Functors for Matching
Kazhdan-Lusztig polynomials for GL(n; F )
Peter E. Trapa
(joint with Dan Ciubotaru)
F : X 7! H0 (X F )[+N]
KLV polynomials
KLV polynomials
GR real reductive group, g = gR R C
KR maximal compact subgroup, K = (KR )C
B variety of Borel subalgebras in g.
LocK (B) set of irreducible K -equivariant locally constant
sheaves supported on single K orbits on B
KLV polynomials
GR real reductive group, g = gR R C
KR maximal compact subgroup, K = (KR )C
B variety of Borel subalgebras in g.
LocK (B) set of irreducible K -equivariant locally constant
sheaves supported on single K orbits on B
Given L; L0 2 LocK (B), we can dene the
Kazhdan-Lusztig-Vogan polynomial
pLL (q):
0
Notation for special case
If L and L0 are the constant sheaves on Q; Q0 2 K nB, instead
write
pQ;Q = pL;L
0
0
Notation for an even more special case
If GR is already a complex Lie group, then the Bruhat
decomposition identies K nB with the Weyl group W of g.
Notation for an even more special case
If GR is already a complex Lie group, then the Bruhat
decomposition identies K nB with the Weyl group W of g. In
fact,
LocK (B) $ W:
Notation for an even more special case
If GR is already a complex Lie group, then the Bruhat
decomposition identies K nB with the Weyl group W of g. In
fact,
LocK (B) $ W:
If Q(x) and Q(y) are two orbits parametrized by x and y, then
we write
px;y = pQ(x);Q(y) :
Another example
GR = U(p; q)
Another example
GR = U(p; q)
KR = U(p) U(q)
Another example
GR = U(p; q)
KR = U(p) U(q)
K = GL(p; C) GL(q; C)
Another example
GR = U(p; q)
KR = U(p) U(q)
K = GL(p; C) GL(q; C)
In fact
LocK (B) $ (GL(p; C) GL(q; C))nB:
What does \matching KLV polynomials" mean?
Let G1R and G2R denote two real groups.
What does \matching KLV polynomials" mean?
Let G1R and G2R denote two real groups.
Suppose we can nd
(1) a subset S 1 LocK 1 (B1 );
What does \matching KLV polynomials" mean?
Let G1R and G2R denote two real groups.
Suppose we can nd
(1) a subset S 1 LocK 1 (B1 );
(2) a subset S 2 LocK 2 (B2 ); and
What does \matching KLV polynomials" mean?
Let G1R and G2R denote two real groups.
Suppose we can nd
(1) a subset S 1 LocK 1 (B1 );
(2) a subset S 2 LocK 2 (B2 ); and
(3) a bijection : S 1 ! S 2
What does \matching KLV polynomials" mean?
Let G1R and G2R denote two real groups.
Suppose we can nd
(1) a subset S 1 LocK 1 (B1 );
(2) a subset S 2 LocK 2 (B2 ); and
(3) a bijection : S 1 ! S 2
such that for any L; L0 2 S 1 ,
pL;L = p(L);(L ) :
0
0
What does \matching KLV polynomials" mean?
Let G1R and G2R denote two real groups.
Suppose we can nd
(1) a subset S 1 LocK 1 (B1 );
(2) a subset S 2 LocK 2 (B2 ); and
(3) a bijection : S 1 ! S 2
such that for any L; L0 2 S 1 ,
pL;L = p(L);(L ) :
0
0
Then we say implements a matching of KLV polynomials.
What does \matching KLV polynomials" mean?
Let G1R and G2R denote two real groups.
Suppose we can nd
(1) a subset S 1 LocK 1 (B1 );
(2) a subset S 2 LocK 2 (B2 ); and
(3) a bijection : S 1 ! S 2
such that for any L; L0 2 S 1 ,
pL;L = p(L);(L ) :
0
0
Then we say implements a matching of KLV polynomials.
Obviously this is more impressive if S 1 is large.
Plan
Plan
Explicitly describe a matching for GL(n; C) with various
U(p; q)'s, p + q = n.
Plan
Explicitly describe a matching for GL(n; C) with various
U(p; q)'s, p + q = n.
The key will be an intermediate combinatorial set (which is
really GL(n; Qp ) in disguise).
Plan
Explicitly describe a matching for GL(n; C) with various
U(p; q)'s, p + q = n.
The key will be an intermediate combinatorial set (which is
really GL(n; Qp ) in disguise).
Indicate briey a general setting to search for such
matchings. (Works in all simple adjoint classical cases, for
instance, but breaks in interesting ways in a handful of
exceptional ones.)
Plan
Explicitly describe a matching for GL(n; C) with various
U(p; q)'s, p + q = n.
The key will be an intermediate combinatorial set (which is
really GL(n; Qp ) in disguise).
Indicate briey a general setting to search for such
matchings. (Works in all simple adjoint classical cases, for
instance, but breaks in interesting ways in a handful of
exceptional ones.)
Hando to Dan: a functor \implementing" the matching.
Intermediate Combinatorial Set
Intermediate Combinatorial Set
Fix = (01 0n ) (an innitesimal character in disguise).
Intermediate Combinatorial Set
Fix = (01 0n ) (an innitesimal character in disguise).
Everything reduces to the case
1 0 2 Z
i
0
2 0
i i+1 2 f0; 1g.
Intermediate Combinatorial Set
Fix = (01 0n ) (an innitesimal character in disguise).
Everything reduces to the case
1 0 2 Z
i
0
2 0
i i+1 2 f0; 1g.
We assume this is the case and rewrite
z
n1
}|
{
z
nk
}|
{
= (1 ; : : : ; 1 > > k ; : : : ; k ):
Intermediate Combinatorial Set
z
n1
}|
{
z
nk
}|
{
= (1 ; : : : ; 1 > > k ; : : : ; k ):
Intermediate Combinatorial Set
z
n1
}|
{
z
nk
}|
{
= (1 ; : : : ; 1 > > k ; : : : ; k ):
Let W () = stabilizer of under the permutation action of
W = Sn .
Intermediate Combinatorial Set
z
n1
}|
{
z
nk
}|
{
= (1 ; : : : ; 1 > > k ; : : : ; k ):
Let W () = stabilizer of under the permutation action of
W = Sn . So
W () = Sn1 Snk :
Intermediate Combinatorial Set
z
n1
}|
{
z
nk
}|
{
= (1 ; : : : ; 1 > > k ; : : : ; k ):
Intermediate Combinatorial Set
z
n1
}|
{
z
nk
}|
{
= (1 ; : : : ; 1 > > k ; : : : ; k ):
Consider a collection of k piles of dots, with the ith pile
consisting of ni dots labeled i .
Intermediate Combinatorial Set
z
n1
}|
{
nk
z
}|
{
= (1 ; : : : ; 1 > > k ; : : : ; k ):
Consider a collection of k piles of dots, with the ith pile
consisting 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 xi
in the i pile and xi+1 in the i+1 pile; and
No two edges originate at the same vertex.
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 xi
in 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 consider
the 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 xi
in 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 consider
the following graph:
5
4
OOOOOO
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 xi
in 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 consider
the following graph:
5
4
3
OOOOOO oooo
oo 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 xi
in 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 consider
the following graph:
5
4
3
2
1
OOOOOO ooooOOOOOO oooo
oo oo
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 xi
in 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 consider
the following graph:
5
4
3
2
1
OOOOOOOOoOoOoOoOOOOOOO oooo
oo oo
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 xi
in 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 consider
the following graph:
5
4
3
2
1
OOOOOOOOoOoOoOoOOOOooOOoOo oooo
oo oo oo
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 xi
in 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 cannot
consider the following graph:
5
4
3
2
1
OOOOOOOOoOoOoOoOOOOoOoOoOo oooo
oo oo oo
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 xi
in 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 consider
the following graph:
5
4
3
2
1
OOOOOOOOoOoOoOoOOOOooOOoOo oooo
oo oo oo
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 xi
in 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 cannot
consider the following graph:
5
4
3
2
1
OOOOOOOOoOoOoOoOOOOoOoOoOo oooo
oo oo oo
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 xi
in 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 consider
the following graph:
5
4
3
2
1
OOOOOOOOoOoOoOoOOOOooOOoOo oooo
oo oo oo
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 xi
in 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 cannot
consider the following graph:
5
4
3
2
1
OOOOOOOOoOoOoOoOOOOoOoOoOojjjojojojo
oo jjjjjoojjjjoo
jj 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 xi
in 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 consider
the following graph:
5
4
3
2
1
OOOOOOOOoOoOoOoOOOOooOOoOo oooo
oo oo oo
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 xi
in 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 consider
the following graphs:
5
4
3
2
1
OOOO OOOoOooOOOoOoo ooo
OOooo OOooo OOooo
Clearly W () ' Sn1 Snk acts on such graphs.
Intermediate Combinatorial Set
For instance,
5
4
3
2
1
OOO
O
O
OOO oOoOoOoOoO OoOoOoOoOo ooooo
o
o
o
and
5
4
are in the same S2 S4 S3 S2 S1 orbit.
3
2
1
Intermediate Combinatorial Set
For instance,
5
4
3
2
1
OOO
O
O
OOO oOoOoOoOoO OoOoOoOoOo ooooo
o
o
o
and
5
4
3
2
are in the same S2 S4 S3 S2 S1 orbit.
The set of equivalence classes, denoted
called
-multisegments.
MS (), are
1
How do multisegments naturally arise?
Let G = GL(n; C), g = gl(n; C), view 2 h , and consider
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.
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
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
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_ .
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-eigenspace of ad_ () on g_
1
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-eigenspace of ad_ () on g_
1
Then G_ () naturally acts on g_1 ().
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-eigenspace of ad_ () on g_
1
Then G_ () naturally acts on g_1 ().
MS () parametrizes these orbits.
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-eigenspace of ad_ () on g_
1
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).
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)
Example of parametrization
Take = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1). So
G_ () '
GL(2; C)n GL(4; C) GL(3; C) GL(2; C)o GL(1; C)
g_1 () ' C2 A! C4 B! C3 C! C2 D! C1
Example of parametrization
Take = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1). So
G_ () '
GL(2; C)n GL(4; C) GL(3; C) GL(2; C)o GL(1; C)
g_1 () ' C2 A! C4 B! C3 C! C2 D! C1
For instance,
5
4
3
2
1
Example of parametrization
Take = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1). So
G_ () '
GL(2; C)n GL(4; C) GL(3; C) GL(2; C)o GL(1; C)
g_1 () ' C2 A! C4 B! C3 C! C2 D! C1
For instance,
5
4
3
2
1
5
a1
a2
4
b1
b2
b3
b4
3
c1
c2
c3
2
d1
d2
1
e1
Example of parametrization
Take = (5; 5; 4; 4; 4; 4; 3; 3; 3; 2; 2; 1). So
G_ () '
GL(2; C)n GL(4; C) GL(3; C) GL(2; C)o GL(1; C)
g_1 () ' C2 A! C4 B! C3 C! C2 D! C1
For instance,
5
4
3
2
1
5
a1 a2
4
b/ 1 b2 b3
b4
3
/c1 /c2 c3
2
d/ 1 /d2
1
/e1
Another kind of KL polynomial
Another kind of KL polynomial
Since MS () parametrizes orbits of G_ () on g_1 (), given
0
s; s 2 MS () we can consider
ps;s (q) := pL;L (q)
0
0
where L; L0 2 LocG () (g_1 ()) are the constant sheaves on the
orbits parametrized by s; s0 .
_
Another kind of KL polynomial
Since MS () parametrizes orbits of G_ () on g_1 (), given
0
s; s 2 MS () we can consider
ps;s (q) := pL;L (q)
0
0
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 compute
these polynomials.
The plan
The plan
We are going to dene injections
1 : MS () ! Sn
The plan
We are going to dene injections
1 : MS () ! Sn
and
2 : MS () !
a
(GL(p; C) GL(q; C))nBn
p+q=n
The plan
We are going to dene injections
1 : MS () ! Sn
and
2 : MS () !
a
(GL(p; C) GL(q; C))nBn
p+q=n
such that all Kazhdan-Lusztig-Vogan polynomials match:
ps;s = p1 (s);1 (s ) = p2 (s);2 (s ) :
0
0
0
The plan
We are going to dene injections
1 : MS () ! Sn
and
2 : MS () !
a
(GL(p; C) GL(q; C))nBn
p+q=n
such that all Kazhdan-Lusztig-Vogan polynomials match:
ps;s = p1 (s);1 (s ) = p2 (s);2 (s ) :
0
0
0
In particular, this gives a matching of KLV polynomials for
GL(n; C) and U(p; q).
Defining 1 : MS () ! Sn
Defining 1 : MS () ! Sn
Defining 1 : MS () ! Sn
1
2
3
4
5
6
7
8
9
10
11
12
Defining 1 : MS () ! Sn
Set = (1 3 4 7 10 12)(4 8 11):
1
2
3
4
5
6
7
8
9
10
11
12
Defining 1 : MS () ! Sn
Set = (1 3 4 7 10 12)(4 8 11): And
1
2
3
4
5
6
7
8
9
1 (s) := longest element in W ():
10
11
12
Dening
2 : MS () !
`
p+q=n
(GL(p; C) GL(q; C))nBn:
Dening
2 : MS () !
`
p+q=n
(GL(p; C) GL(q; C))nBn:
Let
(n) = finvolutions in Sn with signed xed pointsg:
Dening
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
(GL(p; C) GL(q; C))nBn :
p+q=n
A few details on the (n) parametrization.
A few details on the (n) parametrization.
Set G = GL(n; C) and x a torus T .
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:
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; B 0) j T 0 B 0; 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; B 0) j T 0 B 0; z2 = e; 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; B 0) j T 0 B 0; z2 = e; xT 0x 1 = T 0g=G
= fz 2 NG (T ) j z 2 = eg=T
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; B 0) j T 0 B 0; z2 = e; xT 0x 1 = T 0g=G
= fz 2 NG (T ) j z 2 = eg=T $ (n):
A few details on the (n) parametrization.
Recall the reason for introducing X (e):
X (e) $
a
i
Ki nBn
where the union is over all strong real forms (lying over e).
A few details on the (n) parametrization.
Recall the reason for introducing X (e):
X (e) $
a
i
Ki nBn
where the union is over all strong real forms (lying over e).
Our case (G conjugacy classes of elements of order 2):
A few details on the (n) parametrization.
Recall the reason for introducing X (e):
X (e) $
a
i
Ki nBn
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
(GL(p; C) GL(q; C))nBn :
p+q=n
A few details on the (n) parametrization.
Recall the reason for introducing X (e):
X (e) $
a
i
Ki nBn
where the union is over all strong real forms (lying over e).
Our case (G conjugacy classes of elements of order 2):
(n) $
a
(GL(p; C) GL(q; C))nBn :
p+q=n
Notation
Notation
+ s r + +
+ + + , + 2 (12):
Notation
+ s r + +
Read o
p = (# of + signs) +
+ + + , + 2 (12):
1 (# non-xed points)
2
Notation
+ s r + +
Read o
p = (# of + signs) +
+ + + , + 2 (12):
1 (# non-xed points) = 6 + 4 = 8:
2
2
Notation
+ s r + +
Read o
p = (# of + signs) +
q = (# of
+ + + , + 2 (12):
1 (# non-xed points) = 6 + 4 = 8:
2
2
signs) + 12 (# non-xed points)
Notation
+ s r + +
Read o
p = (# of + signs) +
q = (# of
+ + + , + 2 (12):
1 (# non-xed points) = 6 + 4 = 8:
2
2
signs) + 12 (# non-xed points) = 2 + 42 = 4:
Notation
Notation
A convenient short-hand (as in Monty's talk):
Notation
A convenient short-hand (as in Monty's talk): Rewrite
+ s r + +
+ + + , + 2 (12):
Notation
A convenient short-hand (as in Monty's talk): Rewrite
+ s r + +
as
+12 + +
+ + + , + 2 (12):
+ + + 21
A few comments on the (n) parametrization.
A few comments on the (n) parametrization.
General features:
A few comments on the (n) parametrization.
General features:
1 This parametrization is \dual" to the one for
LocO(n;C) (Bn ), i.e. the representation theory of GL(n; R).
A few comments on the (n) parametrization.
General features:
1 This parametrization is \dual" to the one for
LocO(n;C) (Bn ), i.e. the representation theory of GL(n; R).
2 The closed orbits are parametrized by diagrams consisting
of all signs.
A few comments on the (n) parametrization.
General features:
1 This parametrization is \dual" to the one for
LocO(n;C) (Bn ), i.e. the representation theory of GL(n; R).
2 The closed orbits are parametrized by diagrams consisting
of all signs.
3 It's easy to translate the information from the atlas
command kgb in this parametrization.
An example of the parametrization for
(p; q) = (2; 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 .
++
+ +
++
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








++
+ +
++
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?
 ???

??


??
??


??

??



++
+ +
++
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?
 ???
 ???


??
??



??
??

??
??


?
??


??


??


?




++
+ +
++
An example of the parametrization for
(p; q) = (2; 1).
Set = e1 e2 , = e2 e3 , and consider
(GL(2; C) GL(1; C))nB3 .
1+ 1










+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 .
1+?1??

??


?? 

??

??


??


?

+11
11+
??
??


??
??


??
?


?? 

??

?
??
??


?
??


??


??


?




++
+ +
++
Dening 2 :
MS () ! (n)
Consider our running example:
5
4
3
2
1
Dening 2 :
MS () ! (n)
Consider our running example:
5
4
3
2
1
5
+
+
4
3
+
+
+
2
1
+
Dening 2 :
MS () ! (n)
Consider our running example:
5
4
3
2
1
5
+
4
+
3
+
+
2
+
1
Dening 2 :
MS () ! (n)
Consider our running example:
5
4
3
2
1
5
+
4
3
+
+
2
+
1
Dening 2 :
MS () ! (n)
Consider our running example:
5
4
3
Now \atten".
2
1
5
+
4
3
+
+
2
+
1
Flatten...
u +
v
+ + ( + )
Flatten...
u +
v
+ + ( + )
The attening procedure is not well-dened.
Flatten...
u +
v
+ + ( + )
The attening procedure is not well-dened.
Remedy: take largest dimensional orbit obtained this way.
The upshot
The upshot
We have dened injections
1 : MS () ! Sn
The upshot
We have dened injections
1 : MS () ! Sn
and
2 : MS () !
a
p+q=n
(GL(p; C) GL(q; C))nBn :
The upshot
We have dened 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;s = p1 (s);1 (s ) = p2 (s);2 (s ) :
0
0
0
The upshot
We have dened 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;s = p1 (s);1 (s ) = p2 (s);2 (s ) :
0
0
0
In particular, this gives a matching of KLV polynomials for
GL(n; C) and U(p; q).
Where does all this come from?
Where does all this come from?
Suppose G is a complex algebraic group dened over R.
Where does all this come from?
Suppose G is a complex algebraic group dened over R.
Fix an inner class of real forms fG1 ; : : : ; Gk g of G .
Where does all this come from?
Suppose G is a complex algebraic group dened over R.
Fix an inner class of real forms fG1 ; : : : ; Gk g of G . For example
GL(n; R) if n is odd; and
Where does all this come from?
Suppose G is a complex algebraic group dened over R.
Fix an inner class of real forms fG1 ; : : : ; Gk g of G . For example
GL(n; R) if n is odd; and
GL(n; R) and GL(n=2; H) if n is even.
Where does all this come from?
Suppose G is a complex algebraic group dened over R.
Fix an inner class of real forms fG1 ; : : : ; Gk g 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.
Where does all this come from?
Suppose G is a complex algebraic group dened over R.
Fix an inner class of real forms fG1 ; : : : ; Gk g 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
M
i
K HC (Gi ) dual
$ geometry of G _(C) orbits on XR:
Where does all this come from?
Suppose G is a complex algebraic group dened over R.
Fix an inner class of real forms fG1 ; : : : ; Gk g 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
M
i
K HC (Gi ) dual
$ geometry of G _(C) orbits on XR;:
Where does all this come from?
Where does all this come from?
Suppose G is a an adjoint algebraic group dened over F = Qp .
Where does all this come from?
Suppose G is a an adjoint algebraic group dened over F = Qp .
Fix an inner class of real forms fG1 ; : : : ; Gk g of G .
Where does all this come from?
Suppose G is a an adjoint algebraic group dened over F = Qp .
Fix an inner class of real forms fG1 ; : : : ; Gk g of G . For example
GL(n=d; Ed ) djn
Where does all this come from?
Suppose G is a an adjoint algebraic group dened over F = Qp .
Fix an inner class of real forms fG1 ; : : : ; Gk g of G . For example
GL(n=d; Ed ) djn
Then Deligne-Langlands-Lusztig:
M
i
K UN (Gi ) dual
$ geometry of G _(C) orbits on XF :
Where does all this come from?
Suppose G is a an adjoint algebraic group dened over F = Qp .
Fix an inner class of real forms fG1 ; : : : ; Gk g of G . For example
GL(n=d; Ed ) djn
Then Deligne-Langlands-Lusztig:
M
i
K UN (Gi ) dual
$ geometry of G _(C) orbits on XF;:
Where does all this come from?
Where does all this come from?
Write out the spaces. If is real, then there is a natural map
XF;
! XR;:
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 the
natural map XF; ! XR; in much the same way.
Generalizations?
If G is simple, adjoint, and classical, one may unravel the
natural map XF; ! XR; in much the same way.
Find an anologous matching of KLV polynomials for UN (G =F )
and HC (G =R)
Generalizations?
If G is simple, adjoint, and classical, one may unravel the
natural 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 map
XF; ! XR; is less well behaved.
Generalizations?
If G is simple, adjoint, and exceptional, the natural map
XF; ! XR; is less well behaved. (Two orbits can collapse to
one, for instance.)
Back to the nice case of GL(n)
Back to the nice case of GL(n)
The matching of KLV polynomials implies there are nice
relationships between character formuals for
Back to the nice case of GL(n)
The matching of KLV polynomials implies there are nice
relationships between character formuals for
1 Harish-Chandra modules for GL(n; C);
Back to the nice case of GL(n)
The matching of KLV polynomials implies there are nice
relationships between character formuals for
1 Harish-Chandra modules for GL(n; C);
2 Unipotent representations of GL(n; Qp ); and
Back to the nice case of GL(n)
The matching of KLV polynomials implies there are nice
relationships 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).
Back to the nice case of GL(n)
The matching of KLV polynomials implies there are nice
relationships 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?
Existence of Functors?
Since we have maps
1 : XF; ! XC;
2 : XF; ! XR;
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).
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.