developments in alternating sign matricesbressoud/talks/2010/rutgers-1.pdf · developments in...
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Developments in Alternating Sign
Matrices David Bressoud Macalester College St. Paul, MN Rutgers University
New Brunswick, NJ October 14, 2010
PowerPointavailableatwww.macalester.edu/~bressoud/talks
MAA
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1. Developments leading to Doron’s proof of the refined alternating sign matrix conjecture.
2. Counting symmetry classes of ASMs
3. Fully packed loop configurations
4. Fischer’s proof of the refined alternating sign matrix conjecture
5. Fruits of an undergraduate’s exploration.
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DavidRobbins(1942–2003)
Alternating Sign Matrices
Kuperberg’s representation
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Monotone Triangle Strict increase across rows, weak increase along diagonals
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3 2 2 4 3 2 5 4 3 2
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x2 4
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14
x x x 1 4 5
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42 105 135 105 42
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14 7 23 14 26 23 7 14 7 123 124 125 134 135 145 234 235 245 345
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An,k = # of n×n alternating sign matrices with 1 in row n, column k.
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22/3 33/22
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Conjecture:
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Conjecture1:
Conjecture2(corollaryofConjecture1):
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1992: George Andrews proves Robbins conjecture that the number of totally symmetric, self-complementary plane partitions in an n×n×n box is given by
Z, Proof of the Alternating Sign Matrix Conjecture, Elect. J. of Combin., 1996.
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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 had been studying ASM’s for decades, only they called them the six-vertex model.
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Horizontal = 1
Vertical = –1
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southwest
northeast
northwest
southeast
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N = # of vertical vertices (the number of –1’s)
I = inversion number = N + # of SW
N = 2, I = 5
(x2, y3)
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Anatoli Izergin Vladimir Korepin
SUNY Stony Brook
1980’s
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Proof: LHS is symmetric polynomial in x’s and in y’s
Degree n – 1 in x1
By induction, LHS = RHS when x1 = y1
Sufficient to show that RHS is symmetric polynomial in y’s
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Rodney J. Baxter
Australian National University
Proof: LHS is symmetric polynomial in x’s and in y’s
Degree n – 1 in x1
By induction, LHS = RHS when x1 = y1
Sufficient to show that RHS is symmetric polynomial in x’s and in y’s — This follows from Baxter’s triangle-to-triangle relation.
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1996
Doron used this determinant to prove the original conjecture
“Proof of the refined alternating sign matrix conjecture,” New York Journal of Mathematics
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Proofs and Confirmations: The Story of the Alternating Sign Matrix Conjecture
Cambridge University Press & MAA, 1999
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2001, Kuperberg uses the power of the triangle-to-triangle relation to prove some of formulas conjectured by Robbins:
Vertically symmetric ASMs
1800 rotationally symmetric ASMs
900 rotationally symmetric ASMs
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Kuperberg, 2001: proved formulas for counting some new six-vertex models:
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2004, Soichi Okada observes that the number of n × n ASM’s is 3–n(n–1)/2 times the dimension of the irreducible representation of GL2n indexed by
A2n−1+ = ei − ej 1 ≤ i < j ≤ 2n{ }ρ = n − 12,n −
32,…,−n + 12( )
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Okada proves the formula for vertically and horizontally symmetric ASMs:
Okada also proves the comparable formula for AVH (4n–1)
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2006, Razumov and Stroganov prove the conjectured formula for quarter-turn symmetric ASMs of odd order:
There is one remaining unproven conjecture: The number of ASMs of odd order that are symmetric across both diagonals is
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Fully packed loop configuration
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Fully packed loop configuration
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Each ASM defines a pairing of the vertices, π.
Ben Wieland, 2000, proved that the number of ASMs that correspond to a given pairing is unchanged if we rotate the labels.
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Wieland’s theorem is a special case of the Razumov-Stroganoff conjecture:
The number of ASMs associated to a given pairing is a “component of a ground-state wavefunction.”
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The Razumov-Stroganov conjecture was proven this past spring by physicists Luigi Cantini and Andrea Sportiello.
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12 13 14 15 23 24 25 34 35 45
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2006, Ilse Fischer observes that the number of monotone triangles with bottom row k1, k2 is k2 – k1 + 1.
The number with bottom row k1, k2, k3 is
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A(n;k1,…,kn) = the number of monotone triangles with bottom row k1,…,kn.
A(n;k1,…,kn) is a polynomial of total degree n(n–1)/2 and degree n–1 in each of the ki.
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2007, Fischer provides a new proof of the refined ASM conjecture (the value of An,i).
She uses her characterization of A(n;k1,…,kn) to prove that
(An,1,…, An,n) is an eigenvector with eigenvalue 1. Verify that the dimension of the eigenspace is at most one and that the conjectured values satisfy this identify. Check constant.
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2008, Carleton undergraduate Nathan Williams looks at the alternating sums of the rows.
1
0
1
0
9
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1: 1 3: 1 5: 9 7: 676 9: 417 316 11: 21054 33225 13: 8657 65116 22500 15: 28972 58363 89801 95600 17: 7 88319 29114 31396 91797 40176 19: 1742 93613 18275 76565 60875 92718 01924
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AV(2n+1)2 is the number of ASMs of odd order with no 0’s on the central column, ACC(2n+1).
Williams asks: Is there a bijective proof ?
PowerPointavailableatwww.macalester.edu/~bressoud/talks