Tangency and Discriminants FPSAC 2019, Lubljana Sandra Di Rocco, - PowerPoint PPT Presentation

KTH ROYAL INSTITUTE OF TECHNOLOGY Tangency and Discriminants FPSAC 2019, Lubljana Sandra Di Rocco,

Goal ◮ Discriminants: tangency and duality ◮ Discriminants: tangential intersections ◮ Generalized Schäfli decomposition 2/28

Natural concept The discriminant is a concept occurring naturally in connection with the way we grasp 3D objects. Figure: Boundary locally defined by f ( x , y , z ) = 0 . The Discriminant with respect to x is the “plane curve" defined by the equation obtained by eliminating x from { f = 0 , ∂ f ∂ x = 0 } 3/28

The discriminants of univariate polynomials The discriminant: gives information about the nature of the polynomial’s roots. ◮ the discriminant of c 2 x 2 + c 1 x + c 0 is c 2 1 − 4 c 2 c 0 . ◮ for higher degrees the discriminant D d is a ( 2 d − 1 ) × ( 2 d − 1 ) determinant :  c d c d − 1 · · · c 0 0 · · · 0  0 c d c d − 1 · · · 0 · · · 0    . . . . . . .  . . . . . . .   . . . . . . .   D d = ( 1 / c d ) det   dc d ( d − 1 ) c d − 1 · · · 0 0 0 0    . . . . . . .  . . . . . . .   . . . . . . .   0 0 0 0 · · · · · · 2 d 2 c 1 4/28

Algebra vs Geometry Algebra: D d = Res ( p ( x ) , p ′ ( x )) x 2 x 3 Geometry: 1 x P 1 ֒ → P d J 1 = O P 1 ( d − 1 ) ⊕ O P 1 ( d − 1 ) and deg ( c 1 ( J 1 )) = 2 d − 2 5/28

The definition of discriminant Let A ⊂ Z n be a finite subset of lattice points: A = { m 0 , m 1 , . . . , m N } A polynomial p in d variables is supported on A if � c i x m i p ( x 1 , . . . , x n ) = m i ∈A where x m = x k 1 1 x k 2 2 · · · x k n n if m = ( k 1 , . . . , k n ) ∈ A Figure: Quadrics c 0 + c 1 x + c 2 y + c 3 xy 6/28

The definition of discriminant Definition Let A = { m 0 , m 1 , . . . , m N } ⊂ Z n . The discriminant of A is (if it exists!) a polynomial D A ( c 0 , . . . , c N ) in N + 1 variables vanishing whenever the corresponding polynomial m i ∈A c i x m i has some multiple root in ( C ∗ ) n . p ( x ) = � there is x ∈ ( C ∗ ) n s.t. D A ( c 0 , . . . , c N ) = 0 ⇔ p ( x ) = . . . = ∂ p ∂ x j ( x ) = . . . = 0 Otherwise D A = 1 . Existence does not mean an efficient algorithm and hence a formula! 7/28

Example 1 For the configuration A = { ( 0 , 0 ) , ( 1 , 0 ) , ( 0 , 1 ) , ( 1 , 1 ) } ⊂ Z 2 The discriminant is given by an homogeneous polynomial ∆ A ( c 0 , c 1 , c 2 , c 3 ) vanishing whenever the corresponding quadric has a singular point in ( C ∗ ) 2 . I D A ( c 0 , c 1 , c 2 , c 3 ) = det ( M ) = c 0 c 3 − c 1 c 2 . 8/28

Geometry Let Q ⊂ C 2 and p ∈ C 2 ◮ general tangent lines to Q do not contain the point p ◮ exceptional locus: { x ∈ Q | p ∈ T Q , x } has degree 2 . ◮ It gives the degree of the discriminant. 9/28

Geometry ◮ The polar classes P i are codimension i cycles on → P N X ֒ ◮ P 1 on Q is a zero-cycle of degree 2 . 10/28

Projective duality → P n be a smooth embedding of dimension d . X ֒ The dual variety is defined as: X ∗ = { H ∈ ( P n ) ∗ tangent to X at some x ∈ X } 11/28

Projective duality ◮ N ( X ) = { ( x , H ) : H tangent to X at x ∈ X } ⊂ X × ( P N ) ∗ has dimension N − 1 Bertini For general varieties, the restriction of the projection π : N ( X ) → ( P N ) ∗ is generically 1-1. ◮ Im ( π ) = X ∗ , codimension-one irreducible subvariety (generically!) ◮ It is defined by an irreducible polynomial D X , called the discriminant. 12/28

Polar geometry:the degree and dimension of the discriminant Let P 0 ( X ) , . . . , P n ( X ) be the polar classes. Theorem X projective variety of dimension n, then ◮ codim ( X ∗ ) = 1 + n − max { j s.t. P j ( X ) � = 0 } ◮ Let codim ( X ∗ ) = 1 + n − j then deg ( X ∗ ) = deg ( P j ( X )) . Figure: C ∗ is another conic, deg ( P 1 ( X )) = 2 13/28

Toric projective duality= A -discriminants ◮ A = { m 0 , . . . , m N } ⊂ Z n Let P A = Conv ( A ) ◮ ϕ A : ( C ∗ ) n → P N , ϕ ( x ) = ( x m 0 , . . . , x m N ) ◮ X A = Im ( ϕ A ) is a toric embedding ◮ X ∗ A has codimension 1 unless X A is a linear fibration ( P A certain Cayley polytope). ◮ Smooth: codimension 1 if P n ( X A ) = deg ( D A ) = F � P A ( − 1 ) codim ( F ) ( dim ( F ) + 1 )! Vol Z ( F ) � = 0 � ( x , y ) → ( 1 , x , y , xy ) 3 ! · Area − 2 !( perimeter ) + 4 = 6 − 8 + 4 = 2 14/28

Can a discriminant govern multiple roots of sys- tems of polynomials? As before: Let A 1 , . . . , A n be (finite) in Z n and let f 1 , . . . , f n be Laurent polynomials with these support sets and coefficients in an alg. cl. field K , e.g. C : � c i , a x a . p A i ( x ) = a ∈A i If the coefficients c i , a are generic then, by Bernstein’s Theorem , the number of common solutions in the algebraic torus ( C ∗ ) n equals the mixed volume MV ( Q 1 , Q 2 , . . . , Q n ) of the Newton polytopes Q i = conv ( A i ) in R n . 15/28

Example Let n = 2 and A 1 = A 2 = { ( 0 , 0 ) , ( 1 , 0 ) , ( 0 , 1 ) , ( 1 , 1 ) } be the unit square, f 1 = a 00 + a 10 x 1 + a 01 x 2 + a 11 x 1 x 2 , f 2 = b 00 + b 10 x 1 + b 01 x 2 + b 11 x 1 x 2 . 16/28

tangential intersections 17/28

tangential intersections Given p A 1 , p A 2 , we say that x is a tangential solution of the system p A 1 ( u ) = p A 2 ( u ) = 0 if x is a regular point of the hypersurfaces p A i = 0 and their normal lines are dependent. Definition Given a system of type ( A 0 , . . . , A r ) . We call an isolated solution u ∈ ( C ∗ ) n a non-degenerate multiple root if the r + 1 gradient vectors ∇ x p A i ( u ) , i = 0 , . . . , r are linearly dependent. 17/28

The mixed discriminant Given A 0 , . . . , A r ⊂ Z n Definition The mixed discriminant is a (the!) polynomial MD A 0 ,..., A r ( c ) on the c i , a which vanishes whenever the polynomials have tangential roots. MD A 0 , ··· , A r ( c ) is a polynomial in |A 0 | + · · · + |A r | variables When A 0 = · · · = A r = A we denote it by M ( r , A ) . 18/28

Example Let n = 2 and A 1 = A 2 = { ( 0 , 0 ) , ( 1 , 0 ) , ( 0 , 1 ) , ( 1 , 1 ) } be the unit square, f 1 = a 00 + a 10 x 1 + a 01 x 2 + a 11 x 1 x 2 , f 2 = b 00 + b 10 x 1 + b 01 x 2 + b 11 x 1 x 2 . ∆ A 1 , A 2 is the hyperdeterminant of format 2 × 2 × 2: a 2 00 b 2 11 − 2 a 00 a 01 b 10 b 11 − 2 a 00 a 10 b 01 b 11 − 2 a 00 a 11 b 00 b 11 + 4 a 00 a 11 b 01 b 10 + a 2 01 b 2 10 + 4 a 01 a 10 b 00 b 11 − 2 a 01 a 10 b 01 b 10 − 2 a 01 a 11 b 00 b 10 + a 2 10 b 2 01 − 2 a 10 a 11 b 00 b 01 + a 2 11 b 2 00 bidegree = ( 2 , 2 ) 19/28

One more example: The distance to a variety Consider X ⊂ R N . The Euclidian Distance Degree, EDD ( X ) , number of critical points of the algebraic function: u ( X ) where d X ( u ) = min x ∈ X ( d u ( x )) for u ∈ R N generic. u �→ d 2 20/28

Consider now a plane curve. Equivalently one looks at the circles admitting tangent solutions with the curve. C is a conic: 3 x 3 matrix M ( c ij ) and the circle by the the 3 x 3 symmetric matrix M ( u , r ) . The Mixed Discriminant is given by the 2 x 3 x 3 hyperderminant: H ( c ij , u , r ) . 21/28

This proves: Theorem (Cayley) Let C be an irreducible conic, then ◮ EDD(Circle) = 2 ◮ EDD(Parabola) = 3 ◮ EDD = 4 otherwise The key tool is the use of Schläfli decomposition MD ( A 1 , A 2 ) = Hyperdet ([ M 1 , M 2 ]) = Disc t ( det ( M 1 + tM 2 )) . 22/28

Singular intersection of Quadric Surfaces ◮ Brownic[1906], ◮ Salmon [1911] ◮ Farouki [1989] Completely classified singular intersections of quadric surfaces. Key tools: ◮ Classified by the Hyperdeterminant, i.e. discriminant of Segre embeddings ◮ The hyperdeterminant can be computed by iteration 23/28

Two Main Questions: ◮ Question 1 Can the mixed discriminant be computed via iteration? ◮ Question 2 What about singular intersection of higher dimensional quadrics? 24/28

Towards an answer to question 1 Theorem (Dickenstein-DR-Morrison 2019) MD r , A = D Cayley ( r , A ) Definition Let A ⊂ Z d , such that D A � = 1 , deg ( D A ) = δ, and let ( λ 0 , . . . , λ r ) ∈ C r + 1 . Define the iterated discriminant as: ID r , A = D δ ∆ r ( D A ( λ 0 f 0 + . . . + λ r f r )) Abuse of notation: f i = ( c i 0 , . . . , c i N ) deg ( I r , A ) = δ ( δ − 1 )( r + 1 ) 25/28

Answer to question 1 Theorem (Dickenstein-DR-Morrison) A ⊂ Z n , D A � = 1 and 0 � r � n. Then, the mixed discriminant MD r , A � = 1 divides the iterated discriminant ID r , A . Moreover, A ( sing ( X ∗ 1. If codim X ∗ A )) > r, ID r , A = MD r , A . � ℓ i = 1 Ch µ i A ( sing ( X ∗ 2. If codim X ∗ A )) = r, ID r , A = MD r , A Y i , where Y 1 , . . . , Y ℓ are the irreducible components of sing ( X ∗ A ) of maximal dimension r, with respective multiplicities µ i . A ( sing ( X ∗ 3. If codim X ∗ A )) < r, ID r , A = 0 . 26/28

Answer to question 2 Theorem (Dickenstein-DR-Morrison) Let Q 1 , Q 2 be two d-dimensional quadric hypersurfaces then: Q 1 ∩ Q 2 singular if and only if I 1 , 2 ∆ d = MD 1 , 2 ∆ d = 0 27/28

n >> 1 28/28