k

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n

cn−1

cn−k+2 cn−k+1 c1 Fig. 3. Decomposition of π(α). In particular, {(i, j) : i < j, π(α)(i) < π(α)(j)} = {(k, k + 1), . . . , (k, n)}. We deduce that l(π(α)) = n(n − 1)/2 − (n − k). On the other hand, we have l(α) = Pn−1  i=1 (n − i) − (n − k), a contradiction. Since the dimension of Cw equals the length of w, Proposition 8.5 implies P Corollary 8.6. For any σ ∈ Stλ , we have dim Cwσ = i nσ (i). The partition λ is said to be of hook type if λ2 = · · · = λr = 1.

Corollary 8.7. Let σ be a standard Young tableau of shape λ. Then there exists w ∈ Sn such that Fσ is the closure of Cw if and only if λ is of hook type and σ = σmin . Proof. If Fσ is the closure of Cw then w = wσ . Now, there exists w ∈ Sn such that Fσ is the closure of Cw if and only if dim Fσ = l(wσ ). Notice that the inequality dim Fσ ≤ l(wσ ) is obvious. But, by [7], the dimension of Fσ only depends on λ; and, by Propositions 6.9 and 8.5, for any standard Young tableau γ 6= σmin of shape λ, we have l(σmin ) < l(γ). Therefore, if dim Fσ = l(wσ ) then σ = σmin . It remains to prove that dim Fx ≤ l(wσmin ) if and only if λ is of hook type. Pr Pr−1 We have dim Fx = i=2 (i − 1)λi = i=1 nσmin (i). In particular, dim Fx ≤ l(wσmin ) implies that nσmin (r + 1) = 0, and so λ is of hook type. Conversely, if λ Pr−1 Pn−1 is of hook type, one easily checks that i=1 nσmin (i) = i=1 nσmin (i).  8.3. Now, we can prove Theorem 8.1:

Lemma 8.8. Via the map Pa, Γλ identifies with a full subgraph of Γn .

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Proof. The map Stλ → ∆, σ 7→ (nσ (1), . . . , nσ (n−1)), is obviously injective. Now, Lemma 8.3 implies that Pa : Stλ → Sn is injective. Let σ ≻ γ be two k-adjacent standard Young tableaux. One easily checks that nσ (k − 1) > nσ (k) and τk (nσ (1), . . . , nσ (n − 1)) = (nγ (1), . . . , nγ (n − 1)). So, by Lemma 8.4, wσ = wγ sn+1−k . Moreover, by Proposition 8.5, l(wσ ) = l(wγ ) + 1. Finally, via the map Pa, Γλ identifies with a subgraph Γ′λ of Γn . Conversely, consider an edge in Γn labeled by k joining w to w′ with l(w′ ) = l(w) + 1. Consider α = (n1 , . . . , nn−1 ) and β = (n′1 , . . . , n′n−1 ) such that π(α) = w and π(β) = w′ . By Lemma 8.4, either β = τk (α) or α = τk (β). Since l(w′ ) = l(w) + 1, Proposition 8.5 shows that α = τk (β) and nk−1 ≤ nk . We now assume that w = wσ for some σ ∈ Stλ ; and we distinguish three cases: Case A: nk−1 < nk and k > 2. By Lemma 6.10, there exists a standard Young tableau γ which is k-adjacent to σ. Moreover, the first part of the proof shows that wγ = w′ . In particular, the edge considered is an edge of Γ′λ . Case B: nk−1 = nk and k > 2. We assume that w′ = wγ for some γ ∈ Stλ . Lemma 8.4 shows that τk (nγ (1), . . . , nγ (n − 1)) = (nσ (1), . . . , nσ (n − 1)). We have to prove that γ and σ are k-adjacent; by the injectivity of Pa, it is sufficient to prove that there exists a standard Young tableau k-adjacent to γ. Since nσ (k − 1) = nσ (k), by Lemma 6.10 it is sufficient to prove that k − 1 and k are not in the same line in γ. Assume by contradiction that lγ (k − 1) = lγ (k). Since nσ (i) = nγ (i) for all i ≤ k − 2, an immediate induction shows that the integers 1, . . . , k − 2 are in the same boxes in σ and γ. Notice that nγ (k − 1) = nγ (k). Moreover, nσ (k − 1) = nγ (k − 1) − 1 and so lσ (k − 1) > lγ (k − 1). Since nσ (k) ≥ nγ (k), we deduce that lσ (k) < lγ (k). But then nσ (k) > nγ (k), which is a contradiction. Case C: k = 2. Since n1 = n − λ1 , the condition n1 ≤ n2 implies that n2 = n1 and lσ (2) = 1. In particular, n′1 = n1 −1 = n−λ1 −1. But for all γ ∈ Stλ , we have nγ (1) = n−λ1 . Therefore, w′ does not belong to Pa(Stλ ).  8.4. We now give another description of wσ . For i = 2, . . . , n, set nσ (i) := card{j < i : lσ (j) < lσ (i)}. For any j = 1, . . . , n − 1, let cj denote the decreasing cycle of Sn of length nσ (n + 1 − j) ending with j. Theorem 8.9. With the above notation, we have wσ = c1 · · · cn−1 .

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Proof. In this proof, we set wσ = c1 · · · cn−1 . First, we prove by induction on n that the conclusion holds for σmin . Set σ = σmin . By Lemma 5.2, wσ (1) = λ1 + · · · + λlσ (n)−1 + 1 = nσ (n) + 1 = wσ (1). Let σ ′ be the standard Young tableau obtained from σ by deleting the box occupied by n. One can easily check that for all i = 2, . . . , n, ′

wσ (i) = c1 (wσ (i − 1) + 1). Moreover, by Lemma 5.3 (and using its notation), we have, for all i = 2, . . . , n, wσ (i) = φ−1 (wσ′ (i − 1)). Therefore, we have to prove that φ−1 (k) = c1 (k + 1) for all k = 1, . . . , n − 1. This follows easily from nσ (n) + 1 = wσ (1). Let σ and γ be two k-adjacent standard Young tableaux (with some 3 ≤ k ≤ n). One can easily check that wσ = wγ sn+1−k . Then, by Theorem 8.1, wσ satisfies the conclusion of the theorem if and only if wγ does. Now, the conclusion follows from Proposition 6.9 and the first part of this proof.  P σ P Remark 8.10. Theorem 8.9 implies that for any σ we have i n (i) = j nσ (j). This is obvious, since these two quantities are the cardinality of the set of pairs (i, j) such that i < j and lσ (i) < lσ (j). But the equality of the theorem does not seem obvious and remains mysterious to the authors.

References [1] W. Fulton. Young Tableaux. With applications to representation theory and geometry. London Math. Soc. Student Texts 35, Cambridge Univ. Press, Cambridge, 1997. [2] F. Y. C. Fung. On the topology of components of some Springer fibers and their relation to Kazhdan–Lusztig theory. Adv. Math. 178 (2003), 244–276. [3] W. H. Hesselink. Polarizations in the classical groups. Math. Z. 160 (1978), 217–234. [4] H. Kraft. Parametrisierung von Konjugationsklassen in sln . Math. Ann. 234 (1978), 209–220. [5] N. G. J. Pagnon. On the Spaltenstein correspondence. Indag. Math. (N.S.) 15 (2004), 101–114. [6] N. G. J. Pagnon. Localization of the irreducible components of the variety of flags fixed by a nilpotent endomorphism. Preprint, 2004, 10 pp. [7] N. Spaltenstein. The fixed point set of a unipotent transformation on the flag manifold. Nederl. Akad. Wetensch. Proc. Ser. A 79 = Indag. Math. 38 (1976), 452–456. [8] N. Spaltenstein. Classes unipotentes et sous-groupes de Borel. Lecture Notes in Math. 946, Springer, Berlin, 1982. [9] R. Steinberg. An occurrence of the Robinson–Schensted correspondence. J. Algebra 113 (1988), 523–528. [10] M. A. A. van Leeuwen. Flag varieties and interpretations of Young tableau algorithms. J. Algebra 224 (2000), 397–426.

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[11] J. A. Vargas. Fixed points under the action of unipotent elements of SLn in the flag variety. Bol. Soc. Mat. Mexicana (2) 24 (1979), 1–14. N. G. J. Pagnon Chennai Mathematical Institute Plot H1, SIPCOT IT Park Padur PO Siruseri 603103, India e-mail: [email protected] N. Ressayre Universit´e Montpellier II (Math´ematiques) Place Eug`ene Bataillon 34095 Montpellier Cedex 05, France e-mail: [email protected]

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