Proofs

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Proofs Confirmations The story of the
alternating sign matrix conjecture
David M. Bressoud Macalester College
New Jersey Section, MAA November 6, 2005 Monmouth
University
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Bill Mills
IDA/CCR
1 0 0 1 1 0 1 0 1
Howard Rumsey
Dave Robbins
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Charles L. Dodgson aka Lewis Carroll
1 0 0 1 1 0 1 0 1
Condensation of Determinants, Proceedings of
the Royal Society, London 1866
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How many n ? n alternating sign matrices?
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very suspicious
2 ? 3 ? 7
3 ? 11 ? 13
22 ? 11 ? 132
22 ? 132 ? 17 ? 19
23 ? 13 ? 172 ? 192
22 ? 5 ? 172 ? 193 ? 23
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There is exactly one 1 in the first row
1 0 0 1 1 0 1 0 1
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There is exactly one 1 in the first row
1 0 0 1 1 0 1 0 1
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1 1 1 2 3 2 7 14
14 7 42 105 135
105 42 429 1287 2002
2002 1287 429
1 0 0 1 1 0 1 0 1
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1 1 1 2 3 2 7 14
14 7 42 105 135
105 42 429 1287 2002
2002 1287 429
1 0 0 1 1 0 1 0 1



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1 1 1 2 3 2 7 14
14 7 42 105 135
105 42 429 1287 2002
2002 1287 429
1 0 0 1 1 0 1 0 1



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1 1 2/2 1 2 2/3 3 3/2 2 7 2/4 14
14 4/2 7 42 2/5 105 135 105
5/2 42 429 2/6 1287 2002 2002
1287 6/2 429
1 0 0 1 1 0 1 0 1
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1 1 2/2 1 2 2/3 3 3/2 2 7 2/4 14 5/5
14 4/2 7 42 2/5 105 7/9 135 9/7 105 5/2
42 429 2/6 1287 9/14 2002 16/16 2002 14/9 1287
6/2 429
1 0 0 1 1 0 1 0 1
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2/2 2/3 3/2 2/4 5/5 4/2 2/5
7/9 9/7 5/2 2/6 9/14 16/16
14/9 6/2
1 0 0 1 1 0 1 0 1
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Numerators
11 11 12 11 23 13 11
34 36 14 11 45 610 410
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1 0 0 1 1 0 1 0 1
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11 11 12 11 23 13 11
34 36 14 11 45 610 410
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Numerators
1 0 0 1 1 0 1 0 1
Conjecture 1
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Conjecture 1
1 0 0 1 1 0 1 0 1
Conjecture 2 (corollary of Conjecture 1)
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1 0 0 1 1 0 1 0 1
Richard Stanley
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1 0 0 1 1 0 1 0 1
Richard Stanley
Andrews Theorem the number of descending plane
partitions of size n is
George Andrews
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All you have to do is find a 1-to-1
correspondence between n by n alternating sign
matrices and descending plane partitions of size
n, and conjecture 2 will be proven!
1 0 0 1 1 0 1 0 1
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All you have to do is find a 1-to-1
correspondence between n by n alternating sign
matrices and descending plane partitions of size
n, and conjecture 2 will be proven!
1 0 0 1 1 0 1 0 1
What is a descending plane partition?
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Percy A. MacMahon
Plane Partition
1 0 0 1 1 0 1 0 1
Work begun in 1897
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Plane partition of 75
6 5 5 4 3 3
1 0 0 1 1 0 1 0 1
of pps of 75 pp(75)
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Plane partition of 75
6 5 5 4 3 3
1 0 0 1 1 0 1 0 1
of pps of 75 pp(75) 37,745,732,428,153
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Generating function
1 0 0 1 1 0 1 0 1
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1912 MacMahon proves that the generating function
for plane partitions in an n ? n ? n box is
1 0 0 1 1 0 1 0 1
At the same time, he conjectures that the
generating function for symmetric plane
partitions is
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Symmetric Plane Partition
4 3 2 1 1 3 2 2 1 2 2 1 1 1 1
1 0 0 1 1 0 1 0 1
The reader must be warned that, although there
is little doubt that this result is correct,
the result has not been rigorously established.
Further investigations in regard to these matters
would be sure to lead to valuable work. (1916)
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1971 Basil Gordon proves case for n infinity
1 0 0 1 1 0 1 0 1
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1971 Basil Gordon proves case for n infinity
1 0 0 1 1 0 1 0 1
1977 George Andrews and Ian Macdonald
independently prove general case
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1912 MacMahon proves that the generating function
for plane partitions in an n ? n ? n box is
1 0 0 1 1 0 1 0 1
At the same time, he conjectures that the
generating function for symmetric plane
partitions is
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Macdonalds observation both generating
functions are special cases of the following
1 0 0 1 1 0 1 0 1
where G is a group acting on the points in B and
B/G is the set of orbits. If G consists of only
the identity, this gives all plane partitions in
B. If G is the identity and (i,j,k) ? (j,i,k),
then get generating function for symmetric plane
partitions.
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Does this work for other groups of symmetries?
1 0 0 1 1 0 1 0 1
G S3 ?
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Does this work for other groups of symmetries?
1 0 0 1 1 0 1 0 1
G S3 ? No
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Does this work for other groups of symmetries?
1 0 0 1 1 0 1 0 1
G S3 ? No
G C3 ? (i,j,k) ? (j,k,i) ? (k,i,j)
It seems to work.
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Cyclically Symmetric Plane Partition
1 0 0 1 1 0 1 0 1
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Macdonalds Conjecture (1979) The generating
function for cyclically symmetric plane
partitions in B(n,n,n) is
1 0 0 1 1 0 1 0 1
If I had to single out the most interesting open
problem in all of enumerative combinatorics, this
would be it. Richard Stanley, review of
Symmetric Functions and Hall Polynomials,
Bulletin of the AMS, March, 1981.
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1979, Andrews counts cyclically symmetric plane
partitions
1 0 0 1 1 0 1 0 1
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1979, Andrews counts cyclically symmetric plane
partitions
1 0 0 1 1 0 1 0 1
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1979, Andrews counts cyclically symmetric plane
partitions
1 0 0 1 1 0 1 0 1
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1979, Andrews counts cyclically symmetric plane
partitions
1 0 0 1 1 0 1 0 1
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1979, Andrews counts cyclically symmetric plane
partitions
1 0 0 1 1 0 1 0 1
L1 W1 gt L2 W2 gt L3 W3 gt
width
length
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1979, Andrews counts descending plane partitions
1 0 0 1 1 0 1 0 1
L1 gt W1 L2 gt W2 L3 gt W3
6 6 6 4 3 3 3 2
width
length
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6 X 6 ASM ? DPP with largest part 6
What are the corresponding 6 subsets of DPPs?
1 0 0 1 1 0 1 0 1
6 6 6 4 3 3 3 2
width
length
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ASM with 1 at top of first column ?DPP with no
parts of size n.
ASM with 1 at top of last column ?DPP with n1
parts of size n.
1 0 0 1 1 0 1 0 1
6 6 6 4 3 3 3 2
width
length
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Mills, Robbins, Rumsey Conjecture of n ? n
ASMs with 1 at top of column j equals of DPPs
n with exactly j1 parts of size n.
1 0 0 1 1 0 1 0 1
6 6 6 4 3 3 3 2
width
length
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Mills, Robbins, Rumsey proved that of DPPs
n with j parts of size n was given by their
conjectured formula for ASMs.
1 0 0 1 1 0 1 0 1
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Mills, Robbins, Rumsey proved that of DPPs
n with j parts of size n was given by their
conjectured formula for ASMs.
1 0 0 1 1 0 1 0 1
Discovered an easier proof of Andrews formula,
using induction on j and n.
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Mills, Robbins, Rumsey proved that of DPPs
n with j parts of size n was given by their
conjectured formula for ASMs.
1 0 0 1 1 0 1 0 1
Discovered an easier proof of Andrews formula,
using induction on j and n.
Used this inductive argument to prove Macdonalds
conjecture Proof of the Macdonald Conjecture,
Inv. Math., 1982
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Mills, Robbins, Rumsey proved that of DPPs
n with j parts of size n was given by their
conjectured formula for ASMs.
1 0 0 1 1 0 1 0 1
Discovered an easier proof of Andrews formula,
using induction on j and n.
Used this inductive argument to prove Macdonalds
conjecture Proof of the Macdonald Conjecture,
Inv. Math., 1982
But they still didnt have a proof of their
conjecture!
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Totally Symmetric Self-Complementary Plane
Partitions
1 0 0 1 1 0 1 0 1
1983
David Robbins (19422003)
Vertical flip of ASM ? complement of DPP ?
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Totally Symmetric Self-Complementary Plane
Partitions
1 0 0 1 1 0 1 0 1
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1 0 0 1 1 0 1 0 1
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Robbins Conjecture The number of TSSCPPs in a
2n X 2n X 2n box is
1 0 0 1 1 0 1 0 1
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Robbins Conjecture The number of TSSCPPs in a
2n X 2n X 2n box is
1 0 0 1 1 0 1 0 1
1989 William Doran shows equivalent to counting
lattice paths 1990 John Stembridge represents
the counting function as a Pfaffian (built on
insights of Gordon and Okada) 1992 George
Andrews evaluates the Pfaffian, proves Robbins
Conjecture
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December, 1992 Doron Zeilberger announces a proof
that of ASMs of size n equals of TSSCPPs in
box of size 2n.
1 0 0 1 1 0 1 0 1
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December, 1992 Doron Zeilberger announces a proof
that of ASMs of size n equals of TSSCPPs in
box of size 2n.
1 0 0 1 1 0 1 0 1
1995 all gaps removed, published as Proof of the
Alternating Sign Matrix Conjecture, Elect. J. of
Combinatorics, 1996.
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Zeilbergers proof is an 84-page tour de force,
but it still left open the original conjecture
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1996 Kuperberg announces a simple proof
Another proof of the alternating sign matrix
conjecture, International Mathematics Research
Notices
Greg Kuperberg UC Davis
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1996 Kuperberg announces a simple proof
Another proof of the alternating sign matrix
conjecture, International Mathematics Research
Notices
Greg Kuperberg UC Davis
Physicists have been studying ASMs for decades,
only they call them square ice (aka the
six-vertex model ).
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H O H O H O H O H O H H H
H H H H O H O H O H O H O
H H H H H H H O H
O H O H O H O H H H H
H H H O H O H O H O H O H H
H H H H H O H O H O
H O H O H
1 0 0 1 1 0 1 0 1
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1 0 0 1 1 0 1 0 1
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Horizontal ? 1
1 0 0 1 1 0 1 0 1
Vertical ? 1
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southwest
1 0 0 1 1 0 1 0 1
northwest
southeast
northeast
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1 0 0 1 1 0 1 0 1
N of vertical I inversion number
N of NE
x2, y3
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1960s
Rodney Baxters Triangle-to-triangle relation
1 0 0 1 1 0 1 0 1
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1960s
Rodney Baxters Triangle-to-triangle relation
1 0 0 1 1 0 1 0 1
1980s
Vladimir Korepin
Anatoli Izergin
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1996 Doron Zeilberger uses this determinant to
prove the original conjecture
1 0 0 1 1 0 1 0 1
Proof of the refined alternating sign matrix
conjecture, New York Journal of Mathematics
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The End
1 0 0 1 1 0 1 0 1
(which is really just the beginning)
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The End
1 0 0 1 1 0 1 0 1
(which is really just the beginning)
This Power Point presentation can be downloaded
from www.macalester.edu/bressoud/talks