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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
Bumpless Pipe Dreams
Anna Weigandt Bumpless Pipe Dreams and ASMs
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
Anna Weigandt Bumpless Pipe Dreams and ASMs
Example Not an Example
Anna Weigandt Bumpless Pipe Dreams and ASMs
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 .
Anna Weigandt Bumpless Pipe Dreams and ASMs
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
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)
Anna Weigandt Bumpless Pipe Dreams and ASMs
Theorem (Lam-Lee-Shimozono, 2018)
The double Schubert polynomial Sw (x; y) is the weighted sum
Sw (x; y) =∑
P∈Pipes(w)
wt(P).
Anna Weigandt Bumpless Pipe Dreams and ASMs
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
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→
Anna Weigandt Bumpless Pipe Dreams and ASMs
Alternating Sign Matrices
Anna Weigandt Bumpless Pipe Dreams and ASMs
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.
Anna Weigandt Bumpless Pipe Dreams and ASMs
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
Anna Weigandt Bumpless Pipe Dreams and ASMs
Defining Segments are Pipes
Anna Weigandt Bumpless Pipe Dreams and ASMs
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.
Anna Weigandt Bumpless Pipe Dreams and ASMs
Anna Weigandt Bumpless Pipe Dreams and ASMs
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.
Anna Weigandt Bumpless Pipe Dreams and ASMs
A Bijection Through Ice
Anna Weigandt Bumpless Pipe Dreams and ASMs
Theorem (W-, 2018+)
ASM(n) is in bijection with BPD(n).
Call an ASM reduced if its corresponding bumpless pipe dream isreduced.
Anna Weigandt Bumpless Pipe Dreams and ASMs
The Six Vertex Model
Anna Weigandt Bumpless Pipe Dreams and ASMs
Ice and ASMs
0 0 1 00 1 0 01 0 −1 10 0 1 0
7→ 1 7→ −1
Anna Weigandt Bumpless Pipe Dreams and ASMs
Six Vertices and Six Tiles
Anna Weigandt Bumpless Pipe Dreams and ASMs
Lascoux’s Formula for GrothendieckPolynomials
Anna Weigandt Bumpless Pipe Dreams and ASMs
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
Anna Weigandt Bumpless Pipe Dreams and ASMs
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)
Anna Weigandt Bumpless Pipe Dreams and ASMs
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 .
Anna Weigandt Bumpless Pipe Dreams and ASMs
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
Anna Weigandt Bumpless Pipe Dreams and ASMs
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.
Anna Weigandt Bumpless Pipe Dreams and ASMs
Example: w = 2143
Gw (x; y) = (−1)`(w)∑
A∈asm(w)
wt(A).
Gw = z11z33 +z11z21(1−z33)+z11z12(1−z33)−z11z12z21(1−z33).
Anna Weigandt Bumpless Pipe Dreams and ASMs
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].
Anna Weigandt Bumpless Pipe Dreams and ASMs
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).
Anna Weigandt Bumpless Pipe Dreams and ASMs
Transition and Alternating SignMatrices
Anna Weigandt Bumpless Pipe Dreams and ASMs
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).
Anna Weigandt Bumpless Pipe Dreams and ASMs
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.
Anna Weigandt Bumpless Pipe Dreams and ASMs
Example: w = 2143
Gw = Gv − (1− z33)(Gv −Gvt23 −Gvt13 + Gvt13t23)
= z33Gv + (1− z33)(Gvt23 + Gvt13 −Gvt13t23).
Anna Weigandt Bumpless Pipe Dreams and ASMs
Keys from Pipes
Anna Weigandt Bumpless Pipe Dreams and ASMs
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
Anna Weigandt Bumpless Pipe Dreams and ASMs
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
Anna Weigandt Bumpless Pipe Dreams and ASMs
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.
Anna Weigandt Bumpless Pipe Dreams and ASMs
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.
Anna Weigandt Bumpless Pipe Dreams and ASMs
Thank you!
Anna Weigandt Bumpless Pipe Dreams and ASMs