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Nilpotent Cones

Vinoth Nandakumar

An essay submitted in partial fulfillment ofthe requirements for the degree of

B.Sc. (Honours)

Pure MathematicsUniversity of Sydney

August 2010

CONTENTS

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi

Chapter 1. Classifying nilpotent orbits and centralizers of nilpotents inclassical Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1. Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2. Classification of the nilpotent orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3. Describing centralizers of nilpotent elements in type A. . . . . . . . . . . . . . 12

Chapter 2. A resolution of singularities and the closure ordering fornilpotent orbits Oλ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.1. Resolution of singularities in type A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2. Closure ordering in Type A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.3. Resolution of singularities in type C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Chapter 3. The enhanced nilpotent cone and some variations . . . . . . . . . . . 353.1. The enhanced nilpotent cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.2. The exotic nilpotent cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.3. The 2-enhanced nilpotent cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Chapter 4. Springer fibres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524.1. Irreducible components of Springer fibres . . . . . . . . . . . . . . . . . . . . . . . . . 52

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

iii

Introduction

This essay deals with nilpotent cones, in the framework on complex Lie groups.There have been many established connections between the structure theory andrepresentation theory of Lie groups and their nilpotent cones. For instance, theSpringer correspondence relates nilpotent orbits to representations of the Weyl group.More generally, for any representation of a Lie group it is possible to define its null-cone; the nilpotent cone is the nullcone of the adjoint representation.

Chapter 1 begins with a description of the classical Lie algebras in types A, B, Cand D. Then a definition of what the nilpotent cone is, and what nilpotent orbits are,is given. This is followed by a description of the nilpotent orbits in type A, whichis a straightforward consequence of the Jordan form theorem. A description of thenilpotent orbits in type C is then given, which requires more machinery. A descrip-tion of the nilpotent orbits in types B and D are given without proof, since the proofis similar in nature to that of type C. Following this, an explicit description of thecentralizers of nilpotent elements in type A is given.

The second chapter of the essay deals with resolution of singularities of nilpotentorbit closures. Given a nilpotent orbit, its closure will have the structure of a sin-gular algebraic variety. In types A and C, a resolution of singularities is given forthe nilpotent orbit closures; a different technique is required to construct the res-olution in each of these cases. As an application of the resolution of singularitiesin type A, a description of the closure ordering for nilpotent orbits in type A is given.

In the third chapter, we move on to study some variations of nilpotent cones. Intype A, we study the enhanced nilpotent cone, which is the product of a vectorspace with the nilpotent cone of type A. Using a slightly different method to thatoriginally used, we classify the orbits for the action of the general linear group onthe enhanced nilpotent cone. We next examine Kato’s exotic nilpotent cone in typeC, which is the product of a vector space with a variant of the nilpotent cone in typeC. We describe the orbits of the action of the symplectic group on the exotic nilpo-tent cone, and describe part of the proof. We then study the 2-enhanced nilpotentcone in type A, which is the product of two vector spaces with the nilpotent cone oftype A. This is a problem which has not previously been studied, and we describesome partial results about the orbits in this cases.

In the final chapter, we study Springer fibres in type A. Springer fibres are fibresof the resolution of singularities for the full nilpotent cone, which is a special caseof the resolution of singularities of nilpotent orbit closures. Following a paper of

iv

INTRODUCTION v

Spaltenstein, we describe the irreducible components of Springer fibres.

I expect that this essay should be accessible to a fourth year student with a ba-sic knowledge in Lie theory and algebraic geometry. Complete proofs have beengiven, except for the Jacobson-Morozov theorem, and the classification of orbits inthe exotic nilpotent cone.

Acknowledgements

First and foremost, I would like to thank my supervisor, Dr Anthony Henderson.His guidance and help in all facets of my thesis has been priceless. I am sincerelygrateful for the countless hours he has spent, and the patience he has shown, in ex-plaining mathematics to me and checking my work.

My family deserves a profound thank you for all the emotional support that theyhave provided me with throughout this difficult year.

I would also like to thank Rowland Jiang and Clinton Boys for their generous helpwith Latex.

vi

CHAPTER 1

Classifying nilpotent orbits and centralizers of nilpotents inclassical Lie algebras

1.1. Definitions

In the following, we work over the field C for convenience; however, almost all ofthe results obtained will be valid for any algebraically closed field F in characteris-tic 0.

Let g denote a reductive Lie algebra, and let G be a connected Lie group having Liealgebra g. Recall the following definition of the classical Lie groups, and their Liealgebras. In type A, we have the general linear group; in types B and D we havethe special orthogonal group, and in type C we have the symplectic group.

Type A: Here G = GLn(C), g = gln(C). For convenience, we will work withGLn(C) as opposed to SLn(C).

Type B: Here G = SO2n+1(C) = {A ∈ SL2n+1(C) : AtA = I2n+1}, andg = so2n+1(C) = {A ∈ gl2n+1(C) : A+ At = 0}.

Type C: Fix an invertible 2n × 2n matrix J such that J = −J t, so that J spec-ifies a non-degenerate skew-symmetric form 〈·, ·〉 on C2n. Here G = Sp2n(C) ={A ∈ GL2n(C) : AtJA = J}, and g = sp2n(C) = {A ∈ gl2n(C) : JA+AtJ = 0}.

Type D: Here G = SO2n(C) = {A ∈ SL2n(C) : AtA = I2n}, and g = so2n(C) ={A ∈ gl2n(C) : A+ At = 0}.

Proposition 1.1. In each of the above cases, we can define a map, known as theadjoint representation, G→ Aut(g), in which g · x = gxg−1 for g ∈ G, x ∈ g.

Proof. We need to check that this action is well-defined, i.e. given g ∈ G andx ∈ g, gxg−1 ∈ g.

In Type A, the result is clear.

In types B and D, suppose g ∈ SOk(C), so gt = g−1, and x ∈ sok(C).

1

2 1. NILPOTENT ORBITS IN THE CLASSICAL LIE ALGEBRAS

gxg−1 + (gxg−1)t = gxg−1 + (g−1)txtgt = gxg−1 + (gt)−1xtg−1

= gxg−1 + gxtg−1 = g(x+ xt)g−1 = 0

Hence gxg−1 ∈ sok(C), as required.

In Type C, suppose g ∈ Sp2n(C), so gtJg = g, and x ∈ sp2n(C), so Jx+ xtJ = 0.Note gtJg = J implies gtJ = Jg−1, and Jg = (gt)−1J .

J(gxg−1) + (gxg−1)tJ = Jgxg−1 + (gt)−1xtgtJ

= (gt)−1Jxg−1 + (gt)−1xtJg−1

= (gt)−1(Jx+ xtJ)g−1 = 0

Hence gxg−1 ∈ sp2n(C), as required. �

In the definition above of sp2n(C) as {A ∈ gln(C) : JA+AtJ = 0}, an equivalentformulation is {A ∈ gl2n(C) : 〈Av,w〉 = −〈v, Aw〉 ∀v, w ∈ C2n}.

In a Lie algebra g, recall that an element X ∈ g is “nilpotent" if adX is a nilpotentendomorphism of the vector space g. In the case of g being a classical Lie algebra,this is equivalent to X being nilpotent in the sense of matrices.

Definition 1.2. The nilpotent cone of the Lie algebra g, denoted as N , consists ofall nilpotent elements in g.

The group G has a well-defined action on N , since in each of the above cases, Gacts on g by conjugation, and the conjugate of a nilpotent matrix will remain nilpo-tent.

Definition 1.3. The orbits of the group action of G on N are nilpotent orbits. [3]

Thus the nilpotent cone N is a union of a nilpotent orbits, and two nilpotent el-ements are in the same nilpotent orbit iff they are conjugate by an element of G.Alternatively, nilpotent orbits are often referred to as “conjugacy classes", or “ad-joint orbits" in N .

Next, we give an explicit description of nilpotent orbits for the classical Lie alge-bras. Type A (gln) serves as a straightforward example to illustrate the conceptsabove, while type C (sp2n) takes more work. Types B (so2n+1) and type D (so2n)involve similar ideas to type C, and here we state the results without proof.

1.2. Classification of the nilpotent orbits

1.2.1. Type A. In type A, we have that g = gln(C), and the connected Liegroup with Lie algebra g is G = GLn(C). The nilpotent cone N of gln(C) is then

1.2. CLASSIFICATION OF THE NILPOTENT ORBITS 3

simply the set of all nilpotent matrices in gln(C).

The action of the group G on N is by conjugation, so two nilpotent matrices inN lie in the same nilpotent orbit iff they are conjugate under GLn(C). Now theclassification of orbits follows from the Jordan form theorem. Keeping in mind thatnilpotent matrices in GLn(C) have all eigenvalues equal to 0, the Jordan form the-orem says that any matrix x ∈ N is conjugate to precisely one matrix nλ of thefollowing form, where λ1 ≥ λ2 ≥ ... ≥ λk are positive integers whose sum is n:

nλ =

Nλ1

Nλ2

...Nλk

Here the matrixNi denotes the following matrix (where there are i rows and columns):

0 10 1

..0 1

0

Now let us make the following definitions:

Definition 1.4. Let P(n) denote the set of all k-tuples (λ1, ..., λk), with λ1 ≥ λ2 ≥... ≥ λk satisfying

∑ki=1 λi = n.

Definition 1.5. For each partition λ ∈ P(n), let Oλ denote the nilpotent orbitcontaining the matrix nλ.

Using this, it follows that every nilpotent orbit in gln(C) corresponds to a uniqueelement of P(n). Thus we have proven the following classification of nilpotent or-bits in type A:

Proposition 1.6. There is a bijection between the set of nilpotent orbits for the Liealgebra gln, and the set P(n), given by the Jordan canonical form. Specifically, theorbit Oλ corresponding to the partition λ consists of all matrices in gln conjugateto nλ.

If x ∈ Oλ, we refer to λ as the Jordan type of the matrix x. A Jordan basis forx, is a basis {ei,j} of V , with 1 ≤ i ≤ k, 1 ≤ j ≤ λi, such that xei,j = ei,j−1 ifj > 1, and 0 otherwise. The matrix of x with respect to such a basis, ordered sothat ei,j precedes ei′,j′ if i < i′ or i = i′ and j < j′, will then be nλ. It will oftenbe convenient to represent λ as a diagram of boxes, with λ1 boxes in the first row,λ2 boxes in the second row, and so forth. Then the boxes in this diagram can be


thought of as representing the elements of this Jordan basis (with ei,j being the j-thbox in the i-th row), with x moving each box to the box on the left.

1.2.2. Type C. Following Collingwood and McGovern, [3], pg 70-73, we de-scribe the nilpotent orbits in the symplectic Lie algebra sp2n(C). While in the caseof the general linear Lie algebra gln(C), the nilpotent orbits were in 1 − 1 corre-spondence with the set of all partitions of n, here we will find that the nilpotentorbits of the symplectic Lie algebra sp2n(C) are in 1 − 1 correspondence with theset of all partitions of 2n in which every odd part occurs with even multiplicity.In effect, the problem of classifying of nilpotent orbits in sp2n(C) and proving theabove statement is equivalent to proving the following three propositions:

Proposition 1.7. Given any nilpotent element n ∈ sp2n(C), the Jordan type of n isa partition λ ∈ P(2n), satisfying the condition that every odd part in λ occurs witheven multiplicity.

Proposition 1.8. Given any partition λ ∈ P(2n) where every odd part in λ occurswith even multiplicity, there exists a nilpotent element nλ ∈ sp2n(C) such that theJordan type of nλ is λ.

Proposition 1.9. Given any two nilpotent elements nλ, n′λ ∈ sp2n(C) which bothhave Jordan type λ, then we can find g ∈ Sp2n(C) such that g−1nλg = n′λ.

In the proofs of the above propositions, we will use the Jacobson-Morozov Theo-rem, [3], pg 37:

Jacobson-Morozov Theorem: Given a semisimple Lie algebra g, and a nilpotentelementX in g, then there exists a sl2 triple (H,X, Y ) containingX . (Equivalently,we can find H, Y ∈ g such that [X, Y ] = H, [H,X] = 2X, [H,Y ] = −2Y ).

To illustrate the Jacobson-Morozov Theorem, we will prove it for g = gln(C).(Note that here g is not semisimple; nonetheless, the Jacobson-Morozov theorem isstill true in this case). It suffices to find an sl2-triple containing the nilpotent elementnλ. Let µri = i(r + 1− i), and define the following three matrices:

Hr+1 =

r

r − 2· · ·

−r + 2−r

, Xr+1 =

0 1

0 1· · ·

0 10

Yr+1 =

0µr1 0

· · ·0µrr 0


H =

Hλ1

Hλ2

· · ·Hλk

, X =

Xλ1

Xλ2

· · ·Xλk

Y =

Yλ1

Yλ2· · ·

Yλk

It is clear by calculation that {Hr+1, Xr+1, Yr+1} is an sl2-triple, and hence that{H,X, Y } is an sl2-triple, with X = nλ. This proves the Jacobson-Morozov The-orem for gln(C). For future use, we prove the following lemma about sl2-triples insymplectic groups.

Lemma 1.10. Let V be a vector space with even dimension with a chosen symplec-tic form 〈·, ·〉, and suppose {H,X, Y } is an sl2-triple in sp(V ). Let V =

⊕sr=1 Vr

be the decomposition of V into irreducible sl2-modules. If dimVp 6= dimVq, thenVp⊥Vq with respect to the symplectic inner product 〈·, ·〉.

Proof. Let dimVp = m, dimVq = n, and without loss of generality assume thatm < n. Let {v1, v2, · · · , vn} denote a string basis of Vq, so that the actions of Xand Y on Vq are given by the following (see Section 7.2 of [5] for an exposition ofsl2-theory):

Xvi = (n+ 1− i)vi−1 and Xvi = 0; vi =Xn−i

(n− i)!vn

Y vi = ivi+1 and Y vn = 0; vi =Y i−1

(i− 1)!v1

Similarly, let {w1, w2, · · · , wm} denote a string basis of Vp, so that the actions of Xand Y on Vp are given by the following:

Xwi = (m+ 1− i)wi−1 and Xw1 = 0; wi =Xm−i

(m− i)!wm

Y wi = iwi+1 and Xwm = 0; wi =Y i−1

(i− 1)!w1

Note the following two relations:

X iwi =Xm

(m− i)!wm = 0

Y m−i+1wi =1

(i− 1)!Y mw1 = 0


To prove that Vp⊥Vq, it suffices to prove 〈vi, wj〉 = 0 for 1 ≤ i ≤ n, 1 ≤ j ≤ m. Ifn− i ≤ j − 1 and i− 1 ≤ m− j, then n ≤ m, which is a contradiction. So eithern− i > j − 1 or i− 1 > m− j.If n− i > j − 1:

〈vi, wj〉 =⟨

Xn−i

(n− i)!vn, wj

⟩=

(−1)n−i

(n− i)!⟨vn, X

n−iwj⟩

= 0 since n− i ≥ j,Xjwj = 0

If i− 1 > m− j:

〈vi, wj〉 =⟨

Y i−1

(i− 1)!v1, wj

⟩=

(−1)i−1

(i− 1)!

⟨v1, Y

i−1wj⟩

= 0 since i− 1 ≥ m− j + 1, Y m−j+1wj = 0

Thus, in either case, 〈vi, wj〉 = 0, as required. �

Proof. (of Proposition 1.7)Given a nilpotent X of type λ, by the Jacobson-Morozov theorem, we can find ansl2-triple {H,X, Y } containing X . Decompose C2n as the sum of irreducible sl2-modules V1, V2, · · ·Vj , with dim Vi = µi, and without loss of generality, we canassume that µ1 ≥ µ2 ≥ · · · ≥ µj . By sl2-theory, if vi ∈ Vi is a lowest weightvector in Vi, Vi is spanned by {vi, Xvi, · · · , Xµi−1vi}, and Xµivi = 0. Thus the set{Xjvi|0 ≤ j ≤ µi − 1} is a Jordan basis for the nilpotent X , and so X has type µ,and so λ = µ. We have just proven that the dimensions of the irreducible summandsof C2n, considered as an sl2-module, are λ1, · · · , λk (so j = k and dim Vi = λi).Suppose r is odd. Make the following definition:

V r =⊕λi=r

Vi

Let Hr−1 be the highest weight space in V r, i.e. the (r − 1)-weight space. It isclear that dim Hr−1 is the multiplicity of r in the partition λ. Define a form (·, ·) onHr−1 by (v, w) = 〈v, Y r−1w〉. Then the following calculation shows that (·, ·) is askew-symmetric form.

(w, v) =⟨w, Y r−1w

⟩= (−1)r−1

⟨Y r−1w, v

⟩= (−1)r

⟨v, Y r−1w

⟩= −(v, w)


Next we will prove that the form (·, ·) is non-degenerate. Suppose v ∈ Hr−1 and(v, w) = 0 ∀w ∈ Hr−1. As w ranges over Hr−1, Y r−1w will range over the lowestweight space H1−r in V r, by sl2-theory. Thus v is perpendicular to H1−r, with re-spect to 〈·, ·〉.

If s 6= 1 − r, then v will be perpendicular to the s-weight space Hs in V r by thefollowing calculation. Let w ∈ Hs.

〈Hv,w〉 = −〈v,Hw〉〈(r − 1)v, w〉 = −〈v, sw〉

(r − 1 + s) 〈v, w〉 = 0

〈v, w〉 = 0

Since v is perpendicular to Hs for s 6= 1− r, and to H1−r, v is perpendicular to allweight spaces in V r, and so v is perpendicular to V r, which is a direct sum of itsweight spaces. Make the following definition:

V ′r =⊕λi 6=r

Vi

By Lemma 1.10, v is perpendicular to V ′r, and hence v is perpendicular to V r ⊕V ′r = V . Thus v lies in the kernel of the non-degenerate bilinear form 〈·, ·〉, sov = 0.

Hence (·, ·) is a non-degenerate symplectic form on Hr−1, so dim Hr−1 is even.Since dim Hr−1 is equal to multiplicty of r in the partition λ, this proves that everyodd part occurs with even multiplicity in λ. �

Proof. (of Proposition 1.8)First we prove the result for rectangle partitions, λ = (di). Here if d is odd, i mustbe even.

Let V be a vector space of dimension i, with basis {v(1), v(2), · · · , v(i)}. If d is even,define a symmetric form on V via (v(r), v(s)) = δrs. If d is odd and i is even, definea symplectic form on V via (v(r), v(s)) = 0 if |r − s| 6= 1; (v(r), v(r+1)) = 1 if r isodd and 0 otherwise; (v(r+1), v(r)) = −1 if r is odd and 0 otherwise.

We now construct a 2n-dimensional vector space W , an sl2-triple {X, Y,H} forwhich W is a direct summand of i d-dimensional irreducible sl2-submodules, anda symplectic form on W which is invariant under the sl2 action. Let the highestweight space in W have basis {w(1)

d−1, · · · , w(i)d−1}. For 1 ≤ m ≤ d − 1, 1 ≤ j ≤ i,

define w(j)d−1−2m = Ym

m!w

(j)d−1, so that {w(j)

d−1, · · · , w(j)1−d} is a basis for an irreducible

sl2 submodule. The superscript denotes which of the i submodules it lies in, and the


subscript denotes the weight. We adopt the convention that w(i)d+1 = w

(i)−d−1 = 0.

By sl2-theory, the actions of X and Y are given by the following:

Xw(j)d−1−2m = (d−m)w

(j)d+1−2m

Y w(j)d−1−2m = (m+ 1)w

(j)d−3−2m

Define the following form on W :

⟨w

(i)d−2m−1, w

(i′)d−2m′−1

⟩= 0 unless m+m′ = d− 1⟨

w(i)d−2m−1, w

(i′)d−2m′−1

⟩= (−1)m

(d− 1

m

)(v(i), v(i

′)) if m+m′ = d− 1

It is straightforward to check that 〈·, ·〉 is a skew-symmetric form. Ifm+m′ 6= d−1,then

⟨w

(i)d−2m−1, w

(i′)d−2m′−1

⟩= −

⟨wi′

d−2m′−1, wid−2m−1

⟩= 0. If m + m′ = d − 1,

then by construction of the form (·, ·), (v(i), v(i′)) = (−1)d(v(i′), v(i)). Then wecompute:

⟨w

(i)d−2m−1, w

(i′)d−2m′−1

⟩= (−1)m

(d− 1

m

)(v(i), v(i

′))

= (−1)m(d− 1

m′

)(−1)d(v(i′), v(i))

= −(−1)d−1−m(d− 1

m′

)(v(i

′), v(i))

= −(−1)m′(d− 1

m′

)(v(i

′), v(i))

= −⟨w

(i′)d−2m′−1, w

(i)d−2m−1

⟩

Next we check that 〈·, ·〉 is invariant under H .

SinceHw(i′)d−2m−1 = (d−2m−1)w(i′)

d−2m−1, ifm+m′ 6= d−1,⟨Hw

(i)d−2m−1, w

(i′)d−2m′−1

⟩+⟨

w(i)d−2m−1, Hw

(i′)d−2m′−1

⟩= 0 + 0 = 0.

If m+m′ = d− 1:


⟨Hw

(i)d−2m−1, w

(i′)d−2m′−1

⟩+⟨w

(i)d−2m−1, Hw

(i′)d−2m′−1

⟩= (d− 2m− 1 + d− 2m′ − 1)

⟨w

(i)d−2m−1, w

(i′)d−2m′−1

⟩= (2(d− 1)− 2(m+m′))

⟨w

(i)d−2m−1, w

(i′)d−2m′−1

⟩= 0

Next we check that 〈·, ·〉 is invariant under X .⟨Xw

(i)d−2m−3, w

(i′)d−2m′−1

⟩+⟨w

(i)d−2m−3, Xw

(i′)d−2m′−1

⟩=⟨(d−m− 1)w

(i)d−2m−1, w

(i′)d−2m′−1

⟩+⟨w

(i)d−1−2(m+1), (d−m

′)w(i′)d−1−2(m′−1)

⟩If m +m′ 6= d − 1, then both terms in the above sum are clearly 0. Otherwise, ifm+m′ = d− 1:⟨Xw

(i)d−2m−3, w

(i′)d−2m′−1

⟩+⟨w

(i)d−2m−3, Xw

(i′)d−2m′−1

⟩= (d−m− 1)

⟨w

(i)d−2m−1, w

(i′)d−2m′−1

⟩+ (d−m′)

⟨w

(i)d−1−2(m+1), w

(i′)d−1−2(m′−1)

⟩= (d−m− 1)

⟨w

(i)d−2m−1, w

(i′)d−2m′−1

⟩+ (m+ 1)

⟨w

(i)d−1−2(m+1), w

(i′)d−1−2(m′−1)

⟩= (d−m− 1)

(d− 1

m

)(−1)m(v(i), v(i′)) + (m+ 1)

(d− 1

m+ 1

)(−1)m+1(v(i), v(i

′))

= 0, since (d−m− 1)

(d− 1

m

)= (m+ 1)

(d− 1

m+ 1

)

Checking that 〈·, ·〉 is invariant under Y is a similar calculation that we omit.

It is easy to see that the form 〈·, ·〉 is non-degenerate by examining the matrix ofthe form with respect to the basis {w(j)

d−1−2m} of W . For a fixed m, j the expression⟨w

(j)d−1−2m, w

(j′)d−1−2m′

⟩will be non-zero for a unique choice of j′,m′ by construc-

tion of 〈·, ·〉 and (·, ·). The matrix of the form 〈·, ·〉 will have exactly one non-zerovalue in each row and column, and hence be invertible.

Hence X ∈ sp(W ), and X is a nilpotent of type (di) (this follows using the argu-ment at the start of the proof of Proposition 1.7); this proves Proposition 1.9 in thecase of rectangle partitions.

Now suppose λ = (di11 , · · · , dirr ). For 1 ≤ j ≤ r, let Wj be a vector space ofdimension djij , and let Xj be a nilpotent of type dijj in End(Wj) which is invariantfor a non-degenerate symplectic form 〈·, ·〉j on Wj . Let W be the direct sum of


the spaces Wj for 1 ≤ j ≤ r, and let 〈·, ·〉 be the non-degenerate symplectic formobtained as the orthogonal sum of the 〈·, ·〉j . Let X the sum of the matrices {Xj}.Then X ∈ sp(W ), and X will be a nilpotent of type λ, as required. �

Proof. (of Proposition 1.9)Again we first prove the result for the case of rectangle partitions λ = (di). Let Xbe a nilpotent element in sp(W ), for some vector space W of dimension di and asymplectic form 〈·, ·〉. By the Jacobson-Morozov Theorem, we can find an sl2-triple{X, Y,H} in sp(W ). By the argument at the start of the proof of Proposition 1.7,W is the sum of i irreducible sl2-modules of dimension d.

Let Hd−1 denote the highest weight space in W , i.e. the (d − 1)-weight space.Similarly to the proof of Proposition 1.7, define a form (·, ·) on Hd−1 via (v, w) =⟨v, Y

d−1

(d−1)!w⟩

.

(w, v) =

⟨w,

Y d−1

(d− 1)!v

⟩= (−1)d−1

⟨Y d−1

(d− 1)!w, v

⟩= (−1)d

⟨v,

Y d−1

(d− 1)!w

⟩= (−1)d(v, w)

Thus (·, ·) is symmetric if d is even, and symplectic if d is odd. As proved inthe proof of Proposition 1.7, (·, ·) is non-degenerate. In the case of d being even,pick an orthonormal basis of Hd−1 for the form (·, ·), {w(1)

d−1, · · · , w(i)d−1}, so that

(w(i)d−1, w

(j)d−1) = δij . In the case of d being odd, pick a basis ofHd−1, {w(1)

d−1, · · · , w(i)d−1},

so that (w(r)d−1, w

(s)d−1) = 0 if |r − s| 6= 1; (w(r)

d−1, w(r+1)d−1 ) = 1 if r is odd and 0 oth-

erwise; (w(r+1)d−1 , w

(r)d−1) = −1 if r is odd and 0 otherwise.

Definew(j)d−2m−1 =

Ym

m!w

(j)d−1 for 1 ≤ m ≤ d−1, so that {w(j)

d−1, · · · , w(j)1−d} is a string

basis for an irreducible sl2-module. Since the r-weight space is orthogonal under〈·, ·〉 to the s-weight space unless r+s = 0 (proven in the proof of Proposition 1.7):

⟨w

(j)d−2m−1, w

(j)d−2m′−1

⟩= 0 unless m+m′ = d− 1

If m+m′ = d− 1:


⟨w

(j)d−2m−1, w

(j′)d−2m′−1

⟩=

⟨Y m

m!w

(j)d−1, w

(j′)d−2m′−1

⟩=

(−1)m

m!

⟨w

(j)d−1, Y

mw(j′)1−d+2m

⟩=

(−1)m(d−m)(d−m+ 1) · · · (d− 1)

m!

⟨w

(j)d−1, w

(j′)1−d

⟩= (−1)m

(d− 1

m

)(w

(j)d−1, w

(j′)d−1) (since w(j′)

1−d =Y d−1

(d− 1)!w

(j′)d−1)

In the above, Y mw(j′)1−d+2m = (d−m)(d−m+1) · · · (d−1)w(j′)

1−d since Y w(j′)1−d+2m =

(d − m)w(j′)1−d+2m−2, Y w

(j′)1−d+2m−2 = (d − m + 1)w

(j′)1−d+2m−4, and so on until

Y w(j′)1−d+2 = (d− 1)w

(j′)1−d.

Hence we have proven above, that given two nilpotents X and X ′ of type (di) insp(W ), we can construct bases of W , {w(j)

d−2m−1}, {w′(j)d−2m−1}, with 0 ≤ m ≤

d − 1, 1 ≤ j ≤ i, such that the matrix of X with respect to the basis {w(j)d−2m−1} is

the same as the matrix of X ′ with respect to the basis {w′(j)d−2m−1}. Further, we havethat: ⟨

w(j)d−2m−1, w

(j′)d−2m′−1

⟩=⟨w′(j)d−2m−1, w

′(j′)d−2m−1

⟩

Let g be the matrix such that gw(j)d−2m−1 = w

′(j)d−2m−1. By the above condition,

g ∈ Sp(W ); and the matrices X and X ′ are conjugate by g. This proves that anytwo nilpotents X and X ′ of type (d(i)) are conjugate by g, proving the claim in thecase of rectangle partitions.

In the case of an arbitrary partition λ = (di11 , · · · , dirr ), given a nilpotents of typeλ, X ∈ sp(W ), by the Jacobson-Morozov Theorem we can find an sl2-triple{X, Y,H} in sp(W ). Let Wj be the sum of the ij dj-dimensional sl2-submodules.By the Lemma 1.10, Wj is orthogonal under 〈·, ·〉 to Wk if j 6= k. Now we can con-struct bases of the spacesWj , and put them together to construct a basis of the spaceW , such that the matrix of X with respect to the basis, as well as the values of theform 〈·, ·〉 applied to pairs of basis elements, depend solely on the partition λ (andnot on X). Given another nilpotent X ′ of type λ, we can do the same thing. ThenX ′ will be conjugate to X via the matrix g taking the first basis to the second, sincethe matrix of X with respect to the first basis is the same as the matrix of X ′ withrespect to the second, and g ∈ Sp(W ) since the inner products of the correspondingpairs of basis elements is the same. �

Definition 1.11. If λ ∈ P(2n) is a partition in which every odd part occurs witheven multiplicity, then define Oλ to be the set of all nilpotents in sp2n(C) of type λ.


1.2.3. Types B and D. Here we state the results without proofs. The proofs aresimilar in nature to that of type C.

Proposition 1.12. Given any nilpotent X ∈ so2n+1(C), the Jordan type λ of Xsatisfies the condition that every even part in λ occurs with even multiplicity. Con-versely, for any such partition λ of 2n + 1, there exists a nilpotent X ∈ so2n+1(C)with Jordan type λ. Any two such nilpotents X,X ′ ∈ so2n+1(C) are conjugate byan element of the special orthogonal group SO2n+1(C). Thus the nilpotent orbitsin type B are in bijection with partitions of 2n+ 1 in which every even part occurswith even multiplicity.

Proposition 1.13. Given any nilpotent X ∈ so2n(C), the Jordan type λ of X satis-fies the condition that every even part in λ occurs with even multiplicity. Conversely,for any such partition λ of 2n, there exists a nilpotent X ∈ so2n(C) with Jordantype λ. For such a partition λ of 2n, the set of nilpotents of type λ are a single orbitunder the action of SO2n(C); except when λ is a partition containing only evenparts, in which case the set of nilpotents of type λ splits up into two orbits under theaction of SO2n(C).

1.3. Describing centralizers of nilpotent elements in type A

Let λ = (λ1, λ2, ..., λk). Recall that the representative nλ for the nilpotent orbit Oλtakes the following form, where Ni ∈ Matn(C) is defined by Ni =

∑n−1j=1 ej,j+1 (as

in Section 1.2.1).

nλ =

Nλ1

Nλ2

...Nλk

First we compute the centralizer CMn(C)(nλ) of nλ in the matrix algebra Mn(C),then the centralizer CGLn(C)(nλ) in GLn(C).

Proposition 1.14. Given x ∈ Mn(C), let x consist of k2 blocks xi,j , where xi,j hassize λi × λj , as follows.

x =

x11 x1kx21 x2k

...xk1 xkk

Then x ∈ CMn(C)(nλ) precisely when the blocks xi,j take the following form. Ifλi ≥ λj , the entries must satisfy the following (here 1 ≤ k ≤ λi, 1 ≤ l ≤ λj):

1.3. DESCRIBING CENTRALIZERS OF NILPOTENT ELEMENTS IN TYPE A 13

xij,kl = 0 if k > l

xij,kl = xij,1,l−k+1 if k ≤ l

If λi < λj , the entries must satisfy the following (here 1 ≤ k ≤ λi, 1 ≤ l ≤ λj):

xij,kl = 0 if l − k < λj − λixij,kl = xij,1,l−k+1 otherwise

Proof. Since x ∈ CMn(C)(nλ), we have xnλ = nλx:

x11 x1kx21 x2k

...xk1 xkk

Nλ1

Nλ2

...Nλk

=

Nλ1

Nλ2

...Nλk

x11 x1kx21 x2k

...xk1 xkk

Simplifying this expression,

Nλ1x11 Nλ1x1kNλ2x21 Nλ2x2k

· · ·Nλkxk1 Nλkxkk

=

x11Nλ1 x1kNλk

x21Nλ1 x2kNλk

· · ·xk1Nλ1 xkkNλk

Thus we require that Nλixij = xijNλj , for all i, j ∈ {1, · · · , k}.

0 1

0 1...

0 10

xij,11 xij,1λj

...

xij,λi1 xij,λiλj

=

xij,11 xij,1λj

...

xij,λi1 xij,λiλj

0 10 1

...0 1

0

Expanding the above expression gives the following:


xij,21 xij,2λj

...xij,λi1 xij,λiλj0 0

=

0 xij,11 xij,1,λj−1

...

0 xij,λi1 xij,λi,λj−1

(1.15)

Case 1: If λi ≥ λj , equating the entries in (1.15) now gives the following; here1 ≤ k ≤ λi, 1 ≤ l ≤ λj:

xij,kl = 0 if k > l

xij,kl = xij,1,l−k+1 if k ≤ l

To see this, we can illustrate with an example where λi = 4, λj = 3:

xij,21 xij,22 xij,23xij,31 xij,32 xij,33xij,41 xij,42 xij,430 0 0

=

0 xij,11 xij,120 xij,21 xij,220 xij,31 xij,320 xij,41 xij,42

xij,11 = xij,22 = xij,33,xij,12 = xij,23

xij,21 = xij,32 = xij,43 = 0, xij,31 = xij,42 = 0, xij,41 = 0

xij =

xij,11 xij,12 xij,130 xij,11 xij,120 0 xij,110 0 0

Case 2: If λi < λj , equating the entries in (1.15) gives the following (here 1 ≤ k ≤λi, 1 ≤ l ≤ λj):

xij,kl = 0 if l − k < λj − λixij,kl = xij,1,l−k+1 otherwise

To see this, consider the below example where λi = 3, λj = 4:


xij,21 xij,22 xij,23 xij,24xij,31 xij,32 xij,33 xij,340 0 0 0

=

0 xij,11 xij,12 xij,130 xij,21 xij,22 xij,230 xij,31 xij,32 xij,33

xij,11 = xij,22 = xij,33 = 0

xij,21 = xij,32 = 0, xij,31 = 0

xij,12 = xij,23 =xij,34, xij,13 = xij,24

xij =

0 xij,12 xij,13 xij,140 0 xij,12 xij,130 0 0 xij,12

�

Now that we know the structure of the matrices xij which form the matrix x, effec-tively we have computed the centralizer CMn(C)(nλ) of nλ inside the matrix algebraMn(C). As an example of what this centralizer looks like in practice, we examinethe example λ = (3, 22, 1). A typical element x ∈ CMn(C)(nλ) has the followingform:

x11,11 x11,12 x11,13 x12,11 x12,12 x13,11 x13,12 x14,11x11,11 x11,12 x12,11 x13,11

x11,11x21,12 x21,13 x22,11 x22,12 x23,11 x23,12 x24,11

x21,12 x22,11 x23,11x31,12 x31,13 x32,11 x32,12 x33,11 x33,12 x34,11

x31,12 x32,11 x33,11x41,13 x42,12 x43,12 x44,11

Now we will compute the centralizer CGLn(C)(nλ) by describing precisely when amatrix x ∈ CMn(C)(nλ) is invertible.

Proposition 1.16. Let x ∈ CMn(C)(nλ), and let λ = (ab11 , ab22 , · · · , abtt ). Define the

following square matrices, where cs = b1 + · · · + bs, and 0 ≤ s ≤ t − 1. Thenx ∈ CGLn(C)(nλ) precisely when the matrices xs are invertible.

xs+1 =

xcs+1,cs+1,11 · · · xcs+1,cs+bs+1,11...

...xcs+bs+1,cs+1,11 · · · xcs+bs+1,cs+bs+1,11


Proof. We start with the above example λ = (3, 22, 1). We may consider the rowsand columns as being labelled by the set {e1,1, e1,2, e1,3, e2,1, e2,2, e3,1, e3,2, e4,1}.Consider re-labelling the rows and columns instead with the set {e1,1, e2,1, e3,1, e4,1,e1,2, e2,2, e3,2, e1,3}. The matrix now takes the following form:

x11,11 x12,11 x13,11 x14,11 x11,12 x12,12 x13,12 x11,13x22,11 x23,11 x24,11 x21,12 x22,12 x23,12 x21,13x32,11 x33,11 x34,11 x31,12 x32,12 x33,12 x31,13

x44,11 x42,12 x43,12 x41,13x11,11 x12,11 x13,11 x11,12

x22,11 x23,11 x21,12x32,11 x33,11 x31,12

x11,11

The matrix is now block upper-triangular and will be invertible if and only if x11,11 6=0, x44,11 6= 0, and the following matrix is invertible:

(x22,11 x23,11x32,11 x33,11

)

The above example makes it clear that in the general case λ = (ab11 , ab22 , · · · , abtt ),

after re-labeling the rows and columns of the matrix x as indicated above (so thatei,k precedes ej,l either if k < l or if k = l and i < j) to obtain a matrix x′, thefollowing square matrices (as defined above) will occur on the diagonal of x′. Herecs = b1 + · · ·+ bs, where 0 ≤ s ≤ t− 1.

xs+1 =

xcs+1,cs+1,11 · · · xcs+1,cs+bs+1,11...

...xcs+bs+1,cs+1,11 · · · xcs+bs+1,cs+bs+1,11

The matrix xs will occur as times. The entry in the row corresponding to ei,k andcolumn corresponding to ej,l will occur in one of the copies of the matrix xs pre-cisely when k = l and λi = λj . It remains to prove that all the other entries ofthe matrix x′ will lie in the upper half of the matrix, so that x′ will be block upper-triangular with blocks xs; it will then follow that x is invertible precisely when allthe matrices xs are invertible.

First consider an entry xij,kl of the matrix x′ when i ≤ j (so λi ≥ λj). The entryxij,kl lies in the row corresponding to ei,k, and the column corresponding to ej,l. Asshown above, the entry xij,kl can only be non-zero if k ≤ l. If k < l, then ei,k willprecede ej,l in the labelling of the rows and columns in x′, and hence xij,kl will lie inthe upper half of x′. If k = l and i < j, again ei,k will precede ej,l in the labelling ofthe rows and columns, and xij,kl will lie in the upper half of x′. If k = l and i = j,then ei,k = ej,l, and xij,kl will lie on the diagonal of the matrix x′, and will lie in one


of the matrices xs.

Next consider an entry xij,kl of the matrix x′, when i > j (so λi ≤ λj). The entryxij,kl will lie in the row corresponding to ei,k, and the column corresponding to ej,l.As proven above xij,kl can only be nonzero if l − k ≥ λj − λi. If k < l, then ei,kwill precede ej,l in the labelling of the rows and columns in x′, and hence xij,kl willlie in the upper half of x′. If k = l, then 0 ≥ λj − λi ≥ 0, so λi = λj . In this casethe entry xij,kl will occur in one of the matrices xs.

This shows that every non-zero entry in x′ occurs in the upper half, or in of thediagonal blocks xs, proving that x′ is block upper triangular with blocks xs. Itfollows that x is invertible precisely when all the matrices xs are invertible. �

We now compute the dimension of the orbit Oλ. We first compute the dimensionof the centralizer CGLn(C)(nλ) in the following two Lemmas. Given a partition λ,define its transpose partition µ by letting µi be #{j|λj ≥ i}. Diagramatically, thetranspose partition µ is obtained from λ by reflecting λ along the diagonal contain-ing the first box in the first row, the second box in the second row, and so on.

Lemma 1.17. Given a partition λ = (λ1, λ2, · · · , λk) and its transpose partitionµ = (µ1, µ2, · · · , µl), the following identity holds:

k∑j=1

(2j − 1)λj =l∑

j=1

µ2j

Proof. Proof by induction on the number of parts of µ.

Suppose λ1 = λ2 = · · · = λi > λi+1, so µl = i. Consider the partition λ′ =(λ1 − 1, · · · , λi − 1, λi+1, · · · , λk). By induction the result is true for λ′:

(λ1 − 1) + · · ·+ (2i− 1)(λi − 1) + (2i+ 1)λi+1 + · · ·+ (2k − 1)λk =l−1∑j=1

µ2j

Adding µ2l = i2 = 1+3+ · · ·+(2i−1) to each side now gives the required identity

for λ′:

k∑j=1

(2j − 1)λj = λ1 + · · ·+ (2i− 1)λi + · · ·+ (2k − 1)λk

= µ21 + µ2

2 + · · ·+ µ2l

�

Lemma 1.18. The dimension of the centralizer CGLn(C)(nλ) is∑l

j=1 µ2j .


Proof. The dimension of the centralizer is clearly equal to the number of free vari-ables xij,kl in the description of the centralizer. We use the example λ = (3, 22, 1) toillustrate the ideas. In this case, a typical element x of the centralizer CGLn(C)(nλ)takes the following form:

x11,11 x11,12 x11,13 x12,11 x12,12 x13,11 x13,12 x14,11x11,11 x11,12 x12,11 x13,11

x11,11x21,12 x21,13 x22,11 x22,12 x23,11 x23,12 x24,11

x21,12 x22,11 x23,11x31,12 x31,13 x32,11 x32,12 x33,11 x33,12 x34,11

x31,12 x32,11 x33,11x41,13 x42,12 x43,12 x44,11

The matrix x consists of blocks xij for 1 ≤ i, j ≤ k. The number of free variablesin the block xi,j is min(λi, λj). Hence the dimension of the centralizer is equal tothe following sum:

∑1≤i,j≤k

min(λi, λj)

In the above sum, the term λj occurs with multiplicity 2j−1, since min(λi, λj) = λjwhen i ≤ j, so λj = min{λp, λq} when (p, q) ∈ {(1, j), (2, j), · · · , (j, j), (j, j −1), · · · , (j, 1)}. Hence the dimension of the centralizer is equal to

∑kj=1(2j− 1)λj;

the conclusion now follows from Lemma 2.8. �

Given the standard nilpotent nλ ∈ Oλ, clearly Oλ = G.nλ. Hence dim(G.nλ) =dim(G) − dim(Stab(nλ)). Since the stabilizer of nλ in G is simply the centralizerCGLn(C)(nλ), using Lemma 1.18 we now compute:

dim(Oλ) = dim(G)− dim(CGLn(C)(nλ))

= n2 −l∑

i=1

µ2i

CHAPTER 2

A resolution of singularities and the closure ordering fornilpotent orbits Oλ

In this section, given a nilpotent orbit Oλ we describe how to construct a resolu-tion of singularities for the algebraic varietyOλ (hereOλ denotes the closure of thequasi-affine algebraic variety Oλ). As a consequence of this resolution of singu-larities, we derive the closure ordering on the nilpotent orbits Oλ. The resolutionof singularities for the general linear group (type A) and for the symplectic group(type C) are somewhat different in nature.

2.1. Resolution of singularities in type A

Let x ∈ gln(C) be nilpotent, corresponding to a partition λ ∈ P(n). In this section,we will construct a resolution of singularities for the algebraic variety Oλ = Gx,where G = GLn(C).

Consider the following flag of vector spaces inside V ∼= Cn (V being the vectorspace of dimension n on which gln(C) acts):

0 ⊂ ker(x) ⊂ ker(x2) ⊂ · · · ⊂ ker(xm) = V

Define P to be the parabolic subgroup of the Lie group GLn(C) stabilizing theabove flag:

P := {u ∈ GLn(C) : u(kerxi) ⊂ kerxi}

Define p to be the Lie algebra of P , i.e. the parabolic Lie algebra stabilizing theabove flag:

p := {u ∈ gln(C) : u(kerxi) ⊂ kerxi}

Define n to be the Lie algebra moving each vector space in the above flag into theone below:

n := {u ∈ gln(C) : u(kerxi) ⊂ kerxi−1}

One can show that n is the nil-radical of the parabolic Lie algebra P . The belowexample shows the structure of n and p for some specific nilpotent elements:

19

20 2. RESOLUTION OF SINGULARITIES FOR Oλ

Example 2.1. Consider the partition λ = (3, 2, 1). Pick x to be the followingmatrix, of type (3, 2, 1). In this example will compute p and n.

0 1

0 10

0 10

0

Then by calculation, we see the following:

ker(x) =

∗∗∗000

, ker(x2) =

∗∗∗∗∗0

We claim that p and n are given by the following:

p =

∗ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗ ∗

∗ ∗ ∗∗ ∗ ∗

∗

, n =

0 0 0 ∗ ∗ ∗0 0 0 ∗ ∗ ∗0 0 0 ∗ ∗ ∗

0 0 ∗0 0 ∗

0

In the above, the corresponding parabolic subgroup P of GLn(C) will be the inter-section of p with GLn(C), as sets.

To see that p and n have this form, it is clear by inspection that for all u ∈ p,u(ker(x)) ⊂ ker(x), u(ker(x2)) ⊂ ker(x2), and that for all v ∈ n, v(ker(x)) =0, v(ker(x2)) ⊂ ker(x). It remains to see that there are no other matrices in gln(C)satisfying this property. The first column of a matrix u in p is equal to ue1, andhence must lie in ker(x). By a similar argument, the second and third columns ofu must lie in ker(x), the fourth and fifth columns must lie in ker(x2), and the lastcolumn of u must lie in ker(x3). The first column of a matrix v in n is equal tove1 and hence must be 0. By a similar argument, the second and third columns ofv must be 0, the fourth and fifth columns of v must lie in ker(x), and the last col-umn of v must lie in ker(x2). This shows that p and n have the form specified above.

In general, given a nilpotent element x corresponding to a partition λ, the followingresult gives the dimensions of the spaces ker(xi).

Lemma 2.2. If λ = (λ1, · · · , λk), suppose µ = λt is the transpose partition withparts µ = (µ1, · · · , µl). Then for 0 ≤ s ≤ l, dim(ker(xs)) = µ1 + · · ·+µs, and fors > l, dim(ker(xs)) = n.

2.1. RESOLUTION OF SINGULARITIES IN TYPE A 21

Proof. Since x ∈ Oλ, there exists a basis ei,j of V , with 1 ≤ i ≤ k, 1 ≤ j ≤ λi,such that xei,j = ei,j−1 if j ≥ 1, and xei,1 = 0. Then it follows by a quick inductionthat xsei,j = 0 if j ≤ s, and xsei,j = ei,j−s if j > s. We claim that ker(xs) isspanned by {ei,j|j ≤ s}. To prove this, it is clear that the vectors {ei,j|j ≤ s}are linearly independent and lie inside ker(xs). Conversely, suppose a ∈ ker(xs),where a =

∑i,j ai,jei,j . Then:

xsa =∑i,j

ai,jxsei,j

=∑i;j>s

ai,jei,j−s = 0

Hence if j > s, ai,j = 0, so a =∑

i;j≤s ai,jei,j , and a lies in the span of thevectors {ei,j|j ≤ s}. This proves that ker(xs) is spanned by {ei,j|j ≤ s}. It fol-lows that the basis vectors spanning ker(xs) consists of the basis vectors spanningker(xs−1), together with {ei,j|j = s}. The cardinality of the set {ei,j|j = s} isequal to the maximum value of i for which λi ≥ s; this is in turn equal to µs.Thus dim(ker(xs)) − dim(ker(xs−1)) = µs. It then follows by a quick inductionthat dim(ker(xs)) = µ1 + · · · + µs. If s > l, then µl+1 = · · · = µs = 0, sodim(ker(xs)) = µ1 + · · ·+ µl = n.

�

Now we will compute dim p and dim n. From Example 2.1 and Lemma 2.2, it isclear that for a suitable choice of x ∈ Oλ (so that ker(xi) will consist of the vectorsin V with the first µ1 + · · ·+ µi co-ordinates non-zero), p will consists of matricesthat are block upper-triangular, with blocks of sizes µ1, · · · , µl. The subalgebra nof p will consist of those matrices in p with all diagonal blocks equal zero.

dim(p) =∑i≥j

µiµj =1

2

l∑i=1

µ2i +

1

2(

l∑i=1

µ2i + 2

∑i>j

µiµj)

=1

2

l∑i=1

µ2i +

1

2(

l∑i=1

µi)2 =

1

2

l∑i=1

µ2i +

1

2n2

dim(n) = dim(p)−l∑

i=1

µ2i =

1

2n2 − 1

2

l∑i=1

µ2i

Definition 2.3. Let G×P n to be the quotient of the Cartesian product space G× nby the equivalence relation (g, y) ∼ (gp−1, p.y) = (gp−1, pyp−1), where p ∈ P .

Recall that the quotient G/P is the partial flag variety, under the identificationgP → {0 ⊂ g ker(x) ⊂ g ker(x2) ⊂ · · · ⊂ g ker(xl)}:


G/P = {0 ⊂ V1 ⊂ V2 ⊂ · · · ⊂ Vl = V |dim(Vi) = dim(ker(xi))}

The partial flag variety is a projective variety, since there is a natural inclusion ofG/P into the product of the Grassmanians Gra1(V )×Gra2(V )× · · · ×Gral−1

(V )(where ai = dim(ker(xi))), and each Grassmanian is a projective variety.

Proposition 2.4. Define the map φ : G ×P n → G/P × gln(C) by φ(g, y) =(gP, gyg−1). Then the map φ is well-defined and injective, hence giving G×P n thestructure of an algebraic variety (as a closed subvariety ofG/P ×gln(C)). Further,im(φ) = {((Vi), y)|yVi ⊂ Vi−1}.

Proof. To check that φ is well-defined, it suffices to check that φ(g, y) = φ(gp−1, pyp−1).This is a straightforward calculation:

φ(gp−1, pyp−1) = (gp−1P, (gp−1)(pyp−1)(gp−1)−1)

= (gP, (gp−1)(pyp−1)(pg−1))

= (gP, gyg−1) = φ(g, y)

To check that φ is injective, suppose φ(g1, y1) = φ(g2, y2), so that (g1P, g1y1g−11 ) =(g2P, g2y2g

−12 ). Since g1P = g2P , p = g−12 g1 ∈ P . Since g1y1g−11 = g2y2g

−12 ,

g−12 g1y1g−11 g2 = y2 so py1p−1 = y2. Hence (g2, y2) = (g1p

−1, py1p−1) ∼ (g1, y1)

by definition of the equivalence relation in G×P n.

It is clear from the definition of φ that im(φ) = {(gP, y) : g−1yg ∈ n}. Since this isa closed condition, this shows that im(φ) is a closed subvariety of G/P × gln(C).The element gP ∈ G/P corresponds to the partial flag {0 ⊂ g ker(x) ⊂ · · · ⊂g ker(xl)}. Since g−1yg ∈ n, g−1yg ker(xi+1) ⊂ ker(xi), so then yg ker(xi+1) ⊂g ker(xi). Hence any (gP, y) ∈ im(φ) corresponds to an element ((Vi), y) withyVi ⊂ Vi−1. Conversely, given some ((Vi), y) such that yVi ⊂ Vi−1, supposefor each i, Vi = g ker(xi) for some g ∈ G. Then yg ker(xi) ⊂ gker(xi−1), sog−1yg ker(xi) ⊂ ker(xi−1) for each i, and g−1yg ∈ n. Hence ((Vi), y) correspondsto (gP, y) where g−1yg ∈ n, and hence ((Vi), y) lies in im(φ) if yVi ⊂ Vi−1. Thisproves that im(φ) = {((Vi), y)|yVi ⊂ Vi−1}. Since φ is injective, G ×P n can beidentified with this variety.

�

Corollary 2.5. G×P n is a vector bundle over G/P , with fibres isomorphic to n.

Proof. By Proposition 2.4, identifyG×P n with the variety {((Vi), y)| yVi ⊂ Vi−1}.Let p1 denote the projection onto the first factor; p1 is clearly a surjective maponto G/P . Given a fixed element gP = (Vi) ∈ G/P , the fibre p−11 (Vi) is {y ∈gln(C)|yVi ⊂ Vi−1}. This fibre is simply a conjugate of the Lie algebra n, andhence is isomorphic to n as vector spaces. �

2.1. RESOLUTION OF SINGULARITIES IN TYPE A 23

Define the map π : G×Pn→ gln(C), by π(g, y) = gyg−1. As proved in Proposition2.4, π is well-defined. If we instead consider G ×P n as the set {((Vi), y)|yVi ⊂Vi−1}, the map π is projection onto the second factor.

Corollary 2.6. The map π is a projective morphism of varieties.

Proof. By definition, a map π : X → Y is a projective morphism if we can find aprojective variety Z, and a closed embedding φ : X → Z × Y , so that π = p2 ◦ φ,where p2 : Z × Y → Y is the projection onto the second factor. In this case, letZ = G/P be the partial flag variety, and let φ : G ×P n → G/P × gln(C) bethe map from Proposition 2.4. It is clear that Z is a projective variety. Since φ isan injective map (proven in Proposition 2.4), whose image is a closed-subvariety ofG/P ×gln(C) (namely the set {(gP, y)|g−1yg ∈ n}), φ is a closed embedding. It isclear that p2 ◦ φ = π, since p2(φ(g, y)) = p2(gP, gyg

−1) = gyg−1 = π(g, y). Thisproves that π is a projective morphism. �

Now we can state the first main result of this section: the map π is a resolutionof singularities for the orbit closure Oλ. At this point, it is hardly clear the abovemap even surjects onto Oλ; this, among a couple other things are what needs to beverified in order to check that this map is a resolution of singularities.

Theorem 2.7. The map π : G ×P n → gln(C) is a resolution of singularities forthe orbit closure Oλ. Equivalently, the following four statements are true:

(i) The image of the map π in gln(C) is precisely the orbit closure Oλ.

(ii) The map π is injective when restricted to π−1(Oλ). The inverse map, when re-stricted to Oλ, π−1 : Oλ → G×P n is a morphism of algebraic varieties.

(iii) As an algebraic variety, G×P n is smooth and irreducible.

(iv) The map π is a proper morphism of algebraic varieties.

Proof. (of Theorem 2.7 (iii), (iv))The variety G×P n is irreducible, since it is a vector bundle over the partial flag va-riety G/P , and the partial flag variety is an irreducible variety. The variety G×P nis smooth, since it is a vector bundle over the G/P , and G/P is smooth as it is ahomogenous space.

The fact that π is a proper morphism of varieties follows from the fact that it is aprojective morphism of varieties (Corollary 2.6).

�

Proof. (of Theorem 2.7 part (i))The image of the map π contains Oλ, since given n ∈ Oλ, n = gxg−1 for someg ∈ GLn(C), so if we know that x ∈ n, then π(g, x) = n. To verify that x ∈ n,using the definition of n, it suffices to verify that x(ker(xi)) ⊂ (ker(xi−1)). This is


clear, since if u ∈ (ker(xi)), xiu = xi−1(xu) = 0; i.e. xu ∈ (ker(xi−1)).

Since the map π is a projective morphism, the image of π is a closed subvariety ofgln(C). Since the image of π contains Oλ, it follows that the image of π containsOλ. As proven above, the variety G ×P n is an irreducible variety, so its imageunder the morphism φ will be an irreducible variety.

Next, it is clear that the image of the map π is stable under conjugation by elementsof G. If v ∈ im(π), i.e. v = gyg−1 for g ∈ G, y ∈ n, and if w = hvh−1 (i.e. wlies in the same G-orbit as v), it follows that w = h(gyg−1)h−1 = (hg)y(hg)−1, sow ∈ im(π). It is also clear that every element in the image of the map π is nilpotent,since if v = gyg−1 lies in the image of π, y is nilpotent since every element in n isnilpotent, and hence v is nilpotent. This means that the image of the map im(π) isa union of nilpotent orbits of G. However, we already know that im(π) is a closedalgebraic variety, which means that im(π) is a union of some family of nilpotentorbit closures. Since im(π) is an irreducible algebraic variety, it follows that im(π)is a single orbit closure.

Since im(π) contains Oλ, and since the dimension of the image of the morphismπ is at most the dimension of the domain G ×P n, we have the following chain ofinequalities:

dim(Oλ) ≤ dim(im(π)) ≤ dim(G×P n)

Using the fact that the variety G×P n is a vector bundle over the partial flag varietyG/P with fibres isomorphic to n, we can compute its dimension. The dimensionof G/P is equal to dim(G) − dim(P ), where G and P are quasi-affine algebraicvarieties (following the notation of Hartshorne). The dimension of the Lie groupsG and P will be equal to the dimension of the corresponding Lie algebras gln(C)and p. Since we have computed dim(p) and dim(n) previously, we can now computedim(G×P n):

dim(G×P n) = dim(G)− dim(P ) + dim(n)

= dim(gln(C))− dim(p) + dim(n)

= n2 − (1

2n2 +

1

2

l∑i=1

µ2i ) + (

1

2n2 − 1

2

l∑i=1

µ2i )

= n2 −l∑

i=1

µ2i

On the other hand, dim(Oλ) = dim(Oλ) (since the dimension of a closure of analgebraic variety is equal to the dimension of the algebraic variety). As proven in

2.2. CLOSURE ORDERING IN TYPE A 25

Section 1.3, dim(Oλ) = n2 −∑l

i=1 µ2i . Hence dim(Oλ) = n2 −

∑li=1 µ

2i .

Hence dim(G×P n) = dim(Oλ), which means we must have equality in the chainof inequalities dim(Oλ) ≤ dim(im(π)) ≤ dim(G ×P n). Hence im(π) is a singlenilpotent orbit closure containing Oλ, and has the same dimension, so im(π) isequal to Oλ.

�

Proof. (of Theorem 2.7 part (ii))First we need to check that the map π is injective when restriced to π−1(Oλ).Consider G ×P n as being the variety {((Vi), y)|yVi ⊂ Vi−1} and the map π asbeing projection onto the second factor. We are required to prove that given ay ∈ Oλ, there exists a unique flag (Vi) with dim(Vi) = dim(ker(xi)) such thatyVi ⊂ Vi−1. Suppose that (Vi) is such a flag. Since yV1 = 0, V1 ⊂ ker(y); sincey2V2 ⊂ yV1 = 0, V2 ⊂ ker(y2); by a quick induction it follows that Vi ⊂ ker(yi).But dim(Vi) = dim(ker(xi)) = dim(ker(yi)), since x, y ∈ Oλ. Hence Vi = ker(yi).Conversely, it is clear that (ker(yi)) is a flag with the desired properties. This showsthat π is injective when restricted to π−1(Oλ).

Next we need to show that the map π−1 : Oλ → G ×P n is a morphism of vari-eties. From the above paragraph, when restricted to Oλ, π−1(y) = {((ker(yi)), y)}.Hence we must prove that the map α : Oλ → G/P , defined by α(y) = (ker(yi)) isa morphism of varieties.

Let Gx = CGLn(C)(x). We claim that Gx ⊂ P . If v ∈ Gx then vx = xv, and aquick induction shows that vxi = xiv for all i. Then v(ker(xi)) ⊂ ker(xi), sinceif w ∈ ker(xi) for some w ∈ Cn, then xi.w = 0 so xi.(vw) = v(xi.w) = 0, i.e.v.w ∈ ker(xi). Since v(ker(xi)) ⊂ ker(xi) for all i, v ∈ P ; thus Gx ⊂ P . Thismeans that now we can define a morphism of varieties θ : G/Gx → G/P by declar-ing that θ(uGx) = uP .

Since Oλ = Gx, the orbit-stabilizer theorem gives a bijection between Oλ andG/Gx, via uxu−1 → uGx. This bijection is an isomorphism since we are work-ing over C. We claim now that the maps α and θ can be identified under thisbijection. To prove this, under the map α, the element uxu−1 is sent to the flag(ker(uxu−1)i) = (ker(uxiu−1)). Under the map θ, the corresponding element uGx

is sent to uP , which corresponds to the flag (uker(xi)). Thus we need to prove thatker(uxiu−1) = uker(xi). This is a straightforward check: v ∈ ker(uxiu−1) ⇐⇒uxiu−1v = 0 ⇐⇒ xiu−1v = 0 ⇐⇒ u−1v ∈ ker(xi) ⇐⇒ v ∈ uker(xi). Henceα and θ can be identified, proving that α is a morphism of varieties.

�

2.2. Closure ordering in Type A

In this section, given two nilpotent orbitsOλ andOλ′ , we give a criterion forOλ′ ⊂Oλ. Recall the definition of the dominance partial ordering on the set of partitionsP(n) of n:


Definition 2.8. λ ≥ λ′ if for each i, λ1 + · · ·+ λi ≥ λ′1 + · · ·+ λ′i.

Theorem 2.9. Oλ′ ⊂ Oλ precisely when λ′ ≤ λ.

Proof. By Theorem 2.7, Oλ is the image of the π, so y ∈ Oλ if and only if thereexists a flag (Vi) with dimVi = dim(ker(xi)) = µ1 + · · ·+ µi such that yVi ⊂ Vi−1.

Suppose Oλ′ ⊂ Oλ. Pick y ∈ Oλ′; then there exists a flag (Vi) with dimVi =µ1 + · · · + µi such that yVi ⊂ Vi−1. It is clear that yiVi = 0, so Vi ⊂ ker(yi),so dimVi ≤ dim(ker(yi)). If µ′ is the transpose partition to λ′, dim(ker(yi)) =µ′1 + · · · + µ′i, as seen in Proposition 2.2. Hence µ1 + · · · + µi ≤ µ′1 + · · · + µ′ifor each i, so µ ≤ µ′. Since the operation of transposition is an order-reversinginvolution on P(n) (see pg 6-7 of [7]), it follows that λ ≥ λ′.

Conversely, suppose y ∈ Oλ′ , for some λ′ ≤ λ; we will prove that y ∈ Oλ, byconstructing a flag of subspaces (Vi) with dimVi = µ1 + · · · + µi, such that yVi ⊂Vi−1. Let µ′ denote the transpose partition for λ′. To construct this flag of subspaces,pick a Jordan basis for y, and represent it by a diagram for the partition λ′, so thaty moves each box to the box immediately to the left of it. As an example, letλ′ = (42, 22), λ = (5, 3, 2), µ′ = (4, 23), µ = (32, 2, 12). In the below diagram, wehave filled up the partition diagram for λ′ with µ1 1-s, µ2 2-s, µ3 3-s, µ4 4-s and µ5

5-s, so that the numbers strictly increase across the rows. Now let Vi be spanned bythe boxes filled with numbers smaller than or equal to i. Then the spaces Vi willhave the correct dimensions; and will have the property that yVi ⊂ Vi−1, due tostrict increase across rows.

The above example shows that it is sufficient to prove that there exists a way offilling up the diagram for the partition λ′ with µ1 1-s, µ2 2-s, and so on, such thatthe numbers strictly increase across the rows. By transposing the diagram, thisis equivalent to proving that there exists a way of filling up the diagram for thepartition µ′ with µ1 1-s, µ2 2-s, and so on, such that the numbers strictly increasedown the columns. The slightly stronger statement, that there exists a semistandardtableaux of shape µ′ and content µ if µ ≤ µ′ is a standard fact; see pg 26 of Fulton[4]. Here a semistandard tableaux of shape µ′ and content µ is a diagram of thepartition µ′ filled with µ1 1-s, µ2 2-s, and so on, such that the numbers weaklyincrease across rows and strictly increase down columns. �

2.3. RESOLUTION OF SINGULARITIES IN TYPE C 27

2.3. Resolution of singularities in type C

Here we use the theory of the Jacobson-Morozov resolution to construct a resolu-tion of singularities for the nilpotent orbit closure Oλ in the symplectic Lie algebrasp2n(C). Denote G = Sp2n(C), g = sp2n(C). Let X be a nilpotent of type λ inOλ. By the Jacobson-Morozov Theorem, we can find an sl2-triple {H,X, Y ′} ing. Then we can consider g as sl2-module, via the adjoint action. By sl2-theory, wehave the following:

g =⊕a∈Z

ga, where ga = {Y ∈ g|[H,Y ] = aY }

The following calculation shows that [ga, gb] ⊆ ga+b. Let u ∈ ga, v ∈ gb:

[H, [u, v]] = [[H, u], v] + [[v,H], u]

= [au, v] + [−bv, u]= (a+ b)[u, v]

[u, v] ∈ ga+b

Let g≥i =⊕

a≥i ga, and let p = g≥0. Since [ga, gb] ⊆ ga+b, p is closed under the Liebracket, and hence is a Lie sub-algebra of g. Let P be the corresponding connectedLie subgroup of G.

Since there is a natural action of G on a vector space V of dimension 2n, we mayalso consider V as an sl2-module. By the argument at the start of Proposition 1.7,the dimensions of the irreducible sl2-submodules of V are λ1, λ2, · · · , λk. Theweights of V as an sl2-submodule will thus be λ1 − 1, λ1 − 3, · · · , 3 − λ1, 1 −λ1, λ2 − 1, λ2 − 3, · · · , 3− λ2, 1− λ2, · · · . By sl2-theory, we have the following:

V =⊕a∈Z

Va, where Va = {v ∈ V |H · v = av}

The following calculation shows that ga · Vb ⊆ Va+b. Let u ∈ ga, v ∈ Vb, so that[H, u] = au,H · v = bv.

a(u · v) = [H, u] · v = H(u · v)− u(H · v)= H(u · v)− u(bv)= H(u · v)− b(u · v)

H(u · v) = (a+ b)(u · v), so u · v ∈ Va+b


We now construct a partial flag in V as follows:

V≥b =⊕a≥b

Va

0 ⊆ · · · ⊆ V≥2 ⊆ V≥1 ⊆ V≥0 ⊆ V≥−1 ⊆ V≥−2 ⊆ · · · ⊆ V

dimV≥b =∑

1≤i≤k,λi>b

⌈λi − b

2

⌉= db(λ)

We now claim that this partial flag is isotropic, i.e. that V ⊥≥b = V≥1−b. To provethis, first recall from the proof of Proposition 1.7 that the r-weight space Vr willbe orthogonal to the s-weight space Vs unless r + s = 0. This shows that Vr willbe perpendicular to Vs if r ≥ b, s ≥ 1 − b since then r + s ≥ 1 > 0, and henceV≥b is perpendicular to V≥1−b. Further, dimV≥b is equal to the number of weightsin V which are at least b, counted with multiplicity, and dimV≥1−b is equal to thenumber of weights in V which are at least 1 − b, counted with multiplicity. Sincethe weights in V are symmetric about 0, the number of weights in V which are atleast 1 − b is equal to the number of weights in V which are at most b − 1. Sinceevery weight is either at most b − 1 or at least b, and since there are 2n weights intotal, this shows that dimV≥b + dimV≥1−b = 2n. It now follows that V ⊥≥b = V≥1−b.

Proposition 2.10. P is the stabilizer of this partial isotropic flag in Sp2n(C).

Proof. It suffices to prove that the stabilizer of this flag in sp2n(C) is p. To see whythis is true, if A stabilizes this flag, then for each t, exp(tA) will stabilize this flag.Since P is generated is by the one-parameter subgroups exp(tA) as A ranges overp, it follows that P will stabilize this flag.

Conversely we must show that the stabilizer of this flag in Sp2n(C) is not largerthan P ; so suppose that it is P ′, and has Lie algebra p′. Since the stabilizer of anisotropic flag is connected, and the Lie algebra determines the Lie group if the Liegroup is connected, it suffices to prove that p′ ⊆ p, which will imply P ′ ⊆ P andhence P ′ = P . Given any A ∈ p′, for all t, exp(tA) ∈ P ′, so exp(tA) will stabilizethe flag, and so will exp(tA)−1. Defining the norm ||X|| of a matrix to be the max-imum of the absolute values of its entries, recall that the power series for ln(1+X)will converge if ||X|| < 1. Choosing t arbitrarily small so that || exp(tA)− 1|| < 1,then tA = ln(1+exp(tA)−1) will stabilize the flag; henceA will stabilize the flag,so A ∈ p. This shows that p′ ⊆ p; hence it is sufficient to show that the stabilizer ofthe isotropic flag in sp2n(C) is p.

Let q be the stabilizer of the flag in sp2n(C).

q = {Y ∈ g|Y V≥b ⊆ V≥b ∀b}= {Y ∈ g|Y Vb ⊆ V≥b ∀b}


To see the equality of the above two lines, clearly Y V≥b ⊆ V≥b implies Y Vb ⊆ V≥b.Conversely, if Y Vb ⊆ V≥b for all b, then if a ≥ b, Y Va ⊆ V≥a ⊂ V≥b, soY V≥b ⊆ V≥b.

Since giVb ⊆ Vb+i ⊂ V≥b if i ≥ 0, it follows that gi ⊂ q for each i ≥ 0, and thusp ⊆ q. Conversely, suppose x ∈ q, x /∈ p. Let x =

∑i xi, with xi ∈ gi. Since

x /∈ p, for some j < 0, xj 6= 0. Then xjvl 6= 0 for some vl ∈ Vl; we know thatxjvl ∈ Vj+l. Then xvl =

∑i xivl, and xivl ∈ Vi+l; since xjvl 6= 0, and j + l < l, it

follows xvl /∈ V≥l. This contradicts the fact that xVl ⊆ Vl. Hence q ⊆ p, implyingthat p = q, as required. �

Since P is the stabilizer of an isotropic flag, it follows that P is a parabolic sub-group of Sp2n(C). Since [ga, gb] ⊂ ga+b, [p, g≥2] ⊆ g≥2. It follows that P acts byconjugation on g≥2. To see this, given X ∈ p, g≥2 is stable under ad(X), and henceunder exp(ad(X)). But we have that exp(ad(X)) = Ad(exp(X)), so g≥2 is stableunder conjugation by elements of the form exp(X) forX ∈ p. Since P is generatedby elements of the form exp(X) for X ∈ p, it follows that P acts by conjugationon g≥2.

Definition 2.11. Let G×P g≥2 be the quotient of the Cartesian product G× g≥2 bythe equivalence relation (g, y) ∼ (gp−1, p · y) = (gp−1, pyp−1) where p ∈ P .

P is the stabilizer of the isotropic flag {· · · ⊆ V≥1 ⊆ V≥0 ⊆ V≥−1 ⊆ · · · }, andG acts transitively on the set {(Wb)| dimWb = db(λ),W

⊥b = W1−b} (this follows

from the stronger fact that Sp2n(C) acts transitively on the set of flags {0 = V0 ⊂V1 ⊂ · · · ⊂ V2n−1 ⊂ V2n = V |V ⊥i = V2n−i}). Thus under the identificationgP → {· · · ⊆ gV≥1 ⊆ gV≥0 ⊆ gV≥−1 ⊆ · · · }, G/P is the partial flag variety:

G/P = {(Wb)| dimWb = db(λ),W⊥b = W1−b}

The partial flag variety G/P is a projective variety, since there is a natural inclusionofG/P into the product of the Grassmanians

∏bGrdb(λ)(V ), and each Grassmanian

is a projective variety.

Proposition 2.12. Define the map φ : G ×P g≥2 → G/P × g by φ(g, Y ) =(gP, gY g−1). Then φ is well-defined and injective. Then we have that im(φ) ={((Wb), U)|UWb ⊆ Wb+2}. Thus imφ is a closed subvariety of G/P × g, givingG×P g≥2 the structure of an algebraic variety.

Proof. The proof that φ is a well-defined and injective map is the same as in theproof of the Proposition 2.4, and is omitted.

It is clear by definition of φ that im(φ) = {(gP, U)|U ∈ gg≥2g−1}. The conditionthat U ∈ gg≥2g

−1 is a closed condition, and thus im(φ) is a closed subvariety ofG/P × g. Suppose (gP, U) ∈ im(φ), so that U = gY g−1 for some Y ∈ g≥2. Theelement gP ∈ G/P corresponds to the flag (gV≥b), so gY g−1(gV≥b) = gY V≥b ⊆gV≥b+2 (here Y V≥b ⊆ V≥b+2 since Y ∈ g≥2). Hence the (gP, U) corresponds to an


element ((Wb), U) such that UWb ⊆ Wb+2.

Conversely, suppose we have an element (gP, U) = (gP, gY g−1) which corre-sponds to an element ((Wb), U) such that UWb ⊂ Wb+2; we will prove that Y ∈g≥2. Since Wb = gV≥b, we have that gY g−1(gV≥b) ⊂ gV≥b+2, i.e. gY V≥b ⊂gV≥b+2, or Y V≥b ⊂ V≥b+2. In the proof of Proposition 2.10, we proved thatY V≥b ⊆ V≥b for each b implies that Y ∈ g≥0. A similar argument will showthat if Y V≥b ⊂ V≥b+2 for each b, then Y ∈ g≥2. �

Corollary 2.13. G ×P g≥2 is a vector bundle over G/P , with fibres isomorphic tog2.

Proof. By Proposition 2.12, identify G ×P g≥2 with the variety {(gP, U)|U ∈gg≥2g

−1}. Let p1 denote the projection onto the first factor; p1 is clearly a sur-jective map onto G/P . Given a fixed element gP ∈ G/P , the fibre p−11 (gP ) is{U ∈ gg≥2g−1}. This fibre is simply a conjugate of the vector space g≥2, and henceis isomorphic to g≥2 as vector spaces. �

Define the map π : G ×P g≥2 → g by π(g, y) = gyg−1. As proved in Proposition2.12, π is a well-defined map. If we instead considerG×Pg≥2 as {((Wb), U)|UWb ⊆Wb+2}, π is projection onto the second factor. By the exact same argument used inCorollary 2.6, we may prove that π is a projective morphism of varieties. Now westate the main result of this section: the map π is a resolution of singularities for theorbit closure Oλ.

Theorem 2.14. The map π : G ×P g≥2 → g is a resolution of singularities for theorbit closure Oλ. Equivalently, the following four statements are true:

(i) The image of the map π in g is precisely the orbit closure Oλ.

(ii) The map π is injective when restricted to π−1(Oλ). The inverse map, when re-stricted to Oλ, π−1 : Oλ → G×P g≥2 is a morphism of algebraic varieties.

(iii) As an algebraic variety, G×P g≥2 is smooth and irreducible.

(iv) The map π is a proper morphism of algebraic varieties.

Proof. (of Theorem 2.14 (iii), (iv)) The varietyG×P g≥2 is smooth and irreducible,since it is a vector bundle over the partial flag variety G/P , which is smooth andirreducible since it is a homogeneous space.

The fact that π is a proper morphism of varieties follows from the fact that it is aprojective morphism of varieties. �

Proof. (of Theorem 2.14 (i))We first prove that im(π) contains Oλ. Since π is a projective morphism, its imageis a closed subvariety of g, so it suffices to prove that im(π) contains Oλ. Since


X ∈ g2 ⊂ g≥2, X = π(1, X), so X is contained in im(π). If Y ∈ Oλ, thenY = gXg−1 for some g ∈ G, so Y = π(g,X) and so Y is contained in im(π). Thisproves that im(π) contains Oλ.

Next, it is clear that im(π) is stable under conjugation by elements of G. If v ∈im(π), i.e. v = gyg−1 for g ∈ G, y ∈ g≥2, and if w = hvh−1 (i.e. w lies in the sameG-orbit as v), it follows that w = h(gyg−1)h−1 = (hg)y(hg)−1, so w ∈ im(π). Itis also clear that every element of g≥2 is nilpotent. To see this, g = g≥i for some isufficiently small; [g≥2, g≥i] ⊂ [g≥i+2], so if x ∈ g≥2, ad(x)jg≥i ∈ g≥i+2j = 0 if jis picked sufficiently large; this proves that ad(x) is a nilpotent endomorphism of g,and hence x is nilpotent. Since every element in im(π) is conjugate to an element ing≥2, every element in im(π) is nilpotent. Since im(π) is stable under conjugation, itfollows that im(π) is a union of nilpotent orbits. Since im(π) is closed, it is a unionof nilpotent orbit closures. As im(π) is an irreducible variety, it is a single nilpotentorbit closure.

As im(π) contains Oλ, and since the dimension of the image of the morphism πis at most the dimension of its domain G ×P g≥2, we have the following chain ofinequalities:

dim(Oλ) ≤ dim(im(π)) ≤ dim(G×P g≥2)

Since G×P g≥2 is a vector bundle over G/P with fibres isomorphic to g≥2, we maycompute its dimension as follows:

dim(G×P g≥2) = dim(G)− dim(P ) + dim(g≥2)

= dim(g)− dim(p) + dim(g≥2)

= dim(g)− dim(g≥0) + dim(g≥2)

= dim(g)− dim(g0)− dim(g1)

To compute dim(Oλ), first note that dim(Oλ) = dim(Oλ) = dim(G.X) = dim(G)−dim(GX). Here GX denotes the centralizer of X in G, which has the same dimen-sion as its Lie algebra, the centralizer of X in g. To compute this dimension, con-sider g as an sl2-module, under the adjoint action. Then g has a basis of weightvectors for H , with the weights consisting of the strings {µ1 − 1, µ1 − 3, · · · , 3 −µ1, 1 − µ1}, {µ2 − 1, · · · , 1 − µ2}, and so-on (where µi are the dimensions of theirreducible summands of g as an sl2-module). By sl2-theory, X will annihilate thehighest weight vector in each string, and act non-trivially on all other vectors in eachstring. Thus the centralizer of X in g will be spanned by the highest weight vectorsin each string, and its dimension will be equal to the number of weight strings. Eachweight string will either pass through 0 (if the highest weight in the string is even)


or pass through 1 (if the highest weight in the string is odd). Thus the dimension ofthe centralizer of X in g will be equal to dim(g0) + dim(g1). Thus we have:

dim(Oλ) = dim(Oλ)= dim(G)− dim(GX)

= dim(G)− dim(gX)

= dim(G)− dim(g0)− dim(g1)

Hence dim(Oλ) = dim(G ×P g≥2), and we must have equality in the chain ofinequalities dim(Oλ) ≤ dim(im(π)) ≤ dim(G ×P g≥2). Since im(π) is a singlenilpotent orbit closure containing Oλ, and with the same dimension, im(π) = Oλ.

�

Proof. (of Theorem 2.14 (ii))We first need to check that the map π is injective when restricted to π−1(Oλ). Con-sideringG×P g≥2 as the set {((Wb), U)|UWb ⊆ Wb+2} and π as the projection ontothe second factor, we must prove that given U ∈ Oλ, there exists a unique flag (Wb)such that UWb ⊆ Wb+2. It suffices to prove this for U = X; indeed suppose weknow the statement for U = X and suppose for some U = gXg−1 with g ∈ G, wehave two flags (Wb) and (W ′

b) with the required property. Then the flags (g−1Wb)and (g−1W ′

b) both have the required property with respect to X , and must be thesame (since there is only one flag which has the property for X). This then forces(Wb) and (W ′

b) to be the same, proving the statement for U .

So we need to show that if (Wb) is an isotropic flag with XWb ⊆ Wb+2, thenWb = V≥b. We first illustrate with the example λ = (32, 2). The below partitiondiagram shows a basis of V , where X takes each box to the box on its left, and thenumbers represent the weights of the basis vectors. Reading off the diagram, wehave that d3(λ) = 0, d2(λ) = 2, d1(λ) = 3, d0(λ) = 5, d−1(λ) = 6, d−2(λ) = 8.Since XV = XW−2 ⊂ W0, and dimXV = 5, dimW0 = d0(λ) = 5, it follows thatXV = W0; by inspection we also have that XV = V≥0, so W0 = V≥0. Similarly,since X2V = X2W−2 ⊂ W2, but dimX2V = 2 and dimW2 = d2(λ) = 2;by inspection we also have that X2V = V≥2, so W2 = V≥2. Now that we haveW2 = V≥2 and W0 = V≥0, since both flags are isotropic we have W−1 = V≥−1 andW1 = V≥1. This now proves that the two flags are the same, as required.


To show the statement in general, we proceed by induction on the size of the largestpart λ1. The base case is when λ = (1l), for some l even. Here, d0(λ) = l andd1(λ) = 0, so we are dealing with a flag W1 ⊂ W0 with dimW1 = 0, dimW0 = l.This forces W1 = 0 = V≥1,W0 = V = V≥0, proving the statement in this case.

We will first prove that Wλ1−1 = V≥λ1−1 = im(Xλ1−1) (note Wλ1−1 is the small-est non-zero subspace in the flag). Since V = W1−λ1 , applying the fact thatXWb ⊆ Wb+2 λ1 − 1 times, we get that Xλ1−1V ⊆ Wλ1−1. But dimXλ1−1 isequal to the multiplicity of λ1 in λ (this can be seen by looking at how Xλ1−1

acts on a Jordan basis for X); and dimWλ1−1 is equal to the number of weightswhich are greater than or equal to λ1 − 1, which is also equal to the multiplicityof λ1 in λ. This shows that Xλ1−1V = Wλ1−1. The same argument also provesthat Xλ1−1V = V≥λ1−1, and hence Wλ1−1 = V≥λ1−1. Since the flag is isotropic,W⊥λ1−1 = W2−λ1 and V ⊥≥λ1−1 = V≥2−λ1 , and so W2−λ1 = V≥2−λ1 .

Consider V ′ = V≥2−λ1/V≥λ1−1 (as proved in the last paragraph, this is equal toW2−λ1/Wλ1−1). Define a symplectic form on this space as follows:

〈v + V≥λ1−1, v′ + V≥λ1−1〉 = 〈v, v′〉

This form is well-defined since V≥2−λ1 is perpendicular to V≥λ1−1, so v and v′ willbe orthogonal to any vector in V≥λ1−1. This form is clearly skew-symmetric. It isnon-degenerate, since if v + V≥λ1−1 lies in the kernel of the form, then v will be or-thogonal to everything in V≥2−λ1 , so v ∈ V ⊥≥2−λ1 = V≥λ1−1, so then v+V≥λ1−1 = 0.

The endomorphism X will induce a nilpotent endomorphism X ′ in V ′ of type λ′,where λ′ is the partition with parts λ1 − 2, · · · , λr − 2, λr+1, · · · , λk, arranged indecreasing order (where r is the multiplicity of λ1 in λ). To see this, suppose V hasbasis ei,j , where 1 ≤ i ≤ k, 1 ≤ j ≤ λi, such that Xei,j = ei,j−1 if j > 1 and 0otherwise. Then V≥λ1−1 = im(Xλ1−1) will be spanned by {e1,1, · · · , er,1}, whileV≥2−λ1 = im(Xλ1−1)

⊥ will be spanned by all of the ei,j-s excluding {e1,λ1 , · · · , er,λ1}.Thus V ′ is spanned by the images of all ei,j-s satisfying 2 ≤ j ≤ λ1−1 if 1 ≤ i ≤ r,andX acts as before; this shows thatX ′ has type λ′. Consider the two flags inducedby V≥b and Wb in V ′:

0 ⊂ Wλ1−2/V≥λ1−1 ⊂ · · · ⊂ W3−λ1/V≥λ1−1 ⊂ V ′

0 ⊂ V≥λ1−2/V≥λ1−1 ⊂ · · · ⊂ V≥3−λ1/V≥λ1−1 ⊂ V ′

It is easy to check that the above flags are isotropic; we will check that the firstflag is isotropic (the same method will work for the second flag). Given 3 − λ1 ≤b ≤ λ1 − 2, we need to check that (Wb/V≥λ1−1)

⊥ = W1−b/V≥λ1−1. Since W⊥b =

W1−b, the two spaces are orthogonal. It suffices to check that dim(Wb/V≥λ1−1) +dim(W1−b/V≥λ1−1) = dimV ′. This is true because dim(V≥λ1−1) = r, so LHS =


dim(W≥b) + dim(W≥1−b)− 2r = dimV − 2r = dimV ′ = RHS.

Since X(Wb) ⊆ Wb+2, we have X ′(Wb/V≥λ1−1) ⊆ Wb+2/V≥λ1−1, and similarlyX ′(V≥b/V≥λ1−1) ⊆ V≥b+2/V≥λ1−1. By the induction hypothesis applied to X ′, itnow follows that Wb/V≥λ1−1 = V≥b/V≥λ1−1, and hence Wb = V≥b. This gives therequired result, proving that the map π is injective when restricted to π−1(Oλ).

Next we need to prove that the map π−1 : Oλ → G ×P g≥2 is a morphism ofvarieties. Given U = gXg−1 ∈ Oλ, from the above paragraphs, there exists aunique flag (Wb) with UWb ⊆ Wb+2, and it is clear that (gV≥b) is such a flag. Thusπ−1(gXg−1) = ((gV≥b), gXg

−1). To prove that π−1 is a morphism, it suffices toprove that the mapα, defined by α(gXg−1) = (gV≥b) is a morphism of varieties.

Let GX = CSp2n(C)(X). Since Oλ = Gx, the orbit-stabilizer theorem gives a bijec-tion between Oλ and G/GX via uXu−1 → uGX . This bijection is an isomorphismof varieties since we are working over C. We also have that GX ⊂ P ; to see this,given g ∈ GX , then g.(V≥b) will be an isotropic flag withX(gV≥b) ⊂ gV≥b+2. Sincethere is a unique flag with this property, this forces g ∈ P , and henceGX ⊂ P . Thismeans we have a morphism θ : G/GX → G/P , defined via θ(uGX) = uP . Weclaim that θ and α can be identified under the bijection between Oλ and G/GX . Toprove this, under the map α, the element gXg−1 is sent to (gV≥b). Under the map θ,the corresponding element gGX is sent to gP , which corresponds to the flag (gV≥b),as required. This proves that α is a morphism of varieties. �

CHAPTER 3

The enhanced nilpotent cone and some variations

3.1. The enhanced nilpotent cone

The enhanced nilpotent cone for GLn(C) (see Achar-Henderson, [1]), V × N , isdefined to be the product of the vector space V = Cn with the nilpotent cone N ;there is a natural action of GLn(C) on V , which gives a natural action of GLn(C)on V ×N . In this section, we classify the orbits of GLn(C) on V ×N , and provethat the orbits are in 1− 1 correspondence with bi-partitions of n, defined below.

Definition 3.1. An ordered pair of partitions (µ, ν) with |µ| + |ν| = n is called abi-partition of n. The set of all bi-partitions of n will be denoted Q(n). Call (µ, ν)a bi-partition of λ, if µ+ ν = λ.

Definition 3.2. Given a vector v ∈ V , for 1 ≤ j ≤ k, define the j-th “portion"of v to be the vector obtained by considering the λ1 + · · ·+ λj−1 + 1-th, · · · , λ1 +· · ·λj−1 + λj-th coordinates of V .

The below theorem, which re-phrases Proposition 2.3 from [1], is the main Theoremin this section. The proof given here is a variation of the proof given in [1].

Theorem 3.3. The orbits of GLn(C) on V ×N are in one-to-one correspondencewith the set Q(n). The partition µ + ν = λ of n specifies the Jordan type of thenilpotent element n ∈ N . Suppose λ = (λ1, · · · , λk), µ = (µ1, · · · , µk), ν =(ν1, · · · , νk). Then (vµ, nλ) is an orbit representative for the orbit corresponding to(µ, ν), where nλ is as described in section 2.2.1, and vµ is constructed as follows:for each 1 ≤ j ≤ k, the j-th portion has a 1 in the µj-th coordinate, and 0-selsewhere.

The problem of computing the orbits of GLn(C) on V × N is equivalent to theproblem of computing the orbits of CGLn(C)(nλ) on V . To see this, if two elements(v1, n1), (v2, n2) ∈ V ×N are in the same GLn(C)-orbit, then n1 and n2 are nilpo-tent elements with the same Jordan type, say corresponding to the partition λ. Thenit suffices to check when (v1, nλ), (v2, nλ) are in the same GLn(C)-orbit. This hap-pens when there exists a g with gv1 = v2 with gnλg−1 = nλ, i.e. if v1, v2 lie in thesame CGLn(C)(nλ)-orbit.

Example 3.4. Let λ = (3, 2, 2). A typical element g ∈ CGL7(C)(nλ) takes thefollowing form, using the structure of the centralizer CGLn(C)(nλ) as described inSection 2.3.1. In the case of the bi-partition (µ, ν) = ((2, 1, 1), (1, 1, 1)), vµ will beas follows.

35

36 3. THE ENHANCED NILPOTENT CONE AND SOME VARIATIONS

g =

λ1 λ2 λ3 λ4 λ5 λ6 λ7λ1 λ2 λ4 λ6

λ1λ8 λ9 a11 λ12 a12 λ13

λ8 a11 a12λ10 λ11 a21 λ14 a22 λ15

λ10 a21 a22

, vµ =

0101010

gvµ =

λ2 + λ4 + λ6λ10

λ8 + a11 + a120

λ10 + a21 + a220

Definition 3.5. Given a vector v, say that it is of type (α, β), where α = (α1, · · · , αk),β = (β1, · · · , βk) and αi + βi = λi for all 1 ≤ i ≤ k, if the maximal t such thatthe t-th component in the i-th portion is non-zero, is αi for each 1 ≤ i ≤ k. LetWα be the set of vectors of type (α, β). Let vα be constructed as follows: for each1 ≤ j ≤ k, the j-th portion has a 1 in the αj-th coordinate, and 0-s elsewhere.

Definition 3.6. Define the ordering >̇ on pairs of compositions (α, β) such thatα+β = λ, by declaring (α, β)>̇(α′, β′) iff αj ≥ α′j for all 1 ≤ j ≤ k. In particularthis ordering restricts to bi-partitions λ. Define a map φ from pairs of compositionsadding up to λ, to bi-partitions of λ as follows: let φ(α, β) be the unique minimalbi-partition lying above the pair of compositions (α, β).

In the above definition, it is not immediately clear why there must exist a uniqueminimal bi-partition above the pair of compositions (α, β). We justify this in thefollowing proposition, by explicitly constructing φ(α, β) in terms of (α, β).

Proposition 3.7. The unique minimal bi-partitition (γ, δ) = φ(α, β) lying abovethe pair (α, β) is given by the following:

γi = max({αj | j ≥ i}∪{λi−βj | j < i}), δi = min({λi−αj | j ≥ i}∪{βj | j < i})Proof. In order to prove the proposition, we must check that:

a) (γ, δ) ∈ Q(n) and γ + δ = λ,b) (α, β)<̇(γ, δ)c) If (α, β)<̇(γ′, δ′) for some (γ′, δ′) ∈ Q(n) with γ′+δ′ = λ, then (γ, δ)<̇(γ′, δ′).

To check (a), it is clear from the definitions that γ + δ = λ, so it suffices to checkthat γi ≥ γi+1 and δi ≥ δi+1 for all 1 ≤ i ≤ k− 1. To show that γi ≥ γi+1, we mustshow that

max{αi, · · · , αk, λi−β1, · · · , λi−βi−1} ≥ max{αi+1, · · · , αk, λi+1−β1, · · · , λi+1−βi}

3.1. THE ENHANCED NILPOTENT CONE 37

To do this, for every element u in the set {αi+1, · · · , αk, λi+1− β1, · · · , λi+1− βi},we will show that there exists some element u′ in the set {αi, αi+1, · · · , αk, λi −β1, · · · , λi − βi−1} with u′ ≥ u. If u = αj for some i + 1 ≤ j ≤ k, we cansimply pick u′ = u. If u = λi+1 − βj for some 1 ≤ j ≤ i − 1, then we can picku′ = λi−βj . If u = λi+1−βi, then we can pick u′ = αi (since λi+1 ≤ λi = βi+αi).This concludes the proof that γi ≥ γi+1. To show that δi ≥ δi+1, we must show that

min{λi−αi, · · · , λi−αk, β1, · · · , βi−1} ≥ min{λi+1−αi+1, · · · , λi+1−αk, β1, · · · , βi}

To do this we will prove that for any v in the set {λi−αi, · · · , λi−αk, β1, · · · , βi−1},we can find a v′ in the set {λi+1 − αi+1, · · · , λi+1 − αk, β1, · · · , βi} with v ≥ v′.If v = λi − αj with i + 1 ≤ j ≤ k, then we can simply pick v′ = λi+1 − αj . Ifv = λi − αi = βi, we can pick v′ = βi. If v = βj for 1 ≤ j ≤ i − 1, we can pickv′ = βj . This concludes the proof that δi ≥ δi+1, and hence that of (a).

To check (b), it suffices to show that αi ≤ γi for all 1 ≤ i ≤ k. This follows fromthe definition γi = max{αi, αi+1, · · · , αk, λi − β1, · · · , λi − βi−1}.

To check (c), we must show that (γ, δ)<̇(γ′, δ′), i.e. that

γ′i ≥ γi = max{αi, αi+1, · · · , αk, λi − β1, · · · , λi − βi−1}

Thus we must first show that γ′i ≥ αj for i ≤ j ≤ k. This is true because γ′i ≥ γ′j ≥αj . We must also show that γ′i ≥ λi − βj for 1 ≤ j ≤ i − 1. This is equivalent toshowing that βj ≥ λi−γ′i = δ′i. The reason that this is true is because βj ≥ δ′j ≥ δ′i,(where βj ≥ δ′j is true since γ′j ≥ αj and γ′j + δ′j = αj + βj = λj). �

Definition 3.8. Let Vµ be the set of vectors v ∈ V such that the type (α, β) of vsatisfies φ(α, β) = (µ, ν).

It is clear that vµ ∈ Wµ ⊂ Vµ. We claim that Vµ = CGLn(C)(nλ).vµ. This is suf-ficient to prove the Theorem 3.3, since V is the disjoint union of all Vµ, as (µ, ν)ranges overQ(n). The fact Vµ is a single CGLn(C)(nλ)-orbit is a consequence of thefollowing three Lemmas. The reason for this is that Lemma 3.9 implies that Vµ is aunion of CGLn(C)(nλ)-orbits on V , while Lemma 3.10 applied to α = µ, along withLemma 3.11, imply that CGLn(C)(nλ) acts transitively on Vµ.

Lemma 3.9. The set Vµ is stable under the action of CGLn(C)(nλ).

Lemma 3.10. Any vector in Wα can be mapped to any other vector in Wα by someelement of CGLn(C)(nλ).

Lemma 3.11. Given a vector v of type (α, β), there exists another v′ of type φ(α, β)in the same CGLn(C)(nλ)-orbit as v.

Proof. (of Lemma 3.10)It suffices to show that vα can be mapped to any other vector in Wα (since it is clearthat vα ∈ Wα). To illustrate the idea behind this proof, we revisit Example 3.4 withλ = (3, 2, 2), α = (2, 1, 1). Consider the following choice of g ∈ CGLn(C)(nλ):


g =

λ1 λ2 λ3λ1 λ2

λ1a11 λ12

a11a22 λ15

a22

, vα =

0101010

, gvα =

λ2λ10a110a220

The only constraints on the matrix g ∈ CGLn(C)(nλ) above is that λ1, a11, a12 6=0, so the image gvα ranges across all vectors of the form above subject to thisconstraint. But this is precisely how Wα is defined, concluding the proof in the caseof this example. It is clear how to generalize this proof for λ = (3, 2, 2) to arbitraryλ = (λ1, · · · , λk). �

Proof. (of Lemma 3.11)If φ(α, β) = (µ, ν), we shall prove that vα is in the same orbit as some vectorv′ ∈ Vµ; this suffices since by Lemma 3.10 any vector of type (α, β) is in the sameCGLn(C)(nλ)-orbit as vα. We will first prove that there exists g ∈ CMn(C)(nλ) suchthat gvα ∈ Wµ.

By Proposition 3.7, µi = max({αj|j ≥ i} ∪ {λi − βj|j < i}). We will show that,for any s ∈ {αj|j ≥ i} ∪ {λi − βj|j ≤ i}, we can choose g, so that gvα has anon-zero value in the s-th co-ordinate in the i-th portion, and all higher entries inthe i-th portion are 0.

If s = αj , for some j ≥ i, pick any g with a non-zero entry in the αj-th row and αj-th column of the (i, j)-th block of g. Since j ≥ i, λi ≥ λj , so this will be possible,since it permissible to have a non-zero entry in the r-th row and s-th column of the(i, j)-th block of g when r ≤ s in this case, by Proposition 1.14. We stipulate alsothat for all j′, all other entries in the (i, j′)-th blocks of g, apart from those forcedto be equal to this entry, be 0. It follows then that gvα will have a non-zero entry inthe αj-th coordinate of the i-th portion, as required. The fact that all higher entriesin the i-th portion are 0 follows from the fact that all entries higher than αj in thej-th portion of vα are 0.

Otherwise if s = λi − βj for some j ≤ i, pick a g with a non-zero entry in the(λi−βj)-th row and αj-th column of the (i, j)-th block of g. Since j ≤ i, λi ≤ λj , itis permissible to have a non-zero entry in the r-th row and s-th column of the (i, j)-th block only if r−s ≤ λi−λj; thus this is permissible since λi−βj−αj = λi−λj .We stipulate also that all other entries in the (i, j′)-th blocks of g, apart from thosealready forced to be equal to this entry, be 0. It follows then that gvα will have anon-zero entry in the (λi − βj)-th coordinate of the i-th portion, as required. Thefact that all higher entries in the i-th portion are 0 follows from the fact that all en-tries higher than αj in the j-th portion of vα are 0.

3.1. THE ENHANCED NILPOTENT CONE 39

In the above paragraphs, we have shown by imposing certain restrictions on thecoordinates in the (i, j)-th blocks of g, for fixed i, we can ensure that gvα has anon-zero value in the µi-th co-ordinate in the i-th portion, and all higher entries inthe i-th portion are 0. Since there are no dependencies between the entries in the(i, j)-th block and (i′, j′)-th block of g unless i = i′, j = j′, we can impose all ofthese restrictions simultaneously, and find a g so that gvα will have a non-zero valuein the µi-th coordinate of the i-th portion, and zero values in all higher co-ordinatesfor each i. Equivalently, we have found a g so that gvα has type (µ, ν).

Now note that g + aI ∈ CMn(C)(nλ), for all a ∈ C, and (g + aI)(vα) = gvα + avα.It is clear that for all a with a finite number of exceptions, gvα + avα will have type(µ, ν), and that for all a with a finite number of exceptions, g+aI will be invertible.Taking a value of a outside the union of these two finite sets of exceptions, g+aI ∈CGLn(C)(nλ) will have the required property.

�

Proof. (of Lemma 3.9)The proof of Lemma 3.9 uses Lemmas 3.10 and Lemma 3.11. Suppose that Vµ is notstable under the action ofCGLn(C)(nλ); i.e. suppose that for given some v1 ∈ Vµ1 forsome bi-partition (µ1, ν1), we can find some v2 ∈ Vµ2 for some bi-partition (µ2, ν2)with v1, v2 being in the same CGLn(C)(nλ) orbit. Then by Lemmas 3.10 and 3.11,v1 is in the same orbit as vµ1 , and v2 is in the same orbit as vµ2 . Thus vµ1 and vµ2are in the same orbit; we will now show that this is not possible.

We claim that for g ∈ CGLn(C)(nλ), the lth co-ordinate of the ith portion of g.vµ is0 for l > µi. (The computation in Example 3.4 of gvµ illustrates this claim.) Thiswill then imply that vµ1 and vµ2 cannot be in the same CGLn(C)(nλ)-orbit, since forsome i, since µ1 6= µ2, WLOG we must have (µ1)i > (µ2)i. If vµ1 = g.vµ2 , then thevector vµ1 will have a non-zero entry, 1, in the (µ1)i-th position, which contradictsthe above claim since (µ1)i > (µ2)i.

To show that the lth co-ordinate of the ith portion of g.vµ is 0 for l > µi, we mustexamine the structure of the centralizer CGLn(C)(nλ). We are required to prove thatthe (λ1+ · · ·+λi−1+ l)-th coordinate of g.vµ is 0; accordingly, examine the entriesin the (λ1 + · · · + λi−1 + l)-th row of the matrix g ∈ CGLn(C)(nλ). The non-zeroentries in the column vector vµ occur in the co-ordinates (λ1 + · · ·+ λj−1 + µj)-thpositions for j = 1, · · · , k. Thus to prove that the (λ1+ · · ·+λi−1+ l)-th coordinateof g.vµ is 0, it suffices to prove that the (λ1 + · · ·+ λj−1 + µj)-th coordinate in the(λ1 + · · · + λi−1 + l)-th row of g is 0 for all l > µi and all j. To do this, we mustseparately examine the cases i ≤ j and i > j.

If i ≤ j, note that in the (i, j)-th block of the matrix g, which is of size λi × λj , the(u, v)-th coordinate is 0 if u > v, by Proposition 1.14. Here we are interested in theco-ordinate in the (λ1 + · · ·+ λi−1 + l) row and the (λ1 + · · ·+ λj−1 + µj) columnof g. This lies in the l-th row and µj-th column of the (i, j)-th block of the matrix


g. But l > µi ≥ µj since i < j, which concludes the proof of the claim in this case.

If i > j, then note that in the (i, j)-th block of the matrix g, which is of sizeλi×λj , the (u, v)-th coordinate is 0 if v−u < λj−λi, by Proposition 1.14. Again,here we are interested in the co-ordinate in the (λ1 + · · ·+ λi−1 + l)-th row and the(λ1+· · ·+λj−1+µj)-th column of g, which lies in the l-th row and the µj-th columnof the (i, j)-th block of g. But µj− l < µj−µi ≤ λj−λi; where µj−µi ≤ λj−λisince λi− µi ≤ λj − µj (i.e. νi ≤ νj , which is true since i > j). This concludes theproof of the claim in the second case, and hence the proof of Lemma 3.9.

�

3.2. The exotic nilpotent cone

In this section, we study Kato’s exotic nilpotent cone, N. We state the classificationof the orbits in the exotic nilpotent cone, and describe part of the proof of this clas-sification. Let V ∼= C2n be a vector space of dimension 2n, with a symplectic form〈·, ·〉, 〈v, w〉 = vtJw corresponding to the following 2n× 2n matrix J :

1

· · ·1

−1· · ·

−1

Following [2], define S,N as follows (here N denotes N (gl(V )), the set of nilpo-tent matrices in End(V )). N is the exotic nilpotent cone.

S = {x ∈ End(V )| 〈xv, w〉 = 〈v, xw〉 ,∀v, w ∈ V }N = V × (S ∩N )

The following calculation shows that there is a well-defined action of Sp(V ) on Sby conjugation. Let g ∈ Sp(V ), so that 〈gv, gw〉 = 〈v, w〉 for all v, w, and letx ∈ S:

⟨gxg−1v, w

⟩=⟨gxg−1v, gg−1w

⟩=⟨xg−1v, g−1w

⟩=⟨g−1v, xg−1w

⟩=⟨gg−1v, gxg−1w

⟩=⟨v, gxg−1w

⟩Since Sp(V ) clearly acts by conjugation onN , Sp(V ) acts by conjugation on S∩N .There is also a natural action of Sp(V ) on V , and hence there is an action of Sp(V )on the exotic nilpotent cone N = V × (S ∩N ).

3.2. THE EXOTIC NILPOTENT CONE 41

There is a natural inclusion map φ2 : N→ V ×N , of the exotic nilpotent cone intothe enhanced nilpotent cone of size 2n. We next construct a map φ1 : V1×N1 → N,where V1 ∼= Cn is a vector space of dimension n, and N1 = N (gl(V1)) is the set ofnilpotent matrices in End(V1); thus V1×N1 is the enhanced nilpotent cone of size n.

Lemma 3.12. The following map φ1 : V1×N1 → N is well-defined. Here x ∈ N1,and xt denotes its transpose along the skew diagonal, so that xti,j = xn+1−j,n+1−i.

φ1(

v1...vn

, x) = (

v1...vn0...0

,

(x

xt

))

Proof. Let X = ( x xt ). It is clear that if x is nilpotent, then xt is nilpotent, and soX is nilpotent, i.e. X ∈ N . Thus it suffices to prove that X ∈ S, i.e. 〈Xv,w〉 =〈v,Xw〉 for all v, w ∈ V . Letting ei be the vector with a 1 in the i-th co-ordinateand zeroes elsewhere, since {e1, · · · , e2n} is a basis for V , it sufficient to prove that〈Xei, ei′〉 = 〈ei, Xei′〉. By construction of 〈·, ·〉, recall that:

〈ei, ei′〉 =

0 if i+ i′ 6= 2n+ 1,

1 if i < i′, i+ i′ = 2n+ 1,

−1 if i > i′, i+ i′ = 2n+ 1

If 1 ≤ i, i′ ≤ n, then Xei, Xei′ ∈ Span{e1, · · · , en}, and 〈Xei, ei′〉 = 〈ei, Xei′〉 =0. Similarly if n+1 ≤ i, i′ ≤ 2n, 〈Xei, ei′〉 = 〈ei, Xei′〉 = 0. If 1 ≤ i ≤ n, n+1 ≤i′ ≤ 2n, then we compute 〈Xei, ei′〉 , 〈ei, Xei′〉 as follows. Let i′ = j + n, with1 ≤ j ≤ n.

〈Xei, ej+n〉 = 〈x1,ie1 + · · ·+ xn,ien, ej+n〉 = xn+1−j,i

〈ei, Xej+n〉 =⟨ei, x

t1,jen+1 + · · ·+ xtn,je2n

⟩= xtn+1−i,j

By the definition of xt, xn+1−j,i = xtn+1−i,j , so 〈Xei, ej+n〉〈ei, Xej+n〉 in this case.A similar argument works in the case when n + 1 ≤ i ≤ 2n, 1 ≤ i′ ≤ n. HenceX ∈ S, as required. �

Lemma 3.13. Suppose (v, n1) and (w, n2) are in the sameGL(V1) orbit of V1×N1.Then φ(v, n1) and φ(w, n2) are in the same Sp(V )-orbit of N.

Proof. Suppose g(v, n1) = (w, n2). Define g1 = (g(gt)−1 ). We will first check that

g1 ∈ Sp(V ), and then verify that g1(φ(v, n1)) = φ(w, n2).


To check that g1 ∈ Sp(V ), it is necessary to check that gT1 Jg1 = J (gT1 denotes theordinary transpose). Here J = ( I′

−I′ ); where I ′ denotes the n× n matrix with 1’salong the skew diagonal. To help with this computation, note first that gT I ′ = I ′gt,since by inspection both matrices will have gn+1−j,i as their (i, j)-th co-ordinate.Then we obtain that gT I ′(gt)−1 = I ′; and transposing the equation we also haveI ′g = gt

TI ′, so (gt

T)−1I ′g = I ′. Now we compute:(g

(gt)−1

)T (I ′

−I ′)(

g(gt)−1

)=

(gT

(gtT)−1

)(I ′(gt)−1

−I ′g

)=

(gT I ′(gt)−1

−(gtT )−1I ′g

)=

(I ′

−I ′)

Hence g1 ∈ Sp(V ). Next, it is clear that if gv = w, then (g(gt)−1 )( v0 ) = ( w0 ). Since

g(v, n1) = (w, n2), we have gn1g−1 = n2. Since (AB)t = BtAt (one way of seeing

this is that, from above, xT I ′ = I ′xt so xt = I ′−1xT I ′, and this identity holds for theordinary transpose), taking the transpose of gn1g

−1 = n2 gives (gt)−1nt1gt = nt2.

Now we compute:

(g

gt−1

)(n1

nt1

)(g

gt−1

)−1=

(g

gt−1

)(n1

nt1

)(g−1

gt

)=

(gn1g

−1

(gt)−1nt1gt

)=

(n2

nt2

)This proves that g1(φ(v, n1)) = φ(w, n2), concluding the proof. �

From results in section 3.1, we know that the orbits of GL(V1) on V1 × N1 are inbijection with bi-partitions of n; callOµ,ν the orbit corresponding to the bi-partition(µ, ν). By the above Lemma, any two elements in φ(Oµ,ν) will be in the sameSp(V ) orbit of N.

Definition 3.14. Define Oµ,ν to be the Sp(V )-orbit in N containing φ1(Oµ,ν).

Given a partition λ = (λ1, · · · , λk), let λ∪λ denote the partition (λ1, λ1, · · · , λk, λk).

Proposition 3.15. We have that φ2(φ1(Oµ,ν)) ⊆ Oµ∪µ,ν∪ν .

Proof. As in the proof of Theorem 3.3, let (vµ, nλ) be an orbit representative forOµ,ν in V1 ×N1. It suffices to prove that φ2(φ1(vµ, nλ)) ∈ Oµ∪µ,ν∪ν . Consider thefollowing example, where λ = (2, 1), µ = (1, 1), ν = (1, 0).

3.2. THE EXOTIC NILPOTENT CONE 43

φ2(φ1(vµ, nλ)) = (

101000

,

0 1

00

00 1

0

)

This example makes it clear that while nλ consists of the blocks Nλ1 , · · · , Nλk ar-ranged in that order, ntλ will consists of the same blocks arranged in the reverseorder: Nλk , · · · , Nλ1 . Since nλ and ntλ have the same Jordan type, it follows thenthat we can find a matrix g, such that gnλg−1 = ntλ. Let g′ be a 2n × 2n matrixconsisting of the four n× n quadrants, with the top left quadrant being the identitymatrix, the bottom right quadrant being g, and the other two quadrants empty. Act-ing on the above pair by the matrix g′, g′ will stabilize the vector and transform thematrix into a matrix with nλ occurring twice:

(

101000

,

0 1

00

0 10

0

)

Now act on the above pair by the matrix h which will conjugate the matrix above tothe standard nilpotent nλ∪λ. It is clear what effect this will have on the vector.

(

100010

,

0 1

00 1

00

0

)

It is clear how this example generalizes: we have shown that φ2(φ1(vµ, nλ)) is in thesame orbit as (v′µ, nλ∪λ), where v′µ is a vector with its (2i− 1)-st portion consistingof a 1 in the µi-th coordinate and 0-s elsewhere, and its 2i-th portion zero for all1 ≤ i ≤ k. We can now use the results in Section 3.1 to compute the orbit in V ×Nthat it lies in.

The type of the vector v′µ is (α, β), where

α = (µ1, 0, µ2, 0, · · · , µk, 0), β = (ν1, λ1, ν2, λ2, · · · , νk, λk).


Using Proposition 3.7 and Lemmas 3.8 − 3.10, the orbit which (v′µ, nλ∪λ) lies inwill be given by the bi-partition φ(α, β) = (γ, δ). We have that

γ2i = max({αj | j ≥ 2i} ∪ {λi − βj|j < 2i})

Inspecting the values of {αj|j ≥ 2i}, it is clear that max{αj|j ≥ 2i} = µi+1.Inspecting the values of {βj|j < 2i}, it is clear that we have min{βj|j < 2i} =min(ν1, λ1, ν2, λ2, . . . , νi) = νi, so max{λi − βj|j < 2i} = λi − νi = µi. Henceγ2i = max(µi+1, µi) = µi. Next, we have

γ2i−1 = max({αj|j ≥ 2i− 1} ∪ {λi − βj|j < 2i− 1})

Inspecting the values of {αj|j ≥ 2i− 1}, it is clear that max{αj|j ≥ 2i− 1} = µi.Inspecting the values of {βj|j < 2i − 1}, it is clear that min{βj|j < 2i − 1} =min(ν1, λ1, ν2, λ2, . . . , νi−1, λi−1) = νi−1, so max{λi−βj|j ≤ 2i− 1} = λi− νi−1.Hence γ2i−1 = max(µi, λi−νi−1) = µi, since νi−1 ≥ νi = λi−µi, so µi ≥ λi−νi−1.Thus γ2i−1 = γ2i = µi, proving that γ = µ∪µ, and hence δ = ν∪ν. This concludesthe proof that φ2(φ1(vµ, nλ)) ∈ Oµ∪µ,ν∪ν .

�

Corollary 3.16. The orbits Oµ,ν and Oµ′,ν′ are distinct if (µ, ν) 6= (µ′, ν ′) aredistinct bi-partitions.

Proof. This follows from the above Lemma, since φ2(Oµ,ν) and φ2(Oµ′,ν′) belongto Oµ∪µ,ν∪ν and Oµ′∪µ′,ν′∪ν′ , respectively. �

Theorem 3.17. The orbits of Sp(V ) onN ′ are in bijection with the bi-partitions ofn, with the bipartition (µ, ν) corresponding to the orbit Oµ,ν .

Proof. All that remains to be proved is that every point in N lies in one of the orbitsOµ,ν . For this, we refer the reader to Theorem 1.14 in [6]. �

3.3. The 2-enhanced nilpotent cone

Here we describe some partial results regarding the orbits of G = GLn(C) on the2-enhanced nilpotent cone, defined below.

Definition 3.18. The 2-enhanced nilpotent cone is defined to be V × V ×N .

The action of G on the 2-enhanced nilpotent cone is clear, since G acts naturallyon V , and acts by conjugation on N . This is the first instance so far in which wewill find that there are infinitely many orbits. The problem of computing the orbitsof GLn(C) on V × V × N is equivalent to the problem of computing the orbitsof CGLn(C)(nλ) on V × V . To see this, if two elements (v1, w1, n1), (v2, w2, n2) ∈V × V × N are in the same GLn(C)-orbit, then n1 and n2 are nilpotent elementswith the same Jordan form, say corresponding to the partition λ. Thus it suffices tocheck when two vectors (v′1, w

′1, nλ) and (v′2, w

′2, nλ) are in the same GLn(C)-orbit.

This happens iff there exists g with g−1nλg = nλ such that g(v′1, w′1) = (v′2, w

′2);

i.e. iff (v′1, w′1), (v

′2, w

′2) lie in the same CGLn(C)(nλ) orbit.

3.3. THE 2-ENHANCED NILPOTENT CONE 45

Example 3.19. Let λ = (4, 2, 1), and let µ = (2, 2, 1), ν = (2, 0, 0), so that (µ, ν)is a bi-partition of λ. Pick v ∈ V arbitrary, and g ∈ CGLn(C)(nλ). In the followingcalculation, v should be considered as the union of the 1st portion (v1, v2, v3, v4),the 2nd portion (v5, v6), and the 3rd portion (v7). It is clear that the 7 coordinatefunctions specifying the vector gv can be expressed as bilinear functions in thevariables λi specifying the coordinates of g, and the variables vi describing thevector v. For 1 ≤ j ≤ 7, define fj to be the coordinate function describing the j-thcoordinate of the vector gv. Using the below example as a starting point, we willmake an observation about which variables precisely the functions fj depend upon.

g =

λ1 λ2 λ3 λ4 λ12 λ13 λ14λ1 λ2 λ3 λ12

λ1 λ2λ1

λ8 λ9 λ5 λ6 λ15λ8 λ5λ10 λ11 λ7

, v =

∗∗v3v4v5v6v7

gv =

∗∗

λ1v3 + λ2v4λ1v4

λ8v3 + λ9v4 + λ5v5 + λ6v6 + λ15v7λ8v4 + λ5v6

λ10v4 + λ11v6 + λ7v7

In the above example, we note that the functions f3, f4, f5, f6, f7 do not depend onthe variables v1 and v2 (the variables v1 and v2 have been left out to emphasize thisfact). Equivalently, the 3rd, 4th, 5th, 6th and 7th coordinates of gv can be expressedas linear functions of the 3rd, 4th, 5th, 6th and 7th coordinates of v. This leads tothe question: what is so special about those particular co-ordinates which results inthis phenomena? To answer this question, the co-ordinate v3 and v4 are the last 2coordinates in the 1st portion of v, v5 and v6 are the last 2 co-ordinates in the 2ndportion of v, and the coordinate v7 is the last 1 coordinate in the 3rd portion of v;and the triple (2, 2, 1) specifies the bi-partition (µ, ν) of λ. The generalization ofthis phenomenon is stated in the below Proposition.

Proposition 3.20. Let the vector v have portions of sizes (λ1, · · · , λk), let (µ, ν) bea bi-partition of λ, and let Sµ,ν = {µ1+1, · · · , λ1, λ1+µ2+1, · · · , λ1+λ2, · · · , λ1+· · · + λk−1 + µk + 1, · · · , λ1 + · · · + λk} (here |Sµ,ν | = |ν|). If g ∈ CGLn(C(nλ),then for j ∈ Sµ,ν , the j-th coordinate of the vector gv can be expressed as a sum ofproducts of coefficients of g and co-ordinates vi as i ranges over Sµ.

Proof. Since j ∈ Sµ,ν , let j = λ1 + · · · + λs−1 + p, for some µs + 1 ≤ p ≤ λs(here 1 ≤ s ≤ k). We will examine the j-th row of the matrix g. It is sufficient toprove that the entry in the i-th column and j-th row of the matrix g is 0 if i /∈ Sµ,ν .


The required result will then follow since the j-th coordinate of gv, which is theproduct of the row vector formed by the j-th row of g with the column vector v,will only contain terms involving the co-ordinates vi for i ∈ Sµ,ν . Since i /∈ Sµ,ν ,let i = λ1 + · · · + λt−1 + q, for some 1 ≤ q ≤ µt. If we divide up the matrix ginto k2 blocks (with the (u, v)-th block having dimension λu × λv), as describedin Section 2.3.1, the entry in the j-th row and i-th column will be the p-th row andq-th column in the (s, t)-th block. We consider the two cases where s ≤ t and s > t.

If s ≤ t, we have that λs ≥ λt; and the co-ordinate in the p-th row and the q-thcolumn will be zero if p > q. We know that p ≥ µs + 1 > µt ≥ q; here p ≥ µsand q ≤ µt are from the above paragraph, while µs ≥ µt is because s ≤ t. Thiscompletes the proof when s ≤ t.

If s > t, we have that λs ≤ λt; and the co-ordinate in the p-th row and the q-thcolumn will be zero if q−p < λt−λs. Since q ≤ µt, µs+1 ≤ p, q−p < µt−µs ≤λt − λs. The last inequality is because νs ≤ νt (since s > t), so λs − µs ≤ λt − µt,so µt − νs ≤ λt − λs. This completes the proof when s > t. �

Definition 3.21. Given a bi-partition (µ, ν) of λ, let Tµ,ν = {v ∈ V |vi = 0 for all i ∈Sµ,ν}.

Definition 3.22. Let Xµ,a0,a1,··· ,aν1−1 = {(v, w) ∈ V × V |(a0 + a1nλ + · · · +aν1−1n

ν1−1λ )v − w ∈ Tµ,ν}.

Proposition 3.23. The set Xµ,a0,a1,··· ,aν1−1 ⊂ V × V is stable under the action ofCGLn(C)(nλ).

Example 3.24. Let λ = (4, 2, 1). To illustrate the above concepts, the following is adescription of the sets Tµ,ν and Xµ,a0,··· ,aν1−1 , for the bi-partition µ = (2, 0, 0), ν =(2, 2, 1). In the below examples, all vectors should be consisted as consisting ofthe first portion (the first 4 co-ordinates), the second portion (the following 2 co-ordinates), and the third portion (the last co-ordinate).

Tµ,ν = {

∗∗00000

}, a0 + a1nλ =

a0 a1a0 a1

a0 a1a0

a0 a1a0

a0


If v =

∗∗v3v4v5v6v7

, (a0 + a1nλ)v =

∗∗

a0v3 + a1v4a0v4

a0v5 + a1v6a0v6a0v7

Hence (v, w) ∈ Xµ,a0,a1 iff w =

w1

w2

a0v3 + a1v4a0v4

a0v5 + a1v6a0v6a0v7

for some w1, w2.

Now we return to the proof of Proposition 3.23.

Proof. (of Proposition 3.23)Suppose (v, w) ∈ Xµ,a0,a1,··· ,aν1−1; we need to prove that (gv, gw) ∈ Xµ,a0,a1,··· ,aν1−1

for all g ∈ CGLn(C)(nλ). This is equivalent to saying that (a0 + a1nλ + · · · +aµ1−1n

ν1−1λ )gv − gw ∈ Tµ,ν .

Since g commutes with nλ, g commutes with all powers of nλ, so (a0 + a1nλ +· · ·+ aν1−1n

ν1−1λ )gv = g(a0 + a1nλ+ · · ·+ aν1−1n

ν1−1λ )v. Thus we must prove that

g[(a0 + a1nλ + · · · + aν1−1nν1−1λ )v − w] ∈ Tµ,ν . Since (v, w) ∈ Xµ,a0,a1,··· ,aν1−1 ,

(a0 + a1nλ + · · · + aµ1−1nµ1−1λ )v − w ∈ Tµ,ν . Thus it is sufficient for us to show

that Tµ,ν is stable under the action of CGLn(C)(nλ).

The fact that Tµ,ν is stable under the action ofCGLn(C)(nλ) is clear using Proposition3.20 by the following argument. Suppose v ∈ Tµ,ν , i.e. that vi = 0 for all i ∈ Sµ,ν .Given g ∈ G, the i-th co-ordinate of gv is a linear combination of co-ordinatesfrom g and co-ordinates vj for j ∈ Sµ,ν , and since vj = 0 for j ∈ Sµ,ν , the i-thco-ordinate of gv is 0 for i ∈ Sµ,ν . Hence gv ∈ Tµ,ν , completing the proof. �

Thus we have found certain sets which are stable under the action of CGLn(C)(nλ).Using a variant of this method, we next describe the orbits of CGLn(C)(nλ) on V ×Vin the case where λ = (pq) is a rectangle partition. (Here (pq) = (p, p, · · · , p),where p occurs q times.)

Using Theorem 3.3, we can assume that the second vector is in standard form, vµfor some bi-partition (µ, ν) = (rq, (p − r)q) of λ. Then the problem is equiva-lent to classifying the orbits of G(vµ,nλ) on V , where G(vµ,nλ) is the stabilizer inG = GLn(C) of the pair (vµ, nλ). For convenience, let Gr = G(nλ,vµ). Below we


describe the structure of Gr. Recall from Proposition 1.14 that for any g ∈ Gnλ , gcan be divided into q2 p× p blocks gi,j , with 1 ≤ i, j ≤ q, with gi,j upper triangularwith entries on the diagonals equal. Call the entries in the first row of the block gi,j ,gi,j1 , g

i,j2 , · · · , gi,jp . The matrix g must satisfy the condition that the matrix (gi,j1 ) is

invertible.

Lemma 3.25. A matrix g of the form above, will lie in Gr if and only if it satisfiesthe following additional conditions, for 1 ≤ i ≤ q:

q∑j=1

gi,jk = 1 if k = 1,

q∑j=1

gi,jk = 0 if 2 ≤ k ≤ r

If r = 0 there are no additional restrictions on Gr.

Proof. First consider the example p = 4, q = 2, r = 3. For convenience, we usedifferent notation for the entries of g to that above.

a1 b1 c1 d1 a2 b2 c2 d2a1 b1 c1 a2 b2 c2

a1 b1 a2 b2a1 a2

a3 b3 c3 d3 a4 b4 c4 d4a3 b3 c3 a4 b4 c4

a3 b3 a4 b4a3 a4

00100010

=

c1 + c2b1 + b2a1 + a2

0c3 + c4b3 + b4a3 + a4

0

Equating the coefficients gives a1 + a2 = a3 + a4 = 1, b1 + b2 = b3 + b4 =0, c1 + c2 = c3 + c4 = 0. This proves it in the case of this example, and it is clearhow this example will generalize. �

Definition 3.26. Let v ∈ V , v = (v1,1, · · · , v1,p, v2,1, · · · , v2,p, · · · , vq,1, · · · , vq,p).For 1 ≤ i ≤ p, let vi = (v1,i, v2,i, · · · , vq,i). Define the height h of v to be themaximal h such that vh is non-zero (so vh+1 = · · · = vp = 0, vh 6= 0).

It is clear the set of all vectors of a fixed height h will be stable under the action ofGnλ , and hence Gr. To see this, the set of all vectors of a fixed height h is the setTµ1,ν1 −Tµ2,ν2 , where (µ1, ν1) = (hq, (p− h)q), (µ2, ν2) = ((h− 1)q, (p− h+1)q).The result then follows from the fact that Tµ1,ν1 and Tµ2,ν2 are stable under the actionof Gnλ , which is shown in the proof of Proposition 3.23.

Definition 3.27. Let the type of v be a sequence (λ1, · · · , λj) defined as follows. Letj = min(j′, r), where j′ ≤ h is maximal such that vh, vh−1, · · · , vh+1−j′ ∈ Cw0;here w0 = (1, 1, · · · , 1) (with q ‘1’s). For 1 ≤ i ≤ j, let λi be the scalar such thatvh+1−i = λiw0.

Note that in the above definition, j = 0 is allowed, in which case the type will beempty. If j > 0, then λ1 6= 0, since vh = λ1w0 and vh 6= 0 by definition of h.

Lemma 3.28. The set of all vectors of height h and type (λ1, · · · , λj) is stable underthe action of Gr.


Proof. It suffices to show that the set of all vectors with height h and satisfyingvh = λ1w0, · · · , vh+1−j = λjw0 is stable under the action of Gr. The reason whythis suffices is as follows: if we have a vector v1 of type (λ1, · · · , λj) in the sameorbit as a vector v2, then if v2 does not have type (λ1, · · · , λj) then v2 is forced tohave type (λ1, · · · , λj, λj+1, · · · ); but then v1, which is in the same orbit as v2, willnow be forced to have type (λ1, · · · , λj, λj+1, · · · ), which is a contradiction.

Let us consider the example p = 4, q = 2, r = 3, h = 3, j = 2. Here y = λ1, z =λ2.

a1 b1 c1 d1 a2 b2 c2 d2a1 b1 c1 a2 b2 c2

a1 b1 a2 b2a1 a2

a3 b3 c3 d3 a4 b4 c4 d4a3 b3 c3 a4 b4 c4

a3 b3 a4 b4a3 a4

∗zy0∗zy0

=

∗(a1 + a2)z + (b1 + b2)y

(a1 + a2)y0∗

(a3 + a4)z + (b3 + b4)y(a3 + a4)y

0

=

∗zy0∗zy0

Above we have used the relations in Gr, a1+a2 = a3+a4 = 1, b1+ b2 = b3+ b4 =0, c1 + c2 = c3 + c4 = 0. It is fairly clear how this example will generalize. Thereason it is necessary to define j so that j ≤ r, is that by Lemma 3.25 there areprecisely r equations among its entries which govern its structure, and if j > r thenthese equations will run out, and there is no assurance that the conditon vh−r =λr+1w will be preserved by the action of Gr. �

Lemma 3.29. If j < r, then any two vectors of height h and type (λ1, · · · , λj) arein the same Gr-orbit.

Proof. Let v be the vector with height h such that v1 = v2 = · · · = vh−j−1 = 0,vh−j = (0, · · · , 0, 1) (here there are q − 1 0-s), vh+1−j = λjw0, · · · , vh = λ1w0.Then v clearly has height h and type (λ1, · · · , λj). We will prove that any othervector of height h and type (λ1, · · · , λj) lies in the same Gp-orbit as v; this willsuffice, since by Lemma 3.25, Gp ⊂ Gr. Let us consider the example with λ =(42), µ = (32), j = 2, h = 4.


a1 b1 c1 d1 a2 b2 c2 d2a1 b1 c1 a2 b2 c2

a1 b1 a2 b2a1 a2

a3 b3 c3 d3 a4 b4 c4 d4a3 b3 c3 a4 b4 c4

a3 b3 a4 b4a3 a4

00λ2λ101λ2λ1

=

b2a2λ2λ1b4a4λ2λ1

Here we have used the relations inGp, a1+a2 = 1, b1+b2 = c1+c2 = d1+d2 = 0.In the vector on the right above, the value of the pair (a2, a4) can be an arbitraryelement of C2 not lying in Cw0. To see this, the invertible 2× 2 matrix ( a1 a2a3 a4 ) actson C2 × C2, and can take any pair of vectors forming a rank 2 matrix, to any otherpair of vectors forming a rank 2 matrix. In particular, it can take the pair of vectors(0, 1), (1, 1) to (a2, a4), (1, 1) if a2 6= a4. The condition that the matrix stabilizes(1, 1) imposes the conditions that we have on the matrix, a1 + a2 = a3 + a4 = 1,and the matrix will take (0, 1) to (a2, a4). The variables b2 and b4 are free to vary,because the equations relate them to b1 and b3 which do not occur. This then meansthe set of vectors on the right ranges over all vectors of height 4 and type (x, y),proving the statement in the case of the example.

It is clear how this example will generalize. Here we require that j < r; if j = r,in the above example for instance if r = 2 instead, there will be vectors of type(λ1, λ2) with a2 = a4, which is not accounted for in the above argument.

�

Lemma 3.30. If j = r, then any two vectors of height h and type (λ1, · · · , λr) arein the same Gr-orbit.

Proof. Let v be the vector with height h such that v1 = v2 = · · · = vh−r = 0,vh+1−r = λrw0, · · · , vh = λ1w0. Then v is clearly of type (λ1, · · · , λr), so it suf-fices to prove that any other vector of type (λ1, · · · , λr) lies in the same orbit as v.Let us consider the example with λ = (42), µ = (22), h = 4. Here x = λ1, y = λ2.

a1 b1 c1 d1 a2 b2 c2 d2a1 b1 c1 a2 b2 c2

a1 b1 a2 b2a1 a2

a3 b3 c3 d3 a4 b4 c4 d4a3 b3 c3 a4 b4 c4

a3 b3 a4 b4a3 a4

00yx00yx

=

(c1 + c2)y + (d1 + d2)x(c1 + c2)x

yx

(c3 + c4)y + (d3 + d4)x(c3 + c4)x

yx

Here we have used the relations in Gr, a1 + a2 = 1, b1 + b2 = 0. In the above, thevector on the right can be made to be an arbitrary vector of type (x, y). To see this,since x 6= 0, and there are no restrictions on c1, c2, c3 and c4, the values of (c1+c2)x


and (c3 + c4)x can be made arbitrary. Since there are no restrictions on d1, d2, d3and d4, the values of (c1 + c2)y + (d1 + d2)x and (c3 + c4)y + (d3 + d4)x can bemade arbitrary. This proves the statement in the case of the example. It is clear thatthis example will generalize. �

Combining the previous three Lemmas, we can now state the classification of orbitsof Gr on V .

Theorem 3.31. The orbits of Gr on V are given by the height h of the vector v, andits type (λ1, · · · , λj).

As an example to illustrate the theorem, the following are the orbits of Gr on V inthe case where λ = (32), µ = (22).

000000

a00b00

|a 6= b

;

λ00λ00

, λ 6= 0

∗a0∗b0

|a 6= b

;

aλ0bλ0

|a 6= b

,λ 6= 0;

λ2λ10λ2λ10

, λ1 6= 0

∗∗b∗∗a

|a 6= b

;

∗aλ∗bλ

|a 6= b

,λ 6= 0;

∗λ2λ1∗λ2λ1

, λ1 6= 0

CHAPTER 4

Springer fibres

In this section, we will define Springer fibres, and study some properties of theseSpringer fibres. We describe the irreducible components of the Springer fibres as-sociated to a nilpotent of type λ, and prove that they are in bijection with the set ofstandard tableaux of type λ. Here we follow the presentation in Spaltenstein, [8].

4.1. Irreducible components of Springer fibres

Recall that in Section 2.1, we constructed a map π : G ×P n → Oλ. Considerthe case when λ = (n). Using results in Section 2.2, Oλ′ ⊂ Oλ when λ′ ≤ λ.Since λ′ ≤ (n) for any partition λ′, it follows that O(n) = N in our case. The spaceG×P n can be viewed as {(x, (Vi))|xVi ⊆ Vi−1}with dimVi = µ1+ · · ·+µi (whereµ is the transpose partition of (n)); here since µ = (1n), µ1 + · · ·+ µi = i, and theparabolic subgroup P will become a Borel subgroup B. Hence the space G ×B ncan be viewed as {(x, (0 = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V ))|x(Vi) ⊆ Vi−1}, withdimVi = i, and the map π, which is known as the Springer resolution, is projectiononto the first factor.

Definition 4.1. Given x ∈ Oλ, the Springer fibre Fx is defined as follows:

Fx = π−1(x) = {0 = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V |x(Vi) ⊂ Vi−1}Let Sλ denote the set of standard tableaux associated to the partition λ; here a “stan-dard tableau” is a way of filling up the Young diagram for the partition λ with thenumbers 1, 2, 3, · · · , n in such a way that the numbers are strictly increasing alongrows and down columns. Given a σ ∈ Sλ, let σi denote the number of the col-umn which i lies in. It is fairly clear that σ can be reconstructed from the numbersσ1, σ2, · · · , σn, since proceeding inductively, the entry i must be inserted into thehighest entry in the σi-th column which has not yet been occupied by the entries1, 2, · · · , i − 1 (to ensure that entries are increasing down columns). Define a totalordering on Sλ as follows (known as the “reverse lexicographic" ordering): givenσ, σ′ ∈ Sλ, suppose that σ < σ′ if for some i, σi < σ′i but σj = σ′j if i < j ≤ n.

Suppose we have a Jordan basis ei,j for x, with 1 ≤ i ≤ µ1, 1 ≤ j ≤ λi, andxei,j = ei,j−1 if j ≥ 1, and 0 otherwise. Consider an example with λ = (4, 32, 2).The diagram below shows the partition λ, and the basis elements ei,j can be iden-tified with the boxes of λ in an obvious way. Let Wi = span(e1,1, e2,1, · · · , ei,1).In the case of our example, the below diagram shows that kerx = W4, kerx ∩imx3 = W1, kerx ∩ im x2 = W3, kerx ∩ im x = W4. This shows that in generalkerx ∩ im xi−1 = Wµi (here µi is the length of the i-th column of λ). In particular,Wµi is independent of the choice of Jordan basis (although Wi for general i is not).

52

4.1. IRREDUCIBLE COMPONENTS OF SPRINGER FIBRES 53

Lemma 4.2. Consider the flag (V1/V1, V2/V1, · · · , Vn/V1). The nilpotent x clearlyinduces a nilpotent transformation x′ of Vn/V1, and moves each element of this flaginto the previous flag. Then the Jordan type of the nilpotent x′ corresponds to thepartition obtained by deleting the corner at the bottom of the j-th column of λ,where j is such that V1 ∈ Wµj but V1 /∈ Wµj+1

.

Proof. We first consider the example above, λ = (4, 32, 2), with j = 3. ThenV1 ∈ span (e1,1, e2,1, e3,1) but V1 /∈ span (e1,1). The first case is when V1 =span (ae1,1 + e2,1). Then the following set will be a second Jordan basis for x:{e1,1, e1,2, e1,3, e1,4}, {ae1,1+e2,1, ae1,2+e2,2, ae1,3+e2,3}, {e3,1, e3,2, e3,3}, {e4,1, e4,2}.Letting the image of ei,j in V/V1 be e′i,j , note that in V/V1, ae′1,1 + e′2,1 = 0. It fol-lows that a possible Jordan basis for x on V/V1 is given by {e′1,1, e′1,2, e′1,3, e′1,4},{ae′1,2 + e′2,2, ae

′1,3 + e′2,3}, {e′3,1, e′3,2, e′3,3}, {e′4,1, e′4,2}. Thus the Jordan type of x′

on V/V1 is (4, 3, 22), which is obtained from λ by deleting the last box in the thirdcolumn, proving the claim in the first case of this example.

The second case is when V1 = span (ae1,1 + be2,1 + e3,1). Then the follow-ing set will be a second Jordan basis for x: {e1,1, e1,2, e1,3, e1,4}, {e2,1, e2,2, e2,3},{ae1,1 + be2,1 + e3,1, ae1,2 + be2,2 + e3,2, ae1,3 + be2,3 + e3,3}, {e4,1, e4,2}. Lettingthe image of ei,j in V/V1 be e′i,j , note that in V/V1, ae′1,1 + be′2,1 + e′3,1 = 0. It fol-lows that one possible Jordan basis for x′ on V/V1 is given by {e′1,1, e′1,2, e′1,3, e′1,4},{e′2,1, e′2,2, e′2,3}, {ae′1,2 + be′2,2 + e′3,2, ae

′1,3 + be′2,3 + e′3,3}, {e4,1, e4,2}. Thus the

Jordan type of x′ on V/V1 is (4, 3, 22), which is obtained from λ by deleting the lastbox in the third column, proving the claim in the second case of the example, andfinishing the example.

In the general case, it is straightforward to see how to generalize this constructionand construct the necessary Jordan basis for x′ on V/V1. �

By induction on dim V , we will show that there exists a map θ : Fx → Sλ. Sinceby induction this is possible for the smaller flag (V1/V1 ⊂ V2/V1 ⊂ · · · ⊂ Vn/V1),we have a way of associating a standard tableau σ1 of size n − 1 to the action ofx′ on this flag, where σ1 has shape λ′, which is obtained by deleting the corner atthe bottom of the j-th row of λ (by the above Lemma). Then, to the action of x onthe flag (V0 ⊂ V1 ⊂ · · · ⊂ Vn), we will associate the standard tableaux σ of shapeλ, by adding to σ1 the number n in the corner at the bottom of the j-th row (thisis a standard tableaux since this box is a corner). This gives us the required mapθ : Fx → Sλ. By the construction of θ : Fx → Sλ, given F ∈ Fx, the shape of the

54 4. SPRINGER FIBRES

standard tableau containing the entries 1, · · · , n− s and ignoring all higher entriesin θ(F ), corresponds to the Jordan type of x on V/Vs. For σ ∈ Sλ, let Fσ = θ−1(σ).It is clear by induction that Fσ is non-empty. The following theorem is the mainresult of this section.

Theorem 4.3. The irreducible components of the Springer fibre Fx are the closuresFσ, and hence are in bijection with the set of standard tableaux Sλ.

Given a vector space W , let P(W ) denote the set of all one-dimensional subspacesof W ; then P(W ) is a projective algebraic variety. Let Bi = P(Wi) − P(Wi−1)and B1 = P(W1), so that P(kerx) is the disjoint union of Bi with 1 ≤ i ≤ µ1.Define a map p : Fx → P(kerx), which takes a flag (V0 ⊂ V1 ⊂ · · · ⊂ Vn)to V1. Define Xi = p−1(Bi). Then Fx is the disjoint union of X1, X2, · · · , Xµ1 .Let X ′i = p−1(Cei,1). The variety X ′i can be considered as a “n − 1 dimensionalversion” of Fx, since if x′ is the induced nilpotent in V/Cei,1, X ′i can be identifiedwith the set of all complete flags in this quotient space with x′ moving each flaginto the previous flag. Keeping in mind that Xi is the set of all flags with their 1-dimensional component in Bi, and X ′i is the set of all flags with their 1 dimensionalcomponent being a specific element of Bi, the following Proposition is plausible:

Proposition 4.4. We can find an isomorphism of algebraic varieties f : Xi →Bi × X ′i, such that if f(F ) = (bi, F

′) for some flags F, F ′ and bi ∈ Bi, thenθ(F ) = θ(F ′).

Proof. First note that Wi−1 is isomorphic to Bi = P(Wi) − P(Wi−1). To seethis, Wi = Wi−1 ⊕ Cei,1, so every one-dimensional subspace in Bi can be writ-ten as C(ei,1 + wi−1) for some wi−1 ∈ Wi−1, uniquely. It follows that the mapwi−1 → C(ei,1 + wi−1) is an isomorphism from Wi−1 to Bi. This means it sufficesto prove the Proposition with Bi replaced by Wi−1.

Suppose w ∈ Wi−1, so w =∑

1≤s≤i−1 ases,1 for some constants as. For j ≤ λi−1,let wj =

∑1≤s≤i−1 ases,j . Note that xwj = wj−1 (here define w0 = 0 for con-

venience). Define a linear map gw : V → V by letting gw(ei′,j) = ei′,j if i′ 6= i,and gwei,j = ei,j + wj . It is clear that gw is invertible, since the matrix of gw withrespect to the Jordan basis for x is unitriangular. We next check that gwx = xgw.If i′ 6= i, then gwxei′,j = gwei′,j−1 = ei′,j−1, while xgwei′,j = xei′,j = ei′,j−1, sogwxei′,j = xgwei′,j (for all i′, let ei′,0 = 0 to deal with the boundary case). Next,gwxei,j = gwei,j−1 = ei,j−1 + wj−1, while xgwei,j = x(ei,j + wj) = ei,j−1 + wj−1since xwj = wj−1, so gwxei,j = xgwei,j .

Consider the map g : Wi−1×X ′i → Xi defined by g(w,F ) = gw(F ). We first needto show that g is a well-defined map, i.e. that gw(F ) lies in Xi. If F = (V0 ⊂ V1 ⊂· · · ⊂ Vn), then gw(F ) = (V0 ⊂ gw(V1) ⊂ · · · ⊂ gw(Vn−1) ⊂ Vn), so xgw(Vi) =gwx(Vi) ⊂ gw(Vi−1), which means gw(F ) ∈ Fx. If F = (V0 ⊂ V1 ⊂ · · · ⊂ Vn) liesin X ′i, then p(gw(F )) = gw(V1) = gw(Cei,1) = C(ei,1 + w1) = C(ei,1 + w). SinceC(ei,1 + w) ∈ Bi, it follows that gw(F ) ∈ Xi; and hence g is a well defined map.

To check that g is surjective, suppose F ′ = (V ′0 ⊂ V ′1 ⊂ · · · ⊂ V ′n) is some flagin Xi. Suppose V ′1 = C(ei,1 + w). Since gw(C(ei,1)) = C(ei,1 + w) we have


that g−1w (V ′1) = C(ei,1). Since xgw = gwx, xg−1w = g−1w x, and so xg−1w (V ′i ) =g−1w x(V ′i ) = g−1w V ′i−1, which means g−1w (F ′) ∈ X ′i. It then follows that F ′ =g(w, g−1w (F ′)), proving that g is surjective. To prove that g is injective, suppose wehave that gw(F ) = F ′ for some w ∈ Wi−1, F ∈ X ′i, F

′ ∈ Xi. If F ′ = (V ′0 ⊂V ′1 ⊂ · · · ⊂ V ′n), F = (V0 ⊂ V1 ⊂ · · · ⊂ Vn), then gw(V1) = gw(C(ei,1)) =C(ei,1 + w1) = V ′1 . The condition C(ei,1 + w1) = C(ei,1 + w) = V ′1 determinesw uniquely from F ′. Then gw(F ) = F ′, so F = g−1w (F ′), which determines Funiquely from F ′. This shows that g is injective. If we define f to be the inverse ofg, f will an isomorphism between Xi and Wi−1 ×X ′i.

To show the remainder of the proposition, it is enough to show that θ(F ) = θ(gw(F )),for F = (V0 ⊂ V1 ⊂ · · · ⊂ Vn) ∈ X ′i. Recall that the shape of the stan-dard tableau containing the entries 1, · · · , n − s and ignoring all higher entries inθ(F ), corresponds to the Jordan type of x on V/Vs. It thus suffices to prove thatthe Jordan type of x on V/Vs is the same as the Jordan type of x on V/gw(Vs)(here x really means the induced transformation on the quotient). To see this, sup-pose am,n + gw(Vs) is a Jordan basis for x on V/gw(Vs). Then consider the basisg−1w am,n + Vs for V/Vs. Clearly xVs ⊂ Vs, and g−1w xgw = x, so x(g−1w am,n + Vs) =(g−1w xgw)(g

−1w am,n) + Vs = g−1w xam,n + Vs = g−1w am,n−1 + Vs; hence g−1w am,n + Vs

is a Jordan basis for x on V/Vs. This shows that θ(F ) = θ(gw(F )), as required.�

Proposition 4.5. a) The set Fσ is a locally closed subset of Fx, and thus wecan give Fσ the structure of an algebraic variety.

b) The variety Fσ is irreducible, and has dimension∑

iµi(µi−1)

2.

Proof. (a) It suffices to prove that the set ∪σ′≥σFσ′ is closed for each σ. To seethis, since the ordering > is a total ordering, we can find a standard tableauxσ+ which comes immediately before σ, so that σ′ > σ means σ′ ≥ σ+. ThenFσ = ∪σ′≥σFσ′ − ∪σ′≥σ+Fσ′ . It would then follow that Fσ is an intersection of anopen set and a closed set, and is hence locally closed.

Let σn = j. Make the following definitions:

A =⋃

σ′n≥j+1

Fσ′

Z =⋃σ′≥σ

Fσ′

B =⋃σ′n≥j

Fσ′

Then by the way the ordering on Sλ is defined, we have A ⊂ Z ⊂ B. It is clearthat B = p−1(P(Wµj)), and hence B is closed in Fx. Similarly A is closed in Fx.Since B is closed in Fx, to show that Z is closed in Fx it suffices to prove that Z isclosed in B. Equivalently, it is sufficient to prove that B −Z is open in B. Since Ais closed, A is closed in B, so B − A is open in B. Thus to show B − Z is open in

56 4. SPRINGER FIBRES

B, it will suffice to show that B − Z is open in B − A, since B − A is open in B.The spaces B − Z and B − A are the following:

B − Z =⋃

σ′<σ,σ′n=j

Fσ′

B − A =⋃σ′n=j

Fσ′

Pick an element F in B−Z; so suppose F ∈ Fτ for some standard tableaux τ withτ < σ and τn = j. We will construct an open neighbourhood of F in B − A.

Consider the isomorphism Xµj∼= Bµj ×X ′µj defined in Proposition 4.4, and view

X ′µj as F ′x′ , an “n − 1 dimensional version” of Fx, corresponding to the action ofx′ on the flag variety F ′ consisting of quotient spaces (the Jordan type of x′ is thepartition obtained by deleting the last box in the j-th column of λ). If the flag inXµj

lies in Fσ′ for some σ′ < σ, σ′n = j, then by the property stated in Proposition 4.4,the flag inX ′µj will also lie inFσ′ , and hence will lie inF ′σ′1 whenX ′µj is consideredas F ′x′ (here σ′1 is obtained by deleting the n at the end of the j-th column of σ′).We thus have the following:

Under the isomorphism Xµj∼= Bµj ×X ′µj ,

Xµj ∩⋃

σ′<σ,σ′n=j

Fσ′ corresponds to Bµj × (⋃

σ′1<σ1

F ′σ′1)

By induction, we may assume that⋃σ′1<σ1 F ′σ′1 is open in X ′µj = F ′x′ . This will

imply that Bµj ×⋃σ′1<σ1 F ′σ′1 is open in Bµj × X ′µj , so Xµj ∩

⋃σ′<σ,σ′n=j

Fσ′ isopen in Xµj .

Note next that Xµj+1∪ · · · ∪ Xµj =

⋃σ′n=jFσ′ = B − A by considering the pos-

sibilities for the first subspace in the flag. It follows that Xµj+1∪ · · ·Xµj−1 =

p−1(P(Wµj−1)) ∩ (B − A) is closed in B − A, so Xµj is open in B − A.

This now means that Xµj ∩⋃σ′<σ,σ′n=j

Fσ′ is open in Xµj , which is in turn open inB−A; hence it is open in B−A, and it is also clearly contained in B−Z. We willbe done if we can show that it contains F (since this will then be the required openneighbourhood of F in B−A). To show that it contains F , we are required to showthat F is contained in Xµj . F is chosen to be in some Fτ for some τ with τn = j; sothe first subspace in the flag lies in Wµj but not in Wµj+1

. We require that the firstsubspace in the flag F lies in Wµj but not in Wµj−1. To do this, we have freedom inchoosing the Jordan basis for x, so we can simply choose the Jordan basis for x sothat the first subspace in the flag lies in Wµj but not in Wµj−1.


(b) Continue to denote σn by j. Consider the morphism p : Fσ → P(Wµj) −P(Wµj+1

). We have seen that the fibre of this morphism over any point can be iden-tified with F ′σ1 . This means that Fσ is a fibre bundle over P(Wµj)− P(Wµj+1

) withfibres isomorphic to F ′σ1 . We can assume by induction that F ′σ1 is an irreducible va-riety. Since P(Wµj)−P(Wµj+1

) is an open subset of the irreducible variety P(Wµj),it is an irreducible variety. Since a fibre bundle over an irreducible variety with ir-reducible fibres is irreducible, it follows that Fσ is irreducible.

Now assume by induction that dimF ′σ1 = (∑

i 6=j12µi(µi− 1))+ 1

2(µj − 1)(µj − 2).

The base case for this induction is when σ1 is the standard tableaux with one box(containing 1). In this case, the Springer fibre corresponds to a point (the flag 0 =V0 ⊂ V1 = V ), which has dimension 0, as required. For the induction step, clearlydim(P(Wµj) − P(Wµj+1

)) = µj − 1. Since Fσ is a fibre bundle over P(Wµj) −P(Wµj+1

) with fibres isomorphic to F ′σ1 we have:

dimFσ = dimF ′σ1 + dim(P(Wµj)− P(Wµj+1))

= (∑i 6=j

1

2µi(µi − 1)) +

1

2(µj − 1)(µj − 2) + µj − 1

= (∑i 6=j

1

2µi(µi − 1)) +

1

2(µj − 1)(µj)

=∑i

1

2µi(µi − 1)

This now proves the required dimension formula for Fσ. �

Now we are in a position to prove Theorem 4.3.

Proof. It is clear that Fx = ∪Fσ. Since the closure of an irreducible variety isirreducible, (b) of Proposition 4.5 gives us that Fσ is irreducible. Since (b) ofProposition 4.5 tells us that the dimensions of the varieties Fσ are all equal, it isnot possible for any one of these varieties to contain another (since a proper closedsubvariety of an irreducible variety has strictly smaller dimension). Finally, it is notpossible for Fσ = Fσ′ for distinct σ, σ′, since then Fσ and Fσ′ would be disjointnonempty open subsets (since Fσ is locally closed, hence open in its closure), andit is not possible for an irreducible variety to have disjoint nonempty open subsets.This now completes the proof. �

References

[1] Achar, P. N., Henderson, A.: Orbit closures in the enhanced nilpotent cone,Adv. Math. 219 (2008), no. 1, 27-62

[2] Achar, P, N., Henderson, A., Sommers, E.: Pieces of nilpotent cones for clas-sical groups, arXiv:1001.4283v1

[3] Collingwood, D. H., McGovern, W. M.: Nilpotent Orbits in Semisimple LieAlgebras, Van Nostrand Reinhold Mathematics Series, Van Nostrand ReinholdCo., New York (1993)

[4] Fulton, W.: Young Tableaux, with Applications to Representation Theory andGeometry, London Mathematical Society, 35. Cambridge University Press,Cambridge (1997)

[5] Humphreys, J. E.: Introduction to Lie algebras and representation theory,Graduate Texts in Mathematics, Vol. 9. Springer-Verlag, New York-Berlin(1972)

[6] Kato, S.: An exotic Deligne-Langlands correspondence for symplectic groups,Duke Math. J. 148 (2009), no. 2, 305-371

[7] Macdonald, I.G.: Symmetric functions and Hall polynomials, Oxford Mathe-matical Monographs, Oxford University Press, New York, (1995)

[8] Spaltenstein, N.: The fixed point set of a unipotent transformation on the flagmanifold, Indag. Math. 38 (1976), no. 5, 452-456

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