Normality of Closures of Orthogonal Nilpotent Symmetric Orbits

Abstract

We study closures of conjugacy classes in the symmetric matrices of the orthogonal group and we determine which one are normal varieties. In contrast to the result for the symplectic group where all classes have normal closure, there is only a relatively small portion of classes with normal closure. We perform a combinatorial computation on top of the same methods used by Kraft-Procesi and Ohta.

1 Introduction

In a fundamental paper of Kostant [4], the adjoint action on a reductive Lie algebra $\mathfrak {g}$ defined over an algebrically closed field k of characteristic 0 is studied in detail. In the course of his analysis it arose the following problem. Let $A \in \mathfrak {g}$, let C_A be the conjugacy class of A and let $\overline {C_A}$ be the (Zarisky) closure of C_A. Is $\overline {C_A}$ always a normal variety?

In his paper Kostant showed that if A is a regular nilpotent element of $\mathfrak {g}$, so that $\overline {C_A}$ is the nilpotent cone of $\mathfrak {g}$, the normality is always the case.

In [3], the problem of normality of $\overline {C_A}$ is reduced to the case in which A is nilpotent, possibly changing the Lie algebra $\mathfrak {g}$.

Later, in [2, Theorem 4.1] , Broer proved the normality of a handful of nilpotent orbits in $\mathfrak {g}$, including the regular and the subregular orbits (if $\mathfrak {g}$ is simple).

In the fundamental paper [5] of Kraft and Procesi, the normality of $\overline {C_A}$ is proved for every nilpotent A in the case $\mathfrak {g} = \mathfrak {gl}_n$. Their method consists in constructing an auxiliary variety Z which is a normal complete intersection such that $\overline {C_A}$ is a quotient of Z.

Kraft and Procesi extended their method to the case where $\mathfrak {g}$ is the orthogonal or symplectic Lie algebra in [7]. In that case not all nilpotent classes have normal closure. They obtained necessary and sufficient conditions on the partition of A in order to have the normality of $\overline {C_A}$, under suitable hypothesis on A.

In a subsequent paper by Sommers [13], the cases outside of this hypothesis (in [7]) are proved to always have normal closure. Sommers also solved the problem for the exceptional case E₆ in [12].

Kostant and Rallis generalized the study [4] of the adjoint action on $\mathfrak {g}$ to the action on the symmetric space in [8]. We briefly describe their setting. Let 𝜃 be a Lie algebra automorphism of $\mathfrak {g}$ of order 2. Then there exists a decomposition of $\mathfrak {g}$ in 𝜃-eigenspaces given by

$$ \mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p} $$

where $\mathfrak {k} = \{A \in \mathfrak {g} : \theta (A) = A\}$ and $\mathfrak {p} = \{A \in \mathfrak {g} : \theta (A) = -A\}$. Let G be the adjoint group of $\mathfrak {g}$ and $K \subseteq G$ be the subgroup of elements commuting with 𝜃. We notice that $\mathfrak {k}$ is a Lie subalgebra of $\mathfrak {g}$ and $\text {Lie} K = \mathfrak {k}$. We remark that K is not necessarily connected. Moreover the adjoint action of G on $\mathfrak {g}$ induces an action of K on $\mathfrak {g}$ and both $\mathfrak {k}$, $\mathfrak {p}$ are K-stable. The pair $(\mathfrak {g}, \mathfrak {k})$ is called a symmetric pair and $\mathfrak {p}$ is the symmetric space associate to it.

In their study, Kostant and Rallis focused on the action of K on $\mathfrak {p}$. Among other things, they already observed that the nilpotent cone is not irreducible nor normal in general.

In [14], Vinberg further generalized the study of the adjoint action to graded Lie algebras, also called 𝜃-groups. Here 𝜃 is an automorphism of a Lie algebra $\mathfrak {g}$ of finite order n, thus generalizing the case n = 2 of [8]. The graduation induced by 𝜃, $\mathfrak {g} = \mathfrak {g}_0 \oplus {\dots } \oplus \mathfrak {g}_{n-1}$, produces an action of the 𝜃-fixed points of the adjoint group on $\mathfrak {g}_1$.

After the work of Kostant, Rallis and Vinberg, the problem of the normality of $\overline {C_A}$ naturally generalizes to the case of symmetric spaces.

In [11], Sekiguchi studied deeply the geometry of the nilpotent cone in symmetric spaces. In particular he proved several results for the principal nilpotent elements (which are the analogue of the regular nilpotent elements) when the symmetric space is the orthogonal symmetric space or the symplectic symmetric space, that is, the symmetric spaces of the symmetric pairs $(\mathfrak {g}, \mathfrak {k}) = (\mathfrak {gl}(n), \mathfrak {o}(n))$ and $(\mathfrak {g}, \mathfrak {k}) = (\mathfrak {gl}(2m), \mathfrak {sp}(m))$ respectively.

The method of the auxiliary variety Z developed by Kraft and Procesi in [5,6,7] was adapted by Ohta in [10] to the study of the singularities of the orbits in the orthogonal and symplectic symmetric spaces. In particular he proved that $\overline {C_A}$ is always normal in the case of symplectic symmetric space. Moreover he found several orbits C_A in the case of the orthogonal symmetric space which have a non-normal closure.

It is well known that the conjugacy classes of the nilpotent cone of the orthogonal symmetric space are parametrized by the partitions of n (see section 1 for more details). The main purpose of this paper is to give a necessary and sufficient condition on the partition corresponding to the orbit of a nilpotent symmetric element A in order to have the normality of $\overline {C_A}$. The following theorem is the main result of the paper.

Theorem 1

Let $\mathfrak {p}$ be the symmetric space of the symmetric pair $(\mathfrak {gl}(n),\mathfrak {o}(n))$ and let $A \in \mathfrak {p}$ be a nilpotent element. Let $\lambda = (\lambda _1, \dots , \lambda _h)$ be the partition of A. Then $\overline {C_A}$ is normal if and only if

$$ \lambda_{i} - \lambda_{i+1} \leq 1 \quad \forall i = 1, \dots, h $$

(1)

with the convention that λ_h+ 1 = 0.

The proof of Theorem 1 will be carried out in Section 8. The only if part was already proved in [10, 11] (see Section 8 for details).

We remark that, if n is even, C_A is not necessarily connected, as the group K is not connected. Therefore, sometimes, $\overline {C_A}$ is not normal just because it is not irreducible, being the union of two closure of SO(n)-orbits (as an example, if one takes λ = (2), one can shows that $\overline {C_A}$ is the union of two lines which meet at 0). As the referee has kindly pointed out, it would be interesting to investigate the normality of the closure of the SO(n)-orbits.

We summarize the content of the rest of the paper.

In Section 1 we recall some basic facts about symmetric nilpotent orbits for the orthogonal group. In Section 2 we recall from [5] the classification of nilpotent pairs via ab-diagrams and we describe the class of ab-diagrams which parametrize symmetric nilpotent pairs. In Section 3 we recall the construction of the variety Z and some of its properties which are needed for the proof. In Section 4 we define a condition on partitions which plays an important role in our investigation. In Section 5 we introduce some combinatorial description of pairs of partitions. In Section 7, resp. Section 8, we prove a complete intersection, resp. normality, condition for the variety Z using a combinatorial computation carried on in Section 6.

I am grateful to Claudio Procesi for proposing me this problem. I would like to thank Corrado De Concini, Andrea Maffei and Paolo Bravi for many useful comments. I am really grateful to the reviewers for their precise and useful comments. A special thank goes to my PhD advisor Giovanni Cerulli Irelli for many discussions about this problem and his precious help in editing this paper.

2 Symmetric Nilpotent Orbits

In this section we introduce settings and notations for the objects studied in the paper. We follow [7] and [8].

Let V be a vector space over $\mathbb {C}$ together with a symmetric non-degenerate bilinear form (−,−). Even if in this paper we will only be concerned with the orthogonal group O(V ), we will denote the isometry group with respect to (−,−) as G(V ) := O(V ) in order to keep the notation as close as possible with [7, 10]. Let $\mathfrak {gl}(V)$ be the space of linear endomorphisms of V. Clearly G(V ) acts on $\mathfrak {gl}(V)$ by conjugation. We denote by D^∗ the adjoint of an endomorphism D with respect to (−,−). The endomorphism 𝜃 : D↦ − D^∗ of $\mathfrak {gl}(V)$ is involutive, therefore we have a decomposition into eigenspaces for 𝜃 as follows:

$$ \mathfrak{gl}(V) = \mathfrak{k}(V) \oplus \mathfrak{p}(V), $$

where $\mathfrak {k}(V) = \{D: D = \theta (D) = -D^{*}\}$ is the Lie algebra of G(V ). Moreover the action of G(V ) leaves both $\mathfrak {k}(V)$, $\mathfrak {p}(V)$ stable.

In this paper we are concerned with the study of nilpotent orbits of G(V ) in $\mathfrak {p}(V)$. It is well known ([11, Sec. 3.1], [10, Sec. 0]) that the nilpotent orbits in $\mathfrak {p}(V)$ are completely determined by their Jordan form and every partition of $n:=\dim V$ realizes a non-empty nilpotent orbit. We denote by P(n) the set of partitions of n and by $C_{\lambda } \subseteq \mathfrak {p}(V)$ the nilpotent orbit corresponding to λ.

Let λ be a partition. We set |λ| := n if λ ∈ P(n) and we denote the dual partition of λ by $\widehat \lambda$. We frequently identify a partition with its Young diagram. This means that, if $\lambda = (\lambda _{1}, \dots , \lambda _{h})$, $|\lambda | = \lambda _{1}+\dots +\lambda _{h}$, λ_i are the rows of λ and if $\widehat \lambda = (\widehat \lambda _{1}, \dots , \widehat \lambda _{t})$, $\widehat \lambda _{j}$ are the columns of λ.

We recall the dimension formula for the orbit C_λ from [10, Remark 8]:

$$ \dim C_{\lambda} = \frac 1 2 \left( n^{2} - \sum\limits_{i=1}^{t} {\widehat\lambda}_{i}^{ 2}\right). $$

(2)

As the nilpotent cone of $\mathfrak {p}(V)$ is G(V )-stable with only finitely many orbits, we have that orbit closure $\overline {C_{\lambda }}$ is G(V )-stable, and the complement $\overline {C_{\lambda }} \setminus C_{\lambda }$ is a disjoint union of finitely many orbits. The relation $C_{\mu } \subseteq \overline {C_{\lambda }}$ produces a partial order on the partitions, called dominance order and denoted by μ ≤ λ, and it is given by

$$ \sum\limits_{i=1}^{j} \lambda_{i} \geq \sum\limits_{i=1}^{j} \mu_{i}\quad \forall j \in \{1, \dots, h\} $$

or, equivalently,

$$ \sum\limits_{k>j} \widehat\lambda_{k} \geq \sum\limits_{k>j} \widehat\mu_{k} \quad \forall j \in \{1, \dots, t\}. $$

We recall the notion of nilpotent pairs. We follow the same argument of [7]. Let V (resp. U) be a finite dimensional vector space over $\mathbb {C}$ equipped with a non-degenerate symmetric bilinear form (−,−)_V (resp. (−,−)_U). We denote $L(V,U) := \text {Hom}_{\mathbb {C}}(V,U)$, L(V ) := L(V, V ) and we define L_{V, U} := L(V, U) × L(U, V ). We can interpret L_{V, U} as the representation variety of the quiver

with dimension vector $\underline {n}=(\dim V, \dim U)$. This means that

The group G(V ) × G(U) acts on L_{V, U} by change of basis.

For every A ∈ L(V, U) we define the adjoint map A^∗∈ L(U, V ) as the unique map such that

$$ (Av , u )_{U} = (v , A^{*}u )_{V} $$

(3)

for all v ∈ V, u ∈ U.

Lemma 2

Let A ∈ L(V, U) be a linear map. Then

$$ \text{rk} A = \text{rk} A^{*}. $$

Proof

By (3), we get that $\ker A \bot \operatorname {Im} A^{*}$ and $\ker A^{*} \bot \operatorname {Im} A$. □

A pair (A, B) ∈ L_{V, U} is nilpotent if the endomorphism AB (or equivalently BA) is nilpotent. A pair (A, B) ∈ L_{V, U} is symmetric if B = A^∗. We define N_{V, U} as the cone in L_{V, U} of the symmetric nilpotent pairs.

As in [7], we have maps

defined by π(A, B) = AB, ρ(A, B) = BA. We can restrict π, ρ to the subspace of symmetric pairs, i. e. the image of $L(V,U)\rightarrow L_{V,U}$, A↦(A, A^∗):

The subvariety $N_{V,U} \subseteq L_{V,U}$ is G(V ) × G(U)-stable. We call nilpotent symmetric orbit each of the G(V ) × G(U)-orbits in N_{V, U}. We recall [5, Sec. 4.2] that an ab-diagram is a list of ab-strings, i.e., strings with letters a and b occurring on alternate positions. As shown in [5, Sec. 4.3], to each nilpotent pair (A, B) ∈ L_{V, U} corresponds an ab-diagram which determines its GL(V ) × GL(U)-orbit completely; i. e. the GL(V ) × GL(U)-orbits correspond one to one to the ab-diagrams having $\dim V$ a’s and $\dim U$ b’s. The following picture illustrates how to associate an ab-diagram δ = δ_{(A, B)} to a nilpotent pair (A, B):

where {v₁,⋯ ,v₅} is a suitable basis of V and {u₁,⋯ ,u₄} is a suitable basis of U.

If δ is the ab-diagram of an element p ∈ N_{V, U}, we can retrieve the Young diagram of π(p) (resp. ρ(p)) by suppressing the a’s (resp. the b’s) from δ. We denote by π(δ) (resp. ρ(δ)) the Young diagram obtained in this manner.

Let (A, B) ∈ L_{V, U} be a nilpotent pair with ab-diagram δ. For every h ≥ 1, we want to infer the rank of the linear map $(AB)^{h-1}A:V\rightarrow V$ (resp. $(BA)^{h-1}B:U\rightarrow U$) from δ. If γ = c₁⋯c_ℓ is a string, for any 1 ≤ i ≤ j ≤ ℓ, we call c_i⋯c_j a substring of γ. A substring of δ is a substring of one of its ab-strings. For example, for h = 1, the rank of A equals the number of occurrences of the letter a which are not in the last position of a string; this is hence the number of occurrences of the substring (ab) = (ab)^h in δ. With this in mind we prove the following Lemma.

Lemma 3

Let (A, B) be a nilpotent pair with ab-diagram δ. For every h ≥ 1, the rk((AB)^h− 1A) (resp. rk((BA)^h− 1B)) equals the number of occurrences of the substring (ab)^h (resp. (ba)^h) in δ.

Proof

The ab-diagram δ suggests the choice of basis of V and U for which the behavior of A and B is straightforward. For each ab-string $\delta _{i}=c_{i1}c_{i2}\dots c_{i\ell _{i}}$ of δ (where c_ij ∈{a, b}), for each $j=1,\dots ,\ell _{i}$, we pick vectors v_ij ∈ V (resp. u_ij ∈ U) if c_ij = a (resp. c_ij = b), with the following properties: {v_ij}_{i, j} (resp. {u_ij}_{i, j}) is a basis of V (resp. U) and Av_ij = u_{i, j+ 1} (resp. Bu_ij = v_{i, j+ 1}) if j + 1 ≤ ℓ_i or $Av_{i\ell _{i}}=0$ (resp. $Bu_{i\ell _{i}}=0$).

Notice that the map (AB)^h− 1A carries v_ij to u_{i(j+ 2h− 1)} if j + 2h − 1 ≤ ℓ_i or to 0 otherwise. Therefore, rk((AB)^h− 1A) equals the number of pairs (i, j) such that c_ij = a and j + 2h − 1 ≤ ℓ_i. Moreover, every such pair (i, j) corresponds one to one to the substring $c_{ij}{\dots } c_{i(j+2h-1)}$ of δ, which is the substring (ab)^h. A similar argument computes rk((BA)^h− 1B). □

3 Ortho-symmetric ab-Diagrams

In this section we give a combinatorial description of the ab-diagrams of the elements of N_{V, U}.

Let X ∈ N_{V, U} be a nilpotent symmetric pair, let G = G(V ) × G(U) and let G.X be its orbit in N_{V, U}. We also consider $G^{\prime } = \text {GL}(V) \times \text {GL}(U)$ acting on L_{V, U}, and the orbit $G^{\prime }.X \subseteq L_{V,U}$.

The automorphism σ of $G^{\prime }$ defined by

$$ (g_{1}, g_{2})^{\sigma} = ((g_{1}^{*})^{-1},(g_{2}^{*})^{-1}) $$

has G as the set of fixed points. The automorphism (still denoted by σ) of L_{V, U} defined by

$$ \sigma(A,B) = (B^{*}, A^{*}) $$

has the symmetric pairs as the set of fixed points. It is straightforward to check that all hypothesis of [9, Prop. 2.1] hold. Thus we get that:

$$ G^{\prime}.X \cap N_{V,U} = G.X. $$

Let δ be an ab-diagram associated to an orbit $G^{\prime }.Y$. We say that δ is ortho-symmetric if

$$ G^{\prime}.Y \cap N_{V,U} \neq \emptyset. $$

We are left with the following problem: which ab-diagrams are ortho-symmetric?

For any two pairs (A, B) ∈ L_{V, U}, $(A^{\prime }, B^{\prime }) \in L_{V^{\prime }, U^{\prime }}$, the direct sum $(A, B)\oplus (A^{\prime }, B^{\prime }) = (A \oplus A^{\prime }, B \oplus B^{\prime })$ is defined as an element in $L_{V\oplus V^{\prime }, U\oplus U^{\prime }}$. If δ is the ab-diagram of (A, B) and $\delta ^{\prime }$ is the ab-diagram of $(A^{\prime },B^{\prime })$, the ab-diagram of the direct sum $(A,B)\oplus (A^{\prime },B^{\prime })$ is the disjoint union of δ and $\delta ^{\prime }$.

We call a symmetric pair (A, A^∗) indecomposable if it can not be written as a direct sum of two nontrivial symmetric pairs. We claim that the ab-diagrams of indecomposable nilpotent symmetric pairs are given in Table 1. This table closely follows table II in [7, 6.3].

Table 1 Indecomposable ortho-symmetric ab-diagrams

Proposition 4

An ab-diagram δ is ortho-symmetric if and only if it is a disjoint union of finitely many ab-diagrams from Table 1.

Proof

We say that an ab-diagram η has property $\mathcal P$ if, for every h ≥ 1, the number of occurrences of (ab)^h in η equals the number of occurrences of (ba)^h in η.

Every ab-diagrams α_k,β_k,ε_k in Table 1 has property $\mathcal P$. Indeed in the diagrams α_k and β_k, for every h ≥ 1, there are exactly k − h + 1 occurrences of (ab)^h and of (ba)^h. The diagram ε_k has property $\mathcal P$ by symmetry of its two ab-strings.

Let (A, A^∗) ∈ N_{V, U} be a nilpotent symmetric pair with ab-diagram δ, an ortho-symmetric ab-diagram. For every positive integer h, we notice that the pair ((AA^∗)^h− 1A,(A^∗A)^h− 1A^∗) is nilpotent symmetric. By Lemma 2, we have:

$$ \text{rk} (AA^{*})^{h-1} A = \text{rk} (A^{*}A)^{h-1} A^{*}, $$

which, by Lemma 3, implies that δ has property $\mathcal P$.

We claim that an ab-diagram δ with property $\mathcal P$ is a disjoint union of finitely many ab-diagrams from Table 1. We proceed by contradiction: suppose that there exist ab-diagrams with property $\mathcal P$ which are not disjoint union of finitely many ab-diagrams from Table 1 and take δ one with the least number of ab-strings. Let s be one of the ab-strings in δ with maximal length and let l be its length.

If l = 2k + 1 is odd, s = α_k or s = β_k. Let $\delta ^{\prime }$ be the ab-diagram such that $\delta = s \sqcup \delta ^{\prime }$ (we cannot have δ = s, otherwise δ is already in Table 1).

If l = 2k is even, without loss of generality, we assume that s starts with a. Then, in s there exists exactly one occurrence of (ab)^k and no occurrence of (ba)^k. As δ satisfies property $\mathcal P$, by maximality of l, there must be an ab-string $s^{\prime }$ in δ of length l = 2k starting with b. Thus $s \sqcup s^{\prime } = \varepsilon _{k}$, so let $\delta ^{\prime }$ be the ab-diagram such that $\delta = s \sqcup s^{\prime } \sqcup \delta ^{\prime }$ (again, it cannot be empty).

In both cases, δ and $\delta \setminus \delta ^{\prime }$ have property $\mathcal P$, therefore also $\delta ^{\prime }$ has property $\mathcal P$. By minimality of δ, $\delta ^{\prime }$ is disjoint union of finitely many ab-diagrams from Table 1, thus so is δ. This contradicts the existence of such a δ and concludes the only if part of the proposition.

On the other hand, we show that every ab-diagram in Table 1 is ortho-symmetric. In order to show this, we just need to construct a symmetric pair (A, A^∗) associated to each diagram α_k, β_k, ε_k.

α_k::

Let $D: \mathbb {C}^{k+1} \rightarrow \mathbb {C}^{k+1}$ be a symmetric nilpotent endomorphism of nilpotent order exactly k + 1. Let D = I ∘ X be the canonical decomposition of the map D through its image $D(\mathbb {C}^{k+1})$, so that $X: \mathbb {C}^{k+1} \rightarrow D(\mathbb {C}^{k+1})$ and $I: D(\mathbb {C}^{k+1}) \hookrightarrow \mathbb {C}^{k+1}$. Then D induces a non-degenerate bilinear symmetric form in $D(\mathbb {C}^{k+1})$, as shown in [7, 4], and X = I^∗. As $\dim D(\mathbb {C}^{k+1}) = k$ and both X, I have maximal rank, we immediately get that the ab-diagram of the symmetric nilpotent pair (X, I) is α_k.

β_k::

Let (A, A^∗) be a pair with ab-diagram of type α_k. Then (A^∗,A) is a pair with ab-diagram of type β_k.

ε_k::

Let (A, B) be a nilpotent (not symmetric) pair between the spaces $(V, U) = (\mathbb {C}^{k}, \mathbb {C}^{k})$ with ab-diagram ab⋯b. Let (B^∗,A^∗) be the linear dual of (A, B), so that it is a nilpotent pair between the dual spaces (V^∗,U^∗) with ab-diagram ba⋯a. Then we can equip V ⊕ V^∗ and U ⊕ U^∗ with the non-degenerate symmetric bilinear form given by the duality pairing. Therefore (A, B) ⊕ (B^∗,A^∗) is a symmetric nilpotent pair and its ab-diagram has type ε_k.

□

4 The Variety Z

For each partition λ of n we are going to construct a variety Z^(λ) (or just Z, if the partition λ is clear from the context) with the following property: there exists a quotient map $Z \rightarrow \overline {C_{\lambda }}$. In this way, we will be able to assert some properties of $\overline {C_{\lambda }}$ by proving them for the variety Z. The construction of Z is analogous to the one carried out in [7] and it has already been described in [10]; here we just recall its definition.

Let λ ∈ P(n) be a partition and let t = λ₁ be the number of columns of λ. As in [7], we define a variety Z using the representations of the quiver:

Let

$$ n_{i} = \sum\limits_{j>i}^{t} \widehat\lambda_{j} $$

so that n₀ = n and n_t = 0. We fix the dimension vector $\underline {n} = (n_{0}, \dots , n_{t})$ and vector spaces V_i such that $\dim V_{i} = n_{i}$. We also fix a symmetric bilinear non-degenerate form on each V_i.

The variety Z consists of the representations of dimension vector $\underline {n}$ of the quiver Q_t with relations:

• xy = yx;

• y = x^∗ (the adjoint map as in (3)).

Therefore, each point z ∈ Z is a sequence of maps

$$ (A_{1}, B_{1}; A_{2}, B_{2}; \dots; A_{t}, B_{t}) $$

where A_i ∈ L(V_i− 1,V_i), B_i ∈ L(V_i,V_i− 1), A_iB_i = B_i+ 1A_i+ 1 and $B_{i} = A_{i}^{*}$ for each i. As each B_i is adjoint to A_i, we get an inclusion

$$ Z \hookrightarrow L(V_{0}, V_{1}) \times {\dots} \times L(V_{t-1}, V_{t}) $$

mapping

$$ (A_{1}, B_{1}; A_{2}, B_{2}; \dots; A_{t}, B_{t}) \mapsto (A_{1}, \dots, A_{t}). $$

As in [7], we can think of Z as a schematic fiber in the following way. Let

$$ \begin{array}{@{}rcl@{}} M &:=& L(V_{0}, V_{1}) \times {\dots} \times L(V_{t-1},V_{t}),\\ N &:=& \mathfrak{p}(V_{1}) \times {\dots} \times \mathfrak{p}(V_{t-1}),\\ {\Phi} &:& M \rightarrow N\\ && (A_{1}, \dots, A_{t}) \mapsto (A_{1}A_{1}^{*}-A_{2}^{*}A_{2}, \dots, A_{t-1}A_{t-1}^{*} - A_{t}^{*}A_{t}); \end{array} $$

then we put Z := Φ^− 1(0). As M is an affine space, this interpretation will be useful in Section 7 to show that Z is a complete intersection variety (at least for some λ).

As in [7], let $G = G(V_{0}) \times {\dots } \times G(V_{t})$ be the group which acts on Z by change of basis and $H = G(V_{1}) \times {\dots } \times G(V_{t}) \subseteq G$. We have a map

$$ \begin{array}{@{}rcl@{}} {\Theta}:&& Z \rightarrow \overline{C_{\lambda}}\\ &&(A_{1}, B_{1}; \dots; A_{t}, B_{t}) \mapsto B_{1}A_{1}. \end{array} $$

By the same argument as in [7], we have the following:

Proposition 5

The map Θ splits through the quotient by the group H as

$$ {\Theta}: Z \longrightarrow Z/H \tilde\longrightarrow \overline{C_{\lambda}}; $$

moreover Θ is G(V₀) = G/H-equivariant.

(In particular, Θ^− 1(C_μ) is stable under the action of G(V₀) for each partition μ ≤ λ.)

By restricting the quiver Q_t to two consecutive vertices i − 1,i, we get a map $Z \rightarrow N_{V_{i-1},V_{i}}$. As all the orbits in $N_{V_{i-1},V_{i}}$ are defined by their ortho-symmetric ab-diagram, we can map each point z ∈ Z and $i = 1,\dots ,t$ to its i th ab-diagram τ_i. Therefore, for each string of ab-diagrams

$$ \tau = (\tau_{1}, \dots, \tau_{t}) $$

we define

$$ Z_{\tau} = \{z \in Z:\forall i,\text{ the \textit{i}th \textit{ab}-diagram of \textit{z} is $\tau_{i}$ }\}. $$

The following are necessary conditions on τ in order to have Z_τ≠∅:

• (1) τ_i must have exactly n_i− 1 a’s and n_i b’s,

• (2) ρ(τ_i) = π(τ_i+ 1) for $i=1,\dots ,t-1$,

• (3) each τ_i must be ortho-symmetric.

We will denote by Λ^(λ) (or, shortly, by Λ) the set of all strings τ which verify the three conditions above.

As in [5, Lemma 5.4], [7, 8.2], Θ^− 1(C_λ) is given by exactly one stratum $Z_{\tau ^{0}}$. The string of ab-diagrams τ⁰ can be constructed in the following way: for all $i=0,\dots ,t-1$, let λ_i be the partition of n_i given by the columns of λ with indices greater than i; then let τ_i+ 1 be the ab-diagram built by setting one a on each box in the Young diagram of λ_i and by setting one b between each pair of consecutive a’s.

Example 1

Let λ = (3,1) be a partition. Then the string of ab-diagrams τ⁰ is the following:

$$ \tau^{0} = ({\tau^{0}_{1}}, {\tau^{0}_{2}}, {\tau^{0}_{3}}) = \left( \begin{array}{l|l|l} ababa & aba & a \\ a & & \end{array} \right). $$

The stratum $Z_{\tau ^{0}}$ is open in Z as it is the subvariety of the points

$$ (A_{1}, B_{1}, \dots, A_{t}, B_{t}) $$

in Z where each A_i, B_i has maximal rank. We have the following key result after the manner of Kraft, Procesi and Ohta.

Proposition 6 ([7, 5.5] and [10, Prop. 8])

The open set $Z_{\tau ^{0}}$ is contained in the smooth locus of Z.

We recall a dimension formula for the strata Z_τ due to Otha [10, Prop. 6]. First, for each ab-diagram δ, we denote by a_i (resp. b_i) the number of rows of δ starting by a (resp. b) of length i. We define

$$ {\Delta}(\delta) := {\sum}_{i \text{ odd }}a_{i}b_{i} $$

and

$$ o(\delta) := {\sum}_{i \text{ odd }}(a_{i} + b_{i}). $$

Proposition 7

[10, Prop. 6] Let $\tau = (\tau _{1}, \dots , \tau _{t})$. We have

$$ \begin{array}{ll} \dim Z_{\tau} = \frac 1 2 \dim C_{\pi(\tau_{1})} + {\sum}_{i=0}^{t-1}\left( \frac 1 2 n_{i}n_{i+1} - \frac 1 4 (n_{i} + n_{i+1})\right) \\ + {\sum}_{i=1}^{t} \left( \frac 1 4 o(\tau_{i}) -\frac 1 2{\Delta}(\tau_{i})\right). \end{array} $$

(4)

In the following sections we focus on the combinatorics of the τ ∈Λ. We need tools to effectively manipulate formula (4), in particular we aim at proposition 10 and proposition 14 (see Section 6).

5 s-Step Condition

In this section we define a condition which plays a pivotal role in the subsequent study of the nilpotent symmetric orbits.

Definition 1

Given a partition $\lambda = (\lambda _{1}, \dots , \lambda _{h})$, we say that λ satisfies the s-step condition if for every $i = 1, \dots , h$ we have

$$ \lambda_{i} \leq \lambda_{i+1} + s $$

with the convention that λ_h+ 1 = 0.

We want to highlight some simple properties of this definition. First of all, s₁-step condition implies s₂-step condition for each s₂ ≥ s₁. Secondly, if λ satisfies the s-step condition, then if we remove the first row or the first column from λ, the resulting partition still satisfies the s-step condition.

Example 2

The single line partition (n) satisfies the n-step condition, but not the (n − 1)-step condition. The triangular partition $(n, n-1, \dots , 1)$ satisfies the 1-step condition. The 2-skew triangular partition $(2n, 2(n-1), \dots , 2)$ satisfies the 2-step condition (but not the 1-step condition).

6 Differences Between Partitions

In this section we study some quantitative ways to address the difference between two partitions. Let λ, μ ∈ P(n) be two partitions of n such that λ > μ.

Definition 2

If $\lambda = (\lambda _{1}, \dots , \lambda _{k}) \in P(n)$ and $\mu = (\mu _{1}, \dots , \mu _{k}) \in P(n)$, possibly allowing λ_i = 0 for some i, we define

$$ q(\lambda, \mu) = \frac 1 2 {\sum}_{i=1}^{k} |\lambda_{i} - \mu_{i}|. $$

Remark 1

By an easy computation modulo 2, it can be seen that q(λ, μ) is an integer.

Example 3

Let λ = (7,2,2,1) and μ = (5,3,1,1,1,1). We can draw the Young diagram of both partitions and overlay one over the other, so that λ is given by white and gray boxes while μ is given by white and black boxes in the following diagram:

We can compute the q difference:

$$ q(\lambda, \mu) =\frac 1 2 \left( (7-5) + (3-2) + (2-1) + (1-1) + 1 + 1 \right) = 3. $$

We can check that in this case q(λ, μ) is the number of gray boxes or the number of black boxes. This is an easy fact (it is not even needed that λ > μ) that can be derived from the following Lemma.

Lemma 8

Let λ > μ be two partitions of n. Then there exist partitions

$$ \lambda = \lambda^{0} > \lambda^{1} > {\dots} > \lambda^{q(\lambda, \mu)} = \mu $$

such that q(λⁱ,λ^i+ 1) = 1 for each $i = 0, \dots , q(\lambda , \mu )-1$.

Moreover, we can choose λⁱ such that for each row r and for each column c the sequences of integers ${\lambda ^{0}_{r}}, \dots , \lambda ^{q(\lambda , \mu )}_{r}$ and $\widehat {\lambda ^{0}_{c}}, \dots , \widehat \lambda ^{q(\lambda , \mu )}_{c}$ are monotone.

Proof

We proceed by induction on q = q(λ, μ). If q = 1 we trivially have λ¹ = μ. If q > 1, we are going to build λ¹ such that q(λ, λ¹) = 1 and q(λ¹,μ) = q(λ, μ) − 1.

We build λ¹ starting from λ in this way. Let $\tilde \imath$ be the first index such that $\lambda _{\tilde \imath } > \mu _{\tilde \imath }$ and let j be the first index such that λ_j < μ_j. Clearly there must exists such indices because λ≠μ and |λ| = |μ|. Let $i \geq \tilde \imath$ be the last index such that $\lambda _{i} = \lambda _{\tilde \imath }$ (possibly $i = \tilde \imath$). As λ > μ, we must have $\tilde \imath < j$ and therefore i < j. We define

$$ \lambda^{1} = (\lambda_{1}, \dots, \lambda_{i-1}, \lambda_{i}-1, \lambda_{i+1}, \dots, \lambda_{j-1}, \lambda_{j} + 1, \lambda_{j+1}, \dots, \lambda_{k}). $$

The sequence λ¹ is actually a partition of n as |λ¹| = |λ| = n, λ_i − 1 ≥ λ_i+ 1 by the maximality of i and λ_j− 1 ≥ λ_j + 1 by the minimality of j.

Moreover, both the following hold: q(λ, λ¹) = 1 and q(λ¹,μ) = q − 1. We apply induction on the pair (λ¹,μ) and we conclude the proof of the first part of the Lemma.

To conclude the proof, we notice that if, for a row r, λ_r > μ_r (resp. λ_r < μ_r), then we have ${\lambda ^{i}_{r}} \geq \lambda ^{i+1}_{r}$ (resp. ${\lambda ^{i}_{r}} \leq \lambda ^{i+1}_{r}$) by construction. The same holds true for the columns lengths. □

Example 4

Taking the partitions λ, μ defined in the previous example, the construction explained in the Lemma produces the following sequence of partitions:

$$ (7, 2, 2, 1) > (6, 3, 2, 1) > (5, 3, 2, 1, 1) > (5, 3, 1, 1, 1). $$

Remark 2

In view of Lemma 8 we can interpret q(λ, μ) as the minimum number of boxes needed to lower in order to obtain μ from λ.

We consider two other ways of comparing partitions. Given a partition λ, for each box B in the Young diagram of λ we define its column number c(B) (resp. row number r(B)) by counting the columns (resp. rows) starting from the leftmost column (resp. uppermost row). For example, given λ = (7,2,2,1) we have:

Remark 3

By summing up over the rows (resp. the columns) of λ, we have:

$$ {\sum}_{B \in \lambda} c(B) = {\sum}_{i=1}^{h} \binom{\lambda_{i}+1}{2} \quad \left( \text{resp. } {\sum}_{B \in \lambda} r(B) = {\sum}_{i=1}^{t}\binom{\widehat{\lambda}_{i}+1}{2}\right). $$

(5)

Definition 3

Given λ ≥ μ as before, we define:

$$ c(\lambda, \mu) = {\sum}_{B \in \lambda} c(B) - {\sum}_{B \in \mu} c(B), $$

where B is selected among the boxes in the Young diagrams of λ and μ and, similarly,

$$ r(\lambda, \mu) = {\sum}_{B \in \mu} r(B) - {\sum}_{B \in \lambda} r(B). $$

Example 5

Let λ = (7,2,2,1) and μ = (5,3,1,1,1,1). We compute

$$ \begin{array}{@{}rcl@{}} c(\lambda, \mu) & =& (28 + 3 + 3 + 1) - (15 + 6 + 1 + 1 + 1 + 1) = 10, \\ r(\lambda, \mu) & =& (21 + 3 + 3 + 1 + 1) - (10 + 6 + 1 + 1 + 1 + 1 + 1) = 8. \end{array} $$

Remark 4

It is always guaranteed that c(λ, μ) ≥ 0 and r(λ, μ) ≥ 0 for every λ ≥ μ. We even have that

$$ \lambda = \mu \Longleftrightarrow c(\lambda, \mu) = 0 \Longleftrightarrow r(\lambda, \mu) = 0. $$

(6)

In fact we have that c(λ, μ) ≥ 0 is equivalent to

$$ {\sum}_{i} \binom{\lambda_{i}+1}2 \geq {\sum}_{i} \binom{\mu_{i}+1}2. $$

(7)

As the function $n \mapsto \binom {n+1}{2}$ is convex, (7) is given by the Karamata’s inequality [1, Chapter 1, §28].

Remark 5

Given partitions λ ≥ μ ≥ ν, by definition of c and r we have:

$$ \begin{array}{@{}rcl@{}} c(\lambda, \nu) &= c(\lambda, \mu) + c(\mu, \nu), \end{array} $$

(8)

$$ \begin{array}{@{}rcl@{}} r(\lambda, \nu) &= r(\lambda, \mu) + r(\mu, \nu). \end{array} $$

(9)

Remark 6

For every pair of partitions λ ≥ μ we have that

$$ c(\lambda, \mu) \geq q(\lambda, \mu). $$

(10)

In fact, when q(λ, μ) = 1, by (6) we get (10). Otherwise, by Lemma 8, we have a sequence of partitions $\lambda ^{0}, \dots , \lambda ^{q(\lambda , \mu )}$. For each pair (λⁱ,λ^i+ 1), we have c(λⁱ,λ^i+ 1) ≥ 1, therefore the conclusion follows by summing up each inequalities by (8).

Lemma 9

Let λ be a partition satisfying the s-step condition and μ < λ be another partition. Then

$$ s \cdot r(\lambda, \mu) \geq c(\lambda, \mu) + q(\lambda, \mu). $$

(11)

Moreover, (11) holds strictly if there exists a column index c such that $\widehat \mu _{c} > \widehat \lambda _{c} + 1$ or a row index r such that μ_r > λ_r + 1.

Proof

Let q = q(λ, μ). By Lemma 8, we obtain a sequence

$$ \lambda = \lambda^{0} > {\dots} > \lambda^{q} = \mu, $$

with q(λⁱ,λ^i+ 1) = 1 for $i = 0, \dots , q-1$.

We want to prove

$$ s \cdot r(\lambda^{i}, \lambda^{i+1}) \geq c(\lambda^{i}, \lambda^{i+1}) + 1 $$

(12)

for all $i = 0, \dots , q - 1$. Once (12) is proved, the conclusion will follow by summing up each of those inequalities by (9) and (8).

We start by noticing that the s-step condition implies that, for every pair of row indices r₁ < r₂,

$$ \lambda_{r_{1}} - \lambda_{r_{2}} \leq s (r_{2} - r_{1}). $$

Fix r₁ to be an index row such that $\lambda _{r_{1}} \geq \mu _{r_{1}}$ and fix r₂ to be an index row such that $\lambda _{r_{2}} \leq \mu _{r_{2}}$. Lemma 8 asserts that $\lambda ^{i}_{r_{j}}$ is a monotone sequence for both j = 1,2; therefore, for such r₁,r₂, we get $\lambda ^{i}_{r_{1}} - \lambda ^{i}_{r_{2}} \leq \lambda _{r_{1}} - \lambda _{r_{2}}$, so we get

$$ \lambda^{i}_{r_{1}} - \lambda^{i}_{r_{2}} \leq s (r_{2} - r_{1}). $$

(13)

As q(λⁱ,λ^i+ 1) = 1, we already observed that the Young diagrams of λⁱ,λ^i+ 1 have only one box in different positions. This means that there exist exactly two columns c_i,1 < c_i,2 which are not equal between λⁱ,λ^i+ 1. Similarly, there exist exactly two rows indices r_i,1 < r_i,2.

We have c(λⁱ,λ^i+ 1) = c_i,2 − c_i,1 and r(λⁱ,λ^i+ 1) = r_i,2 − r_i,1. Moreover, by definition of λⁱ and λ^i+ 1, we have $\lambda ^{i}_{r_{i,1}} = c_{i,2}$ and $\lambda ^{i}_{r_{i,2}} = c_{i,1} - 1$. Therefore, by (13), we get

$$ \begin{array}{@{}rcl@{}} & s \cdot r(\lambda^{i}, \lambda^{i+1}) = s(r_{i,2} - r_{i,1}) \geq \lambda^{i}_{r_{i,1}} - \lambda^{i}_{r_{i,2}} = c_{i,2} - c_{i,1} + 1 = c(\lambda^{i}, \lambda^{i+1}) + 1. \end{array} $$

Finally, we consider the additional hypothesis of the existence of a column index c such that $\widehat \mu _{c} > \widehat \lambda _{c} + 1$ or a row index r such that μ_r > λ_r + 1. We are going to prove that there exists i such that (13) holds strictly.

Let

$$ I_{\text{cols}}(c) = \left\{ i\in \{0, \dots, q-1\}: \widehat{\lambda^{i+1}_{c}} > \widehat{{\lambda^{i}_{c}}}\right\}. $$

As q(λⁱ,λ^i+ 1) = 1, we have $|\widehat {\lambda ^{i+1}}_{c} - \widehat {\lambda ^{i}}_{c}| \leq 1$ for each i. Therefore, if there exists c such that $\widehat \mu _{c} > \widehat \lambda _{c} + 1$, we have |I_cols(c)|≥ 2. Let j be the minimum in I_cols(c). We show that (13) holds strictly for the rows r_i,1,r_i,2 for every i ∈ I_cols(c) such that i≠j. Indeed, as $\lambda ^{i}_{r_{i,2}} = c = \lambda ^{j}_{r_{j,2}}$, we have

$$ \begin{array}{@{}rcl@{}} && \lambda^{i}_{r_{i,1}} - \lambda^{i}_{r_{i,2}} \leq \lambda^{j}_{r_{i,1}} - \lambda^{j}_{r_{j,2}} \leq \lambda_{r_{i,1}} - \lambda_{r_{j,2}} \\ && \leq s(r_{j,2} - r_{i,1}) \leq s(r_{i,2} - 1 - r_{i,1}) < s(r_{i,2} - r_{i,1}). \end{array} $$

In a similar fashion, let

$$ I_{\text{rows}}(r) = \left\{ i\in \{0, \dots, q-1\}: {\lambda^{i+1}_{r}} > {{\lambda^{i}_{r}}}\right\}. $$

If there exists r such that μ_r > λ_r + 1, then |I_rows(r)|≥ 2 and we take j ∈ I_rows(r) to be the minimum and i ∈ I_rows(r) to be another index. We have $\lambda ^{i}_{r_{i,2}} > \lambda ^{j}_{r_{i,2}} = \lambda _{r_{i,2}}$, so we get

$$ \lambda^{i}_{r_{i,1}} - \lambda^{i}_{r_{i,2}} < \lambda_{r_{i,1}} - \lambda_{r_{i,2}} \leq s(r_{i,2} - r_{i,1}). $$

In both cases (13) holds strictly for at least an i. Therefore (12) holds strictly for this i, and (11) must hold strictly as a consequence. Thus, the Lemma is proven. □

Example 6

Let

$$ \lambda = (6, 4, 2), \quad \mu = (5, 3, 2, 1, 1), \quad \nu = (5, 3, 3, 1) $$

be three partitions of n = 12. We have that μ < λ and ν < λ; moreover, in the pair (λ, μ) the first column differs by 2 boxes, while no column in (λ, ν) differs more than 1 box.

Since λ satisfies the 2-step condition, we can compute each term involved in (11). For (λ, μ) we have: 2 ⋅ 6 ≥ 8 + 2 (which holds strictly), while for (λ, ν) we have: 2 ⋅ 4 ≥ 6 + 2.

7 Inequalities on the Dimensions

In this section we will use the tools introduced in Section 5 to effectively compute the dimensions of the strata in Z.

Let $\lambda = (\lambda _{1},\dots ,\lambda _{k})$ be a partition of n := |λ| with t := λ₁ columns. We want to compare the dimension of $Z_{\tau ^{0}}$ with the dimension of each other stratum Z_τ with τ ∈Λ^(λ). Let μ = π(τ), so that μ ≤ λ. The aim of this section is to give some sufficient combinatorial conditions on the pair of partitions (λ, μ) in order to secure a bound on the difference $\dim Z_{\tau ^{0}} - \dim Z_{\tau }$.

The aim of the current section is to prove proposition 10 and Proposition 14. In order to introduce the stronger inequality in Proposition 14, we need an in-depth study of the combinatorics of Λ, so we prefer to introduce it later.

Proposition 10

Let λ ≥ μ be two partitions and let Z = Z^(λ) be the variety built from λ. Let $Z_{\tau ^{0}}$ be the only stratum of Z with $\pi ({\tau ^{0}_{1}}) = \lambda$ and let Z_τ be a stratum with $\tau =(\tau _{1}, \dots , \tau _{t})$ such that π(τ₁) = μ.

For each real number x, we have that:

$$ 2r(\lambda, \mu) - c(\lambda, \mu) - q(\lambda, \mu) \geq 4x \quad \Longrightarrow \quad \dim Z_{\tau^{0}} - \dim Z_{\tau} \geq x. $$

The proof will proceed as follows: we take the dimension formula of the strata of Z given by (4) then we will estimate each term on the difference. Each of the following lemmas deals with one of the terms.

Lemma 11

Let λ ≥ μ be two partitions of n. Let t = λ₁ be the number of columns of λ. Then

$$ {\sum}_{i=1}^{t} \left( \widehat{\mu}_{i}^{2} - \widehat{\lambda}_{i}^{2}\right) = 2r(\lambda, \mu). $$

(We allow $\widehat {\mu }_{i} = 0$ for each i > μ₁.)

Proof

For every integer m we have: $m^{2} = 2\binom {m+1} 2 - m$. So, by (5), we get:

$$ {\sum}_{i=1}^{t} \left( \widehat{\mu}_{i}^{2} - \widehat{\lambda}_{i}^{2}\right) = {\sum}_{i=1}^{t} \left( 2\binom{\widehat{\mu}_{i}+1} 2 - \widehat{\mu_{i}} - 2\binom{\widehat{\lambda}_{i}+1} 2 + \widehat{\lambda}_{i}\right) = 2r(\lambda, \mu) - n + n. $$

□

In order to carry on the computation, we examine carefully all the strata Z_τ with τ≠τ⁰. We do so by grouping together the τ ∈Λ which can appear in Θ^− 1(C_μ), for fixed μ. We already know that for every $\tau = (\tau _{1}, \dots , \tau _{t})$ such that Θ(Z_τ) = C_μ, we have that π(τ₁) = μ.

The conditions τ ∈Λ and π(τ₁) = μ place important combinatorial restrictions in the choice of the ab-diagrams $\tau _{1}, \dots , \tau _{t}$. We are going to describe them in the following paragraphs.

Let Z^(μ) be the variety Z built from the partition μ (rather than the partition λ). We will denote by σ⁰ the unique stratum in Λ^(μ) such that ${\Theta }(Z^{(\mu )}_{\sigma ^{0}}) = C_{\mu }$. As in the case of τ⁰, every ab-diagram ${\sigma ^{0}_{i}}$ has only rows starting and ending by a, in particular they all have odd length, so:

$$ {\sum}_{i=1}^{t} o({\sigma^{0}_{i}}) = n = {\sum}_{i=1}^{t} o({\tau^{0}_{i}}). $$

(14)

The sequence of ab-diagrams σ⁰ does not belong to Λ^(λ) because the integers n_i computed from μ are different from those computed from λ.

Nonetheless, for every τ ∈Λ^(λ) such that π(τ₁) = μ, the condition ρ(τ_i− 1) = π(τ_i) implies that each τ_i can be obtained from ${\sigma ^{0}_{i}}$ by adding an adequate amount of a’s and b’s letters. For example, if i = 0 we need some b’s, possibly.

For each $i = 1, \dots , t$, let d_i be the list of a’s and b’s we need to add to ${\sigma ^{0}_{i}}$ in order to obtain τ_i. Let ${d_{i}^{a}}$ (resp. ${d_{i}^{b}}$) be the number of a’s (resp. b’s) in d_i. We remark that ${d_{i}^{a}}$ and ${d_{i}^{b}}$ depend only on (λ, μ) and not on the particular τ. In fact, we can compute ${d_{i}^{a}}$ by taking the differences

$$ {d_{i}^{a}} = {\sum}_{j=i}^{t} \left( \widehat{\lambda}_{j} - \widehat{\mu}_{j}\right); $$

(15)

and therefore we have ${d_{i}^{b}} = d_{i+1}^{a}$.

Example 7

Let τ ∈Λ^(λ) and let σ⁰ ∈Λ^(μ). Suppose that the i th diagrams are the following:

$$ {\sigma^{0}_{i}} = \begin{array}{l} aba \\ aba \\ b \end{array}; \qquad \tau_{i} = \begin{array}{l} ababa \\ aba \\ ba \\ ab \end{array}. $$

Then, we have d_i = (a, a, a, b, b), ${d_{i}^{a}} = 3$, ${d_{i}^{b}} = 2$.

Given an ab-diagram δ and a list d of a’s and b’s, we define aug_δ(d) to be the set of ortho-symmetric ab-diagrams obtainable from δ by adding the letters in d.

Example 8

Let d = (a, b) and

$$ \delta = \begin{array}{l} aba \\ aba \\ b \end{array}. $$

Then aug_δ(d) has three elements:

$$ \text{aug}_{\delta}(d) = \left\{ \begin{array}{l} ababa \\ aba \\ b \end{array},\quad \begin{array}{l} aba \\ aba \\ bab \end{array},\quad \begin{array}{l} aba \\ aba \\ a \\ b \\ b \end{array}\right\}. $$

Lemma 12

Let δ⁰ be an ab-diagram with rows starting and ending only with a. Let d be a list of d^a a’s and d^b b’s. Let $\delta \in \text {aug}_{\delta ^{0}}(d)$. Then the following holds:

$$ o(\delta) - 2{\Delta}(\delta) - o(\delta^{0}) \leq \max\{d^{a}, d^{b}\}. $$

Proof

The ab-diagram δ is obtained from δ⁰ by adding d^a a’s and d^b b’s.

Let L (resp. S) be the set of rows of δ longer than 1 (resp. of length 1) built using only letters in d. As o(δ) counts some of the rows in δ, we compare the difference o(δ) − o(δ⁰) with the number of rows built only with letters in d and we get o(δ) − o(δ⁰) ≤|L| + |S|.

Let S_a (resp. S_b) the rows of S starting with a (resp. b), so that S = S_a ⊔ S_b. Recall that a_i (resp. b_i) is the number of rows of length i starting with a (resp. b); in this particular ab-diagram δ we have |S_b| = b₁ and |S_a|≤ a₁.

$$ \begin{array}{@{}rcl@{}} o(\delta) - o(\delta^{0}) - 2{\Delta}(\delta) &\leq & |L| + |S| -2{\sum}_{i \text{ odd}} a_{i} b_{i}\\ &\leq & |L| + |S| -2 a_{1} b_{1} \\ &\leq & |L| + |S_{a}| + |S_{b}| - 2 |S_{a}| |S_{b}|\\ &\leq & |L| + \max\{|S_{a}|, |S_{b}|\}. \end{array} $$

If |S_a|≥|S_b|, then $|L| + |S_{a}| \leq {d_{i}^{a}}$ because each row in L or in S_a contains at least one a. If the opposite holds true, then $|L| + |S_{b}| \leq {d_{i}^{b}}$ for the same reason. Thus the Lemma is proven. □

We can introduce a small, but very important, improvement on Lemma 12.

Lemma 13

In the same setting of Lemma 12, if we furthermore assume that d = (b), that is d^a = 0 and d^b = 1, and that δ⁰ has l rows of length 1, we have the stronger equality:

$$ o(\delta) - 2{\Delta}(\delta) - o(\delta^{0}) = \max\{d^{a}, d^{b}\} - 2l = 1 - 2l. $$

Proof

As all the rows in δ⁰ have odd length, we cannot extend exactly one row with letter b, otherwise the resulting ab-diagram would not be ortho-symmetric. So we must place b on a new row.

In this case, o(δ) − o(δ⁰) = 1 and Δ(δ) = a₁b₁ = l ⋅ 1. □

Using this lemma we will be able to introduce the previously announced stronger version of proposition 10, namely:

Proposition 14

Let λ ≥ μ be two partitions and let Z = Z^(λ) be the variety built from λ. Let $Z_{\tau ^{0}}$ be the only stratum of Z with $\pi ({\tau ^{0}_{1}}) = \lambda$ and let Z_τ be a stratum with $\tau =(\tau _{1}, \dots , \tau _{t})$ such that π(τ₁) = μ. As in proposition 10, we suppose that there exists a real number x such that:

$$ 2r(\lambda, \mu) - c(\lambda, \mu) - q(\lambda, \mu) \geq 4x. $$

If there exists an index i such that d_i = (b) and ${\sigma ^{0}_{i}}$ has l rows of length one, then:

$$ \dim Z_{\tau^{0}} \geq \dim Z_{\tau} + x + l/2. $$

The proof of proposition 14 will be given together with the proof of proposition 10 at the end of this section.

We want an estimate of the sum of the terms $\max \limits \{{d_{i}^{a}}, {d_{i}^{b}}\}$ depending only on λ, μ.

Lemma 15

Let λ ≥ μ be partitions, so that we can define ${d_{i}^{a}}, {d_{i}^{b}}$ for all $i = 1, \dots , t$. Then

$$ {\sum}_{i=1}^{t} \max\{{d_{i}^{a}}, {d_{i}^{b}}\} \leq c(\lambda, \mu) + q(\lambda, \mu). $$

(16)

Proof

First we prove the claim for q(λ, μ) = 1.

In this case, μ is obtained from λ by moving down a single box B. Let us call c₁ (resp. c₂) the column where B lies in μ (resp. λ); so c₁ < c₂. By (15), one gets ${d_{i}^{a}} = 1$ for each c₁ < i ≤ c₂, and 0 otherwise. Moreover one also gets that ${d_{i}^{b}} = 1$ for each c₁ ≤ i < c₂ and 0 otherwise. Therefore $\max \limits \{{d_{i}^{a}}, {d_{i}^{b}}\} = 1$ for all c₁ ≤ i ≤ c₂ and 0 otherwise. Therefore

$$ {\sum}_{i=1}^{t} \max\{{d_{i}^{a}}, {d_{i}^{b}}\} = c_{2} - c_{1} + 1, $$

while

$$ c(\lambda, \mu) = c_{2} - c_{1}, $$

so

$$ {\sum}_{i=1}^{t} \max\{{d_{i}^{a}}, {d_{i}^{b}}\} = c(\lambda, \mu) + 1 $$

(17)

and the conclusion holds in this case.

In the general case, let q := q(λ, μ) so, by Lemma 8, we get a sequence of partitions

$$ \lambda = \lambda^{0} > \lambda^{1} > {\dots} > \lambda^{q} = \mu $$

such that q(λ^j,λ^j+ 1) = 1 for each $j = 0, \dots , q-1$. By (17), for every j, we have

$$ {\sum}_{i=1}^{t} \max\{{d_{i}^{a}}, {d_{i}^{b}}\}(\lambda^{j}, \lambda^{j+1}) = c(\lambda^{j}, \lambda^{j+1}) + 1 $$

and, summing up over j,

$$ {\sum}_{i=1}^{t} {\sum}_{j=0}^{q-1}\max\{{d_{i}^{a}}, {d_{i}^{b}}\}(\lambda^{j}, \lambda^{j+1}) = {\sum}_{j=0}^{q-1}c(\lambda^{j}, \lambda^{j+1}) + q. $$

By (8), the function c is additive on the pairs (λ^j,λ^j+ 1), therefore the right hand side is equal to c(λ, μ) + q. Also ${d_{i}^{a}}$ and ${d_{i}^{b}}$ are additive functions on the pairs (λ^j,λ^j+ 1) (as it follows from (15)), therefore

$$ \begin{array}{@{}rcl@{}} {\sum}_{j=0}^{q-1}\max\{{d_{i}^{a}}, {d_{i}^{b}}\}(\lambda^{j}, \lambda^{j+1}) & =& {\sum}_{j=0}^{q-1}\max\left\{{d_{i}^{a}}(\lambda^{j}, \lambda^{j+1}), {d_{i}^{b}}(\lambda^{j}, \lambda^{j+1})\right\}\\ & \geq& \max\left\{{\sum}_{j=0}^{q-1}{d_{i}^{a}}(\lambda^{j}, \lambda^{j+1}), {\sum}_{j=0}^{q-1}{d_{i}^{b}}(\lambda^{j}, \lambda^{j+1})\right\} \\ & =& \max\left\{{d_{i}^{a}}(\lambda, \mu), {d_{i}^{b}}(\lambda, \mu)\right\} = \max\{{d_{i}^{a}}, {d_{i}^{b}}\} \end{array} $$

and the conclusion follows. □

We are finally able to prove both Proposition 10 and Proposition 14.

Proof of Proposition 10 and Proposition 14

We start from the dimension formula of the strata (4). We recall that Δ(τ⁰) = 0 as no ab-diagram of τ⁰ has rows starting with b. Therefore

$$ \begin{array}{@{}rcl@{}} 4\big(\dim Z_{\tau^{0}} - \dim Z_{\tau} - x\big) &= &\ 2 (\dim C_{\pi({\tau_{1}^{0}})} - \dim C_{\pi(\tau_{1})})+\\ & &+ \left( o(\tau^{0}) - 2{\Delta}(\tau^{0})\right) - \left( o(\tau) - 2{\Delta}(\tau)\right) - 4x\\ &= & \ 2 (\dim C_{\lambda} - \dim C_{\mu}) + o(\tau^{0}) - o(\tau) + 2{\Delta}(\tau) - 4x\\ &= & {\sum}_{i=1}^{t} \left( \widehat{\mu}_{i}^{2}-\widehat{\lambda}_{i}^{2}\right) + {\sum}_{i=1}^{t} \left( o({\tau^{0}_{i}}) - o(\tau_{i}) + 2{\Delta}(\tau_{i})\right) - 4x\\ \text{(by Lemma\nobreakspace 11)} &= &\ 2r(\lambda, \mu) + {\sum}_{i=1}^{t} \left( o({\tau^{0}_{i}}) - o(\tau_{i}) + 2{\Delta}(\tau_{i})\right) - 4x\\ \text{(by (\ref{equ:comb:ots}))} &= &\ 2r(\lambda, \mu) + {\sum}_{i=1}^{t} \left( o({\sigma^{0}_{i}}) - o(\tau_{i}) + 2{\Delta}(\tau_{i})\right) - 4x\\ &\geq &\ 2r(\lambda, \mu) - 4x\ + \\ & \ &-{\sum}_{i=1}^{t} \max_{\tau_{i} \in \text{aug}_{{\sigma^{0}_{i}}}(d_{i})} \big(o(\tau_{i}) - 2{\Delta}(\tau_{i}) - o({\sigma^{0}_{i}}) \big)\\ \text{(by Lemma\nobreakspace 12)} &\geq &\ 2r(\lambda, \mu) - {\sum}_{i=1}^{t} \max\{{d_{i}^{a}}, {d_{i}^{b}}\} - 4x\\ \text{(by Lemma\nobreakspace 15)} &\geq &\ 2r(\lambda, \mu) - c(\lambda, \mu) - q(\lambda, \mu) - 4x. \end{array} $$

Therefore $\dim Z_{\tau ^{0}} - \dim Z_{\tau } \geq x$ is guaranteed as soon as 2r(λ, μ) − c(λ, μ) − q(λ, μ) ≥ 4x.

In order to obtain the sharper result of proposition 14, it is enough to use Lemma 13 in place of Lemma 12 in the second to last step. □

8 Complete Intersection Conditions

In this section we want to give condition under which we can make sure that Z is a complete intersection variety.

Proposition 16

Let λ be a partition and recall that $Z = {\Phi }^{-1}(0) \subseteq M$. We have

$$ \text{codim}_{M}(Z_{\tau^{0}}) = \dim N. $$

Proof

In order to prove the equality of the required dimensions, we will work with the dimension formula given by (4). We are going to prove that

$$ \dim Z_{\tau^{0}} = \dim M - \dim N. $$

(18)

We immediately have:

$$ \dim M = {\sum}_{i=1}^{t} \dim L(V_{i-1},V_{i}) = {\sum}_{i=1}^{t} n_{i-1}n_{i} $$

and

$$ \dim N = {\sum}_{i=1}^{t-1} \dim \mathfrak{p}(V_{i}) = {\sum}_{i=1}^{t-1}\frac 1 2 {n_{i}^{2}} + {\sum}_{i=1}^{t-1}\frac 1 2 n_{i}, $$

so

$$ \dim M - \dim N = {\sum}_{i=1}^{t} n_{i-1}n_{i} - \frac 1 2{\sum}_{i=1}^{t-1} {n_{i}^{2}} - \frac 1 2{\sum}_{i=1}^{t-1} n_{i}. $$

On the other hand, we have:

$$ \begin{array}{ll} \dim Z_{\tau^{0}} = \frac 1 2 \dim C_{\pi({\tau^{0}_{1}})} + {\sum}_{i=0}^{t-1}\left( \frac 1 2 n_{i}n_{i+1} - \frac 1 4 (n_{i} + n_{i+1})\right) + \\ +{\sum}_{i=1}^{t} \left( \frac 1 4 o({\tau^{0}_{i}}) -\frac 1 2{\Delta}({\tau^{0}_{i}})\right). \end{array} $$

We have that $\pi ({\tau ^{0}_{1}}) = \lambda$; we then use formula (2).

We also have that $o({\tau ^{0}_{i}})$ is precisely the number of rows of ${\tau ^{0}_{i}}$ (as they all have odd length), the number of rows of ${\tau ^{0}_{i}}$ is equal to the number of rows of $\lambda ^{i} :=\pi ({\tau ^{0}_{i}})$, the partition λⁱ has clearly $\widehat {\lambda }^{i}_{1}$ rows and $\widehat {\lambda }^{i}_{1} = \widehat {\lambda }_{i}$, that is the i th column of λ.

Finally, we have that ${\Delta }({\tau ^{0}_{i}}) = 0$, because there are no rows in ${\tau ^{0}_{i}}$ starting with b.

Therefore:

$$ \dim Z_{\tau^{0}} = \frac 1 4 \left( {n_{0}^{2}}- {\sum}_{i=1}^{t} {\widehat\lambda_{i}^{2}}\right) + {\sum}_{i=0}^{t-1}\left( \frac 1 2 n_{i}n_{i+1} - \frac 1 4 (n_{i} + n_{i+1})\right) + \frac 1 4 {\sum}_{i=1}^{t}\widehat{\lambda}_{i}; $$

we recall that $\widehat \lambda _{i} = n_{i-1} - n_{i}$ and that n₀ = n = |λ|, n_t = 0; so

$$ \begin{array}{@{}rcl@{}} \dim Z_{\tau^{0}} &=& \frac 1 4 \left( {n_{0}^{2}}- {\sum}_{i=1}^{t} (n_{i-1}-n_{i})^{2}\right) +\\ &&+ \frac 1 2{\sum}_{i=0}^{t-1} n_{i}n_{i+1} - \frac 1 4 n_{0} - \frac 1 2{\sum}_{i=1}^{t-1}n_{i} + \frac 1 4 {\sum}_{i=1}^{t}n_{i-1}-n_{i}\\ &=&- \frac 1 2{\sum}_{i=1}^{t-1} {n_{i}^{2}} + \frac 1 2{\sum}_{i=1}^{t-1} n_{i-1}n_{i} + \frac 1 2 {\sum}_{i=0}^{t-1}n_{i}n_{i+1} - \frac 1 2 {\sum}_{i=1}^{t-1} n_{i}\\ &=& \dim M - \dim N. \end{array} $$

□

We can proceed to the main result of this section.

Proposition 17

If λ satisfies the 2-step condition, then Z is a complete intersection variety.

The argument is basically the same used in [5, Theorem in 3.3] and [7, Theorem in 5.3]. Here we merely put the pieces together. One of these pieces is the following technical lemma.

Lemma 18

If the partition λ satisfies the 2-step condition and τ ∈Λ is a string of ab-diagrams different from τ⁰, then

$$ \dim Z_{\tau^{0}} > \dim Z_{\tau}. $$

Proof of Proposition 17

We start recalling Proposition 6 and Proposition 16.

As $Z \setminus Z_{\tau ^{0}}$ consists of finitely many strata Z_τ and, by Lemma 18, codim_Z(Z_τ) ≥ 1, we deduce that $Z = \overline {Z_{\tau ^{0}}}$. That implies that Z is a complete intersection variety smooth in codimension 0 and Z is reduced as a scheme. □

We can finally use Lemma 9 and Propositions 10 and 14 to prove Lemma 18.

Proof of Lemma 18

Let μ = π(τ), so in particular μ < λ, as τ≠τ⁰. Therefore Lemma 9 implies

$$ 2 r(\lambda, \mu) \geq c(\lambda, \mu) + q(\lambda, \mu) $$

(19)

and, by combining it with proposition 10 we immediately get

$$ \dim Z_{\tau^{0}} \geq \dim Z_{\tau}. $$

So our concern is reduced to obtain a strict inequality. We will gain the strict inequality either by proving that (19) holds strictly for the pair (λ, μ) or by using proposition 14 with l > 0.

Let c be the first column such that $\widehat \mu _{c} > \widehat \lambda _{c}$ and let $r = \widehat \mu _{c}$. This means that the first c − 1 columns of λ and μ are the same and μ_r > λ_r. On the basis of c and r, we distinguish three cases.

$\widehat \mu _{c} - \widehat \lambda _{c} \geq 2$::

in this case we can use Lemma 9 with the column c to obtain that

$$ 2r(\lambda, \mu) > c(\lambda, \mu) + q(\lambda, \mu). $$

$\widehat \mu _{c} - \widehat \lambda _{c} = 1$, μ_r − λ_r ≥ 2::

we can still use Lemma 9, this time with the row r, to obtain that

$$ 2r(\lambda, \mu) > c(\lambda, \mu) + q(\lambda, \mu). $$

$\widehat \mu _{c} - \widehat \lambda _{c} = 1$, μ_r − λ_r = 1::

in this case we can use Proposition 14 with i = c and l ≥ 1. Indeed, the following two facts are immediately seen: by (15), d_c = (b); the ab-diagram ${\sigma ^{0}_{c}}$ has at least one row equal to a single a, namely the $\widehat \mu _{c}$th row.

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9 Normality Conditions

In this section we prove Theorem 1. We start by proving the necessary condition as a consequence of the works by Ohta and Sekiguchi.

Theorem 19 (the only if part of Theorem 1)

Let λ be a partition of n and suppose that λ does not satisfy the 1-step condition. Then $\overline {C_{\lambda }} \subseteq \mathfrak {p}$ is not normal.

Proof

If λ does not satisfy the 1-step condition, then there is an i such that λ_i ≥ λ_i+ 1 + 2. Let

$$ \mu = (\lambda_{1}, \dots, \lambda_{i-1}, \lambda_{i} - 1, \lambda_{i+1} + 1, \dots, \lambda_{h}) $$

be another partition of n, so that μ < λ and μ is a minimal degeneration of λ (in the same sense of [6, 10]). Let m = λ_i − λ_i+ 1 ≥ 2 and let us define the following two partitions of m:

$$ \begin{array}{@{}rcl@{}} &\lambda^{\prime} = (m), \\ &\mu^{\prime} = (m-1, 1). \end{array} $$

The degeneration $\mu ^{\prime } < \lambda ^{\prime }$ is obtained from μ < λ by removing the common rows and columns (of μ and λ). By Theorem [10, Theorem 2], the singularity C_μ of $\overline {C_{\lambda }}$ is normal if and only if the singularity $C_{\mu ^{\prime }}$ of $\overline {C_{\lambda ^{\prime }}}$ is normal.

Moreover, by construction, $C_{\lambda ^{\prime }}$ is the variety of the principal (or regular) nilpotent elements of $\mathfrak {p}$ (in dimension m); therefore we also have that $\overline {C_{\lambda ^{\prime }}} = \mathcal N(\mathfrak {p})$, that is the cone of the nilpotent elements in $\mathfrak {p}$. As $\mu ^{\prime }$ is the minimal degeneration of $\lambda ^{\prime }$, $C_{\mu ^{\prime }}$ is the variety of subregular nilpotent elements of $\mathfrak {p}$. The Theorem [11, §3.1, Theorem 7]. proves that the singularity $C_{\mu ^{\prime }}$ is not normal in $\mathcal N(\mathfrak {p})$. So this concludes the only if part. □

We turn to the sufficient condition. We will make use of the construction of the variety Z.

Theorem 20 (the if part of Theorem 1)

If the partition λ satisfies the 1-step condition, then Z is a normal variety. In particular $\overline {C_{\lambda }}$ is a normal variety.

As λ satisfies the 1-step condition, also λ satisfies the 2-step condition, so, by proposition 17, Z is a complete intersection variety. At this point the proof of Theorem 20 is a direct consequence of the following Lemma, as in [5, Section 3.7].

Lemma 21

If the partition λ satisfies the 1-step condition, then $\dim (Z \setminus Z_{\tau ^{0}}) \leq \dim Z - 2$.

Proof

Let τ ∈Λ such that τ≠τ⁰. Thus μ := π(τ₁) < λ. We need to prove that

$$ \dim Z_{\tau^{0}} - \dim Z_{\tau} \geq 2. $$

By Lemma 9, r(λ, μ) ≥ c(λ, μ) + q(λ, μ) and hence

$$ 2r(\lambda, \mu) - c(\lambda, \mu) - q(\lambda, \mu) \geq 4x $$

where $x = \frac 14\left (c(\lambda , \mu ) + q(\lambda , \mu )\right )$. By proposition 10, we get

$$ \dim Z_{\tau^{0}} - \dim Z_{\tau} \geq x, $$

so we are done if x > 1.

Let us assume that x ≤ 1. For simplicity of notation, we put c := c(λ, μ) and q := q(λ, μ). We claim that there are only four cases left: q = c = 2 and q = 1, c ≤ 3. Indeed, by (6), λ > μ if and only if both c ≥ 1 and q ≥ 1. By (10), we also have c ≥ q. Since 4x = c + q ≤ 4, then q < 3. Indeed, q ≥ 3 ⇒ c ≥ 3 ⇒ c + q ≥ 6. If q = 2, then c ≤ 2 by the same reasoning. If q = 1, then c = 4x − q ≤ 4 − 1.

We analyze the four cases separately and, in each case, we find an index i such that d_i = (b), the ab-diagram ${\sigma ^{0}_{i}}$ has l rows of length one and l is big enough so that we can conclude the proof by proposition 14 as follows:

$$ \dim Z_{\tau^{0}} - \dim Z_{\tau} \geq x + \frac l2 = \frac 14(c+q) + \frac l2 > 1. $$

q = 2, c = 2:

There are two boxes B, $B^{\prime }$ which are lowered from λ in order to obtain μ. Let $i^{\prime }$ be the column of B in λ and i be the column of B in μ. As c = 2, $i = i^{\prime }-1$. By (15), the list d_i = (b). In ${\sigma ^{0}_{i}}$ the rows with indices $\widehat \mu _{i} = \widehat \lambda _{i} + 1$, $\widehat \lambda _{i}$ and $\widehat \lambda _{i}-1$ have length one. Indeed, since λ is 1-step, λ does not have two different columns with the same height. Thus, ${\sigma ^{0}_{i}}$ has l ≥ 3 rows of length one. We found that $\frac 14(c+q) + \frac l2 \geq \frac 52 > 1$.

q = 1, c = 3:

There is a single box B which is lowered from λ in order to obtain μ. Let $i^{\prime }$ be the column of B in λ and i be the column of B in μ. As c = 3, $i = i^{\prime }-3$. By (15), the list d_i = (b). In ${\sigma ^{0}_{i}}$ the rows with indices $\widehat \mu _{i}$, $\widehat \mu _{i}-1$ have length one. Thus, ${\sigma ^{0}_{i}}$ has l ≥ 2 rows of length one. We found that $\frac 14(c+q) + \frac l2 \geq 2 > 1$.

q = 1, c = 2:

Similar to the previous case, there is a single box B moving from column $i^{\prime }=i+2$ to column i; d_i = (b); the rows with indices $\widehat \mu _{i}$, $\widehat \mu _{i}-1$ have length one; so ${\sigma ^{0}_{i}}$ has l ≥ 2 rows of length one. We found that $\frac 14(c+q) + \frac l2 \geq \frac 74 > 1$.

q = 1, c = 1:

There is a single box B moving from column $i^{\prime }=i+1$ to column i. Similar to the case q = 2, c = 2, d_i = (b), the rows with indices $\widehat \mu _{i} = \widehat \lambda _{i} + 1$, $\widehat \lambda _{i}$ and $\widehat \lambda _{i}-1$ have length one and ${\sigma ^{0}_{i}}$ has l ≥ 3 rows of length one. We found that $\frac 14(c+q) + \frac l2 \geq 2 > 1$.

□