An Introduction to Tropical Geometry – Examples Eva-Maria Feichtner Recent Developments on Geometric and Algebraic Methods in Economics August 23, 2014
An Introduction to Tropical Geometry – Examples A degree formula for ∆ A A = 1, w ∈ R n generic. A ∈ Z d × n , codim X ∗ Theorem: [A.Dickenstein, E.M.F., B.Sturmfels] The exponent of x i in the monomial in w (∆ A ) equals the number of intersection points of the halfray w + R > 0 e i with the tropical discriminant τ ( X ∗ A ), counting multiplicities: � � det A T , σ 1 , . . . , σ n − d − 1 , e i � � � � � � in w (∆ A ) = deg x i � . σ ∈B ( kerA ) i , w Eva-Maria Feichtner 2 / 9
An Introduction to Tropical Geometry – Examples Example I 1 1 1 1 1 1 A = 0 1 2 0 1 0 0 0 0 1 1 2 235 5 5 3 2 1 235 456 6 3 456 1 124 6 4 4 2 124 B ( ker A ) τ ( X ∗ A ) Eva-Maria Feichtner 3 / 9
An Introduction to Tropical Geometry – Examples Example I 1 1 1 1 1 1 A = 0 1 2 0 1 0 0 0 0 1 1 2 5 1 3 6 4 2 τ ( X ∗ A ) New (∆ A ) Eva-Maria Feichtner 4 / 9
An Introduction to Tropical Geometry – Examples Example I 1 1 1 1 1 1 A = 0 1 2 0 1 0 0 0 0 1 1 2 x 3 x 2 x 1 x 3 x 6 5 4 x 1 x 2 1 x 2 x 4 x 5 5 3 6 x 2 2 x 6 2 4 τ ( X ∗ A ) New (∆ A ) Eva-Maria Feichtner 5 / 9
An Introduction to Tropical Geometry – Examples Example I 1 1 1 1 1 1 A = 0 1 2 0 1 0 0 0 0 1 1 2 (- 1 ) x 3 x 2 4 x 1 x 3 x 6 5 4 x 1 x 2 1 x 2 x 4 x 5 5 3 6 (- 1 ) x 2 2 x 6 4 2 τ ( X ∗ A ) New (∆ A ) Eva-Maria Feichtner 6 / 9
An Introduction to Tropical Geometry – Examples Example II 1 1 1 1 1 1 A = 1 0 1 2 1 1 0 1 1 1 2 3 b 2 d 2 e 4 b 3 d 3 f 2 24 24 3 1 a 2 e 6 a 4 f 4 6 156 5 5 156 1 6 3 c 4 e 4 c 6 f 2 B ( ker A ) New (∆ A ) τ ( X ∗ A ) Eva-Maria Feichtner 7 / 9
An Introduction to Tropical Geometry – Examples Example II 1 1 1 1 1 1 A = 1 0 1 2 1 1 0 1 1 1 2 3 - 1024 b 3 d 3 f 2 16 b 2 d 2 e 4 24 24 3 1 16 a 2 e 6 729 a 4 f 4 6 156 5 5 156 1 6 3 c 4 e 4 16 c 6 f 2 B ( ker A ) New (∆ A ) τ ( X ∗ A ) Eva-Maria Feichtner 8 / 9
An Introduction to Tropical Geometry – Examples Example II 1 1 1 1 1 1 A = 1 0 1 2 1 1 0 1 1 1 2 3 c 4 e 4 − 8 bc 2 de 4 + 16 b 2 d 2 e 4 − 8 ac 2 e 5 − 32 abde 5 + 16 a 2 e 6 ∆ A = − 8 c 5 e 2 f + 64 bc 3 de 2 f − 128 b 2 cd 2 e 2 f + 68 ac 3 e 3 f +240 abcde 3 f − 144 a 2 ce 4 f + 16 c 6 f 2 − 192 bc 4 df 2 +768 b 2 c 2 d 2 f 2 − 1024 b 3 d 3 f 2 − 144 ac 4 ef 2 + 2304 ab 2 d 2 ef 2 +270 a 2 c 2 e 2 f 2 − 1512 a 2 bde 2 f 2 + 216 a 3 e 3 f 2 + 216 a 2 c 3 f 3 +2592 a 2 bcdf 3 − 972 a 3 cef 3 + 729 a 4 f 4 Eva-Maria Feichtner 9 / 9
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