On caustics by reflection of algebraic surfaces
OON CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES
ALFREDERIC JOSSE AND FRANC¸ OISE P`ENE
Abstract.
Given a point S (the light position) in P and an algebraic surface Z (the mirror)of P , the caustic by reflection Σ S ( Z ) of Z from S is the Zariski closure of the envelope of thereflected lines R m got by reflection of ( Sm ) on Z at m ∈ Z . We use the ramification methodto identify Σ S ( Z ) with the Zariski closure of the image, by a rational map, of an algebraic 2-covering space of Z . We also give a general formula for the degree (with multiplicity) of caustics(by reflection) of algebraic surfaces of P . Introduction
Let S [ x : y : z : t ] ∈ P := P ( W ) (with W a 4-dimensional complex vector space) and let Z = V ( F ) be a surface of P given by some F ∈ Sym d ( W ∨ ) (i.e. F corresponds to a polynomialof degree d in C [ x, y, z, t ]). The caustic by reflection Σ S ( Z ) of Z from S ∈ P is the Zariskiclosure of the envelope of the reflected lines R m of the lines ( mS ) after reflection at m on themirror surface Z .Since the seminal work of von Tschirnhaus [14, 15], caustics by reflection of planar curveshave been studied namely by Chasles [6], Quetelet [12] and Dandelin [7]. Let us also mentionthe work of Bruce, Giblin and Gibson [3, 1, 2] in the real case. A precise computation of thedegree and class of caustics by reflection of planar algebraic curves has been done in [9, 10, 11].The idea was based on the fact that the caustic by reflection of an irreducible algebraic curve C of P from source S ∈ P is the Zariski closure of the image of C by a rational map. Moreover,in the planar case, the generic birationality of the caustic map has been established in [11, 4].The study of caustics by reflection of algebraic surfaces is more delicate. We will see that ageneric point m of Z is associated to two (instead of a single one) points on Σ S ( Z ).A classical way to study envelopes is the ramification theory. Let us mention that thisapproach has been used namely by Trifogli in [16] and by Catanese and Trifogli in [5] for focalloci (which generalize the notion of evolute to higher dimension). We use here the ramificationtheory to construct the caustic by reflection Σ S ( Z ) and to identify it with the Zariski closure ofthe image by some rational map Φ, of an algebraic 2-covering space ˆ Z of Z . We will see that,contrary to the case of caustics by reflection of planar curves, the set of base points of Φ | ˆ Z is never empty . We give a general formula expressing the degree (with multiplicity) mdegof Σ S ( Z ) in terms of intersection numbers of Z with a particular curve (called reflected polarcurve) computed at the projection on Z of the base points of Φ | ˆ Z . As a consequence of ourgeneral result, we prove namely the following generic result (see Theorem 37 for precisions). Theorem 1.
Let d ≥ . For a generic irreducible surface Z ⊂ P of degree d and for a generic S ∈ P , we have mdeg Σ S ( Z ) = d ( d − d − . Date : February 27, 2019.2000
Mathematics Subject Classification.
Key words and phrases. caustic, class, polar, intersection number, pro-branchFran¸coise P`ene is supported by the french ANR project GEODE (ANR-10-JCJC-0108). a r X i v : . [ m a t h . AG ] J un N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 2
We denote by H ∞ the plane at infinity of P : H ∞ = { [ x : y : z : t ] ∈ P : t = 0 } and Z ∞ := Z ∩ H ∞ . In this study the umbilical curve C ∞ plays a particular role. Recall that C ∞ is the intersection of H ∞ with any sphere (see Section 1).In practice, the degree of the caustic will namely depend on the position of S with respect tothe surface Z , to H ∞ , to C ∞ and to the isotropic tangent planes to Z (see Section 1 for thenotion of isotropic planes).We illustrate this by a precise study of the degrees of caustics of a paraboloid Z . In thiscase, Z ∞ is the union of two lines intersecting at the focal point at infinity. We will see that thecaustic of the paraboloid is a surface if the light position is outside Z ∞ and outside the focalpoints of Z .Still in the case of the paraboloid, Z ∩ C ∞ is made of two points I and J and the tangentplanes to Z at these two points are isotropic. Moreover the revolution axis D of the paraboloidis the intersection of these two tangent planes. Proposition 2.
Let Z be the paraboloid V ( x + y − zt ) ⊂ P of axis D = V ( x, y ) and let S ∈ P .If S is a focal point of the paraboloid (either F [0 : 0 : 1 : 0] or F [0 : 0 : 1 : 2] ), then Σ S ( Z ) is reduced to the other focal point.If S ∈ Z ∞ \ { F } , then Σ S ( Z ) is a planar curve of degree 2.If S ∈ D , then the degree of mdeg Σ S ( Z ) = 4 if S [0 : 0 : 0 : 1] and mdeg Σ S ( Z ) = 6 elsewhere.Assume now that S is not on D ∪ Z ∞ ∪ { F } .If S is neither at infinity nor on the paraboloid Z , then: • If x + y = 0 and z (cid:54) = t / (i.e. if S is on T I Z or on T J Z and on two other isotropictangent planes to Z ), then mdeg Σ S ( Z ) = 14 . • If x + y = 0 and z = t / (i.e. if S is on T I Z or on T J Z and on another isotropictangent plane to Z ), then mdeg Σ S ( Z ) = 12 . • If x + y (cid:54) = 0 and x + y + ( z − t / = 0 (i.e. if S is on three isotropic planes butneither on T I Z nor on T J Z ), then mdeg Σ S ( Z ) = 17 . • Otherwise (generic case: S is on four isotropic tangent planes to Z ), then mdeg Σ S ( Z ) =18 .If S is at infinity, then: • If S (cid:54)∈ C ∞ , then mdeg Σ S ( Z ) = 12 . • If S ∈ C ∞ , then mdeg Σ S ( Z ) = 6 .If S is on Z , then: • If x + y + t = 0 (i.e. if the tangent plane to Z at S is isotropic), then mdeg Σ S ( Z ) = 12 . • If x + y = 0 (i.e. if S is on T I Z or on T J Z ), then mdeg Σ S ( Z ) = 14 . • Otherwise mdeg Σ S ( Z ) = 16 . The paper is organized as follows. Section 1 is devoted to the (complex) projectivization oforthogonality in the real euclidean affine 3-space (which plays a crucial role in the present work)and its link with the umbilical curve. In Section 2, we construct the reflected lines. In Section3, we use the reflected lines and the ramification method to define the caustic by reflection. InSection 4, we define the appropriate 2-covering ˆ Z of Z and the rational map Φ. In Section 5, wedetermine precisely the base points of Φ | ˆ Z . We define the reflected polar in section 6 and use it N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 3 in Section 7 to establish a formula for the degree of the caustic by reflection. In Section 8, weprove Theorem 1. In Section 9, we prove Proposition 2. In Section 10, we precise a significativedifference between the caustic by reflection studied in this paper and the focal loci of genericvarieties considered in [16, 5]. In appendix A, we study two families of caustics by reflection ofsurfaces which are related to caustics by reflection of planar curves.1.
Affine and projective perpendicularity, link with umbilical conjugation
Consider the real euclidean affine 3-space E of direction the 3-vector space E (endowedwith some fixed basis). Let W := ( E ⊕ R ) ⊗ C (endowed with the induced basis). Let j : E (cid:44) → P := P ( W ) be the natural map defined on coordinates by j ( x, y, z ) := [ x : y : z : 1]for every m ( x, y, z ) ∈ E . We are interested in the interpretation in the plane at infinity of P of perpendicularity at a point of two affine subvarieties of E . Consider the two followingquadratic forms q ( x, y, z ) = x + y + z on E ⊗ C and Q ( x, y, z, t ) = x + y + z on W . Definition 3.
The umbilical curve of P is the irreducible conic C ∞ := V ( Q |H ∞ ) ∼ = V ( q ) ⊂ P ( E ⊗ C ) . We call cyclic point any point of C ∞ . We recall that every (complex projectivized) sphere contains C ∞ . It is worth noting that, forevery m ∈ E , we have the following classical diagram E j (cid:44) → P ( W ) Π ←− W \{ } ξ m (cid:38) H ∞ where Π is the canonical projection and with ξ m is defined on coordinates by ξ m ( m + ( x, y, z )) =[ x : y : z : 0]. Given any vector subspace V ⊂ E , the projective subspace V := j ( m + V ) of P (where K denotes the Zariski closure of K ) is the complex projectivization of the affine subspace V = m + V of E . We observe that ξ m ( V ) is V ∞ := V ∩ H ∞ .An affine line L (resp. an affine plane H ) containing m ∈ E is defined by m + V (resp. m + V ) with V i an i -dimensional subspace of E . Recall that the (complex) projectivization L of L (resp. H of H ) is the projective line (resp. plane) of P of equations obtained byhomogeneization of the equations of L (resp. H ).Hence, two lines L, L (cid:48) containing m are perpendicular at m if and only if their points atinfinity are conjugated with respect the conic C ∞ .A line L and a plane H containing m are perpendicular if and only if H ∞ is the polar of (cid:96) ∞ with respect to the conic C ∞ in H ∞ ∼ = P . This leads to the following definition of projectivenormal lines to a plane. Definition 4.
Let H = V ( h ) ⊂ P (with h ∈ W ∨ \ { } ) be a projective plane and m ∈ H \ H ∞ .The normal line N m ( H ) to H at m is the line containing m and n ∞ ( H ) := Π( κ ( ∇ h )) with κ : W → W defined on coordinates by κ ( a, b, c, d ) := ( a, b, c, . Remark 5.
Given a projective plane
H ⊂ P ( H (cid:54) = H ∞ ), if n ∞ ( H ) = [ u : v : w : 0] lies onthe umbilical (i.e. ( u, v, w ) lies on the isotropic cone V ( q ) in E ⊗ C ), then the line H ∞ is tangent to C ∞ at n ∞ ( H ) in H ∞ . In this case we have N m ( H ) ⊂ H . Let m = Π( m ) be a non singular point of Z \ H ∞ . We write T m ( Z ) for the projectivetangent plane at m to Z . We also define the projective normal line N m ( Z ) at m to N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 4 Z is the projective normal line to T m ( Z ) at m , i.e. N m ( Z ) is the line containing m and n ∞ ,m ( Z ) = Π( κ ( ∇ F ( m ))).Observe that the line at infinity T ∞ ,m ( Z ) of T m ( Z ) is the polar of the point at infinity n ∞ ,m ( Z ) of N m ( Z ) with respect the conic C ∞ .Later, we will see that the base points of the reflected map can be seen on the geometry onthe normals at infinity with respect to the umbilical. In particular isotropic tangent plane to Z containing S will play some role. Definition 6.
A plane H = V ( h ) (with h ∈ W ∨ \ { } ) is said to be isotropic if ∇ h is anisotropic vector for Q . Remark 7.
A plane
H ⊂ P is isotropic if and only if either it is the plane at infinity H ∞ orif n ∞ ( H ) is in C ∞ (i.e. H contains its normal lines). In particular, the surface Z admits an isotropic tangent plane at one of its nonsingular point m [ x : y : z : 1] if and only if m belongs to V ( Q ( ∇ F ) , F ). We note that the whole curve C ∞ iscontained in every complex projectivized sphere S r and that we have N m ( S r ) ⊂ T m ( S r ) for all m ∈ S r \ H ∞ . This is also true for tori.Consider some particular points on Z , playing a particular role in the construction of thecaustic map. Let B := V ( F, ∆ S F, Q ( ∇ F )) in P , the interpretation in the plane at infinity isthe following one. Let m be a nonsingular point of Z \ H ∞ then m ∈ B ⇐⇒ (cid:26) S ∈ T m ( Z ) n ∞ ,m ( Z ) ∈ C ∞ ⇐⇒ (cid:26) ( mS ) ⊂ T m ( Z ) n ∞ ,m ( Z ) ∈ C ∞ ⇐⇒ (cid:26) ( mS ) ∞ ∈ T ∞ ,m ( Z ) T ∞ ,m ( Z ) = T n ∞ ,m ( Z ) ( C ∞ ) . (1)We observe that B is in general a finite set, but that, for the unit sphere, B is a curve (thecircle apparent contour of Z seen from S ).Let us now specify some additional notations used in this paper. We write S ( x , y , z , t ) ∈ W \ { } . For any m [ x : y : z : t ] ∈ P , we will write m ( x, y, z, t ) ∈ W \ { } . For any d (cid:48) ≥ G ∈ Sym d (cid:48) ( W ∨ ), we write as usual G x , G y , G z , G t ∈ Sym d (cid:48) − ( W ∨ ) for the partialderivatives of in x , y , z and t respectively.2. Reflected lines
The incident lines are the lines (
S m ) with m ∈ Z . We will define the reflected line R m asthe orthogonal symmetric of ( S m ) with respect to the tangent plane to Z at m . To this end,we will define the orthogonal symmetric σ ( m ) of S with respect to the tangent plane to Z at m . Let us first explain how one can give a sense to the notion of orthogonal symmetries in P by complex projectivization of the euclidean affine situation.2.1. Orthogonal symmetric and map σ . To every injective linear map W f → W , corre-sponds a unique morphism P ( W ) P ( f ) → P ( W ). Therefore, to every injective affine map E g → E ,corresponds a unique algebraic map P ( W ) ι ( g ) → P ( W ). This defines an injective groups homomor-phism ι : Af f ( E ) ∼ = E (cid:111) Gl ( E ) → P ( Gl ( W )) such that ι ( Is ( E )) = ι ( E (cid:111) O ( E )) ⊂ P ( O ( ˆ Q )),with ˆ Q = x + y + z + t on W . We apply this to the orthogonal symmetry s H with respectto some affine plane H = V (˜ h ) ⊆ E with ˜ h = ax + by + cz + d . Recall that s H is defined by N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 5 s H ( P ) = P − h ( P ) ∇ ˜ hq ( ∇ ˜ h )) . This leads to the morphism s H := ι ( s h ) : P → P defined by P ( s h )with ∀ P ∈ W , s h ( P ) := Q ( ∇ h ) · P − h ( P ) · κ ( ∇ h ) ∈ W , with H = V ( h ) ⊂ P and with h = ax + by + cz + dt the homogeneized of ˜ h . Now we extendthis definition to any projective plane H ⊂ P as follows. Definition 8.
Consider a plane H = V ( h ) ⊆ P (with h ∈ W ∨ \ { } ). We define the orthogonalsymmetry s H with respect to H as the rational map given by s H = P ( s h ) with ∀ P ∈ W , s h ( P ) := Q ( ∇ h ) · P − h ( P ) · κ ( ∇ h ) ∈ W . We can notice that, when
H (cid:54) = H ∞ , s H ( P ) is well defined in P except if H is an isotropicplane containing P (see Proposition 9). For any non singular m [ x : y : z : t ] ∈ Z , we define σ ( m ) := s T m Z ( S ) = P ( σ )( m ) with σ := Q ( ∇ F ) · S − S F · κ ( ∇ F ) ∈ W (2)on Π − ( Z ) with ∆ S F the equation of the polar hypersurface of P S ( Z ) given by ∆ S F := DF · S (where DF is the differential of F ). We extend the definition of σ ( m ) to any m ∈ W \ { } .Observe that σ defines a unique rational map σ : P → P . Proposition 9.
The base points of the rational map σ |Z are the singular points of Z , the pointsof tangency of Z with H ∞ and the points at which Z has an isotropic tangent plane containing S .Proof. We prove that the base points of σ are the points of P such that F x = F y = F z = 0 orsuch that Q ( ∇ F ) = 0 and ∆ S F = 0. It is easy to see that these points are base points of σ .Now let m = [ x : y : z : t ] be a point of P such that σ ( m ) = 0. • If ∆ S ( F ) = 0, then, since S (cid:54) = 0, we get that Q ( ∇ F ) = 0. • If Q ( ∇ F ) = 0, then either ∆ S F = 0 or κ ( ∇ F ) = 0. • Assume now that Q ( ∇ F ) (cid:54) = 0. We have Q ( ∇ F ) · S = 2∆ S F · κ ( ∇ F ). This impliesthat κ ( ∇ F ) is non zero and proportional to S (which is also non zero), so that t = 0and 0 = y F x − x F y = z F y − y F z = x F z − z F x . Therefore, writing σ ( i ) for the i thcoordinate of σ , we have0 = σ (1) = Q ( ∇ F ) x − x F x + y F x F y + z F x F z )= Q ( ∇ F ) x − x F x + x F y + x F z ) = − Q ( ∇ F ) x . In the same way, we get 0 = σ (2) = − Q ( ∇ F ) y and 0 = σ (3) = − Q ( ∇ F ) z . Thiscontradicts the fact that Q ( ∇ F ) (cid:54) = 0 (since S (cid:54) = 0). (cid:3) Remark 10.
Each σ ( i ) belongs to Sym d − ( W ∨ ) . Moreover, for a general ( Z , S ) , the set V ( F, F x , F y , F z ) is empty and the base points of σ |Z are the d ( d − points of V ( F, Q ( ∇ F ) , ∆ S F ) . Reflected lines.Definition 11.
For any m ∈ Z , the reflected line R m on Z at m is the line ( mσ ( m )) whenit is well defined. Definition 12.
We write M S, Z for the set of points m ∈ P such that m and σ ( m ) areproportional, i.e. M S, Z := { m ∈ P : ∃ [ λ : λ ] ∈ P , λ · m + λ · σ ( m ) = 0 } . Observe that R m is well defined if m ∈ Z \ M S, Z . N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 6
Proposition 13.
We have
Z ∩ M S, Z = Z ∩ (Base( σ ) ∪ { S } ∪ W ) , with W := { m ∈ Z : m = n ∞ ,m ( Z ) , ∆ S F ( m ) (cid:54) = 0 , Q ( m ) = 0 } , with n ∞ ,m ( Z ) := Π( κ ( ∇ F ( m ))) .Proof. We prove
Z ∩ M S, Z ⊆ Z ∩ (Base( σ ) ∪ { S } ∪ W ), the inverse inclusion being clear. Let m ∈ ( Z ∩ M S, Z ) \ Base( σ ). Observe that, due to the Euler identity, we have 0 = DF ( m ) · m and so 0 = DF · σ = − ∆ S F · Q ( ∇ F ) . If ∆ S F = 0, then σ = Q ( ∇ F ) · S , so m = σ ( m ) = S .If Q ( ∇ F ) = 0, then σ = − S F · κ ( ∇ F ). So m = σ ( m ) = n ∞ ,m ( Z ); moreover ∆ S F (cid:54) = 0 and Q = 0. (cid:3) Lemma 14. If dim M S, Z = 3 , then Z = H ∞ or V (∆ S F, Q ( ∇ F )) = P .Proof. Due to Proposition 13, we have
Z ∩ M S, Z ⊆ Z ∩ (Base( σ ) ∪ { S } ∪ C ∞ ). Assume thatdim M S, Z = 3. This implies that Base( σ ) = P . So, due to the proof of Proposition 9, weconclude that P = V ( F x , F y , F z ) ∪ V (∆ S F, Q ( ∇ F )). So, either P = V ( F x , F y , F z ) (whichimplies Z = H ∞ ) or P = V (∆ S F, Q ( ∇ F )). (cid:3) Caustic by reflection
Now, let us introduce some additional notations. We define N S ( m ) as the complexifiedhomogenized square euclidean norm of Sm by N S ( m ) := ( xt − x t ) + ( yt − y t ) + ( zt − z t ) . We will also consider the bilinear Hessian form Hess F of F and its determinant H F . Let ussee how to construct two maps ψ = ψ ± : Z → P such that the surface ψ ( Z ) is tangent to thereflected line R m at ψ ( m ), for a generic m ∈ Z . Observe first that ψ ( m ) is in R m implies that ψ ( m ) can be rewritten ψ ( m ) = λ ( m ) · m + λ ( m ) · σ ( m ) ∈ W \ { } , with [ λ ( m ) : λ ( m )] ∈ P for every m ∈ Z . The main result of this section is the next theoremspecifying the form of λ and λ (belonging to an integral extension of the ring Sym ( W ∨ ))which ensures that, for a generic m ∈ Z , R m is tangent to ψ ( Z ) at ψ ( m ). Theorem 15.
Let ψ : U → P (with U ⊆ Z ) be given by ψ ( m ) = λ ( m ) · m + λ ( m ) · σ ( m ) ∈ W , with λ ( · ) and λ ( · ) in an integral extension of Sym ( W ∨ ) such that α ( m )( λ ( m )) + β ( m ) λ ( m ) λ ( m ) + γ ( m )( λ ( m )) = 0 (3) with α, β, γ ∈ Sym ( W ∨ ) given by α := ∆ S F ∈ Sym d − ( W ∨ ) , (4) β := − (cid:2) Hess F ( S , σ ) + (∆ S F ) ( F xx + F yy + F zz ) (cid:3) ∈ Sym d − ( W ∨ ) (5) and γ := − S F ( d − N S H F ∈ Sym d − ( W ∨ ) . (6) Then, for every m ∈ Z \ V ( tQ ( ∇ F )) , the reflected line R m is tangent to ψ ( Z ) at ψ ( m ) . N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 7
It will be useful to introduce ∀ ( m , λ , λ ) ∈ W × C , Q S ,F ( m , λ , λ ) = α ( m ) λ + β ( m ) λ λ + γ ( m ) λ . One may notice that, for a fixed m , Q S ,F ( m , λ , λ ) is a quadratic form in ( λ , λ ). Roughlyspeaking, Theorem 15 states that the image of Z by ψ ( · ) = λ ( · ) · Id + λ ( · ) · σ ( · ) (for some λ , λ ∈ Sym ( W ∨ )[ (cid:112) β − αγ ]) corresponds to a part of the envelope of the reflected lines R m .More precisely: Definition 16.
The caustic by reflection Σ S ( Z ) of Z from S is the Zariski closure of thefollowing set { P ∈ P : ∃ m ∈ Z , ∃ [ λ : λ ] ∈ P , Q S ,F ( m , λ , λ ) = 0 and P = λ · m + λ · σ ( m ) } . Remark 17. If Z ⊆ V (∆ S F, ( F x + F y + F z ) Hess F ( S , S )) , then (3) becomes on Z and σ S, Z ( Z ) is either { S } or empty. If it is S (i.e. if ∆ S F = 0 in C [ x, y, z, t ] and if Z (cid:54)⊆ V ( F x + F y + F z ) ), we set Σ S ( Z ) = { S } . Theorem 15 states that the points of the caustic Σ S ( Z ) corresponding to m ∈ Z are thepoints of coordinates ψ ± ( m ) with ψ ± ( m ) = (cid:16) ˜ β ( m ) ± (cid:112) ϑ ( m ) (cid:17) · m + ∆ S F ( m ) · σ ( m ) ∈ C , (7)with ˜ β := − β/ ϑ := ˜ β − αγ . Let us observe that if ϑ is a square in C [ x, y, z, t ] / ( F ), then,on Z , (7) corresponds to two rational maps ψ ± : P → P and the caustic by reflection Σ S ( Z )is the union of the Zariski closures of ψ + ( Z ) and of ψ − ( Z ). Let us give some examples. Example 18 (A singular caustic of the saddle surface) . Let us study the caustic by reflectionof Z = V ( xy − zt ) from S = [0 : 0 : 1 : 0] . Observe that α = − t , β = 0 and γ = 4 t . So(3) becomes λ − t λ = 0 (if t (cid:54) = 0 ). Hence Σ S ( Z ) is the union of the Zariski closure of theimages of Z by the two rational maps ψ ± : P (cid:55)→ P defined on coordinates by ψ ± ( x, y, z, t ) =(2 ty ± tx, tx ± ty, x + y − t ± tz, ± t ) . Noting that on Z , tz = xy , we obtain that theZariski closure of ψ ± ( Z ) is the parabola V ( X ∓ Y, ± T Z − ( X − T ) / and so that the caustic Σ S ( Z ) is the union of these two curves (which are parabolas contained in two orthogonal planes). Example 19 (The double-butterfly caustic of the saddle surface) . We are interested in thecaustic by reflection of Z = V ( xy − zt ) from S = [0 : 0 : 1 : 1] . We have α = − z − t , ˜ β = − x + y − zt ) and γ = 4( z + t )( x + y + z + t − tz ) and so ϑ = 4( x + y + z + t − z t + 2 x y + x z + y z + x t + y t ) . Since ϑ ( x, y, xy, is not a square in C [ x, y ] , weconclude that ϑ is not a square in C [ x, y, z, t ] / ( F ) . In this case, the coordinates of ψ ± are inan extension of C [ x, y, z, t ] and do not corresponds to rational maps on P (see Figure 1 for arepresentation of this caustic). Example 20.
Let Z be the paraboloid V (( x + y − zt ) / ⊂ P .The caustic by reflection Z from its focal point F = [0 : 0 : 1 : 0] is its other focal point F = [0 : 0 : 1 : 2] . This can be quickly shown with our Theorem 15 (since α = − t , β = − t and γ = 4 t and so (3) admits a unique solution [ λ : λ ] ∈ P which is [ − t : 1] . The uniqueramification point M m associated to m is then M m = [0 : 0 : − zt + x + y − t : − t ] = F (since x + y = 2 zt ).If light position S is another point of H ∞ , then ϑ = ( x + y )[( x + y + t ) ( x + y ) − x + y + t )( xx + yy − z t ) z t − xx + yy − z t ) t ] N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 8
Figure 1.
The picture on the right represents the caustic by reflection of V ( zt − xy ) from [0 : 0 : 1 : 0] (corresponding to the points of V ( zt − xy ) in thechart t = 1 for x, y ∈ [ − , is not a square in C [ x, y, z, t ] / ( F ) unless if x + y + z = 0 or x + y = 0 (see Proposition30 for the case when x + y = 0 and t = 0 ). The fact that ϑ is not a square means that thecaustic map Φ S,F cannot be decomposed in two rational maps on P . The end of this section is devoted to the proof of Theorem 15.
Proof of Theorem 15.
Let m [ x : y : z : t ] ∈ Z \ V ( t ( Q ( ∇ F ))). We will use several times theEuler identity ( xG x + yG y + zG y + tG t = d G if G is in Sym d ( W ∨ )). We use the idea oframification (used for example in [16, 5]). The points of the caustic corresponding to m are thepoints Π( λ · m + λ · σ ( m )) with [ λ : λ ] ∈ P such that the rank of the Jacobian matrix J of j : ( m , λ , λ ) (cid:55)→ ( λ · m + λ · σ ( m ) , F ( m ))is less than 5. We have J := λ + λ σ (1) x λ σ (1) y λ σ (1) z λ σ (1) t x σ (1) λ σ (2) x λ + λ σ (2) y λ σ (2) z λ σ (2) t y σ (2) λ σ (3) x λ σ (3) y λ + λ σ (3) z λ σ (3) t z σ (3) λ σ (4) x λ σ (4) y λ σ (4) z λ + λ σ (4) t t σ (4) F x F y F z F t , with σ ( i ) the i th coordinates of σ .(1) Let us explain this briefly. Let ψ ( · ) of the form ψ ( m (cid:48) ) = λ ( m (cid:48) ) · m (cid:48) + λ ( m (cid:48) ) · σ ( m (cid:48) ) . We define the following propertythe line ( mσ ( m )) is tangent to ψ ( Z ) at ψ ( m ) . (8)Recall that we have assumed ( Q ( ∇ F ))( m ) (cid:54) = 0. Assume for example F x ( m ) (cid:54) = 0 (theproof is similar if we replace F x by F y or by F z ). Now, Property (8) means that thereexists A ∈ W ∨ \ { } such that A ( m ) = 0 , A ( σ ( m )) = 0 , A (( D ψ ( m ) · F y ( m ) − F x ( m )00 ) = 0 , To prove that δ is not a square in C [ x, y, z, t ] / ( F ), it is enough to see that there exists no polynomial P ∈ C [ x, y ]such that ( P ( x, y )) = δ ( x, y, ( x + y ) / , N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 9 A (( D ψ ( m )) · F z ( m )0 − F x ( m )0 ) = 0 and A (( D ψ ( m )) · F t ( m )00 − F x ( m ) ) = 0 , and so that A ( m ) = 0 , A ( σ ( m )) = 0 , A ( ψ x ( m )) F y ( m ) = F x ( m ) A ( ψ y ( m )) ,A ( ψ x ( m )) F z ( m ) = F x ( m ) A ( ψ z ( m )) and A ( ψ x ( m )) F t ( m ) = F x ( m ) A ( ψ t ( m )) . Therefore, by taking b := A ( ψ x ( m )) /F x ( m ), A ( m ) = 0 , A ( ψ x ( m )) = F x ( m ) b, A ( ψ y ( m )) = F y ( m ) b,A ( ψ z ( m )) = F z ( m ) b and A ( ψ t ( m )) = F t ( m ) b and so that the rank of the following matrix is strictly less than 5ˆ J := (cid:18) ψ x ( m ) ψ y ( m ) ψ z ( m ) ψ t ( m ) m σ ( m ) F x F y F z F t (cid:19) ∈ M at , ( C ) . Let us write C i the i -th column of J . We observe that the four first columns of ˆ J arerespectively equal to C + ( λ ) x C + ( λ ) x C , C + ( λ ) y C + ( λ ) y C , C + ( λ ) z C +( λ ) z C and C + ( λ ) t C + ( λ ) t C . Therefore the J and ˆ J have the same rank and so(8) means that rank ( J ) < Z , xC + yC + zC + tC = λ C + λ C . Since t (cid:54) = 0, C isa linear combination of the other columns and so the rank of J is strictly less than 5 ifand only if the following determinant is null: D ( m , λ , λ ) := (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) λ + λ σ (1) x λ σ (1) y λ σ (1) z x σ (1) λ σ (2) x λ + λ σ (2) y λ σ (2) z y σ (2) λ σ (3) x λ σ (3) y λ + λ σ (3) z z σ (3) λ σ (4) x λ σ (4) y λ σ (4) z t σ (4) F x F y F z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Now let us define τ := Q ( ∇ F ) · S + 2 ( xt − x t ) F x + ( yt − y t ) F y + ( zt − z t ) F z t · κ ( ∇ F ) . Observe that τ = σ + t dFt κ ( ∇ F ) (due to the Euler identity). Therefore, on Z , we have σ = τ . Now we observe that, on Z , we have D ( m , λ , λ ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) λ + λ τ (1) x λ τ (1) y λ τ (1) z x τ (1) λ τ (2) x λ + λ τ (2) y λ τ (2) z y τ (2) λ τ (3) x λ τ (3) y λ + λ τ (3) z z τ (3) λ τ (4) x λ τ (4) y λ τ (4) z t τ (4) F x F y F z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , (9)with τ ( i ) the i th coordinate of τ . Indeed, if we write L i the i -th line of the matrix(with σ ) used in the definition of D and if we write ˜ L i the i -th line of the matrix(with τ ) appearing in the above formula, we obtain (due to the Euler identity) that,on Z , we have ˜ L = L , ˜ L = L and ˜ L = L + λ t dt F x L , ˜ L = L + λ t dt F y L ,˜ L = L + λ t dt F z L . N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 10 (3) On Z , we have D ( m , λ , λ ) = α ( m ) λ + β ( m ) λ λ + γ ( m ) λ , (10)where α , β and γ can be expressed as follows (due to Euler’s identity ensuring that − xF x t − yF y t − zF z t + tx F x + ty F y + tz F z = t ∆ S F on Z ) α := Q ( ∇ F ) t ∆ S F = tQ ( ∇ F ) α (11) β := − t Q ( ∇ F ) B (12) γ := − t − N S .Q ( ∇ F ) . ∆ S F.h F (13)with the following definitions of h F and B . First, on Z , we have h F := (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) F xx F xy F xz F x F xy F yy F yz F y F xz F yz F zz F z F x F y F z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = t ( d − H F , where H F is the Hessian determinant of F . Therefore γ = − t ( d − N S .Q ( ∇ F ) . ∆ S F.H F = tQ ( ∇ F ) γ. (14)Second B := δ x F xx + δ y F yy + δ z F zz + 2( ε x,y F xy + ε x,z F xz + ε y,z F yz ) , with δ x := ( x t − xt ) ( F y + F z ) + (( t y − ty ) F y + ( t z − tz ) F z ) = ( x t − xt ) ( F y + F z ) + ( t ( yF y + zF z ) − t ( y F y + z F z )) = ( x t − xt ) ( F y + F z ) + ( t ( xF x + tF t ) + t ( y F y + z F z )) = ( x t − xt ) ( F y + F z ) + (( t x − x t ) F x + t ∆ S F ) = x t ( F x + F y + F z ) + 2 xt t [ − x ( F x + F y + F z ) + F x ∆ S F ] ++ x t ( F x + F y + F z ) + t (∆ S F ) − x t F x ∆ S F = x t ( F x + F y + F z ) + t ∆ S F (2 xt F x − x tF x ) ++ t ( F x + F y + F z )( x t − xx t ) + t (∆ S F ) , Indeed, if we write ˆ C i for the i -th column of Hess F , due to the Euler formula, on Z , we have ˆ C = d − t ∇ F − ( x ˆ C + y ˆ C + z ˆ C ) /t (where ∇ F is the gradient of F ); therefore H F := (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) F xx F xy F xz F xt F xy F yy F yz F yt F xz F yz F zz F zt F xt F yt F zt F tt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = d − t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) F xx F xy F xz F x F xy F yy F yz F y F xz F yz F zz F z F xt F yt F zt F t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Now, if we write ˆ L i the i -th line of the above matrix, using again the Euler identity, on Z , we have ˆ L = d − t ( F x F y F z − ( x ˆ L + y ˆ L + z ˆ L ) /t and we get H F = ( d − h F /t . N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 11 δ y (resp. δ z ) being obtained from δ x by interverting x and y (resp. x and z ) and ε x,y := − ( t x − x t ) F y (( t x − x t ) F x + ( t z − z t ) F z ) − ( t y − y t ) F x (( t y − y t ) F y + ( t z − z t ) F z ) + ( t x − x t )( t y − y t ) F z = ( t x − x t ) F y [ t yF y + t tF t + x tF x + z tF z ] ++( t y − y t ) F x [ t xF x + t tF t + y tF y + z tF z ] ++( t x − x t )( t y − y t ) F z = ( t x − x t ) F y [ t yF y + t ∆ S F − ty F y ] + ( t y − y t ) F x [ t xF x + t ∆ S F − tx F x ] ++( t x − x t )( t y − y t ) F z = t xy ( F x + F y + F z ) + t ∆ S F (( t x − x t ) F y + ( t y − y t ) F x )+ t ( F x + F y + F z )( tx y − t ( y x + yx )) ε x,z (resp. ε y,z ) being obtained from ε x,y by interverting y and z (resp. x and z ). On Z ,we have0 = xF x + yF y + zF z + tF t and ( d − F w = xF xw + yF yw + zF zw + tF tw , ∀ w ∈ { x, y, z, t } . Therefore0 = x F xx + y F yy + z F zz + t F tt + 2( xyF xy + xzF xz + xtF xt + yzF yz + ytF yt + ztF zt )and so B = ( F x + F y + F z )( b + b + b ) + 2 t ∆ S F b + t (∆ S F ) ( F xx + F yy + F zz ) , with b = − t ( t F tt + 2 t ( xF xt + yF yt + zF zt )) = − t t (2( d − F t − tF tt ) ,b = t ( x F xx + y F yy + z F zz + 2 x y F xy + 2 x z F xz + 2 y z F yz ) ,b = − tt (cid:88) w ∈{ x,y,z } ( w ( xF xw + yF yw + zF zw ))= 2 tt (cid:88) w ∈{ x,y,z } ( w ( tF tw − ( d − F w )) ,b = (cid:88) w ∈{ x,y,z } F w (( t x − tx ) F xw + ( t y − ty ) F yw + ( t z − tz ) F zw )= (cid:88) w ∈{ x,y,z } F w ( t ( d − F w − t ( x F xw + y F yw + z F zw + t F wt )) . Putting all these terms together, we get that B is equal to t (cid:2) Q ( ∇ F ) · Hess F ( S , S ) − S F · Hess F ( S , κ ( ∇ F )) + (∆ S F ) ( F xx + F yy + F zz ) (cid:3) . and so B = t (cid:2) Hess F ( S , σ ) + (∆ S F ) ( F xx + F yy + F zz ) (cid:3) , which leads to β = tQ ( ∇ F ) β. (15)Hence the points of the caustic associated to m are the points Π( λ · m + λ · σ ( m ))where [ λ : λ ] ∈ P satisfies α ( m ) λ + β ( m ) λ λ + γ ( m ) λ = 0 , (16)with α , β and γ given by (11), (15) and (14). Now, since tQ ( ∇ F ) (cid:54) = 0, (16) meansthat α ( m ) λ + β ( m ) λ λ + γ ( m ) λ = 0. (cid:3) N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 12 Covering space ˆ Z and rational map ΦWe consider the algebraic covering space ˆ Z of Z given byˆ Z := { ( m, [ λ : λ ]) ∈ Z × P : Q S ,F ( m , λ , λ ) = 0 } . This set is a subvariety of a particular algebraic variety denoted F (3) ( − d + 3 ,
0) (by extendingthe notations used by Reid in [13, Chapter 2]) which corresponds to the cartesian product of sets P × P endowed with an unusual structure of algebraic variety based on the following definitionof multidegree multideg( P ) for P ∈ Sym ( W ∨ )[ λ , λ ] ∼ = C [ x, y, z, t, λ , λ ]:multideg( x a (cid:48) y b (cid:48) z c (cid:48) t d (cid:48) λ e (cid:48) λ f (cid:48) ) = ( a (cid:48) + b (cid:48) + c (cid:48) + d (cid:48) + (2 d − e (cid:48) , e (cid:48) + f (cid:48) ) . With this notion of multidegree, we have
Sym ( W ∨ )[ λ , λ ] = (cid:76) k,(cid:96) ≥ C k,(cid:96) , where C k,(cid:96) denotesthe homogeneous component of multidegree ( k, (cid:96) ). Now, we define F (3) ( − d +3 ,
0) as the quotientof W × C by the equivalence relation ∼ given by( x, y, z, t, λ , λ ) ∼ ( x (cid:48) , y (cid:48) , z (cid:48) , t (cid:48) , λ (cid:48) , λ (cid:48) ) ⇔ ∃ µ, ν ∈ C ∗ , ( x (cid:48) , y (cid:48) , z (cid:48) , t (cid:48) , λ (cid:48) , λ (cid:48) ) = ( µx, µy, µz, µt, µ d − νλ , νλ ) . We observe that H ( F (3) ( − d + 3 , P ∈ C [ x, y, z, t, λ , λ ] withhomogeneous multidegree multideg defined above.Now, since F ∈ Sym d ( W ∨ ), α ∈ Sym d − ( W ∨ ), β ∈ Sym d − ( W ∨ ) and γ ∈ Sym d − ( W ∨ ),we get that F and Q S ,F are in H ( F (3) ( − d +3 , Z is a subvariety of F (3) ( − d +3 , Z = { ( m, [ λ : λ ]) ∈ F (3) ( − d + 3 ,
0) : F ( m ) = 0 and Q S ,F ( m , λ , λ ) = 0 } . Since each coordinate of σ is in Sym d − ( W ∨ ), the map Φ : W × C → W given by Φ ( m , λ , λ ) := λ · m + λ · σ ( m ) ∈ ( C d − , ) defines a rational map Φ : X → P with X = { ( m, [ λ : λ ]) ∈ F (3) ( − d + 3 ,
0) : Q S ,F ( m , λ , λ ) = 0 } . Let us denote by B Φ | ˆ Z the set of base points of the map Φ | ˆ Z , i.e. B Φ | ˆ Z := { ( m, [ λ : λ ]) ∈ ˆ Z : Φ ( m , λ , λ ) = 0 } . We consider the canonical projection π : F (3) ( − d + 3 , → P (given by π ( m, [ λ : λ ]) = m ). Notation 21.
We write B := π ( B Φ | ˆ Z ) . Observe that, for any m ∈ B , there exists a unique [ λ : λ ] ∈ P such that λ · m + λ · σ ( m ) =0. This gives the following scheme B Φ | ˆ Z (cid:44) → ˆ Z (cid:44) → X Φ −→ P π | B Φ ↓ π | ˆ Z ↓ π |X ↓ B (cid:44) → Z (cid:44) → P Therefore B φ = B . Remark 22.
The caustic by reflection Σ S ( Z ) of Z from S satisfies Σ S ( Z ) = Φ( ˆ Z ) ⊆ P . Note that
B ⊆ M S, Z (with M S,Z defined in Definition 12). Due to the classical blowing-uptheorem, we obtain the following result valid in the general case.
N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 13
Proposition 23.
Assume that the set B is finite and that dim( Z ∩ M S, Z ) ≤ . Then thereexists δ ∈ N ∗ ∪ {∞} such that, for a generic point P ∈ Σ S ( Z ) , we have π (Φ − | ˆ Z ( { P } )) \ B ] = δ .Proof. Observe that, by hypothesis, the set B Φ | ˆ Z is finite. Now, applying the blowing-up resultgiven in [8, Example II-7.17.3], we get the existence of a variety (cid:101) ˆ Z and of two morphisms π : (cid:101) ˆ Z → ˆ Z and (cid:101) Φ : (cid:101) ˆ Z → P such that • π defines an isomorphism from π − ( ˆ Z \ B Φ ) onto ˆ Z \ B Φ , • On π − ( ˆ Z \ B Φ ), we have ˜Φ = Φ ◦ π , • (cid:101) Φ( (cid:101) ˆ Z ) is the Zariski closure of Φ( ˆ Z \ B Φ ), i.e. (cid:101) Φ( (cid:101) ˆ Z ) = Σ S ( Z ), • dim( (cid:101) ˆ Z ) = 2, • E := (cid:101) ˆ Z \ π − ( ˆ Z \ B Φ ) is a variety of dimension at most 1.Let δ be the degree of the morphism (cid:101) Φ. If δ = ∞ , then dim(Σ S ( Z )) ≤ dim( (cid:101) Φ( (cid:101) ˆ Z )) <
2. Assumenow that δ < ∞ . Since (cid:101) Φ is a morphism, every point of (cid:101) Φ( (cid:101) ˆ Z ) has δ preimages by (cid:101) Φ in (cid:101) Φ( (cid:101) ˆ Z ).Now, observe that dim(Σ S ( Z )) = 2 and that dim( (cid:101) Φ( E )) <
2. Therefore, a generic point ofΣ S ( Z ) is in Φ( ˆ Z ) \ [( Z ∩ M S, Z ) ∪ (cid:101) Φ( E )]. Let P in this set. We have δ = − |Z ( { P } )= { ( m, [ λ : λ ]) ∈ Z × P : Q S ,F ( m , λ , λ ) = 0 and Π( λ · m + λ · σ ( m )) = P } . Observe that, for m ∈ Z ∩ M S, Z , Φ( π − ( { m } )) = { m } by Definition 12. Since P (cid:54)∈ Z ∩ M S, Z ,we know that Φ − |Z ( { P } ) ∩ π − ( M S, Z ) = ∅ . So, for any m ∈ π (Φ − |Z ( { P } )), there exists aunique [ λ : λ ] ∈ P such that ( m, [ λ , λ ]) ∈ Φ − |Z ( { P } ). Therefore δ = π (Φ − |Z ( { P } )) = π (Φ − |Z ( { P } )) \ B ] (since π (Φ − |Z ( { P } )) ∩ B ⊂ π (Φ − |Z ( { P } )) ∩ M S, Z = ∅ . (cid:3) Lemma 24. If dim( Z ∩ M S, Z ) = 2 . Then Σ S ( Z ) = Z \ ( V ( β, γ ) \ V ( α )) . So if dim( Z ∩ M S, Z ) = 2, then Σ S ( Z ) = Z except if Z ⊂ V ( β, γ ) \ V ( α ) and in this last caseΣ S ( Z ) = ∅ . Proof of Lemma 24.
Assume that dim( Z ∩M S, Z ) = 2. Then Z ⊂ M S, Z . So, due to Proposition13, we have Z ⊂
Base( σ ). Now, due to Definition 16, we haveΣ S ( Z ) = { m ∈ Z : ∃ [ λ : λ ] ∈ P , Q S ,F ( m , λ , λ ) = 0 , λ (cid:54) = 0 } and the result follows. (cid:3) Base points of Φ Proposition 25.
The base points of Φ | ˆ Z are the points ( m, [ λ : λ ]) ∈ ˆ Z satisfying one of thefollowing conditions: (1) m ∈ V ( F, ∆ S F, Q ( ∇ F )) (i.e. m ∈ Sing ( Z ) or m is a point of tangency of Z with anisotropic plane containing S ) and λ = 0 , (2) t = F x = F y = F z = 0 and x + y + z = 0 (i.e. m is a cyclic point with T m Z = H ∞ )and λ = 0 , (3) t = F x = F y = F z = 0 (i.e. T m Z = H ∞ ) and H F = 0 and λ = 0 , (4) m = S ∈ Z and [ λ : λ ] is the unique element of P such that λ · m + λ Q ( ∇ F ) · S = 0 , N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 14 (5) m is a cyclic point (i.e. m ∈ C ∞ ), [ λ : λ ] = [2∆ S F ( F xx + F yy + F zz ) : 2 d − (cid:54) = [0 : 1] and ( F xx + F yy + F zz ) · m = (2 d − κ ( ∇ F ) .Proof. Let us prove that any base point ( m, [ λ : λ ]) has one of the form announced in thestatement of the proposition (the converse being direct). Let ( m, [ λ : λ ]) ∈ B Φ | ˆ Z . By definitionof Φ, we have 0 = λ · m + λ · σ ( m ). So m ∈ Z ∩ M S, Z and these m have been determined inProposition 13. • Assume first that σ ( m ) = 0. Then the unique [ λ : λ ] ∈ P satisfying λ · m + λ · σ ( m ) =0 is [ λ : λ ] = [0 : 1]. Hence λ = 0, λ (cid:54) = 0 and so Q S ,F ( m , λ , λ ) = γ ( m ) λ . If ∆ S F ( m ) = 0 and Q ( ∇ F ( m )) = 0, then γ ( m ) = 0.Otherwise, according to Proposition 9, we have F x = F y = F z = 0 and so t = 0. Now,we have γ ( m ) = 0 if and only if ( x + y + z ) t H F = 0. • Assume now that m = S and σ ( m ) (cid:54) = 0. Then ∆ S F ( m ) = 0 and so σ ( m ) = Q ( ∇ F ( m )) · S . We consider the unique [ λ , λ ] such that λ · m + λ · σ ( m ) = 0. Observe that λ (cid:54) = 0and that λ (cid:54) = 0. Since ∆ S F ( m ) = 0, we have Q S ,F ( m , λ , λ ) = β ( m ) λ λ . But β ( m ) = − F, m ( S , σ ( m )) = 0 due to m = S = σ ( m ). • Assume finally that m ∈ W . We have ∆ S F ( m ) (cid:54) = 0, x + y + z = 0, m = [ F x ( m ) : F y ( m ) : F z ( m ) : 0], t = 0 and σ ( m ) = − S F ( m ) κ ( ∇ F ( m )). Since t = 0, it followsthat N S ( m ) = ( x + y + z ) t = 0 and so that γ ( m ) = 0. Let [ λ , λ ] ∈ P besuch that λ · m + λ · σ ( m ) = 0. We observe that λ (cid:54) = 0 and λ (cid:54) = 0. We have α ( m ) λ = ∆ S F ( m ) λ and − · Hess F, m ( S , σ ( m )) λ λ = 2 Hess F, m ( S , m ) λ = 2( d − S F ( m ) , since ( d − F w = xF xw + yF yw + zF zw + tF tw for every w ∈ { x, y, z } . Hence we have Q S ,F ( m , λ , λ ) = (2 d − S F ( m ) λ − S F ( m )) ( F xx ( m ) + F yy ( m ) + F zz ( m )) λ λ . Hence Q S ,F ( m , λ , λ ) = 0 if and only if λ /λ = 2∆ S F ( m )( F xx ( m ) + F yy ( m ) + F zz ( m )) / (2 d − λ · m + λ · σ ( m ) = 0 and the formulaobtained for σ ( m ). (cid:3) Corollary 26.
A point m is in B if and only if it satisfies one of the following conditions: (1) m ∈ B = V ( F, ∆ S F, Q ( ∇ F )) , i.e. m is a singular point of Z or m is a point of tangencyof Z with an isotropic plane containing S (see also (1)), (2) m is a point of tangency of Z with H ∞ and m lies on the umbilical curve C ∞ , (3) m is a point of tangency of Z with H ∞ and m lies in the hessian surface of Z , (4) m = S ∈ Z , (5) m lies on C ∞ and ( F xx + F yy + F zz ) · m = (2 d − κ ( ∇ F ) .This can be summarized in the following formula B = Z ∩ [ V (∆ S F, Q ( ∇ F )) ∪ { S } ∪ V ( H F · Q, κ ( ∇ F )) ∞ ∪ G ∞ ] , with G ∞ = { m ∈ C ∞ : ( F xx + F yy + F zz ) · m = (2 d − κ ( ∇ F ) } . Remark 27.
The set B is never empty. Except (iv), the forms of the base points are verysimilar to the base points of the caustic map of planar curves (see [9] ).For a general ( Z , S ) , the set B consists of the points at which Z admits an isotropic tangentplane containing S , i.e. B = B = V ( F, ∆ S F, Q ( ∇ F )) , and in general Z has no singular pointand B ∩ H ∞ = ∅ . In this case B is fully interpreted by (1). N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 15
Let us study the base points when Z is the paraboloid V ( x + y − zt ) (see Example 20). Proposition 28 (Paraboloid) . Let Z = V ( x + y − zt ) and any S ∈ P \ { F , F } . Then B = V ( F, ∆ S P, Q ( ∇ F )) ∪ ( { S } ∩ Z ) and its points are the following ones: (1) the point S if S is in Z , (2) the points [1 : ± i : 0 : 0] if x = y = 0 , (3) the point [ t : it : x + iy : 0] and [ t : − it : x − iy : 0] if t (cid:54) = 0 , (4) the point F [0 : 0 : 1 : 0] if t = 0 and x + y (cid:54) = 0 , (5) the points of the form [ ux : uy : z : 0] (with [ u : z ] ∈ P ) if x + y = 0 , t = 0 , (6) the points m and m − if x + y (cid:54) = 0 , with m ε := [ x ε : y ε : −
12 : 1] ,x ε = x ( z − t ) + iεy (cid:113) x + y + ( z − t ) x + y and y ε = y ( z − t ) − iεx (cid:113) x + y + ( z − t ) x + y . (7) the point (cid:20) z − t x : z − t y : − : 1 (cid:21) if x + y = 0 , x (cid:54) = 0 and z − t (cid:54) = 0 .Proof. We have α = ∆ S F = xx + yy − tz − zt ,˜ β = ( x + y + t )( x + y − t z ) − tz + zt + t t )∆ S F and γ = 4(( x t − xt ) + ( y t − yt ) + ( z t − zt ) )∆ S F . First we observe that there is nopoint of Z satisfying (5) of Corollary 26. Assume that m is such a point. We have t = 0, so x + y = 0 and z = 0 (since x + y + z = 0). Now, using ( F xx + F yy + F zz ) m = (2 d − κ ( ∇ F ),we obtain 2 x = 3 x and 2 y = 3 y , which implies x = y = 0 which contradicts z = t = 0.Now we prove that Z ∩ V ( H F · Q, κ ( ∇ F )) ∩H ∞ = ∅ . Assume that m ∈ Z ∩ V ( H F · Q, κ ( ∇ F )) ∩H ∞ . Due to F x ( m ) = F y ( m ) = F z ( m ) = 0, we obtain x = y = t = 0. Since H F = −
1, we have0 = x + y + z and so z = 0.It remains to identify the points of V ( F, ∆ S F, Q ( ∇ F )). Let m [ x : y : z : t ] be a point of thisset. Then x + y = 2 zt , 0 = x x + y y − z t − t z , 0 = F x + F y + F z = x + y + t . So − t = x + y = 2 zt .If x = y = 0, since t (cid:54) = 0, we obtain x + y + t = 0 = x + y − zt and z = − z t/t .If t = 0, we obtain x + y = 0 and z = 0. This gives (2). If t = 1, from t = − zt , it comes z = − / t = 2 z , which contradicts S (cid:54) = F .From now on, we assume that ( x , y ) (cid:54) = 0.Assume first that t = 0. Then we have x + y = 0 and 0 = ∆ S F = x x + y y − t z .Let ε ∈ {± } such that y = iεx . We have 0 = x x + iεy x − t z . If t = 0, this becomes0 = x ( x + iεy ) and so either x = 0 or x + iεy = 0. This gives (4) and (5). If t (cid:54) = 0 and if x + iεy = 0, we obtain z = 0 and y = iεx , so m = | iε : 0 : 0]. If t (cid:54) = 0 and if x + iεy (cid:54) = 0,we obtain x = t z/ ( x + iεy ), so m = [ t : iεt : x + iεy : 0]. This gives (3).Assume now that t = 1. We have − x + y = 2 z and so z = − /
2. Since x + y = − ε ∈ {± } be such that y = εi √ x . We have0 = x x + y y − z t − t z = x x + y εi (cid:112) x − z + t N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 16 and so − y εi √ x = x x − z + t and we obtain − y (1 + x ) = ( x x − z + t ) and so0 = ( x + y ) x − xx (cid:18) z − t (cid:19) + (cid:18) z − t (cid:19) + y . This gives m ∈ { m , m − } if x + y (cid:54) = 0, with m ε := [ x ε : y ε : − : 1], x ε = x ( z − t ) + ε (cid:113) ( x ( z − t )) − ( x + y )( y + ( z − t ) ) x + y and y ε = y ( z − t ) − ε (cid:113) ( y ( z − t )) − ( x + y )( x + ( z − t ) ) x + y , and so (6). Now, we assume moreover that x + y = 0, we obtain0 = − xx (cid:18) z − t (cid:19) + (cid:18) z − t (cid:19) + y . If x + y = 0 and z − t (cid:54) = 0, then we obtain x = z − t x and y = z − t y , since x (cid:54) = 0 and y (cid:54) = 0. This gives (7). If x + y = 0 and z − t = 0, we obtain y = 0 and so S = F . (cid:3) Remark 29.
Let Z = V (( x + y − zt ) / and S ∈ P \ { F , F } . Observe that B < ∞ exceptif S ∈ Z ∞ . In the particular case where S ∈ Z ∞ , we have the following. Proposition 30.
Let Z = V (( x + y − zt ) / and S = [1 : εi : z : 0] with ε ∈ {± } , then Σ S ( Z ) is the curve of equations z iεx + z y + iεt, ( z − t/ + x + y .Proof. In this case, we have α = x + iεy − tz , ˜ β = − αtz and γ = 4 αz t . So (3) becomes α ( λ + 2 tz λ ) = 0. Hence Σ S ( Z ) is the Zariski closure of the image of Z by the rational mapgiven by ψ ( m ) = − tz · m + σ ( m ) . On Z , using the fact that 2 tz = x + y , this rational mapcan be rewritten ψ ( m ) = − ( x + iεy ) + t iε (( x + iεy ) + t ) − z t + 2( x + iεy ) t − t z . To conclude observe ψ only depends on( x + iεy, t ) and that if ( X, Y, Z, T ) = ψ ( x, y, z, t ), then 2 t = X − εY and 2( x + iεy ) = − X − iεY . (cid:3) Reflected Polar curves
Let H ∈ P ic ( P ) be the hyperplane class. We will identify π , ∗ (Φ ∗ H ) ∈ A ( P ) with theclass of sets P A,B ⊆ P defined as follows. Definition 31.
For any
A, B ∈ W ∨ , we define the set D A,B := V ( A, B ) ⊆ P and the reflectedpolar P A,B by P A,B = π (Φ − ( D A,B )) ∪ π ( Base (Φ)) , i.e. P A,B corresponds to the following set: { m ∈ P : ∃ [ λ , λ ] ∈ P , A ( λ m + λ σ ( m )) = 0 , B ( λ m + λ σ ( m )) = 0 , Q S ,F ( m , λ , λ ) = 0 } . N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 17
Proposition 32.
For generic
A, B in W ∨ , D A,B is a line and P A,B = V ( K , K , K ) , with K ( m ) := A ( σ ( m )) B ( m ) − A ( m ) B ( σ ( m )) , K ( m ) := Q S ,F ( m , − A ( σ ( m )) , A ( m )) ,K ( m ) := Q S ,F ( m , − B ( σ ( m )) , B ( m )) . Proof.
Recall that M S, Z has been defined in Definition 12. Assume that D A,B is a line thatdoes not correspond to any line ( m σ ( m )) for m ∈ P \ M S, Z (this is true for a generic ( A, B )in ( W ∨ ) ). Hence V ( A, B, A ◦ σ , B ◦ σ ) = M S, Z ∩ D A,B . (17)Let m ∈ P A,B and [ λ : λ ] ∈ P be such that A ( Φ ( m , λ , λ )) = 0 = B ( Φ ( m , λ , λ )) = 0 and Q S ,F ( m , λ , λ ) = 0. This implies that λ · A ( m ) + λ · A ( σ ( m )) = 0 = λ · B ( m ) + λ · B ( σ ( m )) . Therefore ( − A ( σ ( m )) , A ( m )) and ( − B ( σ ( m )) , B ( m )) are proportional to ( λ , λ ). But, since Q S ,F ( m , λ , λ ) = 0, we conclude that m is in V ( K , K , K ).Conversely, assume now that m is a point of V ( K , K , K ). Due to (17), we have m (cid:54)∈ V ( A, B, A ◦ σ , B ◦ σ ) or m ∈ M S, Z ∩ D A,B . Assume first that m (cid:54)∈ V ( A, B, A ◦ σ , B ◦ σ ), then ( − A ( σ ( m )) , A ( m )) and ( − B ( σ ( m )) , B ( m ))are proportional and at least one is non null. Let [ λ : λ ] be the corresponding point in P .we have A ( λ · m + λ · σ ( m )) = 0, B ( λ · m + λ · σ ( m )) = 0, and Q S ,F ( m , λ , λ ) = 0. So m ∈ P A,B .Assume finally that m ∈ M S, Z ∩D A,B , then there exists [ λ : µ ] ∈ P such that λ · m + µ · σ ( m ) =0, A ( m ) = B ( m ) = 0. We also have A ( σ ( m )) = B ( σ ( m )) = 0 and so m ∈ P A,B . (cid:3) Notation 33.
We write B S, Z for the set of points m ∈ P for which Q S ,F ( m , λ , λ ) = 0 in C [ λ , λ ] . Observe that, for m ∈ P \ B S, Z , there are at most two [ λ : λ ] ∈ P such that Q S ,F ( m , λ , λ ) =0, and so X ∩ π − ( m )) ≤ Remark 34.
According to the expressions of σ , α , β and γ , we have B S, Z = V ( α, β, γ ) = V (∆ S F, Hess F ( S , S ) · Q ( ∇ F )) . We observe that dim B S, Z ≥
1. Observe that ( M S, Z ∪ B S, Z ) is the set of m ∈ P suchthat Φ( π − ( { m } )) = Π( V ect ( m , σ ( m ))). When m ∈ M S, Z , Φ( π − ( { m } )) = { m } and when m ∈ B S, Z , Φ( π − ( { m } )) = R m . Recall that B = π ( B Φ | ˆ Z ). Proposition 35.
Assume that B < ∞ and Z (cid:54) = H ∞ . Then, for generic A, B ∈ W ∨ , dim P A,B = 1 and deg P A,B = ( d − d − .Proof. As in the proof of the preceding proposition, we consider generic (
A, B ) ∈ ( W ∨ ) suchthat (17) holds. Recall that P A,B = V ( K , K , K ). First, we observe that K ∈ Sym d − ( W ∨ )whereas K , K ∈ Sym d − ( W ∨ ). Now, if m is a point of P \ V ( A, A ◦ σ ), then the followingequivalence holds true m ∈ P A,B ⇔ m ∈ V ( K , K ) and that, if m is a point of P \ V ( B, B ◦ σ ),then m ∈ P A,B ⇔ m ∈ V ( K , K ) . Therefore dim P A,B ∈ { , } . • Let us prove that dim P A,B = 1. Assume first that dim(Σ S ( Z )) ≤
1. Then, for generic(
A, B ) ∈ ( W ∨ ) , we have Σ S ( Z ) ∩ D A,B = ∅ . Therefore, π (Φ − | ˆ Z ( D A,B )) = ∅ and so Z ∩ P
A,B = B is a finite set, which implies that dim P A,B ≤ N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 18
Assume now that dim(Σ S ( Z )) = 2. Let us consider a generic ( A, B ) ∈ ( W ∨ ) suchthat S ( Z ) ∩ D A,B ) < ∞ and such that, for every P ∈ Σ S ( Z ) ∩ D A,B , we have π (Φ − | ˆ Z ( { P } )) \ B ] = δ, (see Proposition 23). This implies that π (Φ − | ˆ Z ( D A,B )) < ∞ and so Z ∩ P
A,B ) < ∞ (since B < ∞ ). Hence dim P A,B ≤ Z = 2. • Let (
A, B ) as above. Since dim P A,B = 1, deg P A,B corresponds to P A,B ∩ H ) for ageneric plane H in P . Due to Corollary 26, we have V ( F, ∆ S F, Q ( ∇ F )) ≤ B < ∞ .So dim V (∆ S F, Q ( ∇ F )) = 1. Moreover we have Z (cid:54) = H ∞ . So, due to Lemma 14, weconclude that dim M S, Z < M S, Z ∩ D A,B ) < ∞ . Now, since(17) holds, we conclude that dim V ( A, A ◦ σ ) = 1 = dim V ( B, B ◦ σ ).Since dim P A,B = 1, we conclude that dim V ( K , K ) = dim V ( K , K ) = 1. More-over, for a generic ( A, B ) ∈ ( W ∨ ) , we have E < ∞ , with E := V ( A, A ◦ σ , K ) ∪ V ( B, B ◦ σ , K ) < ∞ . Let us explain how we get V ( A, A ◦ σ , K ) < ∞ . We recall that, due to (17), V ( A, A ◦ σ, B, B ◦ σ ) < ∞ . We write E := V ( A, A ◦ σ , K ) \ V ( B, B ◦ σ ) to simplify notations. – First we observe that E ∩ M S, Z ⊂ V ( A ) ∩ M S, Z which is finite for a generic A ∈ W ∨ since dim M S, Z < – Second we observe that E ∩ ( B S, Z \ M S, Z ) = ∅ for a generic A ∈ W ∨ . Indeedthis set is contained in V ( A, A ◦ σ , ∆ S F ) \ V ( Q ( ∇ F )). For m in this set, we have σ ( m ) = S . So, we just have to take A such that A ( S ) (cid:54) = 0. – Third we observe that E \ B S, Z < ∞ for generic A, B ∈ W ∨ . Indeed, dim V ( A, A ◦ σ ) = 1 (due to (17)) and, for any m ∈ V ( A, A ◦ σ ) \ B S, Z , there are at most two[ λ : λ ] ∈ P such that Q S ,F ( m, λ , λ ) = 0. So, for a generic B ∈ W ∨ , we have { ( m, [ λ : λ ]) ∈ X : m ∈ V ( A, A ◦ σ ) \ B S, Z , B (Φ( m, [ λ , λ ])) = 0 } < ∞ . Now, if m ∈ E \ B S, Z , [ − B ( σ ( m )) : B ( m )] ∈ P and, due to K ( m ) = 0, we observethat M := ( m, [ − B ( σ ( m )) : B ( m )]) is in X and B (Φ( M )) = 0. So E \ B S, Z < ∞ .Let us prove that deg( P S, Z ,A,B ) = ( d − d −
9) for a generic (
A, B ) ∈ ( W ∨ ) . Weconsider generic A and B such that: deg A = 1, deg A ◦ σ = 2 d − V ( A, A ◦ σ, B, B ◦ σ ) < ∞ and V ( A, A ◦ σ, K ) < ∞ . We observe that the reflected polar curve correspondsto V ( K , K ) outside V ( A, A ◦ σ ) and that the polar curve coincide with V ( K ) on V ( A, A ◦ σ ) (since V ( A, A ◦ σ ) is contained in V ( K , K )). We consider a generic plane H in P , such that H ∩ V ( A, A ◦ σ, B, B ◦ σ ) = ∅ and H ∩ V ( A, A ◦ σ, K ) = ∅ . Since H ∩ V ( A, A ◦ σ, K ) = ∅ , we havedeg( P S, Z ,A,B ) = deg V ( K , K ) − (cid:88) m ∈ V ( A,A ◦ σ ) ∩H i m ( H , V ( K , K ))= 5( d − d − − (cid:88) m ∈ V ( A,A ◦ σ ) ∩H i m ( H , V ( K , K )) . Now, we prove that ∀ m ∈ V ( A, A ◦ σ ) ∩ H , i m ( H , V ( K , K )) = 2 i m ( H , V ( A, A ◦ σ )) . (18)Let m ∈ V ( A, A ◦ σ ) ∩ H . If B ( m ) (cid:54) = 0, then i m ( H , V ( K , K )) = i m ( H , K , K ) = i m ( H , K , B · K )= i m ( H , K , α ( B · A ◦ σ ) − β · A · B · ( B · A ◦ σ ) + γ · A · B )= i m ( H , K , α ( A · B ◦ σ ) − β · A · B · ( A · B ◦ σ ) + γ · A · B ) , N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 19 (since B · A ◦ σ = A · B ◦ σ on V ( K )) and so i m ( H , V ( K , K )) = i m ( H , K , A · K )= i m ( H , K , A ) since K ( m ) (cid:54) = 0= i m ( H , A ◦ σ · B − B ◦ σ · A, A )= 2 i m ( H , A ◦ σ · B − B ◦ σ · A, A )= 2 i m ( H , A ◦ σ · B, A )= 2 i m ( H , A ◦ σ, A ) since B ( m ) (cid:54) = 0 . Analogously, If B ( σ ( m )) (cid:54) = 0, then i m ( H , V ( K , K )) = i m ( H , K , K ) = i m ( H , K , B ◦ σ · K )= i m ( H , K , α · B ◦ σ · A ◦ σ − β · A ◦ σ.B ◦ σ · ( A · B ◦ σ ) + γ · A · B ◦ σ )= i m ( H , K , α ( A ◦ σ · B ◦ σ ) − β · A ◦ σ · B ◦ σ · ( B · A ◦ σ ) + γ · B · A ◦ σ )= i m ( H , K , A ◦ σ · K ) = i m ( H , K , A ◦ σ )= i m ( H , A ◦ σ · B − B ◦ σ · A, A ◦ σ )= 2 i m ( H , A ◦ σ · B − B ◦ σ · A, A ◦ σ ) = 2 i m ( H , A, A ◦ σ ) . Hence we proved (18) and, for a generic plane H , we havedeg( P S, Z ,A,B ) = 5( d − d − − (cid:88) m ∈ V ( A,A ◦ σ ) ∩H i m ( H , V ( K , K ))= 5( d − d − − (cid:88) m ∈ V ( A,A ◦ σ ) ∩H i m ( H , V ( A, A ◦ σ ))= 5( d − d − − A ) deg( A ◦ σ )= 5( d − d − − d −
1) = ( d − d − . (cid:3) A formula for the degree of the caustic
Recall that B has been completely described in Corollary 26 (see also Remark 27 for thegeneral case). We refer to Definition 12 and Proposition 13 for M S, Z and to Notation 33 andRemark 34 for B S, Z . Observe that dim M S, Z ≥ σ ) ⊆ M S, Z . Theorem 36.
We assume that B < ∞ .If dim(Σ S ( Z )) < , then for a generic ( A, B ) ∈ ( W ∨ ) , we have d ( d − d − − (cid:88) m ∈B i m ( Z , P A,B ) . If dim(Σ S ( Z )) = 2 , dim( Z ∩ M S, Z ) ≤ and Z ∩ B S, Z \ M S, Z ) < ∞ , then for a generic ( A, B ) ∈ ( W ∨ ) , we have mdeg(Σ S ( Z )) = d ( d − d − − (cid:88) m ∈B i m ( Z , P A,B ) , where mdeg(Σ S ( Z )) is the degree with multiplicity of (Σ S ( Z )) ( mdeg(Σ S ( Z )) = δ deg(Σ S ( Z )) ,see Proposition 23 for the property satisfied by δ ), where d is the degree of Z and where i m ( Z , P A,B ) denotes the intersection number of Z with P A,B at point m . N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 20
Let us notice, that in this formula, we can replace i m ( Z , P A,B ) by i m ( Z , V ( K , K )), withthe notations of Proposition 32. Indeed, we can take A and B such that V ( A, A ◦ σ, K ) < ∞ (see the proof of Proposition 35) and use P a,b \ V ( A, A ◦ σ ) = V ( K , K ) \ V ( A, A ◦ σ )).Let us recall that the case when dim( Z ∩ M S, Z ) > m ∈ Z ∩ B S, Z \ M S, Z , the reflected line R m is well defined and containedin Σ S ( Z ). Therefore, in the degenerate case when dim( Z ∩ B S, Z \ M S, Z ) ≥
1, the surfaceconstituted by the reflected lines R m for m ∈ Z ∩ B S, Z is contained in Σ S ( Z ). Proof of Theorem 36.
Recall that for a generic (
A, B ) in ( W ∨ ) , we have deg( P A,B ) = ( d − d − S ( Z ) < δ = ∞ ) and that B < ∞ , Taking ( A, B )such that deg( P A,B ) = ( d − d −
9) and D A,B ∩ Σ S ( Z ) = ∅ , we have P A,B ∩ Z = B and so d ( d − d −
9) = deg( Z ) deg( P A,B ) = (cid:88) m ∈Z∩P A,B i m ( Z , P A,B )= (cid:88) m ∈B i m ( Z , P A,B ) = (cid:88) m ∈B i m ( Z , P A,B ) . Assume now that dim Σ S ( Z ) = 2 (i.e. that δ is finite), that dim( Z ∩ M S, Z ) ≤
1, that
Z ∩ B S, Z \ M S, Z ) < ∞ and B < ∞ . We consider ( A, B ) ∈ ( W ∨ ) such that:(a) D A,B is a line containing no reflected line R m = ( m σ ( m )) ( m ∈ Z ),(b) deg( P A,B ) = ( d − d −
9) (this is generic due to Proposition 35),(c) the points P ∈ D A,B ∩ Σ S ( Z ) are such that π (Φ − | ˆ Z ( { P } )) \ B ] = δ (this is generic dueto Proposition 23),(d) For any P ∈ D A,B ∩ Σ S ( Z ), we have i P (Σ S ( Z ) , D A,B ) = 1 (this is true for a generic(
A, B ) since Σ S ( Z ) is a surface),(e) the line D A,B intersects no reflected line R m with m ∈ B S, Z (this is generic since Z ∩ B S, Z \ M S, Z ) < ∞ and dim( Z ∩ M S, Z ) ≤ m ∈ ( P A,B ∩ Z ) \ B , we have i m ( Z , P A,B ) = 1 (this is explained at the end ofthis proof),(g) D A,B does not intersect Φ( π − ( Z ∩ V ( β − αγ ))) if dim( Z ∩ V ( β − αγ )) = 1.Due to (b), we have d ( d − d −
9) = deg( Z ) deg( P A,B ) = (cid:88) m ∈Z∩P A,B i m ( Z , P A,B )= (cid:88) m ∈B i m ( Z , P A,B ) + (cid:88) m ∈ ( Z∩P
A,B ) \B i m ( Z , P A,B ) . Now, we have (cid:88) m ∈ ( Z∩P
A,B ) \B i m ( Z , P A,B ) =
Z ∩ P
A,B ) \ B ) due to ( f )= π (Φ − | ˆ Z ( D A,B )) \ B ] due to Def inition δ S ( Z ) ∩ D A,B ) due to ( c )= δ (cid:88) P i P (Σ S ( Z ) , D A,B ) = δ deg(Σ S ( Z )) due to ( d ) . Let us now explain why (f) is true for a generic (
A, B ) ∈ ( W ∨ ) . Let m ∈ ( P A,B ∩ Z ) \ B .Due to (e), m ∈ π (Φ − | ˆ Z ( D A,B )) \ B S, Z . We consider the cone hypersurface K Z of W associated N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 21 to Z . Since m ∈ Z \ B S, Z , there exist two maps ψ ± : U → P defined on a neighbourhood U of m in P such that, for any m (cid:48) ∈ U , Φ( π − ( { m (cid:48) } )) = { ψ − ( m (cid:48) ) , ψ + ( m (cid:48) ) } . Let ε ∈ { + , −} be such that Φ( π − ( { m } )) ∩ D A,B = { ψ ε ( m ) } ( ψ ε is unique for a generic m ∈ Z according to(a) and to (g)) and the tangent space to P A,B at m is given by V ( A ◦ Dψ ε ( m ) , B ◦ Dψ ε ( m )),where Dψ ± ( m ) are the jacobian matrices of ψ ± taken at m . Now, with these notations, for ageneric m in Z , Dψ ε (cid:48) ( m ) is invertible if dim ψ ε (cid:48) ( Z ) = 2 (if dim ψ ε (cid:48) ( Z ) <
2, take (
A, B ) suchthat D A,B ∩ ψ ε (cid:48) ( Z ) = ∅ ). This combined with (d) gives the result. (cid:3) Proof of Theorem 1
More precisely we prove the following (recall that V ( F, ∆ S F, Q ( ∇ F )) is the number ofisotropic tangent planes to Z passing through S ). Theorem 37.
Let
Z ⊂ P be an irreducible smooth surface and S ∈ P \ ( Z ∪ H ∞ ) be suchthat B = V ( F, ∆ S F, Q ( ∇ F )) (see Corollary 26) and such that this set contains d ( d − d − points. We assume moreover that B ∩ V ( H F N S ) = ∅ . Then mdeg(Σ S ( Z )) = d ( d − d − .Proof. Without any loss of generality, we assume that S [0 : 0 : 0 : 1]. Due to Theorem 36, wehave mdeg(Σ S ( Z )) = d ( d − d − − (cid:80) P ∈B i P ( Z , V ( K , K )) for generic A, B ∈ W ∨ . Sincewe know that B = V ( F, ∆ S F, Q ( ∇ F )) contains d ( d − d −
2) points, we just have to provethat i P ( Z , P A,B ) = 1 for any P ∈ B (for generic A, B ∈ W ∨ ).Let such a point P . Assume that A ( P ) B ( P ) (cid:54) = 0 (this is true for generic A, B ∈ W ∨ ). Since F is smooth, either F x ( P ) F y ( P ) (cid:54) = 0, or F x ( P ) F z ( P ) (cid:54) = 0 or F y ( P ) F z ( P ) (cid:54) = 0. Assume forexample that F x ( P ) F y ( P ) (cid:54) = 0, then there exists a local parametrization h of Z defined on anopen neighbourhood of (0 ,
0) in C such that h (0 ,
0) = P and [ ∂h ( u, v ) /∂u ](0 , (cid:54) = 0. Since V ( F, ∆ S F, Q ( ∇ F )) = d ( d − d − i P ( V ( F, ∆ S F, Q ( ∇ F ))) = 1. Hence weassume that val u,v A ( σ ( h ( u, v ))) = val u,v B ( σ ( h ( u, v ))) = 1. (this is true for generic A, B ∈ W ∨ due to the formula of σ ). Moreover ∆ S F ( h ( u, v )), α ( h ( u, v )), β ( h ( u, v )) and γ ( h ( u, v ))have valuation 1 in ( u, v ). Recall that K and K are given by K ( m ) = A ( σ ( m )) B ( m ) − B ( σ ( m )) A ( m ) and K ( m ) = α ( m )( A ( σ ( m ))) − β ( m ) A ( σ ( m )) A ( m ) + γ ( m )( A ( m )) . On theone hand K ( h ( u, v )) has valuation 1 and its term of degree 1 is the term of degree 1 of γ ( h ( u, v ))( A ( P )) = − A ( σ ( P ))) ∆ S F ( h ( u, v )) N S ( P ) H F ( P ) / ( d − . On the other hand K ( h ( u, v )) = ∆ S F ( h ( u, v ))[ · · · ] + Q ( ∇ F ( h ( u, v )))[ A ( S ) B ( h ( u, v )) − B ( S ) A ( h ( u, v ))] . Hence, for generic
A, B ∈ W ∨ , A ( S ) B ( P ) (cid:54) = B ( S ) A ( P ) and so i P ( V ( F, K , K )) = 1. (cid:3) Degree of caustics of a paraboloid
This section is devoted to the proof of Proposition 2. We consider again the case when Z is the paraboloid V ( F ) with F = ( x + y − zt ) /
2. We recall that the base points have beenstudied in Proposition 28 and that we have written F [0 : 0 : 1 : 0] and F [0 : 0 : 1 : 2] for thetwo focal points of this paraboloid. Let S ∈ P \ { F , F } . We have F = ( x + y − zt ) / , ∆ S F = x x + y y − z t − t z, σ = ( x + y + t ) · S − S F · xy − t ,K ( m ) = A ( σ ( m )) B ( m ) − B ( σ ( m )) A ( m ) ,K ( m ) = α ( m )( A ( σ ( m ))) − β ( m ) A ( σ ( m )) A ( m ) + γ ( m )( A ( m ))
2N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 22 with α = ∆ S F, γ = − S F N S ,β = − x + y − z t )( x + y + t ) + 4∆ S F ( t z + z t + t t ) , and N S = ( x t − xt ) + ( y t − yt ) + ( z t − zt ) . It will be useful to observe that Z is invariant by composition by [ x : y : z : t ] (cid:55)→ [¯ x : ¯ y : ¯ z : ¯ t ]and by [ x : y : z : t ] (cid:55)→ [ ax + by : − bx + ay : cz : t/c ] with a + b = c = 1. • If x = y = 0, then, due to Theorem 42, the degree of Σ S ( Z ) corresponds to the degreeof a caustic of the parabola. Hence, we have mdeg Σ S ( Z ) = 4 if S [0 : 0 : 0 : 1] andmdeg Σ S ( Z ) = 6 elsewhere (see [9]).We assume now that x (cid:54) = 0. • [Generic case] If S (cid:54)∈ ( Z ∪ H ∞ ), if x + y (cid:54) = 0 and if x + y + ( z − ( t / (cid:54) = 0,then Theorem 37 applies and mdeg Σ S ( Z ) = 18. Indeed B = V ( F, ∆ S F, Q ( ∇ F )) = { C, D, m , m − } with C [ t : it : x + iy : 0], C [ t : − it : x − iy : 0] and m ε [ x ε ; y ε : − / x ε and y ε defined in Proposition 28. Moreover we have N S ( C ) = ( x + iy ) t (cid:54) = 0, N S ( D ) = ( x − iy ) t (cid:54) = 0 and N S ( m ε ) = ( x + y ) + ( z − t ) (cid:54) = 0.In order to apply our Theorem 36, we will have to verify its assumptions on B S, Z and M S, Z .This is the aim of the next proposition. Proposition 38.
Let S ∈ P \ { F , F } be such that B < ∞ . Then Z ∩ M S, Z ) < ∞ .Moreover if Hess F ( S , S ) (cid:54) = 0 (i.e. if S (cid:54)∈ Z ), then Z ∩ B S, Z ) < ∞ and so Theorem 36applies.Proof. Let us prove that
Z∩M S, Z ) < ∞ . Since B < ∞ , we already know that V ( F, ∆ S F, Q ( ∇ F )) < ∞ . Moreover we have V ( F, F x , F y , F z ) = { [0 : 0 : 1 : 0] } and W = { [1 : ± i : 0 : 0] } (indeed for m ∈ W , we have ( x, y ) (cid:54) = 0, so z = t and − z = x + y = 2 zt = 2 z ).If Hess F ( S , S ) (cid:54) = 0, then Z ∩ B S, Z = V ( F, ∆ S F, Q ( ∇ F )) ⊆ B which is finite. (cid:3) • Let S (cid:54)∈ ( Z ∪ H ∞ ) such that x + y (cid:54) = 0 and x + y + ( z − ( t / = 0. Withoutloss of generality we assume that x = 0 and y = 1 and z − ( t /
2) = i . The factthat S (cid:54)∈ Z implies that t (cid:54) = − i . We have B = V ( F, ∆ S F, Q ( ∇ F )) = { C, D, E } with C [ t : it : i : 0], D [ t : − it : − i : 0] and E [0 : i : − : 1].Around E , we parametrize Z by h ( x, y ) = ( x, i + y, x +( i + y ) , S F ◦ h ( x, y ) = y (1 − it ) − t x + y and Q ( ∇ F ) ◦ h ( x, y ) = 2 iy + x + y . Hence i E ( Z , V (∆ S F, Q ( ∇ F ))) =2 and so i C ( Z , V (∆ S F, Q ( ∇ F ))) = 1 and i D ( Z , V (∆ S F, Q ( ∇ F ))) = 1. Since more-over N S ( C ) = ( x + iy ) t (cid:54) = 0 and N S ( D ) = ( x − iy ) t (cid:54) = 0, due to the proofof Theorem 37, we have i C ( Z , V (∆ S F, Q ( ∇ F ))) = i D ( Z , V (∆ S F, Q ( ∇ F ))) = 1. Ob-serve that for a generic A ∈ W ∨ , A ◦ σ ◦ h has valuation 1 with dominating term2[ a y ( i − it ) + a y (1 − t i ) + ia yt ]. Hence val x,y K ◦ h ( x, y ) = 1 and its dominatingterm is proportional to y . Using the fact that z + t = i + t = i (1 − it ), we have N S ◦ h ( x, y ) = − it x − it ( i + t ) y + ... , val x,y α ◦ h ( x, y ) = 1, val x,y γ ◦ h ( x, y ) = 3.Moreover x + y − z t = (1 − it ) so β ◦ h ( x, y ) = − x (1 − it ) − y (1 − it ) + ... .Therefore val x,y K ◦ h ( x, y ) = 3 and we conclude that i E ( Z , V ( K , K )) = 3 and thatmdeg Σ S ( Z ) = 22 − − − • If S (cid:54)∈ ( Z ∪ H ∞ ), if x + y = 0 ( x (cid:54) = 0) and z (cid:54) = t /
2. We assume without lossof generality that x = 1 and y = i . B = V ( F, ∆ S F, Q ( ∇ F )) = { C, D, E } with C [1 : i : 0 : 0], D [ t : − it : 2 : 0] and E [ z − ( t / x : z − ( t / y : − : 1].We use the parametrization h ( z, t ) = (1 , i √ − zt, z, t ) at a neighbourhood of Z around C . We have ∆ S F ( h ( z, t )) = − z t − t z + (1 − √ − zt ) and Q ( ∇ F )( h ( z, t )) = N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 23 (2 z + t ) t and so i C ( Z , V (∆ S F, Q ( ∇ F ))) = 2 and so the intersection numbers of Z with V (∆ S F, Q ( ∇ F )) is equal to 1 at D and E . Since N S ( D ) = ( x − iy ) t (cid:54) = 0 and N S ( E ) =( z − t ) (cid:54) = 0. Hence, due to the Proof of Theorem 37, we have i D ( Z , V (∆ S F, Q ( ∇ F ))) = i E ( Z , V (∆ S F, Q ( ∇ F ))) = 1. It remains to estimate i C ( Z , V (∆ S F, Q ( ∇ F ))). We have σ ◦ h ( z, t ) = t + 2( z t + t z + √ − zt − − zt )2( z t + t z + √ − zt − i √ − zt + 2 izt + it z + t ) tz − z t − √ − zt − t (2 z + t ) t t . Hence, for generic
A, B ∈ W ∨ , K ◦ h ( z, t ) has valuation 2 with dominating terms[ z (2 z + t ) + ( t − z )2 z ( z + t )](( a + ia ) b − ( b + ib ) a )++ t ( t − z )[( b + ib ) a − b ( a + ia )] . Moreover val z,t α ◦ h ( z, t ) = 1, val z,t β ◦ h ( z, t ) = 2 and val z,t γ ◦ h ( z, t ) = 2. Hence K ◦ h ( z, t ) = (4( z t − t z ) − z − t ) zt )( z t + t z )( a + ia ) + ... for a generic A ∈ W ∨ , so i C ( Z , V ( K , K )) = 6 and mdeg Σ S ( Z ) = 22 − − − • If S (cid:54)∈ ( Z ∪ H ∞ ), if x + y = 0 ( x (cid:54) = 0) and z = t /
2. We assume without loss ofgenerality that x = 1 and y = i . B = V ( F, ∆ S F, Q ( ∇ F )) = { C, D } with C [1 : i : 0 : 0], D [ t : − it : 2 : 0]. We use again the parametrization h ( z, t ) = (1 , i √ − zt, z, t ) ata neighbourhood of Z around C . We observe that i C ( Z , V (∆ S F, Q ( ∇ F ))) = 3 andso i D ( Z , V (∆ S F, Q ( ∇ F ))) = 1. Moreover N S ( D ) = ( x − iy ) t (cid:54) = 0. Hence, dueto the proof of Theorem 37, we have i D ( Z , V ( K , K )) = 1 . Moreover, we prove that i C ( Z , V ( K , K )) = 9 (probranches of K ◦ h and K ◦ h have intersection number 3/2)and so mdeg Σ S ( Z ) = 22 − − Proposition 39.
Let S ∈ Z \ ( D ∪ H ∞ ) . Then mdeg Σ S ( Z ) = 12 if x + y + t = 0 , mdeg Σ S ( Z ) = 14 if x + y = 0 and mdeg Σ S ( Z ) = 16 otherwise.Proof. Observe that ∆ S F divides α , β and γ . In this case we define Σ S ( Z ) by replacing Q S ,F by ˜ Q S ,F := Q/ ∆ S F , we define analogously ˜ α := 1, ˜ β := 4( t z + z t + t t ) and ˜ γ := − N S .Following our argument above, we define ˜ B S, Z := V ( ˜ α, ˜ β, ˜ γ ) = ∅ and the new reflected polarcurve ˜ P A,B = V ( K , ˜ K , ˜ K ) by using ˜ Q S ,F instead of Q S ,F . Following the proof of Proposition35, we obtain that deg P A,B = deg K deg ˜ K − A deg( A ◦ σ ) = 8 and the correspondingset of base points ˜ B is contained in ( { S } ∩ Z ) ∪ V ( F, ∆ S F, Q ( ∇ F ) , N S )) (recall that B = ( { S } ∩Z ) ∪ V ( F, ∆ S F, Q ( ∇ F ))). • If S ∈ Z \H ∞ , x + y (cid:54) = 0 and x + y + t (cid:54) = 0 (so z + t (cid:54) = 0), then V ( F, ∆ S F, Q ( ∇ F )) = { C, D, m , m − } with C [ t : − it : x − iy : 0], D [ t : it : x + iy : 0] and m ε [ x ε : y ε : − : 1] with x ε := x z t ( z − t )+ iε y t z ( z + t ) and y ε := y z t ( z − t ) − iε x t z ( z + t ).Hence the intersection number of Z with V (∆ S F, Q ( ∇ F )) is 1 at these four points. Weobserve that N S ( C ) = t ( x + iy ) (cid:54) = 0, N S ( D ) = t ( x − iy ) (cid:54) = 0 and N S ( m ε ) =( z + t ) (cid:54) = 0. Hence i C ( V ( F, K , ˜ K )) = i D ( V ( F, K , ˜ K )) = i m ε ( V ( F, K , ˜ K )) = 0.It remains to compute i S ( V ( F, K , ˜ K )). We have σ ( S ) = ( x + y + t ) · S and˜ K ( S ) = − x + y + t ) ( A ( S )) (cid:54) = 0. We conclude that mdeg Σ S ( Z ) = 2 × • If S ∈ Z \ H ∞ and x + y + t = 0, then z = − t / B = V ( F, ∆ S F, Q ( ∇ F )) = { S, C, D } with C [ t : it : x + iy : 0] and D [ t : − it : x − iy : 0]. Assume t = 1.Observe that N S ( C ) (cid:54) = 0 and so that i C ( V ( F, K , ˜ K )) = 0. Analogously we have i D ( V ( F, K , ˜ K )) = 0. Using the parametrization h ( x, y ) = ( x + x, y + y, ( x + x ) +( y + y ) , Z around S and the fact that ˜ α = 1, ˜ β ◦ h ( x, y ) = 4( xx + yy + x + y ), that N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 24 ˜ γ ◦ h ( x, y ) = − x + y + ( xx + yy + ( x + y ) / ]. Moreover σ ◦ h ( x, y ) = (2 xx + 2 yy + x + y ) · S + x + y · x + xy + y − . Hence ˜ K ◦ h ( x, y ) = x + y [( a x + a y − a ) B ( S ) − ( b x + b y − b ) A ( S )] and ˜ K :=16( xx + yy ) ( A ( S )) . So i S ( V ( F, ˜ K , ˜ K )) = 4 and mdeg Σ S ( Z ) = 2 × − • If S ∈ Z \ H ∞ and x + y = 0, then we assume without loss of generality that x = 1and y = i (so z = 0). We have B = { S, C, D, E } with C [1 : i : 0 : 0], D [ t : − it : 2 : 0]and E [ − t x : − t y : − : 1]. We have N S ( D ) = 4 t and N S ( E ) = 5 t / i D ( V ( F, K , ˜ K )) = i E ( V ( F, K , ˜ K )) = 0. Observe that ∆ S F ( S ) = 0, N S ( S ) = 0, σ ( S ) = t · S and ˜ β ( S ) = 4 t (cid:54) = 0. So ˜ K ( S ) (cid:54) = 0 and i S ( Z , V ( K , ˜ K )) = 0.Around C , we parametrize Z by h ( z, t ) = (1 , i √ − zt, z, t ). We have ∆ S F ◦ h ( z, t ) =1 − √ − zt − t z and σ ◦ h ( z, t ) = t z + 2 zt + t + 2( √ − zt − it z √ − zt + 2 izt + it + 2 i ( √ − zt − √ − zt t tz − t ( √ − zt − z + t ) t t . Hence A ◦ σ ◦ h ( z, t ) has valuation 1. Moreover ˜ α = 1, ˜ β ◦ h ( z, t ) = 4( t z + t t ) and˜ γ ◦ h ( z, t ) = 8 t t + ... . Hence, for a generic A ∈ W ∨ , ˜ K ◦ h ( z, t ) = 8 t tA ( C ) + ... .Moreover we have K ◦ h ( z, t ) = 2 t { ( a tz + a t − a z )( b + ib ) − ( b tz + b t − b z )( a + ia ) } + ... Hence i C ( Z , V ( K , ˜ K )) = 2 and so mdeg Σ S ( Z ) = 16 − (cid:3) We assume now that S is at infinity. In this case N S = ( x + y + z ) t . • If S ∈ H ∞ \ ( Z ∪ C ∞ ), then x + y (cid:54) = 0 and B = { F , m , m − } with m ε [ x z + iεy (cid:112) x + y + z : y z − iεx (cid:112) x + y + z : − x + y : x + y ]. Then we will provethat the intersection number of V ( F, K , K ) is 8 at F and 1 at the two other basepoints and so mdeg Σ S ( Z ) = 22 − − − N S ( m ε ) (cid:54) = 0, to prove that i m ε ( Z , V ( K , K )) =1 it is enough to prove that i m ε ( V ( F, ∆ S F, Q ( ∇ F )) = 1. To see this, we use theparametrization h ( x, y ) = ( x + x, y + y, (( x + x ) + ( y + y ) ) / ,
1) of Z around m ε = [ x : y : − / S F ◦ h and Q ( ∇ F ) ◦ h arerespectively x x + y y and 2( xx + yy ) which are not proportional since x + y (cid:54) = 0.For F , we use the parametrization h ( x, y ) = ( x, y, , ( x + y ) /
2) of Z around F .We observe that, for generic A, B ∈ W ∨ , A ◦ σ ◦ h and B ◦ σ ◦ h have valuation 2 withrespective dominating terms: θ A := a [ − x x + x y − y xy ] + a [ − y y + y x − x xy ] + a ( x + y ) z ,θ B := b [ − x x + y y − y xy ] + b [ − y y + x x − x xy ] + b ( x + y ) z . Therefore the lowest degree terms of K ◦ h are given by( b a − a b )[ − x x + x y − y xy ] + ( b a − a b )[ − y y + y x − x xy ] . N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 25
Moreover the valuations of α ◦ h , β ◦ h and γ ◦ h are respectively 1, 2 and 5. Hence K ◦ h has valuation 4 and its dominating term is a x + y )( x + y ) θ A . Hence the curves ofequations K ◦ h and K ◦ h are transverse and we conclude that i F ( Z , V ( K , K )) = 8. • If S ∈ C ∞ \ Z , then x + y (cid:54) = 0, z (cid:54) = 0 and B = { F , m } with m [ x z : y z : − x + y : x + y ]. Then N S = 0 in C [ x, y, z, t ] (so γ = 0 in C [ x, y, z, t ] ). We prove that i F ( Z , V ( K , K )) = 8 as in the previous case. We compute i m ( Z , V ( K , K )). Weassume without loss of generality that x = 0, y = 1 and z = i . Around m [0 : z : − / Z by h ( x, y ) = ( x, z + y, (( x + x ) + ( y + y ) ) / , α ◦ h ( x, y ) = ∆ S F ◦ h ( x, y ) = y , Q ( ∇ F ) ◦ h ( x, y ) = x + 2 iy + y and σ ◦ h ( x, y ) = ( x + 2 iy + y ) i − y xi + y − = − xyx − y z ( x + y )0 . Hence, for generic
A, B ∈ W ∨ , K ◦ h ( x, y ) has valuation 2. Moreover α ( h ( x, y )) = y , β ( h ( x, y )) = − x + y ) and γ ( h ( x, y )) = 0. Hence K ◦ h ( x, y ) has valuation 4 and wehave i m ( Z , V ( K , K )) = 8, so mdeg Σ S ( Z ) = 22 − − • The case when S ∈ Z ∩ H ∞ has been studied in Proposition 30.10. About a reflected bundle
Recall that O Z ( −
1) = { ( m, v ) ∈ Z × W : v ∈ m } . Observe that the set R ( −
1) of ( m, v ) inthe trivial bundle
Z × W such that v corresponds to a point of P on the reflected line R m is: R ( −
1) = O Z ( −
1) + { ( m, v ) ∈ Z × W : v ∈ σ ( m ) } . Observe that this sum is direct in the generic case (when S (cid:54)∈ Z and when W = ∅ , see Proposition13). But, contrarily to the normal bundle considered in [16, 5] to study the evolute, R ( −
1) doesnot define a bundle since its rank is not constant. Indeed, the dimension of
V ect ( m , σ ( m ))equals 2 in general but not at every point m ∈ Z (it is strictly less than 2 when m is a basepoint of σ |Z and, as seen in Proposition 9, such points always exist). Appendix A. Caustics of surfaces linked with caustics of curves
For the classes of examples studied in this section, caustics of surfaces are linked with ofcaustics of planar curves. We start with some facts on caustics of planar curves.A.1.
Caustic of a planar curve.
Let S [ x : y : t ] ∈ P and an irreducible algebraiccurve C = V ( G ) ⊂ P with G ∈ C [ x, y, t ] homogeneous of degree d ≥
2. We write ∆ S G := x G x + y G y + z G z , N S = ( x t − xt ) + ( y t − yt ) , Hess G for the Hessian form of G and H G for its determinant and σ S ,G = ( G x + G y ) · S − S G · ( G x , G y , Definition 40 ([9]) . The caustic map of C from S is the rational map Φ S , C : P → P corresponding to Φ S ,G : C → C given by Φ S ,G = − H G N S ( d − · Id +∆ S G · σ S ,G . The causticby reflection Σ S ( C ) is the Zariski closure of Φ S , C ( C ) . We start with a technical lemma making a link between the formulas involved in Theorems 40and 15. Inspired by the three dimensional case, let us define the following quantities: α S ,G :=∆ S G , β S ,G := − (cid:2) Hess G ( S , σ S ,G ) + (∆ S G ) ( G xx + G yy ) (cid:3) and γ S ,G := − S G ( d − N S H G . N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 26
Lemma 41.
Let S ∈ P and C = V ( G ) ⊂ P be an irreducible algebraic curve with G ∈ C [ x, y, t ] being homogeneous of degree d ≥ . We have β S ,G = d − N S H G = γ S ,G α S ,G and so Φ S ,G ( m ) = λ · m + λ · σ S ,G ( m ) , with [ λ : λ ] = [ − β S ,G : α S ,G ]( m ) (i.e. α S ,G ( m ) λ + β S ,G ( m ) λ = 0 if ( α S ,G , β S ,G )( m ) (cid:54) = ). We omit the straightforward proof of this lemma.A.2.
Caustic of a surface from a light position on a revolution axis.
To simplify, weconsider the case of a surface Z = V ( F ) with axis of revolution V ( x, y ). We will use the fact that F = G ◦ h where h ( x, y, z, t ) = ( (cid:112) x + y , z, t ) for some homogeneous polynomial G ∈ C [ r, z, t ]with monomials of even degree in r . Such a surface Z is written R ( G ) and is called surface ofrevolution of axis V ( x, y ) of the curve V ( G ) ⊂ P . Theorem 42.
Let Z = R ( G ) with G ∈ C [ r, z, t ] irreducible homogeneous of degree d ≥ (the monomials of G being of even degree in r ). Assume that Z (cid:54)⊆ V (∆ S F, ( F x + F y + F z ) Hess F ( S , S )) . Let S [0 : 0 : z : t ] ∈ P and S [0 : z : t ] ∈ P .If d = 2 and if S is a focal point of Z , then Σ S ( Z ) is reduced to another other focal point.Otherwise, we have Σ S ( Z ) = V ( x, y ) ∪ R (Σ S ( V ( G ))) .Proof. We have α = [ z F z + t F t ] = ∆ S G ◦ h. (19)Now let us prove that γ = (cid:20) G r r γ S ,G (cid:21) ◦ h, (20)Since ∆ S F = ∆ S G ◦ h and N S = N S ◦ h , we just have to prove that H F = (cid:2) G r r H G (cid:3) ◦ h on Z .Recall that t ( d − H F = h F on Z and that t ( d − H G = h G on V ( G ) with h F := (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) F xx F xy F xz F x F xy F yy F yz F y F xz F yz F zz F z F x F y F z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) and h G := (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) G rr G rz G r G rz G zz G z G r G z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . We will write as usual G r , G z , G t for the first order derivatives of G and G rr , G rz , G rt , G zt and G tt for the second order derivatives. We also write F r := G r ◦ h and we define analogously F xr , F yr , F zr and F rr . Due to the particular form of F , we immediately obtain that F x = xr F r , F y = yr F r , F xt = xr F rt , F yt = yr F rt ,F xx = F r r + x r F rr − x r F r , F yy = F r r + y r F rr − y r F r ,F xy = xyr F rr − xyr F r , F xz = xr F rz , F yz = yr F rz . with r := (cid:112) x + y . Due to these relations and to the above formula of h F , we have h F = 1 ry (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) rF r − xF r r y r F rr yF rz yF r yr F rz F zz F z yr F r F z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = F r r (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) F rr F rz F r F rz F zz F z F r F z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) writing respectively L i and C i for the i-th line and for the i-th row, we make successively the following linearchanges: L ← yL − xL , C ← yC − xC and L ← rL + xr L N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 27 and (20) follows. Now let us prove that β = (cid:20) β S ,G − S G ) G r r (cid:21) ◦ h. (21)Using the above formulas, we obtain F xx + F yy + F zz = F rr + F zz + F r r . Now (21) comes fromthe definition of β , and from the above expressions of F x , F y , F xz , F yz F zz , F xt , F yt and F tt .Let m [ x : y : z : t ] ∈ Z \ V ( α, β ). To prove the result, it is enough to prove that the twosolutions [ λ : λ ] of (3) are [ λ (1)0 : λ (1)1 ] = [2∆ S G.G r : r ] ◦ h and [ λ (2)0 : λ (2)1 ] = [ − h G N S : t ∆ S G ] ◦ h = Φ S ,G ◦ h . Indeed, since r = (cid:112) x + y , x = y = 0, F x = xr F r and F y = yr F r , wehave M := λ (1)0 · Id + λ (1)1 · σ ∈ V ( x, y ) and M := λ (2)0 · Id + λ (2)1 · σ = (cid:2) x R r : y R r : Z : T (cid:3) ◦ h (using Lemma 41) with Φ S ,G = ( R , Z , T ). Observe that R /r ∈ C [ r, z, t ]. Due to Theorem40, the Zariski closure of Φ S ,V ( G ) ( V ( G )) is the caustic Σ S ( V ( G )). Observe that, for every( r, z, t ), the set [ x : y ] goes along P when ( x, y ) moves in { ( x, y ) ∈ C : x + y = r } . So theZariski closure of M ( Z ) is the revolution surface R (Σ S ( V ( G ))). Observe moreover that theZariski closure of M ( Z ) is V ( x, y ) unless it is a single point A of V ( x, y ), which would meanthat every reflected line contains this point A , this would imply that Σ S ( V ( G )) is reduced toa point and so that V ( G ) is a conic and S one of its focal point (see e.g. [10]). To prove that[ λ (1)0 : λ (1)1 ] and [ λ (2)0 : λ (2)1 ] are the solutions of (3), it is enough to prove that[ γ : α ] = [ λ (1)0 λ (2)0 : λ (1)1 λ (2)1 ] (22)and [ − β : α ] = [ λ (1)0 λ (2)1 + λ (2)0 λ (1)1 : λ (1)1 λ (2)1 ] . (23)Now (22) comes from [ λ (1)0 λ (2)0 : λ (1)1 λ (2)1 ] = [ − h G N S G r : rt ] ◦ h , (19) and (20). Now (23) isequivalent to β = (cid:104) − S G ) G r r + h G N S t (cid:105) ◦ h . So (23) comes from (21) and Lemma 41. (cid:3) Remark 43.
Due to Theorem 42, the caustic by reflection of a sphere S of center A from S (cid:54) = A is the union of the line ( A S ) and of the revolution surface of axis ( A S ) obtained fromthe caustic curve of the circle S ∩ P where P is any plane containing ( A S ) . We consider the case where Z = V ( x + y − zt ) with S = [0 : 0 : z : 1] (with S [0 : z : 1])with z (cid:54) = 1 / S ∈ V ( x, y ) and S is not a focal point of P ). Observe that Z = R ( V ( G ))with G ( r, z, t ) = r / − zt . Due to [9], Σ S ( V ( G )) has degree 6 except if z = 0 (correspondingto S ∈ Z ) and, in this last case, Σ S ( V ( G )) has degree 4. More precisely: Proposition 44.
Let Z = V ( x + y − zt ) ⊂ P and S [0 : 0 : z : 1] ∈ P with z (cid:54) = 1 / . Then Σ S ( Z ) = V ( x, y ) ∪ R ( V ( H )) , where • if z (cid:54) = 0 , the curve V ( H ) is the sextic given by H ( r, z, t ) := 27 r z − z t + 288 z r t + 108 r t z +(3072 zt − r zt − t − z t +4992 zr t +4096 z t − z r t − r t − r t − r ) z +( − zt − zr t − z t +6144 z t +288 r t +108 r z − r zt +3195 r t +72 r +2688 z r t ) z + ( − z t + 3072 z t + 90 r zt − r z − r t − z r t + 4032 zr t − r ) z . • if z = 0 , the curve V ( H ) is the cuartic given by H ( r, z, t ) := 27 r − zt + 288 r t .Proof. We have Z = R ( V ( G )) with G ( r, z, t ) = ( r − zt ) /
2. Due to Theorem 42, we know thatΣ S ( Z ) = V ( x, y ) ∪ R (Σ S ( C )) with C = V ( G ) ⊆ P and S [0 : z : 1] ∈ P . Observe that C N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 28
Figure 2.
Caustics C (cid:48) of V ( r − zt ) from [0 : z : 1] for z = − , − , , , , z , the two-dimensional part of the caustic by reflection of V ( x + y − zt ) from [0 : 0 : z : 1] is the revolution surface of C (cid:48) around V ( x, y ).admits the rational parametrization ( u, v ) (cid:55)→ [ uv : ( u /
2) : v ]. Due to Theorem 40, Σ S ( C ) hasthe rational parametrization ( u, v ) (cid:55)→ Φ S ,G ([ uv : ( u /
2) : v ]) and Φ S ,G (cid:18) uv, u , v (cid:19) = (cid:18) u v (1 − z ) ,
14 (4 z v + 6 z v u ( v − u ) + u ( u + 6 v )) ,
12 (2 z − v (2 z v − u ) (cid:19) , which parametrizes Σ S ( C ). So Σ S ( C ) = V ( H ). (cid:3) A.3.
Caustic of a cylinder.
To simplify, we restrict ourselves to the study of a cylindricalsurface Z = V ( F ) with axis V ( x, y ). We will use the fact that F ( x, y, z, t ) = G ( x, y, t ) for somehomogeneous polynomial G ∈ C [ x, y, t ]. Such a surface Z is called the cylinder of axis V ( x, y )and of basis V ( G ) ⊂ P . We then write Z = Cyl ( G ). Observe that, in this particular case, thetangent plane to Z at m = [ x : y : z : t ] does not depend on z . Remark 45. If Z = Cyl ( G ) (with G as above) and if S [0 : 0 : 1 : 0] ∈ P , then Z ⊆ V (∆ S F, ( F x + F y + F z ) Hess F ( S , S )) . Theorem 46.
Let S [ x : y : z : t ] ∈ P \{ [0 : 0 : 1 : 0] } and let Z = Cyl ( G ) with G ∈ C [ x, y, t ] an irreducible homogeneous polynomial of degree d ≥ . Assume that Z (cid:54)⊆ V (∆ S F, ( F x + F y + F z ) Hess F ( S , S )) . We set S [ x : y : t ] ∈ P .If V ( G ) (cid:54)⊆ V ( H G , N S ) , then Σ S ( Z ) = σ ( Z ) ∪ Cyl (Σ S ( V ( G ))) , where σ ( Z ) is the algebraiccurve corresponding to the Zariski closure of the sets of orthogonal symmetrics of S with respectto the tangent planes to Z . Otherwise Σ S ( Z ) = σ ( Z ) .Proof. Observe that, since F xz = F yz = F zz = F zt = 0, we have H F = 0 and so γ = 0. Let m [ x : y : z : t ] ∈ Z \ V ( α, β ). We have α = ∆ S F = ∆ S G ◦ h and β = β S ,G ◦ h with h ( x, y, z, t ) = ( x, y, t ). So (3) becomes λ ( α ( m ) λ + β ( m ) λ ) = 0 and its solutions [ λ : λ ] ∈ P are [0 : 1] and [ − β ( m ) : α ( m )]. The corresponding points on Σ S ( Z ) are M ( m ) := σ ( m ) and N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 29 M ( m )[ X ( m ) : Y ( m ) : Z ( m ) : T ( m )] with [ X : Y : T ] = Φ S ,V ( G ) (due to Lemma 41) and Z ( m ) = (cid:104) − H G ( h ( m )) N S ( h ( m ))( d − z + ∆ S G ( h ( m ))( G x ( h ( m )) + G y ( h ( m )) z (cid:105) . Due to Theorem 40,the Zariski closure of Φ S ,V ( G ) ( V ( G )) is Σ S ( V ( G )). If V ( G ) (cid:54)⊆ V ( H G N S ), then, for every[ x : y : t ] ∈ V ( G ), Z ( x, y, z, t ) goes all over C when z describes C . If V ( G ) ⊆ V ( H G N S ), then,due to Lemma 41, β = 0 on Z , which implies that M = M on Z . (cid:3) Proposition 47 (parabolic cylinder with light at infinity) . The caustic of reflection of Z = V ( y − xt ) ⊂ P from S [1 − v : 2 v : z : 0] with v (cid:54) = 0 is σ ( Z ) ∪ V ( H ) , with H ( x, y, z, t ) = 4 y (1 − v ) + ( − t + 108 xt − y t + 24 xy − x t )( v + v )+ 12 y (2 x + 3 t + y )(2 x + 3 t − y )( v − v )+ 2(16 x + 27 t − x t − xy + 216 xt − y t ) v , (geometrically, v corresponds to the tangent of the half-angle of (1 , with the direction of S ). Figure 3.
Caustics of V ( H ) for t = 0 and x + iy = e iθ for θ = π , π , π , π .Such surfaces are used in practice to concentrate sunrays on a tube (put along the line madeof the focal points of the parabols) in order to heat the water circulating in it. Figure 3 is atransverse representation of this solar heater, the tube being at the focal point (1 / , Proof.
Let S [1 − v : 2 v : 0] and C := V ( G ) ⊂ P with G ( x, y ) := ( y − xt ) /
2. Due to Theorem46, the caustic by reflection of V ( y − xt ) ⊂ P from S is V ( x, y ) ∪ Cyl (Σ S ( C )). Moreover weknow from [9] that deg Σ S ( C ) = 3. Using the parametrization ψ : ( a, b ) (cid:55)→ (cid:104) a : ab : b (cid:105) of C together with Theorem 40, we conclude that Φ S ,V ( G ) ◦ ψ ( a, b ) = [ X ( a, b ) : Y ( a, b ) : Z ( a, b )] isa parametrization of Σ S (cid:48) ( C ). We obtain σ S ,G = (( y − t ) x + 2 ty y, ( t − y ) y + 2 tx y and so X = 2 v (1 − v ) a b + 12 v a b − v (1 − v ) ab + (1 − v ) b Y = − v a b + 6 v (1 − v ) a b + 12 v ab − v (1 − v ) b , Z = 2(1 + v ) b . (cid:3) To complete the study of this example, let us specify σ ( Z ). Proposition 48.
Under assumptions of the previous result, we have • if v (cid:54) = − , then σ ( Z ) = H ∞ ∩ V (( x + y ) z − ( v + 1) z ) ; • if z (cid:54) = 0 and v = − , then σ ( Z ) = H ∞ ∩ V ( y + 2 x ) , • if z = 0 and v = − , then σ ( Z ) = { [2 : − v : 0 : 0] } . N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 30
Proof.
We have σ ( x, a, z, b ) = a (1 − v ) + 4 vab − b (1 − v ) =: g ( a, b ) a ( − v ) + 2(1 − v ) ab + b v =: g ( a, b )( a + b ) z =: g ( a, b )0 . First observe that σ ( Z ) is included in H ∞ . Second, we compute the resultant in a of( xg ( a, − g ( a, , yg ( a, − g ( a, v ) (( x + y ) z − ( v + 1) ). If v (cid:54) = −
1, this resultant gives the result (by homogeneization with z ).Assume now that v = 1. We have σ ( x, a, z, b ) = ( a + vb ) a + vb ) − v ( a + vb )( a − vb ) z . If z = 0, we have σ ( x, a, z, b ) = ( a + vb ) (2 , − v, ,
0) and so σ ( Z ) = { [2 : − v : 0 : 0] } . Finally,if z (cid:54) = 0 (still with v = 1), using the fact that [ a − vb : a + vb ] describes P when [ a : b ] movesin P , we obtain σ ( Z ) = H ∞ ∩ V ( y + 2 x ). (cid:3) References [1] J. W. Bruce, P. J. Giblin and C. G. Gibson.
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N CAUSTICS BY REFLECTION OF ALGEBRAIC SURFACES 31
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