Topology of Positive Zero Sets of Bivariate Pentanomials Malachi - PowerPoint PPT Presentation

Topology of Positive Zero Sets of Bivariate Pentanomials Malachi Alexander 1 , Ashley De Luna 2 Christian McRoberts 3 1 California State University, Monterey Bay 2 California State Polytechnic University, Pomona 3 Morehouse College MSRI-UP 2017, 08/04/2017

Motivation 1876: Carl Gustav Axel Harnack investigated the number of connected components of real algebraic curves.

Motivation 1876: Carl Gustav Axel Harnack investigated the number of connected components of real algebraic curves. 1900: David Hilbert posed his sixteenth problem about determining arrangements of algebraic curves.

Motivation Previous work has shown, For n -variate ( n + 1 ) -nomials: OR

Motivation Previous work has shown, For n -variate ( n + 1 ) -nomials: For n -variate ( n + 2 ) -nomials: OR OR

Motivation Previous work has shown, For n -variate ( n + 1 ) -nomials: For n -variate ( n + 2 ) -nomials: OR OR Our work is on n -variate ( n + 3 ) -nomials, We focus on the first unknown case: bivariate pentanomials!

Motivation Automate computing the topology of positive zero sets. Graph the A -discriminant curve of sparse polynomials.

Motivation 3 4 2 1 5 6 12 7 8 11 10 9 Automate computing the topology of positive zero sets. Graph the A -discriminant curve of sparse polynomials. Observe number of connected components.

Motivation Automate computing the topology of positive zero sets. Graph the A -discriminant curve of sparse polynomials. Observe number of connected components. Visualize the topology of the positive zero sets.

Motivation Example 1: f ( x 1 , x 2 ) = x 100 x 100 + 1 + x 2001 + x 2011 − 10 x 2 1 x 2 1 2 1 2 2 1.2 0.2 1.0 0 0.8 -0.2 -0.4 0.6 -0.6 0.4 -0.8 -1 0.2 -1.2 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 Trop + ( f ) Z + ( f )

Motivation Example 2 : f ( x 1 , x 2 ) = 1 + x 1 + x 2 − 6 5 x 4 1 x 2 − 6 5 x 1 x 4 2 3.0 0.1 2.5 0 -0.1 2.0 -0.2 -0.3 1.5 -0.4 -0.5 1.0 -0.6 -0.7 0.5 -0.8 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Trop + ( f ) Z + ( f )

Project Overview Step 1: Plot the A -Discriminant Variety Step 2: Compute Positive Tropical Variety Step 3: Find the Topology of Inner Chambers

Step 1: Plot the A -Discriminant Variety A -Discriminant Variety Let A ∈ Z 2 × 5 , with columns a 1 , . . . , a 5 and x = ( x 1 , x 2 ) s.t. f ( x ) = c 1 x a 1 + c 2 x a 2 + c 3 x a 3 + c 4 x a 4 + c 5 x a 5 ∈ C [ x 1 , x 2 ] ∇ A := { [ c 1 · · · c 5 ] ∈ P 4 C | f ( x ) has a degenerate root in ( C \{ 0 } ) 2 } [ c 1 , . . . , c 5 ] ∈ ∇ A [ c 1 , . . . , c 5 ] �∈ ∇ A

Step 1: Plot the A -Discriminant Variety ∇ A has simple ( C \{ 0 } ) 2 -fibers over a two-dimensional base.

Step 1: Plot the A -Discriminant Variety ∇ A has simple ( C \{ 0 } ) 2 -fibers over a two-dimensional base. We take a two-dimensional slice to get the reduced A -discriminant variety.

Step 1: Plot the A -Discriminant Variety Given any polynomial f with x = ( x 1 , x 2 ) , a i = ( a 1 i , a 2 i ) , f ( x ) = c 1 x a 1 + c 2 x a 2 + c 3 x a 3 + c 4 x a 4 + c 5 x a 5

Step 1: Plot the A -Discriminant Variety Given any polynomial f with x = ( x 1 , x 2 ) , a i = ( a 1 i , a 2 i ) , f ( x ) = c 1 x a 1 + c 2 x a 2 + c 3 x a 3 + c 4 x a 4 + c 5 x a 5 � � � 2 � 1 − 1 5 4 9 1 5 8 4 A = A = 4 2 3 1 4 2 8 8 4 1

Step 1: Plot the A -Discriminant Variety f ( x 1 , x 2 ) = 10 x 2 1 x 2 2 − 8 x 1 x 8 2 − 11 x 5 1 x 8 2 + 8 x 8 1 x 4 2 + 3 x 4 1 x 2

Step 1: Plot the A -Discriminant Variety f ( x 1 , x 2 ) = 10 x 2 1 x 2 2 − 8 x 1 x 8 2 − 11 x 5 1 x 8 2 + 8 x 8 1 x 4 2 + 3 x 4 1 x 2 � 2 � 1 5 8 4 A = 2 8 8 4 1

Step 1: Plot the A -Discriminant Variety f ( x 1 , x 2 ) = 10 x 2 1 x 2 2 − 8 x 1 x 8 2 − 11 x 5 1 x 8 2 + 8 x 8 1 x 4 2 + 3 x 4 1 x 2 � 2 � 1 5 8 4 A = 2 8 8 4 1   1 1 1 1 1 ˆ A = 2 1 5 8 4   2 8 8 4 1

Step 1: Plot the A -Discriminant Variety f ( x 1 , x 2 ) = 10 x 2 1 x 2 2 − 8 x 1 x 8 2 − 11 x 5 1 x 8 2 + 8 x 8 1 x 4 2 + 3 x 4 1 x 2 � 2 � 1 5 8 4 A = 2 8 8 4 1   1 1 1 1 1 ˆ A = 2 1 5 8 4   2 8 8 4 1 We define B to be the right null space of ˆ A : ˆ A B = O where O is the zero vector.

θ π 0 Step 1: Plot the A -Discriminant Variety Use the Horn-Kapranov Uniformization .

θ π 0 Step 1: Plot the A -Discriminant Variety Use the Horn-Kapranov Uniformization . ϕ A ( λ ) := ( Log | λ B T | ) B

θ π 0 Step 1: Plot the A -Discriminant Variety Use the Horn-Kapranov Uniformization . ϕ A ( λ ) := ( Log | λ B T | ) B λ = ( λ 1 , λ 2 ) λ = ( sin ( θ ) , cos ( θ ))

Step 1: Plot the A -Discriminant Variety Use the Horn-Kapranov Uniformization . ϕ A ( λ ) := ( Log | λ B T | ) B λ = ( λ 1 , λ 2 ) λ = ( sin ( θ ) , cos ( θ )) θ π 0 0 ≤ θ < π

Step 1: Plot the A -Discriminant Variety f ( x 1 , x 2 ) = 10 x 2 1 x 2 2 − 8 x 1 x 8 2 − 11 x 5 1 x 8 2 + 8 x 8 1 x 4 2 + 3 x 4 1 x 2

Project Overview Step 1: Plot the A -Discriminant Variety Step 2: Compute Positive Tropical Variety Step 3: Find the Topology of Inner Chambers

Step 2: Compute the Positive Tropical Variety Theorem 1. If f ∈ R [ x 1 , x 2 ] has coefficient vector c and Log | c | B lying in an outer chamber, then Z + ( f ) and Trop + ( f ) are isotopic.

Step 2: Compute the Positive Tropical Variety Theorem 1. If f ∈ R [ x 1 , x 2 ] has coefficient vector c and Log | c | B lying in an outer chamber, then Z + ( f ) and Trop + ( f ) are isotopic. f ( x 1 , x 2 ) = 10 x 2 1 x 2 2 − 8 x 1 x 8 2 − 11 x 5 1 x 8 2 + 8 x 8 1 x 4 2 + 3 x 4 1 x 2 � � c f = 10 − 8 − 11 8 3 g ( x 1 , x 2 ) = 2 x 2 1 x 2 2 − 5 x 1 x 8 2 − 11 x 5 1 x 8 2 + 8 x 8 1 x 4 2 + 3 x 4 1 x 2 � � c g = 2 − 5 − 11 8 3 h ( x 1 , x 2 ) = 11 x 2 1 x 2 2 − 8 x 1 x 8 2 − 10 x 5 1 x 8 2 + 7 x 8 1 x 4 2 + 2 x 4 1 x 2 � � c h = 11 − 8 − 10 7 2

Step 2: Compute the Positive Tropical Variety 50 40 30 20 10 Log | c | B = ( α, β ) 0 -10 -20 -30 -40 -50 -50 -40 -30 -20 -10 0 10 20 30 40 50

Step 2: Compute the Positive Tropical Variety 50 40 30 20 10 Log | c | B = ( α, β ) 0 f ( x 1 , x 2 ) -10 -20 -30 -40 -50 -50 -40 -30 -20 -10 0 10 20 30 40 50

Step 2: Compute the Positive Tropical Variety 50 40 30 20 10 Log | c | B = ( α, β ) 0 f ( x 1 , x 2 ) -10 g ( x 1 , x 2 ) -20 -30 -40 -50 -50 -40 -30 -20 -10 0 10 20 30 40 50

Step 2: Compute the Positive Tropical Variety 50 40 30 20 10 Log | c | B = ( α, β ) 0 f ( x 1 , x 2 ) -10 g ( x 1 , x 2 ) -20 h ( x 1 , x 2 ) -30 -40 -50 -50 -40 -30 -20 -10 0 10 20 30 40 50

Step 2: Compute the Positive Tropical Variety Theorem 1. If f ∈ R [ x 1 , x 2 ] has coefficient vector c and Log | c | B lying in an outer chamber, then Z + ( f ) and Trop + ( f ) are isotopic. f ( x 1 , x 2 ) = 10 x 2 1 x 2 2 − 8 x 1 x 8 2 − 11 x 5 1 x 8 2 + 8 x 8 1 x 4 2 + 3 x 4 1 x 2 � � c f = 10 − 8 − 11 8 3 g ( x 1 , x 2 ) = 2 x 2 1 x 2 2 − 5 x 1 x 8 2 − 11 x 5 1 x 8 2 + 8 x 8 1 x 4 2 + 3 x 4 1 x 2 � � c g = 2 − 5 − 11 8 3 h ( x 1 , x 2 ) = 11 x 2 1 x 2 2 − 8 x 1 x 8 2 − 10 x 5 1 x 8 2 + 7 x 8 1 x 4 2 + 2 x 4 1 x 2 � � c h = 11 − 8 − 10 7 2

Step 2: Compute the Positive Tropical Variety Isotopy : Given two sets X , Y ∈ R 2 , a homeomorphism ι : [ 0 , 1 ] × R 2 → R 2 is an isotopy if ι is a continuous map such that ι ( 0 , X ) = X and ι ( 1 , X ) = Y

Step 2: Compute the Positive Tropical Variety Isotopy : Given two sets X , Y ∈ R 2 , a homeomorphism ι : [ 0 , 1 ] × R 2 → R 2 is an isotopy if ι is a continuous map such that ι ( 0 , X ) = X and ι ( 1 , X ) = Y

Step 2: Compute the Positive Tropical Variety Theorem 1. If f ∈ R [ x 1 , x 2 ] has coefficient vector c and Log | c | B lying in an outer chamber, then Z + ( f ) and Trop + ( f ) are isotopic. f ( x 1 , x 2 ) = 10 x 2 1 x 2 2 − 8 x 1 x 8 2 − 11 x 5 1 x 8 2 + 8 x 8 1 x 4 2 + 3 x 4 1 x 2 � � c f = 10 − 8 − 11 8 3 g ( x 1 , x 2 ) = 2 x 2 1 x 2 2 − 5 x 1 x 8 2 − 11 x 5 1 x 8 2 + 8 x 8 1 x 4 2 + 3 x 4 1 x 2 � � c g = 2 − 5 − 11 8 3 h ( x 1 , x 2 ) = 11 x 2 1 x 2 2 − 8 x 1 x 8 2 − 10 x 5 1 x 8 2 + 7 x 8 1 x 4 2 + 2 x 4 1 x 2 � � c h = 11 − 8 − 10 7 2

Step 2: Compute the Positive Tropical Variety f ( x 1 , x 2 ) = 10 x 2 1 x 2 2 − 8 x 1 x 8 2 − 11 x 5 1 x 8 2 + 8 x 8 1 x 4 2 + 3 x 4 1 x 2

Step 2: Compute the Positive Tropical Variety f ( x 1 , x 2 ) = 10 x 2 1 x 2 2 − 8 x 1 x 8 2 − 11 x 5 1 x 8 2 + 8 x 8 1 x 4 2 + 3 x 4 1 x 2 � 2 � 1 5 8 4 A = 2 8 8 4 1 � � c = 10 − 8 − 11 8 3

Step 2: Compute the Positive Tropical Variety f ( x 1 , x 2 ) = 10 x 2 1 x 2 2 − 8 x 1 x 8 2 − 11 x 5 1 x 8 2 + 8 x 8 1 x 4 2 + 3 x 4 1 x 2 � 2 � 1 5 8 4 A = 2 8 8 4 1 � � c = 10 − 8 − 11 8 3 Newt ( f ) := Conv ( { a i } ) 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8