Contact in algebraic and tropical geometry Damiano Testa University of Warwick S´ eminaire d´ ematerialis´ e de “Alg` ebre and G´ eom´ etrie” Universit´ e de Versailles March 31, 2020 @ 11am (GMT+2) Stream here Damiano Testa (Warwick) Social distancing March 31, 2020 1 / 22
Outline 1 Overview Background Inflection points and bitangent lines of plane curves 2 An elementary approach 3 Inflection points and inflection lines 4 Bitangent lines 5 Further directions Damiano Testa (Warwick) Social distancing March 31, 2020 2 / 22
Goal Reconcile classical constructions in recent results in algebraic geometry over and tropical geometry the complex numbers via positive characteristic . Based on joint work with Marco Pacini (UFF, Rio de Janeiro). Damiano Testa (Warwick) Social distancing March 31, 2020 3 / 22
Conventions Curve usually means a projective, plane curve C ⊂ P 2 k over a field k . Often, a curve C is general among plane curves of the same degree as C . In particular, there is no loss in thinking that curves are smooth. The field k can be taken to be algebraically closed. Important characteristics of fields in this talk are 0, 3 and 2. Damiano Testa (Warwick) Social distancing March 31, 2020 4 / 22
Inflection points of plane curves An inflection point of a curve C is a point a ∈ C at which the tangent line to C meets the curve C with multiplicity at least 3. The Fermat cubic x 3 + y 3 + z 3 = 0 and its 9 inflection points: with ζ 3 + 1 = 0 . [0 , 1 , ζ ] , [1 , 0 , ζ ] , [1 , ζ, 0] , Damiano Testa (Warwick) Social distancing March 31, 2020 5 / 22
Bitangent lines of plane curves A bitangent line of a curve C is a line ℓ ⊂ P 2 k that is tangent to C at two distinct points. The quartic ( x 2 − z 2 ) 2 = y (2 z 3 − xz 2 + yz 2 − xy 2 ) and its bitangent line y = 0. Damiano Testa (Warwick) Social distancing March 31, 2020 6 / 22
Complex and tropical geometry Two starting points. Theorem. � plane curve over C � 3 d ( d − 2) � � A of degree d has inflection tropical plane curve d ( d − 2) points. Theorem. � plane quartic over C � 28 � � A has bitangent lines. tropical plane quartic 7 (Recall: curve means general curve .) Damiano Testa (Warwick) Social distancing March 31, 2020 7 / 22
Algebraic and tropical geometry Let k and l be algebraically closed fields of characteristics 3 and 2. Our observations are underlined. Theorem. � plane curve over C � 3 d ( d − 2) � � A plane curve over k of degree d has d ( d − 2) inflection tropical plane curve d ( d − 2) points. Theorem. � plane quartic over C � 28 � � A plane quartic over l has 7 bitangent lines. tropical plane quartic 7 Damiano Testa (Warwick) Social distancing March 31, 2020 8 / 22
Inflection points Arguing via the real numbers, Klein, Ronga, Schuh, Viro, Brugall´ e, L´ opez de Medrano,. . . address the factor 3 between the 3 d ( d − 2) complex and the d ( d − 2) tropical inflection points. Theorem. � � plane curve over R A general of degree d has at most d ( d − 2) tropical plane curve � � real inflection points. The upper bound is achieved. distinct tropical Takeaway For each real inflection point, there are two further complex conjugate inflection points. Reading off real multiplicities in tropical geometry is hard! Damiano Testa (Warwick) Social distancing March 31, 2020 9 / 22
An elementary approach We propose a local approach in positive characteristic to explain geometrically the discrepancy between the complex and the tropical counts. The intuition is that the contact multiplicities interact with the characteristic of the field. For instance, working with inflection points, we reduce modulo 3; working with bitangent lines, we reduce modulo 2. The method has the potential for broader applications. Damiano Testa (Warwick) Social distancing March 31, 2020 10 / 22
Basic computation: multiple roots Lemma. Let k be a field and let f ( x ) ∈ k [ x ] a polynomial with a root α of multiplicity m . The gcd( f, f ′ ) is divisible by ( x − α ) m − 1 ; divisible by ( x − α ) m if and only if char k | m . Proof. Write f ( x ) = ( x − α ) m g ( x ) , with g ( x ) ∈ k [ x ], and g ( α ) � = 0. Compute f ′ ( x ) ( x − α ) m − 1 ( mg ( x ) + ( x − α ) g ′ ( x )) . = Thus ( x − α ) m − 1 divides f ′ ( x ) and ( x − α ) m divides f ′ ( x ) ⇐ ⇒ ( x − α ) divides mg ( x ) ⇐ ⇒ m = 0 in k . Damiano Testa (Warwick) Social distancing March 31, 2020 11 / 22
Basic computation: multiple roots Conclusion Let k be a field, let p be a prime number and let n ≥ p be an integer. There is a rational function r p ( f 0 , . . . , f n ) in ( n + 1) variables with the following property. Assume that the polynomial f = f 0 x n + f 1 x n − 1 + · · · + f n has a unique root α of multiplicity at least 2 and that the multiplicity of α is p . If char k � = p , then r p ( f ) = α . If char k = p , then r p ( f ) = α p . Proof. Use the previous lemma and induction to show � x − α, if char k � = p ; � f, f ′ , . . . , f ( p − 1) � gcd = ( x − α ) p = x p − α p , if char k = p. (Derivatives = Hasse derivatives) Compute gcd via Euclid’s Algorithm. Damiano Testa (Warwick) Social distancing March 31, 2020 12 / 22
Inflection points and inflection lines Back to inflection points of curves in P 2 k . Fix a curve C ⊂ P 2 k . Define the incidence correspondence: � � � x is an inflection point of C , � ( x, ℓ ) ∈ P 2 k × ( P 2 k ) ∨ F C = � � ℓ is the tangent line to C at x . � k ) ∨ is projective plane dual to P 2 where ( P 2 k . k ) ∨ correspond to lines in P 2 (Points of ( P 2 k .) Question. Can we reconstruct F C from either the set of inflection points or the set of inflection lines of C ? Damiano Testa (Warwick) Social distancing March 31, 2020 13 / 22
Inflection points and inflection lines (= ⇒ ) From inflection points to inflection lines. � � � x is an inflection point of C , � ( x, ℓ ) ∈ P 2 k × ( P 2 k ) ∨ F C = � � ℓ is the tangent line to C at x . � Fix an inflection point x . We can reconstruct the corresponding inflection line , by computing the tangent line to C at x . Thus, F C has as many elements as C has inflection points : # F C = # { inflection points of C } . Damiano Testa (Warwick) Social distancing March 31, 2020 14 / 22
Inflection points and inflection lines ( ⇐ =) From inflection lines to inflection points. � � � x is an inflection point of C , � ( x, ℓ ) ∈ P 2 k × ( P 2 k ) ∨ F C = � ℓ is the tangent line to C at x . � � Fix an inflection line ℓ to the curve C . As C is general , the inflection line ℓ is tangent to C at just one point x and the intersection multiplicity between ℓ and C at x is exactly 3. Thus, a polynomial F vanishing on C restricts to a polynomial F | ℓ on ℓ ≃ P 1 k with the following properties: F | ℓ has a unique repeated root, corresponding to the inflection point of C on ℓ ; the multiplicity of the repeated root is 3. Damiano Testa (Warwick) Social distancing March 31, 2020 15 / 22
Inflection points and inflection lines Inflection lines to inflection points. Fix an inflection line ℓ to C . As C is general [. . . ], we find a polynomial F | ℓ satisfying: F | ℓ has a unique repeated root α , corresponding to the inflection point of C on ℓ ; the multiplicity of the repeated root α is 3. We have seen earlier to what extent we can reconstruct α from F ! If char k � = 3, then we can reconstruct α from F | ℓ and we deduce that # F C = # { inflection lines of C } . If char k = 3, then we can reconstruct α 3 from F | ℓ and we deduce that # F C = 3 · # { inflection lines of C } . Damiano Testa (Warwick) Social distancing March 31, 2020 16 / 22
� � Summary for inflection points � � x is an inflection point of C, � ( x, ℓ ) ∈ P 2 k × ( P 2 k ) ∨ � F C = � ℓ is the tangent line to C at x. � char k � =3 , birational char k =3 , purely inseparable birational of degree 3 � � � x inflection � “Gauss map” � � k ) ∨ � � � ℓ inflection x ∈ P 2 ℓ ∈ ( P 2 � � k point of C line of C Amusing consequence. (If you happen to like imperfect fields) Let k be a separably closed field of characteristic 3. Let C be a general plane curve defined over k . The coordinates of the inflection lines are contained in k . 1 3 . The coordinates of the inflection points are contained in k Damiano Testa (Warwick) Social distancing March 31, 2020 17 / 22
� � Bitangent lines Similarly for bitangent lines. Fix a plane quartic C ⊂ P 2 k . � � � x is point of bitangency of C, ( x, ℓ ) ∈ P 2 k × ( P 2 k ) ∨ � � ℓ is the (bi)tangent line to C at x. � char k � =2 , double cover char k =2 , purely inseparable birational of degree 4 “Gauss map” � � � � x bitangent � � k ) ∨ � � � ℓ bitangent x ∈ P 2 ℓ ∈ ( P 2 � � k point of C line of C In char 2, contact points contribute 2 each to the inseparable degree. Thus, bi tangents give a 4 : 1 degree ratio. The 28 bitangents of plane quartics over C , correspond to the 7 = 28 4 bitangents in characteristic 2. Damiano Testa (Warwick) Social distancing March 31, 2020 18 / 22
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