bumpless pipe dreams and alternating sign matricesweigandt/bumplessupload.pdf · bumpless pipe...

Bumpless Pipe Dreams and Alternating SignMatrices

Anna Weigandt

University of Michigan

[email protected]

September 21st, 2018

Anna Weigandt Bumpless Pipe Dreams and ASMs

Bumpless Pipe Dreams


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


Example Not an Example


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 .


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


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)


Theorem (Lam-Lee-Shimozono, 2018)

The double Schubert polynomial Sw (x; y) is the weighted sum

Sw (x; y) =∑

P∈Pipes(w)

wt(P).


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).


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→


Alternating Sign Matrices


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.


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


Defining Segments are Pipes


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.


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.


A Bijection Through Ice


Theorem (W-, 2018+)

ASM(n) is in bijection with BPD(n).

Call an ASM reduced if its corresponding bumpless pipe dream isreduced.


The Six Vertex Model


Ice and ASMs

0 0 1 00 1 0 01 0 −1 10 0 1 0

7→ 1 7→ −1


Six Vertices and Six Tiles


Lascoux’s Formula for GrothendieckPolynomials


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.



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)


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 .


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).



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.


Example: w = 2143

Gw (x; y) = (−1)`(w)∑

A∈asm(w)

wt(A).

Gw = z11z33 +z11z21(1−z33)+z11z12(1−z33)−z11z12z21(1−z33).


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].


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).


Transition and Alternating SignMatrices


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).


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.


Example: w = 2143

Gw = Gv − (1− z33)(Gv −Gvt23 −Gvt13 + Gvt13t23)

= z33Gv + (1− z33)(Gvt23 + Gvt13 −Gvt13t23).


Keys from Pipes


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


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


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.


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.


Thank you!


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