July 22, 2005

THETA CORRESPONDENCE FOR UNITARY GROUPS

1. Notation

  [sec:1]

1.1. [sec:1.1] For a reductive Lie algebra 𝔤, the complexification of a real Lie algebra 𝔤_{0}, let 𝔤 = 𝔨 + 𝔰 be the Cartan decomposition. In the case of u(p, q), we think of the Lie algebra as Hom_{ℂ}[V, V], where V = V^{+} + V^{−}. Then the Cartan decomposition is

𝔨 =(Hom[V^{+}, V^{+}] + Hom[V^{−}, V^{−}])
𝔰 =(Hom[V^{+}, V^{−}] + Hom[V^{−}, V^{+}]). (1.1.1)

The action of K_{c} = GL(V^{+}) × GL(V^{−}) on 𝔰 is given by

(g^{+}, g^{−}) · (A, B) := (g^{+}A(g^{−})^{−1}, g^{−}B(g^{+})^{−1}).  (1.1.2)

A nilpotent orbit is represented by a pair η := (A, B) which satisfies AB (equivalently BA) is nilpotent. We will view η as an element of Hom[V, V] whenever convenient. It can also be represented by a partition (p_{1}, …, p_{k}) and corresponding Young diagram where each entry is labelled ± alternating on the rows. The signs in the first column give the decomposition of the kernel of η according to V^{±}, the signs in the first two columns the decomposition of the kernel of η^{2} according to V^{±}, and so on.

The transpose partition is denoted (p̂_{1}, …, p̂_{t}).

Write C_{η} for the orbit of η under the group K_{c}. The following lemma describes closure relations.

Lemma. [l:1.1] Let η and τ be nilpotent orbits, and assume the partitions have the same number of columns by padding one with zeroes if necessary. Then C_{τ} ⊂ \overline{C}_{η}, if and only if for every i, the number of ± in the largest i columns of τ is greater than or equal to the corresponding number of ± for η.

Alternatively, C_{τ} ⊂ \overline{C}_{η} if and only if there is a sequence of (η_{i}, η_{i+1}) such that

• η_{1} = η, η_{n} = τ, and

• η_{i+1} is obtained from η_{i} as follows. Remove a chunk from the beginning (respectively) end of one row, of η_{i}, and add it to the beginning (end respectively) of the next smaller row for which the result is still the tableau of a nilpotent orbit.

Proof. The necessity of the condition follows from the fact that

dim ker τ_{V^{±}}^{i} ≥ dim ker η_{V^{±}}^{i}.  (1.1.3)

The sufficiency of the condition follows by direct calculation, using the alternate description of the condition. □

1.2. [sec:1.2] We define a sequence of partitions

η_{t}, …, η_{1},  (1.2.1)

where η_{t} = η, and η_{i} is obtained from η_{i+1} by removing the largest column. We construct spaces

(U_{1}^{+}, U_{1}^{−}), …, (U_{t}^{+}, U_{t}^{−}),  (1.2.2)

where each U_{i}^{±} has dimension the number of ± in the tableau of η_{i}. Note that in this situation,

dim U_{i}^{+} ≥ (dim U_{i−1}^{−} − dim U_{i−2}^{−}) + dim U_{i−1}^{+},
dim U_{i}^{−} ≥ (dim U_{i−1}^{+} − dim U_{i−2}^{+}) + dim U_{i−1}^{−},

(1.2.3)

1.3. [sec:1.3] Write U_{i} := U_{i}^{+} + U_{i}^{−}. Write L for Hom . The space

W := L(U_{1}, U_{2}) × L(U_{2}, U_{1}) × ⋯ × L(U_{t−1}, U_{t}) × L(U_{t}, U_{t−1}) (1.3.1)

is symplectic. We choose a Lagrangian subspace

\begin{matrix}L := & L(U_{1}^{+}, U_{2}^{+}) + L(U_{2}^{+}, U_{1}^{−}) + L(U_{1}^{−}, U_{2}^{−}) + L(U_{2}^{−}, U_{1}^{+}) + … \\ + L(U_{t−1}^{+}, U_{t}^{+}) + L(U_{t}^{+}, U_{t−1}^{−}) + L(U_{t−1}^{−}, U_{t}^{−}) + L(U_{t}^{−}, U_{t−1}^{+}) \end{matrix} (1.3.2)

A typical element of this space is

α := (A_{1}^{+}, B_{1}^{+}, A_{1}^{−}, B_{1}^{−}, …, A_{t−1}^{+}, B_{t−1}^{+}, A_{t−1}^{−}, B_{t−1}^{−})  (1.3.3)

1.4. [sec:1.4] Let

N := L(U_{1}^{+}, U_{1}^{−})×L(U_{1}^{−}, U_{1}^{+})×⋯×L(U_{t−1}^{+}, U_{t−1}^{−})×L(U_{t−1}^{−}, U_{t−1}^{+}). (1.4.1)

Then the restriction of the map Φ in [KP] is given by

\begin{matrix}Φ(α) := & (B_{1}^{+}A_{1}^{+}, B_{1}^{−}A_{1}^{−}, B_{2}^{+}A_{2}^{+} − A_{1}^{−}B_{1}^{+}, B_{2}^{−}A_{2}^{−} − A_{1}^{+}B_{1}^{−}, …, \\ B_{t−1}^{+}A_{t−1}^{+} − A_{t−2}^{+}B_{t−2}^{−}, B_{t−1}^{−}A_{t−1}^{−} − A_{t−2}^{−}B_{t−2}^{+}). \end{matrix} (1.4.2)

The variety Z is defined as Φ^{−1}(0). The group K_{c} acts in the obvious way, extending (1.1.2). Consider the map

\begin{matrix}Θ : L ⟶ L(U_{t}^{+}, U_{t}^{−}) × L(U_{t}^{−}, U_{t}^{+}), \\ Θ(α) := (A_{t−1}^{+}B_{t−1}^{−}, A_{t−1}^{−}B_{t−1}^{+}). \end{matrix}  (1.4.3)

2. The variety Z

  [sec:2]

2.1. [sec:2.1] We consider the case when there are just two vector spaces U_{1} = U_{1}^{+} + U_{1}^{−} and U_{2} = U_{2}^{+} + U_{2}^{−}. An element ϵ ∈ L can be viewed as a cyclic quiver

U_{1}^{+} ⟶^{A_{1}^{+}} U_{2}^{+} ⟶^{B_{1}^{+}} U_{1}^{−} ⟶^{A_{1}^{−}} U_{2}^{−} ⟶^{B_{1}^{−}} U_{1}^{+},  (2.1.1)

and we can talk about the powers ϵ^{i}. The orbit structure of L under the group K_{c} is given by partitions/tableaus T where each row is labelled

… |a|b|c|d|a| ….

The number of a’s in the largest i rows is the dimension of the kernel of ϵ^{i} on U_{1}^{+}, and so on.

This is the analogue of the Jordan canonical form for a single nilpotent transformation.

The pairs

\begin{matrix}(B_{1}^{+}A_{1}^{+}, B_{1}^{−}A_{1}^{−}) ∈ L(U_{1}^{+}, U_{1}^{−}) × L(U_{1}^{−}, U_{1}^{+}), \\ (A_{1}^{+}B_{1}^{−}, A_{1}^{−}B_{1}^{+}) ∈ L(U_{2}^{+}, U_{2}^{−}) × L(U_{2}^{−}, U_{2}^{+}) \end{matrix}  (2.1.2)

correspond to tableaus T_{ac}, and T_{bd} respectively, obtained from T by removing the b, d’s, and a, c’s respectively.

2.2. [sec:2.2] In view of (1.2.3), we consider the special case when

dim U_{1}^{+} + dim U_{1}^{−} ≤ dim U_{2}^{+}, dim U_{2}^{−}.  (2.2.1)

Let T ∈ L, and write η_{1} := Φ_{1}(T), η_{2} := Φ_{2}(T). Then

η_{1} = (B_{1}^{+}A_{1}^{+}, B_{1}^{−}A_{1}^{−}),  η_{2} = (B_{2}^{+}A_{2}^{+} − A_{1}^{−}B_{1}^{+}, B_{2}^{−}A_{2}^{−} − A_{1}^{+}B_{1}^{+}). (2.2.2)

The tableau corresponding to η_{1} is denoted T_{bd}, the one corresponding to η_{2}, is denoted T_{ac}. They are obtained by removing b, c or a, d respectively from T. The correspondence with the ± tableaus is a, d ↦ +, and b, c ↦ −.

Proposition. [p:1.4] The image of Z under Θ is \overline{C}_{η}, the closure of the orbit of η.

Proof. We do an induction on t. For t = 1, consider the case (A_{1}^{+}, B_{1}^{+}, A_{1}^{−}, B_{1}^{−}). Let T_{bd} correspond to the case (B_{1}^{+}A_{1}^{+}, B_{1}^{−}A_{1}^{−}) = (0, 0). It follows that T_{ac} is a single column with the number of a’s equal to dim U_{1}^{+}, and number of c’s equal to dim U_{1}^{−}. Let T map to this T_{ac}. It is obtained from T_{ac} by

(1) adding b’s to the beginning of some rows starting with c, or the the end of rows finishing in a,

(2) adding d’s to the beginning of some rows starting with a, or the the end of rows finishing in c,

(3) adding the rest of the b, d’s as rows each formed of a single b, and rows each formed of a single d at the bottom of the tableau.

Let T^{max} be the tableau obtaind from T_{ac} by replacing each row equal to a by dab, each row of the form c by bcd, and adding the rest of the b’s and d’s as single rows at the bottom. This is possible because of (2.2.1). Then lemma 1.1 implies that the orbit given by T_{bd} corresponding to T^{max} contains all others in its closure. Thus (A_{1}^{−}B_{1}^{+}, A_{1}^{+}B_{1}^{−}) is in \overline{C}_{η_{2}}.

Now assume by induction that (A_{i−1}^{−}B_{i−1}^{+}, A_{i−1}^{+}B_{i−1}^{−}) ∈ \overline{C}_{η_{i}}. Let T map to a T_{ac} corresponding to this. Then

(B_{i}^{+}A_{i}^{+}, B_{i}^{−}A_{i}^{−}) = (A_{i−1}^{−}B_{i−1}^{+}, A_{i−1}^{+}B_{i−1}^{−}) ∈ \overline{C}_{η_{i}}.  (2.2.3)

The tableau T is obtained from T_{ac} by (1) inserting a b between every ac, and a d between every ca,

(2) adding b’s to some rows starting with c, and some rows ending in a,

(3) adding d’s to some rows starting with a, and some rows ending in c,

(4) adding the remainig a’s and c’s at the bottom as single rows. Fix a T_{ac}. As before we can construct a T^{max} by applying (1)-(4) above, but in (2), and (3), adding b, d to the beginning and end of all rows. Let η^{max} be the orbit corresponding to T_{bd}^{max}. By lemma 1.1 any other tableau T_{bd} obtained from T mapping to the fixed T_{ac} corresponds to an orbit in the closure of the one for eta^{max}. In turn the largest η^{max} corresponds to η_{i+1}. This means,

(A_{i}^{−}B_{i}^{+}, A_{i}^{+}B_{i}^{−}) ∈ \overline{C}_{η_{i+1}}.  (2.2.4)
□

2.3. [sec:2.2a]

Proposition. [p:2.2a] The dimension of the orbit of ϵ equals the sum of

(1) for each row, x| … |y , the number of x in the later/smaller rows plus the number of y in the later/smaller rows. If a row has size 1, then count twice,

(2) for each row x| … |y , the number of x in the row itself.

Proof. Let U = U_{1}^{+} + U_{2}^{+} + U_{1}^{−} + U_{2}^{−}, and view ϵ as an endomorphism of U. Let ζ be a primitive fourth root of unity, and define a diagonal element

ρ := (1, …, 1, ζ, …, ζ, ζ^{2}, …, ζ^{2}, ζ^{3}, …, ζ^{3})  (2.3.1)

such that its centralizer is K_{c}. Then

[ρ, ϵ] = ζϵ.  (2.3.2)

Conjugate ϵ so that it is in Jordan canonical form, with the blocks in decreasing order, and ρ is still diagonal. Then on each block, ρ is of the form

(r, rζ, rζ^{2}, rζ^{3}, r, …).  (2.3.3)

If r = 1, the corresponding row in the tableau starts with a, if r = ζ it starts with b, and so on. Now ρ acts on the centralizer of ϵ, and the centralizer in 𝔨_{c} is the fixed space under ρ.

Write

U = W_{1} + ⋯ + W_{k},  (2.3.4)

for the block decomposition of ϵ. Suppose i < j and therefore dim W_{i} ≥ dim W_{j}. Then the contributions of Hom[W_{i}, W_{j}] and Hom[W_{j}, W_{i}] to the centralizer are given by the matrices

\begin{matrix}[\begin{matrix}I \\ 0 \end{matrix}] ,  [0 I] \\ I = [\begin{matrix}x_{1} & x_{2} & … & x_{ℓ−1} & x_{ℓ} \\ 0 & ... & ... & ... & x_{ℓ−1} \\ ⋮ & & ... & ... & ⋮ \\ 0 & & & ... & x_{2} \\ 0 & & … & … & x_{1} \end{matrix}] \end{matrix}  (2.3.5)

The first sum comes from counting the off-diagonals centralized by ρ. For i = j, the same argument applies, but there is only one Hom[W_{i}, W_{i}].

□

2.4. [sec:2.3] We fix T_{bd}, and consider all the T ∈ L that map to it.

Definition. We call T_{max} the tableau obtained from T_{bd} as follows:

Insert a, c between all the b, d, and add one a, c to the beginning and to the end of each row. If there are any a, c left, add them as single rows to the bottom.

Proposition. [p:2.3] Suppose T maps to T_{bd}. Then dim C_{T} ≤ dim C_{T_{max}}.

Proof. Any such tableau is obtained from T_{max} by moving some a, c from the beginning or end of rows, and moving them down to the bottom of the tableau. Then in fact

C_{T} ⊂ \overline{C_{T_{max}}}.  (2.4.1)

We also check the dimension statement directly. Number the rows in decreasing order, so that the largest one is the first (equal size rows in arbitrary order). Then the number of a and c in the rows higher than i can only increase compared to the corresponding number for T_{max}. Consider the number n_{i} contributed by row i.

• If row i starts with a, c, then n_{i} can only incresase. Similarly if it ends in a or c.

• Suppose row i for T starts with b (d is similar).

– Each b but the first one in row i is preceded by an a. Thus the number for i is either the same as for T_{max}, or has decreased by – The number of b in the lower rows of T are at most one less than the number of a’s in the corresponding row of T_{max}.

– Each row of T later than i that has fewer b’s than a’s must end in an a in T_{max}. Thus there is an extra count of 1 from the a that has moved from row i. This offsets the drop of 1 from counting the b’s.

– The a that has moved down contributes another 1.

□

2.5. [sec:2.4]

Proposition. Assume that Φ is such that Φ_{bd} satisfies C_{Φ_{bd}} ⊂ \overline{C_{T_{bd}}}. Then

dim C_{Φ} ≤ dim C_{T_{max}}.

Proof. It is enough to consider the case when Φ_{bd} is obtained from T_{bd} by removing the edge (beginning or end) of one row and moving it down to a lower row. Given such a Φ_{bd}, we can construct Φ_{max} the same way as we constructed T_{max} from T_{bd}. By proposition 2.4, dim C_{Φ} ≤ dim C_{Φ_{max}}. Then Φ_{max} is obtained from T_{max} by moving the corresponding edge lower. Thus Φ_{max} ⊂ \overline{C_{T_{max}}}, and the claim follows. □

2.6. [sec:2.6] We prove the analogue of lemma 1.1 for the case of cyclic quivers as in (2.1.1). Undoubtedly a general version is known.

Given a tableau T, denote by A_{i}, B_{i}, C_{i}, D_{i} the number of a, b, c, d respectively, in the largest i rows. We say T′ ≤ T, if T can be obtained from T′ by applying the following procedure repeatedly:

Remove a chunk from the beginning (or end) of a row, and add it to the beginning (or end) of a smaller row.

(2.6.1)

Theorem. C_{T} ⊂ \overline{C_{T′}} if and only if

A_{i} ≥ A′_{i},  B_{i} ≥ B′_{i},  C_{i} ≥ C′_{i},  D_{i} ≥ D′_{i}, for all i.

The condition is clearly necessary, so we will concentrate on the sufficiency. The proof will be given over several sections. It is by induction on

D(T, T′) := ∑(A′_{i} − A_{i}) + (B′_{i} − B_{i}) + (C′_{i} − C_{i}) + (D′_{i} − D_{i}).

If D(T, T′) = 0, then T′ = T, and there is nothing to prove. Otherwise we will find T′′ so that D(T, T′′) < D(T, T′), and T′ is obtained from T′′ by applying (2.6.1). Section 2.6 shows that if T′ and T are related by (2.6.1), then indeed C_{T′} ⊂ \overline{C_{T}} . sections 2.6-2.11 are devoted to constructing T′′.

We first observe that for all i,

\begin{matrix}A_{i+1} − A_{i} ≤ D_{i} − D_{i−1}, \\ B_{i+1} − B_{i} ≤ C_{i} − C_{i−1}, \\ C_{i+1} − C_{i} ≤ B_{i} − B_{i−1}, \\ D_{i+1} − D_{i} ≤ A_{i} − A_{i−1}. \end{matrix}  (2.6.2)

This is just the fact that in each row, a d must be followed by an a, and so on.

2.7. [sec:2.7] We consider the case when ϵ has a T which is formed of just two rows. We choose a Jordan canonical basis, and write the corresponding maps as

\begin{matrix}x_{1} ↦ y_{1} ↦ z_{1} ↦ w_{1} ↦ x_{2} ↦ … ↦ x_{m} ↦ y_{m} ↦ z_{m} ↦ w_{m} ↦ 0 \\ α_{1} ↦ β_{1} ↦ γ_{1} ↦ δ_{1} ↦ … ↦ ρ_{n} ↦ τ_{n} ↦ η_{n} ↦ μ_{n} ↦ 0, \end{matrix} (2.7.1)

where x_{1}, … is a basis of U_{1}, and α_{1}, … a basis of U_{2}. As in section 2.3, let ρ be the semisimple element so that [ρ, ϵ] = ζϵ. The notation x_{i}, y_{i} … reflects the eigenvalue of ρ on the corresponding basis element.

Let

x = [\begin{matrix}I & X \\ 0 & I \end{matrix}]  (2.7.2)

be an element of Hom[U_{1} + U_{2}, U_{1} + U_{2}]. It belongs to K_{c} precisely when X is fixed by ρ. Assume m ≥ n, and let n ≤ k ≤ m. Then there exists x such that xT x^{−1} satisfies

μ_{n} ↦ r_{m−k−1},  (2.7.3)

instead of 0, and the rest the same as before. There exists a diagonal element a_{t} such that

lim_{t→0} a_{t} · r_{m−k} = 0,  a_{t} · s_{l} = s_{l},  (2.7.4)

for all other basis elements s_{l}. Then lim_{t→0} a_{t}T a_{t}^{−1} exists, and is a tableau T′ as in (2.6.2), where a chunk from the end of the larger row has been added to the end of a smaller row. If T′ is in the ζ eigenspace of ρ, x can be chosen to be in K_{c}. If T consists of more than two rows, choose a Jordan canonical basis, and apply the above argument to just two of the rows.

The argument for when a chunk is added from the beginning of the larger row to the beginning of the smaller row is similar; basically interchange m and n.

2.8. [sec:2.8] Assume that the first k columns of T and T′ are the same. Then we can remove them from both, and argue on the remainders. Thus assume that the first column of T′ is longer than that of T, and in particular,

A′_{1} > A_{1}.  (2.8.1)

Let k be such that A′_{i} > A_{i} for i ≤ k, and A′_{k+1} = A_{k+1}. We claim that there is j ≤ k such that

D′_{j} − D′_{j−1} > A′_{j+1} − A′_{j}.  (2.8.2)

If equality were to hold for all j ≤ k, we would conclude

\begin{matrix}D′_{k} = & (D′_{k} − D′_{k−1}) + ⋯ + D′_{1} = \\ (A′_{k+1} − A′_{k}) + ⋯ + (A′_{2} − A′_{1}) = A′_{k+1} − A′_{1} < \\ A_{k+1} − A_{1} ≤ D_{k}, \end{matrix}  (2.8.3)

a contradiction.

2.9. [sec:2.9] Assume there is an “a” in the 1-rows. If there is also a “d”, then let T′′ be the tableau obtained from T′ by

\begin{matrix}a \\ d \end{matrix}  ⟶  ^{a d}  (2.9.1)

Then T′′ is between T and T′, and we can apply induction.

If there is no “d” in the 1-rows, then

D′_{1} = A′_{2} − A′_{1}.  (2.9.2)

If a 2-row ends in a “d”, then make T′′ by

\begin{matrix}c & d \\ a \end{matrix}  ⟶  ^{c d a}  (2.9.3)

Then

A′′_{1} = A′_{1} − 1, A′′_{2} = A′_{2} − 1. (2.9.4)

For T′′ to be a tableau, it is necessary that A′_{2} > A_{2}. But if A′_{2} = A_{2}, then

D′_{1} = A′_{2} − A′_{1} = A_{2} − A′_{1} < A_{2} − A_{1} ≤ D_{1},  (2.9.5)

a contradiction. Thus T′′ is a tableau between T and T′ and so we can apply induction. Suppose no 1-row or 2-row ends in a “d”, but a 3-row does. Then

D′_{1} = A′_{2} − A′_{1},  D′_{2} − D′_{1} = A′_{3} − A′_{2}.  (2.9.6)

We have already seen that A′_{2} > A2. Suppose A′_{3} = A_{3}. Then

D′_{2} = D′_{1} + (A′_{3} − A′_{2}) = A′_{3} − A′_{1} = A_{3} − A′_{1} < A_{3} − A_{1} ≤ D_{2}, (2.9.7)

a contradiction. Thus A′_{3} > A_{3}, and we can form T′′ between T and T′ by moving

\begin{matrix}b & c & d \\ a \end{matrix}  ⟶  ^{b c d a}  (2.9.8)

2.10. [sec:2.10] Assume that A′_{1} > A_{1}, and no 1-row of T′ starts with an “a”. Then

B′_{2} − B′_{1} = A′_{1}.  (2.10.1)

Thus

B′_{2} = B′_{1} + A′_{1} > B′_{1} + A_{1} ≥ B_{1} + A_{1} ≥ B_{2}.  (2.10.2)

Assume that T′ has a 2-row “ab”. If one of the 1-rows of T′ is a “d”, then T′′ obtained by

\begin{matrix}a & b \\ d \end{matrix}  ⟶  ^{d a b}  (2.10.3)

is a tableau between T and T′′, so we can apply induction. If no 1-rows have a “d”, then

D′_{1} = A′_{2} − A′_{1}.  (2.10.4)

¿From (2.10.2), B′_{2} > B_{2}, and A′_{2} > A_{2} (2.9.4 and 2.9.5).

Assume that B′_{3} = B_{3}. We claim that A′_{2} − A′_{1} > B′_{3} − B′_{2}, so that T′ has a 2-row “da”. Indeed, otherwise

D′_{1} = A′_{2} − A′_{1} = B′_{3} − B′_{2} = B_{3} − B′_{2} < B_{3} − B_{2} ≤ A_{2} − A_{1} ≤ D_{1}, (2.10.5)

a contradiction. Then we can form T′′ between T and T′ by

\begin{matrix}d & a \\ a & b \end{matrix}  ⟶  \begin{matrix}d & a & b \\ a \end{matrix}  (2.10.6)

and apply induction.

So assume B′_{3} > B_{3}, and T′ has no 2-row “da”. Thus

B′_{3} − B′_{2} = A′_{2} − A′_{1}.  (2.10.7)

If B′_{4} = B_{4}, then

\begin{aligned}D′_{2} − D′_{1} & = A′_{3} − A′_{2} = B′_{4} − B′_{3} = B_{4} − B′_{3} \\ & < B_{4} − B_{3} ≤ A_{3} − A_{2} ≤ D_{2} − D_{1} ≤ C_{1} \end{aligned}  (2.10.8)

Since also C_{1} ≤ C′_{1}, we conclude that T′ has a 1-row “c”, because

D′_{2} − D′_{1} < C′_{1},  (2.10.9)

and also

D′_{1} > D_{1},  (2.10.10)

because otherwise

D′_{2} = D′_{1} + (D′_{2} − D′_{1}) < D_{1} + (D_{2} − D_{1}) = D_{2}.  (2.10.11)

If T′ has a 3-row “dab”, then we can form T′′ between T and T′ by

\begin{matrix}d & a & b \\ c \end{matrix}  ⟶  ^{c d a b}  (2.10.12)

If not, then B′_{3} − B′_{2} = C′_{4} − C′_{3}. Then C′_{4} > C_{4}, or else the analogue of (2.10.6) starting with C′_{4} − C′_{3} implies D′_{1} < D_{1}, a contradiction. We can continue this argument with D′_{5} and so on, to conclude that we can form T′′ between T and T′ by

\begin{matrix}d & a & … \\ c \end{matrix}  ⟶  ^{c d a …}  (2.10.13)

2.11. [sec:2.11] We continue to assume A′_{1} > A_{1}. We have dealt with the cases when T′ has a 1-row “a”, and a 2-row “ab”. Assume that T′ has no such rows, the first row starting with an “a” is “abc”. This implies

\begin{aligned}A′_{1} & = B′_{2} − B′_{1},  no “a”, \\ B′_{2} − B′_{1} & = C′_{3} − C′_{2},  no “ab”. \end{aligned}  (2.11.1)

We conclude

A′_{1} > A_{1}, B′_{2} > B_{2}, C′_{3} > C_{3}. (2.11.2)

Otherwise, if say C′_{3} = C_{3}, then

A′_{1} = B′_{2} − B′_{1} = C′_{3} − C′_{2} ≤ C_{3} − C_{2} ≤ B_{2} − B_{1} ≤ A_{1},  (2.11.3)

a contradiction. Thus if T′ has a row “d”, “da” or “dab”, we can form T′′ between T and T′ by

\begin{matrix}\begin{matrix}a & b & c \\ d \end{matrix} ⟶ ^{d a b c}, \\ \begin{matrix}a & b & c \\ d & a \end{matrix} ⟶ \begin{matrix}d & a & b & c \\ a \end{matrix}, \\ \begin{matrix}a & b & c \\ d & a & b \end{matrix} ⟶ \begin{matrix}d & a & b & c \\ a & b \end{matrix}. \end{matrix}  (2.11.4)

They give

\begin{matrix}C′′_{3} = C′_{3} − 1, & B′′_{2} = B′_{2} − 1, & A′′_{1} = A′_{1} − 1, \\ C′′_{3} = C′_{3} − 1, & B′′_{2} = B′_{2} − 1, \\ C′′_{3} = C′_{3} − 1, \end{matrix}  (2.11.5)

which are satisfied by (2.11.2). The conclusion follows by induction.

So assume

D′_{1} = A′_{2} − A′_{1},  no “d”,
A′_{2} − A′_{1} = B′_{3} − B′_{2},  no “da”,
B′_{3} − B′_{2} = C′_{4} − C′_{3},  no “dab”.

(2.11.6)

As for (2.11.2), we conclude that

A′_{2} > A_{2},  B′_{3} > B_{3},  C′_{4} > C_{4}.  (2.11.7)

Otherwise, if say C′_{4} = C_{4},

D′_{1} = A′_{2} − A′_{1} = B′_{3} − B′_{2} = C′_{4} − C′_{3} < C_{4} − C_{3} ≤ B_{3} − B_{2} ≤ A_{2} − A_{1} ≤ D_{1},

(2.11.8)

a contradiction (since D′_{1} ≥ D_{1}). We conclude that if T′ has a row “cd”, “cda”, “cdab”, we can form a tableau T′′ between T and T′ by

\begin{matrix}a & b & c \\ c & d \end{matrix} ⟶ ^{c d a b c} ,

\begin{matrix}a & b & c \\ c & d & a \end{matrix} ⟶ \begin{matrix}c & d & a & b & c \\ a \end{matrix} ,  (2.11.9)

\begin{matrix}c & d & a & b \\ a & b & c \end{matrix} ⟶ \begin{matrix}c & d & a & b & c \\ a & b \end{matrix}.

This gives

\begin{matrix}C′′_{3} = C′_{3} − 1,  C′′_{4} = C′_{4} − 1, \\ B′′_{2} = B′_{2} − 1,  B′′_{3} = B′_{3} − 1, \\ A′′_{1} = A′_{1} − 1,  A′′_{2} = A′_{2} − 1, \end{matrix}  (2.11.10)

so (2.11.2) and (2.11.7) imply that T′′ is between T and T′. The conclusion follows by induction.

Thus assume that T′ satisfies

D′_{2} − D′_{1} = A′_{3} − A′_{2},  no “cd”,
A′_{3} − A′_{2} = B′_{4} − B′_{3},  no “cda”,
B′_{4} − B′_{3} = C′_{5} − C′_{4},  no “cdab”.

(2.11.11)

This implies that

C′_{5} > C_{5},  B′_{4} > B_{4},  A′_{3} > A_{3},  (2.11.12)

or

D′_{2} − D′_{1} < C′_{1}  and  D′_{1} > D_{1}.  (2.11.13)

Indeed, if say C′_{5} = C_{5}, then

\begin{aligned}D′_{2} − D′_{1} & = A′_{3} − A′_{2} = B′_{4} − B′_{3} = C′_{5} − C′_{4} < \\ & C_{5} − C_{4} ≤ B_{4} − B_{3} ≤ A_{3} − A_{2} ≤ D_{2} − D_{1} ≤ C_{1} ≤ C′_{1}, \end{aligned} (2.11.14)

and if D′_{1} = D_{1}, we would conclude

D′_{2} = (D′_{2} − D′_{1}) + D′_{1} < (D_{2} − D_{1}) + D_{1} = D_{2},  (2.11.15)

contradicting the assumption D′_{2} ≥ D_{2}.

In case (2.11.13), T′ has rows

\begin{matrix}d & a & … & x_{m} \\ c \end{matrix} .  (2.11.16)

We can form T′′ by

\begin{matrix}d & a & … & x_{m} \\ c \end{matrix}  ⟶  ^{c d a … x_{m}}  (2.11.17)

with dab … x_{m} the smallest row starting with a “d”. Then

D′_{1} = A′_{2} − A′_{1} = ⋯ = X′_{j} − X′_{j−1} = ⋯ = X′_{m} − X′_{m−1}.  (2.11.18)

To have T′′ between T ad T′, we need X′_{j} > X_{j} for all j ≤ m. As for (2.11.7), any equality X′_{j} = X_{j} implies D′_{1} ≤ D_{1}, a contradiction.

In case (2.11.12), if T′ has any row of the form “bcd”, “bcda” or “bcdab”, we can form T′′ by

\begin{matrix}b & c & d \\ a & b & c \end{matrix} ⟶ ^{b c d a b c}

\begin{matrix}b & c & d & a \\ a & b & c \end{matrix} ⟶ \begin{matrix}b & c & d & a & b & c \\ a \end{matrix}  (2.11.19)

\begin{matrix}b & c & d & a & b \\ a & b & c \end{matrix} ⟶ \begin{matrix}b & c & d & a & b & c \\ a & b \end{matrix} .

Then (2.11.12) implies that T′′ is between T and T′, and we can apply induction.

Thus assume that T′ satisfies

D′_{3} − D′_{2} = A′_{4} − A′_{3},  no “bcd”,
A′_{4} − A′_{3} = B′_{5} − B′_{4},  no “bcda”,
B′_{5} − B′_{4} = C′_{6} − C′_{5},  no “bcdab”.

(2.11.20)

Suppose

C′_{6} = C_{6}, B′_{5} = B_{5}, or A′_{4} = A_{4}. (2.11.21)

Then

B′_{1} > D′_{3} − D′_{2} ≥ 0, and D′_{2} > D_{2}.  (2.11.22)

Indeed, if say C′_{6} = C_{6},

D′_{3} − D′_{2} = ⋯ = C′_{6}−C′_{5} <  (2.11.23)

C_{6} − C_{5} ≤ ⋯ ≤ D_{3} − D_{2} ≤ C_{2} − C_{1} ≤ B_{1} ≤ B′_{1}.

Condition D′_{2} > D_{2}, comes from the fact that if D′_{2} = D_{2}, then

D′_{3} = (D′_{3} − D′_{2}) + D′_{2} < (D_{3} − D_{2}) + D_{2} = D_{3}.  (2.11.24)

Condition B′_{1} > D′_{3} −D′_{2} implies that T′ has a row “b” or “bc”. If C′_{2} −C′_{1} = D′_{3} − D′_{2}, there is no “bc”. In addition, C′_{2} − C′_{1} < C_{2} − C_{1}, so C′_{1} > C_{1}. Otherwise

C′_{2} = (C′_{2} − C′_{1}) + C′_{1} < (C_{2} − C_{1}) + C_{1} = C_{2},  (2.11.25)

a contradiction. Thus (still assuming (2.11.21) ), in addition to D′_{2} > D_{2} and B′_{1} > 0,

C′_{2} − C′_{1} = D′_{3} − D′_{2}, so also C′_{1} > C_{1}, B′_{1} > C′_{2} − C′_{1}, (2.11.26)

or

C′_{2} − C′_{1} > D′_{3} − D′_{2}.  (2.11.27)

In case (2.11.26), T′ has rows

\begin{matrix}c & d & a & … & x_{m} \\ b \end{matrix} .  (2.11.28)

THETA CORRESPONDENCE FOR UNITARY GROUPS 13

The argument in case (2.11.13) applies, so that T′′ obtained by

\begin{matrix}c & d & a & b & … & x_{m} \\ b \end{matrix} ⟶ ^{b c d a b … x_{m}}  (2.11.29)

is between T and T′.

In case (2.11.27), T′ has a 2-row “bc”. Suppose D′_{1} = D_{1}, then C′_{1} > 0. Otherwise C_{1} = 0 as well, and

0 = C′_{1} ≥ D′_{2} − D′_{1}, 0 = C_{1} ≥ D_{2} − D_{1} = 0. (2.11.30)

Then

D′_{2} = D′_{1} = D_{1} = D_{2},  (2.11.31)

a contradiction. Thus T′ has two rows

\begin{matrix}c & d & a & b & … & x_{m} \\ b & c \end{matrix} ,  (2.11.32)

and we can apply the argument for (2.11.26).

Thus assume (2.11.27) and D′_{1} > D_{1}. Consider T′′ obtained by

\begin{matrix}d & a & … & x_{m} \\ b & c \end{matrix}  ⟶  ^{b c d a … x_{m}} .  (2.11.33)

Then T′′ satisfies

\begin{matrix}D′′_{1} = D′_{1} − 1, …, X′′_{i} = X′_{i} − 1, …, X′′_{m} = X′_{m} − 1, \\ D′′_{2} = D′_{2} − 1, …, Y′′_{j} = Y′_{j} − 1, …, Y′′_{m+1} = Y′_{m+1} − 1. \end{matrix}  (2.11.34)

If any X′_{i} = X_{i}, then as in (2.11.18), we get D′_{1} ≤ D_{1}, a contradiction to the hypothesis. If on the other hand Y′_{j} = Y_{j}, we get (2.11.14). Thus T′′ is between T and T′.

We have reduced to the case

C′_{6} > C_{6}, B′_{5} > B_{5}, A′_{4} > A_{4}. (2.11.35)

??? The argument continues in a similar fashion. ???

3. Symplectic and Orthogonal groups

  [sec:3]

3.1. [sec:3.1] We consider the pair sp(R) × o(S). Write R = L + L^{⊥}, for the symplectic space, and S = V + W for the orthogonal space. The basis for L is denoted by e’s, the dual basis in L^{⊥} by f. A nilpotent element in 𝔰 ∩ sp is formed of orthogonal blocks. An even sized block is of the form

\begin{matrix}e_{1} ↦ f_{2} ↦ e_{3} ↦ … ↦ e_{n} ↦ f_{n} ↦ e_{n−1} … ↦ f_{1} ↦ 0, \\ e_{1} ↦ f_{2} ↦ e_{3} ↦ … ↦ f_{n} ↦ e_{n} ↦ e_{n−1} … ↦ f_{1} ↦ 0, \end{matrix}  (3.1.1)

depending whether n is odd or even, or the reverse order.

An odd sized block has to come in pairs,

\begin{matrix}e_{1} ↦ f_{2} ↦ e_{3} ↦ … ↦ e_{n} ↦ 0, \\ f_{n} ↦ e_{n−1} ↦ … ↦ f_{1} ↦ 0, \end{matrix}  (3.1.2)

or

\begin{matrix}e_{1} ↦ f_{2} ↦ e_{3} ↦ … ↦ f_{n} ↦ 0, \\ e_{n} ↦ e_{n−1} ↦ … ↦ f_{1} ↦ 0, \end{matrix}  (3.1.3)

depending whether n is even or odd. Take a pair (A, B) ∈ Z mapping to a nilpotent orbit in sp(R). This coincides with the case of the unitary groups via taking the quadruple

A ∈ Hom(L, V), B ∈ Hom( L^{⊥}, W), A^{∗} ∈ Hom(V, L^{⊥}), B^{∗} ∈ Hom(W, L).

(3.1.4)

We write

A : e_{i} ↦ x_{i},  B′ : f_{j} ↦ y_{i},  (3.1.5)

with x_{i} ∈ V, y_{i} ∈ W. Then for an even block, we get

e_{1} ↦ x_{1} ↦ f_{2} ↦ y_{2} … ↦ f_{1} ↦ y_{1} ↦ 0,  (3.1.6)

or the reverse. In case n is even, we find that |x_{2i−1}| = |x_{2i}| = 0, and (x_{2i−1}, x_{2i}) = 1, while |x_{n}| = 1. The rest of the inner products are 0. Similarly, |y_{2i}| = |y_{2i+1}| = 0, (y_{2i}, y_{2i+1}) = 1, the rest of the inner products are 0. There are two cases, y_{1} = 0 and y_{1} ≠ 0, and in either situation, |y_{1}| = 0. In the first case, the string is (3.1.6), and we write it as

abcd … c,  or  cdab … a  (3.1.7)

where (contrary to upq) the last entry on the right represents the kernel. When y_{1} ≠ 0, there is a dual vector y′_{1}, which maps to e_{1}. This corresponds to the string

dabc … d,  or  bcda … b.  (3.1.8)

For the odd blocks, we get pairs of strings

\begin{matrix}a & b & c & d & … & a \\ c & d & a & b & … & c \end{matrix}, ; \begin{matrix}a & b & c & d & … & a & b \\ b & c & d & a & … & d & a \end{matrix} ; \begin{matrix}d & a & b & c & … & d & a \\ c & d & a & b & … & c & d \end{matrix} ; \begin{matrix}d & a & b & c & … & a & b \\ b & c & d & a & … & c & d \end{matrix}  (3.1.9)

References

[KP] H. Kraft, C. Procesi, Closures of conjugacy classes of matrices are normal, Inv. Math., vol. 53, (1979), pp. 227-247