R -systems Pavel Galashin MIT galashin@mit.edu AMS Joint Meeting, - PowerPoint PPT Presentation

R -systems Pavel Galashin MIT galashin@mit.edu AMS Joint Meeting, San Diego, CA, January 13, 2018 Joint work with Pavlo Pylyavskyy Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 1 / 23

Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 2 / 23

Part 1: Definition

A system of equations Let G = ( V , E ) be a strongly connected digraph . c ( c + d ) d ( c + d ) b c a ac d bc X X ′ Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 4 / 23

A system of equations Let G = ( V , E ) be a strongly connected digraph . c ( c + d ) d ( c + d ) b c a ac d bc X X ′ Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 4 / 23

A system of equations Let G = ( V , E ) be a strongly connected digraph . c ( c + d ) d ( c + d ) b c b ′ c ′ a ac d a ′ bc d ′ X X ′ Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 4 / 23

A system of equations Let G = ( V , E ) be a strongly connected digraph . c ( c + d ) d ( c + d ) b c b ′ c ′ a ac d bc a ′ d ′ X X ′ � − 1 � � � �� 1 X v X ′ ∀ v ∈ V , v = X w X ′ u v → w u → v Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 4 / 23

A system of equations Let G = ( V , E ) be a strongly connected digraph . c ( c + d ) d ( c + d ) b c a ac d bc X X ′ � − 1 � � � �� 1 ∀ v ∈ V , X v X ′ v = X w X ′ u v → w u → v Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 4 / 23

A system of equations Let G = ( V , E ) be a strongly connected digraph . c ( c + d ) d ( c + d ) b c a ac d bc X X ′ � − 1 � � � �� 1 X v X ′ ∀ v ∈ V , v = X w X ′ u v → w u → v � − 1 � 1 1 d · ac = ( a ) c ( c + d ) + d ( c + d ) Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 4 / 23

Solution Theorem (G.-Pylyavskyy, 2017) Let G = ( V , E ) be a strongly connected digraph. Then there exists a birational map φ : P V ��� P V such that X , X ′ ∈ P V give a solution X ′ = φ ( X ) . ⇐ ⇒ Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 5 / 23

Periodic examples (exercise) b a c b g c a f d e e f d Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 6 / 23

Arborescence formula b c b c b c a a a d d d wt = ad 2 wt = acd wt = abd b c b c b c a a a d d d wt = bd 2 wt = abc wt = bcd Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 7 / 23

Map Cluster Cluster Integrable Integrable LP algebras algebras algebras systems systems Zamolodchikov Superpotential & R-systems R-systems periodicity Mirror symmetry Birational Birational Birational Geometric rowmotion rowmotion toggling RSK Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 8 / 23

Map Cluster Cluster Integrable Integrable LP algebras algebras algebras systems systems Zamolodchikov Zamolodchikov Superpotential & R-systems R-systems periodicity periodicity Mirror symmetry Birational Birational Birational Birational Geometric Geometric rowmotion rowmotion toggling toggling RSK RSK Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 9 / 23

Birational rowmotion ⊆ R -systems ˆ ˆ ˆ 1 1 1 ˆ ˆ ˆ 0 0 0 P Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 10 / 23

Birational rowmotion ⊆ R -systems ˆ ˆ ˆ ˆ 1 1 1 1 ˆ ˆ ˆ ˆ 0 0 0 0 ˆ P P Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 10 / 23

Birational rowmotion ⊆ R -systems ˆ ˆ ˆ ˆ 1 1 1 1 ˆ ˆ ˆ ˆ 0 0 0 0 ˆ P P G ( P ) Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 10 / 23

Birational rowmotion ⊆ R -systems ˆ ˆ ˆ 1 1 1 ˆ ˆ ˆ 0 0 0 ˆ P P G ( P ) Proposition (G.-Pylyavskyy, 2017) Birational rowmotion on P = R-system associated with G ( P ) . Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 10 / 23

Part 2: Singularity confinement

Map Cluster Cluster Cluster Integrable Integrable Integrable LP algebras LP algebras algebras algebras algebras systems systems systems Zamolodchikov Superpotential & R-systems R-systems periodicity Mirror symmetry Birational Birational Birational Geometric rowmotion rowmotion toggling RSK Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 12 / 23

The Laurent phenomenon ατ n +1 τ n +3 + βτ 2 Somos-4 sequence: τ n +4 = n +2 . τ n Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 13 / 23

The Laurent phenomenon ατ n +1 τ n +3 + βτ 2 Somos-4 sequence: τ n +4 = n +2 . τ n Theorem (Fomin-Zelevinsky, 2002) For each n > 4 , τ n is a Laurent polynomial in α, β, τ 1 , τ 2 , τ 3 , τ 4 . Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 13 / 23

The Laurent phenomenon ατ n +1 τ n +3 + βτ 2 Somos-4 sequence: τ n +4 = n +2 . τ n Theorem (Fomin-Zelevinsky, 2002) For each n > 4 , τ n is a Laurent polynomial in α, β, τ 1 , τ 2 , τ 3 , τ 4 . Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 13 / 23

Singularity confinement Consider a mapping of the plane ( x n − 1 , x n ) �→ ( x n , x n +1 ) given by substitute x n = τ n +1 τ n − 1 x n +1 = α x n + β n . τ 2 x n − 1 x 2 n Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 14 / 23

Singularity confinement Consider a mapping of the plane ( x n − 1 , x n ) �→ ( x n , x n +1 ) given by substitute x n = τ n +1 τ n − 1 x n +1 = α x n + β n . τ 2 x n − 1 x 2 n x 3 = α x 2 + β x 1 x 2 2 Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 14 / 23

Singularity confinement Consider a mapping of the plane ( x n − 1 , x n ) �→ ( x n , x n +1 ) given by substitute x n = τ n +1 τ n − 1 x n +1 = α x n + β n . τ 2 x n − 1 x 2 n x 3 = α x 2 + β x 1 x 2 2 x 4 = ( β x 1 x 2 2 + α 2 x 2 + αβ ) x 1 x 2 ( α x 2 + β ) 2 Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 14 / 23

Singularity confinement Consider a mapping of the plane ( x n − 1 , x n ) �→ ( x n , x n +1 ) given by substitute x n = τ n +1 τ n − 1 x n +1 = α x n + β n . τ 2 x n − 1 x 2 n x 3 = α x 2 + β x 1 x 2 2 x 4 = ( β x 1 x 2 2 + α 2 x 2 + αβ ) x 1 x 2 ( α x 2 + β ) 2 x 5 = ( αβ x 2 1 x 3 2 + ··· + β 3 )( α x 2 + β ) ( β x 1 x 2 2 + α 2 x 2 + αβ ) 2 x 1 Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 14 / 23

Singularity confinement Consider a mapping of the plane ( x n − 1 , x n ) �→ ( x n , x n +1 ) given by substitute x n = τ n +1 τ n − 1 x n +1 = α x n + β n . τ 2 x n − 1 x 2 n x 3 = α x 2 + β x 1 x 2 2 x 4 = ( β x 1 x 2 2 + α 2 x 2 + αβ ) x 1 x 2 ( α x 2 + β ) 2 x 5 = ( αβ x 2 1 x 3 2 + ··· + β 3 )( α x 2 + β ) ( β x 1 x 2 2 + α 2 x 2 + αβ ) 2 x 1 x 6 = ( α 3 β x 2 1 x 4 2 + ··· + αβ 4 )( β x 1 x 2 2 + α 2 x 2 + αβ ) ( αβ x 2 1 x 3 2 + ··· + β 3 ) 2 x 2 Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 14 / 23

Singularity confinement Consider a mapping of the plane ( x n − 1 , x n ) �→ ( x n , x n +1 ) given by substitute x n = τ n +1 τ n − 1 x n +1 = α x n + β n . τ 2 x n − 1 x 2 n x 3 = α x 2 + β x 1 x 2 2 x 4 = ( β x 1 x 2 2 + α 2 x 2 + αβ ) x 1 x 2 ( α x 2 + β ) 2 x 5 = ( αβ x 2 1 x 3 2 + ··· + β 3 )( α x 2 + β ) ( β x 1 x 2 2 + α 2 x 2 + αβ ) 2 x 1 x 6 = ( α 3 β x 2 1 x 4 2 + ··· + αβ 4 )( β x 1 x 2 2 + α 2 x 2 + αβ ) ( αβ x 2 1 x 3 2 + ··· + β 3 ) 2 x 2 x 7 = ( αβ 3 x 4 1 x 6 2 + ··· + β 6 x 2 )( αβ x 2 1 x 3 2 + ··· + β 3 ) x 1 x 2 ( α 3 β x 2 1 x 4 2 + ··· + αβ 4 ) 2 Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 14 / 23

Singularity confinement Consider a mapping of the plane ( x n − 1 , x n ) �→ ( x n , x n +1 ) given by substitute x n = τ n +1 τ n − 1 x n +1 = α x n + β n . τ 2 x n − 1 x 2 n x 3 = α x 2 + β x 1 x 2 2 x 4 = ( β x 1 x 2 2 + α 2 x 2 + αβ ) x 1 x 2 ( α x 2 + β ) 2 x 5 = ( αβ x 2 1 x 3 2 + ··· + β 3 )( α x 2 + β ) ( β x 1 x 2 2 + α 2 x 2 + αβ ) 2 x 1 x 6 = ( α 3 β x 2 1 x 4 2 + ··· + αβ 4 )( β x 1 x 2 2 + α 2 x 2 + αβ ) ( αβ x 2 1 x 3 2 + ··· + β 3 ) 2 x 2 x 7 = ( αβ 3 x 4 1 x 6 2 + ··· + β 6 x 2 )( αβ x 2 1 x 3 2 + ··· + β 3 ) x 1 x 2 ( α 3 β x 2 1 x 4 2 + ··· + αβ 4 ) 2 x 8 = ( α 3 β 3 x 6 1 x 8 2 + ··· + αβ 8 )( α 3 β x 2 1 x 4 2 + ··· + αβ 4 ) ( αβ 3 x 4 1 x 6 2 + ··· + β 6 x 2 ) 2 x 2 1 x 2 Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 14 / 23

Singularity confinement Consider a mapping of the plane ( x n − 1 , x n ) �→ ( x n , x n +1 ) given by substitute x n = τ n +1 τ n − 1 x n +1 = α x n + β n . τ 2 x n − 1 x 2 n x 3 = α x 2 + β x 1 x 2 2 x 4 = ( β x 1 x 2 2 + α 2 x 2 + αβ ) x 1 x 2 ( α x 2 + β ) 2 x 5 = ( αβ x 2 1 x 3 2 + ··· + β 3 )( α x 2 + β ) ( β x 1 x 2 2 + α 2 x 2 + αβ ) 2 x 1 x 6 = ( α 3 β x 2 1 x 4 2 + ··· + αβ 4 )( β x 1 x 2 2 + α 2 x 2 + αβ ) ( αβ x 2 1 x 3 2 + ··· + β 3 ) 2 x 2 x 7 = ( αβ 3 x 4 1 x 6 2 + ··· + β 6 x 2 )( αβ x 2 1 x 3 2 + ··· + β 3 ) x 1 x 2 ( α 3 β x 2 1 x 4 2 + ··· + αβ 4 ) 2 x 8 = ( α 3 β 3 x 6 1 x 8 2 + ··· + αβ 8 )( α 3 β x 2 1 x 4 2 + ··· + αβ 4 ) ( αβ 3 x 4 1 x 6 2 + ··· + β 6 x 2 ) 2 x 2 1 x 2 Pavel Galashin (MIT) R -systems San Diego, CA, 01/13/2018 14 / 23