Functors for MatchingKazhdan-Lusztig polynomials for GL(n; F )

Peter E. Trapa(joint with Dan Ciubotaru)

F� : X 7! H0(X F )[�+N�]

KLV polynomials

GR real reductive group, g = gR R CKR maximal compact subgroup, K = (KR)CB variety of Borel subalgebras in g.LocK(B) set of irreducible K-equivariant locally constantsheaves supported on single K orbits on B

Given L;L0 2 LocK(B), we can de�ne theKazhdan-Lusztig-Vogan polynomial

pLL0(q):

KLV polynomials

GR real reductive group, g = gR R CKR maximal compact subgroup, K = (KR)CB variety of Borel subalgebras in g.LocK(B) set of irreducible K-equivariant locally constantsheaves supported on single K orbits on B

Given L;L0 2 LocK(B), we can de�ne theKazhdan-Lusztig-Vogan polynomial

pLL0(q):

KLV polynomials

GR real reductive group, g = gR R CKR maximal compact subgroup, K = (KR)CB variety of Borel subalgebras in g.LocK(B) set of irreducible K-equivariant locally constantsheaves supported on single K orbits on B

Given L;L0 2 LocK(B), we can de�ne theKazhdan-Lusztig-Vogan polynomial

pLL0(q):

Notation for special case

If L and L0 are the constant sheaves on Q;Q0 2 KnB, insteadwrite

pQ;Q0 = pL;L0

Notation for an even more special case

If GR is already a complex Lie group, then the Bruhatdecomposition identi�es KnB with the Weyl group W of g.

Infact,

LocK(B) $ W:

If Q(x) and Q(y) are two orbits parametrized by x and y, thenwe write

px;y = pQ(x);Q(y):

Notation for an even more special case

If GR is already a complex Lie group, then the Bruhatdecomposition identi�es KnB with the Weyl group W of g. Infact,

LocK(B) $ W:

If Q(x) and Q(y) are two orbits parametrized by x and y, thenwe write

px;y = pQ(x);Q(y):

Notation for an even more special case

If GR is already a complex Lie group, then the Bruhatdecomposition identi�es KnB with the Weyl group W of g. Infact,

LocK(B) $ W:

If Q(x) and Q(y) are two orbits parametrized by x and y, thenwe write

px;y = pQ(x);Q(y):

Another example

GR = U(p; q)

KR = U(p)�U(q)K = GL(p;C)�GL(q;C)

In factLocK(B) $ (GL(p;C)�GL(q;C))nB:

Another example

GR = U(p; q)KR = U(p)�U(q)

K = GL(p;C)�GL(q;C)

In factLocK(B) $ (GL(p;C)�GL(q;C))nB:

Another example

GR = U(p; q)KR = U(p)�U(q)K = GL(p;C)�GL(q;C)

In factLocK(B) $ (GL(p;C)�GL(q;C))nB:

Another example

GR = U(p; q)KR = U(p)�U(q)K = GL(p;C)�GL(q;C)

In factLocK(B) $ (GL(p;C)�GL(q;C))nB:

What does \matching KLV polynomials" mean?

Let G1R and G2

R denote two real groups.

Suppose we can �nd

(1) a subset S1 � LocK1(B1);

(2) a subset S2 � LocK2(B2); and

(3) a bijection � : S1 ! S2

such that for any L;L0 2 S1,

pL;L0 = p�(L);�(L0):

Then we say � implements a matching of KLV polynomials.

Obviously this is more impressive if S1 is large.

What does \matching KLV polynomials" mean?

Let G1R and G2

R denote two real groups.Suppose we can �nd

(1) a subset S1 � LocK1(B1);

(2) a subset S2 � LocK2(B2); and

(3) a bijection � : S1 ! S2

such that for any L;L0 2 S1,

pL;L0 = p�(L);�(L0):

Then we say � implements a matching of KLV polynomials.

Obviously this is more impressive if S1 is large.

What does \matching KLV polynomials" mean?

Let G1R and G2

R denote two real groups.Suppose we can �nd

(1) a subset S1 � LocK1(B1);

(2) a subset S2 � LocK2(B2); and

(3) a bijection � : S1 ! S2

such that for any L;L0 2 S1,

pL;L0 = p�(L);�(L0):

Then we say � implements a matching of KLV polynomials.

Obviously this is more impressive if S1 is large.

What does \matching KLV polynomials" mean?

Let G1R and G2

R denote two real groups.Suppose we can �nd

(1) a subset S1 � LocK1(B1);

(2) a subset S2 � LocK2(B2); and

(3) a bijection � : S1 ! S2

such that for any L;L0 2 S1,

pL;L0 = p�(L);�(L0):

Then we say � implements a matching of KLV polynomials.

Obviously this is more impressive if S1 is large.

What does \matching KLV polynomials" mean?

Let G1R and G2

R denote two real groups.Suppose we can �nd

(1) a subset S1 � LocK1(B1);

(2) a subset S2 � LocK2(B2); and

(3) a bijection � : S1 ! S2

such that for any L;L0 2 S1,

pL;L0 = p�(L);�(L0):

Then we say � implements a matching of KLV polynomials.

Obviously this is more impressive if S1 is large.

What does \matching KLV polynomials" mean?

Let G1R and G2

R denote two real groups.Suppose we can �nd

(1) a subset S1 � LocK1(B1);

(2) a subset S2 � LocK2(B2); and

(3) a bijection � : S1 ! S2

such that for any L;L0 2 S1,

pL;L0 = p�(L);�(L0):

Then we say � implements a matching of KLV polynomials.

Obviously this is more impressive if S1 is large.

What does \matching KLV polynomials" mean?

Let G1R and G2

R denote two real groups.Suppose we can �nd

(1) a subset S1 � LocK1(B1);

(2) a subset S2 � LocK2(B2); and

(3) a bijection � : S1 ! S2

such that for any L;L0 2 S1,

pL;L0 = p�(L);�(L0):

Then we say � implements a matching of KLV polynomials.

Obviously this is more impressive if S1 is large.

Plan

� Explicitly describe a matching for GL(n;C) with variousU(p; q)'s, p+ q = n.

� The key will be an intermediate combinatorial set (which isreally GL(n;Qp) in disguise).

� Indicate brie y a general setting to search for suchmatchings. (Works in all simple adjoint classical cases, forinstance, but breaks in interesting ways in a handful ofexceptional ones.)

� Hando� to Dan: a functor \implementing" the matching.

Plan

� Explicitly describe a matching for GL(n;C) with variousU(p; q)'s, p+ q = n.

� The key will be an intermediate combinatorial set (which isreally GL(n;Qp) in disguise).

� Indicate brie y a general setting to search for suchmatchings. (Works in all simple adjoint classical cases, forinstance, but breaks in interesting ways in a handful ofexceptional ones.)

� Hando� to Dan: a functor \implementing" the matching.

Plan

� Explicitly describe a matching for GL(n;C) with variousU(p; q)'s, p+ q = n.

� The key will be an intermediate combinatorial set (which isreally GL(n;Qp) in disguise).

� Indicate brie y a general setting to search for suchmatchings. (Works in all simple adjoint classical cases, forinstance, but breaks in interesting ways in a handful ofexceptional ones.)

� Hando� to Dan: a functor \implementing" the matching.

Plan

� Explicitly describe a matching for GL(n;C) with variousU(p; q)'s, p+ q = n.

� The key will be an intermediate combinatorial set (which isreally GL(n;Qp) in disguise).

� Indicate brie y a general setting to search for suchmatchings. (Works in all simple adjoint classical cases, forinstance, but breaks in interesting ways in a handful ofexceptional ones.)

� Hando� to Dan: a functor \implementing" the matching.

Plan

� Explicitly describe a matching for GL(n;C) with variousU(p; q)'s, p+ q = n.

� The key will be an intermediate combinatorial set (which isreally GL(n;Qp) in disguise).

� Indicate brie y a general setting to search for suchmatchings. (Works in all simple adjoint classical cases, forinstance, but breaks in interesting ways in a handful ofexceptional ones.)

� Hando� to Dan: a functor \implementing" the matching.

Intermediate Combinatorial Set

Fix � = (�01 � � � � � �0n) (an in�nitesimal character in disguise).

Everything reduces to the case

1 �0i 2 Z

2 �0i � �0i+1 2 f0; 1g.

We assume this is the case and rewrite

� = (

n1z }| {�1; : : : ; �1 > � � � � � � � � � >

nkz }| {�k; : : : ; �k):

Intermediate Combinatorial Set

Fix � = (�01 � � � � � �0n) (an in�nitesimal character in disguise).

Everything reduces to the case

1 �0i 2 Z

2 �0i � �0i+1 2 f0; 1g.

We assume this is the case and rewrite

� = (

n1z }| {�1; : : : ; �1 > � � � � � � � � � >

nkz }| {�k; : : : ; �k):

Intermediate Combinatorial Set

Fix � = (�01 � � � � � �0n) (an in�nitesimal character in disguise).

Everything reduces to the case

1 �0i 2 Z

2 �0i � �0i+1 2 f0; 1g.

We assume this is the case and rewrite

� = (

n1z }| {�1; : : : ; �1 > � � � � � � � � � >

nkz }| {�k; : : : ; �k):

Intermediate Combinatorial Set

Fix � = (�01 � � � � � �0n) (an in�nitesimal character in disguise).

Everything reduces to the case

1 �0i 2 Z

2 �0i � �0i+1 2 f0; 1g.

We assume this is the case and rewrite

� = (

n1z }| {�1; : : : ; �1 > � � � � � � � � � >

nkz }| {�k; : : : ; �k):

Intermediate Combinatorial Set

� = (

n1z }| {�1; : : : ; �1 > � � � � � � � � � >

nkz }| {�k; : : : ; �k):

Let W (�) = stabilizer of � under the permutation action ofW = Sn. So

W (�) = Sn1 � � � � � � � � � � � Snk :

Intermediate Combinatorial Set

� = (

n1z }| {�1; : : : ; �1 > � � � � � � � � � >

nkz }| {�k; : : : ; �k):

Let W (�) = stabilizer of � under the permutation action ofW = Sn.

SoW (�) = Sn1 � � � � � � � � � � � Snk :

Intermediate Combinatorial Set

� = (

n1z }| {�1; : : : ; �1 > � � � � � � � � � >

nkz }| {�k; : : : ; �k):

Let W (�) = stabilizer of � under the permutation action ofW = Sn. So

W (�) = Sn1 � � � � � � � � � � � Snk :

Intermediate Combinatorial Set

� = (

n1z }| {�1; : : : ; �1 > � � � � � � � � � >

nkz }| {�k; : : : ; �k):

Consider a collection of k piles of dots, with the ith pileconsisting of ni dots labeled �i.For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), consider:

5 4 3 2 1�

��

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�

�

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�

Intermediate Combinatorial Set

� = (

n1z }| {�1; : : : ; �1 > � � � � � � � � � >

nkz }| {�k; : : : ; �k):

Consider a collection of k piles of dots, with the ith pileconsisting of ni dots labeled �i.

For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), consider:

5 4 3 2 1�

��

��

��

��

��

�

�

�

�

�

Intermediate Combinatorial Set

� = (

n1z }| {�1; : : : ; �1 > � � � � � � � � � >

nkz }| {�k; : : : ; �k):

Consider a collection of k piles of dots, with the ith pileconsisting of ni dots labeled �i.For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), consider:

5 4 3 2 1�

��

��

��

��

��

�

�

�

�

�

Intermediate Combinatorial Set

Consider a directed graph such that on this vertex set such that

� The only allowable edges are of the form xi ! xi+1 with xiin the �i pile and xi+1 in the �i+1 pile; and

� No two edges originate at the same vertex.

For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), we can considerthe following graph:

5 4 3 2 1�

��

��

��

��

��

�

�

�

�

�

Intermediate Combinatorial Set

Consider a directed graph such that on this vertex set such that

� The only allowable edges are of the form xi ! xi+1 with xiin the �i pile and xi+1 in the �i+1 pile; and

� No two edges originate at the same vertex.

For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), we can considerthe following graph:

5 4 3 2 1�

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��

��

��

��

�

�

�

�

�

Intermediate Combinatorial Set

Consider a directed graph such that on this vertex set such that

� The only allowable edges are of the form xi ! xi+1 with xiin the �i pile and xi+1 in the �i+1 pile; and

� No two edges originate at the same vertex.

For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), we can considerthe following graph:

5 4 3 2 1�

�

OOOOOO�

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�

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�

�

�

Intermediate Combinatorial Set

Consider a directed graph such that on this vertex set such that

� The only allowable edges are of the form xi ! xi+1 with xiin the �i pile and xi+1 in the �i+1 pile; and

� No two edges originate at the same vertex.

For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), we can considerthe following graph:

5 4 3 2 1�

�

OOOOOO�

�ooo

ooo�

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�

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Intermediate Combinatorial Set

Consider a directed graph such that on this vertex set such that

� The only allowable edges are of the form xi ! xi+1 with xiin the �i pile and xi+1 in the �i+1 pile; and

� No two edges originate at the same vertex.

For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), we can considerthe following graph:

5 4 3 2 1�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo�

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Intermediate Combinatorial Set

Consider a directed graph such that on this vertex set such that

� The only allowable edges are of the form xi ! xi+1 with xiin the �i pile and xi+1 in the �i+1 pile; and

� No two edges originate at the same vertex.

For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), we can considerthe following graph:

5 4 3 2 1�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

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ooo�

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OOOOOO�

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Intermediate Combinatorial Set

Consider a directed graph such that on this vertex set such that

� The only allowable edges are of the form xi ! xi+1 with xiin the �i pile and xi+1 in the �i+1 pile; and

� No two edges originate at the same vertex.

For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), we can considerthe following graph:

5 4 3 2 1�

�

OOOOOO�

�ooo

ooo�

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OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo

�

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Intermediate Combinatorial Set

Consider a directed graph such that on this vertex set such that

� The only allowable edges are of the form xi ! xi+1 with xiin the �i pile and xi+1 in the �i+1 pile; and

� No two edges originate at the same vertex.

For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), we cannotconsider the following graph:

5 4 3 2 1�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

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ooo�

�

OOOOOO�

�ooo

ooo

� ��

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Intermediate Combinatorial Set

Consider a directed graph such that on this vertex set such that

� The only allowable edges are of the form xi ! xi+1 with xiin the �i pile and xi+1 in the �i+1 pile; and

� No two edges originate at the same vertex.

For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), we can considerthe following graph:

5 4 3 2 1�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo

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Intermediate Combinatorial Set

Consider a directed graph such that on this vertex set such that

� The only allowable edges are of the form xi ! xi+1 with xiin the �i pile and xi+1 in the �i+1 pile; and

� No two edges originate at the same vertex.

For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), we cannotconsider the following graph:

5 4 3 2 1�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo�

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OOOOOO�

�ooo

ooo

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Intermediate Combinatorial Set

Consider a directed graph such that on this vertex set such that

� The only allowable edges are of the form xi ! xi+1 with xiin the �i pile and xi+1 in the �i+1 pile; and

� No two edges originate at the same vertex.

For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), we can considerthe following graph:

5 4 3 2 1�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo

�

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Intermediate Combinatorial Set

Consider a directed graph such that on this vertex set such that

� The only allowable edges are of the form xi ! xi+1 with xiin the �i pile and xi+1 in the �i+1 pile; and

� No two edges originate at the same vertex.

For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), we cannotconsider the following graph:

5 4 3 2 1�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

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ooo

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Intermediate Combinatorial Set

Consider a directed graph such that on this vertex set such that

� The only allowable edges are of the form xi ! xi+1 with xiin the �i pile and xi+1 in the �i+1 pile; and

� No two edges originate at the same vertex.

For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), we can considerthe following graph:

5 4 3 2 1�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo

�

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Intermediate Combinatorial Set

Consider a directed graph such that on this vertex set such that

� The only allowable edges are of the form xi ! xi+1 with xiin the �i pile and xi+1 in the �i+1 pile; and

� No two edges originate at the same vertex.

For instance, if � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1), we can considerthe following graphs:

5 4 3 2 1�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo

�

�

�

�

Clearly W (�) ' Sn1 � � � � � Snk acts on such graphs.

Intermediate Combinatorial Set

For instance,

5 4 3 2 1�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo

�

�

�

�

and 5 4 3 2 1� �� �� �� �

� �� ��

�

�

�

are in the same S2 � S4 � S3 � S2 � S1 orbit.

The set of equivalence classes, denoted MS(�), arecalled �-multisegments.

Intermediate Combinatorial Set

For instance,

5 4 3 2 1�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo�

�

OOOOOO�

�ooo

ooo

�

�

�

�

and 5 4 3 2 1� �� �� �� �

� �� ��

�

�

�

are in the same S2 � S4 � S3 � S2 � S1 orbit.

The set of equivalence classes, denoted MS(�), arecalled �-multisegments.

How do multisegments naturally arise?

Let G = GL(n;C), g = gl(n;C), view � 2 h�, and consider

1 G(�) = Ad-centralizer of � in G.

2 g1(�) = +1-eigenspace of ad(�) on g

Except this doesn't make sense.

How do multisegments naturally arise?

Let G = GL(n;C), g = gl(n;C), view � 2 h�, and consider

1 G(�) = Ad-centralizer of � in G.

2 g1(�) = +1-eigenspace of ad(�) on g

Except this doesn't make sense.

How do multisegments naturally arise?

Let G = GL(n;C), g = gl(n;C), view � 2 h�, and consider

1 G(�) = Ad-centralizer of � in G.

2 g1(�) = +1-eigenspace of ad(�) on g

Except this doesn't make sense.

How do multisegments naturally arise?

Let G = GL(n;C), g = gl(n;C), view � 2 h�, and consider

1 G(�) = Ad-centralizer of � in G.

2 g1(�) = +1-eigenspace of ad(�) on g

Except this doesn't make sense.

Try again: How do multisegments naturally

arise?

Let G = GL(n;C), g = gl(n;C), view � 2 h� ' h_, and consider

1 G_(�) = Ad-centralizer of � in G_.

2 g_1 (�) = +1-eigenspace of ad_(�) on g_

Then G_(�) naturally acts on g_1 (�).

MS(�) parametrizes these orbits.

There is a beautiful generalization to all types (Kawanaka,Lusztig, Vinberg).

Try again: How do multisegments naturally

arise?

Let G = GL(n;C), g = gl(n;C), view � 2 h� ' h_, and consider

1 G_(�) = Ad-centralizer of � in G_.

2 g_1 (�) = +1-eigenspace of ad_(�) on g_

Then G_(�) naturally acts on g_1 (�).

MS(�) parametrizes these orbits.

There is a beautiful generalization to all types (Kawanaka,Lusztig, Vinberg).

Try again: How do multisegments naturally

arise?

Let G = GL(n;C), g = gl(n;C), view � 2 h� ' h_, and consider

1 G_(�) = Ad-centralizer of � in G_.

2 g_1 (�) = +1-eigenspace of ad_(�) on g_

Then G_(�) naturally acts on g_1 (�).

MS(�) parametrizes these orbits.

There is a beautiful generalization to all types (Kawanaka,Lusztig, Vinberg).

Try again: How do multisegments naturally

arise?

Let G = GL(n;C), g = gl(n;C), view � 2 h� ' h_, and consider

1 G_(�) = Ad-centralizer of � in G_.

2 g_1 (�) = +1-eigenspace of ad_(�) on g_

Then G_(�) naturally acts on g_1 (�).

MS(�) parametrizes these orbits.

There is a beautiful generalization to all types (Kawanaka,Lusztig, Vinberg).

Try again: How do multisegments naturally

arise?

Let G = GL(n;C), g = gl(n;C), view � 2 h� ' h_, and consider

1 G_(�) = Ad-centralizer of � in G_.

2 g_1 (�) = +1-eigenspace of ad_(�) on g_

Then G_(�) naturally acts on g_1 (�).

MS(�) parametrizes these orbits.

There is a beautiful generalization to all types (Kawanaka,Lusztig, Vinberg).

Try again: How do multisegments naturally

arise?

Let G = GL(n;C), g = gl(n;C), view � 2 h� ' h_, and consider

1 G_(�) = Ad-centralizer of � in G_.

2 g_1 (�) = +1-eigenspace of ad_(�) on g_

Then G_(�) naturally acts on g_1 (�).

MS(�) parametrizes these orbits.

There is a beautiful generalization to all types (Kawanaka,Lusztig, Vinberg).

Example of parametrization

Take � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1).

So

G_(�)'GL(2;C)�GL(4;C)�GL(3;C)�GL(2;C)�GL(1;C)

g_1 (�) 'nC2 A

�! C4 B�! C3 C

�! C2 D�! C1

o

For instance,

5 4 3 2 1� �� �� �� �

� �� ��

�

�

�

5 4 3 2 1a1 b1b1 c1c1 d1d1 e1

b2 c2c2 d2a2

b3b4

c3

Example of parametrization

Take � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1). So

G_(�)'GL(2;C)�GL(4;C)�GL(3;C)�GL(2;C)�GL(1;C)

g_1 (�) 'nC2 A

�! C4 B�! C3 C

�! C2 D�! C1

o

For instance,

5 4 3 2 1� �� �� �� �

� �� ��

�

�

�

5 4 3 2 1a1 b1b1 c1c1 d1d1 e1

b2 c2c2 d2a2

b3b4

c3

Example of parametrization

Take � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1). So

G_(�)'GL(2;C)�GL(4;C)�GL(3;C)�GL(2;C)�GL(1;C)

g_1 (�) 'nC2 A

�! C4 B�! C3 C

�! C2 D�! C1

o

For instance,

5 4 3 2 1� �� �� �� �

� �� ��

�

�

�

5 4 3 2 1a1 b1b1 c1c1 d1d1 e1

b2 c2c2 d2a2

b3b4

c3

Example of parametrization

Take � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1). So

G_(�)'GL(2;C)�GL(4;C)�GL(3;C)�GL(2;C)�GL(1;C)

g_1 (�) 'nC2 A

�! C4 B�! C3 C

�! C2 D�! C1

o

For instance,

5 4 3 2 1� �� �� �� �

� �� ��

�

�

�

5 4 3 2 1a1 b1b1 c1c1 d1d1 e1

b2 c2c2 d2a2

b3b4

c3

Example of parametrization

Take � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1). So

G_(�)'GL(2;C)�GL(4;C)�GL(3;C)�GL(2;C)�GL(1;C)

g_1 (�) 'nC2 A

�! C4 B�! C3 C

�! C2 D�! C1

o

For instance,

5 4 3 2 1� �� �� �� �

� �� ��

�

�

�

5 4 3 2 1a1 b1b1 c1c1 d1d1 e1

b2 c2c2 d2a2

b3b4

c3

Example of parametrization

Take � = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1). So

G_(�)'GL(2;C)�GL(4;C)�GL(3;C)�GL(2;C)�GL(1;C)

g_1 (�) 'nC2 A

�! C4 B�! C3 C

�! C2 D�! C1

o

For instance,

5 4 3 2 1� �� �� �� �

� �� ��

�

�

�

5 4 3 2 1a1 b1

� //b1 c1� //c1 d1

� //d1 e1� //

b2 c2� //c2 d2

� //a2

b3b4

c3

Another kind of KL polynomial

Since MS(�) parametrizes orbits of G_(�) on g_1 (�), givens; s0 2MS(�) we can consider

ps;s0(q) := pL;L0(q)

where L;L0 2 LocG_(�)(g_1 (�)) are the constant sheaves on the

orbits parametrized by s; s0.

In general, Lusztig (2006) has given an algorithm to computethese polynomials.

Another kind of KL polynomial

Since MS(�) parametrizes orbits of G_(�) on g_1 (�), givens; s0 2MS(�) we can consider

ps;s0(q) := pL;L0(q)

where L;L0 2 LocG_(�)(g_1 (�)) are the constant sheaves on the

orbits parametrized by s; s0.

In general, Lusztig (2006) has given an algorithm to computethese polynomials.

Another kind of KL polynomial

Since MS(�) parametrizes orbits of G_(�) on g_1 (�), givens; s0 2MS(�) we can consider

ps;s0(q) := pL;L0(q)

where L;L0 2 LocG_(�)(g_1 (�)) are the constant sheaves on the

orbits parametrized by s; s0.

In general, Lusztig (2006) has given an algorithm to computethese polynomials.

The plan

We are going to de�ne injections

�1 : MS(�) �! Sn

and

�2 : MS(�) �!a

p+q=n

(GL(p;C)�GL(q;C))nBn

such that all Kazhdan-Lusztig-Vogan polynomials match:

ps;s0 = p�1(s);�1(s0) = p�2(s);�2(s0):

In particular, this gives a matching of KLV polynomials forGL(n;C) and U(p; q).

The plan

We are going to de�ne injections

�1 : MS(�) �! Sn

and

�2 : MS(�) �!a

p+q=n

(GL(p;C)�GL(q;C))nBn

such that all Kazhdan-Lusztig-Vogan polynomials match:

ps;s0 = p�1(s);�1(s0) = p�2(s);�2(s0):

In particular, this gives a matching of KLV polynomials forGL(n;C) and U(p; q).

The plan

We are going to de�ne injections

�1 : MS(�) �! Sn

and

�2 : MS(�) �!a

p+q=n

(GL(p;C)�GL(q;C))nBn

such that all Kazhdan-Lusztig-Vogan polynomials match:

ps;s0 = p�1(s);�1(s0) = p�2(s);�2(s0):

In particular, this gives a matching of KLV polynomials forGL(n;C) and U(p; q).

The plan

We are going to de�ne injections

�1 : MS(�) �! Sn

and

�2 : MS(�) �!a

p+q=n

(GL(p;C)�GL(q;C))nBn

such that all Kazhdan-Lusztig-Vogan polynomials match:

ps;s0 = p�1(s);�1(s0) = p�2(s);�2(s0):

In particular, this gives a matching of KLV polynomials forGL(n;C) and U(p; q).

The plan

We are going to de�ne injections

�1 : MS(�) �! Sn

and

�2 : MS(�) �!a

p+q=n

(GL(p;C)�GL(q;C))nBn

such that all Kazhdan-Lusztig-Vogan polynomials match:

ps;s0 = p�1(s);�1(s0) = p�2(s);�2(s0):

In particular, this gives a matching of KLV polynomials forGL(n;C) and U(p; q).

Defining �1 : MS(�) �! Sn

� �� �� �� �

� �� ��

�

�

�

1 33 77 1010 12

4 88 112

5

6

9

Set � = (1 3 4 7 10 12)(4 8 11): And

�1(s) := longest element in �W (�):

Defining �1 : MS(�) �! Sn

� �� �� �� �

� �� ��

�

�

�

1 33 77 1010 12

4 88 112

5

6

9

Set � = (1 3 4 7 10 12)(4 8 11): And

�1(s) := longest element in �W (�):

Defining �1 : MS(�) �! Sn

� �� �� �� �

� �� ��

�

�

�

1 33 77 1010 12

4 88 112

5

6

9

Set � = (1 3 4 7 10 12)(4 8 11): And

�1(s) := longest element in �W (�):

Defining �1 : MS(�) �! Sn

� �� �� �� �

� �� ��

�

�

�

1 33 77 1010 12

4 88 112

5

6

9

Set � = (1 3 4 7 10 12)(4 8 11):

And

�1(s) := longest element in �W (�):

Defining �1 : MS(�) �! Sn

� �� �� �� �

� �� ��

�

�

�

1 33 77 1010 12

4 88 112

5

6

9

Set � = (1 3 4 7 10 12)(4 8 11): And

�1(s) := longest element in �W (�):

De�ning

�2 : MS(�) �!`

p+q=n(GL(p;C)�GL(q;C))nBn:

Let

��(n) = finvolutions in Sn with signed �xed pointsg:

Then there is a bijection

��(n)$a

p+q=n

(GL(p;C)�GL(q;C))nBn:

De�ning

�2 : MS(�) �!`

p+q=n(GL(p;C)�GL(q;C))nBn:

Let

��(n) = finvolutions in Sn with signed �xed pointsg:

Then there is a bijection

��(n)$a

p+q=n

(GL(p;C)�GL(q;C))nBn:

De�ning

�2 : MS(�) �!`

p+q=n(GL(p;C)�GL(q;C))nBn:

Let

��(n) = finvolutions in Sn with signed �xed pointsg:

Then there is a bijection

��(n)$a

p+q=n

(GL(p;C)�GL(q;C))nBn:

A few details on the ��(n) parametrization.

Set G = GL(n;C) and �x a torus T .Fix the inner class of (classical) real forms represented by

fU(p; q) j p+ q = n; p � qg:

Consider its one-sided parameter space

X = f(z; T 0; B0) j T 0 � B0; z2 2 Z(G); xT 0x�1 = T 0g=G

A few details on the ��(n) parametrization.

Set G = GL(n;C) and �x a torus T .

Fix the inner class of (classical) real forms represented by

fU(p; q) j p+ q = n; p � qg:

Consider its one-sided parameter space

X = f(z; T 0; B0) j T 0 � B0; z2 2 Z(G); xT 0x�1 = T 0g=G

A few details on the ��(n) parametrization.

Set G = GL(n;C) and �x a torus T .Fix the inner class of (classical) real forms represented by

fU(p; q) j p+ q = n; p � qg:

Consider its one-sided parameter space

X = f(z; T 0; B0) j T 0 � B0; z2 2 Z(G); xT 0x�1 = T 0g=G

A few details on the ��(n) parametrization.

Set G = GL(n;C) and �x a torus T .Fix the inner class of (classical) real forms represented by

fU(p; q) j p+ q = n; p � qg:

Consider its one-sided parameter space

X = f(z; T 0; B0) j T 0 � B0; z2 2 Z(G); xT 0x�1 = T 0g=G

A few details on the ��(n) parametrization.

Set G = GL(n;C) and �x a torus T .Fix the inner class of (classical) real forms

fU(p; q) j p+ q = n; p � qg:

Consider its one-sided parameter space

X (e) = f(z; T 0; B0) j T 0 � B0; z2 = e; xT 0x�1 = T 0g=G

= fz 2 NG(T ) j z2 = eg=T $ ��(n):

A few details on the ��(n) parametrization.

Set G = GL(n;C) and �x a torus T .Fix the inner class of (classical) real forms

fU(p; q) j p+ q = n; p � qg:

Consider its one-sided parameter space

X (e) = f(z; T 0; B0) j T 0 � B0; z2 = e; xT 0x�1 = T 0g=G

= fz 2 NG(T ) j z2 = eg=T

$ ��(n):

A few details on the ��(n) parametrization.

Set G = GL(n;C) and �x a torus T .Fix the inner class of (classical) real forms

fU(p; q) j p+ q = n; p � qg:

Consider its one-sided parameter space

X (e) = f(z; T 0; B0) j T 0 � B0; z2 = e; xT 0x�1 = T 0g=G

= fz 2 NG(T ) j z2 = eg=T $ ��(n):

A few details on the ��(n) parametrization.

Recall the reason for introducing X (e):

X (e) $ai

KinBn

where the union is over all strong real forms (lying over e).

Our case (G conjugacy classes of elements of order 2):

X (e)$a

p+q=n

(GL(p;C)�GL(q;C))nBn:

A few details on the ��(n) parametrization.

Recall the reason for introducing X (e):

X (e) $ai

KinBn

where the union is over all strong real forms (lying over e).

Our case (G conjugacy classes of elements of order 2):

X (e)$a

p+q=n

(GL(p;C)�GL(q;C))nBn:

A few details on the ��(n) parametrization.

Recall the reason for introducing X (e):

X (e) $ai

KinBn

where the union is over all strong real forms (lying over e).

Our case (G conjugacy classes of elements of order 2):

X (e)$a

p+q=n

(GL(p;C)�GL(q;C))nBn:

A few details on the ��(n) parametrization.

Recall the reason for introducing X (e):

X (e) $ai

KinBn

where the union is over all strong real forms (lying over e).

Our case (G conjugacy classes of elements of order 2):

��(n)$a

p+q=n

(GL(p;C)�GL(q;C))nBn:

Notation

+ � ++� ,,+ + � � + + + �rr �ss 2 ��(12):

Read o�

p = (# of + signs) +1

2(# non-�xed points) = 6 +

4

2= 8:

q = (# of � signs) +1

2(# non-�xed points) = 2 +

4

2= 4:

Notation

+ � ++� ,,+ + � � + + + �rr �ss 2 ��(12):

Read o�

p = (# of + signs) +1

2(# non-�xed points) = 6 +

4

2= 8:

q = (# of � signs) +1

2(# non-�xed points) = 2 +

4

2= 4:

Notation

+ � ++� ,,+ + � � + + + �rr �ss 2 ��(12):

Read o�

p = (# of + signs) +1

2(# non-�xed points)

= 6 +4

2= 8:

q = (# of � signs) +1

2(# non-�xed points) = 2 +

4

2= 4:

Notation

+ � ++� ,,+ + � � + + + �rr �ss 2 ��(12):

Read o�

p = (# of + signs) +1

2(# non-�xed points) = 6 +

4

2= 8:

q = (# of � signs) +1

2(# non-�xed points) = 2 +

4

2= 4:

Notation

+ � ++� ,,+ + � � + + + �rr �ss 2 ��(12):

Read o�

p = (# of + signs) +1

2(# non-�xed points) = 6 +

4

2= 8:

q = (# of � signs) +1

2(# non-�xed points)

= 2 +4

2= 4:

Notation

+ � ++� ,,+ + � � + + + �rr �ss 2 ��(12):

Read o�

p = (# of + signs) +1

2(# non-�xed points) = 6 +

4

2= 8:

q = (# of � signs) +1

2(# non-�xed points) = 2 +

4

2= 4:

Notation

A convenient short-hand (as in Monty's talk): Rewrite

+ � ++� ,,+ + � � + + + �rr �ss 2 ��(12):

as+ 1 2 + + � � + + + 2 1

Notation

A convenient short-hand (as in Monty's talk):

Rewrite

+ � ++� ,,+ + � � + + + �rr �ss 2 ��(12):

as+ 1 2 + + � � + + + 2 1

Notation

A convenient short-hand (as in Monty's talk): Rewrite

+ � ++� ,,+ + � � + + + �rr �ss 2 ��(12):

as+ 1 2 + + � � + + + 2 1

Notation

A convenient short-hand (as in Monty's talk): Rewrite

+ � ++� ,,+ + � � + + + �rr �ss 2 ��(12):

as+ 1 2 + + � � + + + 2 1

A few comments on the ��(n) parametrization.

General features:

1 This parametrization is \dual" to the one forLocO(n;C)(Bn), i.e. the representation theory of GL(n;R).

2 The closed orbits are parametrized by diagrams consistingof all signs.

3 It's easy to translate the information from the atlascommand kgb in this parametrization.

A few comments on the ��(n) parametrization.

General features:

1 This parametrization is \dual" to the one forLocO(n;C)(Bn), i.e. the representation theory of GL(n;R).

2 The closed orbits are parametrized by diagrams consistingof all signs.

3 It's easy to translate the information from the atlascommand kgb in this parametrization.

A few comments on the ��(n) parametrization.

General features:

1 This parametrization is \dual" to the one forLocO(n;C)(Bn), i.e. the representation theory of GL(n;R).

2 The closed orbits are parametrized by diagrams consistingof all signs.

3 It's easy to translate the information from the atlascommand kgb in this parametrization.

A few comments on the ��(n) parametrization.

General features:

1 This parametrization is \dual" to the one forLocO(n;C)(Bn), i.e. the representation theory of GL(n;R).

2 The closed orbits are parametrized by diagrams consistingof all signs.

3 It's easy to translate the information from the atlascommand kgb in this parametrization.

A few comments on the ��(n) parametrization.

General features:

1 This parametrization is \dual" to the one forLocO(n;C)(Bn), i.e. the representation theory of GL(n;R).

2 The closed orbits are parametrized by diagrams consistingof all signs.

3 It's easy to translate the information from the atlascommand kgb in this parametrization.

An example of the parametrization for

(p; q) = (2; 1).

Set � = e1 � e2, � = e2 � e3, and consider(GL(2;C)�GL(1;C))nB3.

+ +� +�+ �++

An example of the parametrization for

(p; q) = (2; 1).

Set � = e1 � e2, � = e2 � e3, and consider(GL(2;C)�GL(1;C))nB3.

+ +� +�+ �++

An example of the parametrization for

(p; q) = (2; 1).

Set � = e1 � e2, � = e2 � e3, and consider(GL(2;C)�GL(1;C))nB3.

+ +� +�+ �++

+11

+ +�

+11

�

��������������

An example of the parametrization for

(p; q) = (2; 1).

Set � = e1 � e2, � = e2 � e3, and consider(GL(2;C)�GL(1;C))nB3.

+ +� +�+ �++

+11

+ +�

+11

�

��������������+�+

+11

�

??????????????

An example of the parametrization for

(p; q) = (2; 1).

Set � = e1 � e2, � = e2 � e3, and consider(GL(2;C)�GL(1;C))nB3.

+ +� +�+ �++

+11

+ +�

+11

�

��������������+�+

+11

�

??????????????

11+

�++

11+

�

??????????????

An example of the parametrization for

(p; q) = (2; 1).

Set � = e1 � e2, � = e2 � e3, and consider(GL(2;C)�GL(1;C))nB3.

+ +� +�+ �++

+11

+ +�

+11

�

��������������+�+

+11

�

??????????????

11+

�++

11+

�

??????????????+�+

11+

�

��������������

An example of the parametrization for

(p; q) = (2; 1).

Set � = e1 � e2, � = e2 � e3, and consider(GL(2;C)�GL(1;C))nB3.

+ +� +�+ �++

+11

+ +�

+11

�

��������������+�+

+11

�

??????????????

11+

�++

11+

�

??????????????+�+

11+

�

��������������

1 + 1

+11

1 + 1

�

��������������

An example of the parametrization for

(p; q) = (2; 1).

Set � = e1 � e2, � = e2 � e3, and consider(GL(2;C)�GL(1;C))nB3.

+ +� +�+ �++

+11

+ +�

+11

�

��������������+�+

+11

�

??????????????

11+

�++

11+

�

??????????????+�+

11+

�

��������������

1 + 1

+11

1 + 1

�

��������������11+

1 + 1

�

??????????????

De�ning �2 : MS(�) �! ��(n)

Consider our running example:

5 4 3 2 1� �� �� �� �

� �� ��

�

�

�

5 4 3 2 1+ �� ++ �� +

� ++ �+

�

�

+

De�ning �2 : MS(�) �! ��(n)

Consider our running example:

5 4 3 2 1� �� �� �� �

� �� ��

�

�

�

5 4 3 2 1+ �� ++ �� +

� ++ �+

�

�

+

De�ning �2 : MS(�) �! ��(n)

Consider our running example:

5 4 3 2 1� �� �� �� �

� �� ��

�

�

�

5 4 3 2 1� + � + �

� ++ �+

�

�

+

De�ning �2 : MS(�) �! ��(n)

Consider our running example:

5 4 3 2 1� �� �� �� �

� �� ��

�

�

�

5 4 3 2 1� � � + �

� ++ �+

�

�

+

Now \ atten".

De�ning �2 : MS(�) �! ��(n)

Consider our running example:

5 4 3 2 1� �� �� �� �

� �� ��

�

�

�

5 4 3 2 1� � � + �

� ++ �+

�

�

+

Now \ atten".

Flatten...

� ))+ � � � � ((� + + �vv+ �uu

The attening procedure is not well-de�ned.

Remedy: take largest dimensional orbit obtained this way.

Flatten...

� ))+ � � � � ((� + + �vv+ �uu

The attening procedure is not well-de�ned.

Remedy: take largest dimensional orbit obtained this way.

Flatten...

� ))+ � � � � ((� + + �vv+ �uu

The attening procedure is not well-de�ned.

Remedy: take largest dimensional orbit obtained this way.

The upshot

We have de�ned injections

�1 : MS(�) �! Sn

and

�2 : MS(�) �!a

p+q=n

(GL(p;C)�GL(q;C))nBn:

Theorem (Zelevinsky, CT)

All Kazhdan-Lusztig-Vogan polynomials match:

ps;s0 = p�1(s);�1(s0) = p�2(s);�2(s0):

In particular, this gives a matching of KLV polynomials for

GL(n;C) and U(p; q).

The upshot

We have de�ned injections

�1 : MS(�) �! Sn

and

�2 : MS(�) �!a

p+q=n

(GL(p;C)�GL(q;C))nBn:

Theorem (Zelevinsky, CT)

All Kazhdan-Lusztig-Vogan polynomials match:

ps;s0 = p�1(s);�1(s0) = p�2(s);�2(s0):

In particular, this gives a matching of KLV polynomials for

GL(n;C) and U(p; q).

The upshot

We have de�ned injections

�1 : MS(�) �! Sn

and

�2 : MS(�) �!a

p+q=n

(GL(p;C)�GL(q;C))nBn:

Theorem (Zelevinsky, CT)

All Kazhdan-Lusztig-Vogan polynomials match:

ps;s0 = p�1(s);�1(s0) = p�2(s);�2(s0):

In particular, this gives a matching of KLV polynomials for

GL(n;C) and U(p; q).

The upshot

We have de�ned injections

�1 : MS(�) �! Sn

and

�2 : MS(�) �!a

p+q=n

(GL(p;C)�GL(q;C))nBn:

Theorem (Zelevinsky, CT)

All Kazhdan-Lusztig-Vogan polynomials match:

ps;s0 = p�1(s);�1(s0) = p�2(s);�2(s0):

In particular, this gives a matching of KLV polynomials for

GL(n;C) and U(p; q).

The upshot

We have de�ned injections

�1 : MS(�) �! Sn

and

�2 : MS(�) �!a

p+q=n

(GL(p;C)�GL(q;C))nBn:

Theorem (Zelevinsky, CT)

All Kazhdan-Lusztig-Vogan polynomials match:

ps;s0 = p�1(s);�1(s0) = p�2(s);�2(s0):

In particular, this gives a matching of KLV polynomials for

GL(n;C) and U(p; q).

Where does all this come from?

Suppose G is a complex algebraic group de�ned over R.Fix an inner class of real forms fG1; : : : ; Gkg of G. For example

GL(n;R) if n is odd; and

GL(n;R) and GL(n=2;H) if n is even.

Simplify: G adjoint. Then main result of ABV

Mi

KHC(Gi)dual$ geometry of G_(C) orbits on XR:

Where does all this come from?

Suppose G is a complex algebraic group de�ned over R.

Fix an inner class of real forms fG1; : : : ; Gkg of G. For example

GL(n;R) if n is odd; and

GL(n;R) and GL(n=2;H) if n is even.

Simplify: G adjoint. Then main result of ABV

Mi

KHC(Gi)dual$ geometry of G_(C) orbits on XR:

Where does all this come from?

Suppose G is a complex algebraic group de�ned over R.Fix an inner class of real forms fG1; : : : ; Gkg of G.

For example

GL(n;R) if n is odd; and

GL(n;R) and GL(n=2;H) if n is even.

Simplify: G adjoint. Then main result of ABV

Mi

KHC(Gi)dual$ geometry of G_(C) orbits on XR:

Where does all this come from?

Suppose G is a complex algebraic group de�ned over R.Fix an inner class of real forms fG1; : : : ; Gkg of G. For example

GL(n;R) if n is odd; and

GL(n;R) and GL(n=2;H) if n is even.

Simplify: G adjoint. Then main result of ABV

Mi

KHC(Gi)dual$ geometry of G_(C) orbits on XR:

Where does all this come from?

Suppose G is a complex algebraic group de�ned over R.Fix an inner class of real forms fG1; : : : ; Gkg of G. For example

GL(n;R) if n is odd; and

GL(n;R) and GL(n=2;H) if n is even.

Simplify: G adjoint. Then main result of ABV

Mi

KHC(Gi)dual$ geometry of G_(C) orbits on XR:

Where does all this come from?

Suppose G is a complex algebraic group de�ned over R.Fix an inner class of real forms fG1; : : : ; Gkg of G. For example

GL(n;R) if n is odd; and

GL(n;R) and GL(n=2;H) if n is even.

Simplify: G adjoint.

Then main result of ABV

Mi

KHC(Gi)dual$ geometry of G_(C) orbits on XR:

Where does all this come from?

Suppose G is a complex algebraic group de�ned over R.Fix an inner class of real forms fG1; : : : ; Gkg of G. For example

GL(n;R) if n is odd; and

GL(n;R) and GL(n=2;H) if n is even.

Simplify: G adjoint. Then main result of ABV

Mi

KHC(Gi)dual$ geometry of G_(C) orbits on XR:

Where does all this come from?

Suppose G is a complex algebraic group de�ned over R.Fix an inner class of real forms fG1; : : : ; Gkg of G. For example

GL(n;R) if n is odd; and

GL(n;R) and GL(n=2;H) if n is even.

Simplify: G adjoint. Then main result of ABV

Mi

KHC(Gi)�dual$ geometry of G_(C) orbits on XR;�:

Where does all this come from?

Suppose G is a an adjoint algebraic group de�ned over F = Qp.Fix an inner class of real forms fG1; : : : ; Gkg of G. For example�

GL(n=d;Ed)

���� djn�

Then Deligne-Langlands-Lusztig:

Mi

KUN (Gi)dual$ geometry of G_(C) orbits on XF :

Where does all this come from?

Suppose G is a an adjoint algebraic group de�ned over F = Qp.

Fix an inner class of real forms fG1; : : : ; Gkg of G. For example�GL(n=d;Ed)

���� djn�

Then Deligne-Langlands-Lusztig:

Mi

KUN (Gi)dual$ geometry of G_(C) orbits on XF :

Where does all this come from?

Suppose G is a an adjoint algebraic group de�ned over F = Qp.Fix an inner class of real forms fG1; : : : ; Gkg of G.

For example

�GL(n=d;Ed)

���� djn�

Then Deligne-Langlands-Lusztig:

Mi

KUN (Gi)dual$ geometry of G_(C) orbits on XF :

Where does all this come from?

Suppose G is a an adjoint algebraic group de�ned over F = Qp.Fix an inner class of real forms fG1; : : : ; Gkg of G. For example�

GL(n=d;Ed)

���� djn�

Then Deligne-Langlands-Lusztig:

Mi

KUN (Gi)dual$ geometry of G_(C) orbits on XF :

Where does all this come from?

Suppose G is a an adjoint algebraic group de�ned over F = Qp.Fix an inner class of real forms fG1; : : : ; Gkg of G. For example�

GL(n=d;Ed)

���� djn�

Then Deligne-Langlands-Lusztig:

Mi

KUN (Gi)dual$ geometry of G_(C) orbits on XF :

Where does all this come from?

Suppose G is a an adjoint algebraic group de�ned over F = Qp.Fix an inner class of real forms fG1; : : : ; Gkg of G. For example�

GL(n=d;Ed)

���� djn�

Then Deligne-Langlands-Lusztig:

Mi

KUN (Gi)�dual$ geometry of G_(C) orbits on XF;�:

Where does all this come from?

Write out the spaces. If � is real, then there is a natural map

XF;� �! XR;�:

In the case of G = GL(n), this unravels (on the level of orbits)to give the map

�2 : MS(�) �!a

p+q=n

(GL(p;C)�GL(q;C))nBn:

Where does all this come from?

Write out the spaces. If � is real, then there is a natural map

XF;� �! XR;�:

In the case of G = GL(n), this unravels (on the level of orbits)to give the map

�2 : MS(�) �!a

p+q=n

(GL(p;C)�GL(q;C))nBn:

Where does all this come from?

Write out the spaces. If � is real, then there is a natural map

XF;� �! XR;�:

In the case of G = GL(n), this unravels (on the level of orbits)to give the map

�2 : MS(�) �!a

p+q=n

(GL(p;C)�GL(q;C))nBn:

Generalizations?

If G is simple, adjoint, and classical, one may unravel thenatural map XF;� �! XR;� in much the same way.

Find an anologous matching of KLV polynomials for UN (G=F )and HC(G=R) (and a weaker one for HC(G=C)).

Generalizations?

If G is simple, adjoint, and classical, one may unravel thenatural map XF;� �! XR;� in much the same way.

Find an anologous matching of KLV polynomials for UN (G=F )and HC(G=R)

(and a weaker one for HC(G=C)).

Generalizations?

If G is simple, adjoint, and classical, one may unravel thenatural map XF;� �! XR;� in much the same way.

Find an anologous matching of KLV polynomials for UN (G=F )and HC(G=R) (and a weaker one for HC(G=C)).

Generalizations?

If G is simple, adjoint, and exceptional, the natural mapXF;� �! XR;� is less well behaved.

(Two orbits can collapse toone, for instance.)

Generalizations?

If G is simple, adjoint, and exceptional, the natural mapXF;� �! XR;� is less well behaved. (Two orbits can collapse toone, for instance.)

Back to the nice case of GL(n)

The matching of KLV polynomials implies there are nicerelationships between character formuals for

1 Harish-Chandra modules for GL(n;C);

2 Unipotent representations of GL(n;Qp); and

3 Harish Chandra modules for GL(n;R).

Are there functors explaining these relationships?

Back to the nice case of GL(n)

The matching of KLV polynomials implies there are nicerelationships between character formuals for

1 Harish-Chandra modules for GL(n;C);

2 Unipotent representations of GL(n;Qp); and

3 Harish Chandra modules for GL(n;R).

Are there functors explaining these relationships?

Back to the nice case of GL(n)

The matching of KLV polynomials implies there are nicerelationships between character formuals for

1 Harish-Chandra modules for GL(n;C);

2 Unipotent representations of GL(n;Qp); and

3 Harish Chandra modules for GL(n;R).

Are there functors explaining these relationships?

Back to the nice case of GL(n)

The matching of KLV polynomials implies there are nicerelationships between character formuals for

1 Harish-Chandra modules for GL(n;C);

2 Unipotent representations of GL(n;Qp); and

3 Harish Chandra modules for GL(n;R).

Are there functors explaining these relationships?

Back to the nice case of GL(n)

The matching of KLV polynomials implies there are nicerelationships between character formuals for

1 Harish-Chandra modules for GL(n;C);

2 Unipotent representations of GL(n;Qp); and

3 Harish Chandra modules for GL(n;R).

Are there functors explaining these relationships?

Back to the nice case of GL(n)

The matching of KLV polynomials implies there are nicerelationships between character formuals for

1 Harish-Chandra modules for GL(n;C);

2 Unipotent representations of GL(n;Qp); and

3 Harish Chandra modules for GL(n;R).

Are there functors explaining these relationships?

Existence of Functors?

Since we have maps

�1 : XF;� �! XC;�

�2 : XF;� �! XR;�

we anticipate functors

HC(GL(n;C))� �! UN (GL(n;Qp))

HC(GL(n;R))� �! UN (GL(n;Qp))

Existence of Functors?

Since we have maps

�1 : XF;� �! XC;�

�2 : XF;� �! XR;�

we anticipate functors

HC(GL(n;C))� �! UN (GL(n;Qp))

HC(GL(n;R))� �! UN (GL(n;Qp))

Existence of Functors?

Since we have maps

�1 : XF;� �! XC;�

�2 : XF;� �! XR;�

we anticipate functors

HC(GL(n;C))� �! UN (GL(n;Qp))

HC(GL(n;R))� �! UN (GL(n;Qp))

Existence of Functors?

Since we have maps

�1 : XF;� �! XC;�

�2 : XF;� �! XR;�

we anticipate functors

O� �! UN (GL(n;Qp))

HC(GL(n;R))� �! UN (GL(n;Qp))

Existence of Functors?

We expect that the functors

O� �! UN (GL(n;Qp))

HC(GL(n;R))� �! UN (GL(n;Qp))

should take standard modules to standard modules (or zero)and irreducibles to irreducibles (or zero).

See Dan Ciubotaru's talk.

Existence of Functors?

We expect that the functors

O� �! UN (GL(n;Qp))

HC(GL(n;R))� �! UN (GL(n;Qp))

should take standard modules to standard modules (or zero)and irreducibles to irreducibles (or zero).

See Dan Ciubotaru's talk.