Shadow of a Cubic Surface

Home , Cubic surface, Quartic plane curve

Faculteit B`etawetenschappen

Shadow of a cubic surface

Bachelor Thesis

Rein Janssen Groesbeek Wiskunde en Natuurkunde

Supervisors: Dr. Martijn Kool Departement Wiskunde Dr. Thomas Grimm ITF

June 2020 Abstract 3 For a smooth cubic surface S in P we can cast a shadow from a point P ∈ S that does not lie on one of the 27 lines of S onto a hyperplane H. The closure of this shadow is a smooth quartic curve. Conversely, from every smooth quartic curve we can reconstruct a smooth cubic surface whose closure of the shadow is this quartic curve. We will also present an algorithm to reconstruct the cubic surface from the bitangents of a quartic curve. The 27 lines of S together with the tangent space TP S at P are in correspondence with the 28 bitangents or hyperﬂexes of the smooth quartic shadow curve. Then a short discussion on F-theory is given to relate this geometry to physics.

Acknowledgements I would like to thank Martijn Kool for suggesting the topic of the shadow of a cubic surface to me and for the discussions on this topic. Also I would like to thank Thomas Grimm for the suggestions on the applications in physics of these cubic surfaces. Finally I would like to thank the developers of Singular, Sagemath and PovRay for making their software available for free.

i Contents

1 Introduction 1

2 The shadow of a smooth cubic surface 1 2.1 Projection of the ﬁrst polar ...... 1 2.2 Reconstructing a cubic from the shadow ...... 5

3 The 27 lines and the 28 bitangents 9 3.1 Theorem of the apparent boundary ...... 9 3.2 The last bitangent ...... 11 3.3 Injectivity ...... 13

4 Recovering the cubic from a quartic with computer algebra 14 4.1 Calculating the bitangents ...... 15 4.2 Calculating the points of contact ...... 16 4.3 Finding a syzygetic quadruple of bitangents ...... 17 4.4 Finding a fourth bitangent ...... 18 4.5 Recovering the cubic surface ...... 18

5 Algebraic geometry in physics: F-theory 19 5.1 D7-branes ...... 19 5.2 Elliptic ﬁbration ...... 19 5.3 Intersecting D7-branes ...... 20 5.4 Born infeld action ...... 20

6 Appendix: lemmas and notation 23 6.1 Notation ...... 23 6.2 Lemmas ...... 23 1 Introduction

Two classical results in algebraic geometry are that a smooth cubic surface in P3 contains 27 lines and a smooth quartic curve in P2 contains 28 contact lines.

Figure 2: A smooth quartic curve with 28 Figure 1: A smooth cubic surface containing contact lines. 27 lines.

In the following paper we will show how by taking the shadow of the smooth cubic surface, also known as the projection from a point, one can obtain a smooth quartic curve and relate the 27 lines to the 28 contact lines. The paper is structured as follows:

• In the second chapter we will show how the (closure of the) shadow of a smooth cubic surface is a smooth quartic curve. To show smoothness, we use ideas from section 5.8.12 of [Bel09]. Then for a smooth quartic curve we will give a construction of a smooth cubic surface that has this curve as its shadow. The idea of this construction is from exercise 5.7.11 of [Bel09] and the worked out details in coordinates is new.

• In the third chapter we will show how the 27 lines get mapped to the 28 contact lines. Here we use ideas from section 5.8.9 of [Bel09].

• In the fourth chapter we will describe an algorithm that reconstructs a smooth cubic surface from a smooth quartic curve. This is largely based on the paper and code samples from [PSV11].

• In the ﬁfth chapter we will give a short sketch of how algebraic surfaces are used in F-theory, this is unrelated to the previous chapters so can be read separately. The main reference for this chapter is [Wei18].

2 The shadow of a smooth cubic surface 2.1 Projection of the ﬁrst polar We will work in 3 = 3 , projective 3-dimensional space over , but the results also hold over P PC C any algebraically closed ﬁeld of characteristic zero.

1 For a smooth cubic surface S in P3 we can create its shadow by projecting it from a point P ∈ S. Because S is an algebraic variety, we expect that the shadow also is algebraic. In this section we will show that if we choose P to not lie on a line of S, then the shadow (or the closure thereof) is a smooth quartic curve. After that we will also show that any smooth quartic can be considered the projection of some smooth cubic surface.

Definition 1. For S a smooth hypersurface in P3 we define the shadow Sh(S, P, H) from a point P ∈ S onto a hyperplane H not containing P , to be the locus of points PQ ∩ H where PQ ⊂ TQS is a line tangent to S at some Q ∈ S \{P }. ♦ This definition is close to our intuition of a shadow, but it is cumbersome to work with. We can reformulate it using the notion of a first polar.

n Definition 2. Let X be a smooth hypersurface in P given by the polynomial f ∈ k[x0, . . . , xn] of degree r. Let P be a point in X. The first polar X1(P ) of X at P is defined as the hypersurface given by the degree r − 1 polynomial

n X ∂f ∆1 f := P . (1) P i ∂x i=0 i So this is similar to the equation for the tangent space, except that we now leave the argument of the derivatives as variables, and we ﬁx the point in the tangent space. This can be generalized to higher polars, for 0 ≤ k ≤ r we deﬁne the k-th polar Xk(P ) as the hypersurface given by the degree r − k polynomial

k k X ∂ f ∆P f := Pi1 ····· Pik . (2) ∂xi1 . . . ∂xik i1,...,ik=0,...,n

♦

Now we will give an equivalent definition of a shadow using the first polar S1(P ). Note that S was chosen to be smooth, so this construction indeed works. Let π : P3 \{P } → H be the projection that maps Q to the intersection PQ ∩ H. Let S1(P ) be the first polar of S at P . If S is given as the zero set of the cubic polynomial f, 1 P3 ∂f then S1(P ) is the quadric hypersurface with equation ∆ f := Pi . P i=0 ∂xi If a point Q 6= P lies in S and in S1(P ), then P lies in the tangent space at Q, which means the line PQ is tangent at Q. For every point R ∈ Sh(S, P, H) ⊂ H there exists a point Q ∈ S ∩S1(P ) with Q 6= P such that π(Q) = PQ ∩ H = R, which means we have Sh(S, P, H) ⊂ π(S ∩ S1(P ) \{P }). Conversely, for Q ∈ S ∩ S1(P ) \{P }, the line PQ intersects H in π(Q), and this intersection lies in Sh(S, P, H), so π(S ∩ S1(P ) \{P }) ⊂ Sh(S, P, H). This means we get an alternative definition of Sh(S, P, H) = π(S ∩ S1(P ) \{P }) as the projection of the intersection of S with its first polar at P . Note that the previous discussion applies to any smooth hypersurface in Pn and not just for smooth cubic surfaces in P3. We have that S ∩ S1(P ) \{P } is not closed in S because it is missing the point P . Similarly Sh(S, P, H) is not closed in H because we are missing the limiting line obtained by taking the limit of the lines PQ as Q approaches P , thus the tangent lines of S at P . Thus if we take the closure of Sh(S, P, H) in H we add these missing points. In the following theorem we will prove that π induces an isomorphism of varieties between S ∩ S1(P ) \{P } and Sh(S, P, H) and that the closure Sh(S, P, H) in H is a smooth plane quartic.

2 Theorem 1. Let S be a non-singular cubic hypersurface in P3. Let P ∈ S a point lying outside the 27 lines of S, and H a hyperplane not containing P . Then Sh(S, P, H), the closure of its shadow in H, is a smooth quartic curve. Proof. We ﬁrst choose a change of coordinates g ∈ GL(4, k) such that g([0 : 0 : 0 : 1]) = P and g(Z(x3)) = H, for the details on the construction of g see Lemma 6. From now on we will use these new coordinates. This means that S is given as the zero set of the homogeneous polynomial

2 f = x3f1 + x3f2 + f3 (3) with f1, f2, f3 homogeneous polynomials of degree 1,2,3. Also we can identify points in H with 2 points in P by the identification [x0 : x1 : x2 : 0] ↔ [x0 : x1 : x2], which means we can consider the shadow Sh(S, P, H) to be a subset of P2. 3 2 Let π : P \{P } → P be the projection from a point defined as π([x0 : x1 : x2 : x3]) = [x0 : x1 : x2]. Let C := S ∩ S1(P ) be the intersection of S with S1(P ), its first polar at P . The first polar at P has equation 3 X ∂f ∂f ∆1 f = P = 1 · = 2x f + f . (4) P i ∂x ∂x 3 1 2 i=0 i 3 2 This means C = Z(x3f1 + x3f2 + f3, 2x3f1 + f2). 2 We claim that π restricts to an isomorphism of varieties between C \{P } and Z(f2 − 4f1f3) \ −1 f2(a0,a1,a2) Z(f1), where the inverse map is given by π :[a0 : a1 : a2] 7→ [a0 : a1 : a2 : − ]. 2f1(a0,a1,a2)

• Let A = [a0 : a1 : a2 : a3] ∈ C \{P }. If f1(a0, a1, a2) = 0, then this would imply that 1 0 = ∆P f(A) = 0 + f2(a0, a1, a2) and 0 = f(A) = 0 + 0 + f3(a0, a1, a2), which means the entire line λA + µP is contained in S, and we presumed that P does not lie on one of the 27 lines of S. 1 Thus f1(a0, a1, a2) 6= 0. This means that from 0 = ∆P f(a0, a1, a2, a3) = 2a3f1(a0, a1, a2) + f2(a0,a1,a2) f2(a0, a1, a2) we can get that a3 = − . 2f1(a0,a1,a2) 2 2 2 2 f2 f2 f2 2 This means 0 = a f1 + a3f2 + f3 = − + f3 = − + f3 which implies (f − 3 4f1 2f1 4f1 2 2 4f1f3)(a0, a1, a2) = 0. Thus π(A) = [a0 : a1 : a2] ∈ Z(f2 − 4f1f3) and π(A) 6∈ Z(f1). 2 • Conversely, if [a0 : a1 : a2] ∈ Z(f2 − 4f1f3) \ Z(f1), we have

−1 f2(a0, a1, a2) f(π [a0 : a1 : a2]) = f(a0, a1, a2, − ) 2f1(a0, a1, a2) − 1 f 2(a , a , a ) + f (a , a , a )f (a , a , a ) = 4 2 0 1 2 1 0 1 2 3 0 1 2 = 0 (5) f1(a0, a1, a2)

1 f2(a0, a1, a2) and ∆P f(a0, a1, a2, − ) = 0 2f1(a0, a1, a2)

−1 −f2(a0,a1,a2) −1 so π [a0 : a1 : a2] = [a0 : a1 : a2 : ] ∈ S∩S1(P ), and because π [a0 : a1 : a2] 6= P 2f1(a0,a1,a2) −1 because (a0, a1, a2) 6= (0, 0, 0), we have that π [a0 : a1 : a2] ∈ S ∩ S1(P ) \{P }.

−1 f2(a0,a1,a2) Also we have π π([a0 : a1 : a2 : a3]) = [a0 : a1 : a2 : − ] = [a0 : a1 : a2 : a3] and • 2f1(a0,a1,a2) −1 ππ ([a0 : a1 : a2]) = [a0 : a1 : a2].

3 2 This proves that π is an isomorphism between C \{P } and Z(f2 −4f1f3)\Z(f1) and Sh(S, P, H) = 2 π(S ∩ S1(P ) \{P }) = Z(f2 − 4f1f3) \ Z(f1). 2 2 We will now prove smoothness of Z(f2 −4f1f3) by ﬁrst showing smoothness at Z(f2 −4f1f3)\ 2 Z(f1) and then smoothness at Z(f2 − 4f1f3) ∩ Z(f1). • We will use the following identity: 2 2 2 1 2 f2 − 4f1f3 = (2x3f1 + f2) − 4f1(x3f1 + x3f2 + f3) = (∆P f) − 4f1f. (6)

2 −1 Let A = [a0 : a1 : a2] ∈ Z(f2 − 4f1f3) \ Z(f1), so π A ∈ C \ P . Then by diﬀerentiating (6) in π−1A, for i = 0, 1, 2 we have

1 2 2 ∂f ∂(∆P f) ∂(f2 − 4f1f3) ∂f1 4f1(A) = − − 4f(A) ∂xi π−1A ∂xi π−1A ∂xi π−1A ∂xi π−1A 1 2 1 ∂∆P f ∂(f2 − 4f1f3) = 2∆P f(A) − − 0 (7) ∂xi π−1A ∂xi π−1A ∂(f 2 − 4f f ) = 2 1 3 . ∂xi π−1A −1 Thus by smoothness of f at π A ∈ S and because f1(A) 6= 0, we have that there is an index 2 ∂f ∂(f2 −4f1f3) 0 ≤ j ≤ 2 such that 4f1(A) | −1 6= 0, so | −1 6= 0. ∂xj π A ∂xj π A 2 Thus Z(f2 − 4f1f3) \ Z(f1) is smooth. 2 2 • Next we will prove smoothness at Z(f2 − 4f1f3) ∩ Z(f1). Let A = [a0 : a1 : a2] ∈ Z(f2 − 4f1f3) ∩ Z(f1). 2 2 Then f1(a0, a1, a2) = 0 and 0 = (f2 − 4f1f3)(a0, a1, a2) = f2 (a0, a1, a2) so f2(a0, a1, a2) = 0. 1 If f3(A) = 0, then we have for [λ : µ] ∈ P that 2 2 3 f(λ[a0 : a1 : a2 : 0] + µP ) = µ λf1(a0, a1, a2) + µλ f2(a0, a1, a2) + λ f3(a0, a1, a2) = 0 + 0 + 0 (8) which means the line λ[a0 : a1 : a2 : 0]+µP is contained in S, which contradicts the assumption that P does not lie on one of the 27 lines contained in S. Thus f3(A) 6= 0. 3 The set Z(f1) in P is the tangent space TP S of S, because by Lemma 3 we have 1 1 ∂2f 1 ∆1 f(P ) = ∆3−1f(Q) = = f (Q) (9) Q P 2 1 (3 − 1)! 2 ∂x3 Q 2 for all Q ∈ P3. So by smoothness of S at P the tangent space is a plane, thus there exists an index 0 ≤ j ≤ 2 such that ∂f1 6= 0. ∂xj

∂f1 This means 4 ∂x f3(A) 6= 0 which means we have j A 2 ∂(f2 − 4f1f3) ∂f2 ∂f3 ∂f1 ∂f1 = 2f2(A) −4f1(A) −4 f3(A) = −4 f3(A) 6= 0. (10) ∂xj A ∂xj A ∂xj A ∂xj A ∂xj A 2 2 So f2 − 4f1f3 is non-singular at A which means Z(f2 − 4f1f3) ∩ Z(f1) is smooth. 2 2 Thus the entire curve Z(f2 − 4f1f3) is smooth. Then because a smooth curve in P is irreducible, 2 2 we have that Z(f2 − 4f1f3) is irreducible, and Sh(S, P, H) = Z(f2 − 4f1f3) \Z(f1) is a non-empty 2 2 open subset of Z(f2 −4f1f3), so by irreducibility we get Sh(S, P, H) = Z(f2 −4f1f3) (a non-empty open subset is dense in an irreducible space).

4 2.2 Reconstructing a cubic from the shadow 4+2 A quartic homogeneous polynomial in 3 variables has 2 = 15 variable coefficients, while a cubic 3+3 polynomial in 4 variables that passes through P like in equation (3) has 3 − 1 = 20 − 1 = 19 variable coefficients. This means we have a projective space P14 of coefficients of quartic polynomials and a projective space P18 of coefficients of cubic polynomials that pass through a point P , so by a naive comparison of dimensions, we expect that every smooth quartic plane curve should be able to be written as the shadow of a smooth cubic surface. This indeed turns out to be the case; in the following theorem we will see that a quartic plane curve can be written as such a shadow if it has at least one bitangent, which is true if the quartic curve is smooth. We will now define the bitangent and the closely related hyperflex. Before we do that, we need to define the concept of intersection multiplicity in A2. Definition 1. Let X,Y ⊂ A2 be two curves given by polynomials f, g ∈ k[x, y] and suppose P ∈ X ∩ Y . We define the intersection multiplicity of X and Y at P = (Px,Py), denoted by i(X,Y ; P ) as the non-negative integer

i(X,Y ; P ) := dimk(k[[x, y]]/(f(x − Px, y − Py)g(x − Px, y − Py))). (11)

Here k[[x, y]] is the ring of formal power series in x and y and dimk is the vector space dimension.

If we want to calculate the intersection multiplicity at an intersection point P ∈ P2, we can choose an affine chart A2 around P and use the definition above. We can now define bitangents and hyperflexes.

Deﬁnition 2. Let C ⊂ P2 be a smooth curve.

• A line ` ⊂ P2 is a bitangent of C if there exist two distinct points P,Q ∈ C, P 6= Q such that i(`, C; P ) = i(`, C; Q) ≥ 2.

• A line ` ⊂ P2 is a hyperﬂex of C if there exists a point P ∈ C such that i(`, C; P ) = 4. • A line ` ⊂ P2 is a contact line of C if for each point P ∈ ` ∩ C we have i(`, C; P ) ≥ 2. Thus a hyperﬂex is the limiting case of the bitangent when the two points P,Q approach each other. The following theorem is useful in calculations with intersection multiplicites:

Theorem 2 (B´ezout). Let f, g ∈ k[x0, x1, x2], f =6 g be two distinct homogeneous polynomials without common factors, deﬁning two curves in P2. Then X i(Z(f),Z(g); P ) = deg f · deg g. (12) P ∈Z(f,g)

Proof. See [Har77], Theorem II.7.8. The following result is an application of Bezout.

Lemma 1. Let C ⊂ P2 be a smooth quartic curve. Let ` ⊂ P2 be a contact line of C. Then ` is a bitangent or a hyperflex of C. P Proof. We have by Bezout’s theorem that P ∈`∩C i(C, `; P ) = 1 · 4, and because ` is a contact line of C we have i(C, `; P ) ≥ 2 for all P ∈ ` ∩ C. Then because we can split 4 into 2+2 or 4, we either have 2 points P1,P2 with i(C, `; Pi) = 2 for i = 1, 2 or a single point P1 with i(C, `; P1) = 4. So ` is a bitangent in the first case and a hyperflex in the second.

5 We will use the following Black Box, which is a statement that we will not prove.

Black Box 1. Let C ⊂ P2 be a smooth quartic curve. Then C has a bitangent. Proof. From section 6 of [Dol12] we know that C has 28 contact lines, thus bitangents or hyperflexes, and from [KK77] we know that C has at most 12 hyperflexes, where the only quartics having 12 hyperflexes are x4 + y4 + z4 = 0, the Fermat quartic, and x4 + y4 + z4 + 3(x2y2 + y2z2 + z2x2) = 0. Thus C has at least 28 − 12 = 16 bitangents. We can now use this bitangent to reconstruct the cubic surface. The following idea is based on Exercise 5.7.11 of [Bel09]

Theorem 3. Let C be a quartic curve with a bitangent. Then there exist homogeneous polynomials 2 f1, f2, f3 ∈ k[x0, x1, x2] such that C = Z(f2 − 4f1f3).

Proof. Let C be given by the quartic polynomial f ∈ k[x0, x1, x2] and let its bitangent ` be given by the linear polynomial f1 ∈ k[x0, x1, x2]. Let P 6= Q be the two distinct points of contact of ` with C. 1 1 We have that P ∈ TQC and Q ∈ TP C, which means ∆P f(Q) = ∆Qf(P ) = 0. By the double 3 3 reading method of polars (Lemma 3), this implies that also ∆P f(Q) = ∆Qf(P ) = 0. Also note 0 0 that ∆Qf(P ) = f(P ) = 0 and ∆Qf(P ) = f(Q) = 0 and by the double reading method we get 4 4 ∆Qf(P ) = ∆P f(Q) = 0. Thus if we Taylor expand f restricted to the line ` in terms of its polars, we can write

4 X λ4−iµi λ2µ2 f(λP + µQ) = ∆i f(Q) = 0 + 0 + ∆2 f(Q) + 0 + 0. (13) i! P 2 P i=0 Of course because f deﬁnes a smooth quartic, it is irreducible, so it cannot contain `, which 2 means ∆P f(Q) 6= 0. λ2µ2 2 To be able to write f in the desired form, we want to eliminate the 2 ∆pf(Q) term by 2 adding a doubled conic f2 onto it. Choose a degree two polynomial f2 ∈ k[x0, x1, x2] that deﬁnes an irreducible conic passing 2 through P and Q. Then square it to get a quartic curve f2 , for this curve we have for i = 0, 1, 2 ∂f 2 ∂f 2 2 ∂f2 2 1 2 1 2 that ∂x = 2f2(P ) · ∂x = 0 and similarly ∂x = 0, which means ∆Af2 (P ) = ∆Af2 (Q) = 0 i P i P i Q for all A ∈ P2. 3 2 3 2 By the double reading method of polars (Lemma 3), we also get ∆P f2 (A) = ∆Qf2 (A) = 0 2 0 2 2 for all A ∈ P . Also because the conic passes through P and Q, we have ∆Qf2 (P ) = f2 (P ) = 0 2 4 2 4 2 f2(P ) = 0 and ∆P f2 (Q) = 0, and ∆Qf2 (P ) = ∆P f2 (Q) = 0 by the double reading method. 2 Then if we Taylor expand f2 restricted to the line ` in terms of its polars, we get

4 X λ4−iµi λ2µ2 f 2(λP + µQ) = ∆i f 2(Q) = 0 + 0 + ∆2 f 2(Q) + 0 + 0. (14) 2 i! P 2 2 P 2 i=0

2 2 2 Note that if ∆P f2 (Q) = 0, then ` is contained in Z(f2 ) = Z(f2), which means Z(f2) is reducible, 2 2 which is false. Thus ∆P f2 (Q) 6= 0.

6 2 1 Now consider the family of quartics af + bf2 for [a : b] ∈ P and restrict this family on the line `. We then get the equation

4 4 X λ4−iµi X λ4−iµi (af + bf 2)(λP + µQ) = a ∆i f(Q) + b ∆i f 2(Q) 2 i! P i! P 2 i=0 i=0 (15) λ2µ2 = (a∆2 f(Q) + b∆2 f 2(Q)). 2 P P 2

2 ∆P f(Q) 2 We can then choose a = 1 and b = − 2 2 to get (af + bf2 )(λP + µQ) = 0 for all ∆P f2 (Q) [λ : µ] ∈ P1. 2 Thus we get that f + bf2 splits into the line ` = Z(f1) and some cubic f3 ∈ k[x0, x1, x2], so 1 f + bf 2 = f · f or f = −bf 2 + f · f = −b(f 2 − 4 · ( · f ) · f ). (16) 2 1 3 2 1 3 2 4b 1 3

1 Thus if we remove the scalar factor −b and insert the factor 4b inside f1, then we can write 2 C = Z(f) = Z(f2 − 4f1f3) as we wanted.

3 3

2 2

1 1

0 0

-1 -1

-2 -2

-3 -3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3

Figure 3: Quartic plane curve splitting in a bitangent line and a cubic. On the left we have the quartic f, one of its bitangents ` = Z(f1) and a conic f2 passing through its bitangent points. 2 On the right we have in brown the quartic f + bf2 splitting into a line and a cubic. The equation of the quartic f is xy3 + y − x 4 + x2 + x3 = 0.

Remark 1. The previous theorem also holds if we replace the bitangent ` by a hyperﬂex, and it can be proven in a similar way after making the following amendments: Let Q be the unique point of contact of C with `. We choose f2 to be an irreducible conic passing through Q and tangent to ` at Q. If we then 1 choose a point P ∈ ` with P 6= Q, then we have ∆P f2(Q) = 0 by tangency of ` with f2 at Q.

7 This means we have we have 0 2 2 ∆P f2 (Q) = f2 (Q) = 0 1 2 1 ∆P f2 (Q) = 2f2(Q)∆P f2(Q) = 0 2 2 2 2 2 2 2 X ∂ f2 X ∂f2 ∂f2 ∂ f2 ∆P f2 (Q) = PiPj = PiPj 2 + 2f2(Q) ∂x ∂x Q ∂x Q ∂x Q ∂x ∂x Q i,j=0 i j i,j=0 i i i j (17) 1 2 = 2(∆P f2(Q)) = 0 3 2 1 2 1 1 ∆P f2 (Q) = 6∆Qf2 (P ) = 6f2(P )∆Qf2(P ) = 6f2(P )∆P f2(Q) = 0 4 2 0 2 2 ∆P f2 (Q) = 24∆Qf2 (P ) = 24f2 (P ) 6= 0. Also because ` is a hyperflex of C at Q, which means ` touches C at Q with degree 4, we have 0 1 2 3 4 ∆P f(Q) = ∆P f(Q) = ∆P f(Q) = ∆P f(Q) = 0 and ∆P f(Q) 6= 0. 2 µ4 4 4 2 So we can look at the polar expansion (af + bf2 )(λP + µQ) = 4! (a∆P f(Q) + b∆P f2 (Q)) 4 ∆P f(Q) f(P ) again, and choose a = 1, b = − 4 2 = − 2 . ∆P f2 (Q) f2 (P ) Note that in both cases (` bitangent or hyperflex) we have some degree of freedom in choosing f2. In general we have five degrees of freedom in choosing the 6 coefficients of a degree 2 homogeneous polynomial in 3 variables. However in the case of a bitangent we enforce f2(Q) = f2(P ) = 0 so we 1 lose two degrees of freedom and in the case of a hyperflex we enforce f2(Q) = 0 and ∆P f2(Q) = 0 which also removes two degrees of freedom. So the space of conics f2 for which this construction works is 3 dimensional. These 3 dimensions correspond to the difference of 3 dimensions of the map P18 → P15 which sends the 19 coefficients 2 2 4 of the cubic x3f1 + x3f2 + f3 to the 15+1 coefficients of the quartic f2 − 4f1f3 + λx3. Corollary 1. If C ⊂ P2 is a smooth quartic curve, then there exists a smooth cubic surface S such that Sh(S, [0 : 0 : 0 : 1], {x3 = 0}) = C. 2 Proof. By the previous theorem and lemma, we can write C = Z(f2 − 4f1f3). 2 This means we can choose the cubic surface S given as the zero-set of f = x3f1 + x3f2 + f3. Now we check smoothness of S. Let P = [0 : 0 : 0 : 1].

2 • Smoothness at P : We have f1 =6 0, because if f1 is the zero polynomial, then C = Z(f2 ) is non-reduced thus not smooth.

At P = [0 : 0 : 0 : 1] we can look at f in the aﬃne chart {x3 =6 0}, and the dehomogeneization of f is f1 + f2 + f3, from here we can see that the tangent space TP S of S at P is given by the degree 1 part, thus TP S = Z(f1). Then because f1 6= 0, this implies that S is smooth at P . • Smoothness at S \{P }: We can check smoothness on S \{P } again using equation (6). 1 ∂f If S has a singularity A = [a0 : a1 : a2 : a3], with A 6= P , then ∆ f(A) = |A = 0, which P ∂x3 means A ∈ S1(P ). 2 So this means A ∈ S ∩ S1(P ) \{P }, thus [a0 : a1 : a2] = π(A) ∈ Z(f2 − 4f1f2). For i = 0, 1, 2 we have 2 1 ∂(f2 − 4f1f3) 1 ∂∆P f ∂f ∂f1 |A = 2∆P f(A) |A − 4f1(A) |A − 4f(A) |A ∂xi ∂xi ∂xi ∂xi (18) = 0 − 4f1(A) · 0 − 0.

2 This means A is a singularity of f2 − 4f1f3 which contradicts its smoothness. Thus S cannot have singularities, so it is smooth.

8 3 The 27 lines and the 28 bitangents

A famous result in classical algebraic geometry is that a smooth cubic surface in P3 contains exactly 27 lines and that a smooth quartic plane curve has 28 bitangents or hyperﬂexes. In this section we will show that for a smooth cubic surface S and a point P not lying on one of the lines of S, the 27 lines of S and the tangent plane TP S correspond to the 28 bitangents or hyperﬂexes on the shadow quartic curve.

3.1 Theorem of the apparent boundary We will ﬁrst prove a special case of the “theorem of the apparent boundary” from [Bel09, p. 156]. For a non-singular cubic surface S ⊂ P3 and a point P lying outside the 27 lines of S we recall the curve C := S1(P ) ∩ S, which we call the apparent boundary with respect to P . We also denote 3 by Γ := π(S1(P ) ∩ S \ P ) = Sh(S, P, H) the closure of the shadow in H, where π : P \{P } → H is the projection map Q 7→ PQ ∩ H from the point P onto a hyperplane H.

` C

S P2

TP1 S = TP2 S

π(P2)

π(P1)

π(`) Γ H

Figure 4: Idea of the theorem of the apparent boundary: The green plane TP1 S = TP2 S is simultaneously tangent to S at P1,P2 and to Γ at π(P1), π(P2).

The following theorem shows that lines ` ⊂ S become contact lines of Γ after you project them with π. The proof relies again on the identity of equation (6). Theorem 4. Let S ⊂ P3 be a non-singular cubic hypersurface. Let P be a point lying outside the 27 lines of S and H a hyperplane not containing P . For Q ∈ C \{P } we have TQS ∩ H = Tπ(Q)Γ. Proof. By a change of coordinates (Lemma 6) we may assume that P = [0 : 0 : 0 : 1] and 2 H = {x3 = 0}. We can identify H with P by ignoring the coordinate x3. 3 The projection map π : P \{P } → H can now be written in coordinates as [x0 : x1 : x2 : 2 x3] 7→ [x0 : x1 : x2]. Also we can write S = Z(x3f1 + x3f2 + f3) for f1, f2, f3 ∈ k[x0, x1, x2] homogeneous polynomials of degree respectively 1,2,3.

9 We make the following two observations.

1. For A, B ∈ P3 and µ, λ ∈ k, we have

3 2 1 2 X ∂(f2 − 4f1f3) ∆A+λP (f2 − 4f1f3)(B + µP ) = (Ai + λPi) ∂x B+µP i=0 i 3 2 X ∂(f2 − 4f1f3) = Ai (19) ∂x B+µP i=0 i 3 2 X ∂(f2 − 4f1f3) 1 2 = Ai = ∆A(f2 − 4f1f3)(B). ∂x B i=0 i

2 ∂(f2 −4f1f3) 2 The second equality follows from the fact that = 0 because f − 4f1f3 does not ∂x3 2 2 ∂(f2 −4f1f3) depend on x3, and the third equality follows from the fact that for i = 0, 1, 2 ∂xi does not depend on x3, which means it gives the same value when evaluated on B or B +µP .

2. For A, B ∈ P3 we have that 1 2 1 1 2 ∆A(f2 − 4f1f3)(B) = ∆A((∆P f) − 4f1f)(B) 1 1 1 1 1 (20) = 2∆P f(B) · ∆A(∆P f)(B) − 4f1(B) · ∆Af(B) − 4f(B) · ∆Af1(B). The ﬁrst equality follows from equation (6) and the second equality follows from the product rule of polars, see Lemma 4.

Let R ∈ TQS ∩ H. We then calculate

1 2 1 1 2 ∆ (f2 − 4f1f3)(π(Q)) = ∆R(f2 − 4f1f3)(Q) π(R) (21) 2 1 1 1 1 1 = 2∆P f(Q) · ∆R(∆P f)(Q) − 4f1(Q) · ∆Rf(Q) − 4f(Q) · ∆Rf1(Q).

Here =1 follows from observation 1 and =2 follows from observation 2. 1 Then note that because Q ∈ S1(P ) and Q ∈ S, we have ∆P f(Q) = 0 and f(Q) = 0. Also 1 because R ∈ TQS, we have ∆Rf(Q) = 0. Thus the right hand side of equation (21) is equal to zero. This means π(R) ∈ Tπ(Q)Γ. Furthermore because R ∈ H, we have R = π(R), so R ∈ Tπ(Q)Γ. This implies TQS ∩ H ⊂ Tπ(Q)Γ. Then because we have that one line is contained in the other, we conclude that the two lines are equal, so we conclude that Tπ(Q)Γ = TQS ∩ H.

Corollary 2. Let S and Γ be as in the previous theorem and let ` ⊂ S. Then π(`) is a contact line of Γ. Proof. Note that π(`) ∩ Γ has 4 points of intersection counted with multiplicity by B´ezout’s theorem (Theorem 2), so π(`) ∩ Γ 6= ∅. Let Q ∈ π(`) ∩ Γ, Then π−1Q ∈ ` ∩ C. Because ` ⊂ S, we have for R ∈ `, R 6= π−1Q that f(λR + µπ−1Q) = 0 for all [λ : µ] ∈ P1, thus by Taylor expansion we get

−1 3 0 −1 2 1 −1 2 2 −1 3 3 −1 0 = f(λR+µπ Q) = µ ∆Rf(π Q)+λµ ∆Rf(π Q)+λ µ∆Rf(π Q)+λ ∆Rf(π Q). (22)

10 Then because this equality must hold for all choices of [λ : µ] ∈ P1, we conclude that all coefficients 0 −1 1 −1 2 −1 3 −1 must be zero, so ∆Rf(π Q) = ∆Rf(π Q) = ∆Rf(π Q) = ∆Rf(π Q) = 0. 1 −1 So in particular we have ∆Rf(π Q) = 0 which means R ∈ Tπ−1QS, so ` ⊂ Tπ−1QS. This means π(`) = Tπ−1Q ∩ H. Then by the previous theorem we have Tπ−1Q ∩ H = Tππ−1QΓ = TQΓ which means π(`) is tangent to Γ at Q. Thus we see that π(`) is a bitangent or hyperflex of Γ by Lemma 1. Then depending on whether ` ∩ C consists of two points, so ` intersects C transversely, or ` ∩ C is one point, thus ` is tangent to C, we get that π(`) is a bitangent or a hyperflex. These two cases are illustrated in Figure 5.

` S ∩ S1(P )

S ∩ S1(P )

π(`) π(`)

Figure 5: Illustration of a line ` projecting to a bitangent in the left image and to a hyperﬂex on the right image. Note that on the left, ` is tangent to S and S1(P ) but not to S ∩ S1(P ), while on the right ` is tangent to S ∩ S1(P ).

3.2 The last bitangent 2 We have 27 lines in S and 28 bitangents or hyperflexes on Z(f2 − 4f1f3), so because each line of 4 S is mapped to a bitangent of Z(f2 − 4f1f3), then one bitangent (or hyperflex) is not the image of a line on S. We will show that this last bitangent is equal to the line TP S ∩ H. It will turn out that an interesting feature of this last bitangent/hyperflex is that the type of singularity of the curve S ∩ S1(P ) at P (node or cusp), corresponds to the type of tangency of the last bitangent in the projection (bitangent or hyperflex). We will first define what a node or cusp is.

Definition 3. Let C ⊂ P3 be a curve and P a of C. Choose an affine chart U of P3 containing P . Suppose that C in the affine chart U is given as the zero set of the ideal I. Then the tangent cone of TCP is the variety given as the zero set of the ideal in(I) which consists of all minimal degree terms in(f) for f ∈ I.

11 For example if I = (x2 + z3, x2 + y4) then x2 ∈ in(I) and z3 = in(z3 + x2 − (x2 + y4)) ∈ in(I).

Definition 4. Let C ⊂ P3 be a curve and P a double point of C. • We define P to be a node (or A1 singularity) of C if the tangent cone of C at P consists of two distinct lines `1, `2. • We define P to be a cusp (or A2 singularity) of C if the tangent cone of C at P consists of one line `1. We can now state the theorem.

Theorem 5. Let TP S be the tangent space of S at P . Then the line TP S ∩ H is a contact line of Γ.

• If TP S ∩ H is a bitangent of Γ, then S ∩ S1(P ) has a node singularity at P ,

• If TP S ∩ H is a hyperﬂex of Γ, then S ∩ S1(P ) has a cusp singularity at P .

2 Proof. We choose coordinates such that S = Z(f) with f = x3f1 + x3f2 + f3 and H = Z(x3) and P = [0 : 0 : 0 : 1] as in the previous theorem. Then by the previous chapter we have 2 Γ = Z(f2 − 4f1f3). Also we have shown earlier that TP S = Z(f1). Denote ` := TP S ∩ H. Let Q ∈ ` ∩ Γ, then we have to show that ` ⊂ TQΓ. For this it is suﬃcient to show for a point 1 2 R ∈ `, R 6= Q that R ∈ TQΓ, thus that ∆R(f2 − 4f1f3)(Q) = 0. We will use equation (6) again. We have

1 2 1 1 2 ∆R(f2 − 4f1f3)(Q) = ∆R((∆P f) − 4f1f)(Q) 1 1 1 1 1 (23) = 2∆P f(Q) · ∆R(∆P f)(Q) − 4f1(Q) · ∆Rf(Q) − 4f(Q) · ∆Rf1(Q) by using equation (6) and the product rule of polars (Lemma 4). 2 We have Q ∈ TP S = Z(f1) so f1(Q) = 0, and because Q ∈ Γ we have 0 = (f2 − 4f1f3)(Q) = 2 1 f2 (Q) so f2(Q) = 0. This means ∆P f(Q) = 2Q3f1(Q) + f2(Q) = 0 + 0. 1 Also we have by Lemma 5 that ∆Rf1(Q) = f1(R) because f1 is a linear polynomial. Then 1 because R ∈ ` = TP S ∩ H, we have R ∈ TP S = Z(f1) so f1(R) = 0, which means ∆Rf1(Q) = 0. 1 2 Thus the right hand side of equation (23) is zero, which means ∆R(f2 − 4f1f3)(Q) = 0. Thus R ∈ TQΓ, which proves ` ⊂ TQΓ. Then because the choice of Q was arbitrary, this proves that ` is a contact line of Γ. We will now analyze the singularity of S ∩ S1(P ) at P depending on whether ` is a bitangent 1 or a hyperﬂex. We have that C = S ∩ S1(P ) = Z(f, ∆P f) and in the aﬃne chart x3 = 1 we have

C|U3 = Z(f1 + f2 + f3, 2f1 + f2). Then in(f1 + f2 + f3, 2f1 + f2) = (f1, f2). Note that Z(f1, f2) considered as subset of H is equal to the set of points Γ ∩ `.

• ` is a bitangent of Γ: Then we have Z(f1, f2) = Γ ∩ ` = {Q1,Q2} two distinct points. 1 Then the distinct lines λP + µQ1, λP + µQ2 for [λ : µ] ∈ P are equal to the tangent cone 3 Z(f1, f2) of C at P considered as a subset of P . Thus C has a node singularity at P .

• ` is a hyperﬂex of Γ: Then we have Z(f1, f2) = Γ ∩ ` = {Q1} considered as a subset of H, 3 so the tangent cone Z(f1, f2) considered as a subset of P consists of the line λP + µQ1 for [λ : µ] ∈ P1. Thus C has a cusp singularity at P .

12 Example 1. We will now give two examples illustrating the two diﬀerent singularity types and the resulting bitangents.

2 1. The ﬁrst example is the cubic surface S given by the equation f = x3f1 + x3f2 + f3 with 2 2 3 2 f1 = x2, f2 = x0 − x1, f3 = x0 + x0x2. 2 2 The tangent cone Z(f1, f2) = Z(x2, x0 − x1) consists of the two lines λ[0 : 0 : 0 : 1] + µ[−1 : 1 : 0 : 0] and λ[0 : 0 : 0 : 1] + µ[1 : 1 : 0 : 0] for [λ : µ] ∈ P1.

We have TP S ∩ H = Z(f1, x3) = Z(x2, x3) so we get the bitangent line λ[1 : 1 : 0 : 0] + µ[−1 : 2 4 2 2 4 3 3 1 : 0 : 0] of Γ = Z(f2 − 4f1f3) = Z(x0 − 2x0y0 + y0 − 4x0x2 − 4x0x2). In Figure 6 an illustration of this case is given.

2 3 2 2. The second example is the cubic surface S given by f1 = x2, f2 = x0, f3 = x0 + x0x2. The 2 tangent cone is Z(f1, f2) = Z(x2, x0) which consists of the single line λ[0 : 0 : 0 : 1] + µ[0 : 1 : 0 : 0].

We have TP S ∩H = Z(f1, x3) = Z(x2, x3) so we get the hyperﬂex λ[0 : 1 : 0 : 0]+µ[1 : 0 : 0 : 0] 2 4 3 3 of Γ = Z(f2 − 4f1f3) = Z(x1 − 4x0x2 − 4x0x2). In ﬁgure 7 an illustration of this case is given.

3.3 Injectivity 2 We have shown that each line of S is projected to a bitangent of Z(f2 − 4f1f3), but we have not yet excluded that two lines of S can be mapped to the same bitangent. We will now show the injectivity of this correspondence.

Lemma 2. Let `1 =6 `2 be two distinct lines in S. Then π(`1) 6= π(`2), the projections are distinct bitangents.

Proof. Suppose `1 =6 `2 are two lines in S with π(`1) = π(`2). Let Q be an intersection point of `1 and Γ. Then P 6= Q as P does not lie on `1. Also the line PQ is tangent to S at Q, so mQ(S, P Q) ≥ 2. The line PQ then intersects `2 in a point R, thus the line PQ intersects S in three distinct points P, Q, R with total multiplicity at least 1 + 2 + 1 = 4. Then because S has degree 3, this implies that PQ ⊂ S. But this contradicts the assumption that P does not lie on any line. Thus by contradiction we have π(`1) 6= π(`2).

Also because the range of π is Γ \ TP S, it does not contain the last bitangent TP S ∩ H, so it cannot happen that the last bitangent is mapped to one of the 27 bitangents corresponding to one of the 27 lines of S. Thus we conclude that we have a bijective correspondence

{27 lines on S} ∪ {TP S} ↔ {28 bitangent lines of π(Γ)}. (24) between the 27 lines on S and TP S with the 28 bitangent lines of π(Γ).

13 C S

S1(P )

TP S

Γ TP S ∩ H

Figure 6: A node singularity on S ∩ S1(P ) at P , which corresponds to a bitangent of Γ.

4 Recovering the cubic from a quartic with computer algebra

In this section we will describe an algorithm that recovers a cubic surface from a plane quartic such that the shadow of the cubic surface is the quartic. An example implementation of this algorithm is provided in the attached SageMath ﬁle shadow_final.sage. 2 The strategy is to rewrite a quartic polynomial f in the form f2 −4f1f3 by using the bitangents of f, like in the theorem of chapter 1. The algorithm will perform the following steps: 1. Calculate the 28 bitangents of a quartic. 2. Calculate the points of contact of the bitangents with the quartic. 3. Select a quadruple of bitangents with eight distinct points of contact through which a conic g passes (such a quadruple is called syzygetic).

2 4. Rewrite the quartic in the form f = f2 − 4f1f3 where f2 is the conic passing through the eight contact points and f1f3 is the product of the four bitangents.

14 S C

TP S

S1(P ) Γ

TP S ∩ H

Figure 7: A cusp singularity on S ∩ S1(P ) at P , which corresponds to a hyperﬂex of Γ.

In the following sections we will describe these steps one by one.

4.1 Calculating the bitangents We will start by calculating the 28 bitangents of a smooth plane quartic Γ0. The algorithm described in this section is taken from [PSV11]. If ` : ax+by +cz is a general line in P2, and f ∈ k[x, y, z] is a homogeneous quartic polynomial deﬁning a smooth plane quartic Γ0, the condition ` is bitangent to Γ0 can be expressed as an algebraic equation as follows: Without loss of generality we suppose c 6= 0, so we can take c = 1, so z = −ax − by on `, then the condition ` is bitangent to S is then

2 2 2 f|`(x, y) = f(x, y, −ax − by) = (k0x + k1xy + k2y ) (25)

For some k0, k1, k2. 2 2 We can then write k0x +k1xy+k2y = (C1x+D1y)(C2x+D2y) for some C1,D1,C2,D2. This equation expresses that f restricted to the line ` looks like two parabolae (See Figure 8), which

15 means ` is a bitangent. Note that we allow the two contact points [D1 : −C1 : −aD1 + bC1], [D2 : −C2 : −aD2 + bC2] to be equal, in which case the bitangent becomes a hyperﬂex.

Z(f)

] + bC2 −aD2 −C2 : [D2 : ] + bC1 −aD1 −C1 : [D1 : ` = Z(ax + by + c)

Figure 8: Locally on `, a bitangent looks like two parabolae.

The five monomials x4, x3y, x2y2, xy3, y4 form a basis of the homogeneous degree 4 polynomials in k[x, y], so if we expand equation (25) into the equations for each monomial, we get a system of 5 equations in the variables a, b, k0, k1, k2. This gives rise to an ideal I ⊂ k[a, b, k0, k1, k2] generated by 5 elements (the five equations for the five monomials). Then we can use a Gröbnerbasis algorithm (from the computer algebra library Singular) to calculate the elimination ideal J = I ∩ k[a, b], which eliminates the variables k0, k1, k2. Now the condition that ` : ax + by + z is a bitangent of f is equivalent to a, b being roots of all polynomials of J. We can then use computer algebra to calculate the roots a, b. This procedure can then be repeated for the case ` : ax + y, thus assuming that c = 0 and b = 1, so [a : b : c] = [a : 1 : 0] and the case ` : x, thus [a : b : c] = [1 : 0 : 0]. The set of all solutions then consists of 28 triples [a : b : c] which correspond to the 28 bitangents of a smooth plane quartic.

4.2 Calculating the points of contact To calculate the points of contact of each bitangent ` with the quartic Γ0, we want to ﬁnd common roots of f and `, or equivalently the common roots of f|` and `, where f|` is the restriction of f to `. We can separate the line ` in three cases ` = [a : b : 1], [a : 1, 0], [1 : 0 : 0] again. So for example if ` = [a : b : 1], then f|` ∈ k[x, y, a, b] and J + (f|`) ⊂ k[x, y, a, b], so we can then calculate the 0 elimination ideal J = (J + (f|`)) ∩ k[x, y]. 0 The roots of J are then precisely all zeros of f|` which are also zeros of J, thus all roots of f|` such that ` is a bitangent of f. The roots of J 0 can be approximated by computer algebra (using the solve subroutine of Singular). These roots are the union of the roots of each bitangent, thus after calculating the roots, the algorithm should match each bitangent to its roots.

16 4.3 Finding a syzygetic quadruple of bitangents Using the contact points of the previous section, we now want to ﬁnd a quadruple of bitangents such that the eight contact points of the bitangents with f lie on a conic. Such a quadruple is called syzygetic. This is illustrated in Figure 9.

f f

g `3

`1 `4 f

Figure 9: Syzygetic quadruple of bitangents. Here `3 is a hyperﬂex, while `1, `2, `4 are normal bitangents.

If P1,P2,P3,P4,P5,P6 are the six contact points of bitangent triple `1, `2, `3, then these six points lie on a conic (thus `1, `2, `3 is a syzygetic triple) if and only if

 2 2 2  P1,x P1,xP1,y P1,y P1,yP1,z P1,xP1,z P1,z 2 2 2 P2,x P2,xP2,y P2,y P2,yP2,z P2,xP2,z P2,z  2 2 2  P3,x P3,xP3,y P3,y P3,yP3,z P3,xP3,z P3,z det  2 2 2  = 0. (26) P4,x P4,xP4,y P4,y P4,yP4,z P4,xP4,z P4,z  2 2 2  P5,x P5,xP5,y P5,y P5,yP5,z P5,xP5,z P5,z 2 2 2 P6,x P6,xP6,y P6,y P6,yP6,z P6,xP6,z P6,z Using the computer we can try out all triples of bitangents.

17 4.4 Finding a fourth bitangent Then after we find all syzygetic triple, we can find syzygetic quadruples, by trying for a fixed syzygetic triple `1, `2, `3 all bitangents `4 and testing whether a contact point Q of `4 lies on the conic spanned by the contact points of `1, `2, `3. Thus if P1,...,P5 are 5 of the 6 contact points of `1, `2, `3, then we want to test whether a contact points Q of `4 is the root of the conic

 2 2 2  P1,x P1,xP1,y P1,y P1,yP1,z P1,xP1,z P1,z 2 2 2 P2,x P2,xP2,y P2,y P2,yP2,z P2,xP2,z P2,z  2 2 2  P3,x P3,xP3,y P3,y P3,yP3,z P3,xP3,z P3,z g(x, y, z) = det  2 2 2  . (27) P4,x P4,xP4,y P4,y P4,yP4,z P4,xP4,z P4,z  2 2 2  P5,x P5,xP5,y P5,y P5,yP5,z P5,xP5,z P5,z x2 xy y2 yz xz z2

Then we want to find the subset of syzygetic quadruples of bitangents that are not hyperflexes, thus whose two contact points are distinct. This is because if we have a hyperflex, the determinant of the conic is automatically zero as we have two identical rows. This could give a false positive of a syzygetic quadruple.

4.5 Recovering the cubic surface

By the procedure of the previous section we find a quadruple `1, `2, `3, `4 of bitangents that are not hyperflexes and whose contact points line on a conic g. We can multiply the bitangents to get a quartic q = `1 · `2 · `3 · `4. Also we can square the conic to also get a quartic g2. Then the 8 contact points of the quadruple of bitangents are roots of f, g2, q. The following theorem, called the AF + BG-theorem, gives the existence of coefficients λ, µ such that f = λg2 + µq. Theorem 6 (AF + BG-theorem). Let F, G, H be projective plane curves. Assume F and G have no common components. Then there is an equation H = AF + BG (with A, B forms of degree deg(H) − deg(F ), deg(H) − deg(G) respectively) if and only if at every P ∈ F ∩ G,

2 H∗ ∈ (F∗,G∗) ⊂ OP ,P , thus if H is generated by F and G in the local ring at P . For a proof, see [Ful89, p. 61]. In our case, we have F = g2, G = q and H = f. By equation 25, 2 2 2 0 2 2 we see that f = A`+B(k0x +k1xy +k2y ) near a contact point, thus f∗ = A q∗ +Bg∗ ∈ (q∗, g∗). This means by the AF + BG-theorem there exist coefficients λ, µ of degree deg(f) − deg(g2) = deg(f) − deg(q) = 0 such that f = λg2 + µq. To calculate λ, µ, choose two coefficients ai, aj of two monomials of f and let bi, bj be the 2 corresponding monomial coefficients of g and ci, cj the corresponding monomial coefficients of q. Then we can solve the system of equations

b c λ a i i = i (28) bj cj µ aj for λ, µ. This choice of λ, µ then gives that λg2 + µq = f. Thus we take a linear combination 2 2 λg + µ`1`2`3`4 such that√ f = λg + µ`1`2`3`4. 2 −µ 2 2 −µ Then we have f = ( λg) − 4`1( 4 )(`2`3`4) so the polynomial t f1 + tf2 + f3 = t ( 4 `2) + √ √ µ t λg + `2`3`4 with f1 = `1, f2 = λg, f3 = − 4 `2`3`4 gives a cubic polynomial such that when projected from P = [0 : 0 : 0 : 1] gives the quartic polynomial f.

18 5 Algebraic geometry in physics: F-theory

In this section we will give a short impression of how complex algebraic surfaces are used in F-theory in physics. It is not strongly related to the previous sections so it can be read separately. The material in this section is based on section 2 of [Wei18].

5.1 D7-branes F-theory is an extension of Type IIB superstring theory where we model the universe as a 10 real dimensional manifold of the form 1,3 M = R × B3 (29) 1,3 where R represents the 1+3 dimensions of time and space and B3 is a compact manifold of complex dimension 3. The interpretation of the extra dimensions introduced by B3 are dimensions that cannot be observed yet because they are very small or very big. In this space-time M we have 8 real dimensional hypersurfaces living inside, which are called D7-branes (see Figure 11), where the 7 in D7-branes refers to it having 7 real spatial dimensions. These D7-branes are the locations where a 1-dimensional object called the fundamental string can end (see Figure 10). The fundamental string or F-string is a generalisation of a classical point particle.

Figure 10: F-string ending on a D7-brane.

The D7-branes are magnetic sources for a field called C0, which is a generalisation of the electromagnetic field of Maxwell theory, and a fundamental string gains magnetic charge from a D7-brane by having its endpoints on this brane. The C0 field together with another field called the dilaton φ, which describes how strongly −φ strings couple to one another, leads to a complex field τ = C0 + ie on the space M, called the axio-dilaton field.

5.2 Elliptic fibration The field τ on M turns out to define a structure of an elliptic curve on each point x ∈ M, which is smooth away from the D7-branes and singular on the D7-branes. So by assigning to each point p ∈ B3 this complex field τ(p), this leads to an elliptic fibration Eτ → Y4 → B3. This means we have a map π : Y4 → B3 for some 4-complex dimensional −1 manifold Y4 to B3 such that if p ∈ B3 lies outside of a D7-brane, the fiber π (p) is a smooth elliptic curve or a complex torus Eτ , while if p lies on a D7-brane the fiber is singular. Suppose we have a section of the map π, then it turns out that we can write this section in a standard Weierstrass equation form y2 − (x3 + fxz4 + gz6) = 0 in weighted projective space P2,3,1, where f, g ∈ C[x, y, z] are respectively degree 8 and degree 6 polynomials, then the locus of the D7-branes corresponds to the singular locus of this Weierstrass equation, which turns out to be the vanishing locus of the determinant ∆ = 4f 3 + 27g2.

19 By resolving the singularity of this elliptic curve by blow-ups, the singularity can be classiﬁed by a graph called a Dynkin diagram, and it turns out that the type of singularity is related to the gauge symmetry group of the ﬁeld induced by the D7-brane (see [Wei18]).

1,3 R z B3

D7 x y (x, y, z, t)

Figure 11: The 10 dimensional space M with some intersecting D7-branes.

5.3 Intersecting D7-branes When two D7-branes intersect, they lead to a particle living on the intersection, as an F-string can then have endpoints on both intersecting D7-branes (see Figure 12). The singularity of the elliptic ﬁbration on the intersection can then become more complicated than the singularities on the two D7-branes and lead to a fundamental particle with more structure. The type of this singularity determines the type of fundamental particle and its gauge symmetry group.

U(1)

SO(2)

U(1)

Figure 12: Intersecting D7-branes with two F-strings between them.

5.4 Born infeld action To give an example of one of the actions of string theory, we will discuss in this subsection the 1 µν Born-infeld action, which is a generalisation of the action L = 4 F Fµν of Maxwell-theory, which describes electromagnetic ﬁelds in ﬂat Minkowski space R1,3.

20 We know that Maxwell-theory is invariant under the Lorentz group of coordinate transformations of special relativity in flat Minkowski space [Gri13], but because the equations of Maxwell are formulated on flat Minkowski space R1,3, we cannot use these to describe electromagnetic fields in curved spacetime. By postulating that the electromagnetic Lagrangian must be invariant under general space- time transformations, we arrive at the Born-Infeld Lagrangian density as one of the possible amendments. This derivation is explained in [BI34]. The Born-Infeld Lagrangian density is defined as 1/2 1/2 LBI = (− det(ηµν + Fµν )) − (− det(ηµν )) . (30)

Here ηµν is the pseudo-Riemannian metric on some space-time manifold M and Fµν is the Electromagnetic tensor, which depend both on the geometry of the space M. We will now check that to ﬁrst order, the Born-Infeld action reduces properly to the standard 1,3 Maxwell-equations if we take M = R and ηµν = diag(−1, 1, 1, 1) and Fµν equal to

 Ex Ey Ez  0 c c c Ex − c 0 Bz −By F :=  Ey  . (31) − c −Bz 0 Bx  Ez − c By −Bx 0 We can expand the determinant [nLa20] using Laplace expansion to get 1 det(η + F ) = a1a2a3a4 b1b2b3b4 (η + F )(η + F )(η + F )(η + F ) µν µν 4! a1b1 a1b1 a2b2 a2b2 a3b3 a3b3 a4b4 a4b4 1 = a1a2a3a4 b1b2b3b4 η η η η 4! a1b1 a2b2 a3b3 a4b4 3 + a1a2a3a4 b1b2b3b4 η η η F 4! a1b1 a2b2 a3b3 a4b4 6 + a1a2a3a4 b1b2b3b4 η η F F 4! a1b1 a2b2 a3b3 a4b4 3 + a1a2a3a4 b1b2b3b4 η F F F 4! a1b1 a2b2 a3b3 a4b4 1 + a1a2a3a4 b1b2b3b4 F F F F 4! a1b1 a2b2 a3b3 a4b4 (32) where a1a2a3a4 is the Levi-Civita symbol. Then the ﬁrst summand is 1 a1a2a3a4 b1b2b3b4 η η η η = det(η) = −1 (33) 4! a1b1 a2b2 a3b3 a4b4 and the ﬁfth summand is 1 a1a2a3a4 b1b2b3b4 F F F F = det(F ). (34) 4! a1b1 a2b2 a3b3 a4b4

We have Fa4b4 = −Fb4a4 because F is an antisymmetric matrix, and ηa1b1 = ηb1a1 because η is a symmetric matrix. So for the second summand 3 3 a1a2a3a4 b1b2b3b4 b1b2b3b4 a1a2a3a4 ηa1b1 ηa2b2 ηa3b3 Fa4b4 = − ηb1a1 ηb2a2 ηb3a3 Fb4a4 4! 4! (35) 3 = − a1a2a3a4 b1b2b3b4 η η η F 4! a1b1 a2b2 a3b3 a4b4

21 where the last equality follows from renaming the summation indices. This implies for the second summand 3 a1a2a3a4 b1b2b3b4 η η η F = 0. (36) 4! a1b1 a2b2 a3b3 a4b4

By the same argument, because Fa2b2 Fa3b3 Fa4b4 = −Fb2a2 Fb3a3 Fb4a4 , we have for the fourth summand 3 a1a2a3a4 b1b2b3b4 η F F F = 0. (37) 4! a1b1 a2b2 a3b3 a4b4 For the third summand we have  1 · −1 if a3 = b3, a4 = b4, a3 6= a4, a1 = 0 xor a2 = 0  1 · 1 if a = b , a = b , a 6= a , a 6= 0 and a 6= 0  3 3 4 4 3 4 1 2 a1a2a3a4 b1b2b3b4 ηa1b1 ηa2b2 = −1 · −1 if a3 = b4, a4 = b3, a3 6= a4, a1 = 0 xor a2 = 0 (38)  −1 · 1 if a3 = b4, a4 = b3, a3 6= a4, a1 6= 0 and a2 6= 0  0 otherwise.

a3b3 Then Fa3b3 Fa4b4 = −F Fa4b4 if a3 = 0 xor b3 = 0, thus if a1 6= 0 and a2 =6 0. This means that we get 6 1 a1a2a3a4 b1b2b3b4 η η F F = − · 2 · F a3b3 F . (39) 4! a1b1 a2b2 a3b3 a4b4 4 a4b4 So we get 1 det(η + F ) = −1 − F µν F + det(F ) (40) µν µν 2 µν which means we can Taylor expand LBI to second order in the coeﬃcients of F to get 1 1 (− det(η +F ))1/2 −(− det(η ))1/2 = (1+ F µν F −det(F ))1/2 −1 = F µν F +O((F )4). µν µν µν 2 µν 4 µν uν (41) 1 µν Thus if we neglect the higher order terms, we get exactly the Lagrangian 4 F Fµν of the Electromagnetic ﬁeld in Minkowski space.

22 6 Appendix: lemmas and notation 6.1 Notation In the following table we collect some symbols that are used often:

Symbol Meaning k an algebraically closed ﬁeld of characteristic 0, used implicitly everywhere in this text. You may replace this with C without loss of generality. k[x0, . . . , xn]d the ring of homogeneous polynomials of degree d in n + 1 variables over k. Pn n-dimensional projective space over k. Sk(P ) the k-th polar of a hypersurface S from the point P . k ∆P f the equation for the k-th polar of a polynomial f from the point P . Z(f) the set of points Q ∈ Pn such that f(Q) = 0. TP S the tangent space of a variety S at a point P . QP for Q 6= P , the unique line in Pn passing through Q and P . Can be parametrised by λQ + µP for [λ : µ] ∈ P1.

6.2 Lemmas

Lemma 3 (Double reading method of polars). Let f ∈ k[x0, . . . , xn] be a degree r homogeneous polynomial and P , Q two points. 1 s 1 r−s Then for 0 ≤ s ≤ r we have s! ∆P f(Q) = (r−s)! ∆Q f(P ). Proof. We can taylor expand f(λP + µQ) in two ways:

r X λr−iµi f(λP + µQ) = ∆i f(Q) i! P i=0 r (42) X µr−iλi f(λP + µQ) = ∆i f(P ). i! Q i=0 Now if we match the coeﬃcients of λiµj of both expansions, we get 1 1 ∆s f(Q) = ∆r−sf(P ). (43) s! P (r − s)! Q

Lemma 4 (Product rule of polars). Let f, g ∈ k[x0, . . . , xn] be homogeneous polynomials and P , Q two points. 1 1 1 Then ∆P (fg)(Q) = g(Q)∆P f(Q) + f(Q)∆P g(Q). Proof. We expand out in components: n n 1 X ∂(fg) X ∂f ∂g ∆P (fg)(Q) = Pi = Pi g(Q) + f(Q) ∂x Q ∂x Q ∂x Q i=0 i i=0 i i (44) 1 1 = g(Q)∆P f(Q) + f(Q)∆P g(Q).

23 Lemma 5 (Polar of a linear polynomial). Let f ∈ k[x0, . . . , xn] be a homogeneous polynomial of degree 1 and P , Q two points. 1 Then ∆P f(Q) = f(P ). Proof. We expand out in components:

n n 1 X ∂f X ∂f ∆P f(Q) = Pi = Pi = deg f · f(P ) = f(P ). (45) ∂x Q ∂x P i=0 i i=0 i

Here in the second equality we used that ∂f is constant for all i = 0, . . . , n and in the third ∂xi Pn ∂g equality we used the Euler identity xi = deg g · g for homogeneous polynomials. i=0 ∂xi

Lemma 6 (Choosing suitable coordinates). Let S = Z(f) ⊂ P3 be a cubic hypersurface, P a point in S and H a hyperplane not containing P . Then there exists a change of coordinates 2 matrix g such that g([0 : 0 : 0 : 1]) = P , g(Z(x3)) = H and f ◦ g = x3f1 + x3f2 + f3 for some homogeneous polynomials f1, f2, f3 ∈ k[x0, x1, x2] of degree 1,2,3 respectively. Proof. We choose three points A, B, C ∈ H that span the whole plane, thus choose A, B, C such that there is no line containing all three. P3 The hyperplane H is given as the zero-set of some linear polynomial i=0 hixi with [h0 : h1 : 3 h2 : h3] ∈ P the coefficients of this polynomial. P3 We have P 6∈ H, which means i=0 hiPi 6= 0. Then with the coefficients of the points A, B, C, P we can define the following matrix   A0 B0 C0 P0 A1 B1 C1 P1 [g] =   . (46) A2 B2 C2 P2 A3 B3 C3 P3

Because P 6∈ H and A, B, C do not lie on a line, we have that the columns of the matrix are linearly independent, so g is an invertible matrix. Furthermore g([0 : 0 : 0 : 1]) = P and 3 g(Z(x3)) = {µ1A + µ2B + µ3C :[µ1 : µ2 : µ3] ∈ P } = H. 3 Also we have (f ◦ g)([0 : 0 : 0 : 1]) = f(P ) = 0 which means the monomial x3 is not present 2 in f ◦ g, so we can write f ◦ g = x3f1 + x3f2 + f3 for f1, f2, f3 ∈ k[x0, x1, x2] homogeneous polynomials of degree 1,2,3.

24 References

[Bel09] Mauro Beltrametti. Lectures on curves, surfaces and projective varieties : a classical view of algebraic geometry. Zürich: European Mathematical Society, 2009. isbn: 978- 3037190647. [BI34] Max Born and Leopold Infeld. “Foundations of the new field theory”. In: Proceedings of the Royal Society of London. Series A, Containing Papers of a Mathematical and Physical Character 144.852 (1934-03), pp. 425–451. doi: 10.1098/rspa.1934.0059. url: https://doi.org/10.1098/rspa.1934.0059. [Dol12] I Dolgachev. Classical algebraic geometry : a modern view. Cambridge New York: Cambridge University Press, 2012. isbn: 9781107017658. [Ful89] W. Fulton. Algebraic curves: an introduction to algebraic geometry. Advanced book clas- sics. Addison-Wesley Pub. Co., Advanced Book Program, 1989. isbn: 9780201510102. [Gri13] David J Griffiths. Introduction to electrodynamics; 4th ed. Re-published by Cambridge University Press in 2017. Boston, MA: Pearson, 2013. doi: 1108420419. url: https: //cds.cern.ch/record/1492149. [Har77] Robin Hartshorne. Algebraic Geometry. Springer New York, 1977. [KK77] Akikazu Kuribayashi and Kaname Komiya. “On Weierstrass points of non-hyperelliptic compact Riemann surfaces of genus three”. In: Hiroshima Math. J. 7.3 (1977), pp. 743– 768. doi: 10.32917/hmj/1206135658. url: https://doi.org/10.32917/hmj/ 1206135658. [nLa20] nLab authors. Dirac-Born-Infeld action. http://ncatlab.org/nlab/show/Dirac- Born-Infeld%20action. Revision 37. 2020-04. [PSV11] Daniel Plaumann, Bernd Sturmfels, and Cynthia Vinzant. “Quartic curves and their bitangents”. In: Journal of Symbolic Computation 46.6 (2011-06), pp. 712–733. issn: 0747-7171. doi: 10.1016/j.jsc.2011.01.007. url: http://dx.doi.org/10.1016/j. jsc.2011.01.007. [Wei18] Timo Weigand. TASI Lectures on F-theory. 2018. arXiv: 1806.01854 [hep-th].