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Bumpless Pipe Dreams and Alternating SignMatrices
Anna Weigandt
University of Michigan
September 21st, 2018
Anna Weigandt Bumpless Pipe Dreams and ASMs
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Bumpless Pipe Dreams
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Bumpless Pipe Dreams
A bumpless pipe dream is a tiling of the n × n grid with the sixtiles pictured above so that there are n pipes which
1 start at the right edge of the grid,
2 end at the bottom edge, and
3 pairwise cross at most one time.
Lam, T., Lee, S. J., and Shimozono, M. (2018). Back stable Schubert calculus. arXiv:1806.11233
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Example Not an Example
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The Permutation of a Bumpless Pipe Dream
1 2 3 4 5
1
4
3
2
5
Write Pipes(w) for the set of bumpless pipe dreams which traceout the permutation w .
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Why Bumpless?
Ordinary pipe dreams use the tiles:
Example:
2 1 4 3
1
2
3
4
Fomin, S. and Kirillov, A. N. (1996). The Yang-Baxter equation, symmetric functions, and Schubert polynomials.Discrete Mathematics, 153(1):123–143
Bergeron, N. and Billey, S. (1993). RC-graphs and Schubert polynomials. Experimental Mathematics, 2(4):257–269
Knutson, A. and Miller, E. (2005). Grobner geometry of Schubert polynomials. Annals of Mathematics, pages1245–1318
Anna Weigandt Bumpless Pipe Dreams and ASMs
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The Weight of a Bumpless Pipe Dream
If a blank tile sits in row i and column j , assign it the weight(xi − yj). The weight of a bumpless pipe dream is the product ofthe weights of its blank tiles.
y1 y2 y3 y4 y5
x1
x2
x3
x4
x5
wt(P) = (x1 − y1)(x1 − y2)(x3 − y2)
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Theorem (Lam-Lee-Shimozono, 2018)
The double Schubert polynomial Sw (x; y) is the weighted sum
Sw (x; y) =∑
P∈Pipes(w)
wt(P).
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Example: w = 2143
Sw = (x1 − y1)(x3 − y3) + (x1 − y1)(x2 − y1) + (x1 − y1)(x1 − y2).
Sw = (x1 − y1)(x3 − y1) + (x1 − y1)(x2 − y2) + (x1 − y1)(x1 − y3).
Anna Weigandt Bumpless Pipe Dreams and ASMs
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Upshot: There is not a weight preserving bijection from bumplesspipe dreams to ordinary pipe dreams for w .
Problem: Specializing the yi ’s to 0, find a weight preservingbijection between bumpless and ordinary pipe dreams for w .
7→
7→
7→
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Alternating Sign Matrices
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Alternating Sign Matrices
A matrix A is an alternating sign matrix (ASM) if:
A has entries in {−1, 0, 1}Rows and columns sum to 1
Non-zero entries alternate in sign along rows and columns0 0 1 00 1 −1 11 −1 1 00 1 0 0
Let ASM(n) be the set of n × n ASMs.
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The Rothe Diagram of an ASM
Plot the 1’s of A as black dots and the −1’s as white dots:
Write D(A) for the positions of boxes in the diagram and N(A) forthe positions of the negative entries in A.
Lascoux, A. Chern and Yang through ice
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Defining Segments are Pipes
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A Quick Sanity Check
n #ASM(n) =n−1∏k=0
(3k + 1)!
(n + k)!
∑w∈Sn
Sw (1; 0)
1 1 1
2 2 2
3 7 7
4 42 41
5 429 393
The formula which enumerates ASMs was famously conjectured byMills-Robbins-Rumsey in 1983. It was proved independently, firstby Zeilberger and then Kuperberg.
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Upshot: We can extend the definition of bumpless pipe dreams toallow multiple crossings. Say a bumpless pipe dream is reduced ifeach pair of pipes crosses at most one time.
Write BPD(n) for the set of all n × n bumpless pipe dreams.
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A Bijection Through Ice
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Theorem (W-, 2018+)
ASM(n) is in bijection with BPD(n).
Call an ASM reduced if its corresponding bumpless pipe dream isreduced.
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The Six Vertex Model
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Ice and ASMs
0 0 1 00 1 0 01 0 −1 10 0 1 0
7→ 1 7→ −1
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Six Vertices and Six Tiles
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Lascoux’s Formula for GrothendieckPolynomials
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The weight of an ASM
Write zij = xi + yj − xiyj .
Fix A ∈ ASM(n).
Assign each (i , j) ∈ D(A) the weight −zij .Assign each (i , j) ∈ N(A) the weight 1− zij .
All other entries are given the weight 1.
Define wt(A) to be the product of the weights over all entries in A.
Lascoux, A. Chern and Yang through ice
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The weight of an ASM
Example:
A =
0 0 1 00 1 0 01 0 −1 10 0 1 0
D(A) =
wt(A) = −z11z12z21(1− z33)
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Lascoux’s Formula
Theorem (Lascoux)
Fix w ∈ Sn. The double Grothendieck polynomial Gw (x; y) is aweighted sum over ASMs:
Gw (x; y) = (−1)`(w)∑
A∈asm(w)
wt(A).
Here, asm(w) is the set of ASMs whose key is w .
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Inflation on Alternating Sign Matrices
If Aij = 1, we say it is a neighbor of Ars if
1 (i , j) sits weakly northwest of (r , s),
2 (i , j) 6= (r , s), and
3 there are no other nonzero entries in A which sit weaklybetween (i , j) and (r , s).
Lascoux, A. Chern and Yang through ice
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Inflation on Alternating Sign Matrices
A −1 in A ∈ ASM(n) is removable if there are no negative entriesweakly northwest of its position in A. The region enclosed by aremovable entry and its neighbors forms a (reverse) partition shape.
7→
The key of an ASM is the permutation matrix obtained byapplying inflation iteratively until no removable entries remain.
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Example: w = 2143
Gw (x; y) = (−1)`(w)∑
A∈asm(w)
wt(A).
Gw = z11z33 +z11z21(1−z33)+z11z12(1−z33)−z11z12z21(1−z33).
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Applying Lascoux’s Formula to Schubert Polynomials
Lemma
A ∈ ASM(n) is reduced if and only if #D(A) = `(key(A)). Inparticular, if A is not reduced, then #D(A) > `(key(A)).
To get Sw (x; y) from Gw (x; y), replace each yi with −yi and thentake the lowest degree terms. This agrees with the weight onbumpless pipe dreams given in [Lam et al., 2018].
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Example: w = 2143
zij = xi + yj − xiyj 7→ (xi − yj)
Gw = z11z33 +z11z21(1−z33)+z11z12(1−z33)−z11z12z21(1−z33).
Sw = (x1 − y1)(x3 − y3) + (x1 − y1)(x2 − y1) + (x1 − y1)(x1 − y2).
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Transition and Alternating SignMatrices
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Transition for Double Grothendieck Polynomials
Take w ∈ Sn and let (r , s) be the position of the last box in thelast row of its diagram. Let v be the (unique) permutation whosediagram is obtained by removing this box.
Let i1 < i2 < · · · < ik be the list of rows of the neighbors of (r , s).
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Transition for Double Grothendieck Polynomials
Theorem (Lascoux)
Gw = Gv − (1− zrs)Gv · (1− ti1r ) · · · (1− tik r ).
Here Gv · u = Gvu.
Lascoux proved the ASM formula for Grothendieck polynomials byshowing the weight on ASMs is compatible with transition.
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Example: w = 2143
Gw = Gv − (1− z33)(Gv −Gvt23 −Gvt13 + Gvt13t23)
= z33Gv + (1− z33)(Gvt23 + Gvt13 −Gvt13t23).
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Keys from Pipes
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Resolving Multiple Crossings
Given P ∈ BPD(n), we can read off a permutation δ(P) byreplacing any crossing tiles with bumping tiles whenever two pipeshave previously crossed.
1 2 3 4 5 6 7
5
2
4
1
7
6
3
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Theorem (W-, 2018+)
If A ∈ ASM(n) corresponds to the bumpless pipe dream P thenkey(A) = δ(P).
1 2 3 4 5 6 7
5
2
4
1
7
6
3
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References I
Bergeron, N. and Billey, S. (1993). RC-graphs and Schubert polynomials.
Experimental Mathematics, 2(4):257–269.
Fomin, S. and Kirillov, A. N. (1996). The Yang-Baxter equation, symmetricfunctions, and Schubert polynomials.
Discrete Mathematics, 153(1):123–143.
Knutson, A. and Miller, E. (2005). Grobner geometry of Schubertpolynomials.
Annals of Mathematics, pages 1245–1318.
Kuperberg, G. (1996). Another proof of the alternating-sign matrixconjecture.
International Mathematics Research Notices, 1996(3):139–150.
Lam, T., Lee, S. J., and Shimozono, M. (2018). Back stable Schubertcalculus.
arXiv:1806.11233.
Lascoux, A. Chern and Yang through ice.
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References II
Mills, W. H., Robbins, D. P., and Rumsey, H. (1983). Alternating signmatrices and descending plane partitions.
Journal of Combinatorial Theory, Series A, 34(3):340–359.
Zeilberger, D. (1996). Proof of the alternating sign matrix conjecture.
Electron. J. Combin, 3(2):R13.
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Thank you!
Anna Weigandt Bumpless Pipe Dreams and ASMs