Symmetries of stochastic colored vertex models Pavel Galashin (UCLA) - - PowerPoint PPT Presentation

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Symmetries of stochastic colored vertex models Pavel Galashin (UCLA) - - PowerPoint PPT Presentation

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Symmetries of stochastic colored vertex models Pavel Galashin (UCLA) Dimers in Combinatorics and Cluster Algebras 2020 arXiv:2003.06330 Q 11 Q 10 Q 9 Q 8 Q 11 Q 10 Q 9 Q 8 Q 11 Q 10 Q 9 Q 8 Q 11 Q 10 Q 9 Q 8 y 7 y 1 P 11 Q 7 P 11 Q 7 P 11 Q 7 P

slide-1
SLIDE 1

Symmetries of stochastic colored vertex models

Pavel Galashin (UCLA) Dimers in Combinatorics and Cluster Algebras 2020 arXiv:2003.06330

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 y1 y2 y3 y4 y5 y6 y7

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 y7 y6 y5 y4 y3 y2 y1

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Hπ = H Vπ = V

= Z H,V(x, y) Z 180◦(H),V(x, rev(y))

Hπ′ = 180◦(H) Vπ′ = V

slide-2
SLIDE 2

Stochastic colored six-vertex model

  • Introduced in 2016:

[KMMO16] A. Kuniba, V. V. Mangazeev, S. Maruyama, and M. Okado. Stochastic R matrix for Uq(A(1)

n ). Nuclear Phys. B, 913:248–277, 2016.

slide-3
SLIDE 3

Stochastic colored six-vertex model

  • Introduced in 2016:

[KMMO16] A. Kuniba, V. V. Mangazeev, S. Maruyama, and M. Okado. Stochastic R matrix for Uq(A(1)

n ). Nuclear Phys. B, 913:248–277, 2016.

  • Limiting cases include many other interesting probabilistic models
slide-4
SLIDE 4

Stochastic colored six-vertex model

  • Introduced in 2016:

[KMMO16] A. Kuniba, V. V. Mangazeev, S. Maruyama, and M. Okado. Stochastic R matrix for Uq(A(1)

n ). Nuclear Phys. B, 913:248–277, 2016.

  • Limiting cases include many other interesting probabilistic models
slide-5
SLIDE 5

Stochastic colored six-vertex model

  • Introduced in 2016:

[KMMO16] A. Kuniba, V. V. Mangazeev, S. Maruyama, and M. Okado. Stochastic R matrix for Uq(A(1)

n ). Nuclear Phys. B, 913:248–277, 2016.

  • Limiting cases include many other interesting probabilistic models
slide-6
SLIDE 6

n lattice paths of colors 1, 2, . . . , n move up/right on Z2

4 3 2 1 5 6 7 8 9 10 n = 11

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SLIDE 7

n lattice paths of colors 1, 2, . . . , n move up/right on Z2 When two paths of colors c1 < c2 enter a square from the bottom/left, they form either a crossing or an elbow

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

4 3 2 1 5 6 7 8 9 10 n = 11

slide-8
SLIDE 8

n lattice paths of colors 1, 2, . . . , n move up/right on Z2 When two paths of colors c1 < c2 enter a square from the bottom/left, they form either a crossing or an elbow

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

4 3 2 1 5 6 7 8 9 10 n = 11

slide-9
SLIDE 9

n lattice paths of colors 1, 2, . . . , n move up/right on Z2 When two paths of colors c1 < c2 enter a square from the bottom/left, they form either a crossing or an elbow

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

4 3 2 1 5 6 7 8 9 10 n = 11

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SLIDE 10

n lattice paths of colors 1, 2, . . . , n move up/right on Z2 When two paths of colors c1 < c2 enter a square from the bottom/left, they form either a crossing or an elbow

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

4 3 2 1 5 6 7 8 9 10 n = 11

slide-11
SLIDE 11

n lattice paths of colors 1, 2, . . . , n move up/right on Z2 When two paths of colors c1 < c2 enter a square from the bottom/left, they form either a crossing or an elbow

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

4 3 2 1 5 6 7 8 9 10 n = 11

slide-12
SLIDE 12

n lattice paths of colors 1, 2, . . . , n move up/right on Z2 When two paths of colors c1 < c2 enter a square from the bottom/left, they form either a crossing or an elbow

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

4 3 2 1 5 6 7 8 9 10 n = 11

slide-13
SLIDE 13

n lattice paths of colors 1, 2, . . . , n move up/right on Z2 When two paths of colors c1 < c2 enter a square from the bottom/left, they form either a crossing or an elbow

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp 0 < q < 1 is fixed

4 3 2 1 5 6 7 8 9 10 n = 11

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SLIDE 14

n lattice paths of colors 1, 2, . . . , n move up/right on Z2 When two paths of colors c1 < c2 enter a square from the bottom/left, they form either a crossing or an elbow

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp 0 < q < 1 is fixed spectral parameter p depends on the square

4 3 2 1 5 6 7 8 9 10 n = 11

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SLIDE 15

n lattice paths of colors 1, 2, . . . , n move up/right on Z2 When two paths of colors c1 < c2 enter a square from the bottom/left, they form either a crossing or an elbow

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp 0 < q < 1 is fixed spectral parameter p depends on the square pi,j = yj − xi yj − qxi

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 y1 y2 y3 y4 y5 y6 y7

slide-16
SLIDE 16

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp pi,j =

yj−xi yj−qxi

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 y1 y2 y3 y4 y5 y6 y7

slide-17
SLIDE 17

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp pi,j =

yj−xi yj−qxi

2MN pipe dreams − → n! permutations

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 y1 y2 y3 y4 y5 y6 y7

π =   1 2 3 4 5 6 7 8 9 10 11 ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ 2 1 3 10 8 6 4 11 5 7 9  

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SLIDE 18

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp pi,j =

yj−xi yj−qxi

2MN pipe dreams − → n! permutations For each π ∈ Sn, let Hπ and Vπ record the endpoints of all “horizontal” and “vertical” pipes

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 y1 y2 y3 y4 y5 y6 y7

π =   1 2 3 4 5 6 7 8 9 10 11 ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ 2 1 3 10 8 6 4 11 5 7 9  

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SLIDE 19

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp pi,j =

yj−xi yj−qxi

2MN pipe dreams − → n! permutations For each π ∈ Sn, let Hπ and Vπ record the endpoints of all “horizontal” and “vertical” pipes

x1 x2 x3 x4 y1 y2 y3 y4 y5 y6 y7

P1 P2 P3 P4 P5 P8 P6 P7 P9 P10 P6 P7 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

π =   1 2 3 4 5 6 7 8 9 10 11 ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ 2 1 3 10 8 6 4 11 5 7 9  

Hπ = {(6, 6), (7, 4), (9, 5), (10, 7)}

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SLIDE 20

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp pi,j =

yj−xi yj−qxi

2MN pipe dreams − → n! permutations For each π ∈ Sn, let Hπ and Vπ record the endpoints of all “horizontal” and “vertical” pipes

x1 x2 x3 x4 y1 y2 y3 y4 y5 y6 y7

P1 P2 P3 P4 P4 P5 P8 P6 P7 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q10 Q11

π =   1 2 3 4 5 6 7 8 9 10 11 ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ 2 1 3 10 8 6 4 11 5 7 9  

Hπ = {(6, 6), (7, 4), (9, 5), (10, 7)} Vπ = {(4, 10)}

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SLIDE 21

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp pi,j =

yj−xi yj−qxi

2MN pipe dreams − → n! permutations For each π ∈ Sn, let Hπ and Vπ record the endpoints of all “horizontal” and “vertical” pipes Given H, V, let Z H,V(x, y) =probability of observing π ∈ Sn with Hπ = H and Vπ = V.

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 y1 y2 y3 y4 y5 y6 y7

π =   1 2 3 4 5 6 7 8 9 10 11 ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ 2 1 3 10 8 6 4 11 5 7 9  

Hπ = {(6, 6), (7, 4), (9, 5), (10, 7)} Vπ = {(4, 10)}

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SLIDE 22

Flip theorem (G., 2020)

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 y1 y2 y3 y4 y5 y6 y7

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 y7 y6 y5 y4 y3 y2 y1

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

Hπ = H Vπ = V

= Z H,V(x, y) Z 180◦(H),V(x, rev(y))

Hπ′ = 180◦(H) Vπ′ = V

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SLIDE 23

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

pi,j =

yj−xi yj−qxi

Flip theorem: Z H,V(x, y) = Z 180◦(H),V(x, rev(y))

slide-24
SLIDE 24

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

pi,j =

yj−xi yj−qxi

Flip theorem: Z H,V(x, y) = Z 180◦(H),V(x, rev(y))

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y1 y2 y3

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y1 y2 y3

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y3 y2 y1

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y3 y2 y1

p1,3 p2,3 1 − p1,2 1 − qp2,2 p1,1 1 − p2,1 p1,3 p2,3 1 − p1,2 p2,2 1 − p1,1 1 − p2,1 1 − p1,1 1 − qp2,1 p1,2 1 − p2,2 p1,3 p2,3 1 − p1,1 p2,1 1 − p1,2 1 − p2,2 p1,3 p2,3

slide-25
SLIDE 25

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

pi,j =

yj−xi yj−qxi

Flip theorem: Z H,V(x, y) = Z 180◦(H),V(x, rev(y))

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y1 y2 y3

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y1 y2 y3

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y3 y2 y1

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y3 y2 y1

p1,3 p2,3 1 − p1,2 1 − qp2,2 p1,1 1 − p2,1 p1,3 p2,3 1 − p1,2 p2,2 1 − p1,1 1 − p2,1 1 − p1,1 1 − qp2,1 p1,2 1 − p2,2 p1,3 p2,3 1 − p1,1 p2,1 1 − p1,2 1 − p2,2 p1,3 p2,3

All 4 probabilities are different!

slide-26
SLIDE 26

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

pi,j =

yj−xi yj−qxi

Flip theorem: Z H,V(x, y) = Z 180◦(H),V(x, rev(y))

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y1 y2 y3

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y1 y2 y3

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y3 y2 y1

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y3 y2 y1

p1,3 p2,3 1 − p1,2 1 − qp2,2 p1,1 1 − p2,1 p1,3 p2,3 1 − p1,2 p2,2 1 − p1,1 1 − p2,1 1 − p1,1 1 − qp2,1 p1,2 1 − p2,2 p1,3 p2,3 1 − p1,1 p2,1 1 − p1,2 1 − p2,2 p1,3 p2,3

All 4 probabilities are different! = ⇒ there is no weight-preserving bijection

slide-27
SLIDE 27

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

pi,j =

yj−xi yj−qxi

Flip theorem: Z H,V(x, y) = Z 180◦(H),V(x, rev(y))

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y1 y2 y3

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y1 y2 y3

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y3 y2 y1

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y3 y2 y1

p1,3 p2,3 1 − p1,2 1 − qp2,2 p1,1 1 − p2,1 p1,3 p2,3 1 − p1,2 p2,2 1 − p1,1 1 − p2,1 1 − p1,1 1 − qp2,1 p1,2 1 − p2,2 p1,3 p2,3 1 − p1,1 p2,1 1 − p1,2 1 − p2,2 p1,3 p2,3

All 4 probabilities are different! = ⇒ there is no weight-preserving bijection Sum of the first two = sum of the second two

slide-28
SLIDE 28

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

pi,j =

yj−xi yj−qxi

Flip theorem: Z H,V(x, y) = Z 180◦(H),V(x, rev(y))

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y1 y2 y3

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y1 y2 y3

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y3 y2 y1

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y3 y2 y1

p1,3 p2,3 1 − p1,2 1 − qp2,2 p1,1 1 − p2,1 p1,3 p2,3 1 − p1,2 p2,2 1 − p1,1 1 − p2,1 1 − p1,1 1 − qp2,1 p1,2 1 − p2,2 p1,3 p2,3 1 − p1,1 p2,1 1 − p1,2 1 − p2,2 p1,3 p2,3

All 4 probabilities are different! = ⇒ there is no weight-preserving bijection Sum of the first two = sum of the second two

Corollary

The number of pipe dreams for Z H,V equals the number of pipe dreams for Z 180◦(H),V.

slide-29
SLIDE 29

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

pi,j =

yj−xi yj−qxi

Flip theorem: Z H,V(x, y) = Z 180◦(H),V(x, rev(y))

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y1 y2 y3

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y1 y2 y3

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y3 y2 y1

P1 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5

x1 x2 y3 y2 y1

p1,3 p2,3 1 − p1,2 1 − qp2,2 p1,1 1 − p2,1 p1,3 p2,3 1 − p1,2 p2,2 1 − p1,1 1 − p2,1 1 − p1,1 1 − qp2,1 p1,2 1 − p2,2 p1,3 p2,3 1 − p1,1 p2,1 1 − p1,2 1 − p2,2 p1,3 p2,3

All 4 probabilities are different! = ⇒ there is no weight-preserving bijection Sum of the first two = sum of the second two

Corollary

The number of pipe dreams for Z H,V equals the number of pipe dreams for Z 180◦(H),V.

Problem

Find a bijective proof.

slide-30
SLIDE 30

Generalized flip theorem

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 x5 y1 y2 y3 y4 y5 y6

slide-31
SLIDE 31

Generalized flip theorem

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 x5 y1 y2 y3 y4 y5 y6

C

Cut C

slide-32
SLIDE 32

Generalized flip theorem

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 x5 y1 y2 y3 y4 y5 y6

C Ht(C; x, y) =

Cut C → random variable Ht(C; x, y)

slide-33
SLIDE 33

Generalized flip theorem

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 x5 y1 y2 y3 y4 y5 y6

C Ht(C; x, y) = 2

Cut C → random variable Ht(C; x, y) Ht(C; x, y) measures the number of pipes that “cross” C from left to right

slide-34
SLIDE 34

Generalized flip theorem

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 x5 y1 y2 y3 y4 y5 y6

C Ht(C; x, y) = 2

Cut C → random variable Ht(C; x, y) Ht(C; x, y) measures the number of pipes that “cross” C from left to right Fact: distribution of Ht(C; x, y) is a symmetric function in suppH(C; x) := {xℓ, xℓ+1, . . . , xr} and suppV (C; y) := {yd, yd+1, . . . , yu}

slide-35
SLIDE 35

Generalized flip theorem

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 x5 y1 y2 y3 y4 y5 y6 xℓ xr yd yu

C Ht(C; x, y) = 2

Cut C → random variable Ht(C; x, y) Ht(C; x, y) measures the number of pipes that “cross” C from left to right Fact: distribution of Ht(C; x, y) is a symmetric function in suppH(C; x) := {xℓ, xℓ+1, . . . , xr} and suppV (C; y) := {yd, yd+1, . . . , yu}

slide-36
SLIDE 36

Theorem 2 (G., 2020)

slide-37
SLIDE 37

Theorem 2 (G., 2020)

Given any cuts C1, . . . , Cm and C ′

1, . . . , C ′ m, we have

slide-38
SLIDE 38

Theorem 2 (G., 2020)

Given any cuts C1, . . . , Cm and C ′

1, . . . , C ′ m, we have

suppH(Ci; x) = suppH(C ′

i ; x′)

and suppV (Ci; y) = suppV (C ′

i ; y′)

for all i

slide-39
SLIDE 39

Theorem 2 (G., 2020)

Given any cuts C1, . . . , Cm and C ′

1, . . . , C ′ m, we have

suppH(Ci; x) = suppH(C ′

i ; x′)

and suppV (Ci; y) = suppV (C ′

i ; y′)

for all i if and only if

slide-40
SLIDE 40

Theorem 2 (G., 2020)

Given any cuts C1, . . . , Cm and C ′

1, . . . , C ′ m, we have

suppH(Ci; x) = suppH(C ′

i ; x′)

and suppV (Ci; y) = suppV (C ′

i ; y′)

for all i if and only if

  • Ht(C1; x, y), . . . , Ht(Cm; x, y)

d =

  • Ht(C ′

1; x′, y′), . . . , Ht(C ′ m; x′, y′)

  • .
slide-41
SLIDE 41

Theorem 2 (G., 2020)

In other words, if a transformation preserves individual distributions of Ht(Ci; x, y)-s then it preserves their joint distribution. Given any cuts C1, . . . , Cm and C ′

1, . . . , C ′ m, we have

suppH(Ci; x) = suppH(C ′

i ; x′)

and suppV (Ci; y) = suppV (C ′

i ; y′)

for all i if and only if

  • Ht(C1; x, y), . . . , Ht(Cm; x, y)

d =

  • Ht(C ′

1; x′, y′), . . . , Ht(C ′ m; x′, y′)

  • .
slide-42
SLIDE 42

Theorem 2 (G., 2020)

In other words, if a transformation preserves individual distributions of Ht(Ci; x, y)-s then it preserves their joint distribution. Unexpected behavior – these random variables are far from independent! Given any cuts C1, . . . , Cm and C ′

1, . . . , C ′ m, we have

suppH(Ci; x) = suppH(C ′

i ; x′)

and suppV (Ci; y) = suppV (C ′

i ; y′)

for all i if and only if

  • Ht(C1; x, y), . . . , Ht(Cm; x, y)

d =

  • Ht(C ′

1; x′, y′), . . . , Ht(C ′ m; x′, y′)

  • .
slide-43
SLIDE 43

Theorem 2 (G., 2020)

Given any cuts C1, . . . , Cm and C ′

1, . . . , C ′ m, we have

suppH(Ci; x) = suppH(C ′

i ; x′)

and suppV (Ci; y) = suppV (C ′

i ; y′)

for all i if and only if

  • Ht(C1; x, y), . . . , Ht(Cm; x, y)

d =

  • Ht(C ′

1; x′, y′), . . . , Ht(C ′ m; x′, y′)

  • .
slide-44
SLIDE 44

Theorem 2 (G., 2020)

slide-45
SLIDE 45

Theorem 1

Z H,V(x, y) = Z 180◦(H),V(x, rev(y))

Theorem 2

suppH(Ci; x) = suppH(C ′

i ; x′)

suppV (Ci; y) = suppV (C ′

i ; y′)

if and only if

  • Ht(C1; x, y), . . . , Ht(Cm; x, y)
  • d

=

  • Ht(C ′

1; x′, y′), . . . , Ht(C ′ m; x′, y′)

slide-46
SLIDE 46

Theorem 1

Z H,V(x, y) = Z 180◦(H),V(x, rev(y))

Theorem 2

suppH(Ci; x) = suppH(C ′

i ; x′)

suppV (Ci; y) = suppV (C ′

i ; y′)

if and only if

  • Ht(C1; x, y), . . . , Ht(Cm; x, y)
  • d

=

  • Ht(C ′

1; x′, y′), . . . , Ht(C ′ m; x′, y′)

  • Theorem 2 =

⇒ Theorem 1

slide-47
SLIDE 47

Theorem 1

Z H,V(x, y) = Z 180◦(H),V(x, rev(y))

Theorem 2

suppH(Ci; x) = suppH(C ′

i ; x′)

suppV (Ci; y) = suppV (C ′

i ; y′)

if and only if

  • Ht(C1; x, y), . . . , Ht(Cm; x, y)
  • d

=

  • Ht(C ′

1; x′, y′), . . . , Ht(C ′ m; x′, y′)

  • Theorem 2 =

⇒ Theorem 1 Theorem 2 generalizes the results and confirms a conjecture of [BGW19] Alexei Borodin, Vadim Gorin, and Michael

  • Wheeler. Shift-invariance for vertex models and
  • polymers. arXiv:1912.02957, 2019.
slide-48
SLIDE 48

Theorem 1

Z H,V(x, y) = Z 180◦(H),V(x, rev(y))

Theorem 2

suppH(Ci; x) = suppH(C ′

i ; x′)

suppV (Ci; y) = suppV (C ′

i ; y′)

if and only if

  • Ht(C1; x, y), . . . , Ht(Cm; x, y)
  • d

=

  • Ht(C ′

1; x′, y′), . . . , Ht(C ′ m; x′, y′)

  • Theorem 2 =

⇒ Theorem 1 Theorem 2 generalizes the results and confirms a conjecture of [BGW19] Alexei Borodin, Vadim Gorin, and Michael

  • Wheeler. Shift-invariance for vertex models and
  • polymers. arXiv:1912.02957, 2019.

Shift of [BGW19] = double flip

slide-49
SLIDE 49

Theorem 1

Z H,V(x, y) = Z 180◦(H),V(x, rev(y))

Theorem 2

suppH(Ci; x) = suppH(C ′

i ; x′)

suppV (Ci; y) = suppV (C ′

i ; y′)

if and only if

  • Ht(C1; x, y), . . . , Ht(Cm; x, y)
  • d

=

  • Ht(C ′

1; x′, y′), . . . , Ht(C ′ m; x′, y′)

  • Theorem 2 =

⇒ Theorem 1 Theorem 2 generalizes the results and confirms a conjecture of [BGW19] Alexei Borodin, Vadim Gorin, and Michael

  • Wheeler. Shift-invariance for vertex models and
  • polymers. arXiv:1912.02957, 2019.

Shift of [BGW19] = double flip Proof of Theorem 1: Yang–Baxter relation combined with a Hecke algebra interpretation of the model [LLT97] Alain Lascoux, Bernard Leclerc, and Jean-Yves Thibon. Flag varieties and the Yang-Baxter equation. Lett. Math. Phys., 40(1):75–90, 1997.

slide-50
SLIDE 50

Theorem 1

Z H,V(x, y) = Z 180◦(H),V(x, rev(y))

Theorem 2

suppH(Ci; x) = suppH(C ′

i ; x′)

suppV (Ci; y) = suppV (C ′

i ; y′)

if and only if

  • Ht(C1; x, y), . . . , Ht(Cm; x, y)
  • d

=

  • Ht(C ′

1; x′, y′), . . . , Ht(C ′ m; x′, y′)

  • Theorem 2 =

⇒ Theorem 1 Theorem 2 generalizes the results and confirms a conjecture of [BGW19] Alexei Borodin, Vadim Gorin, and Michael

  • Wheeler. Shift-invariance for vertex models and
  • polymers. arXiv:1912.02957, 2019.

Shift of [BGW19] = double flip Proof of Theorem 1: Yang–Baxter relation combined with a Hecke algebra interpretation of the model [LLT97] Alain Lascoux, Bernard Leclerc, and Jean-Yves Thibon. Flag varieties and the Yang-Baxter equation. Lett. Math. Phys., 40(1):75–90, 1997. Hecke algebra approach also gives a one-line proof of [BB19] Alexei Borodin and Alexey Bufetov. Color-position symmetry in interacting particle

  • systems. arXiv:1905.04692.
slide-51
SLIDE 51

Positroid varieties

Gr(k, n) := {V ⊆ Fn | dim(V ) = k}

slide-52
SLIDE 52

Positroid varieties

Gr(k, n) := {V ⊆ Fn | dim(V ) = k} = {full rank k × n matrices}/row operations.

slide-53
SLIDE 53

Positroid varieties

Gr(k, n) := {V ⊆ Fn | dim(V ) = k} = {full rank k × n matrices}/row operations. Gr(k, n) is stratified into positroid varieties. Here’s the most interesting one:

slide-54
SLIDE 54

Positroid varieties

Gr(k, n) := {V ⊆ Fn | dim(V ) = k} = {full rank k × n matrices}/row operations. Gr(k, n) is stratified into positroid varieties. Here’s the most interesting one: Π◦

k,n := {X ∈ Gr(k, n) | ∆1,...,k(X), ∆2,...,k+1(X), . . . , ∆n,...,k−1(X) = 0},

where ∆I(X) =maximal minor of X with column set I.

slide-55
SLIDE 55

Positroid varieties

Gr(k, n) := {V ⊆ Fn | dim(V ) = k} = {full rank k × n matrices}/row operations. Gr(k, n) is stratified into positroid varieties. Here’s the most interesting one: Π◦

k,n := {X ∈ Gr(k, n) | ∆1,...,k(X), ∆2,...,k+1(X), . . . , ∆n,...,k−1(X) = 0},

where ∆I(X) =maximal minor of X with column set I.

Example

k = 2, n = 4: Π◦

k,n ∼

= 1 a b 1 c d

  • a = 0, d = 0, ad − bc = 0
  • .
slide-56
SLIDE 56

Positroid varieties

Gr(k, n) := {V ⊆ Fn | dim(V ) = k} = {full rank k × n matrices}/row operations. Gr(k, n) is stratified into positroid varieties. Here’s the most interesting one: Π◦

k,n := {X ∈ Gr(k, n) | ∆1,...,k(X), ∆2,...,k+1(X), . . . , ∆n,...,k−1(X) = 0},

where ∆I(X) =maximal minor of X with column set I.

Example

k = 2, n = 4: Π◦

k,n ∼

= 1 a b 1 c d

  • a = 0, d = 0, ad − bc = 0
  • .

Number of such matrices over Fq: #Π◦

k,n(Fq) = (q − 1)2(q2 − (q − 1)) = (q − 1)4 + q(q − 1)2.

slide-57
SLIDE 57

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

y→0

− − → pi,j =

yj−xi yj−qxi

slide-58
SLIDE 58

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

y→0

− − → pi,j =

yj−xi yj−qxi

pi,j = 1/q

slide-59
SLIDE 59

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

y→0

− − →

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: 1/q 1 − 1/q

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: 1

pi,j =

yj−xi yj−qxi

pi,j = 1/q

slide-60
SLIDE 60

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

y→0

− − →

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: 1/q 1 − 1/q

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: 1

pi,j =

yj−xi yj−qxi

pi,j = 1/q

Definition

Take a k × (n − k) rectangle and let Z id

k,n(x, 0) := the probability of observing id ∈ Sn when y → 0.

slide-61
SLIDE 61

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

y→0

− − →

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: 1/q 1 − 1/q

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: 1

pi,j =

yj−xi yj−qxi

pi,j = 1/q

Definition

Take a k × (n − k) rectangle and let Z id

k,n(x, 0) := the probability of observing id ∈ Sn when y → 0.

Example

k = 2, n = 4:

Probability:

(1 − 1/q)4 1/q · (1 − 1/q)2

slide-62
SLIDE 62

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: p 1 − p

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: qp 1 − qp

y→0

− − →

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: 1/q 1 − 1/q

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: 1

pi,j =

yj−xi yj−qxi

pi,j = 1/q

Definition

Take a k × (n − k) rectangle and let Z id

k,n(x, 0) := the probability of observing id ∈ Sn when y → 0.

Example

k = 2, n = 4:

Probability:

(1 − 1/q)4 1/q · (1 − 1/q)2

Z id

k,n(x, 0) = (1 − 1/q)4 + 1/q · (1 − 1/q)2

slide-63
SLIDE 63

Coincidence?

Example

k = 2, n = 4: #Π◦

k,n(Fq) = (q − 1)2(q2 − (q − 1)) = (q − 1)4 + q(q − 1)2.

slide-64
SLIDE 64

Coincidence?

Example

k = 2, n = 4: #Π◦

k,n(Fq) = (q − 1)2(q2 − (q − 1)) = (q − 1)4 + q(q − 1)2.

Example

k = 2, n = 4:

Probability:

(1 − 1/q)4 1/q · (1 − 1/q)2

Z id

k,n(x, 0) = (1 − 1/q)4 + 1/q · (1 − 1/q)2

slide-65
SLIDE 65

Coincidence?

Example

k = 2, n = 4: #Π◦

k,n(Fq) = (q − 1)2(q2 − (q − 1)) = (q − 1)4 + q(q − 1)2.

Example

k = 2, n = 4:

Probability:

(1 − 1/q)4 1/q · (1 − 1/q)2

Z id

k,n(x, 0) = (1 − 1/q)4 + 1/q · (1 − 1/q)2

Proposition (G., 2020)

#Π◦

k,n(Fq) = qk(n−k)Z id k,n(x, 0)

slide-66
SLIDE 66

Proposition (G., 2020)

#Π◦

k,n(Fq) = qk(n−k)Z id k,n(x, 0)

slide-67
SLIDE 67

Proposition (G., 2020)

#Π◦

k,n(Fq) = qk(n−k)Z id k,n(x, 0)

Proof.

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: 1/q 1 − 1/q

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: 1

slide-68
SLIDE 68

Proposition (G., 2020)

#Π◦

k,n(Fq) = qk(n−k)Z id k,n(x, 0)

Proof.

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: 1/q 1 − 1/q

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: 1

[Deo85] Vinay V. Deodhar. On some geometric aspects of Bruhat orderings. I. A finer decomposition

  • f Bruhat cells. Invent. Math., 79(3):499–511, 1985.
slide-69
SLIDE 69

Proposition (G., 2020)

#Π◦

k,n(Fq) = qk(n−k)Z id k,n(x, 0)

Proof.

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: 1/q 1 − 1/q

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: 1

[Deo85] Vinay V. Deodhar. On some geometric aspects of Bruhat orderings. I. A finer decomposition

  • f Bruhat cells. Invent. Math., 79(3):499–511, 1985.

#Π◦

k,n(Fq) is a Kazhdan–Lusztig R-polynomial.

slide-70
SLIDE 70

Proposition (G., 2020)

#Π◦

k,n(Fq) = qk(n−k)Z id k,n(x, 0)

Proof.

c1 c2 p

→

c1 c2

  • r

c1 c2

Probability: 1/q 1 − 1/q

c2 c1 p

→

c2 c1

  • r

c2 c1

Probability: 1

[Deo85] Vinay V. Deodhar. On some geometric aspects of Bruhat orderings. I. A finer decomposition

  • f Bruhat cells. Invent. Math., 79(3):499–511, 1985.

#Π◦

k,n(Fq) is a Kazhdan–Lusztig R-polynomial.

The whole story generalizes to arbitrary positroid varieties.

slide-71
SLIDE 71

Theorem (G.–Lam, 2020+)

Assume that gcd(k, n) = 1. Then #Π◦

k,n(Fq) = (q − 1)n−1 · Catk,n(q),

where Catk,n(q) = 1 [n]q n k

  • q

is the rational q-Catalan number.

slide-72
SLIDE 72

Theorem (G.–Lam, 2020+)

Assume that gcd(k, n) = 1. Then #Π◦

k,n(Fq) = (q − 1)n−1 · Catk,n(q),

where Catk,n(q) = 1 [n]q n k

  • q

is the rational q-Catalan number. Alternatively: the probability that a random point of Gr(k, n; Fq) belongs to Π◦

k,n(Fq) is

(q − 1)n qn − 1 .

slide-73
SLIDE 73

Theorem (G.–Lam, 2020+)

Assume that gcd(k, n) = 1. Then #Π◦

k,n(Fq) = (q − 1)n−1 · Catk,n(q),

where Catk,n(q) = 1 [n]q n k

  • q

is the rational q-Catalan number. Alternatively: the probability that a random point of Gr(k, n; Fq) belongs to Π◦

k,n(Fq) is

(q − 1)n qn − 1 .

Proof: knot theory.

slide-74
SLIDE 74

Theorem (G.–Lam, 2020+)

Assume that gcd(k, n) = 1. Then #Π◦

k,n(Fq) = (q − 1)n−1 · Catk,n(q),

where Catk,n(q) = 1 [n]q n k

  • q

is the rational q-Catalan number. Alternatively: the probability that a random point of Gr(k, n; Fq) belongs to Π◦

k,n(Fq) is

(q − 1)n qn − 1 .

Proof: knot theory. For q = 1, the Catalan (n − k = k + 1) and Fuss-Catalan (n − k = mk ± 1) cases are due to David Speyer.

slide-75
SLIDE 75

Theorem (G.–Lam, 2020+)

Assume that gcd(k, n) = 1. Then #Π◦

k,n(Fq) = (q − 1)n−1 · Catk,n(q),

where Catk,n(q) = 1 [n]q n k

  • q

is the rational q-Catalan number. Alternatively: the probability that a random point of Gr(k, n; Fq) belongs to Π◦

k,n(Fq) is

(q − 1)n qn − 1 .

Proof: knot theory. For q = 1, the Catalan (n − k = k + 1) and Fuss-Catalan (n − k = mk ± 1) cases are due to David Speyer. Forthcoming (G.–Lam, 2020+): a q, t-version.

slide-76
SLIDE 76

Theorem (G.–Lam, 2020+)

Assume that gcd(k, n) = 1. Then #Π◦

k,n(Fq) = (q − 1)n−1 · Catk,n(q),

where Catk,n(q) = 1 [n]q n k

  • q

is the rational q-Catalan number. Alternatively: the probability that a random point of Gr(k, n; Fq) belongs to Π◦

k,n(Fq) is

(q − 1)n qn − 1 .

Proof: knot theory. For q = 1, the Catalan (n − k = k + 1) and Fuss-Catalan (n − k = mk ± 1) cases are due to David Speyer. Forthcoming (G.–Lam, 2020+): a q, t-version.

[LS16] Thomas Lam and David Speyer. Cohomology of cluster varieties. I. Locally acyclic case. arXiv:1604.06843.

slide-77
SLIDE 77

Theorem (G.–Lam, 2020+)

Assume that gcd(k, n) = 1. Then #Π◦

k,n(Fq) = (q − 1)n−1 · Catk,n(q),

where Catk,n(q) = 1 [n]q n k

  • q

is the rational q-Catalan number. Alternatively: the probability that a random point of Gr(k, n; Fq) belongs to Π◦

k,n(Fq) is

(q − 1)n qn − 1 .

Proof: knot theory. For q = 1, the Catalan (n − k = k + 1) and Fuss-Catalan (n − k = mk ± 1) cases are due to David Speyer. Forthcoming (G.–Lam, 2020+): a q, t-version.

[LS16] Thomas Lam and David Speyer. Cohomology of cluster varieties. I. Locally acyclic case. arXiv:1604.06843. [GL19] Pavel Galashin and Thomas Lam. Positroid varieties and cluster algebras. arXiv:1906.03501.

slide-78
SLIDE 78

Philosophy

Stochastic colored 6-vertex model

slide-79
SLIDE 79

Philosophy

Stochastic colored 6-vertex model qk(n−k)Z id

k,n(x, 0) = #Π◦ k,n(Fq)

y → 0

slide-80
SLIDE 80

Philosophy

Stochastic colored 6-vertex model qk(n−k)Z id

k,n(x, 0) = #Π◦ k,n(Fq)

y → 0 Π◦

k,n ⊆ Gr(k, n)

count points over Fq

slide-81
SLIDE 81

Philosophy

Stochastic colored 6-vertex model qk(n−k)Z id

k,n(x, 0) = #Π◦ k,n(Fq)

y → 0 Π◦

k,n ⊆ Gr(k, n)

count points over Fq Common generalization?

slide-82
SLIDE 82

Philosophy

Stochastic colored 6-vertex model qk(n−k)Z id

k,n(x, 0) = #Π◦ k,n(Fq)

y → 0 Π◦

k,n ⊆ Gr(k, n)

count points over Fq Common generalization?

Flip theorem as y → 0: Z H,V(x, 0) = Z 180◦(H),V(x, 0)

slide-83
SLIDE 83

Philosophy

Stochastic colored 6-vertex model qk(n−k)Z id

k,n(x, 0) = #Π◦ k,n(Fq)

y → 0 Π◦

k,n ⊆ Gr(k, n)

count points over Fq Common generalization?

Flip theorem as y → 0: Z H,V(x, 0) = Z 180◦(H),V(x, 0) – lifts to the Gr(k, n) level

slide-84
SLIDE 84

Philosophy

Stochastic colored 6-vertex model qk(n−k)Z id

k,n(x, 0) = #Π◦ k,n(Fq)

y → 0 Π◦

k,n ⊆ Gr(k, n)

count points over Fq Common generalization?

Flip theorem as y → 0: Z H,V(x, 0) = Z 180◦(H),V(x, 0) – lifts to the Gr(k, n) level Proof of flip theorem (Yang–Baxter and Hecke algebra)

slide-85
SLIDE 85

Philosophy

Stochastic colored 6-vertex model qk(n−k)Z id

k,n(x, 0) = #Π◦ k,n(Fq)

y → 0 Π◦

k,n ⊆ Gr(k, n)

count points over Fq Common generalization?

Flip theorem as y → 0: Z H,V(x, 0) = Z 180◦(H),V(x, 0) – lifts to the Gr(k, n) level Proof of flip theorem (Yang–Baxter and Hecke algebra) – also lifts to Gr(k, n):

slide-86
SLIDE 86

Philosophy

Stochastic colored 6-vertex model qk(n−k)Z id

k,n(x, 0) = #Π◦ k,n(Fq)

y → 0 Π◦

k,n ⊆ Gr(k, n)

count points over Fq Common generalization?

Flip theorem as y → 0: Z H,V(x, 0) = Z 180◦(H),V(x, 0) – lifts to the Gr(k, n) level Proof of flip theorem (Yang–Baxter and Hecke algebra) – also lifts to Gr(k, n):

[MS16] Greg Muller and David E. Speyer. Cluster algebras of Grassmannians are locally

  • acyclic. Proc. Amer. Math. Soc., 144(8):3267–3281, 2016.
slide-87
SLIDE 87

Same recurrence

[MS16] Greg Muller and David E. Speyer. Cluster algebras

  • f Grassmannians are locally acyclic. Proc. Amer.
  • Math. Soc., 144(8):3267–3281, 2016.
slide-88
SLIDE 88

Same recurrence

[MS16] Greg Muller and David E. Speyer. Cluster algebras

  • f Grassmannians are locally acyclic. Proc. Amer.
  • Math. Soc., 144(8):3267–3281, 2016.

Pl Pl+1 Qr Qr+1

  • Pl

Pl+1 Qr Qr+1

  • Pl

Pl+1 Qr Qr+1

  • H

H′ = H · sr H′′ = sl · H · sr

[G., 2020]

slide-89
SLIDE 89

Open problems

Find a common generalization of the 6-vertex model and positroid varieties.

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SLIDE 90

Open problems

Find a common generalization of the 6-vertex model and positroid varieties. Clarify the relationship with (bumpless, etc) pipe dreams from Schubert calculus.

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SLIDE 91

Open problems

Find a common generalization of the 6-vertex model and positroid varieties. Clarify the relationship with (bumpless, etc) pipe dreams from Schubert calculus. How is the flip theorem related to the geometric RSK? [Dau20] Duncan Dauvergne. Hidden invariance of last passage percolation and directed polymers. arXiv:2002.09459.

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SLIDE 92

Thank you!

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 y1 y2 y3 y4 y5 y6 y7

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

x1 x2 x3 x4 y7 y6 y5 y4 y3 y2 y1

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

Hπ = H Vπ = V

= Z H,V(x, y) Z 180◦(H),V(x, rev(y))

Hπ′ = 180◦(H) Vπ′ = V