ABSTRACT
Title of dissertation: EQUIVARIANT GIAMBELLI FORMULAE
FOR GRASSMANNIANS
Elizabeth V. McLaughlin (Wilson),
Doctor of Philosophy, 2010
Dissertation directed by: Professor Harry Tamvakis
Department of Mathematics
In this thesis we use Young?s raising operators to deflne and study polyno-
mials which represent the Schubert classes in the equivariant cohomology ring of
Grassmannians. For the type A and maximal isotropic Grassmannians, we show
that our expressions coincide with the factorial Schur S, P, and Q functions. We
deflne factorial theta polynomials, and conjecture that these represent the Schubert
classes in the equivariant cohomology of non-maximal symplectic Grassmannians.
We prove that the factorial theta polynomials satisfy the equivariant Chevalley for-
mula, and that they agree with the type C double Schubert polynomials of [IMN]
in some cases.
Equivariant Giambelli Formulae for Grassmannians
by
Elizabeth V. McLaughlin (Wilson)
Dissertation submitted to the Faculty of the Graduate School of the
University of Maryland, College Park in partial fulflllment
of the requirements for the degree of
Doctor of Philosophy
2010
Advisory Committee:
Professor Harry Tamvakis, Chair/Advisor
Professor Jonathan Rosenberg
Professor Niranjan Ramachandran
Professor Larry Washington
Professor Luis Orozco
c Copyright by
Elizabeth V. McLaughlin (Wilson)
2010
Acknowledgments
This thesis would not have been possible without my advisor Professor Harry
Tamvakis. I am grateful to him for introducing me to this beautiful area of mathe-
matics, and directing me while I worked on this thesis problem.
I am also very grateful for my friends. Without their support I would not have
had the strength to push myself through this process. I would like to especially thank
Edward Clifiord, Walter Ray-Dulany, and Ben Lauser for their help and support.
LastlyIwouldliketothankmyfamily. Myparents?encouragementthroughout
the years has been amazing. I have always felt loved and supported in everything
I?ve done, and I can?t describe to them how much I appreciate it. Finally to Kevin
Wilson, my husband, I wish I had timed this process a little better, but if anyone
could make wedding planning and thesis writing at the same time possible, it is
Kevin. When I was stressed, he always made me feel better. He is my rock.
ii
Table of Contents
List of Figures v
1 Introduction 1
1.1 Classical Schubert Calculus . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Equivariant Cohomology . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Equivariant Schubert Calculus . . . . . . . . . . . . . . . . . . . . . . 4
1.4 Raising Operators as a Tool for Schubert Calculus . . . . . . . . . . . 5
1.5 Outline of Thesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2 Preliminaries 7
2.1 Schubert Cells, Varieties, and Classes . . . . . . . . . . . . . . . . . . 7
2.2 Equivariant Cohomology Rings for Complete Flags in Types A and C 8
2.2.1 The Presentations in Terms of Chern Classes . . . . . . . . . . 9
2.3 The Vanishing Theorem and Ramiflcations . . . . . . . . . . . . . . . 12
2.4 The Chevalley Formula . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3 The Type A Equivariant Giambelli Rule 16
3.1 Preliminaries for Type A Grassmannians . . . . . . . . . . . . . . . . 16
3.1.1 The Classical Schubert Classes and Schur S functions . . . . . 16
3.1.2 Torus Fixed Points and the Equivariant Cohomology . . . . . 17
3.1.3 The Type A Grassmann Permutation for a Partition ? . . . . 19
3.1.4 Restriction to Torus Fixed Points in Type A . . . . . . . . . . 21
3.2 The Raising Operator Expression . . . . . . . . . . . . . . . . . . . . 22
3.3 The Vanishing Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.4 The Chevalley Formula . . . . . . . . . . . . . . . . . . . . . . . . . . 25
4 The Giambelli Formula for the Maximal Isotropic Grassmannians 31
4.1 Preliminaries for the Lagrangian Grassmannian . . . . . . . . . . . . 31
4.2 The Raising Operator Expression . . . . . . . . . . . . . . . . . . . . 33
4.3 The Chevalley Formula . . . . . . . . . . . . . . . . . . . . . . . . . . 35
4.4 Giambelli Revisited for Maximal Orthogonal Grassmannians . . . . . 39
5 The Giambelli Formula for General Isotropic Grassmannians 42
5.1 Preliminaries for IG(n?k;2n) . . . . . . . . . . . . . . . . . . . . . 42
5.2 The Theta Polynomials of [BKT] and the Classical Chevalley Formula 43
5.2.1 The Substitution Rule . . . . . . . . . . . . . . . . . . . . . . 46
5.3 The k-Grassmannian Signed Permutation of a Partition ? . . . . . . . 59
5.4 Restriction to Torus Fixed Points in Type C . . . . . . . . . . . . . . 60
5.5 Factorial Theta Polynomials . . . . . . . . . . . . . . . . . . . . . . . 61
5.6 The Equivariant Chevalley Formula . . . . . . . . . . . . . . . . . . . 69
5.7 The Equivariant Substitution Rule . . . . . . . . . . . . . . . . . . . 70
5.8 The Vanishing Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 80
5.9 Generalizations and Further Work . . . . . . . . . . . . . . . . . . . . 81
iii
A Example of the Substitution Rule 82
A.1 Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
A.2 Algorithm With Corrections . . . . . . . . . . . . . . . . . . . . . . . 84
Bibliography 93
iv
List of Figures
5.1 k-related: In this example the box [r;c] is k-related to [r0;c0]. . . . . . 44
5.2 The set C near the pair x = (h;gh) 2 @C . . . . . . . . . . . . . . . . 48
5.3 The diagram of ? with marked k-related diagonals. . . . . . . . . . . 60
A.1 The Young diagram of ?. . . . . . . . . . . . . . . . . . . . . . . . . . 83
A.2 The Substitution Rule applied to (C;?;;;?1) . . . . . . . . . . . . . . 84
A.3 The Substitution Rule applied to (C;?;;;?2) . . . . . . . . . . . . . . 86
A.4 The Substitution Rule applied to (C;?;;;?3) . . . . . . . . . . . . . . 87
A.5 The Substitution Rule applied to (C;?;;;?4) . . . . . . . . . . . . . . 88
A.6 The Substitution Rule applied to (C;?;;;?5) . . . . . . . . . . . . . . 90
A.7 The Substitution Rule applied to (C;?;;;?6) . . . . . . . . . . . . . . 91
A.8 The Substitution Rule applied to (C;?;;;?7) . . . . . . . . . . . . . . 92
v
Chapter 1
Introduction
1.1 Classical Schubert Calculus
Schubert Calculus was invented in the late nineteenth century by Hermann
Schubert in order to solve various counting problems in projective geometry.
We let the Grassmannian G(k;n) be the set of k-dimensional subspaces of
some flxed n-dimensional complex vector space V. The goal of Schubert Calculus
in this setting is to describe the intersection theory, or equivalently, the cohomology
ring of the Grassmannian as a projective complex manifold.
We flx a complete  ag F? = 0 ? F1 ????? Fn = V where dimC(Fj) = j. For
the Grassmannian G(k;n) it is well known that the Schubert cells/varieties/classes
are indexed by partitions ? whose Young diagrams are contained in a k by n ? k
rectangle. The Schubert cell, X?? corresponding to a partition ? is given by the
following formula:
X?? = f? 2 G(k;n)jdim(?\Fn?k+i??i) = i for i ? kg:
These cells are a?ne spaces of dimension k(n ? k) ?j?j, where j?j = Pi ?i. The
Schubert variety corresponding to the partition ? is the closure of this Schubert cell,
and the class of this variety in the cohomology ring, denoted  ?; is independent of
the choice of  ag F?.
1
The Grassmannian described above is the standard Lie type A Grassmannian.
We can similarly deflne symplectic/orthogonal Grassmannians for the other Lie
types as follows. Let V be a complex vector space of dimension 2n for types C and
D, or of dimension 2n + 1 for type B. Let h; i be an non-degenerate bilinear form
on V, where for type C it is skew symmetric, and for types B and D it is symmetric.
Then we deflne the symplectic/orthogonal Grassmannians as follows. For type C we
have isotropic Grassmannians, IG(n?k;2n) = f?n?k ? Vj8v;w 2 ?n?khv;wi = 0g,
where 0 ? k ? n. Where if k = 0 then we obtain the Lagrangian Grassmannian
LG(n;2n). For type B (resp. type D) we have orthogonal Grassmannians, denoted
OG(n?k;2n+1) (resp. OG(n?k;2n)); which are deflned in the same way.
We can similarly deflne Schubert cells/varieties/classes in these types by flxing
a complete  ag and providing incidence conditions for an isotropic subspace with
respect to this  ag. The indexing set for Schubert varieties will be discussed in
further detail later in the thesis.
Two main problems arise in the Schubert calculus of the Grassmannian. The
flrst problem is the Giambelli problem. In the Grassmannian the Schubert classes
correspondingtopartitionswithonepartarecalledspecialclasses, andaretheChern
classes of the tautological quotient vector bundle over G(k;n). It is known that these
special classes are generators for the cohomology ring overZ. The Giambelli problem
is to flnd a formula which gives any Schubert class in terms of these special classes.
The answer, due to Giambelli, is well known for G(k;n) and is usually given as the
determinant below
 ? = det[ ?i+j?i]1?i;j?k:
2
The other main problem in classical Schubert calculus is determining what happens
when we multiply an arbitrary Schubert class  ? by a special class  p. This is called
the Pieri rule. We note that the Pieri rule equivalently tells one how the Schubert
variety X? intersects the variety Xp. In all of the classical settings if one solves the
Giambelli problem then the Pieri rule follows formally, and visa versa.
1.2 Equivariant Cohomology
In equivariant cohomology, we are considering both the intersection theory of
some homogeneous space and the action of a torus. The general setup is as follows.
Let G be a classical Lie group, flx a Borel subgroup B, let P ? B be a parabolic
subgroup, and let T ? B be a maximal torus. There is an action of T on the
homogeneous space G=P. We flnd a contractible space ET on which T acts freely
and then form
ET ?T G=P := (ET ?G=P)=[(e?t;x)s(e;t?x)]:
This is otherwise known as the quotient stack [TnG=P]. The equivariant cohomology
H?T(G=P) = H?(ET ?T G=P). Since ET ?G=P is homotopy equivalent to G=P we
are in efiect taking the cohomology of G=P modulo the action of T.
For the Grassmannians G=P, where P is usually a maximal parabolic sub-
group, the equivariant cohomology ring is similarly generated by special classes
corresponding to partitions of length one. The difierence is now they generate the
equivariant cohomology ring of the Grassmannian over Z[t1;:::tn], where n is the
rank of G and the ti?s are Chern classes corresponding to characters of the torus.
3
1.3 Equivariant Schubert Calculus
Similarly to the classical situation, equivariant Schubert calculus studies the
intersection theory of the equivariant Schubert varieties. A main problem of equiv-
ariant Schubert calculus is to represent the equivariant Schubert classes by poly-
nomials whose multiplicative structure coincides with the intersection theory of the
Schubert varieties.
Equivariant Giambelli formulas for the usual type A Grassmannian have been
obtained by various authors (cf. [MS], [F2], [FP], etc.). Most of these authors
express a Schubert class as a Schur determinant and note that the classes can be
represented by factorial Schur S-functions where the variables are Chern classes.
There are also nice polynomial representatives for the equivariant Schubert classes
of Grassmannians of maximal isotropic subspaces in types B,C, and D. In these
cases the equivariant Schubert class is usually represented as a Pfa?an and can
be recognized as factorial Schur Q- orP- functions. One reference which includes
both formulas is chapter 3 and chapter 7 of [FP]. We will describe these polynomi-
als in chapters 3 and 4 of the thesis. In all cases the polynomials representing the
equivariant Schubert classes will coincide with the double Schubert polynomials of
Lascoux and Sch?utzenberger for type A, and the double Schubert polynomials of
Ikeda, Mihalcea, and Naruse for types B, C, and D. To recognize these polynomi-
als as representatives of the equivariant Schubert classes for the Grassmannian we
consider the projection of the complete  ag variety into the Grassmannian, and use
the induced injection of the cohomology rings. It is well known that the factorial
4
Schur S functions coincide with the double Schubert polynomials of Lascoux and
Sch?utzenberger. Also Ikeda, Mihalcea, and Naruse show that their double Schubert
polynomials for types B, C, and D coincide with the factorial Schur P functions for
types B and D and factorial Schur Q functions for type C whenever the indexing
permutation of their double Schubert polynomial corresponds to a strict partition.
In the equivariant setting the Pieri rule is signiflcantly more complicated. Thus
one flrst tries to describe the product of a general equivariant Schubert class  ?
with the equivariant class of a Schubert divisor  1. This is known as the equivariant
Chevalley Formula.
1.4 Raising Operators as a Tool for Schubert Calculus
Recently Buch, Kresch, and Tamvakis [BKT] used Young?s raising operators
to solve the Giambelli problem for all Grassmannians in classical Lie types. As
mentioned in the flrst section, the solution to the Giambelli problem for the standard
Lie type A Grassmannian is a determinant. In the case of the maximal isotropic
Grassmannian in Lie type C, otherwise known as the Lagrangian Grassmannian,
the Giambelli problem was solved using a Pfa?an by Pragacz [FP]. The raising
operators give a beautiful way to interpolate between the two solutions and in the
process give a solution for general isotropic Grassmannians which was not known
before.
5
1.5 Outline of Thesis
In my thesis I aim to use Young?s raising operators to give a polynomial rep-
resentation for an equivariant Schubert class in the equivariant cohomology ring of
a Grassmannian of any classical Lie type. In chapter 3, this is done for the type A
Grassmannian G(k;n), where I give a raising operator expression which coincides
with the factorial Schur S functions. I show that my expression does indeed coincide
with the corresponding Schur S function, and then prove the Chevalley formula for
my expression independent of previous results. In chapter 4, I give expressions which
coincide with the Schur Q functions for the Lagrangian Grassmannian, along with
the Schur P functions for the maximal orthogonal Grassmannians. Again I prove
the Chevalley formula independently. Finally, in chapter 5, I give a conjecture for a
Giambelli type formula for a Schubert class in a general symplectic isotropic Grass-
mannian, prove the corresponding Chevalley formula, and prove that it represents
a Schubert class in cases where the indexing partition is su?ciently small.
6
Chapter 2
Preliminaries
2.1 Schubert Cells, Varieties, and Classes
Let G be a complex reductive algebraic group. Fix a Borel subgroup B, and
a maximal torus T ? B. Then if we consider a  ag variety G=P for a parabolic
subgroup P  B, we can index the Schubert varieties of this  ag variety by a
particular subset of the Weyl group W = NG(T)=T. The Schubert cells of this  ag
variety are the B-orbits of G=P, and so are indexed by W=WP, where WP is the
subgroup of W which flxes P. Let w0 be the element of longest length in W. Each
coset vWP has a unique element of shortest length in W. Let WP be the set of
minimal length coset representatives, and let w0 be the element of longest length in
W. Then the Schubert varieties are indexed by the set fw0v 2 W : v 2 WPg. We
denote the Schubert cell as X?w = Bw0wP=P, and its closure, the Schubert variety,
as Xw. The class [Xw] is the Schubert class in the classical cohomology of G=P.
Each Schubert cell contains exactly one torus flxed point which we will denote as
ew 2 X?w.
In the equivariant setting we need to also consider the action of a maximal
torus. Backinsection1.2wedescribedthequotientstack[TnG=P]. Wecansimilarly
deflne ET ?T Xw this will be the equivariant Schubert variety, and its class in the
cohomology ring will be its equivariant Schubert class, which is usually denoted as
7
 Tw. Since the rest of the thesis will be about equivariant Schubert classes, we will
shorten the notation to  w for an equivariant Schubert class corresponding to a Weyl
group element w.
For Weyl group elements u;v;w let cwuv be such that  u ? v = Pw cwuv w. Also
the restriction of  w to the torus flxed point ev will be denoted  wflflv. For a given
classical Lie type, these restrictions will be stable as the rank of G increases. In other
words, let n and m be the ranks of classical Lie groups Gn and Gm of a flxed Lie
type with Weyl groups Wn and Wm respectively. Then if n < m we have Wn ? Wm
and for any v;w 2 Wn, we have that  wflflv 2 H?Tm(ev) is constant as m varies, where
Tm is the corresponding maximal torus for Gm. These descriptions of the Schubert
classes are referred to in [IMN, x1.1] as stable Schubert classes. As in [IMN] we will
let H1 be the span of the stable Schubert classes. We will describe the restriction
map in greater detail for the Grassmannians in the later chapters of the thesis.
2.2 Equivariant Cohomology Rings for Complete Flags in Types A
and C
We note that in every case we can consider the projection G=B ! G=P of a
complete  ag variety G=B onto a Grassmannian G=P. This projection induces an
inclusion of the cohomology ring of the Grassmannian into the cohomology ring of
the complete  ag. In my thesis all of the results, including the conjecture made in
chapter 5, match the descriptions of double Schubert polynomials which describe
the equivariant Schubert classes for a complete  ag; the reader can refer to chapter
8
one of [FP] for the type A double Schubert polynomials, and to chapter eight of
[IMN] for the double Schubert polynomials in other types. I will be using the same
conventions as these sources.
2.2.1 The Presentations in Terms of Chern Classes
Let F? = 0 = F0 ? F1 ? ??? ? Fr = V be a  ag representing a point in
G=B for a given classical Lie group G of rank n and a Borel subgroup B. We
note that if G is type A then r = n, if G is type B then r = 2n + 1 and if G is
types C or D then r = 2n. Also if G is types B,C, or D, then Fi is isotropic for
i ? n and Fi = F?r?i+1. Given a Lie type we can describe the tautological sub- and
quotient bundles fSigri=1 and fQigri=1 of the corresponding  ag variety. The flber
above the point F? for Si is Fi and the flber for Qi is V=Fi. We can make all of these
bundles equivariant by instead considering the point ET ?T F? inside of [TnG=B].
Let e1;:::;er be the standard set of basis vectors for V, then we will also flx a  ag
E? = 0 ? heri ? her;er?1i ? ??? ? her;:::;e1i = V. Then we can deflne a new set
of vector bundles fTigri=1 where the flber above every point of G=B for Ti is Ei.
For all Lie types, let ti = c1(Ti); we note that these will correspond to charac-
ters of the torus. In the following we let ?T =Z[t1;:::;tn].
For type A, the variable xi = c1(ker(Qi ! Qi?1)). Then for the equivariant
cohomology ring we consider the projection
p : G=B ! G(k;n):
This map induces an injection of the equivariant cohomology rings so that we can
9
realize H?T(G(k;n);Z) as a subring of
H?T(G=B;Z) ?= ?T[x1;:::;xn]=hei(x)?ei(t)i;
using the Borel presentation of the equivariant cohomology of the complete  ag
variety as described in [F2] and [LS]. In fact, for the Grassmannian G(k;n), we will
only need the variables x1;:::;xn?k and we will denote this (n?k)-tuple as simply x,
as in case of the ordinary cohomology when the Schubert classes are realized as Schur
polynomials in [F3, 10.6]. In the equivariant case it has been shown [cf. F] that the
double Schubert polynomial Sw?(x;t) of [LS] which corresponds to an equivariant
Schubert class in the Grassmannian is the factorial Schur function s?(xjt), which we
deflne in chapter 3.
For types B, C, and D, x will denote a countably inflnite set of formal variables
x1;x2;:::. Also in types B, C, and D we will need an additional set of variables
zi = c1(Sn?i+1=Sn?i) where 1 ? i ? n. Let T = 'ni=1T ?i where T ?i is the dual
bundle to Ti and recall that for vector bundles A and B the total Chern class of the
formal difierence A?B is deflned by
c(A?B) = c(A)=c(B) = 1+c1(A)+c2(A)+:::1+c
1(B)+c2(B)+:::
so that
1X
i=0
ci(A?B)ui =
P1
i=0 ci(A)u
i
P1
i=0 ci(B)ui
:
As in [IMN, x10] we deflne fli = ci(S?n ? T ) where S?n is the dual bundle to the
tautological subbundle Sn. We note that T and S?n have Chern roots ?t1;:::;?tn
10
and ?z1;:::;?zn respectively so that
1X
i=0
fliui =
Qn
i=1 1?ziuQn
i=1 1?tiu
:
As noted in [IMN, x1.1] we are considering a limit as n ! 1, so for the
stable Schubert classes of the complete  ag we will have inflnitely many z vari-
ables to apply the restriction map to. Recall from x2.1 that H1 is the span of
the stable Schubert classes. In [IMN, x5.2, 6.1] the authors show that H1 in-
jects into lim??H?Tn(Gn=Bn;Z) and, in type C, produce an isomorphism from H1 to
?T[z1;z2;:::]?ZZ[Q1(x);Q2(x);:::] where Qi(x) is the ith Schur Q-function which
is deflned in chapter 4. In [IMN, x10] the authors produce a ring homomorphism
?n : ?T[z1;z2;:::]?ZZ[Q1(x);Q2(x);:::] ! H?Tn(Gn=Bn;Z)
such that ?n(Qi(x)) = fli where ?n(ti) = ti, and ?n(zi) = zi for i ? n while
?n(zi) = 0 for i > n. For any Schubert class  w 2 H?Tn(Gn=Bn;Z), the authors of
[IMN] deflne the type C double Schubert polynomial Cw(x;z;t) 2 ?T[z1;z2;:::]?Z
Z[Q1(x);Q2(x);:::] and the type B and type D double Schubert polynomials
Bw(x;z;t);Dw(x;z;t) 2 ?T[z1;z2;:::] ?Z Z[P1(x);P2(x);:::] where Pi(x) is the
Schur P-function which is deflned in chapter 4. These double Schubert polynomials
have the property that ?n(Sw(x;z;t)) =  w where Sw(x;z;t) = Bw(x;z;t) if Gn =
SO2n+1(C), Sw(x;z;t) = Cw(x;z;t) if Gn = Sp2n(C), and Sw(x;z;t) = Dw(x;z;t)
if Gn = SO2n(C).
Then for the equivariant cohomology ring of the isotropic Grassmannian we
consider the projection p : Gn=Bn ! IG(n?k;2n). Again this induces an injection
of the equivariant cohomology rings, so that we can realize H?Tn(IG(n ?k;2n);Z)
11
as a subring of H?Tn(Gn=Bn;Z). The Borel presentation of the complete  ag in type
C [cf. FP] is given by
H?Tn(Gn=Bn;Z) = ?Tn[z1;z2;:::;zn]=hei(z21;z22;:::;z2n)?ei(t21;t22;:::;t2n)i:
Then for a Schubert class  ? 2 H?Tn(IG(n?k;2n);Z) we have that
p?( ?) =  w? = ?n(Cw?(x;z;t)):
Our goal is to deflne a raising operator expression which will coincide with these
double Schubert polynomials.
2.3 The Vanishing Theorem and Ramiflcations
A major ingredient in proving that a polynomial expression represents a Schu-
bert class is a vanishing theorem for the restriction of a Schubert class to the flxed
points of the torus (cf. [MS] for type A, and [IMN] for other Lie types).
Theorem 1 (Vanishing).  wflflv = 0 unless w ? v.
One notes that we expect this since we should only expect  wflflv 6= 0 if ev 2 Xw,
which only happens if  v   w which implies w ? v in the Bruhat order.
If the vanishing theorem holds then we immediately get several nice equations.
cwww =  wflflw (2.1)
This equation holds since if look at  w ? w = Pcvww v and restrict both sides
to the flxed point corresponding to w we get that the only non-zero term on the
12
right will be when v = w, thus canceling  wflflw from each side we get the desired
equation.
Similarly we obtain
cvwv =  wflflv: (2.2)
From this we can use that our expressions given in chapters 3, 4 and 5 are
homogeneous polynomials in each separate set of variables to show that we have
Schubert classes.
It is well known that the equivariant Schubert classes form a basis for the
equivariant cohomology ring for Grassmannians of all types, where in each type the
indexing set for the Schubert classes in the Grassmannian can be recognized as a set
of partitions. In my thesis I give a raising operator expression in each Lie type for
a general equivariant Schubert class. In each type, my raising operator expression
specializes to a known solution to the classical Giambelli formula when each ti is set
to 0 (cf. [F3] for type A, [BKT] for other classical Lie types). Therefore using that
my expressions are homogeneous polynomials, which are indexed by the same set as
the Schubert classes, one can show that in each case my set of expressions will form
a basis for the given cohomology ring.
To show that the raising operator expression represents an equivariant Schu-
bert class we just need to show that it is the same basis as the Schubert basis for
the equivariant cohomology ring over ?T. Let my basis be denoted as fT?g and let
the Schubert basis be denoted f ?g which we can realize as a set of polynomials
13
by considering the projection of the complete  ag onto the Grassmannian in each
case, and using the induced map on the equivariant cohomology rings to express
each Schubert class as a double Schubert polynomial. Then we expand my basis in
terms of the Schubert basis as follows:
T? =
X
?
a?? ?:
If we assume the vanishing theorem for both bases, then we note that if a?? 6= 0
then ?  ? since otherwise we could restrict to ? and get 0 on the left hand side
and a non-zero value on the right hand side. Since both bases are homogenous
polynomials, we know that j?j ? j?j. Therefore a?? = 0 unless ? = ?. Hence both
bases are the same.
2.4 The Chevalley Formula
For the Grassmannian the Chevalley Formula is a special case of the Pieri rule.
In general the Chevalley formula on the  ag variety G=B tells us how to multiply any
Schubert class by the class of a divisor, denoted by  si, where si is a simple re ection
which is a Weyl group element of length one. Each simple re ection corresponds to
a simple root. The set of all positive roots will be the set of all linear combinations
of simple roots where the coe?cients on the simple roots are all non-negative.
For any positive root fi set cfi;si = (wi;fi_) where (; ) is the standard inner
product for V, wi is the ith fundamental weight corresponding to si and fi_ is the
coroot of fi.
14
Let R+ be the a set of positive roots. The general equivariant Chevalley
Formula (cf. [IMN, Lemma 6.8]) is
 w ? si =  siflflw w +
X
fi2R+;?(wsfi)=1+?(w)
cfi;si wsfi:
In order to prove the Chevalley formula, one simply notes that we are only
multiplying the general class by a polynomial of degree one. One knows that the
classical terms will still appear (i.e. the  wsfi) since when all ti are set to zero we
have the classical expression. Hence we only need to flnd the coe?cient of  w, and
this was shown to be  siflflw already using the vanishing theorem.
In each of the following chapters of the thesis we will prove using a raising
operator approach that the given raising operator expression satisfles the Chevalley
formula, and note that this gives strong evidence that the given expression is a solu-
tion to the Giambelli problem. The author hopes that in the future this information
can be used to prove the conjecture stated in chapter 5.
15
Chapter 3
The Type A Equivariant Giambelli Rule
3.1 Preliminaries for Type A Grassmannians
Consider the Grassmannian G(k;n) of k-planes in n-dimensional complex
space Cn (i.e. G(k;n) ?= GLn(C)=P, where P is the parabolic subgroup of GLn(C)
which consists of block matrices where there are zeros in the bottom left k by n?k
entries and the rest of the entries can be arbitrary so long as the resulting matrix is
invertible).
3.1.1 The Classical Schubert Classes and Schur S functions
In section 9.4 of [F3], Fulton describes how the Schubert classes of G(k;n) are
indexed by partitions ? whose Young diagram flt in a k by (n ? k) rectangle. As
discussed in the introduction, once we flx a complete  ag in type A F? = 0 ? F1 ?
???? Fn = V where the dimC(Fj) = j, then the Schubert cell corresponding to ? is
X?? = X??(F?) = f? 2 G(k;n)jdim(?\Fn?k+i??i) = i for i ? kg:
Also the Schubert variety corresponding to ? is the closure of the cell, so it is
X? = X?(F?) = f? 2 G(k;n)jdim(?\Fn?k+i??i) ? i for i ? kg:
Also in 9.4 of [F3], Fulton describes how one can obtain a cohomology class  ? from
the above variety using Poincar?e duality. The Schubert class,  ?, does not depend
16
on the choice of  ag F?.
Fulton also describes in [F3] how one can represent the Schubert class  ? as
the Schur S polynomial s?(x). In Chapter 6 of [F3], Fulton gives the Jacobi-Trudy
identity for the Schur S functions below.
s?(x) = det[e?i+j?i(x)]1?i;j?k = det[h?0i+j?i(x)]1?i;j?n?k
where er(x) and hr(x) are the elementary and complete symmetric polynomials and
x = (x1;:::xn?k) are the variables described inx2.2.1. For an inflnite list of variables
a = (a1;a2;:::) the generating functions for these polynomials are
E(a;u) =
1Y
i=1
(1+aiu) =
1X
i=0
ei(a)ui H(a;u) =
1Y
i=1
(1?ait)?1 =
1X
i=0
hi(a)ui
We note that if x is a flnite set of k variables then ei(x) = 0 for i > k. The
elementary and complete symmetric polynomials also have the following very nice
property which we will use later.
Let a = (a1;a2;:::) and deflne a(j) to be (a1;:::;aj?1;aj+1;:::) so that aj is
removed. Then
ei(a) = ei(a(j))+ajei?1(a(j)) and hi(a) = hi(a(j))+ajhi?1(a): (3.1)
3.1.2 Torus Fixed Points and the Equivariant Cohomology
Let
E? = 0 ?he1i?he1;e2i?????he1;:::;eni
17
be the complete  ag given by the standard basis vectors. Let T be the usual maximal
torus of diagonal matrices. Then we note that a subspace ? of dimension k will be
flxed by the action of T only when exactly k standard basis vectors give a basis for
the subspace. In other words we have ? = hei1;:::;eiki where 1 ? i1 < i2 < ??? <
ik ? n. One notices that each Schubert cell in G(k;n) will contain exactly one torus
flxed point. For a partition ? whose Young diagram is contained in the k by n?k
rectangle, the torus flxed point is
e? = hen?k??1+1;en?k??2+2;:::;en??ki2 X??(E?):
Let H?T(G(k;n);Z) be the equivariant cohomology of this Grassmannian with
respect to the maximal torus T ?= (C?)n of diagonal matrices in GLn(C). It is
well known that the equivariant Schubert classes form a basis of this ring over
?T = Z[t1;:::;tn] where the ti?s are flrst equivariant Chern classes of line bundles
over the Grassmannian (namely ti is the flrst Chern class of the line bundle whose
flber is the complex line generated by the ith standard basis vector as described
in x2.2.1). To get a presentation of the ring in terms of the variables described
we consider the projection p : Gln(C)=B ! G(k;n). As described in x2.2.1, this
map induces an injection of the equivariant cohomology rings so that we can realize
H?T(G(k;n);Z) as a subring of
H?T(Gln(C)=B;Z) ?= ?T[x1;:::;xn]=hei(x)?ei(t)i1?i?n (3.2)
using the Borel presentation of the equivariant cohomology of the complete  ag
variety. The goal of this section is to use raising operators to express the equivariant
Schubert classes as a polynomial in the presentation (3.2).
18
There is a known representation of Schubert classes as polynomials in these
variables given by factorial Schur S-polynomials. The factorial Schur S-polynomial
s?(xjt) is deflned as one of 2 expressions in Equations (3.3) and (3.4) below:
s?(xjt) = det[(xjjt)
n?k?i+?0i]1?i;j?n?k
Q
i<j(xi ?xj)
(3.3)
where (xjjt)m = Qmr=1(xj ?tr) and ?0 is the conjugate partition to ? (i.e. ?0 has ??s
rows as it?s columns and ??s columns as it?s rows); or
s?(xjt) = det[en?k+i??i?i+j?i ]1?i;j?k (3.4)
where
eqp = ep(x1;:::;xn?kjt1;:::;tq) =
pX
i=0
(?1)iep?i(x)hi(t):
We note that the usual deflnition of s?(xjt) has the conjugate (or transpose)
of the partitions indicated in these equations. We use the transpose so that we can
apply these results directly for the classes indexed by \small" partitions in Chapter
5. We note that the two deflnitions above where shown to be equivalent in [M, I.3];
see also [MS] and [FP,F].
3.1.3 The Type A Grassmann Permutation for a Partition ?
Let ? be a partition whose Young diagram flts in a k by l box where l = n?k.
We recall that all such partitions will index the Schubert classes in the Grassmannian
G(k;n). As in section 7 of [F] we denote the set of jumping numbers for ? to be
I(?) = l +1??1;l +2??2;:::;l +k ??k. Also we similarly set J(?) to be the
19
complement of I(?) in the set f1;:::;ng.
The Weyl group element corresponding to ? is w? where w?(i) = ?0l?i+1+i for
i ? l where ?0 is the conjugate partition for ? and is a strictly increasing sequence
for the remaining l +1 ? i ? n.
Notice that w? has the property that w?(i) < w?(i+1) for all i 6= l.
Proposition 1. We have fw?(1);:::;w?(l)g = J(?).
Proof. Assume for a contradiction that w?(l ?i + 1) 2 I(?) for some i ? l. Then
there exists a j ? k such that l + j ??j = ?0i +l?i + 1. If this were true then we
would have
0 = ?j ?j +?0i ?i+1: (3.5)
Consider cases.
1. If there is a box in the ijth place of the Young diagram of ? then we know that
?j ? i and ?0i ? j. Therefore the right hand side of Equation (3.5) is positive
and we have a contradiction.
2. If there is not a box in the ijth place of the Young diagram of ? then we know
that ?j < i and ?0i < j. Therefore the right hand side of Equation (3.5) is
negative and we have a contradiction.
Hence fw?(1);:::;w?(l)g = J(?) as desired.
20
3.1.4 Restriction to Torus Fixed Points in Type A
We have shown that the flxed points of the torus action on G(k;n) will corre-
spond to partitions, and we call the permutation w? corresponding to the partition
? a Grassmann permutation. In order to get the restriction map we look at the
inclusion ? : pt ,! G(k;n), where pt denotes a torus flxed point. This induces a ring
homomorphism on the equivariant cohomology rings ?? : H?T(G(k;n)) ! H?T(pt) =
Z[t1;:::;tn]. This homomorphism takes the equivariant Chern class xi to the equiv-
ariant Chern class tw?(i) if the point we are restricting to corresponds to ?.
So when restricting to a flxed point e? we have the following:
xi ! tw?(i)
and as l-tuples, where tJ(?) = ftjgj2J(?), we have
x ! tJ(?):
When we restrict a Schubert class  ? to the flxed point e? we will denote the
image as  ?flfl?. In particular the above gives us that
 1flfl? =
X
j2J(?)
tj ?
lX
i=1
ti (3.6)
since in G(k;n),  1 is represented by the polynomial Pli=1(xi ? ti). We will use
Equation (3.6) in x3.4.
21
3.2 The Raising Operator Expression
The raising operator, Rij for i < j, as deflned by A. Young, acts on a tuple
of integers by raising the ith entry by 1, and lowering the jth entry by 1. So for
example R2;4(5;4;3;2;1) = (5;5;3;1;1). For integer sequences ? and  , we deflne
the monomial m?; = Q?i=1 e i?i where ? is the length of the sequence ? (i.e. ? =
maxfi : ?i 6= 0g). In this monomial we agree that if any number in either ? or  
is negative then m?; = 0. We deflne Rijm?; = mRij(?); . From here on out all of
our integer sequences will have flnite length, however we may still regard the integer
sequence as an inflnite sequence with only a flnite number of nonzero entries.
Let
T?; =
Y
1?i<j??
(1?Rij)m?; (3.7)
For a partition ? we deflne  (?) to be the integer sequence such that  (?)i =
n?k +i??i and deflne T? = T?; (?).
Theorem 2. The equivariant Schubert class  ? is represented by T?.
We will prove the above theorem by showing that the raising operator expres-
sion (3.7) satisfles the vanishing theorem.
3.3 The Vanishing Theorem
We show below that our raising expression is equivalent to the expression given
in (3.4).
Let w 2 Sn be a permutation. Let Iw = f(i;j) : i < j;w(i) > w(j)g be the set
22
of inversions of w. Deflne
Rw :=
Y
(i;j)2Iw
Rij:
Then we claim that the following raising operators are equivalent.
Y
1?i<j?n
(1?Rij) =
X
w2Sn
(?1)?(w)Rw: (3.8)
Note that we can think of Rij as acting on a polynomial in the variables
x1;:::;xn by raising the power of the xi by one and lowering the power of xj by
one. In this setting the raising operator Rij is simply multiplication by xi=xj. Let
A =
0
BB
BB
BB
BB
BB
@
1 x1 x21 ::: xn?11
1 x2 x22 ::: xn?12
... ... ... ...
1 xn x2n ::: xn?1n
1
CC
CC
CC
CC
CC
A
:
Then
det(A) =
X
w2Sn
(?1)?(w)xw(1)?11 ???xw(n)?1n =
? nY
j=1
xi?1i
! X
w2Sn
(?1)?(w)xw(1)?11 ???xw(n)?nn
and from Vandermonde we also have
det(A) =
Y
1?i<j
(xj ?xi) =
? nY
j=1
xi?1i
! Y
1?i<j
(1? xix
j
):
Then we simply note that Qni=1 xw(i)?ii = Q(i;j)2Iw xixj which can be proven by in-
duction on ?(w). So Equation (3.8) amounts to two difierent calculations of the
Vandermonde determinant.
Theorem 3. T? = s?(xjt)
23
Proof.
T? =
Y
1?i<j??
(1?Rij)m?; (?)
=
X
w2S?
(?1)l(w)Rw
?Y
i=1
en?k+i??i?i
=
X
w2S?
(?1)l(w)
?Y
i=1
en?k+i??i?i+(w(i)?i)
= det[en?k+i??i?i+(j?i) ]i;j??
= s?(xjt) by Equation (3.4)
Below we include a proof of the vanishing theorem similar to the one for
Theorem 2.1 of [MS] for completeness.
Theorem 4. [MS] s?(t?jt) = 0 unless ?  ?, where t? is the image of the x variables
under the localization map ??? : H?T(G(k;n)) ! H?T(e?) where ???(xi) = tw?(i).
Proof. In this setting we will use Equation (3.3), so recall
s?(xjt) = det[(xjjt)
n?k?i+?0i]1?i;j?n?k
Q
i<j(xi ?xj)
:
Then the numerator of the expression when restricted to ? will be
det[(tw?(j)jt)n?k+i??i]1?i;j?n?k. If ? is not contained in ? this means that there
exists an index r ? k such that ?r > ?r. Similarly there is a part in their conjugate
partitions s ? n?k such that ?0s > ?0s. Then xn?k?s+1 is mapped to t?0s+n?k?s+1.
Then since ?0s > ?0s we have that ?0s +n?k?s ? ?0s +n?k?s+1. For i ? s ? j
24
we have the following inequality:
1 ? ?0j +n?k ?j +1 ? ?0s +n?k ?s+1 ? ?0s +n?k ?s ? ?0i +n?k ?i:
Note that the matrix given by A = [(tw?(j)jt)n?k+i??i]1?i;j?n?k = [aij] will have the
property that
aij = (t?0j+n?k?j+1 ?t1)???(t?0j+n?k?j+1 ?t?0i+n?k?i)
so aij = 0 for all i ? s ? j. This gives us a block of zeros which implies that
det(A) = 0, and thus proves the theorem.
Combining Theorems 3 and 4 we have the following Corollary.
Corollary 1. T? satisfles the vanishing theorem so that if T?flfl? 6= 0 then ? ? ?.
Proof. This is a consequence of the above theorems and Macdonald?s proof of the
equivalence of Equations (3.4) and (3.3).
3.4 The Chevalley Formula
We will show that our raising operator expression when multiplied by  1 has
the same outcome as when we multiply the corresponding equivariant Schubert
class by  1. We note that this is true as a consequence of x3.3, but we will give an
independent raising operator proof.
Recall that the equivariant Chevalley Formula for G(k;n) is
 ? ? 1 =  1flfl? ? +
X
?+
 ?+
25
where  1flfl? is the restriction of the divisor class to a flxed point of the torus, and
where ?+ is a partition containing ? such that j?+j = j?j+1.
To prove the analogue of this statement for our T? expressions we will use a
series of lemmas. The flrst lemma gives a relation between two difierent T expres-
sions.
Lemma 1. For any integer sequences ?;?; and  , and integers r and s we have:
T(?1;:::;?j?1;r;s;?); = ?T(?1;:::;?j?1;s?1;r+1;?);sj :
where sj is the transposition which switches the jth and (j +1)th entries.
Proof. We prove this statement via an induction argument similar to that of Buch,
Kresch and Tamvakis in [BKT, Lemma 1.1] for isotropic Grassmannians.
Note that
T(?1;:::;?j?1;r); =
X
fi2f0;1gj?1
(?1)jfijT?+fi; Tr?jfij; j (3.9)
where Tr?jfij; j = e jr?jfij
Using equation 3:9 once we prove the statement is true for ? = ; we will have
our result by induction on ?(?). For the base case we assume that the ? in our
lemma is empty and prove the statement T(?1;:::;?j?1;r;s); = ?T(?1;:::;?j?1;s?1;r+1);sj :
Here we apply the above recursion twice to get
T(?1;:::;?j?1;r;s); =
X
fi;fl2Zf0;1gj?1
(?1)jfij+jfljT?+fi+fl; (e jr?jfije j+1s?jflj ?e jr+1?jfije j+1s?1?jflj):
Similarly we get
T(?1;:::;?j?1;s?1;r+1);sj =
X
fi;fl2Zf0;1gj?1
(?1)jfij+jfljT?+fi+fl; (e j+1s?1?jflje jr+1?jfij ?e j+1s?jflje jr?jfij):
26
Hence T(?1;:::;?j?1;r;s); = ?T(?1;:::;?j?1;s?1;r+1);sj and so the base case is proven. Then
we use Equation (3.9) to perform the inductive step and we have proven the lemma.
Corollary 2. Let ? be a partition such that ? = (?;r;r;?). Also let  be such that
 i =
8
>><
>>:
 (?)i if i 6= l(?)+1
 (?)i +1 if i = l(?)+1
Then T(?;r+1;r;?); = 0.
Proof. Note that  (?)l(?)+1 = n ? k ? r + l(?) + 1 and  (?)l(?)+2 = n ? k ? r +
l(?) + 2. Hence  = sl(?)+1 and by Lemma 1 we get the expression T(?;r+1;r;?); =
?T(?;r+1;r;?); so T(?;r+1;r;?); = 0.
The next lemma is a general relation between factorial elementary symmetric
polynomials which is necessary later in showing that the T??s produce the correct
structure constants.
Lemma 2. er+1p+1 = erp+1 ?tr+1 ?er+1p
Proof.
er+1p+1 =
pX
i=0
(?1)iep+1?i(x)hi(t1;:::;tr+1)
=
pX
i=0
(?1)iep+1?i(x)(hi(t1;:::;tr)+tr+1hi?1(t1;:::;tr+1))
= erp+1 ?er+1p (tr+1)
via a shifting of indices.
27
The next lemma gives us the Chevalley formula for multiplying a T? by T1.
We note that
(e1(x)?h1(t))flfl? =
X
j2J(?)
tj ?
n?kX
i=1
ti =  1flfl?
so that the expression below does indeed coincide with the Chevalley formula dis-
cussed in x3.1.4 Equation (3.6).
We recall that the J-set, J(?), for a partition ? is the set of numbers less than
or equal to n which are not jumping numbers for the partition (i.e. not equal to
n?k +i??i for some i).
Lemma 3.
T? ?T1 =
X
?+
T?+ +
0
@ X
j2J(?)
tj ?
n?kX
i=1
ti
1
AT?:
where ?+ is such that j?+j = ?+1 and ? ? ?+.
Proof. Note that T1 = en?k1 = en?k+r1 +Pri=1 tn?k+i; and ?+ must be a partition of
length at most one more than the length of ?. We enable raising operator expressions
of length ? + 1 by multiplying the expression by Q?i=1 1?Ri;?+11?Ri;?+1. We note that since
there is only one box added to the diagram of ?, so (1 ? Ri;?+1)?1 = 1 + Ri;?+1 +
R2i;?+1+R3i;?+1 ::: will act the same as simply (1+Ri;?+1) since Rri;?+1 will act on the
28
monomial and result in an expression which is zero. Therefore
T? ?T1
=
Y
1?i<j??
(1?Rij)m?; (?) ?en?k1
=
Y
1?i<j??
(1?Rij)
?
1+
?X
r=1
Rr;?+1
!? ?Y
i=1
e (?)i?i
!
?en?k+1+??11 +
? ?X
i=1
tn?k+i
!
T?; (?)
=
Y
1?i<j??+1
(1?Rij)
?
m?;1; (?;1) +
?X
r=1
?r?1Y
i=1
en?k??i+i?i ?en?k??r+r?r+1 ?
?Y
i=r+1
en?k??i+i?i
!!
+
? ?X
i=1
tn?k+i
!
T?; (?)
by the previous lemma en?k??r+r?r+1 = en?k??r?1+r?r+1 +en?k??r+r?r (?tn?k+r??r) so the above
=
Y
1?i<j??+1
(1?Rij)
?
m?;1;? +
?X
r=1
?r?1Y
i=1
e (?)i?i ?e (?)?1?r+1 ?
?Y
i=r+1
e (?)i?i
!!
+
? ?X
i=1
(?tn?k+i??i +tn?k+i)
!
T?; (?)
=
X
?
T?; (?) +
? ?X
i=1
(tn?k+i ?tn?k+i??i)
!
T?
where ? is obtained by adding one box to any row of ?
=
X
?+
T?+ +
0
@ X
j2J(?)
tj ?
n?kX
i=1
ti
1
AT?
since Lemma 1 implies that T? = 0 for any ? which is not a partition and since
fn?k +1;:::;ngrI(?) = I(?)crf1;:::;n?kg = J(?)rf1;:::;n?kg.
Hence we have shown the given raising operator expression satisfles the Cheval-
ley formula independent of the previous results for factorial Schur S functions. In
29
the following chapters we will look at the other Lie types.
30
Chapter 4
The Giambelli Formula for the Maximal Isotropic Grassmannians
4.1 Preliminaries for the Lagrangian Grassmannian
Let V be a 2n-dimensional complex vector space and h; i be a skew-symmetric
bilinear form on V. Consider the Lagrangian Grassmannian
LG(n;2n) = f? ? Vjdim(?) = n;8v;w 2 ?hv;wi = 0g:
We call a subspace W of V isotropic if 8v;w 2 W;hv;wi = 0.
Notice that LG(n;2n) is the set of maximal isotropic subspaces of V.
Let F? = 0 ? F1 ? ??? ? F2n = V be a type C complete  ag so that for
i ? n;Fi is an isotropic subspace of V, and F2n?i = F?i . Then Schubert cells and
varieties are deflned similarly to the type A case. The Schubert varieties for LG
are indexed by strict partitions ? with ?1 ? n, where a partition is called strict
if ?i > ?i+1 for all i such that ?i > 0. For such a strict partition ? the Schubert
variety corresponding to ? is given by
X? = X?(F?) = f? 2 LG(n;2n)jdim(?\Fn+1??j) ? j for j ? l(?)g:
The corresponding Schubert cell is
X?? = f? 2 LG(n;2n)jdim(?\Fn+1??j) = j for j ? l(?)g:
Like the type A case, from any Schubert variety we can obtain a class in the coho-
mology ring which does not depend on the choice of  ag.
31
For LG(n;2n) we similarly have that the torus flxed points are n dimensional
subspaces generated by exactly n basis vectors. We recall that for a vector space V 2n
the standard basis vectors are e1;:::;e2n with the standard antidiagonal symplectic
form h; i where hei;e2n?i+1i = 1 and hei;eji = 0 for all j 6= 2n?i + 1. Let T be
the maximal torus with respect to this basis. Let E? = 0 ? he1i ? he1;e2i ? ??? ?
he1;:::;e2ni. Similarly, any Schubert cell will contain exactly one torus flxed point.
For any strict partition ? of length ? ? n, the torus flxed point is
e? = hen+1??1;:::;en+1???;en+i1;:::en+in??i2 X??(E?)
where i1 < i2 < ??? < in?? is such that ?j 6= ir for any j;r.
It is well known that the classical Schubert classes on LG(n;2n) can be rep-
resented by Schur Q functions. Let x = (x1;x2;:::). Then the functions qi(x) are
deflned by the generating function:
1Y
i=1
1?xiu
1+xiu =
1X
i=0
qi(x)ui:
For a pair of positive integers r;s, we set
Qr;s(x) = qr(x)qs(x)+2
X
i?1
(?1)iqr+i(x)qs?i(x) =: 1?R1;21+R
1;2
qr(x)qs(x):
For any strict partition ? of we can regard it has having even length ? by setting
the last part of ? be zero if the actual length of ? is odd, and leaving it unchanged
otherwise. Let
Q?(x) = Pf[Q?i;?j]i<j??:
It is well known that
Q?(x) =
Y
1?i<j?n
1?Rij
1+Rijm?
32
where m? = Qi q?i(x). In the above the operatorQ1?i<j?n 1?Rij1+Rij acts on a monomial
m? just as in Chapter 3 where we expand each factor 1?Rij1+Rij := 1 ? 2Rij + 2R2ij ?
2R3ij +::: as a formal sum.
4.2 The Raising Operator Expression
We deflne the factorial Schur Q-functions in terms of the usual Schur Q func-
tions and elementary symmetric polynomials as follows in a similar manner as in
section 4 of [IMN]. For a positive integer k, the one part factorial Schur Q function
is
Qrk(xjt) :=
k?1X
i=0
(?1)iQk?i(x1;x2;:::)ei(t1;:::;tr):
Then the 2 part factorial Schur Q function is
Qk;l(xjt) := Qk?1k (xjt)Ql?1l (xjt)+2
1X
i=1
(?1)iQk?1k+i(xjt)Ql?1l?i(xjt):
Finally the factorial Schur Q function for a strict partition ? of even length ? is
deflned as
Q?(xjt) = Pf(Q?i;?j(xjt))i<j??: (4.1)
Proposition 2. [IMN, Prop. 4.2] For a strict partition ?, Q?(xjt) satisfles the
vanishing theorem.
We set the following notation:
Qrk[j] :=
k?1X
i=0
(?1)iQk+j?i(x1;x2 :::)ei(t1;:::;tr)
= 2
1X
i=1
xk?r+ji (xi ?t1)???(xi ?tr):
33
For integer sequences ?; ;? where the length of ? is ? we let m?; ;? = Q?i=1 Q i?i[?i].
Then a raising operator R acts on m?; ;? by acting on only ?. We notice
Qk;l(xjt) = 1?R1;21+R
1;2
Qk?1k [0]Ql?1l [0]: (4.2)
We deflne
T?; ;? :=
Y
1?i<j??
1?Rij
1+Rijm?; ;?:
In LG(n;2n) we deflne  (?) such that  (?)i = ?i?1. For the remainder of the
thesis when we call an integer sequence 0 we mean the inflnite sequence of zeros.
Deflne T? = T?; (?);0. We wish to show that T? is the same as the factorial
Schur-Q polynomials described in section 4 of [IMN]. We note that since  (?)i =
?i?1 we have that Q (?)i?i [j] = Q?i?1?i+j[0] since for j < 0 this is true for any  and for
j > 0 we will have that er(t1;:::;t?i?1) = 0 for r > ?i ?1 so that
Q?i?1?i [j] =
?i?1X
i=1
(?1)iQ?i+j?i+1(x1;x2 :::)ei(t1;:::;t?i?1)
=
?i+j?1X
i=1
(?1)iQ?i+j?i+1(x1;x2 :::)ei(t1;:::;t?i?1)
= Q?i?1?i+j[0]:
Hence for any shift in the index ? is equivalent to the same shift in the index ?.
From this we can assume that our raising operator is acting on the sequence ? rather
than the sequence ? in the special case of T?.
Note that
Y
1?i<j??
 1?R
ij
1+Rij
?
= Pf
 1?R
ij
1+Rij
?
i<j??
due to an identity of Schur.
34
We have
T? =
Y
1?i<j??
 1?R
ij
1+Rij
? ?Y
k=1
Q?k?1?k
= Pf
 1?R
ij
1+Rij
? ?Y
k=1
Q?k?1?k
= 12?=2(?=2)!
X
 2S?
(?1)l( )
?=2Y
i=1
 1?R
 (2i?1); (2i)
1+R (2i?1); (2i)
? ?Y
k=1
Q?k?1?k
= 12?=2(?=2)!
X
 2S?
(?1)l( )
?=2Y
i=1
Q? (2i?1);? (2i)
= Pf(Q?i;?j)i<j??
= Q?
From this and Proposition 2 we have that T? will satisfy the same vanishing
theorem as in [IMN] and x2.3, and thus represents the equivariant Schubert class
 ?.
4.3 The Chevalley Formula
The equivariant Chevalley formula for LG(n;2n) states
 ? ? 1 =  ?;1 +2
X
?+
 ?+ +2
?X
i=1
t?i ?
where again ?+ is a strict partition obtained from adding a box to ? and  1flfl? =
P
i 2t?i as in [I, x4.5]. We show that the T? satisfy the equivariant Chevalley formula
for LG(n;2n) independently. First we will prove a lemma similar to Lemma 1.
35
Lemma 4. For any integer sequences ?;?; and  , where the length of ? is j?1 and
any integers r and s we have that T(?1;:::;?j?1;r;s;?); ;0 = ?T(?1;:::;?j?1;s;r;?);sj ;0.
Proof. We will prove this using the following recursion: For any partition ? of length
?
T?;r; =
X
fi2Z??0
(?1)jfij2#fi:fii>0gT?; ;fiQ ?+1r [?jfij]:
Then similarly to the proof of Lemma 1 we proceed by induction on the length
of ? and using the recursion above we need only prove the base case where ? is empty.
So applying the above recursion twice we get
T(?1;:::;?j?1;r;s); ;0 =
X
fi;fl2Zj?0
(?1)jfij+jfljT?; ;fi+fl?
0
@Q jr [?jfij]Q j+1s [?jflj]+2 X
i2Z?0
(?1)iQ jr [i?jfij]Q j+1s [?i?jflj]
1
A:
We similarly get
T(?1;:::;?j?1;s;r);sj ;0 =
X
fi;fl2Zj?0
(?1)jfij+jfljT?; ;fi+fl?
0
@Q j+1s [?jflj]Q jr [?jfij]+2 X
i2Z?0
(?1)iQ js [i?jflj]Q j+1r [?i?jfij]
1
A:
The fact that Qr;s(x) = ?Qs;r(x) is a well known relation of Q functions. We
36
notice in the above that
0
@Q jr [?jfij]Q j+1s [?jflj]+2 X
i2Z?0
(?1)iQ jr [i?jfij]Q j+1s [?i?jflj]
1
A
=
?r?1X
k=0
(?1)kQr?jfij?k(x)ek(t1;:::;t j)
!?s?1X
k=0
(?1)kQs?jflj?k(x)ek(t1;:::;t j+1)
!
+2
X
i2Z?0
(?1)i
?r?1X
k=0
(?1)kQr?jfij?k+i(x)ek(t1;:::;t j)
!
?
?s?1X
k=0
(?1)kQs?jflj?k?i(x)ek(t1;:::;t j+1)
!
=
X
p<r;q<s
(?1)p+qep(t1;:::;t j)eq(t1;:::;t j+1)?
0
@Qr?jfij?p(x)Qs?jflj?q(x)+2 X
i2Z?0
(?1)iQr?jfij?p+i(x)Qs?jflj?q?i(x)
1
A
=
X
p<r;q<s
(?1)p+qep(t1;:::;t j)eq(t1;:::;t j+1)Qa;b(x)
where a+p = r?jfij and b+q = s?jflj. Similarly
0
@Q j+1s (?jflj)Q jr (?jfij)?2 X
i2Z?0
Q js (i?jflj)Q j+1r (?i?jfij)
1
A
=
X
p<r;q<s
(?1)p+qep(t1;:::;t j)eq(t1;:::;t j+1)Qb;a(x)
where a+p = r?jfij and b+q = s?jflj. Hence using the relation Qr;s(x) = ?Qs;r(x)
we get that T?1;:::;?j?1;r;s; = ?T?1;:::;?j?1;s;r;sj and the lemma follows.
Corollary 3. If ? = (?;r;r;?) then T? = 0.
Proof. This is immediate from the last lemma since  (?)j =  (?)j+1 if the r is the
jth and (j +1)th row of ?.
37
Now we will give a relation amongst the factorial Schur Q polynomials which
will be useful in proving the Chevalley formula.
Lemma 5. For integers p and r, Qrp+1[0] = Qr+1p+1[0]?tr+1Qrp[0].
Proof.
Qr+1p+1[0] =
pX
i=0
(?1)iQp+1?i(x)ei(t1;:::;tr+1)
=
pX
i=0
(?1)iQp+1?i(x)(ei(t1;:::;tr)+tr+1ei?1(t1;:::;tr))
= Qrp+1[0]?tr+1Qrp[0] once we shift indices:
Theorem 5. For a strict partition ? of length ?, we have that T? satisfles the
Chevalley formula for LG(n;2n).
38
Proof. Note that T1 = Q01[0] so we have the following:
T? ?T1
=
Y
1?i<j??
 1?R
ij
1+Rij
? ?Y
i=1
Q?i?1?i [0]?Q01[0]
=
Y
1?i<j??+1=l
 1?R
ij
1+Rij
?
(1+2
?X
r=1
Ri;l)
?Y
i=1
Q?i?1?i [0]?Q01[0]
= T?;1; (?;1);0 +2
?X
r=1
Y
1?i<j??
 1?R
ij
1+Rij
?r?1Y
i=1
Q?i?1?i [0]?Q?r?1?r [1]?
?Y
i=r+1
Q?i?1?i [0]
= T?;1; (?;1);0 +2
?X
r=1
Y
1?i<j??
 1?R
ij
1+Rij
?r?1Y
i=1
Q?i?1?i [0]?Q?r?1?r+1[0]?
?Y
i=r+1
Q?i?1?i [0]
= T?;1 +2
?X
r=1
Y
1?i<j??
 1?R
ij
1+Rij
?r?1Y
i=1
Q?i?1?i [0]?(Q?r?r+1[0]+t?rQ?r?1?r )[0]?
?Y
i=r+1
Q?i?1?i [0]
= T?;1 +2
X
?
T?; (?) +2
?X
r=1
t?rT?:
Where ? is obtained from ? by adding one box. By Lemma 4 the only ? where T?; (?)
is non-zero will be strict partitions. Hence we have proven the Chevalley formula
for LG(n;2n).
4.4 Giambelli Revisited for Maximal Orthogonal Grassmannians
We will now consider the maximal orthogonal Grassmannians OG(n;2n) and
OG(n;2n + 1). These are isomorphic as varieties, but the Lie groups which act
on them are difierent. The Lagrangian Grassmanian and the maximal orthogonal
Grassmannians are deflned similarly; the difierence is that the Lagrangian Grass-
39
mannian is a set of subspaces which are isotropic with respect to a skew-symmetric
form, while the orthogonal Grassmannians are sets of subspaces which are isotropic
with respect to a symmetric form. Given the similarity between the two spaces it is
not surprising that the Schubert classes will have similar representatives.
Here we deflne factorial Schur P functions as section 4 of in [IMN].
Prk(j) = Pk+j(xjt1;:::;tr)
=
k?1X
i=0
(?1)iPk+j?i(x1;x2 :::)ei(t1;:::;tr)
=
1X
i=1
xk?r+ji (xi ?t1)???(xi ?tr):
As in the case of the Lagrangian Grassmannian, the 2-part factorial Schur P func-
tions are deflned as:
Pk;l(xjt) = Pk(xjt)Pl(xjt)+2
X
i?1
(?1)iPk+i(xjt)Pl?i(xjt) =
 1?R
1;2
1+R1;2
?
Pk?1k [0]Pl?1l [0]
as in [IMN]. In fact the reader will notice that the only difierence between the P
functions and the Q functions is a factor of two. Also similar to the Lagrangian case,
for any strict partition ? of even length ?, where again we set every strict partition
to have even length by making the last part 0 if the strict partition is actually of
odd length, we have
P?(xjt) = Pf[P?i;?j]1?i;j??:
Thus we can take our Giambelli Formula for the Lagrangian Grassmannian, multiply
it by the appropriate factor of 12, and appropriately shift the t variables as in [IMN,
Theorem 6.6] for whether we are in the even or odd orthogonal case to obtain an
equivariant Giambelli formula for the orthogonal Grassmannians.
40
Theorem 6. Set
T? = 2??
Y
1?i<j??
 1?R
ij
1+Rij
? ?Y
i=1
Q?i?i:
Then T? represents the Schubert class corresponding to ? in a maximal orthog-
onal Grassmannian, where we flx t1 = 0 in the odd orthogonal case.
We can show that these formulas coincide with the formulas in [IMN, Theorem
6.6] similarly to the case of LG(n;2n) where we note that we are using the factorial
Q functions with the appropriate factor of 2 removed instead of using the factorial
P functions for the theorem. This is to ensure that the equivariant correction term
T1flfl? in the Chevalley formula will not have the factor of 2 which is inherited from
the expansion of the raising operator to include Ri;?+1.
41
Chapter 5
The Giambelli Formula for General Isotropic Grassmannians
5.1 Preliminaries for IG(n?k;2n)
We now consider the non-maximal symplectic Grassmannian IG(n ? k;2n)
where k ? n. IG(n?k;2n) = f?n?k ? Vj8v;w 2 ?n?k hv;wi = 0g, where h ; i is
a symplectic form on V ?C2n.
In [BKT2], the authors describe how any Schubert variety in the symplectic
Grassmannian, IG(n?k;2n), is indexed byak-strict partition whose Young diagram
flts inside a (n ? k) by (n + k) rectangle. A partition ? is called k-strict if for
all i such that ?i > k we have that ?i > ?i+1. Fix a complete  ag in type C,
F? = 0 ? F1 ? ??? ? F2n = V so that for i ? n;Fi is an isotropic subspace of V,
and F2n?i = F?i . Then Schubert cells and varieties are deflned similarly to the type
A case. For any k-strict partition ?, the Schubert variety is described below as it is
in section 1 of [BKT2]
X? = X?(F?) = f? 2 IG(n?k;2n)jdim(?\Fpj(?)) ? j for 1 ? j ? ?(?)g
where for any k-strict partition ?,
pj(?) = n+k +1??j +#fi < j : ?i +?j ? 2k +j ?ig: (5.1)
The corresponding Schubert cell is
X?? = f? 2 IG(n?k;2n)jdim(?\Fpj(?)) = j for 1 ? j ? ?(?)g:
42
Again from any Schubert variety we can obtain a class in the cohomology ring of
IG(n ? k;2n) which does not depend on the choice of  ag F?. When k = 0, we
denote IG(n;2n) by LG(n;2n), as it is the Lagrangian Grassmannian of maximal
isotropic subspaces described in Chapter 4.
For IG(n ? k;2n) we similarly have that all torus flxed points for the stan-
dard maximal torus are n ? k dimensional subspaces generated by exactly n ? k
standard basis vectors. We recall that for a vector space V 2n with an antidiago-
nal skew symmetric non-degenerate bilinear form h; i, the standard basis vectors
are e1;:::;e2n where hei;e2n?i+1i = 1 and hei;eji = 0 for all j 6= 2n ? i + 1. Let
E? = 0 ?he1i?he1;e2i?????he1;:::;e2ni. Similarly, we have that any Schubert
cell, X??(E?), will contain exactly one torus flxed point. For any k-strict partition
?, the torus flxed point is given by:
e? = hep1(?);:::;epn?k(?)i2 X??(E?)
where pj(?) is as deflned in Equation (5.1).
5.2 The Theta Polynomials of [BKT] and the Classical Chevalley
Formula
We begin by recalling a few deflnitions from [BKT].
Deflnition 1. [BKT2] Let ? be a Young diagram. Then the box [r;c] and the box
[r0;c0]are called k-related if c + c0 = 2k + 2 + r ?r0 where c ? k < c0. For example
in Figure 5.1 below the two marked boxes are k-related.
43
Figure 5.1: k-related: In this example the box [r;c] is k-related to [r0;c0].
For an integer sequence ?, we deflne C(?) = f(i;j)ji < j ? ?(?);?i + ?j >
2k + j ?ig as in [BKT], and we let ?? = f(i;j) 2N?Nj1 ? i < jg. We deflne a
partial order on this set by (i;j) ? (i0;j0) if i ? i0 and j ? j0. We call a flnite subset
of ?? valid if it is an order ideal. We recall from [BKT] that a subset is valid ifi it
is equal to C(?) for some k-strict partition ?.
Deflnition 2. For any valid set of pairs D the raising operator RD is deflned as
RD =
Y
i<j
(1?Rij)
Y
(i;j)2D
(1+Rij)?1:
We set RC(?) = R?, and notice that
R? =
Y
1?i<j??
(1?Rij)
Y
(i;j)2C(?)
(1+Rij)?1
where ? is the length of the partition ?.
Deflnition 3. [BKT, x5.1] Let x = (x1;x2;:::) and z = (z1;:::;zk) be the variables
deflned for type C in section x2.2.1. For a positive integer r, the standard Theta
polynomial  r(x;z) := Pri=0 qr?i(x)e(z) where qi(x) is the usual Schur q function of
degree i. For an integer sequence fi, set  fi = Q?i=1  fii.
44
Then for a k-strict partition ?, the theta polynomial ?? is deflned by
?? = R? ?:
In [BKT], the authors prove that the above theta polynomials satisfy the Pieri
rule for IG(n?k;2n), where n is su?ciently large. Below we specialize their proof
to the Chevalley formula.
For any two k-strict partitions ? and ?, we write ? ! ? if ? may be obtained
by adding a box to ?, or removing r boxes from a single column of the flrst k columns
of ? and adding r+1 boxes in a single row of the result, so that the removed boxes
and the bottom box of ? in that column are each k-related to one of the added
boxes. If ? ! ?, let e?? = 2 if ?  ? and the added box in ? is not k-related to
a bottom box in one of the flrst k columns of ?, and otherwise set e?? = 1. The
Chevalley rule for IG states that
 1 ? ? =
X
?!?
e??  ?: (5.2)
For any valid set D and any integer sequence fi we deflne T(D;fi) = RD fi so
that ?? = T(C(?);?).
Theorem 7.
 1 ?T(C(?);?) =
X
?!?
e?? T(C(?);?):
In [BKT] this is proven via a series of Lemmas and a substitution rule. The
following Lemmas are taken from [BKT].
Lemma 6. [BKT, Lemma 1.2] Let ? = (?1;:::;?j?1) and ? = (?j+2;:::;??) be
integer vectors. Assume that (j;j + 1) =2 D and that for each h < j, (h;j) =2 D ifi
45
(h;j +1) =2 D. Then for any integers r and s, where D = C(?;r;s;?) we have
T(D;(?;r;s;?)) = ?T(D;(?;s?1;r +1;?)):
In particular, T(D;(?;r;r +1;?)) = 0.
Lemma 7. [BKT, Lemma 1.3] Let ? = (?1;:::;?j?1) and ? = (?j+2;:::;??) be
integer vectors, assume (j;j + 1) 2 D, and that for each h > j + 1, (j;h) 2 D ifi
(j +1;h) 2 D. If r;s 2Z are such that r+s > 2k, where D = C(?;r;s;?), then we
have
T(D;(?;r;s;?)) = ?T(D;(?;s;r;?)):
In particular, T(D;?;r;r;?) = 0 for any r > k.
Below we give a simplifled version of the substitution rule of [BKT] which
su?ces in proving the Chevalley formula above holds.
5.2.1 The Substitution Rule
We flx the k-strict partition ? of length ?, let C = C(?), and choose n su?-
ciently large so that all Pieri terms that can possibly appear in the Chevalley formula
do not vanish. For any d ? 1 deflne the raising operator R?d by
R?d =
Y
1?i<j?d
(1?Rij)
Y
i<j:(i;j)2C
(1+Rij)?1:
We compute that
 1 ??? =  1 ?R??  ? = R??+1 ?
?Y
i=1
(1?Ri;?+1)?1  ?;1 = R??+1 ?
?Y
i=1
(1+Ri;?+1) ?;1
46
and therefore
 1 ?T(C;?) =
?+1X
j=1
T(C;?(j)); (5.3)
where ?(j) is the integer sequence obtained from ? such that ?(j)i = ?i for all i 6= j
and ?(j)j = ?j + 1. We aim to show that the right hand side of Equation (5.3) is
equal to the right hand side of the Chevalley rule.
Let m ? 1 be minimal such that ?m ? k; we call m the middle row of ?.
Deflnition 4. A valid triple is a triple ? = (D;?;S), such that D is a valid set of
pairs containing C, all pairs (i;j) in D satisfy i ? m and j ? ? + 1, ? is an integer
sequence of length at most ? + 1, and S is a subset of DrC. The evaluation of ?
is deflned by ev(?) = T(D;?) 2 H?(IG;Z).
All valid triples encountered here will also satisfy that D ? C [ @C, but for
technical reasons we do not require this in the deflnition. We will represent the set
?? as the positions above the main diagonal of a matrix, and the various sets of
pairs in D as sets of entries in this matrix, as in [BKT].
For 1 ? h ? m, we let bh = minfj ? m j (h;j) 62 Cg and gh = bh?1 (by
convention we set g1 = ?+1); the next result follows from [BKT, Lemma 3.5].
Lemma 8. If 2 ? h ? m then we have ?h?1 ??h ? gh ?bh +1.
Let (i;j) 2 ?? be arbitrary. We deflne a weight condition W(i;j) on an integer
sequence ? as follows.
W(i;j) : ?i +?j > 2k +j ?i:
47
Figure 5.2: The set C near the pair x = (h;gh) 2 @C
Also we will call (i;j) an outer corner of a valid set D if D [ (i;j) is also a
valid set. We note that if there is no outer corner in column j then this means that
(i;j ?1) =2 D ifi (i;j) =2 D, and similarly if there is no outer corner in row i then
this means that (i?1;j) 2 D ifi (i;j) 2 D. This gives us a new way to view the
hypotheses of Lemmas 6 and 7.
The following substitution rule will be applied iteratively to rewrite the right
hand side of (5.3). It may be applied to any valid triple and will result in either a
REPLACE statement, indicating that the triple should be replaced by one or two
new triples, or a STOP statement, indicating that the triple should not be replaced.
Substitution Rule
Let (D;?;S) be a valid triple. Let h ? ? + 1 be largest such that one of the
following four conditions is true (if none hold for any h, then STOP).
(i) (h ? 1;h) =2 D and there is an outer corner (i;h) of D with i ? m such
that W(i;h) holds;
(ii) (h?1;h) =2 D, D has no outer corner in column h, and ?h = ?h?1 +1;
48
(iii) (h ? 1;h) 2 D and there is an outer corner (h;j) of D with j ? ? + 1
such that W(h;j) holds;
(iv) (h?1;h) 2 D and ?h = ?h?1.
If condition (i) holds, then REPLACE (D;?;S) with (D [ (i;h);?;S) and (D [
(i;h);Rih?;S[(i;h)). If(iii)holds, thenREPLACE(D;?;S)with(D[(h;j);Rhj?;S[
(h;j)) if ?j > ?j?1, or REPLACE (D;?;S) with (D [(h;j);Rhj?;S [(h;j)) and
(D[(h;j);?;S) if ?h ? ?h?1. If (ii) or (iv) holds, then STOP.
Deflnition 5. [BKT, Deflnition 3.9] Let (x) be one of the conditions (i){(iv) of
the Substitution Rule. We say that a valid triple ? meets condition (x) if ? reaches
condition (x) in the Substitution Rule, and condition (x) is satisfled. The corre-
sponding integer h ? 1 is called the level of (D;?;S). Whenever the Substitution
Rule REPLACES ? by one or two triples ?i, we refer to ? as the parent term and
the ?i are its children.
Initially we let ? = f(C;?(j);;) : 1 ? j ? ? + 1g so that P?2? ev(?) agrees
with the right hand side of Equation (5.3). We then apply the above substitution
rule which will change this set by replacing some triples with one or two new valid
triples. Whenever the substitution rule results in a REPLACE statement, then
the set is changed accordingly; otherwise the substitution rule results in a STOP
statement, in which case the triple (D;?;S) is left unchanged. These substitutions
are iterated until no further elements can be REPLACED, i.e., until the substitution
rule results in a STOP statement when applied to any remaining triple. Since the set
of pairs D is not allowed to grow beyond column ?+1, this algorithm will terminate
49
after a flnite number of steps.
Suppose that the triple ? = (D;?;S) occurs in the algorithm. If ? is RE-
PLACED by two triples ?1 and ?2, we deduce from the identity
1?Rij = 1?Rij1+R
ij
+ 1?Rij1+R
ij
Rij
that ev(?) = ev(?1) + ev(?2). Moreover, if ? meets (iii) and is REPLACED by
?0 = (D[(h;j);Rhj?;S[(h;j)), one can show using Lemma 6 that ev(?) = ev(?0)
since we must have that ?j?1 = ?j ?1 and that D [(h;j) has no outer corner in
column j, so ev(D[(h;j);?) = 0.
When the algorithm terminates, let ?0 denote the collection of all triples
(D;?;S) in the flnal set such that (i)-(iv) fail for all h, and ?1 the remaining
triples (i.e. the ones which satisfy conditions (ii) or (iv)). We say that a triple ?
survives the algorithm if at least one of its successors lies in ?0 as in [BKT]. The
above analysis implies that
?+1X
j=1
T(C;?(j)) =
X
?2?0
ev(?)+
X
?2?1
ev(?):
We remark that the triples ? 2 ?0 with ??+1 < 0 evaluate to zero trivially.
Claim 1. For any ? 2 ?1 we have ev(?) = 0.
Proof. If (h ? 1;h) =2 D, then ? meets (ii) and ?h?1 = ?h ? 1. Since there is no
outer corner in column h by Lemma 6 we have that ev(?) = 0. If ? meets (iv),
where (h ? 1;h) 2 D and ?h?1 = ?h, then Lemma 8 implies that D has no outer
corner in row h since in this case we must have that gh = bh. Hence, Lemma 7
shows that ev(?) = 0. The next assertion will therefore prove that the Chevalley
rule holds.
50
Claim 2. For each triple ? = (D;?;S) in ?0 with ??+1 ? 0, ? is a k-strict partition
with ? ! ? and ev(?) = T(C(?);?). Furthermore, for each such partition ?, there
are exactly e?? such triples ?, in agreement with the Chevalley rule.
In the appendix we provide an example of the equivariant substitution rule.
One can also view the example as an example of the classical substitution rule by
making the adjustments described in Remark 2.
Recall the flxed choices of ?, ?, C, and m. Let ? = (D;?;S) denote a triple
which occurs at some step in the algorithm. The symbols D, ?, and S will refer to
components of the triple ?. Throughout the algorithm, observe that ? is obtained
from the initial composition ? by removing boxes from rows weakly below the middle
row m of ? and adding them to rows weakly above the middle row. The set D is
initially equal to C and grows when REPLACE statements are encountered. All
pairs added to D come from the outer rim @C:
Lemma 9. If (j ? 1;j) 62 D and ? does not meet (i) or (ii) at any level h ? j,
then ?j ? ?j?1.
Proof. Assume that ?j > ?j?1. Since ? does not meet (ii), at level j, it follows
that D has an outer corner (i;j) in column j. Since (i;j ? 1) 2 C we obtain
?j + ?i > ?j?1 + ?i > 2k + (j ?1)?i. Hence ? satisfles W(i;j) and meets (i) at
level j, which is a contradiction.
Proposition 3. Let ? = (D;?;S) be a valid triple and suppose that (i;j);(i0;j0) 2
DrC. Then either i = i0 or j = j0.
51
The above proposition is special to the Chevalley formula, as it tells us that
not only is D ? C [@C, but that the set difierence DrC will only be a row or a
column.
Proof. Let ? be as in the statement of the proposition. Recall that initially ? =
f(C;?(j);;) : 1 ? j ? ?g. Let r be such that ? is a descendent of (C;?(r);;) := ?0.
This proof will trace through the family tree of ?0 to ?. We flrst note that if ? = ?0
then D rC is empty. So we consider cases for the children of ?0. Let h be the
level of ?0. Since ?(r) is obtained by only adding a box to the rth row of ? we must
have that any change in the pairs which satisfy the weight condition must contain
r. We also notice that all pairs in ? which have non-trivial descendants will satisfy
condition (i), since if (i;j) =2C but W(i;j) is satisfled for i < j, then we must have
(j ?1;j) =2C.
1. If (r ? 1;r) 2 C, then h > r since ?0 satisfles condition (i), and we have
W(r;h) holds. Hence ?0 has children ?1 = (C [ (r;h);?(r);;) and ?2 =
(C[(r;h);Rrh?(r);f(r;h)g).
We note ?1 cannot have any children since condition (i)cannot be satisfled
since h is the largest such that ?r + 1 + ?h > 2k + h?r, and (r;h) is in D
for ?1. Also condition (iii) cannot be satisfled for ?1 since r would have to be
the level and again h is the largest such that ?r + 1 + ?h > 2k + h?r, and
(r;h) is in D for ?1. If ? = ?1 then DrC is only a singleton.
We notice that ?2 may have children. The sequence ? for ?2 has the
property that ?i = ?i for i 6= h;r, where ?r = ?r + 2 and ?h = ?h ? 1.
52
Thus any change in the pairs which satisfy the weight condition must again
contain r, since the rth position is the only position which was increased. We
note ?2 cannot satisfy condition (i) since (r;h) 2 D and so the level of ?2 is
less than h, as we have progressed past h in the algorithm. If ? = ?2 then
DrC is again only a singleton. If ?2 satisfles condition (iii) then W(r;h+1)
is satisfled. We note this happens when ?h = ?h+1. Then ?2 has children
?3 = (C [ (r;h) [ (r;h + 1);Rrh?(r);f(r;h)g) and ?4 = (C [ (r;h) [ (r;h +
1);Rr;hRr;h+1?(r);f(r;h);(r;h + 1)g). Similarly to ?1, ?3 cannot have any
children, and similarly ?4 can have children as ?2 did if ?4 satisfles condition
(iii). We note that if ?4, or its descendants have children then their D sets
will only increase along row r since in each case r is the only position where
? is increased, and hence the only position which can take part in a change in
pairs which satisfy the weight condition.
Thus in the case where (r ?1;r) 2 C we have that ?, as a descendent of ?0,
will have the property that if (i;j);(i0;j0) 2 DrC, then i = i0 = r.
2. If(r?1;r) =2C, thenh = r, andthereexistsani ? msuchthat(i;r)isanouter
corner of C and W(i;r) holds. Hence ?0 has children ?1 = (C [(i;r);?(r);;)
and ?2 = (C[(i;r);?(i);f(i;r)g).
We note that in this case ?2 cannot have any children since condition (i)
cannot be satisfled since r is the largest such that ?r +1+?i > 2k+r?i, and
(i;r) is in D for ?2. Also condition (iii) cannot be satisfled for ?2 since i would
have to be the level and again r is the largest such that ?r+1+?i > 2k+r?i,
53
and (i;r) is in D for ?2. If ? = ?2 then DrC is only a singleton.
We notice that ?1 could possibly have children since if ?i = ?i+1 +1 then
?1 satisfles condition (i) again with level r. If this were the case then ?1 would
have children ?3 = (C [ (i;r) [ (i + 1;r);?(r);;) and ?4 = (C [ (i;r) [ (i +
1;r);?(i+1);f(i;r);(i+1;r)g). Similarly we have that ?4 cannot have children,
while ?3 can have descendants if ?i = ?i+2 +2.
Thus in the case where (r ?1;r) =2 C we have that ?, as a descendent of ?0,
will have the property that if (i;j);(i0;j0) 2 DrC, then j = j0 = r.
From the above 2 cases the proposition is proved.
Proposition 4. Any (i;j) which is an outer corner of D such that W(i;j) holds
for ?, has the property that ?i +?j = 2k +j ?i+1.
Proof. If ? = ?(i) for some i then this is clear, since C  D and each row of ? only
difiers from ? by one box. Otherwise, assume not. Then ?i + ?j > 2k + j ?i + 1,
while ?i +?j ? 2k +j?i. This can only happen if S is non-empty and contains at
least one pair (i;r) where r < j. In this case we must have that DrC is a row. Let r
be minimal such that (i;r) 2 DrC. Then we must have that ?r = ?r+1 = ??? = ?j
since from the proof of the previous proposition this is the only way to make DrC
be a row. The ancestor of such a progression will be (C;?;;) where ? = ?(i). This ?
has the property that ?r+?i > 2k+r?i while we know ?r+?i ? 2k+r?i. Therefore
?r + ?i = 2k + r?i + 1. The triple (C;?;;) satisfles condition (i) for h = r and so
it has children. As we continue along row i we note that we are increasing part i of
54
each descendent by looking at the child, which will be a product of condition (iii),
that adds (i;h) to S whenever (i;h) is an outer corner at that point in the algorithm.
Thus at most ?i = ?i+j?r. We also note that ?j will remain unchanged during this
process, so that ?j = ?j = ?r = ?r. Therefore 2k+j?i+1 < ?i+?j ? ?i+?r+j?r.
This would imply that ?i+?r > 2k+r?i+1 which is a contradiction. So any (i;j)
which is an outer corner of D such that W(i;j) holds for ?, has the property that
?i +?j = 2k +j ?i+1.
We next study the triples ? = (D;?;S) 2 ?0 with ??+1 ? 0.
Proposition 5. Suppose that ? = (D;?;S) 2 ?0 and ??+1 ? 0. Then ? is a k-
strict partition with j?j = j?j+1, satisfying ?j ?1 ? ?j ? ?j?1 for every j ? 1, and
?j ? ?j when ?j > k. Furthermore, we have D = C(?) and ? ! ?.
Proof. First note that ? = (D;?;S) 2 ?0 means that ? does not meet conditions
(ii) or (iv). Also if ?0 is part of the initial set of triples ? then we know ?0 =
(C;?(j);;) for some j where j?(j)j = j?j+ 1, and so since every REPLACE step of
the substitution rule does not change the weight of the integer sequence we have
that every ? = (D;?;S) 2 ?0 has the property that j?j = j?j+1. Also since DrC
is either a row or a column we only have a couple possibilities for a descendent
? 2 ?0 of ?0 2 ?. Let ? = (D;?;S) be a descendent of ?0 = (C;?(j);;). Consider
cases:
1. If DrC = ; then ?0 does not meet conditions (i) - (iv). So ? = ?0 and we
claim that ?(j) is a k-strict partition since ? is. Indeed, the only way ?(j) will
not be a k-strict partition is if either k ? ?j?1 = ?j in which case since ?0
55
does not satisfy condition (i) it would satisfy condition (ii) which would be
a contradiction, or if k < ?j?1 = ?j + 1 in which case since does not satisfy
condition (iii) it would satisfy condition (iv) which would be a contradiction.
Also since ?0 does not meet conditions (i) - (iv), there are no outer corners
of C such that the weight condition holds so C = C(?(j)) in this case. It is clear
that in this case if we set ? = ?(j) then ?j ?1 ? ?j ? ?j?1 for every j ? 1,
?j ? ?j when ?j > k, and ? ! ?.
2. If DrC is in the hth column, then we have either j = h or j 6= h.
If j = h then we must have j > m. In this case ?0 must meet condition
(i). Since DrC is the jth column we must have that in the family tree of
?0, ? is always a descendent of the terms whose integer sequence ?(j) remains
unchanged. Let (r;j) be the last new addition to D, so that ??s parent is
(Dr(r;j);?(j);;). Then? = (D;?;S)iseither (D;?(j);;)or(D;?(r);f(r;j)g).
In both case ? is k-strict partition with j?j = j?j+ 1, satisfying ?j ? ?j and
? ! ?. To see that D = C(?) note that D is a valid set such that the weight
condition is only met for pairs in D, and that every pair of D meets the weight
condition for ? since ? ? ? and the only pairs which were added to C to form
D satisfled the weight condition for ?.
If j 6= h then we must have that ?0 meets condition (i) at position h since we
have only changed ? in the jth row and so the fact that only the hth column is
changed means that (j;h) must be an outer corner ofC. In this case ?0 must be
the parent of ? since DrC is a column. So ? = (D;?;S) = (C[(j;h);?(j);;)
56
or ? = (D;?;S) = (C [ (j;h);Rjh?(j);(f(j;h)g). Since ? does not meet
conditions (iii) or (iv) we must have that ? is a k-strict partition satisfying
the desired properties. We note that in the case ? = Rjh?(j) we have ? ! ?
since we would have removed a box from the ?thh column and added two boxes
to the jth row.
3. If DrC is in the hth row, then again we have either j = h or j 6= h.
If j = h then we must have j ? m. Let c be the smallest such that (j;c) 2
DrC. Then ?0 satisfles condition(i)at position c. IfjDrCj > 1 then we must
have that ? is a descendent of (C[(j;c);Rjc?(j);f(j;c)g). We remark that it
is necessary for a triple to satisfy condition (iii) in order to add to row j of
C[(j;c) for descendants of ?0 and the triple (C[(j;c);?(j);;) will not satisfy
condition (iii) since ?(j)c+1 = ?c+1 ? ?c = ?(j)c . Then if d is the greatest such
that (j;c+d) 2 DrC wehavethat? = (C[di=0(j;c+i);Qdi=0(Rj;c+i)?(j);f(j;c+
i)gdi=0). We note that if ? = (C[di=0 (j;c+i);Qd?1i=0(Rj;c+i)?(j);f(j;c+i)gd?1i=0)
then this would have to vanish by Lemma 6. The fact that condition (iv) is
not satisfled guarantees that ? is k-strict. We have that ? ! ? since in order
to meet condition (iii) at each branch in the family tree of ?0 we must have
had that ?c = ?c+1 = ??? = ?c+d so that ? is obtained from ? by removing
d+1 boxes from column ?c of ? and adding d+2 boxes to row j. We similarly
argue that C(?) = D in this case.
If j 6= h then we must have (h;j) is an outer corner of C and that ?0 satisfles
condition (i) at position j. We note that ? is a k-strict partition so it is not
57
possible that ?(j)j = ?(j)j+1 hence jDrCj = 1 and we have that ? is either ?(j)
or ?(h) and since none of the conditions hold for ? this will guarantee that ?
and D have the desired properties.
Proposition 5 tells us that if ? = (D;?;S) is any triple in ?0 with ??+1 ? 0,
then ? is a k-strict partition with ? ! ?, D = C(?) is uniquely determined by ?,
and ev(?) = T(C(?);?) is a term appearing in the Chevalley rule.
To account for the multiplicities, flx an arbitrary k-strict partition ? such that
? ! ?, and suppose that e?? = 2. Then the unique box B = [i;c] of ?r? is
not k-related to a bottom box in one of the flrst k columns of ?. We associate a
pair (i;j) to B as follows: i is equal to the row number of B and j such that B is
k-related to the box [j ?1;d?1] in the flrst k columns of ? and [j;d] =2 ?; if there
is no such box then let j = ? + 1. We then have that the two triples in ?0 which
contribute to T(C(?);?) are (C(?);?;;) and ((C(?);?;f(i;j)g).
This concludes the summary of the argument that the Theta polynomials sat-
isfy the classical Chevalley formula. In x5.5 we deflne factorial Theta polynomials
and attempt to show that they satisfy the equivariant Chevalley formula. In or-
der to understand the equivariant Chevalley formula, we flrst need to understand
Grassmann permutations and the restriction map to a torus flxed point.
58
5.3 The k-Grassmannian Signed Permutation of a Partition ?
We let ? be a k-strict partition which flts inside the n?k by n+k rectangle.
We let ?0 be the partition whose rows are the columns of ?. Note ?0 is not going to
be k-strict.
The length of a k-related diagonal for ? is given by the length of a diagonal,
where flrst we draw a line from the each of bottom boxes of the flrst k columns of
? to the k-line and then flnd the length of the diagonal going up from the vertical
k-line to the flrst place where it intersects ?. The lengths of these diagonals are
given by
nj(?) = # of f(i;j0)jj0 > max(j;?j);?i < j0;(?0j;j) is k-related to (i;j0)g
for 1 ? j ? k.
Then we deflne the signed Grassmann permutation corresponding to ? as in
[BKT2]. Let m be the middle row of ? as deflned in x5.2.1. The signed permutation
corresponding to ? is w? where w? has the property that w?(i) < w?(i + 1) for all
i 6= k, w?(i) > 0 for all i ? k +m, and
w?(i) =
8
>><
>>:
nk?i+1(?) if 1 ? i ? k
k ??i?k if k +1 ? i < k +m
This is perhaps easier to see in a picture, taken from [BKT2]. Let n = 7;k =
3 and let ? be the partition (8,5,2,1). Below the diagram for ? drawn with the
appropriate k-related diagonals fllled with dots.
We note that the w?(i) will be the length of a k-related diagonal. So in Figure
59
Figure 5.3: The diagram of ? with marked k-related diagonals.
5.3 we have that the flrst k-related diagonal is given by nk(?) where in this case
since k = 3 we have ?03 = 2 and we notice that the boxes 3-related to the (2;3)
box are (3;4);(2;5); and (1;6), but since ?1 ? 6 and ?2 ? 5 we see that n3(?) = 1
and we see in the flgure that the smallest k-related diagonal has length 1. Similarly
the other k-related diagonals have length 4 and 7 respectively. Also we note that
k??1 = ?5 and k??2 = ?2 but k??i ? 0 for i > 2 so in the above example, for
n = 7;k = 3 and ? = (8;5;2;1) we have w? = (1;4;7;?5;?2;3;6).
5.4 Restriction to Torus Fixed Points in Type C
We have an inclusion ? : pt ,! IG(n?k;2n). This induces a ring homomor-
phism on the equivariant cohomology rings ?? : H?T(IG(n ? k;2n)) ! H?T(pt) =
Z[t1;:::;tn]. When the point we are restricting to corresponds to ?, this homo-
morphism takes the equivariant Chern class zi deflned in x2.2.1 to the equivariant
Chern class tw?(i) and takes xi to t?i?k or 0 if ?i ?k ? 0, where we cite [IMN] for
the geometric details for the localization map.
Remark 1. We note that as a tuple the z variables are mapped to the tuple of
60
t variables indexed by the lengths of the related k-diagonals, and that for the x
variables xi is mapped to the t variable indexed by the absolute value of w?(k+i) if
w?(k + i) < 0 and is mapped to zero otherwise. We also note that if w? 2 Sn then
all x variables are mapped to zero!
5.5 Factorial Theta Polynomials
We will flrst deflne the factorial Theta polynomials indexed by partitions of
length one, use these to deflne ?? for an arbitrary k-strict partition ?, and then we
will give a conjectural Giambelli formula for the general Schubert classes. Note that
it is necessary that when we consider the non-equivariant cohomology for IG(n?
k;2n) it should coincide with the known results of Buch, Kresch, and Tamvakis.
Thus we have the following preliminary deflnition:
Deflnition 6. For p > k
 p(x;zjt) =
pX
i=k+1
(?1)p?i i(x1;x2;:::jz1;:::;zk)ep?i(t1;:::;tp?k?1)
and for p ? k
 p(x;zjt) =
pX
i=1
(?1)i p?i(x1;x2;:::jz1;:::;zk)hi(t1;:::;tk?p+1)
It should be noted that the above factorial theta polynomials coincide with
the double Schubert functions described in [IMN]. We prove this in a proposition
below.
Proposition 6. The  p(x;zjt) deflned above represent the special Schubert classes
for the partition p.
61
Proof. Let Cw(x;zjt) for a Weyl group element w denote the type C double Schubert
polynomial described in [IMN], let Cv(x;z) represent the type C Billey-Haiman
Schubert polynomial for a Weyl group element v, which are known to represent the
Schubert class corresponding to v in a presentation of the classical cohomology of
the complete  ag variety in type C, and let Su(t) be the classical type A Schubert
polynomial for a permutation u which is known to represent the Schubert class
corresponding to u in the classical cohomology of the complete  ag variety in type
A. Then by [IMN, Cor. 8.10] we have
Cw(x;zjt) =
X
u;v
Su?1(?t)Cv(x;z)
where l(w) = l(u)+l(v);uv = w; and v;w 2 W, u a permutation.
Then we note in type C, if the rank of the group is n, W is generated by
simple re ections fsigni=0, where si = (i;i + 1) for i ? 1, and s0(i) = ?i for all i.
Then we can write the Weyl group elements corresponding to the special classes as
sk?p+1 :::sk for  p when p ? k, and sp?k?1 :::s0s1 :::sk for  p when n+k ? p > k.
In general we call such a listing of simple re ections whose product is a given
Weyl group element a word. Observe that u is a permutation only when we can write
it as a combination of simple re ections which do not contain s0. We notice that
for p 2 f1:::kg, any u which is a subword of wp, will be a permutation. However
for p 2 fk + 1;:::;n + kg, we notice that only a small subset of subwords do not
contain the re ection s0. To get our result we only need to look at cases.
Consider the case when p ? k. Then we get all possibilities of u;v such that
uv = w since s0 is not part of the word for wp. Then we note that in each case
62
u?1 is the permutation corresponding to the partition 1r in type A, where r is the
length of the subword. Therefore the Schubert polynomial in type A for u?1 will be
hr(t1;:::;tk?p+1). We also notice that for each u the remaining v will be the Weyl
group element for q ? p. Then we get the desired term by summing all possibilities,
and noting that the Billey-Haiman type C Schubert polynomials coincide with the
theta polynomials in this case.
Now consider the case when p > k. Then we notice that the only possible
subwords u are those that don?t contain s0. Therefore we are left with u?1 which
correspond to a one part partition of length r < p ? k ? 1 in type A, for which
the Schubert polynomial will be er(t1;:::;tp?k?1). Again the remaining v will be
the Weyl group element for k + 1 ? q ? p. And again we get the desired term by
summing all possibilities.
Deflnition 7. For any integers p;r;j we deflne
 rp[j] =
8>
><
>>:
Pp?k?1
i=0 (?1)
i p+j?i(x1;x2;:::jz1;:::;zk)ei(t1;:::;tr) if k < p
Pp+j
i=0(?1)
i p+j?i(x1;x2;:::jz1;:::;zk)hi(t1;:::;tr) if k ? p
where if either p + j or r are negative we take the corresponding factorial theta
polynomial to be zero and if r = 0 then we recover the standard theta polynomial
 p+j(x;z) of Deflnition 3. We also note that if p > k, r = p ? k ? 1, and j > 0
then  rp[j] =  rp+j[0] and similarly if p ? k and p + j ? k for an integer j then
 rp[j] =  rp+j[0].
Deflnition 8. For a valid set D deflne the following two functions:
1. aj(D) = #fiji < j;(i;j) =2 Dg
63
2. cj(D) = #fiji < j;(i;j) 2 Dg = j ?1?aj(D)
For simplicity we deflne aj(?) := aj(C(?)) and cj(?) := cj(C(?)).
Deflnition 9. For integer sequences ?; ;? let  ?; ;? = Q?i=1   i?i[?i]:
For any k-strict partition ?, set  (?) to be the ?-tuple such that  (?)i =
jk ??i +1+ai(?)j.
Deflnition 10. Let ?, ?, and  be any integer sequences, and let D be any valid
set. Deflne
?D?; ;? = RD ?; ;?
where for any i < j, Rij ?; ;? :=  ?; ;Rij(?).
Proposition 7. If p;q > k then
 1?R
1;2
1+R1;2
?
 rp[0] sq[0] = ?
 1?R
1;2
1+R1;2
?
 sq[0] rp[0]
for any non-negative integers r;s.
Proof. Let t(r) = (t1;:::;tr) and similarly let t(s) = (t1;:::;ts). Then
 1?R
1;2
1+R1;2
?
 rp[0] sq[0]
=  rp sq +2
1X
j=1
(?1)j rp[j] sq[?j]
=
?p?k?1X
i=0
(?1)i p?i(x;z)ei(t(r))
!?q?k?1X
i=0
(?1)i q?i(x;z)ei(t(s))
!
+2
1X
j=1
(?1)j
?p?k?1X
i=0
(?1)i p+j?i(x;z)ei(t(r))
!?q?k?1X
i=0
(?1)i q?j?i(x;z)ei(t(s))
!
=
pX
a=k+1
qX
b=k+1
(?1)p+q?a?bep?a(t(r))eq?b(t(s))?a;b
64
Then by Lemma 7 ?a;b(x;z) = ??b;a(x;z) since a+b > 2k. Note that by a similar
argument
 1?R
1;2
1+R1;2
?
 sq[0] rp[0] =
pX
a=k+1
qX
b=k+1
(?1)p+q?a?bep?a(t(r))eq?b(t(s))?b;a:
Therefore
 1?R
1;2
1+R1;2
?
 rp[0] sq[0] = ?
 1?R
1;2
1+R1;2
?
 sq[0] rp[0]:
Deflnition 11. For a k-strict partition ?, deflne the factorial Theta polynomial
corresponding to ? as
?? = ?C(?)?; (?);0 = R? ?; (?);0:
Conjecture 1. Let ? be a k-strict partition. Then the Schubert class corresponding
to ? in the equivariant cohomology of IG(n?k;2n) is represented by ??.
We?ll now prove a series of lemmas similar to Lemmas 6 and 7 which the
factorial Theta polynomials satisfy.
Lemma 10. Let ? = (?1;:::;?j?2) and ? = (?j+1;:::;??) be integer vectors. As-
sume that (j?1;j) =2 D and that for each h < j, (h;j?1) =2 D ifi (h;j) =2 D. Then
for any integers r and s, and any integer sequences ? and ^? such that ?j?1 +1 = ^?j
and ^?j?1 +1 = ?j we have
?D(?;r;s;?); ;? = ??D(?;s;r;?);sj( );^? :
To prove this we will use a similar recursion and induction as in the type A
case. As in [BKT; 1.2] we set for a sequence of nonnegative integers fi and a positive
65
integer ? we set m(D;fi;?) = #fi ? c?(D) : fii > 0g and call fi(D;?)-compatible if
fii 2 f0;1g for i > c?(D). Note that for any integer sequences ?;? of length ??1
we have
?D(?;r); ;(?;s)
=
X
fi
(?1)jfij2m(D;fi;?)?D?; ;fi+?  ?+1r [s?jfij]
where fi is (D;?)-compatible.
Proof. From the above recursion we can assume that ? is empty. Then applying the
recursion twice to ?D(?;r;s); ;? we have:
X
fi;fl
(?1)jfij+jflj2m(D;fi;j)+m(D;fl;j?1)(?D?; ;fi+fl+?0)?
(  j?1r [?j?1 ?jfij]  js [?j ?jflj]?  j?1r [?j?1 +1?jfij]  js [?j ?1?jflj])
where ?0 = ?1;:::;?j?2, fi is (D;j)-compatible, and fl is (D;j ? 1)-compatible.
Similarly applying the recursion twice to ?(?;s;r);sj( );^? we have:
X
fi;fl
(?1)jfij+jflj2m(D;fi;j?1)+m(D;fl;j)(?D?; ;fi+fl+^?0)?
(  js [^?j?1 ?jflj]  j?1r [^?j ?jfij]?  js [^?j?1 +1?jfij]  j?1r [^?j ?1?jflj])
where ^?0 = ^?1;:::; ^?j?2, fi is (D;j ?1)-compatible, and fl is (D;j)-compatible. We
note that since for each h < j, (h;j ? 1) =2 D ifi (h;j) =2 D the set of sequences
which are (D;j ?1)-compatible is the same as those which are (D;j)-compatible,
so that ?(?;s;r);sj( );^? is the negative of the expression for ?D;(?;r;s); ;? since we have
?j?1 +1 = ^?j and ^?j?1 +1 = ?j so the lemma is proven.
66
Corollary 4. Suppose that we have the same situation as in the previous lemma
where ?j?1 = ?j ? k and for each h < j, (h;j ?1) =2 C(?) ifi (h;j) =2 C(?). Then
we have the following
R?Rj;?+1 (?;1); (?);0 = ?tk??j+1+aj(?)??:
Proof. Let ?(j) be the result of adding one box to the jth row of ? (note that this
might no longer be a partition), and let  =  (?) for this proof. Note that since for
each h < j, (h;j) =2C(?) ifi (h;j +1) =2C(?) we have that C(?) = C(?(j)).
R?Rj;?+1 (?;1); ;0
= R?
j?1Y
i=1
 jk+1??i+ai(?)j?i [0] k+1??j+aj(?)?j [1]
?Y
i=j+1
 k+1??i+ai(?)?i [0]
= R?
j?1Y
i=1
 jk+1??i+ai(?)j?i [0] k??j+aj(?)?j [1]
?Y
i=j+1
 k+1??i+ai(?)?i [0]
+R?
j?1Y
i=1
 jk+1??i+ai(?)j?i [0](?tk??j+1+aj(?)) k+1??j+aj(?)?j [0]
?Y
i=j+1
 k+1??i+ai(?)?i [0]
by equation (3.1) for complete symmetric polynomials
= ?C(?)?; (j?);?
j
?tk??j+1+aj(?)??; 
= ?tk??j+1+aj(?)??
Where here we set  (j?) =  everywhere except the jth spot where it is one less.
So since aj?1(?)+1 = aj(?) since for each h < j, (h;j) =2C(?) ifi (h;j +1) =2C(?),
we have sj( (j?)) =  (j?). Hence by the previous lemma ?C(?)?; (j?);?
j
= ??C(?)?; (j?);?
j
where ?j is the sequence which is zero everywhere except the jth position where it
is 1. This means it must be zero, and we are left with the desired result.
67
Lemma 11. Let ? = (?1;:::;?j?1) and ? = (?j+2;:::;??) be integer vectors, as-
sume (j;j + 1) 2 D, and that for each h > j + 1, (j;h) 2 D ifi (j + 1;h) 2 D. If
r;s 2Z are such that r;s > k and D = C(?;r;s;?), then we have
?D(?;r;s;?); ;0 = ??D(?;s;r;?);sj( );0 :
Proof. We will again proceed by using our recursion formula to assume that ? is
empty. Following the proof of Lemma 7 in [BKT] we set ? = j + 1, and note that
(h;h0) 2 D for all h < h0 ? ?. If m > 0 is the least integer such that 2m ? ?, we
claim that ?Dfi; ;0 = ?D(?;r;s); ;0 satisfles the relation
?Dfi; ;0 =
2mX
i=2
(?1)i ?D(fi1;fii);( 1; i);0 ?D(fi2;:::;cfii;:::;fi2m);( 2;:::;b i;:::; 2m);0: (5.4)
Equation (5.4) follows from the formal identity of raising operators
Y
1?h<h0?2m
1?Rhh0
1+Rhh0 =
2mX
i=2
(?1)i 1?R1i1+R
1i
Y
2?h<h0?2m
h6=i6=h0
1?Rhh0
1+Rhh0;
which is equivalent to the classical formula
Y
1?h<h0?2m
xh ?xh0
xh +xh0 = Pfa?an
 x
h ?xh0
xh +xh0
?
1?h;h0?2m
due to Schur. The proof is completed using induction, starting from the base case
of j = 1, which was obtained in Proposition 7.
Corollary 5. If ? is a k-strict partition, ?j?1 > ?j > k and ?j + 1 = ?j?1 then if
for each h > j +1, (j;h) 2C(?) ifi (j +1;h) 2C(?) we have
R?Rj;?+1 (?;1); (?);0 = t?j?k??; (?);0
68
Proof. For this proof we let  (j+) =  (?) everywhere except the jth place where it
is one more. Then since ?j + 1 = ?j?1 > k + 1 we have sj( (j+)) =  (j+). We also
note that since ?j > k we have that  ?j?k?1?j [1] =  ?j?k?1?j+1 [0]. Hence we have:
R?Rj;?+1 (?;1); ;0
= R?
j?1Y
i=1
 (?i?k?1)?i [0] (?j?k?1)?j [1]
?Y
i=j+1
 jk+1??i+ai(?)j?i [0]
= R?
j?1Y
i=1
 (?i?k?1)?i [0] (?j?k)?j+1 [0]
?Y
i=j+1
 jk+1??i+ai(?)j?i [0]
+R?
j?1Y
i=1
 (?i?k?1)?i [0](t?j?k) (?j?k?1)?j [0]
?Y
i=j+1
 jk+1??i+ai(?)j?i [0]
by equation (3.1) for elementary symmetric polynomials
= ?C(?)?(j); (j+);0 +t?j?k??
= t?j?k??:
Since C(?) = C(?(j)) since for each h > j+1, (j;h) 2C(?) ifi (j+1;h) 2C(?)
and since ?C(?)?(j); (j+);0 = 0 by the above lemma.
5.6 The Equivariant Chevalley Formula
We will recall the equivariant Chevalley formula for IG(n ? k;2n) will be
indexed by the same partitions as the classical Chevalley formula, and will have
an additional equivariant correction given by  1flfl? ?. Therefore the equivariant
69
Chevalley formula is given by
 1 ? ? =  1flfl? ? +
X
?!?
e??  ? (5.5)
where e?? is as it is deflned in the classical case. Also we note that when we consider
the special Schubert class  1 as a Theta polynomial we have the following coe?cient
on our equivariant correction term:
 1flfl? =  k1flfl? = 2
?X
i=1
t?i?k +
kX
j=1
tw?(j) ?
kX
l=1
tl
where we set ti = 0 whenever i ? 0.
The way we prove the equivariant Chevalley formula will be very similar to
how the classical Chevalley formula is proven. The difierence here will be that in
our substitution algorithm we need to also keep track of any equivariant corrections.
5.7 The Equivariant Substitution Rule
We flx a k-strict partition ? of length ?, set m to be the middle row of ? deflned
in x5.2.1, let C = C(?), and choose n su?ciently large so that all Pieri terms that
can possibly appear in the Chevalley formula do not vanish just as we did in the
classical case. For the equivariant substitution rule we will need to keep track of the
changes necessary to transform  (?) into  (?) as well as the usual transforming of
C into C(?) and ? to 0. To account for this we will need to make our triple into a
4-tuple.
Deflnition 12. For any integer sequence ? the 4-tuple (D;?;S;?) will be called valid
if the triple (D;? + ?;S) was valid in the classical case. A 4-tuple will correspond
70
to a Schubert class if it is of the form (C(?);?;S;0) for some k-strict partition ?.
For any valid 4-tuple ? = (D;?;S;?) we let
ev(?) = RD ?; (D;?;?);?
where  (D;?;?)i = jk + 1??i ??i + ai(D)j: We note that if D = C(?) and ? = 0
then ev(?) = ?? since  (C(?);?;0) =  (?).
For a 4-tuple ? = (D;?;S;?) we deflne a new weight condition. We say the
4-tuple ? satisfles the weight condition W(i;j) for j > i if ?i+?i+?j+?j > 2k+j?i.
Let  =  (?;1) which is the same as  (?) in the flrst ? components, let ?(j)
be such that ?(j)i = ?i for all i 6= j and ?(j)j = ?j + 1, and let  (j+) =  + ?j and
 (j?) =  ? ?j, where we recall that ?j is the sequence which has a 1 in the jth
position and zeros elsewhere.
We compute that
?? ??1 = R? (?;1); ;0 +
a?+1(?;1)X
i=1
tk+i??
since ?1 = ?;1;k;0 and since  (?;1)?+1 = k + a?+1(?;1). Then the above can be
expanded as in the classical case to
R??+1 ?
?Y
i=1
(1+Ri;?+1) (?;1); ;0 +
a?+1(?;1)X
i=1
tk+i??:
At this point we notice this is the same as
?? ??1 =
?+1X
j=1
?C(?; (?);?j) +
a?+1(?;1)X
i=1
tk+i??; (5.6)
where we note  r0[1] =  r1[0].
71
We aim to show that the right hand side of Equation (5.6) is equal to the right
hand side of the equivariant Chevalley rule. We will do this using the an equivariant
version of the substitution rule from the classical case.
Thus in the above the 4-tuples corresponding to terms on the right hand side
of Equation (5.6) will be f(C;?;;;?j) : 1 ? j ? ?+1g := ?.
The following substitution rule will be applied iteratively to rewrite the right
hand side of (5.6). It may be applied to any valid 4-tuple and will result in either
a REPLACE statement, indicating that the 4-tuple should be replaced by one or
two new 4-tuples, or a STOP statement, indicating that the 4-tuple should not be
replaced.
Equivariant Substitution Rule
Let (D;?;S;?) be a valid 4-tuple. Let h ? ? + 1 be largest such that one of
the following flve conditions is true (if none hold for any h, then STOP).
(i) (h ? 1;h) =2 D and there is an outer corner (i;h) of D with i ? m such
that W(i;h) holds;
(ii) (h?1;h) =2 D, D has no outer corner in column h, and ?h+?h = ?h?1+1;
(iii) (h ? 1;h) 2 D and there is an outer corner (h;j) of D with j ? ? + 1
such that W(h;j) holds;
72
(iv) (h?1;h) 2 D and ?h +?h = ?h?1 +?h?1.
(v) ? 6= 0.
If condition (i) holds, then REPLACE (D;?;S;?) with (D [ (i;h);?;S;?)
and (D [(i;h);?;S [(i;h);Rih?). If (iii) holds, then REPLACE (D;?;S;?) with
(D [ (h;j);?;S [ (h;j);Rhj?) if ?h > ?h?1, or REPLACE (D;?;S;?) with (D [
(h;j);?;S [ (h;j);Rhj?) and (D [ (h;j);?;S;?). If (v) holds then REPLACE
(D;?;S;?) with (D;?+?;S;0) If (ii) or (iv) holds, then STOP.
We begin by performing the Substitution rule on terms at level h = ?+1. First
note that (?;?+1) =2C and of the flrst 4 conditions, only condition (i) has any chance
of holding. If condition (i) holds then we note that it must hold for i = 1, and in fact
will hold for i = 1;2;:::;c?+1(?;1). Thus we will have c?+1(?;1) additional terms,
corresponding to the valid 4-tuples (C;?;f(j;? + 1)g;?j) for j = 1;2;:::;c?+1(?;1),
and this will correspond to the following in our expansion in the Chevalley formula
73
for the factorial Theta polynomials.
?? ??1
= R?;1
c?+1(?;1)Y
i=1
(1+2Ri;?+1) (?;1); 
?Y
i=c?+1(?;1)+1
(1+Ri;?+1) ?;1; +
a?+1(?;1)X
i=1
tk+i??; 
= ??;1 +2
X
j?c?+1(?;1)
R? ?; ;?j +
X
j>c?+1(?;1)
R? ?; ;?j +
a?+1(?;1)X
i=1
tk+i??; ;0
= ??;1 +2
X
j?c?+1(?;1)
R? ?; (j+);?j +
X
j>c?+1(?;1);j<m
R? ?; (j+);?j
+
X
m?j>c?+1(?;1)
R? ?; (j?);?j
+
0
@ X
j?cl?+1(?;1)
2t?j?k +
X
j>c?+1(?;1)
(t?j?k ?tk??j+1+aj(?))+
al+1(?;1)X
i=1
tk+i
1
A??
where we note that if the index on t is negative, then we take it to be zero. The
equivariant correction terms on the last line are gathered using the same relations
amongst elementary and complete symmetric functions as in the proofs of Corollaries
4 and 5. Thus our new goals are to show that the coe?cient on ?? is ?1flfl?, and
that the descendants of the 4-tuples:
1. (C;?;f(j;?+1)g;?j) for j ? c?+1(?;1)
2. (C;?;;;?j) for j < m
3. (C;?;;;?j) for j ? m
which evaluate to the right hand side of the above equation correspond to Schubert
classes appearing the right hand side of Equation (5.5).
74
To do this we will apply the equivariant substitution rule, and show the algo-
rithm will give us the correct results.
First note that in input for the algorithm, our 4-tuples correspond to the triples
in the classical case in that our equivariant algorithm acts on (D;?;S;?) the same
way as the classical algorithm acts on (D;?+?;S), where we use Lemma 10 in place
of Lemma 6 and Lemma 11 in place of Lemma 7. Hence we need only show that
after the algorithm has terminated, the remaining (C(?);?;S;?) have the property
that that ? = 0. We also must show that our replace steps match up algebraically
with correct manipulations of the factorial theta polynomials.
We work through this algorithm with the example ? = (7;4;3;2;1;1) for
IG(9?2;18) in the appendix.
Recall from the classical setting that the REPLACE step for conditions (i)
and (iii) corresponds to equality of the raising operators below
1?Rij = 1?Rij1+R
ij
+ 1?Rij1+R
ij
Rij: (5.7)
We have ev(D;?;S;?) = RD ?; (D;?;?);?. For simplicity let  =  (D;?;?).
Then when we apply Equation (5.7) we would a priori have
ev(D;?;S;?) = RfD[(i;j)g ?; ;? +RfD[(i;j)g ?; ;Rij?:
Then we note that in the algorithm if we perform this REPLACE step then we were
in one of two possible cases, where both cases will have the property that W(i;j)
holds for ?+? but (i;j) is an outer corner of D. This means that  i = ?i+?i?k?1
while  j = k + 1??j ??j + #fr : r < j;(r;j) =2 Dg. So ( (i+))i = ?i + ?i ?k and
75
( (j?))j = k ??j ??j + #fr : r < j;(r;j) =2 Dg. Since (i;j) is an outer corner, we
have that
#fr : r < j;(r;j) =2 Dg = j ?i:
W(i;j) being satisfled by ?+? means that ?i+?i+?j +?j > 2k+j?i. Combining
these relations we must have that
( (i+))i ? j = ?i?i +?j +?j ?2k ?j +i?1:
Then byProposition 4 weknow?j+?j+?i+?i = 2k+j?i+1so wehave( (i+))i =  j.
We notice that  (i+) =  (D[(i;j);?;Rij?) and  (j?) =  (D[(i;j);?;Rij?).
Hence our REPLACE step corresponds to the usual application of Equation
5.7 along with the relations
 ?; ;? =  ?; (j?);? ?t j ?; ;???j
from Equation (3.1) for the complete symmetric functions, and
 ?; ;Rij? =  ?; (i+);Rij? +t( (i+))i Rij?; ;??ei
from Equation (3.1) for the elementary symmetric functions. Note that ? ? ?j =
Rij? ? ei so these correction terms will cancel by the above argument. Hence we
have
ev(D;?;S;?) = ev(D[(i;j);?;S;?)+ev(D[(i;j);?;S [(i;j);Rij?)
and our REPLACE step for conditions (i) and (iii) correspond to correct manipu-
lation of the Theta polynomials.
76
For condition (v) we note that for a 4-tuple ? = (D;?;S;?) if Rij is applied
to ? in a REPLACE step then i ? m and j > m, then as long as ?i ? 0 for i ? m
and ?i ? 0 for i > m we have that ev(?) = ev(?0) where ?0 = (D;?+?;S;0). Thus
we need only check what happens for our initial 4-tuples ? = (C;?;S;?j) for j > m.
Then we note that if ?j < k then ?j +1 ? k so again we have ev(?) = ev(?0) where
?0 = (C;?(j);S;0). So at this point we know that as long as ?j 6= k then we have
that the REPLACE from condition (v) corresponds to correct manipulations of the
indices.
Assume ?j = k and consider the 4-tuple ? = (C;?;S;?j). We consider the
following cases:
1. If ?j?1 > k + 1, then aj(C) = 0 so that  (C;?;?j)j = 0. So we note that
 0k[1] =  k+1(x;z) =  0k+1[0] and so applying the REPLACE step for condition
(v) is an algebraically correct operation.
2. If ?j?1 = k + 1, then (C;?;S;?j) satisfles condition (i) so it has children
?1 = (C [(j ?1;j);?;S [f(j ?1;j)g;ej?1) and ?2 = (C [(j ?1;j);?;S [
f(j ? 1;j)g;ej). Then ?1 has the property that ?i = 0 except for when
i = j?1 and there we have that ?j?1 > k so that ev(?1) = ev(?01) where ?01 =
(C[(j?1;j);?+ej?1;S;0) which amounts to applying the REPLACE step for
condition(v). Wethenconsider?2. If (?)j = 2then (C[(j?1;j);?;?j)j = 0
and we note that since  0k[1] =  k+1(x;z) =  0k+1[0] and  (?)j?1 = 0 so that
ev(?2) is zero by Lemma 11. If  (C[(j ?1;j);?;?j)j 6= 0 then we must have
that ?j?r = k +r for some r > 1, in which case ?2 satisfles condition (i) and
77
has children ?21 = (C [ f(j ? 1;j);(j ? 2;j)g;?;S [ f(j ? 2;j)g;ej?2) and
?22 = (C[f(j ?1;j);(j ?2;j)g;?;S [f(j ?1;j)g;ej) where here ?21 can be
evaluated similarly to ?1 by applying the REPLACE step for condition (v)
and ?22 is either zero by Lemma 11 or satisfles condition (i). Continuing in
this fashion we are able to make all terms vanish except those where ? = 0.
3. If ?j?1 = k, then we notice that ? will either satisfy condition (i) or condition
(ii). If it satisfles condition (ii) then it will vanish by Lemma 10. If it satisfles
condition (i) then it will have a descendant which will satisfy condition (ii)
and all other descendants will have the property that ?i > 0 only when i ? m.
In such a case we know that condition (v) will be satisfled and the REPLACE
step is a correct manipulation of indices.
Then the fact that all of the surviving 4-tuples are terms in the right hand
side of the equivariant Chevalley formula follows from the classical case.
Hence to prove the Chevalley formula it remains to show that our initial cor-
rection of
0
@ X
j?c?+1(?;1)
2t?j?k +
X
??j>c?+1(?;1)
(t?j?k ?tk??j+1+aj(?))+
a?+1(?;1)X
i=1
tk+i
1
A
is the correct coe?cient for  ?. We know the correct coe?cient of  ? is
 1j? = 2
?X
j=1
t?j?k +
kX
i=1
(tw?(i) ?ti)
where w? is the signed permutation corresponding to ?.
We notice that these are equivalent coe?cients only if
k+a?+1(?;1)X
i=1
ti =
X
j>c?+1(?;1)
(t?j?k +tk??j+1+aj(?))+
kX
i=1
tw?(i):
78
Observe flrst that if the indices on the right hand side are distinct then there
are precisely a?+1(?;1) of them. This is because the set f? ? j > c?+1(?;1)g has
cardinality ??c?+1(?;1) = al+1(?;1), and only one of ?j ?k or k ??j + 1 + aj(?)
will be positive. Then we also note that so long as j ? ? we have that
k ??j +1+aj(?) ? k +aj(?) < k +a?+1(?;1):
Similarly so long as j > c?+1(?;1) we know that ?j ? 2k +??j so
?j ?k ? k +??j < k +a?+1(?;1):
It is known that w? will have a unique jump after i = k and is increasing before then,
so we calculate w?(k). This will be equal to ?+k?c?+1(?;1) = k+a?+1(?;1). This
is because if we look at the length of the k related diagonal coming from the flrst
column of ? it will be ?+k?#fjj?j > 2k+??jg, since we note that the k related
diagonal will \hit" the jth row only if ?j > k+??j. Therefore w?(k) = k+a?+1(?;1),
and all other terms on the right hand side are less than it.
Thus to show we have the correct coe?cient we need only show that the terms
X
j>cl+1(?;1)
(t?j?k +tk??j+1+aj(?))+
kX
i=1
tw?(i)
are distinct. To do this we simply notice that the indices are all the absolute value
of w?(i) for some 1 ? i ? n. This is obvious for the tw?(i). Then we also know that
for all j such that ?j ?k > 0, w?(k+j) = ?(?j ?k) so jw?(k+j)j = ?j ?k. Lastly
we know that if we have ?j ? k ? 0, then w?(k + j) corresponds to the length of
a non-related diagonal, which is k ??j + 1 + aj(?) as explained in x5.3. So these
indices all correspond to w? and are thus distinct.
79
Hence we have the correct coe?cient for  ? and have proven the Chevalley
Formula for our raising operator expression.
5.8 The Vanishing Theorem
Now in order to prove the Conjecture 1 we still need to show that the raising
operator expressions satisfles the vanishing theorem described in x2.3.
Currently I have only proven this in the case that the Schubert class is indexed
by a k-strict partition ?; for which ?i ? k for all i. From here until the end of the
chapter we will use the word \small" to describe any such partition ?
Theorem 8. Let ?;? be small partitions such that ?*?. Then ??flfl? = 0.
Proof. We will use the localization map for restricting to the torus flxed point e?
described in x5.4. We note that in this map the variable xi will go to t?i?k which
in the case of ? being small will always be 0. Also the localization map sends zi to
tw?(i). Thus we have the following:
??flfl? = R? ?; (?);0flfl?
=
Y
1?i<j??
(1?Rij)
?Y
i=1
 k??i+1+ai(?)?i (0)flfl?
since ? is small. We also note that in general w?(i) for i ? k will be the k related
diagonals as described in x5.3. Let ?0 be the partition given by the columns of
?. Then the k related diagonals are exactly ?0k?i+1 + i. We note that in this case
80
w? in Lie type C is the same as w? in Lie type A when we are considering the
Grassmannian G(n?k;n). Thus we note that tw?(i) is the image of both xi when
considering G(n?k;n) and zi when considering IG(n?k;2n). Also since ? is small
we have that ai(?) = i?1 for all i. Then note that
 (k??i+i)?i (0)flfl? =
X
r;s
r+s??i
(?1)sq?i?r?s(x)er(z)hs(t)flfl?
=
?iX
j=1
(?1)je?i?j(t?)hj(t):
Hence
??flfl? = s?(t?jt):
Thus the Vanishing theorem in this case follows from the type A result.
5.9 Generalizations and Further Work
There are several directions I can go from here. The most obvious of which
is to prove my conjecture. Once I am able to prove this conjecture one could also
generalize these results to the Quantum Equivariant Cohomology ring of a Grass-
mannian. I may also try to generalize to the Quantum Equivariant Cohomology
ring in order to prove my conjecture, since a recent result of Mihalcea in [Mi] shows
that satisfying the equivariant quantum Chevalley formula is enough to prove the
Giambelli formula in this setting.
81
Chapter A
Example of the Substitution Rule
A.1 Setup
Let us consider the 2-strict partition ? = (7;4;3;2;1;1) indexing a Schubert
cell in IG(9?2;18). The Chevalley formula corresponding to this partition is
 ? ? 1 =
2 (8;4;3;2;1;1) + (10;4;3;2) +2 (7;5;3;2;1;1) + (7;6;3;1;1;1) + (7;4;3;2;2;1) + (7;4;3;2;1;1;1)
+(2t5 +t2 +t1 +t4 +t8) ?:
In this example C = C(?) = f(1;2);(1;3);(1;4);(2;3)g. Our initial set of 4-tuples is
?0 = f(C;?;;;?1);(C;?;;;?2);(C;?;;;?3);(C;?;;;?4);(C;?;;;?5);(C;?;;;?6),
(C;?;;;?7)g: We note that for this example  (?) = (4;1;0;3;6;7). Below we will
follow the substitution algorithm for each 4-tuple (D;?;S;?), including the changes
in the integer sequences ? and ? in the Young diagram and the set D in the flgure
of circles just underneath each diagram. In the pictures below the gray boxes will
represent the positive parts of the integer sequence ? and the dashed boxes are the
negative parts of the integer sequence ?. The survivors of the algorithm will match
up with the Chevalley formula given above.
Remark 2. We note that the solid outline of the resulting diagram is ? + ? and
thus will match up with the steps of the classical substitution rule.
82
        ? =
Figure A.1: The Young diagram of ?.
In the flgures below the REPLACE steps for conditions (i) and (iii) will result
in a new generation of 2 children (indicated by a double arrow in the flgure) while we
will resolve the REPLACE step for condition (v) (indicated by a single downward
arrow) within the description for the parent generation.
We also include the equivariant corrections in the flgure. A ?ti in the last box
of the jth row of the diagram indicates that for the corresponding 4-tuple (D;?;S;?),
? makes it necessary to include or remove ti from the jth part of the monomial
 ?; (D;?;?);? while it was or wasn?t included in the previous generation?s monomial.
We perform this correction using equation (3.1) as described in x5.5.
83
A.2 Algorithm With Corrections
Figure A.2: The Substitution Rule applied to (C;?;;;?1)
000
000
111
111
000
000
111
111
000
000
111
111
00
00
11
11
000
000
111
111
00
00
11
11 000000111111
00
00
11
11
000
000
111
111
(1)?  =
+t 5
7
 6
7
+t
?t
 6+t
?t
First generation:
?(C;?;;;?1) satisfles condition (i) at level h = 5 since 7+1+1 > 4+5?1. We
note that  (C;?;?1) = (5;1;0;3;6;7) =  +?1 so we must add t5?ev(C;?;;;?1??1) =
t5?? in order for ev(C;?;;;?1) to be correct.
84
Second generation:
?(C[(1;5);?;;;?1)meetscondition(v)andisreplacedwith(C[(1;5);?(1);;;0)
which does not meet any conditions and survives the algorithm. We note that
 (C[(1;5);?;?1) = (5;1;0;3;5;7) =  (C;?;?1)??5 so we must subtract t6 ?ev(C[
(1;5);?;;;?1 ??5).
?(C [ (1;5);?;f(1;5)g;2?1 ? ?5) satisfles condition (iii) at level h = 1 since
7 + 2 + 1 > 4 + 6 ? 1. We note that  (C [ (1;5);?;2?1 ? ?5) = (6;1;0;3;6;7) =
 (C;?;?1)+?1 so we must add t6?ev(C[(1;5);?;f(1;5)g;2?1??5??1). We note that
this term will cancel with the correction term from this 4-tuple?s second generation
brother.
Third generation:
?(C [(1;5)[(1;6);?;f(1;5)g;2?1 ??5) satisfles condition (ii) at level h = 6
and its evaluation vanishes by Lemma 10. We note that  (C[(1;5)[(1;6);?;2?1?
?5) = (6;1;0;3;6;6) =  (C;?;2?1 ??5)??6 so we must subtract t7 ?ev(C [(1;5)[
(1;6);?;f(1;5)g;2?1 ??5 ??6).
?(C [ (1;5) [ (1;6);?;f(1;5);(1;6)g;3?1 ? ?5 ? ?6) meets condition (v) and
is replaced with (C [ (1;5);(10;4;3;2);;;0) which does not meet any conditions
and survives the algorithm. We note that  (C [ (1;5) [ (1;6);?;3?1 ? ?5 ? ?6) =
(7;1;0;3;6;7) =  (C;?;2?1??5)+?1 sowemustaddt7?ev(C[(1;5)[(1;6);?;f(1;5);(1;6)g;3?1?
?5 ??6 ??1). We note that this term will cancel with the correction term from this
4-tuple?s third generation brother.
We notice that in the end the descendants (D;?;S) of (C;?(1);;) have the
property that DrC is a row.
85
Figure A.3: The Substitution Rule applied to (C;?;;;?2)
00
00
11
11
000
000
111
111
00
00
11
11 000
000
111
111
?  =(2)
+t
+t
 3 ?t
 3
 2
First generation:
?(C;?;;;?2) satisfles condition (i) at level h = 4 since 4+1+2 > 4+4?2. We
note that  (C;?;?2) = (4;2;0;3;6;7) =  (?)+?2) so we must add t2?ev(C;?;;;?2?
?2) = t2??.
Second generation:
?(C[(2;4);?;;;?2)meetscondition(v)andisreplacedwith(C[(2;4);?(2);;;0)
which does not meet any conditions and survives the algorithm. We note that
 (C[(2;4);?;?2) = (4;2;0;2;6;7) =  (C;?;?2)??4) so we must subtract t3?ev(C[
86
(2;4);?;;;?2 ??4).
?(C[(2;4);?;f(2;4)g;2?2??4) meets condition (v) and is replaced with (C[
(2;4);(7;6;3;1;1;1);f(2;4)g;0) which does not meet any conditions and survives
the algorithm. We note that  (C[(2;4);?;2?2??4) = (4;3;0;3;6;7) =  (C;?;?2)+
?2) so we must add t3?ev(C[(2;4);?;f(2;4)g;2?2??4??2). We note that this term
will cancel with the correction term from this 4-tuple?s second generation brother.
Figure A.4: The Substitution Rule applied to (C;?;;;?3)
(3)?  =
+t1
First generation:
?(C;?;;;?3) satisfles condition (iv) at level h = 3 and its evaluation vanishes
by Lemma 11. We note that  (C;?;?3) = (4;1;1;3;6;7) =  (?) + ?3) so we must
add t1 ?ev(C;?;;;?3 ??3) = t1??.
87
Figure A.5: The Substitution Rule applied to (C;?;;;?4)
?t
000
000
111
111000000111111
00
00
11
11
00
00
11
11
00
00
11
11
0011
?t
?(4)=
 ?t 1
 +t
 +t
2
  2
?t 3
    1
First generation:
?(C;?;;;?4) satisfles condition (i) at level h = 4 since 4 + 2 + 1 > 4 + 4?2.
We note that  (C;?;?4) = (4;1;0;2;6;7) =  (?) ? ?4) so we must subtract t3 ?
ev(C;?;;;?4 ??4) = ?t3??.
Second generation:
?(C[(2;4);?;;;?4) satisfles condition(i)at level h = 4 since 3+2+1 > 4+4?3.
We note that  (C [ (2;4);?;?4) = (4;2;0;1;6;7) =  (C;?;?4) ? ?4) so we must
subtract t2 ?ev(C[(2;4);?;;;?4 ??4).
88
?(C [ (2;4);?;f(2;4)g;?2) satisfles condition (v) and is replaced with (C [
(2;4);?(2);f(2;4)g;0) which does not meet any conditions and survives the algo-
rithm. We note that  (C [ (2;4);?;?2) = (4;3;0;2;6;7) =  (C;?;?4) + ?2) so we
must add t2 ?ev(C[(2;4);?;;;?2 ??2). We note that this term will cancel with the
correction term from this 4-tuple?s second generation brother.
Third generation:
?(C [ (2;4) [ (3;4);?;;;?4) satisfles condition (iv) at level h = 4 and its
evaluation vanishes by Lemma 11. We note that  (C [ (2;4) [ (3;4);?;?4) =
(4;2;0;0;6;7) =  (C [ (2;4);?;?4) ? ?4) so we must subtract t1 ? ev(C [ (2;4) [
(3;4);?;;;?4 ??4).
?(C [ (2;4) [ (3;4);?;f(3;4)g;?3) satisfles condition (iv) at level h = 3 and
its evaluation vanishes by Lemma 11. We note that  (C [ (2;4) [ (3;4);?;?3) =
(4;2;1;1;6;7) =  (C[(2;4);?;?4)+?3)sowemustaddt1?ev(C[(2;4)[(3;4);?;;;?3?
?3). We note that this term will cancel with the correction term from this 4-tuple?s
third generation brother.
We notice that in the end the descendants (D;?;S) of (C;?(4);;) have the
property that DrC is a column.
89
Figure A.6: The Substitution Rule applied to (C;?;;;?5)
00
00
11
11
000
000
111
111
00
00
11
11
000
000
111
111
(5)?   =
?t
?t
?t
 6
 5
 5
First generation:
?(C;?;;;?5) satisfles condition (i) at level h = 5 since 7 + 1 + 1 > 4 + 5?1.
We note that  (C;?;?5) = (4;1;0;3;5;7) =  (?) ? ?5) so we must subtract t6 ?
ev(C;?;;;?5 ??5) = ?t6??.
Second generation:
?(C[(1;5);?;;;?5)meetscondition(v)andisreplacedwith(C[(1;5);?(5);;;0)
which does not meet any conditions and survives the algorithm. We note that
 (C[(1;5);?;?5) = (4;1;0;3;4;7) =  (C;?;?5)??5) so we must subtract t5?ev(C[
90
(1;5);?;;;?5 ??5).
?(C[(1;5);?;;;?1)meetscondition(v)andisreplacedwith(C[(1;5);?(1);;;0)
which does not meet any conditions and survives the algorithm. We note that
 (C [ (1;5);?;?1) = (5;1;0;3;5;7) =  (C;?;?5) ? ?5) so we must add t5 ? ev(C [
(1;5);?;;;?1??1). We note that this term will cancel with the correction term from
this 4-tuple?s second generation brother.
Figure A.7: The Substitution Rule applied to (C;?;;;?6)
  7 ?t
?  =(6)
First generation:
?(C;?;;;?6) satisfles condition (ii) at level h = 6 and its evaluation vanishes
by Lemma 10. We note that  (C;?;?6) = (4;1;0;3;6;6) =  (?) ? ?6) so we must
subtract t7 ?ev(C;?;;;?6 ??6) = ?t7??.
91
Figure A.8: The Substitution Rule applied to (C;?;;;?7)
(7)?  =
First generation:
?(C;?;;;?7) meets condition (v) and is replaced with (C;?(7);;;0) which does
not meet any conditions and survives the algorithm. We note that  (C;?;?7) =
(4;1;0;3;6;7;8) and that in the Chevalley product, ?1 = ?1(x;z)?t1 ?t2 so we
must add (P8i=3 ti)?? which is not included as a correction on our flgure.
Remark 3. In the end when we apply the evaluation map we notice the survivors
match with the terms in the equivariant Chevalley formula for ? given at the begin-
ning of the appendix.
92
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