Galois Groups of Enumerative Problems

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Galois groups of enumerative problems

Yongquan Zhang Advisor: Prof. Joe Harris

Minor thesis

Department of Mathematics Harvard University

January 2019 Contents

1 Introduction 2

2 Preliminaries 3

3 Galois groups of ﬂexes of plane curves 11

4 Galois groups of bitangents of plane curves 15

5 Galois group of 27 lines on a cubic surface 19

6 Galois group of the problem of ﬁve conics 26

7 Resolvent degree 30

References 33

1 1 Introduction

In this report, we discuss the solvability of some enumerative problems following [Har1]. To illustrate the type of problems at hand, take the famous example of 27 lines on a cubic surface 3 3 in P . Let S ⊂ P be a cubic surface given as the locus of a degree 3 homogenous polynomial P I 3 F (X) = #I=3 aI X in some homogenous coordinate X = [X0 : X1 : X2 : X3] on P . A line 3 in P is given by the span of two linearly independent vectors, say A = (a0, a1, a2, a3) and B = (b0, b1, b2, b3); for a generic cubic surface, we may even assume a0 = 0 = b1, a1 = 1 = b0. 1 Then the line lies on F (X) if and only if F (sA + tB) = 0 for all [s : t] ∈ P . This gives 4 degree 3 (nonhomogenous) polynomials in a2, a3, b2, b3, with coefficients as polynomials in aI ’s. Hence we do expect finitely many lines lying on the cubic surface. The question is then how many, and given coefficients aI , whether we can solve for these lines algebraically.

To interpret solvability in terms of Galois theory, note that the solutions to a system of polynomial equations have algebraic coordinates over the base field, which is, in our case above, the field of rational functions C(aI ) in the coefficients aI . In particular, one may ask about the solvability of the Galois group of the algebraic extension obtained by adjoining the coordinates of all the solutions to C(aI ). More generally, one may ask about the solvability of this algebraic extension over C(aI ) adjoining the coordinates of some of the solutions, which in our case is equivalent to ask if we can solve for the remaining lines given the coordinates of some of them.

Note that C(aI ) is essentially the function field of the complete linear system of cubic 3 surfaces in P . Consider the incidence correspondence in the space of pairs of a cubic surface and a line, cut out by the condition that the line lies on the cubic surface. Clearly the algebraic extension mentioned in the previous paragraph gives, in essence, the function field of this algebraic variety. There is also a natural projection from the incidence correspondence to the complete linear system. This projection is of finite degree (27 in this case), and hence we can define the associated monodromy group (for details, see Section 2.6). The key observation is that the Galois group and the monodromy group coincide, and hence we may study the Galois group by looking at the monodromy group and the rich geometry there. The Galois group then reveals information about the solutions; in particular, determining the solvability when the coordinates of some of the lines are given is the same as determining the solvability of the corresponding subgroup.

This approach enables us to study this type of problems systematically. Instead of looking at solutions to ﬁnd additional structures, we look at the group directly. This is in direct contrast to more classical explorations of these problems, where one obtain solvability by study the structure of the solutions, and then determine the Galois group from the information. For example, the Galois group acts on the 27 lines on the cubic surface preserving the intersection conﬁguration, and hence cannot be the full symmetry group. We will comment on this classical approach when we can, drawing from references such as [Jor].

The structure of this report is as follows: some preliminary facts are reviewed in Section 2, many of which come from a basic course in algebraic geometry, mostly about plane curves, and are included for the beneﬁt of the author, whose knowledge in the subject has been somewhat

2 rusty. The remaining sections are then each devoted to a different enumerative problem. Our focus will be the problems of flexes and bitangents on plane curves. In order to solve the problem of bitangents on quartics completely, we also need the solution to the problem of 27 lines on a cubic surface. After that, we include a short exposition on the problem of five conics to showcase the variety of problems that may be settled using this approach. Finally, we give an account of resolvent degree, including some more recent related research.

2 Preliminaries

Some references for this section include [Har2, Ful, GH].

2.1 Flexes and bitangents of plane curves

A plane curve of degree d is the vanishing locus of a homogeneous polynomial F (X0,X1,X2) 2 of degree d in P . The (arithmetic) genus of a plane curve is given by g = (d − 1)(d − 2)/2 by the genus-degree formula.

2 Fix some homogenous coordinate [X0 : X1 : X2] on P , and corresponding dual coordinate 2∗ 2 2 [Y0 : Y1 : Y2] on P , the projective space of lines in P . That is, a point [X0 : X1 : X2] on P P lies on a line corresponding to [Y0 : Y1 : Y2] if and only if i XiYi = 0. Given a smooth plane 2∗ curve C of degree d, we can deﬁne a map f : C → P by sending p ∈ C to the tangent line of C at p. This map is algebraic: indeed, it is the restriction of the map " # ∂F ∂F ∂F p = [X0 : X1 : X2] 7→ : : ∂X0 p ∂X1 p ∂X2 p

to the curve C. Thus the image of C is also a curve, called the dual curve and denoted by C∗. Moreover the map f is generically one-to-one, as easily veriﬁed from its expression given above.

The degree d∗ of C∗ is readily determined: it is the number of intersection of a generic 2∗ ∗ 2∗ 2 line in P with C . A line in P corresponds to a point in P , and thus the degree is also 2 the number of tangent lines to C passing through a generic point in P . Given a point q, assume none of the lines passing through q is multiply tangent to C. Consider the projection of C from q onto a generic line. This is a map of degree d, ramiﬁed at points of tangency. By Riemann-Hurwitz, the number of branch points is given by

2g − 2 + 2d = (d − 1)(d − 2) − 2 + 2d = d(d − 1). P This is as expected, as these tangent points are solutions of F (X) = i ai∂F/∂Xi = 0 where ∗ ∗ [a0 : a1 : a2] are the coordinate of q. Hence the degree d of the curve C is d(d − 1).

The discrepancy in degree of C and C∗ is accounted for by the fact that C∗ is not smooth even if C is. We now try to determine the singularities of C∗. Let p ∈ C be a point such that

3 the tangent line to C at p has contact order m ≥ 2. For simplicity, we choose homogeneous 2 coordinates on P and (holomorphic) local coordinate t on C, so that p = [1 : 0 : 0] and C is parametrically given locally (in an analytic small neighborhood) as

t 7→ [1 : t : X2(t)]

m where X2(t) vanishes at t = 0 to degree m. Write X2(t) = t v(t) for some nonzero holomorphic function v in a neighborhood of zero. Then one can calculate, locally, the expression of f:

0 0 m m+1 0 m−1 m 0 f(t) = [tX2(t)−X2(t): −X2(t) : 1] = (m − 1)t v(t) + t v (t): −mt v(t) − t v (t) : 1

m m−1 Therefore, in some local coordinate et, f is given as t˜ 7→ [t˜ u(t˜): t˜ : 1] for some nonzero holomorphic function u in a neighborhood of zero. In particular, we see that C∗ has a cuspidal singularity of multiplicity m − 1 at f(p) and the map f is smooth at p if and only if m = 2. Moreover, from the expression for f, which gives a parametrization of C∗ locally, we see that at point f(p) of C∗, the tangent line to C∗ corresponds to p.

As we have discussed, the map f is generically one-to-one over a point l ∈ C∗, unless the ∗ corresponding line l is tangent to C at several points p1, . . . , pe. For each branch of C at l, ∗ the tangents are distinct, as these tangents correspond to points pi respectively. Hence C has a e-fold ordinary singularity at l.

We will always assume there are only two types of singularity at a point l ∈ C∗: ordinary double point and cusps of degree 2. In particular, these correspond to lines tangent to C at two distinct points, each with contact order 2, and lines tangent to C at a unique point with contact order 3. The former will be referred to as bitangents, and the latter ordinary flexes (or simply flexes when no confusion arises; in general, a flex is a point at which the tangent line has contact order at least 3). This assumption holds for generic curves of degree 2 d. Indeed, let Wd be the complete linear system of plane curves of degree d in P . For any 2∗ curve C ∈ Wd and a line l ∈ P , consider the divisor C ∩ l on the line. This is a divisor P of degree d. Write C ∩ l = i mipi, where mi is the order of contact at pi of C and l, we P P define mC (l) = i(mi − 1). The space of divisors having the form i mipi has codimension P 2∗ i(mi − 1) in the space of all divisors of degree d. Therefore, the subvariety Jm ⊂ Wd × P defined by Jm := {(C, l): mC (l) ≥ m} has codimension m has well. Therefore dim Jm = dim Wd +2−m. In particular, when m ≥ 3, Jm cannot surject onto Wd and hence has image of codimension at least 1. Thus a generic curve C has mC (l) ≤ 2 for all l. In particular, the only singular possibilities are two tangent points of contact order 1 (i.e. bitangents) or a tangent point of contact order 3 (i.e. flexes).

To obtain a much more symmetric picture, we now relax the condition on C to allow ∗ ordinary double points and cusps. The map f : C → P is well defined away from the singularities. At an ordinary double point, along different branches of C the map f has different limits; the tangent line of C∗ at these two limit point coincide. At a cusp, we may define the tangent line by taking limit of nearby points, and check that this corresponds to a flex point of C∗. Moreover, we have seen that C∗∗ = C. In summary, flexes of C corresponds

4 to cusps of C∗ and vice versa, double tangents of C corresponds to ordinary double points of C∗ and vice versa. These singularities are clearly resolved under blowup at these points in 2 2∗ P or P . Now under the assumption specified in the discussions above, we determine the relation- ships between the number of double points, cusps, flexes and bitangents, which we denote by δ, κ, f and b respectively, and the degree d and genus g of C. First we determine the degree of C∗. As before, this corresponds to the number of tangent lines to C passing through a generic 2 point p of P . Assume the tangent lines passing through p do not intersect C at any singular points, and are not multiply tangent to C. Assume moreover p = [1 : 0 : 0] up to change of 2 coordinates on P . Then the tangent points correspond to points on C where ∂F/∂X0 = 0. 0 Let C be the curve defined by ∂F/∂X0 = 0. This curve has degree d − 1, and passes through ordinary double points of C with multiplicity 2 and cusps of C with multiplicity 3. Hence the number of smooth intersections of C and C0 is

d(d − 1) − 2δ − 3κ, and this is the degree d∗ of C∗. Note that this formula agrees with the one obtained in the case C is nonsingular.

On the other hand, consider the projection from p of C to a generic line (technically, we need a smooth curve to apply Riemann-Hurwitz, and thus we need to compose p with the map from the proper transformation of C under the blowup at singular points). This is a map of degree d. Branch points come from tangents passing through p, and also cusps of C. Hence by Riemann-Hurwitz,

2g − 2 = −2d + d∗ + κ = d2 − 3d − 2δ − 2κ and therefore g = (d − 1)(d − 2)/2 − δ − κ, which agrees the classical genus-degree formula when C is nonsingular. We have similar formulae for C∗, in particular,

(d∗ − 1)(d∗ − 2) d = d∗(d∗ − 1) − 2b − 3f, g = − b − f 2 These formulae are often referred to as the Pl¨uckerformulae. Setting δ = κ = 0 and solve for b and f, we have for a generic curve C 1 f = 3d(d − 2), b = d(d − 2)(d − 3)(d + 3). 2

2.2 Plane cubics and their ﬂexes

Plane cubics are curves of genus 1. In particular, they are elliptic curves, or complex tori. On the other hand, given a complex torus C/Λ, where Λ = Zω1 + Zω2 for some ω1, ω2 ∈ C that are R-linearly independent . We can exhibit it as a cubic curve by the embedding

z 7→ [1 : ℘(z): ℘0(z)]

5 where ℘(z) = ℘Λ(z) is the Weierstarss function with respect to the lattice Λ. The group law apparent on a complex torus can also be defined on the cubic: for any two points p, q ∈ C, let r be the third point of intersection of C and the line pq. Define ϕ : C ×C → C by ϕ(p, q) = r. Choose a point O as the origin, and then define addition on C as p + q = ϕ(O, ϕ(p, q)). For details, see [Ful, §5.6, Prop. 4].

Now the Weierstrass ℘-function has a unique pole of order 2 viewed as a meromorphic 0 function on C/Λ, and ℘ has a unique pole of order 3. Together with the constant function 1, they form a basis of a degree 3 line bundle. As they becomes coordinates in the projective embedding, this line bundle is OC (1). In particular, if H = p1 + p2 + p3 is the divisor of this 2 line bundle, p1, p2, p3 are collinear in P . By Plücker’s formula, there are 9 flexes on a generic cubic C. As a matter of fact, any nonsingular cubic has only ordinary flexes, as any other types of flexes would imply an intersection multiplicity at least 4 between the tangent line and the cubic curve. Flexes are intersections of a line and the cubic C of order 3; therefore p ∈ C is a flex point then the divisor 3p is linearly equivalent to p1 + p2 + p3. Conversely, given a divisor 3p linearly equivalent to p1 + p2 + p3, and as C is projectively normal, 3p is cut out by a line. Suppose p, p1, p2, p3 corresponds to points z, z1, z2, z3 ∈ C/Λ under the embedding noted above. Then by Abel’s theorem, 3z is linearly equivalent to z1 + z2 + z3 if and only if

3z = z1 + z2 + z3 mod Λ. In particular, any two flexes z, z0 satisfy 1 z − z0 ∈ Λ. 3 Therefore if we fix any flex point as the origin, then the 9 torsion points of order 3 are 1 ∼ exactly all the flexes. As 3 Λ/Λ = Z/3 × Z/3. This gives the set Γ of 9 flexes the structure of a 2-dimensional affine space over Z/3. Note that for any three points in Γ whether or not z1 + z2 + z3 = 0 is independent of the choice of origin; indeed, any change of origin is tantamount to adjusting each zi by a fixed order 3 torsion point. As 0 is chosen to be one of the flex, this is equivalent to the fact that the corresponding point p1, p2, p3 on C is collinear. Hence this affine space structure is perfectly visible geometrically: z1 + z2 + z3 = 0 if and only if p1, p2, p3 are collinear.

2.3 Plane quartics and their bitangents

The genus of a smooth plane quartic C is g = 3, and the degree of the canonical divisor is 2g − 2 = 4, precisely the degree of the curve. By adjunction, the canonical line bundle of C is linear equivalent to the restriction of the class of a line. On the other hand, given a smooth curve of genus 3, the dimension of its canonical sections is g = 3, and unless the curve is 2 hyperelliptic, the associated map to P is an embedding. That is, a smooth plane quartic is canonically embedded, and non-hyperelliptic.

Now a bitangent l = p1p2 with tangency points p1, p2 intersects the curve in a divisor 2p1 + 2p2. This divisor of degree 4 is hence linearly equivalent to the canonical divisor KC .

6 Conversely, given a divisor of the form 2p1 +2p2 linearly equivalent to KC , as C is projectively normal, it is then cut out by a line and hence corresponds to a bitangent. Therefore, bitangents of a smooth plane quartic are in one-to-one correspondence with eﬀective semicanonical divisors, that is, those divisors D = p1 + p2 satisfying 2D ∼ KC .

Now given any semicanonical divisor κ, we then have 2(κ − p1 − p2) ∼ 0. Moreover, p1 + p2 6∼ q1 + q2 if p1, p2 6= q1, q2, as otherwise C would be hyperelliptic. Thus up to a choice of κ, bitangents p1 + p2 corresponds to torsion points κ − p1 − p2 of order 2 in the Jacobian 0 ∗ ∼ 3 6 J (C). Let V be the group of order 2 points in J (C) = H (C,KC ) /H1(C, Z) = C /Z . Then this group is given as 1 V = Λ/Λ =∼ ( /2)6 2 Z where Λ = H1(C, Z). Now, as in the case of ﬂexes, the additive structure can be viewed geometrically: no matter what origin κ we choose, four bitangents p1p2, q1q2, r1r2, s1s2 adds up to zero if p1 + p2 + q1 + q2 + r1 + r2 + s1 + s2 ∼ 2KC , and this is true if and only if the 8 point all lie on a conic, as C is projectively normal.

2.4 Riemann theta-function and Riemann’s theorem

We follow [GH, §2.7]. Given a compact Riemann surface S of genus g, choose a basis {α1, . . . , αg, β1, . . . , βg} of H1(S, Z) so that with respect to the intersection form (·, ·) we have (αi, αj) = (βi, βj) = 0, (αi, βj) = δij. We call such a basis a normalized basis with re- 1 1 ∗ spect to (·, ·). We may embed H1(S, Z) into H (S, ΩS) by integrating 1-forms along 1-cycles. It turns out that the image of this embedding is a lattice Λ. Choose a basis {ω1, . . . , ωg} for H1(S, Ω1 ) so that S Z ωj = δij. αi 1 1 ∗ The Jacobian J (S) is deﬁned as H (S, ΩS) /H1(S, Z). With respect to the basis we choose above, the lattice is generated by Z Z Z Z λi = ω1,..., ωg = ei, λi+g = ω1,..., ωg αi αi βi βi

g where ei is the i-th factor in the standard basis of C . The period matrix Ω, which is a g ×2g matrix with columns λ1, . . . , λ2g, is then

Ω = IZ

By Riemann bilinear relations, we have Z is symmetric, and Y = Im Z is positive deﬁnite. Consider the theta function deﬁned by X θ(z) = eπihl,Zli+2πihl,zi, l∈Zn g where h·, ·i is the standard inner product on C . Clearly θ(z) = θ(−z). Then θ gives a section of a line bundle on J (S). In fact θ generates the one-dimensional vector space of sections

7 of this line bundle. Let Θ be its divisor. For any κ ∈ J (S), we denote by Θκ the divisor of the section determined by the translated theta function θ(z − κ). With respect to the group structure, this is simply a translation of Θ by κ, i.e. Θκ = Θ + κ.

There is a map µ from S to its Jacobian J (S) deﬁned as follows. Fix a base point z0 on S, then for any z ∈ S, we have Z z Z z µ(z) = ω1,..., ωg . z0 z0

This is only well defined up to a choice of a path from z0 to z1, but any two such paths differ by a loop and those map to 0 in the Jacobian. Given an effective divisor of degree d

D = p1 + ··· + pd, (d) we deﬁne µ(D) = µ(p1)+···+µ(pd). Let S be the space of all eﬀective divisors of degree d, (d) and set Wd = µ(S ). It turns out that for d ≤ g, Wd is an analytic subvariety of dimension d. The following theorem relates the divisor Θ with Wg−1, both of which are of dimension g − 1: Theorem 2.1 (Riemann’s theorem). For some constant κ ∈ J (S), we have

Θκ = Wg−1.

By Riemann-Roch theorem, given an eﬀective divisor D of degree g − 1, K − D is also eﬀective. Hence µ(K) − Wg−1 = Wg−1. Therefore Wg−1 − κ = Θ = −Θ = −Wg−1 + κ = Wg−1 − µ(K) + κ. Thus Wg−1 is invariant under translation by µ(K) − 2κ, and so does Θ. If µ(K) − 2κ is nonzero, then θ(z − µ(K) + 2κ) is another theta function associated to the same line bundle, which is impossible. Hence µ(K) = 2κ, that is κ corresponds to a semicanonical divisor.

One may wonder how do properties of a divisor D of degree g − 1 predict singularities of Wg−1 at µ(D). We have the following: Theorem 2.2 (Riemann-Kempf Singularity Theorem for d = g − 1). 0 multµ(D)(Θκ) = h (D)

One important consequence of these theorems is that, there exists a semicanonical divisor κ (of degree g − 1) so that for any E ∈ J (S), viewed as both a point on a complex abelian variety and a degree 0 line bundle, θ vanishes at E if and only if κ + E is linearly equivalent to an eﬀective divisor, and this is true if and only if h0(κ + E) > 0; more accurately, the multiplicity of θ at E is equal to h0(κ + E).

2.5 Bilinear forms over a ﬁeld of characteristic 2

We now introduce some terminologies regarding bilinear forms and quadratic forms over a ﬁeld k of characteristic 2. First of all, the usual distinctions between symmetric and skew- symmetric bilinear forms break down. We deﬁne a bilinear form Q on a vector space V over

8 k to be strictly skew-symmetric if Q(v, v) = 0 for all v ∈ V . Given a strictly skew-symmetric bilinear form on V , a quadratic reﬁnement is deﬁned to be a function q satisfying

Q(v, w) = q(v + w) − q(v) − q(w).

Finally a quadratic form on V is a bilinear form with quadratic reﬁnement. Consider the strictly skew-symmetric form Q on a vector space V of dimension 2n as follows:

n X Q(x, y) = (xiyi+n + xi+nyi). i=1 Then it has two quadratic reﬁnements:

n + X q (x) = xixi+n, i=1 and n − X 2 2 q (x) = xixi+n + xn + x2n. i=1 + n−1 n − n−1 n For k = Z/2, q has 2 (2 +1) zeros and q has 2 (2 −1) (as can be shown by induction on n). The the group of linear automorphisms of V preserving Q, q+, q− will be denoted as + − O(2n, Z/2),O (2n, Z/2),O (2n, Z/2).

2.6 Galois and monodromy groups

In this section, we detail the general framework of the concrete examples we will consider in the following sections. Throughout the section, let X,Y be two irreducible algebraic varieties of the same dimension over the complex number C, and π : Y → X a generically finite map of −1 degree d. Let p ∈ X be a generic point so that π (p) consists of d distinct points q1, . . . , qd. We also fix a numbering of these points. We define two groups associated to this setup:

Monodromy group. This is similarly defined as monodromy groups arising from covering maps in many topological and geometric situations. Let U ⊂ X be a sufficiently small Zariski open set so that π is an unbranched covering map of degree d restricted to V = π−1(U). We may also assume p ∈ U. For any loop γ : [0, 1] → U based at p, and any lift qi of p, there exists a unique lift of γ, denoted by γei : [0, 1] → V so that γei(0) = qi. The endpoint of γei is well defined up to homotopy of γ. Therefore, we have an action of π1(U, p) on the set {q1, . . . , qd} so that the equivalence class of homotopic loops [γ] ∈ π1(U, p) sends pi to the endpoint of the lifted arc γei. With respect to the fixed numbering, this gives a homomorphism

π1(U, p) → Σd,

sometimes referred to as the monodromy representation. The image of this homomorphism is called the monodromy group of the covering map π : V → U. Note that, a priori, the monodromy group may depend on the choice of the Zariski open subset U, although we will show that it does not.

9 Galois group. Let π∗ : K(X) → K(Y ) be the inclusion of the function ﬁelds induced by π. By primitive element theorem, K(Y ) is generated over K(X) by a single element f ∈ K(Y ) which satisﬁes a degree d polynomial irreducible over K(X):

d d−1 P (f) = f + g1 · f + ··· + gd = 0.

Let ∆ be the field of germs of meromorphic functions near p, and ∆i the field of germs of meromorphic functions near qi. Then K(X) injects into ∆ by restriction, and we denote ∼ the image by K. Similarly, K(Y ) injects into each ∆i by restriction. Moreover, ∆ = ∆i as π is a (unbranched, holomorphic) covering map over a small neighborhood of p. Let L be ∼ the subfield of ∆ generated by the image of K(Y ) in ∆i = ∆. Let fi be the image of f via ∼ K(Y ) → ∆i = ∆. Then each fi satisfies the polynomial

d d−1 Pe(fi) = fi + ge1 · fi + ··· + ged = 0, where gek is the image of gk via K(X) → K ⊂ ∆. Clearly L is generated over K by {fi}. Moreover, fi’s are different: f has different values at {q1, . . . , qd} as it generates K(Y ) over K(X). Therefore L is normal over K (and is actually the normalization of K(Y ) over ∼ K(X) = K) and hence Galois. The Galois group of L/K acts on {fi} and with respect to the fixed numbering we have an inclusion

Gal(L/K) ,→ Σd.

We will identify Gal(L/K) with its image under this inclusion, and refer to it as the Galois group of π. Note that unlike the monodromy group, the deﬁnition of the Galois group does not refer to any Zariski open subset.

In the deﬁnitions of both, note that a change of numbering results in a conjugation in the permutation group Σd. As promised, we have the following: Proposition 2.3. For π : Y → X as above, the monodromy group equals the Galois group. In particular, the monodromy group does not depend on the choice of Zariski open subset U.

Proof. We first show that the monodromy group is included in the Galois group. Let γ : [0, 1] → U be any loop based at p. Suppose γ maps to σ under the monodromy representation. Then the lifted arc γei : [0, 1] → V with γei(0) = qi satisfies γei(1) = qσ(i). ∼ Consider fi ∈ ∆ = ∆i, regarded as a meromorphic function defined in a small neighborhood of qi. We may extend fi by analytic continuation along γei; indeed, fi is obtained by restriction of f to a small neighborhood of qi. Hence the resulting germ we obtain at the endpoint qσ(i) is fσ(i). This gives a way to define analytic continuation of any elements in L ⊂ ∆ along γ(t): any element in K = K(X) is fixed, and fi is mapped to fσ(i); an arbitrary element h ∈ L is a polynomial in fi’s with coefficients in K(X) and hence can be analytically continued accordingly. Thus we obtain an automorphism of L fixing K, and hence an element of Gal(L/K). It clearly induces σ in the permutation group. Therefore the monodromy group is included in the Galois group.

10 To show that they are equal, it suffices to show that the fixed field of the monodromy group in L is exactly K. Let h ∈ L be an element fixed by the monodromy group. In particular, h is fixed under analytic continuation along any loops γ in U based at p. We may thus analytically continue h to all of U: indeed, for any r ∈ U, define h at r by analytically continue it along any arc from p to r. This is well defined by our assumption that h is fixed under analytic continuation along loops. On the other hand, h is a polynomial in fi’s with coefficients in K(X). As fi’s come from meromorphic functions on Y , we conclude that h cannot have essential singularities. Therefore h extends to all of X as a meromorphic function, i.e. h ∈ K. This finishes our proof.

It turns out, unsurprisingly, in most cases the monodormy group is the full symmetric group. The general strategy we adopt is the following. First, we show that the monodromy group is twice transitive. Second, we exhibit a simple transposition in the monodromy group. It then follows that the monodromy group contains all transpositions, and is therefore the full symmetric group. The following lemma comes in handy to ﬁnd a simple transposition:

Lemma 2.4. Let π : Y → X be a holomorphic map of degree n. Suppose there exists a point p ∈ X such that the ﬁber of Y over p consists of exactly d − 1 distinct points, i.e., simple points q1, . . . , qd−2 and a double point qd−1 = qd. Suppose furthermore Y is locally irreducible at qd−1. Then the monodromy group M of π contains a simple transposition.

Proof. Take a small neighborhood ∆ of X such that π−1(∆) consists of d − 1 disjoint irreducible components ∆i, where ∆i is a small neighborhood of qi, and π restricted to ∆i gives an isometry for i = 1, . . . , d − 2 and a double cover branched only at qd−1 for i = d − 1. Take −1 a pointp ˜ ∈ ∆ withp ˜ 6= p. Then π (˜p) consists of d distinct pointsq ˜i so thatq ˜i ∈ ∆i for i = 1, . . . , d − 2, andq ˜d, q˜d−1 ∈ ∆d−1. Letγ ˜ be any arc connectingq ˜d, q˜d−1 in ∆d−1. Then the monodromy representation of γ = π(˜γ) gives a simple transposition exchanging d and d − 1.

3 Galois groups of ﬂexes of plane curves

As discussed in Section 2.1, for a generic plane curve C of degree d, there are f = 3d(d − 2) flexes, and each of these flexes is tangent to C at a unique point with contact order of exactly 3. Let us set up the problem in the general framework discussed in the previous section. Let ∼ (d+2)−1 Wd = P 2 be the complete linear system of plane curves of degree d. Fix homogenous 2 2∗ 2 2∗ coordinates [X0 : X1 : X2] on P and correspondingly [Y0 : Y1 : Y2] on P . Let I0 ⊂ P × P be the subvariety defined by I0 := {(p, l): p ∈ l}. P In the coordinates above, this is simply the locus cut out by i XiYi = 0, and hence is a smooth irreducible 3-fold. Let Id ⊂ Wd × I0 be the subvariety defined by

Id := {(C, p, l): mp(C · l) ≥ 3}.

11 This is indeed a subvariety as it is defined as the locus of triples (C, p, l) where the polynomial defining C restricted to the line l has a root of order at least 3 at p. All of these can be written down as algebraic conditions in some affine local charts.

Let π : Id → Wd and η : Id → I0 be projections. Then π is generically ﬁnite when d ≥ 3, with degree f = 3d(d − 2). The ﬁbers of η : Id → I0 are linear subspaces of Wd of codimension 3, and I0 is irreducible, and hence Id is irreducible [Har2, Thm 11.14]. Hence we have the setup as in the general framework with Y = Id,X = Wd and π this projection map we conveniently denote by the same letter. We adopt the same notations as in the general framework, with U ⊂ Wd being a Zariski open set over which π is a unbranched covering −1 map, and V = π (U) ⊂ Id. The open set V is connected as Id is irreducible. We denote the monodromy group by Md.

We first show that Md is transitive. Indeed, as V is connected, for any generic curve C and flexes (p, l) and (p0, l0) there exists an arcγ ˜ in V connecting (C, p, l) to (C, p0, l0). The monodromy action corresponding to γ = π(˜γ) then transports (C, p, l) to (C, p0, l0). In a similar manner we show that Md is doubly transitive. It suffices to show that the stabilizer of any point in Md acts transitively on the remaining. Let C0 be a generic curve with a flex (p0, l0). Set 0 W := {C : mp0 (l0 · C) ≥ 3} ⊂ Wd 0 −1 as W = π(η (p0, l0)), this is a linear subspace of Wd. Moreover, define 0 0 I := {(C, p, l): mp(l · C) ≥ 3, p 6= p0, l 6= l0} ⊂ W × I0

The projection onto the second factor maps onto a Zariski open subset of I0 defined by 0 p 6= p0 and l 6= l0, and the fibers are linear subspaces of W of codimension 3. Therefore, 0 0 I is irreducible. Moreover, as C0 is generic, all its other flexes lie in I . Connect (C0, p, l) 0 0 0 0 and (C0, p , l ) in I by an arcγ ˜ in V . This is possible as V ∩ I is connected. Then the 0 0 monodromy action associated to π(˜γ) fixes (C0, p0, l0) and transports (C0, p, l) to (C0, p , l ). Therefore Md is doubly transitive, as desired.

To show that Md is the full symmetric group Σ3d(d−2), it then suﬃces to exhibit a simple transposition. It turns out that. this is not true in the case d = 3. Thus we discuss the cases d ≥ 4 and d = 3 separately.

3.1 Flexes of a plane curve of degree ≥ 4

To apply Lemma 2.4, we need to find a plane curve of degree d with exactly 3d(d − 2) − 1 flexes. For this, we need to slightly generalize the Plücker formulae obtained in Section 2.1. Let C be a smooth plane curve of degree d with a simple hyperflex (i.e., a point at which the tangent line has contact order 4 with the curve, but intersects the curve transversely elsewhere) in addition to ordinary flexes and bitangents. This corresponds to C∗ having a singularity locally modeled on y3 = x4. In the calculation of branch points when projecting onto a generic line from a generic point, such a singularity of C∗ contributes a branch point of ramification m − 1 instead of 2, and hence 2g − 2 = −2d∗ + d + f + 2,

12 where f is the number of ordinary flexes. Solving for f, we have f = 3d(d − 2) − 2. Together with the simple hyperflex, there are thus 3d(d − 2) − 1 flexes.

−1 We need to show such a curve C exists, and Id is irreducible at each point of π (C). Clearly we need d ≥ 4. We might as well fix the hyperflex (p0, l0). Consider 00 W := {C : mp0 (l0 · C) ≥ 4} ⊂ Wd. 00 This is a linear subspace of Wd of codimension 4. A generic curve C ∈ W is smooth (this follows from Bertini’s theorem, see [GH, §1.1], as the linear system does not have any base point away from p0 and clearly smooth at p0) and has a simple hyperflex at p0. Consider the incidence correspondence 00 00 I := {(C, p, l): mp(l · C) ≥ 3, l 6= l0, p 6= p0} ⊂ W × I0 Again, the projection onto the second factor surject onto the Zariski open set defined by 00 00 p 6= p0, l 6= l0, and fibers are linear subspaces of W of codimension 3. Hence I is irreducible. 00 00 For a generic C ∈ W , for all flexes (p, l) aside from (p0, l0) we have (C, p, l) ∈ I . Consider the subvariety J 00 ⊂ I00 defined by 00 J := {(C, p, l): mC (l) ≥ 3} P P where we recall mC (l) = i(mi − 1) if the divisor C ∩ l on the line l is given by i mipi. Note that this is a closed subvariety of I00 (we have considered something similar in Section 2.1) and as I00 is irreducible, we have either J 00 = I00 or J 00 is a subvariety of strictly less dimension. On the other hand, for any (p1, l1) with l1 6= l0 and p1 6= p0, the linear system of plane curves C of degree d with mp0 (l0 · C) ≥ 4 and mp1 (l1 · C) ≥ 3 cuts out the complete linear system on l1 of effective divisors with order at least 3 at p1 and total degree d. Hence for a generic C in this linear system, the point p1 is an ordinary flex point. Hence it cannot be the case that J 00 = I00. Therefore, for a generic C ∈ W 00, all the other tangent lines are either regular tangent lines, bitangents or ordinary flexes. As discussed above, such a curve will have exactly 3d(d − 2) − 1 flexes.

Finally, as the ﬁbers of η : Id → I0 are all linear subspaces, Id is smooth and hence irreducible at every point. Now we can apply Lemma 2.4 to conclude that Md is the full symmetry group Σ3d(d−2).

3.2 Flexes of a plane cubic

The classical approach. In [Jor], Jordan commented on previous works by Hess that the nine ﬂexes on a plane cubic have additional structures: there are 12 lines passing through triples among them, and each ﬂex lies on 4 lines. Denote by (xy) where x, y = 0, 1, 2 the nine points, and (xy)(x0y0)(x00y00) the line passing through them, he observed that the Galois group has to preserve the expression ϕ =(00)(01)(02) + (10)(11)(12) + (20)(21)(22) +(00)(10)(20) + (01)(11)(21) + (02)(12)(22) +(00)(11)(22) + (01)(20)(12) + (02)(10)(21) +(00)(12)(21) + (01)(10)(22) + (02)(20)(11)

13 He then asserted that maps of the form (xy) 7→ (ax + by + α, a0x + by0 + α0) are exactly those preserving ϕ, and noted that these maps form a solvable group. In modern language, his expression is a way to specify the linear conditions on the affine space structure of the set of flexes, as discussed in Section 2.2, and the group he determined is the affine linear group AGL(2, Z/3). This group is indeed solvable (see below). Monodromy group. We now determine the Galois group using the monodromy group. Recall from Section 2.2, the set Γ of flexes has the structure of an affine space of dimension 2 over Z/3. First note that the structure of affine space has to be preserved under monodromy. Indeed, let p1, p2, p3 be the flexes of a generic cubic C so that the corresponding points z1, z2, z3 satisfy z1 + z2 + z3 = 0. In other words, p1, p2, p3 are collinear. Let γ(t) = C(t) be a loop in U ⊂ W3, and letγ ˜i(t) = (C(t), pi(t), li(t)) be the lift based at (C, pi, li). For each t, let p(t) be the third point of intersection of C(t) with p1(t)p2(t) and l the tangent line at p(t). Then α(t) = (C(t), p(t), l(t))

is also a lift of γ based at (C, p3, l3). Therefore α(t) =γ ˜3(t), and in particular, the end points p1(1), p2(1), p3(1) are also collinear. Therefore the monodromy group is a priori a subgroup of the aﬃne general linear group AGL(2, Z/3). We have, in fact: Proposition 3.1. The monodromy group is ASL(2, Z/3).

Proof. We claim that for any flex of a generic cubic C, with respect to the structure of affine space on the set of flexes, the stabilizer in the monodromy is SL(2, Z/3). Assuming this is true, and let ψ be any element of ASL(2, Z/3) and fix p ∈ Γ. The monodromy group acts transitively, as discussed above, and hence there exists an element ϕ in the monodromy group so that ϕ(p) = ψ(p) and hence ψ−1ϕ fixes p. Note the stabilizer of the point p in either group −1 equals ASL(2, Z/3), and hence ψ ϕ lies in the monodromy group, and so does ψ. Thus it suffices to prove the claim. Let C be identified with some unit area complex torus C/Λ via z 7→ [1, ℘(z), ℘0(z)]. In this identification, the set of flexes are identified with torsion points of order 3 on the a¯ ¯b torus. The origin is in particular a flex point. Let Λ = ω + ω . Let A = be any Z 1 Z 2 c¯ d¯ a b element in SL(2, /3) and A = reducing to A mod 3. As SL(2, ) is connected, Z c d R choose an arc A(t) with A(0) = I and A(1) = A. Set ω (t) ω 1 = A(t) 1 ω2(t) ω2 and Λt := Zω1(t) + Zω2(t) Then C/Λt is also a unit area complex torus (we need det A(t) = 1 to preserve area). Let ℘t be the associated Weierstrass function, and set C(t) be the image under the embedding:

0 z 7→ [1, ℘t(z), ℘t(z)].

14 The origin maps to a flex point p(t) of C(t) with tangent line l(t). Hence this gives an arc in V , which maps down to a loop based at C = C(0), as A(1) = A fixes the original lattice Λ. Moreover the monodromy corresponding to this loop fixes the flex corresponding to the origin, and acts on the other 8 flexes by the matrix A with respect to the affine structure, as clear from the construction. Hence we are done.

Finally, note that the group ASL(2, Z/3) is solvable. Indeed, the derivative map is a homomorphism from ASL(2, Z/3) to SL(2, Z/3), with kernel the group of translations. SL(2, Z/3) acts on 1 (as lines on a vector space of dimension 2 over /3). As 1 consists of 4 PZ/3 Z PZ/3 points, we have SL(2, Z/3) → Σ4, with kernel ±I. Hence SL(2, Z/3) is solvable. The group of translations is abelian, and hence solvable. Therefore ASL(2, Z/3) is solvable. P i j k This then suggests that given the defining equation F (X) = i+j+k=3 aijkX Y Z of a generic plane cubic C, it is possible to express the coordinates of the 9 flexes using radicals and rational functions in the coefficients aijk. Indeed, first note that there are 12 lines each passing through 3 flexes, and at each flex there are 4 such lines. Moreover, there are four i i i triples of lines {L1,L2,L3} such that they contain all 9 flex points. (In terms of 9 torsion 3 points forming a 3 by 3 matrix, these triples come from triples of horizontal, vertical, diagonal i i i i and anti-diagonal lines.) Then C = L1 + L2 + L3 is a singular cubic containing all the flex points of C. Note that we have a pencil of cubics passing through these 9 points, defined by

Gλ(X) = F (X) + λH(X),

where H(X) is the Hessian of F (X) whose coeﬃcients are expressible as rational functions i of aijk’s. The singular member of this pencil are exactly C ’s. Now consider the discriminant

∆(λ) := disc(Gλ(X)) whose coefficients are also expressible as rational functions of aijk. The discriminant vanishes when Gλ(X) is reducible. As a degree 12 polynomial, ∆(λ) has 4 triple roots ωi. The reduced polynomial 4 Y ∆(Λ)˜ := (λ − ωi) i=1 satisfies ∆˜ 3 = ∆, and hence its coefficients are expressible as those of ∆. Solve the polynomial ˜ i i ∆, and we obtain the roots ωi and also the defining equations Gi = Gωi for C . Since C are collection of three lines, its singular points are pairwise intersections of these lines. Solving the resultant of {∂Gi(X)/∂X, ∂Gi(X)/∂Y, ∂Gi(X)/∂Z}, which is cubic, gives these singular points, and hence the lines. The flexes points are simply suitable intersections of these lines.

4 Galois groups of bitangents of plane curves

As discussed in Section 2.1, for a generic plane curve C of degree d, there are b = d(d − 2)(d − 3)(d + 3)/2 bitangents, and each of these bitangents intersect the curve C transversely elsewhere. Again, we setup the problem so that we can apply the general framework. As

15 2 2 before, Wd denotes the complete system of plane curves of degree d. Let J0 = P × P − ∆ 2 2 where ∆ ⊂ P × P is the diagonal. Then J0 is a Zariski open subset of an irreducible projective variety, and hence irreducible. For d ≥ 4, let Jd ⊂ Wd × J0 be the incidence correspondence

Jd := {(C, p1, p2): mp1 (C · p1p2) ≥ 2, mp2 (C · p1p2) ≥ 2}.

Again, Jd is a subvariety. Let π : Jd → Wd and η : Jd → J0 be projections. The projection π is generically ﬁnite of degree b. For the projection η, its ﬁbers are linear subspaces of codimension 4. Hence Jd is irreducible. We again have the set up of the general framework, and we ask for the monodromy group Md associated to π : Jd → Wd. Again, U denotes a −1 Zariski open subset of Wd over which π is an unbranched cover and V = π (U).

We ﬁrst show that Md is transitive. As Jd is irreducible, V is connected. Proceed in the same manner as we did in Section 3 to conclude that Md is indeed transitive. Second we show that Md is twice transitive. Again, it suﬃces to show that the stabilizer of any point in Md acts transitively on the remaining. Let C0 be a generic curve with a bitangent p1p2. Set

0 W := {C : mp1 (C · p1p2) ≥ 2, mp2 (C · p1p2) ≥ 2} ⊂ Wd

Again, this is a linear subspace of Wd (of codimension 4). Moreover, deﬁne

0 0 I := {(C, q1, q2): mq1 (C · q1q2) ≥ 2, mq2 (C · q1q2) ≥ 2, qi 6= pj, q1q2 6= p1p2} ⊂ W × J0

The projection onto the second factor maps onto a Zariski open subset of J0 deﬁned by 0 0 qi 6= pj, q1q2 6= p1p2, with ﬁbers linear subspaces of W of codimension 4. Hence I is irreducible. Moreover it contains all bitangents to C0 except p1p2. Proceed as before and we conclude Md is doubly transitive.

Finally, to show that Md is the full symmetric group Σb, it suﬃces to exhibit a simple transposition. Again, this is not true in the lowest degree case d = 4. Thus we discuss the cases d ≥ 5 and d = 4 separately.

4.1 Bitangents of a plane curve of degree ≥ 5

To apply Lemma 2.4, we need to find a plane curve of degree d with exactly d(d − 2)(d − 3)(d+3)/2−1 bitangents. Again, for this we need to slightly generalize the Plücker formulae. Let C be a smooth plane curve of degree d with a simple flex bitangent, i.e., a bitangent line ∗ p1p2 such that mp1 (C · p1p2) = 2, mp2 (C · p1p2) = 3. This corresponds to a singularity of C locally modeled by (x − y)(x3 − y2) = 0. Still, let b be the number of ordinary bitangents, f be the number of ordinary flexes. Clearly f = 3d(d − 2) − 1. When calculating the degree of dual curves, a singularity of C∗ of type modeled by (x − y)(x3 − y2) = 0 contributes an intersection of multiplicity 7, that is,

d = d∗(d∗ − 1) − 2b − 3f − 7,

which gives b = d(d − 2)(d − 3)(d + 3)/2 − 2. Together with the simple ﬂex bitangent, there are then exactly d(d − 2)(d − 3)(d + 3)/2 − 1 bitangents.

16 We need to show such a curve C exists. We might as well fix the tangency points p1 6= p2. Let 00 W := {C : mp1 (C · p1p2) ≥ 2, mp2 (C · p1p2) ≥ 3} ⊂ Wd. 00 00 Then W is a linear subspace of Wd of codimension 5. A generic element C ∈ W is smooth (again by Bertini’s theorem) and has a simple flex bitangent p1p2. Consider the incidence correspondence 00 00 I := {(C, p, l): mp(C · l) ≥ 3, p 6= p1, p2, l 6= p1p2} ⊂ W × I0. 00 00 Since the fibers of the projection I → I0 are codimension 3 linear subspaces of W and this 00 projection surjects onto a Zariski open subset of I0, I is irreducible. Consider the subvariety of I00 defined by 00 J := {(C, p, l): mC (l) ≥ 3} Then again either J 00 = I00 or J 00 is a subvariety of strictly less dimension. On the other hand, given (p, l) with p 6= p1, p2 and l 6= p1p2, the linear system of plane curves of degree d

with mp1 (C · p1p2) ≥ 2, mp2 (C · p1p2) ≥ 3, mp(C · l) ≥ 3 cut out the complete linear system of divisors on l with order at least 3 at p1 and total degree d. Hence for a generic C in this linear system, the point p is an ordinary flex point. Hence it cannot be the case that J 00 = I00. Therefore for a generic C ∈ W 00, all the other tangent lines except the flex bitangent are either regular tangent lines, simple bitangents or ordinary flexes. As discussed above, such a curve has d(d + 2)(d + 3)(d − 3)/2 − 1 bitangent lines.

Thus Md is the full symmetry group.

4.2 Bitangents of a plane quartic

Recall that for a generic plane quartic C the set of bitangents Γ can be identiﬁed with a subset 1 ∼ 6 of the group of elements of order 2 in J (C), which is a vector space V = 2 Λ/Λ = (Z/2) . ∼ On Λ = H1(C, Z), there is a skew-symmetric, nondegenerate pairing (·, ·) coming from intersections of cycles. The form 4(·, ·) then induces a bilinear form Q on V and it is strictly skew-symmetric. Choose normalized basis αi, . . . , α3, β1, . . . , β3 with respect to (·, ·), and 0 correspondingly choose a basis for ω1, . . . , ω3 for H (C,KC ). As in Section 2.4, we then have P3 the lattice is generated by λ1, . . . , λ6, with λj = ej and λg+j = i=1 Zijei where Z = (Zij) is a symmetric matrix. Let θ be the associated Riemann theta function. Since g = 3, by Riemann’s theorem, there exists a semicanonical divisor κ (of degree 2) on C such that for any divisor class E ∈ J (C) of degree 0, we have

κ + E is linearly equivalent to an eﬀective one if and only if θ vanishes at E, with vanishing order h0(κ + E).

Now consider the order 2 points, i.e., elements of V in the Jacobian:

3 3 3  3  X λi X λ3+i X X ei a Zb v = a + b = a + Z b = + i 2 i 2  i ij j 2 2 2 i=1 i=1 i=1 j=1

17 where ai, bi ∈ {0, 1} (or we may extend to ai, bi ∈ Z; this does not matter as we are only interested whether θ vanishes or not), and a = (ai), b = (bi) are column vectors. Then

3 0 0 X 0 0 Q(v, v ) = 4(v, v ) = (aibi + biai) i=1

+ + P3 Let q be the quadratic reﬁnement of Q given by q (v) = i=1 aibi = ha, bi. Also, for v ∈ V we have X θ(v) = eπihl,Zli+πihl,ai+πihl,Zbi l∈Z3 Now replace l by −b − l in the expression, we have X θ(v) = eπih−b−l,−Zl−Zbi+πih−b−l,ai+πih−b−l,Zbi l∈Z3 X = e−πihb,ai eπihl,Zli+πihl,Zbi−πihl,ai l∈Z3 X = e−πihb,ai eπihl,Zli+πihl,Zbi+πihl,ai−2πihl,ai l∈Z3 X + = e−πihb,ai eπihl,Zli+πihl,Zbi+πihl,ai = e−πiq (v)θ(v) l∈Z3 Clearly, when q+(v) = 1 then θ(v) = 0. Similarly, as

∂θ X πihl,Zli+2πihl,vi X πihl,Zli+πihl,ai+πihl,Zbi (v) = 2πi lje = 2πi lje ∂zj l∈Z3 l∈Z3 replacing l by −b − l renders

∂θ −πihb,ai X πihl,Zli+πihl,Zbi+πihl,ai (v) = 2πie (−bj − lj)e ∂zj l∈Z3 −πiq+(v) ∂θ = −e 2πibjθ(v) + (v) ∂zj

If q+(v) = 0, then for all j we have

∂θ (v) = −πibjθ(v). ∂zj

Suppose θ(v) = 0, then θ vanishes at v up to order 2, and thus h0(κ + v) ≥ 2. This is impossible, as otherwise C would be hyperelliptic. hence θ(v) 6= 0. In summary, we have

κ + v is (linearly equivalent to) an eﬀective semicanonical divisor if and only if q+(v) = 1.

18 + + Let κ0 = κ+v0 be an eﬀective semicanonical divisor on C, and set q(v) = q (v0 +v)−q (v0). + + 6 2 3 Since q (v0) = 1, we have q(v) = 0 if and only if q (v0+v) = 1. Thus q(v) has 2 −2 (2 +1) = 22(23 − 1) zeros. Moreover

+ + + + q(v + w) + q(v) + q(w) = q (v0 + v + w) + q (v0 + v) + q (v0 + w) + q (v0) + + + = q (v0 + v + w) + q (v0 + v) + q (w) + + + + q (v0 + w) + q (v0) + q (w)

= Q(v0 + v, w) + Q(v0, w) = Q(v, w) Hence q(v) is also a quadratic reﬁnement; in fact, it is (equivalent to) the odd one. Now we have + q(v) = 0 ⇐⇒ q (v + v0) = 0 ⇐⇒ κ0 + v = κ + v0 + v is eﬀective. Thus bitangents of the plane quartic corresponds to the zeros of q in V .

Note that all these structures are preserved under the monodromy group. As in the case of flexes, given four bitangents with 8 tangency points add up to 0 in the affine structure, any five tangent points of three bitangents will determine a conic, and the last bitangent has tangency points the remaining intersection of the conic and the quartic. Hence throughout the loop, the affine structure is preserved. Similarly the quadratic form and the zeros of q. Hence the monodromy group is restricted to a subgroup of

H = {v 7→ Av + b : Q(Av, Aw) = Q(v, w)∀v, w ∈ V, q(Av + b) = q(v)∀v ∈ V }

This group is called the Steiner group. Note that the stabilizer of the origin in H is exactly − the odd orthogonal group O6 (Z/2). If we can show that the stabilizer of the monodromy − group ﬁxing κ0 is O6 (Z/2), then following the same argument as in the case of ﬂexes, we can conclude that the monodromy group is precisely H. We claim this is true, but defer the arguments until the next section, after we have calculated the Galois group of 27 lines on a cubic surface.

5 Galois group of 27 lines on a cubic surface

5.1 Cubic surfaces and their 27 lines

We refer to [Bea] for basic facts about complex algebraic surfaces.

3 2 A cubic surface in P is the blowup of P at 6 points in general position. Write S = 2 2 blp1,...,p6 P −→ P with exceptional curves Ei coming from blowup at pi. Then NS(S), the N´eron-Severi group of S is generated by Ei and the pullback of the hyperplane class L on 2 ∼ 2 2 P . As NS(S) = H (S, Z) for rational surfaces, we have H (S, Z) is 7 dimensional. The 3 ∗ ∗ hyperplane class H on P restricts to 3 L − E1 − · · · − E6. As KS = KP2 + E1 + ... + E6 and KP2 = −3L, we conclude that H = −KS. We claim that lines on S are exactly the exceptional curves. Indeed, let E be an exceptional curve on S. Then g(E) = 0 and E · E = −1. By genus formula g(E) =

19 1 + (E · E + E · KS)/2, we have E · KS = −1 and therefore E · H = 1, that is, E is a P line on S. Similarly the converse is true. Set E = mL − i miEi and assume E 6= Ei. Since mi = E · Ei, and both E and Ei are lines, mi = 0 or 1. Then 2 X 2 X 1 = E · (−KS) = 3mL + miE = 3m − mi i i Hence we have the following cases:

1. m = 1, and two of mi’s are 1 while the rest are 0. Let L˜ij be the line passing through 2 −1 ˜ pi, pj on P . Then (Lij) consists of two exceptional lines Ei,Ej and the strict −1 transform Lij of L˜ij. As L = (L˜ij) = Ei + Ej + Lij, we have Lij = L − Ei − Ej is precisely this exceptional divisor.

2. m = 2, and ﬁve of mi’s are 1 and the rest are 0. Let C˜i be the conic passing through −1 ˜ P ﬁve points of blowup except pi. Then 2L = (Ci) = Ci + j6=i Ej where Ci is the strict transform of C˜i. Thus again Ci is exactly this exceptional divisor.

6 Therefore in total we have 6 + 2 + 6 = 6 + 15 + 6 = 27 lines on a cubic surface. 1 If we look at H (S, Z/2), then any exceptional curve E · H = 1 mod 2. Let V be the 1 6-dimensional subspace of H (S, Z/2) of elements D with D · H = 0 mod 2. Then the set of exceptional curves is a subset of the affine space E + V for any fixed exceptional curve 1 E. The intersection pairing on H (S, Z/2) gives a pairing on V . For any D ∈ V we have D · H = 0 mod 2; moreover as D · D + D · K D · D − D · H g(D) = 1 + S = 1 + ∈ 2 2 Z we have D · H = 0 mod 2. Thus the pairing on V , denoted by Q, is strictly skew-symmetric. Define D · D q(D) = mod 2 2 where we abuse the notation and denote D ∈ V and a choice of its lift in integral cohomology using the same letter. First note as D · D is even, this is well defined. Moreover (D + D0) · (D + D0) + D · D + D0 · D0 q(D + D0) + q(D) + q(D0) = = D · D0 = Q(D,D0) 2 modulo 2, and thus q gives a quadratic refinement of Q.

Now Let E be any exceptional curve. Then (E − H) · H = 1 − 3 = 0 mod 2. Hence E − H ∈ V . Moreover −1 + 3 − 2 q(E − H) = (E − H) · (E − H)/2 = = 0 2 hence exceptional curves correspond to zeros of q. On the other hand, by our analysis above, the map E → E −H ∈ V is injective and avoids 0 (all exceptional curves are diﬀerent modulo 2 and not equal to H). On the other hand,

2 X 6 (xL + y1E1 + ··· + y6E6) · (xL + y1E1 + ··· + y6E6) = x − yi . i

20 It is easily veriﬁed that there are 27 nontrivial choices of x, yi ∈ {0, 1} so that this expression is a multiple of 4. Hence q has 27 + 1 zeros whose nontrivial zeros are in correspondence with exceptional curves. Note that q is then equivalent to q−, an odd quadratic reﬁnement.

1 Now, let Γ be the set of lines (i.e. exceptional curves) on S. As Γ span H (S, Z) = NS(S), any permutation of Γ respecting the intersection form extends to a linear automorphism 1 of H (S, Z) preserving the intersection form. As H is equal to the sum of there mutually intersecting lines, this linear automorphism must preserve H. Hence this linear automorphism gives a linear automorphism in V preserving the form Q and zeros of q. On the other hand, looking at the nontrivial zeros of q we recover the permutation. Hence the conﬁguration − group of these lines is equal to O (6, Z/2).

1 Moreover, any set of 6 pairwisely disjoint lines together with H span H (S, Z) and more- − over have the same intersection pattern. Thus an element of O (6, Z/2) is determined by sending a ﬁxed collection of 6 disjoint lines to another collection together the permutation of these 6 lines. Let Ω be the set of collections of 6 disjoint lines. Then #Ω = 72. Indeed, it contains the following collections:

A = {Ei}i

Bijk = {Ei,Ej,Ek} ∪ {Lmn}m,n6=i,j,k

Cij = {Ci,Ei} ∪ {Ljk}k6=i

Dijk = {Ci,Cj,Ck} ∪ {Lmn}m,n6=i,j,k

E = {Ci}i

The key to obtain these is that since we want, say Ci and Ljk, disjoint, their projections in 2 P can only intersect at the blowup points; since deg(Ci · Ljk) = 2, we must have Ci passes through pj, pk and hence i 6= j, k. Now an elementary counting gives there is one of A, 20 of Bijk’s, 30 of Cij’s, 20 of Dijk’s and one of E.

− Set G = O (6, Z/2). Then #G = 6!·#Ω = 720·72. We say an element of G is elementary if it ﬁxes an element of Ω, and set G0 to be the group generated by elementary elements. Let 0 0 0 Ω be the orbit of A under G . Then as permutations of Eis ﬁxes A and rearrange elements of A, we have #G0 = 6! · #Ω0 = 720 · #Ω0, 0 and in particular Ω divides 72. On the other hand, permutations of Ei acts transitively on 0 each of {A}, {Bijk}, {Cij}, {Dijk}, {E},Ω must be the union of some of them. The only union of some of these sets that has cardinality dividing 72 is that of all of them. Hence G0 = G, and in particular G is generated by elementary elements.

5.2 Galois group of 27 lines on cubic surfaces

We describe ﬁrst the generalities of the enumerative problem of lines on a generic hypersurface n S ⊂ P of degree d = 2n − 3. In the case n = 3, we have d = 3 and this is the case of lines n on a cubic surface. Let W be the linear system of hypersurfaces of degree d = 2n − 3 in P ,

21 and let I ⊂ W × Gr(2, n + 1) deﬁned by

I := {(S, l): l ⊂ S}.

We can then deﬁne the monodromy group associated to the projection π : I → W . It turns out for n ≥ 4, the answer is the full symmetric group. The approach is very much the same as in the cases of ﬂexes and bitangents, modulo some machineries. See [Har1] for details.

We now focus on the case n = 3. The monodromy group M clearly preserves the incidence relations of the 27 lines, and hence by the discussions in the last section, M is a subgroup of − − O (6, Z/2). We claim it is the full group G = O (6, Z/2). Again by the discussions in the last section, it suﬃces to prove:

Proposition 5.1. M contains all the elementary elements in G.

Proof. Let σ be a permutation of 6 disjoint lines l1, . . . , l6 on S. Blow down the lines to get 2 P , with image of the lines being pi. Drawing arcs pi(t) such that pi(0) = pi and pi(t) = pσ(i). 2 2 6 As the set U of six points of P in general position is a Zariski open subset of (P ) , it is connected. Therefore we may assume (pi(t)) remain in general position for all t ∈ [0, 1]. Now 2 blow up P at the six points pi(t), and let li(t) be the exceptional divisors. Denote the resulting 3 surface S(t). S(t) is embedded into P by sections of the anticanonical bundle −KS(t). Let 3 φ1, φ2, φ3, φ4 be the pullback of homogeneous coordinates Xi on P . As GL(4, C) is connected, 0 we can choose φ1(t), φ2(t), φ3(t), φ4(t) ∈ H (S(t), −KS(t)) so that they vary continuous with 3 respect to t, gives embeddings to P and φi(t) = φi(0). Then the monodromy action of the loop given by the embedded surfaces S(t) is then the desired permutation.

− Therefore we can conclude that M = O (6, Z/2). This group is not solvable (see [Jor]). Using the monodromy group and its action on the set of 27 lines, we can conclude the following:

1. Given two lines on S with no intersection, the problem is still not solvable. Clearly the 0 − Galois group G corresponding to this problem is the subgroup of O (6, Z/2) ﬁxing a − pair of skew lines. Since there are 27 · 16 pairs of skew lines and the group O (6, Z/2) 72 × 720 acts transitively on them, we have #G = = 120. On the other hand, there 27 × 16 are 5 diﬀerent ways to extend this pair to a collection of 6 disjoint lines, and G0 acts 0 ∼ on them as permutations. Therefore G = Σ5 is not solvable.

− 2. Given three skew lines on S, the problem is solvable. Again, O (6, Z/2) acts transitively on triples of skew lines. As there are 27 · 16 · 10 triples, the subgroup G00 ﬁxing a triple 00 then has order 12 and therefore G is solvable. Let E1,E2,E3 be a triple of skew lines. The quadric surface Q containing them can be determined by rational functions in points on E1,E2,E3. Q has two rulings {Mλ} and {Nµ}. As one of the intersect in the cubic surface in three lines, the other must also. These three lines are disjoint, and each intersect all of E1,E2,E3, and hence must be C4,C5,C6. Let λ, µ be the coordinates on Q with respect to the rulings. Write down the restriction of the deﬁning equation

22 of S to Q in terms of λ and µ, factor out the ones corresponding to the three known lines, and solve the cubic polynomial, we can then obtain the new lines.

The plane spanned by Ei,Cj intersects the cubic surface in a third line Lij, and hence we obtain the coordinates for Lij for i = 1, 2, 3 and j = 4, 5, 6. Using Plücker coordinates and Plücker relations (see [Har2, Example 6.1]), we can then solve for an additional line (Plücker relations are quadratic, and hence this is solvable). The remaining lines are successively residuals to the existing pair of lines in the plane they span, and hence we can solve for them one by one. 3. Finally, if we are given two incident lines, the problem is again solvable. Indeed, let 000 − e1, e2 be the corresponding elements in V , and G the stabilizer of them in O (6, Z/2). Again, we calculate the order of G000 to be 192. Now Q and q are both identically zero on the subspace spanned by e1 and e2. The orthogonal complement of V1 := span{e1, e2} 000 ⊥ ∼ 2 with respect to Q must also be preserved. Thus G acts on V1 /V1 = (Z/2) as ∼ ⊥ GL(2, Z/2) = Σ3 (the nontrivial elements in V1 /V1 are not zeros of q). The kernel of this action has order 32 and hence solvable. In practice, given two incident lines, say E1 and L12, we can look at the pencil {Ht} of planes passing through E1. There are five planes intersecting S residually in two lines, one of which we already know. Restrict the defining equation of S to Ht and factor out the equation for E1, we get a quadric polynomial, whose discriminant will be a quintic in t. Factor out the one root we already know, we then need to solve a quartic. After we obtain these 5 planes, solving a quadratic equation gives us the individual lines on each plane. Then we can proceed using Plücker coordinates as in the previous case. One also realize in this process given just one line won’t work, as one need to solve a quintic polynomial otherwise.

In conclusion, given a generic cubic surface, and together with the coordinates of either 1) three skew lines or 2) two incident lines, we can solve for the others explicitly, but no smaller set of lines suﬃces.

5.3 Galois group of bitangents of a plane quartic

We first discuss the relations between 27 lines on a cubic surface and 28 bitangents of a plane quartic. Let S be a generic smooth cubic surface. Let p ∈ S be a point not lying on any lines ∼ 2 of S. One can then look at the projection of S from p to a generic plane H = P . This is of course not well defined at p, but well defined on the blowup S˜ = blp S. Each point on the exceptional divisor gives rise to a tangent direction, and the projection of that point is then the intersection of the tangent line at p in that direction with H.

Now the projection map is generically 2 to 1, and branched over a locus C in H. Note that C is the image of the set of points of tangency on a line emanating from p. a Let l be a generic line on H, and let Hl be the plane spanned by l and p. Then Hl intersects S in a cubic curve Cl passing through p. The number of intersection of l and C is then the number of tangent lines passing through p (but excluding the tangent lines at p). This is easily seen ∗ as deg(Cl )−2 = 4. Hence the branching locus is a curve of degree 4. It is generically smooth. Another way to see this is that to ﬁrst note that the preimage of the branching locus gives

23 points on the cubic where there exists a tangent line passing through p. This is the case if and only if the tangent plane there passing through p. Let F (X) be the deﬁning equation for the cubic surface. Then the tangent plane at a point corresponds to

∂F ∂F ∂F ∂F : : : ∂X0 ∂X1 ∂X2 ∂X3

in the dual projective space. Hence this plane passing through p = [P0 : P1 : P2 : P3] if and only if ∂F ∂F ∂F ∂F P0 + P1 + P2 + P3 = 0 ∂X0 ∂X1 ∂X2 ∂X3 3 and this gives a quadric surface in P , called the polar quadric of the cubic surface S at p. The intersection of this quadric surface and S is the preimage of the branching locus. Now this quadric surface passes through p (as the tangent plane of the cubic surface at p passes through p), and its tangent plane at p corresponds to

" 2 # X ∂ F Pi . ∂XiXj i p j

P ∂G Given a homogeneous polynomial G(X) of degree d, we have i Xi = dG. Apply this ∂Xi to ∂F/∂Xj, we have X ∂2F ∂F X = 2 i ∂X X ∂X i i j j and hence at p, the tangent planes at p to the polar quadric and the cubic coincide. Now the polar quadric and the cubic surface intersect in a curve of degree 6, and has a double point at p (see the next paragraph). Projecting from p to the plane H hence gives a curve of degree 6 − 2 = 4.

As quadric surfaces are ruled, let m1, m2 be the two lines passing though p. Then m1, m2 span the tangent plane to the quadric at p and hence also span the tangent plane to the cubic S at p. Near p, Consider the curve Tp(S) ∩ S. The projection restricted to this curve is unramiﬁed away from p, and hence this curve cannot intersect m1 and m2 away from p. Thus m1 and m2 are also tangent lines to Tp(S) ∩ S at p (as we have discussed above, this is a double point of the curve).

Let L be a line on S. Then the intersection of the plane HL spanned by L and p has a conic component CL residual to L. The image of the line L only intersect C at two points, the image of the two points in L·CL. Hence the image of L is a bitangent of C. This accounts for 27 bitangents. The remaining one D is the intersection of the tangent plane TpS with H. Note that TpS intersects with S in a cubic with a double point at p. The intersection of the two tangents with H will then be the tangency points of this bitangent.

In this way, we identify the 27 lines on S realized as a subset of a vector subspace V of 1 H (S, Z/2), with 27 of the 28 bitangents of C, realized as the nontrivial element in a subset of a vector subspace W of order 2 points in J (C), with the remaining bitangent D at origin.

24 We also remark that any quartic plane cubic can be realized as the branching locus of a projection from a point on a cubic surface as above, and we may choose any of its bitangents to be the distinguished one. There is a very explicit way of doing this as explained brieﬂy in [Har1].

Now recall that in order to show that the monodromy group of bitangents of a plane quartic is the Steiner group, we need to show that the stabilizer of a bitangent D in the − group is O (6, Z/2). In order to do this, we are going to show: 1) the identiﬁcation above extends to a linear isomorphism V =∼ W identifying all the structures. 2) the stabilizer corresponds to the monodromy group of 27 lines on a cubic surface via this identiﬁcation.

First of all, we have

Lemma 5.2. Let L1,L2,L3 be three lines on S, and p1, p2, q1, q2, r1, r2, s1, s2 be points of tangency on the bitangents corresponding to L1,L2,L3 (we will simply use the same symbols to denote the bitangents as well) and the bitangent D. Then L1 + L2 + L3 − 3H is divisible 1 by 2 in H (S, Z) then {pi, qi, ri, si} all lie on a conic.

Proof. We ﬁrst show that L1 + L2 + L3 − 3H is divisible by 2 if and only if L1,L2,L3 are coplanar. For one direction, if L1,L2,L3 is coplanar, then L1 + L2 + L3 − 3H = −2H is divisible by 2. For the converse, assume L1 + L2 + L3 − 3H is divisible by 2. Then P P Li · (L1 + L2 + L3 − 3H) = Li · j6=i Lj − 4 is even, and hence Li · j6=i Lj is even for i = 1, 2, 3. In particular, either Li · Lj all equal to 0 or all equal to 1 for i 6= j. On the other 2 P hand, we have (L1 + L2 + L3 − 3H) = 6 + 2 i6=j Li · Lj is divisible by 4. It then follows that Li · Lj all equal to 1 for i 6= j, and hence L1,L2,L3 are coplanar.

Next, we show that L1,L2,L3 lie on a plane H0 then {pi, qi, ri, si} all lie on a conic. Indeed, consider the polar quadric Qp at p. Then H0 ∩ Qp is a quadric and projects down to a quadric curve containing {pi, qi, ri}. Next, note that the tangent lines m1 and m2 to the curve TpS ∩ S at p lie in Qp. Thus mi ∩ H0 lies on the conic, and so does its projection. But this projection is none other than si. Hence all 8 points lie on a conic.

A direct consequence of this lemma is that the identification of 27 lines and 28 bitangents with one distinguished as the origin extends to an identification of V and W , as the linear conditions are precisely the two described in the lemma. Moreover, this is an identification of vector spaces with quadratic forms, as it maps the zeros of one on V to another on W . We have

− Proposition 5.3. The stabilizer of D in the monodromy group of 28 lines is O (6, Z/2).

Proof. This follows from the fact any quadric may be realized as the branching locus with any bitangent ﬁxed as origin, as described above. It is easy to see the conﬁgurations can be chosen to vary continuously. Hence this stabilizer is the same as the monodromy group of 27 lines on a cubic surface. The proposition then follows from the analysis in the previous section.

25 In particular, the problem of determining 28 bitangents on a quartic is not solvable. But reading from the analysis we have done in the case of 27 lines, we have, given a generic quartic plane curve, and together with either 1) three lines whose points of tangency lie on a conic or 2) four lines whose points of tangency do not lie on a conic, we can solve for the others explicitly, but no smaller collection suﬃces.

6 Galois group of the problem of ﬁve conics

Now we have discussed the main problems in a somewhat self-contained manner, we finally briefly describe another classical enumerative problem, albeit with fewer details: determine the number of conics tangent to all given 5 conics in general position. This problem is sometimes referred to as the Steiner’s conic problem. We will first briefly explain why there are 3264 of them, and then determine the Galois group to be the full symmetric group.

6.1 Counting the number of conics

∼ 5 We will follow [GH, §6.1]. Let W = P be the complete system of plane conics. Let C be a smooth plane conic, and VC ⊂ W the set of conics tangent to C. This gives a hypersurface of some degree d in W . To calculate the degree of VC , let L be a generic line in W , and let C[λ:µ] = λC1 + µC2 be the corresponding pencil of conics. Then C[λ:µ] cuts out a linear 1 system of degree 4 on C without base point, which exhibits C as a 4-sheeted cover to P , with branch points corresponding to elements in the pencil tangent to C. By Riemann-Hurwitz, the number of branch points is 1 2g(C) − 2 − 4(2g(P ) − 2) = 6, 0 0 and therefore VC has degree 6. Moreover, let C be a smooth point of VC and suppose C and C is simply tangent at a point p with all other intersections transverse. Let Hp ⊂ W be the hyperplane consisting of conics passing though p, and L a generic line passing though 0 C . Let C[λ:µ] = λC1 + µC2 be the complete linear system corresponding to that line. Then C[λ:µ] cuts out a linear system of degree 4 on C with base point p, which then exhibits C as 1 a 3-sheeted cover to P , again, with branch points corresponding to elements in the pencil tangent to C other than C0. By Riemann-Hurwitz, the number of branch points is 1 2g(C) − 2 − 3(2g(P ) − 2) = 4, and hence there are generically 4 conics other than C0 tangent to C. Hence the line L is 0 0 tangent to VC at C . Thus Hp is the tangent plane to VC at C .

Steiner [Ste] originally thought that, having determined the degree of VC , the number of conics tangent to 5 given ones in general position is then 65. The problem is that for any conic C, VC contains the subvariety of double lines W1 = {2L}L∈P2∗ (i.e. rank 1 conics), even though they do not all contain the subvariety of two (distinct) lines (i.e. rank 2 conics). To account for this, we consider the blowup of W along the subvariety W1:

π : Wf = blW1 W → W,

26 and set VeC to be the strict transform of VC . We ﬁrst determine the multiplicity of W1 in VC . Let 2L be a generic double line, and C[λ:µ] be a generic pencil containing 2L. Again, this 1 exhibits C as a 4-sheeted map over P with 6 branch points. But two of these are simply the intersection L ∩ C. Hence there are 4 other conics tangent to C. That is, the line passing through the double line 2L in W intersect W1 with degree 2. Hence the multiplicity of W1 in VC is 2.

Let ω be the hyperplane class on W , andω ˜ its pullback via π, and e be the class of the exceptional divisor E. Then the class of VeC is given by 6˜ω − 2e. The answer we seek is then 5 2 ∼ 2 (6˜ω − 2e) . Now let l and p = l be the classes of a line and a point on W1 = P . As W1 is ∼ 5 2∗ the image of Veronese embedding in W = P (indeed, the line [a0 : a1 : a2] ∈ P corresponds 2 2 2 to the conic [a0 : a1 : a2 : 2a0a1 : 2a0a2 : 2a1a2]), we have

ω| = 2l W1 As the Chern class of a projective space of dimension n is given by (1 + ω)n+1, we have 6 c(W ) = (1 + ω) . Its restriction to W1 is hence

c TW | = 1 + 6(2l) + 15(2l)2 = 1 + 12l + 60l2 W1

3 2 On the other hand, we have c(W1) = (1+l) = 1+3l+3l . Consider the smooth decomposition TW | = TW ⊕ N, where N is the normal bundle of W in W . This gives c TW | = W1 1 1 W1 c(TW1)c(N). Performing a long division, it is easily checked that

c(N) = 1 + 9l + 30l2. ∼ The exceptional divisor is given as the projectivised normal bundle E = PN. Let ζ be the ﬁrst Chern class of the tautological line bundle over E. We quote the following theorem [GH, p. 606]:

Proposition 6.1. For X any compact oriented smooth manifold, E → X any complex vector ∗ ∗ bundle of rank r, the cohomology ring H (P(E)) is generated as an H (X) algebra by the ﬁrst Chern class ζ of the tautological line bundle, with the single relation

r r−1 r−1 r ζ − c1(E)ζ + ··· + (−1) cr−1(E)ζ + (−1) cr(E) = 0.

Applied to N → W1 we have

ζ3 − 9˜lζ2 + 30˜l2ζ = 0

2 where ˜l is the pullback of l to H (E, Z). As the tautological line bundle restricts to the universal line bundle on each ﬁber Ep of E → W1. Hence

ζ2 · ˜l2 = 1.

Now we have 0 = ˜l · ζ3 − 9˜l2 · ζ2 + 30˜l3ζ = ˜l · ζ3 − 9 as ˜l3 = 0. And therefore

˜l · ζ3 = 9.

27 Finally ζ4 = 9˜l · ζ3 − 30˜l2 · ζ2 = 81 − 30 = 51. ˜ Now asω ˜|E = 2l, and e|E = ζ, we can do the following calculations as follows:

ω˜5 = ω5 = 1, 4 4 ˜ 4 ω˜ · e = (ω ˜|E) = (2l) = 0, ω˜3 · e2 = (2˜l)3 · ζ = 0, ω˜2 · e3 = (2˜l)2 · ζ2 = 4, ω˜ · e4 = 2l · ζ3 = 18, e5 = ζ4 = 51, and hence (6˜ω − 2e)5 = 32 (3˜ω)5 − 5(3˜ω)4 · e + 10(3˜ω)3 · e2 − 10(3˜ω)2 · e3 + 5 · 3˜ω · e4 − e5 = 32(243 − 360 + 270 − 51) = 3264.

Note that in order for the calculation to be correct, we need that for ﬁve conics in general position, VeCi ’s satisfy some transversality conditions. We omit the rigorous proof, but instead quote [FM]:

Theorem 6.2. If no two of the ﬁve conics C1,...,C5 are tangent, no three are concurrent and no three are tangent to any line, then the intersection VC1 ∩· · ·∩VC5 ∩(W −W1) consists of isolated points with X mC (VC1 ··· VC5 ) = 3264,

C∈VC1 ∩···∩VC5 ∩(W −W1) and the intersection multiplicity of VCi at a point C ∈ W − W1 is given by   Y X mC (VC1 ··· VC5 ) =  (mp(C · Ci) − 1) . i p∈C∩Ci

Note that this theorem not only gives us a count, but also tells us about the cases when some of the intersection points have multiplicities. This is important for proving that the Galois group in this case is the full symmetric group, in the spirit of Lemma 2.4, as in the cases of ﬂexes and bitangents of plane curves.

Before going into the Galois group of this problem, we ﬁrst prove some facts about VC . 0 00 We also denote by VC the subvariety of conics bitangent to C, and VC the subvariety of conics having a point of contact of order ≥ 3 with C. We have

0 00 Proposition 6.3. Given a smooth conic C, VC ,VC ,VC are irreducible. For D not tangent 00 to C, VC ∩ VD and VC ∩ VD are irreducible.

28 0 00 Proof. Consider the incidence correspondences IC ⊂ W × C,IC ⊂ W × C × C,IC ⊂ W × C deﬁned by

IC := {(E, p): mp(E · C) ≥ 2}, 0 IC := {(E, p, q): mp(E · C) ≥ 2, mq(E · C) ≥ 2}, 00 IC := {(E, p): mp(E · C) ≥ 3}.

0 0 00 00 The projection η : IC → C, η : IC → C ×C, η : IC → C are surjective, with ﬁbers projective 3 1 2 0 00 spaces P , P , P respectively. Therefore IC ,IC ,IC are irreducible, and their projections onto 0 W are irreducible. Note moreover that as π : IC → VC is one-to-one outside of VC (where it 0 is 2-to-1), VC is locally irreducible away from VC . For the second part of the assertion, similarly consider

IC,D := {(E, p, q): mp(E · C) ≥ 2, mq(E · D) ≥ 2} ⊂ W × C × D, 0 IC,D := {(E, p, q): mp(E · C) ≥ 2, mq(E · D) ≥ 3} ⊂ W × C × D.

1 The projection IC,D → C ×D is surjective, whose fibers are P , except at 4 intersection points of C and D and 4 points (p, q) with TpC = TqD (there are four as being tangent to a quadric 2 2 defines a conic in the space of lines in P ), where the fibers are P . Hence IC,D is irreducible 2 away from eight P ’s. The same is true for VC ∩ VD. But VC ∩ VD is of pure dimension 3, 5 being the intersection of two degree 6 hypersurface in P , so VC ∩VD is irreducible. Similarly, 0 the projection IC,D → C ×D is surjective, and one-to-one except over the eight points above, 1 00 where the fibers are P . Hence VC ∩ VD is irreducible outside of a collection of 8 lines and hence irreducible.

6.2 Galois group of the problem of ﬁve conics

5 The setup is similar to all the cases we have discussed. Let X = W and Y ⊂ X × (W − W1) be the incidence correspondence given by

Y := {(C1,...,C5,C): C is tangent to Ci∀i}.

The projection Y → X is generically finite of degree 3264 and we want to study the Ga- lois/monodromy group M associated to this map. Note that we cannot exclude rank 2 conics from the second factor, as they could arise: the two lines are tangent to two disjoint pairs of conics and the fifth conic passes through the double point. This makes the arguments slightly more complicated, as fibers of Y → W over rank 2 conics will be reducible. Nevertheless, we have

Proposition 6.4. M is the full symmetric group Σ3264.

Proof. Again we proceed by proving that M is doubly transitive and contains a simple transposition. First, M is transitive. Indeed, consider the projection η : Y → W − W1. Over a 5 smooth conic C, the ﬁbers are (VC ) and hence irreducible (and of dimension 4 · 5 = 20). Over a rank 2 conic the ﬁbers are reducible (the components are conics tangent to each of

29 the lines and conics passing through the double point), but they are still of dimension 20. Hence Y has only one connected component of dimension 25 = dim X. Thus a generic ﬁber of π is completely contained in this irreducible component. Proceed as in the cases of ﬂexes and bitangents, we conclude that M is transitive.

Second, M is doubly transitive, or equivalently, the stabilizer of a point in a generic fiber 5 acts transitively on the remaining. Suppose C is one point in the fiber. Let XC = (VC ) ⊂ X be the set of quintuples of conics tangent to X, and set YC ⊂ XC × (W − W1) be defined by

0 0 0 YC := {(C1,...,C5,C ): C is tangent to Ci∀i, C 6= C}.

The projection of YC to W − W1 maps onto an affine open of W − W1. Moreover, the 0 0 5 fibers over C where C is smooth and transverse to C is (VC0 ∩ VC ) (which is of dimension 15), which is irreducible by Prop. 6.3. Otherwise the fibers may be reducible, but they are all of dimension 15. In particular YC only has one irreducible component of dimension 20 = dim XC . As above, this implies that M is doubly transitive.

Next, we want to exhibit a simple transposition in M. For this we show that there exists a ﬁber with exactly one double point. Let C be a smooth conic, and C1,...,C4 be generic conics tangent to C and C5 have a contact point of order 3 with C. That is, let (C1,...,C5) 4 00 be a generic point of JC := (VC ) × VC . Then we have mC (VC1 ··· VC5 ) = 2 by Theorem 6.2. To show that a generically chosen conﬁguration like this have all the other points simple, we set KC ⊂ JC × (W − W1) as

0 0 0 KC := {(C1,...,C5,C ): C is tangent to Ci∀i, C 6= C}, and its closed subvariety

0 0 0 KC := {(C1,...,C5,C ) ∈ KC : mC (VC1 ··· VC5 ) ≥ 2}. 0 We want to show that KC cannot dominate JC . Now the projection of KC to W − W1 0 4 00 surjects onto an affine open set, and its fiber over a point C is (VC ∩ VC0 ) × (VC ∩ VC0 ). When C0 is smooth and intersect transversely with C, this is irreducible by Prop. 6.3 (and 0 00 of dimension 14). When C does not lie in VC the fiber is reducible, but still of dimension 14. 00 When C ∈ VC the fibers are of dimension 15. We see that there is only one 19 dimensional 0 irreducible component of KC dominating JC . But this cannot be contained in KC , as for C0 smooth and intersect transversely with C, a generic fiber over C0 will be conics simply tangent to C0.

Finally, we need some local irreducibility to apply Lemma 2.4. As we have discussed, we have irreducibility away from bitangent loci, and this is an open condition. Therefore M is indeed the full symmetric group.

7 Resolvent degree

We base this section on a recent preprint [FW], which partly motivated the author to work on this particular topic for the minor thesis.

30 The classical Galois theory provides a dichotomy of solvability of algebraic problems: they can either solved by radicals or not. Resolvent degree measures complexity of solving an algebraic problem in a related but slightly diﬀerent manner: how large is the space of problems one have to solve using operations other than basic algebraic operations? For solving by radicals, each time we take a radical, we are essentially solving xd + a = 0 for a ∈ k, where k is the ﬁeld we are working in at that step. This is a one parameter of equations we need to solve. For solving quintics, one can show that the problem can be reduced via radicals to solving x5 + ax + 1 = 0 which is again a one parameter of equations. Indeed, if we further allow an operation called Bring radical that takes in the parameter and outputs one solution of x5 + ax + 1 = 0, quintics are solvable.

We will attempt to formulate this in mathematical terms. In the context of solving degree d polynomials with complex coefficients, we work in K = C(a0, . . . , ad−1), where ai’s are the d−1 d coefficients of a monic complex polynomial. Let L = K[λ]/(a0 + a1λ + ··· + ad−1λ + λ ), which is a degree d extension of K. Finding a general formula for solving polynomials is tantamount to finding a sequence of finite field extensions

K = L0 ⊂ L1 ⊂ · · · ⊂ Lr

so that Lr contains L as a subﬁeld, and for all i = 0, . . . , r − 1,

dpi Li+1 = Li ⊗C(Xi) C(Xi, Xi) where Xi ∈ Li, and di is an integer ≥ 2. Note that in general Lr is larger than√ L, as it also d contains the conjugates of λ in an algebraic closure of K. Note that C(Xi, i Xi) is a finite extension over C(Xi). To solve quintics, we allow for each i the extension as given above, or 5 Li+1 = Li ⊗C(Xi) C(Xi, µ)/(µ + Xiµ + 1). 5 Again, note that C(Xi, µ)/(µ + Xiµ + 1) is of finite degree over C(Xi). The fact that we only need to solve one parameter of equations is reflected by the fact that C(Xi) is of transcendental degree 1 over C. Generalizing this formulation, we have Definition 7.1 (Algebraic version). Let K,L be fields over a base field k. Suppose K ⊂ L is a finite extension. The resolvent degree RDk(L/K) of L/K is defined to be the smallest integer d such that there exists a sequence of finite field extensions

K = L0 ⊂ L1 ⊂ · · · ⊂ Lr where Lr contains L as a subfield, and for all i = 0, . . . , r − 1, ˜ Li+1 = Li ⊗Fi Fi where Fi is a subfield of Li and has transcendental degree at most d over k, and F˜i is a finite extension over Fi.

Note that the resolvent degree remains the same if we replace L with its Galois closure L. Indeed, at each step of the sequence required in the deﬁnition of L, if we replace F˜i by its Galois closure, and modify the sequence correspondingly, we then have a sequence for L.

31 One might already notice that if we set Pd be the complete linear system of degree d 1 1 homogeneous polynomials on P , and Id ⊂ Pd × P be the incidence correspondence {(P, λ): P (λ) = 0}, and π : Id → Pd, then the function field on an affine open set U of Pd is precisely −1 K = C(a0, . . . , ad−1), and the function field on the affine open cover π (U) is precisely K(λ). Translating the above definition in a geometric context, we then have Definition 7.1 (Geometric version). Let X,Y be algebraic varieties over a base field k. Suppose π : Y → X be a generically finite dominant rational map. The resolvent degree RDk(Y → X) of the map π : Y → X is defined to be the smallest integer d such that there exists a sequence of dominant generically finite dominant rational maps

Yr → Yr−1 → · · · → Y0 = X

where Yr → X factor through Y → X, and for all i = 0, . . . , r − 1, we have the pullback diagram

Yi+1 Z˜i

Yi Zi

where Z˜i → Zi is a generically ﬁnite dominant rational map of varieties of dimension at most d over k.

It is not hard to show the equivalence of these two definitions: Proposition 7.1. If π : Y → X is a generically finite dominant rational map of irreducible algebraic varieties over the field k, then

RDk(Y → X) = RDk(k(Y )/k(X)).

This equivalence is in the same spirit as the equality of monodromy and Galois groups. Finally, in the context of the enumerative problems we have discussed above, we have

1. For solvable problems, the resolvent degree is 1. In particular, the following problems have resolvent degree 1: locating the 9 flexes on a generic plane cubic; locating the remaining 25 bitangents of a plane quartic given 3 bitangents with the 6 points of tangency on a conic; locating the remaining 24 bitangents of a plane quartic given 4 bitangents with the 8 points of tangency not on a conic; finding the remaining 25 lines on a generic cubic surface given 2 incident lines; finding the remaining 24 lines on a generic cubic surface given 3 skew lines.

2. As discussed above, the problem of solving a quintic has resolvent degree 1. In our discussion of 27 lines on a cubic surface, if we are given a single line, then we have noted that we need to solve a quintic. Hence the problem of ﬁnding the remaining 26 lines on a generic cubic surface given one has resolvent degree 1.

32 3. It is shown in [FW] that the problem of locating 27 lines without any given has resolvent degree ≤ 3. The problem of ﬁnding the remaining 27 bitangents on a generic plane quartic given one is equivalent to this problem, and hence also has resolvent degree ≤ 3.

4. We know very little about the resolvent degree of the other problems, most of which have full symmetric groups as their monodromy group.

References

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[Ful] W. Fulton. Algebraic Curves: An Introduction to Algebraic Geometry. Addison- Wesley, 1989.

[FM] W. Fulton and R. MacPherson. Deﬁning algebraic intersection. In Algebraic Geometry, pages 1–30. Springer-Verlag, 1977. Proceedings of the Tromsø Symposium.

[GH] P. Griﬃths and J. Harris. Principles of Algebraic Geometry. Wiley & Sons, 1994.

[Har1] J. Harris. Galois group of enumerative problems. Duke Mathematical Journal 46(1979), 685–724.

[Har2] J. Harris. Algebraic Geometry: A First Course. Number 133 in Graduate Texts in Mathematics. Springer-Verlag, 1992.

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