sign in sign up
<<
Home , Cubic surface

Vol. 46, No. 4 DUKE MATHEMATICAL JOURNAL (C) December 1979

GALOIS GROUPS OF ENUMERATIVE PROBLEMS JOE HARRIS

Contents. Oo Introduction 685 I. Galois and groups 688 II. Flexes and bitangents of plane curves 690 1. Pliicker formulas 690 2. Flexes of a plane cubic 693 3. Flexes of a plane curve of degree d > 4 697 4. Bitangents to a smooth plane quartic 700 5. Bitangents of a plane curve, degree d > 5 706 III. Lines on hypersurfaces 707 1. The Fano scheme 707 2. Lines on a hypersurface in P", n > 4 712 3. Lines on a cubic surface 715 4. Bitangents of a plane quartic II 719 IV. The problem of five conics 721 References 724 0. Introduction. This paper is concerned with the solvability of certain enumerative problems in . To illustrate the questions raised, consider one of the problems dealt with belowmthat of locating the flexes of a plane curve. We let C c P be a complex plane curve of degree d, given as the locus of the homogeneous F(Xo,X,Xa)= ZagjX-i-Jx[xJ2 (or in j. euclidean coordinates x Xi/Xo, as the locus of f(x,x2) ,aijxlxa), we will take coefficients ai. to be general complex numbers. At a generic point of C, then, the tangent lne Tp(C) intersects C with multiplicity mp(C. I)= 2; we say that p is a flex point of C if mp(l. C)> 3. An elementary count of parameters leads us to expect that C will have a finite number of flex points p; and accordingly we may ask two questions: first, how many flexes does C possess? and second, can we find them?--that is, is it possible to give a formula for coordinates xi(p) of the flex points p of C in terms of the coefficients There are a number of ways of answering the first of these questions; historically the first was to note that the flexes of C comprise the locus F(X) H(X)= O, where H(X) is the Hessian

det H(X) OXiOXj Received February 13, 1979. Research partially supported by NSF Grant MCS 77-01964.

685 686 JOE HARRIS since H(X) has degree 3(d-2), we expect 3d(d-2) flexes. To set up the second problem, we note that since the coordinates x(p,) are roots of the ith resultant Ri(f, h), (xi)a polynomial of degree 3d(d- 2) in x whose coefficients are elements of the field G(aj) of rational functions in the ao.the field C,(ai,x(p,O) obtained from G(a0.) by adjoining the xi(p,0 is a finite algebraic extension of C(aj); and there will exist a formula for xi(p,,) in terms of the a/j exactly when the Galois G of this extension is solvable. We may thus replace our second question with the broader, What is the Galois group G of the extension C,(ai,xi(pa))/C,(aij) 9. We note that determining G will also answer some other questions in the same vein. For example, knowing G and the action of G on the roots of Ri, we may ascertain whether the subgroup G’ GalG(ao.,xi(p))/G(aij,xi(Pl),..., xi(pk)) of G fixing k roots is solvable-- that is whether, given the coordinates of k flexes of C, we may solve explicitly for the remaining 3d(d- 2)- k. In similar fashion, there is associated to most enumerative problems a Galois group, whose structure reflects the difficulty of obtaining explicit solutions to the problem. Historically, these Galois groups were first studied systematically by Camille Jordan in his book Traitb des Substitutions (1870) where he discusses, among others, the Galois groups of the flexes of a plane cubic, the bitangents to a plane quartic curve, and the lines on a cubic surface in [a3. Jordan’s work, however--as well as later refinements by Dickson [4] and Weber [13]--was in a direction opposite to that suggested above: the equations defining the coordinates of the flexes of a plane cubic, bitangents to a plane quartic and lines in a cubic surface had already been extensively studied by Cayley, Clebsch, Hesse, Steiner and others ([2], [3], [12]) some time before the advent of Galois theory. What Jordan did was to show how this earlier work would be expressed and interpreted in terms of Galois theory. Thus, for example, in considering the flexes of a plane cubic, Jordan takes Hesse’s solution of the problem and analyzes it to determine the group. In the present paper we take the opposite tack, and determine the solvability of four enumerative problems--the first three of which are generalizations of the ones mentioned above--by computing their Galois groups. There are, in general, two steps in solving each of these problems: First, we look to see what, if any, intrinsic structure is carried by the set of solutions to the problem (e.g., among the nine flexes of a plane cubic, there are 12 sets of three collinear points). This structure must be preserved under the action of the Galois group G (e.g., if Pl, P2, P3 are three collinear flexes of a plane cubic F(X)= 0, the coordinates of P3 may be rationally expressed in terms of those of P and P2 by restricting F to the line PiP2 and factoring out the roots corresponding to _Pl and P2; and so must be preserved under any 0 G fixing xi(Pl) and xg(p2)). This then gives us an "upper bound" on the group. The second half of the problem--showing that G must be at least as large as some subgroup H of the on the set of solutionsseems harder. Certainly it is not practical to analyze the equation in generalfor d large one GALOIS GROUPS OF ENUMERATIVE PROBLEMS 687 could hardly hope, for example, to determine whether the group G of the flexes of a plane d- ic was contained in the alternating group by writing down the discriminant of the degree 3d(d- 2) polynomial R and determining whether it was a square in G(aj). Our technique, rather, is to equate the group G with the monodromy group M of a generically finite map of algebraic varieties, as described in chapter of this paper. This allows us to exhibit elements of the Galois/monodromy group simply by drawing arcs in the parameter space for the problem. One situation in which this approach is particularly effective is in showing that G is the full symmetric group. Here, after showing that G is twice transitive, we want to exhibit a simple transposition in G; and by the lemma of section 11.3 below, this amounts--modulo some secondary argumentsmto locating a special case of the problem which admits a "solution of multiplicity tWO." The four problems considered here are the flexes and bitangents of a general plane curve of any degree d; the lines on a general hypersurface 1/" c an of degree 2n 3, for any n; and the conics tangent to each of five given conics in the plane. For each of the first three, we consider separately the case studied by Jordan, giving a new proofmusing monodromy argumentsmof his theorem, and say what this means in terms of solutions to the problem. We then consider the general case. The arguments in all four cases follow pretty much the same lines; as indicated above, the essential ingredient is the determination of the multiplicity of a solution. For the flexes and bitangents of plane curves, this is implicit in the general form of the classical Pliicker formulas; in the case of lines in hypersurfaces it is a straightforward computation made in section III.1; and for the conics tangent to five conics the answer was recently supplied by Fulton and MacPherson in [5]. The findings may be stated readily: the Galois groups are (i) For the flexes of a plane d- ic, the affine special linear group ASL2(Z/3) in case d 3; and the full symmetric group in case d > 4; (ii) For the bitangents to a plane d- ic, the orthogonal group O6(Z/2) in case d 4 and the full symmetric group in case d > 5; (iii) For the lines on a hypersurface l/’2n_31a: the odd orthogonal group 06- (Z/2) in case n 3, and the full symmetric group in case n > 4; and (iv) For the conics tangent to five conics, the full symmetric group. One remark, finally, about the significance of the results, and in particular the fact that, in every case not considered classically, the group is shown to be the full symmetric group. On one hand, it is of course a disappointment that no explicit solutions may be given for the problems mentioned. On the other hand, however, the results represent an affirmation of one understanding of the geometry underlying each of these problems, in the following sense: in every case dealt with here, the actual structure on the set of solutions of the enumerative problem as determined by the Galois group of the problems, is readily described in terms of standard algebrao-geometric constructions. In particular, in every case in which current theory had failed to discern any 688 JO HARRIS intrinsic structure on the set of solutionsit is proved herethere is in fact none. I would like to thank Steven Kleiman, who introduced this problem to me, and whose ideas have guided this work throughout.

I. Galois and monodromy groups. Suppose X and Y are irreducible algebraic varieties of the same dimension over the complex numbers G, and r: YX a map of degree d > 0. Let p X be a general point of X and F r:-(p)= {q,..., qa) the fiber of r over X. We may then define two subgroups of the group Xa of F, as follows: 1. Let r* K(X)--) K( Y) be the inclusion of function fields induced by rr. By the primitive element theorem, K(Y) is generated over K(X) by a single f K(Y) satisfying a polynomial of degree d e(f) fa + g fa- + + ga 0 where g,..., ga K(X). Now let A be the field of germs of meromorphic functions around p, A the field of germs of meromorphic functions around q, and the induced by the map r restricted to a small neighborhood of q. Let , :K(X)-A be the inclusion obtained by restriction, K q(K(X)), and . K(r)- the inclusion obtained by restriction to A composed with r; let L itself denote the subfield of A generated by the subfields K O,,(K(Y)). Finally, let i @( gi) f (f). Now, each function element f satisfies the polynomial and since the f are distinctindeed, since f generates K(Y) over K(X), it must have all distinct values at the points (q)we see that these are all the roots of ft. The field L c A is then the normalization of the extension K.(Y)/K.(X) K,,/K, and the Galois group G Gal(L/K) acts on the roots {f) of P; so we have an inclusion GALOIS GROUPS OF ENUMERATIVE PROBLEMS 689

2. If we let U c X be a suitable small Zariski open subset of X, V--- r-I(U) c Y, then the restricted map r" V U will be an unbranched covering space. Now, for any loop ," [0, 114 U in U with base point p and any point q r-(p), there is a unique lifting of 3’ to a loop in 1/with "7 (0) q. We may accordingly define a permutation qr of F by sending each point q to the endpoint of the lifted arc Clearly qr depends only on the homotopy class of "r, and so we have a homomorphism. r( U, p) a. The image of this homomorphism is called the monodromy group of the map r" V- U and denoted M (we will see from the next proposition that in fact M does not depend on the choice of Zariski open U cX as long as r" V r-(U)- U is unramified). We have now the basic PROPOSITION. For Y- X as above, the monodromy group equals the Galois group. Proof. We first show that M c G" let y(t) be any loop in U with base point p, -/’ the lifting of , to V with q(O) q, and o Ea the induced permutation of r- (p), i.e.,

For any germ h L c A of a meromorphic function at p in the field L, analytic continuation of h along the path y(t) is well-defined: this is certainly true for any germ h K,, of meromorphic function on Y at a point q, r-l(p); and any h EL is a polynomial in such germs. Analytic continuation along 7 thus defines an automorphism of the field L fixing K, and sending the function element f into fo,,), so that the permutation o of F is indeed in the Galois group G. Conversely, we claim that any automorphism of the field L over K is obtained by analytic continuation along some arc y in U. To see this, we have to show that the fixed field of L under the subgroup M c G is just K, i.e., that any function element h L fixed under analytic continuation along every loop y in U with base point p is in fact the ge.rm of a meromorphic function on X. We may define a meromorphic function h on U by choosing for every r U an arc 6 from p to r and letting the germ of h at r be the analytic continuation of h along 6; choosing a different arc 6’ from p to r must yield the same germ since the continuation of h along the loop 6- 6’ is again h. Writing h-" O(hl,...,ha) where h is the germ in A -A,, of a meromorphic function h on Y and Q is a polynomial we see that h cannot have essential singularities and so extends to a meromorphic function on X with germ h at p. Q.E.D. 690 JOE HARRIS

Note. In case the varieties Y and X are defined over a subfield k of C, the Galois group Gk of the extension K(Y)/K(X) of function fields over k also acts on the general fiber of r; the group G will be in general a subgroup of G. Finally, note that the monodromy group M of a map r: Y--> X is well-defined even in case Y is reducible. In this case we simply define the Galois group of r to be M, so that we can discuss G without first checking the irreducibility of Y. G is a subgroup of, but not always equal to, the product of the Galois groups G of the irreducible components Y of Y dominating X; the’action of G on the fibers of Y, however, is the same as that of G. II. Flexes and bitangents of plane curves 1. Plicker formula. The first problem we shall consider is the flexes of a general plane curve of degree d. To set up, we denote by W Wd P(a5)- the 2 linear system of all plane curves of degree d. Let I0 c 2, be the locus

I0 ((p,/): p l). In terms of homogeneous coordinates [Xo,X1,X2] on a2 and dual coordinates [Yo, Y1, Y2] on p2,, i0 is simply the locus of the equation Si Yi 0; in particular we see that I0 is a smooth irreducible threefold. Now let I=IaC Wa Io be given by

Id {(C; p,l)’mp(C, l) > 3) i.e., Id is the set of triples (C; p,l) such that p is a flex of C with tangent line l. Let r:IdoWd and l:IdoIo be the projection maps. If d>3, r will be generically finite; our problem is to determine the monodromy group of this map. We first determine the degree of rmi.e., find the number of flexes on a general plane curve of degree d--by deriving the classical Pliicker formulas for plane curves. We will actually prove these formulas in somewhat greater generality than is necessary to solve the simple enumerative problem; we will require the additional generality later in answering the monodromy question. Let C now be a smooth plane curve of degree d and let C* c [a2, be the dual curve of C, that is, the image of C under the map f sending a point p C to the tangent line Tp (C). The degree of C*, given as the number of tangent lines to C passing through a general "point q a2, is readily found: assuming that no line through q is multiply tangent to C, projection of C from q onto a line expresses C as a d-sheeted cover of [1 with a simple branch point for every tangent line to C through q; by Reimann-Hurwitz the number of branch points is b 2g- 2 + 2d GALOIS GROUPS OF ENUMERATIVE PROBLEMS 691 and since the genus g of C is (d- 1)(d- 2) this yields d* deg C* d(d 1). Now, at a point p C such that the tangent line to C has contact of order m with C at p, C may be given parametrically in terms of a suitable local coordinate on C centered around p and homogeneous coordinates on p2 by t--v(t)=[1, t+ "’’,t m+ ]. The map f is then given, in terms of dual coordinates on 2,, by the 2 2 minor determinants of the matrix

v’(t) O, 1 + ,mtm-l +

f(t) [1 + ,m. tm-l + ,(m- 1)t + ]. The point f(p) C* is thus an (m- 1)-fold point of the image f(A) of a neighborhood A of p; in particular the map f is smooth at p if and only if me(C. Tt,(C))= 2. Of course, f will be one-to-one over a point C* unless the line is tangent to C at several points pl,..., Pe; in this case, we note that the e branches of the curve C* at have distinct tangent lines, these being the pencils of lines through Pl, ’Pe" We see then that if a line has e(l) points p,..., Pe of tangency with C, with multiplicities m, m then will be an m(/)-fold point of C*, where m(l) Z(m,, 1); and the proper transform of C* in the blow-up of W at will be smooth. We thus obtain one relation on the multiplicities m(/): by the genus formula, (d- 1)(d- 2) (d* 1)(d* 2) 2 =g= 2 m(l)(m(l)- 1) ] EP2. 2 (a(d-1)- 1)(d(a-1)- 2) m(l)(m(l)- l) 692 JOE HARRIS

m(l)(m(l)- 1) (d(d-1)- 1)(d(d- 1)- 2) (d-1)(d-2) Z. p2, 2 2 2 (1) A second relation comes from the fact that the tangent line to an arc f(A) c C* at a point p is the line in p2. dual to p, i.e., the dual curve of C* is C. In particular, we see that through a general point q p2, there pass d tangent lines to C*. Assuming that q does not lie on any of the tangent lines to C* at singular points, or on any multiply tangent line to C*, the branch locus of the composition of f with projection from q onto a line will be the d tangent lines to C* through q, plus a point of multiplicity Y(m- a)= m(l)- e(l) at a line l with contacts of order m as above. Applying Riemann-Hurwitz, then, we have d+ (m(l) e(l)) 2g- 2 + 2d* Ep2, d(d- 3) + 2d(d- 1)

(re(l) e(l))= 3d(d- 2). (2) p2, The relations (1) and (2) are the classical Pliicker formula for plane curves. To apply the Pliicker formulas to a general plane curve, we simply note that since divisors of the form X moap, have Y.(m 1) among all divisors on a line c p2 the subvariety of Wd of curves C for which m(l)> m has codimension m in Wd, i.e., the dimension of the incidence correspondence p2, Jm C Wd defined by Jm ((C,l)’mc(l) > m) has dimension dim Jm dim Wd + 2 m. We see accordingly that for m > 3, Jm cannot surject onto Wd so that for a general curve C in p2, m(l) < 2 for all lines l. The only multiply tangent lines to C are then either (1) bitangent lines, for which m(l)= 2 and e(l)= 2; or (2) flexes, for which m(l)= 2 and e(l)= 1. Denoting the number of bitangents and flexes by b and f respectively, we have by the Pliicker formulas f= 3d(d- 2) GALOIS GROUPS OF ENUMERATIVE PROBLEMS 693 and (d(d- 1)- 1)(d(d- 1)- 2) (d- 2)(d- 2) b 2 2 3d(d- 2).

2. Flexes of a plane cubic. We will consider first the Galois group of the nine flexes on a plane cubic curve C c p2. To see what this group is, recall that such a curve C may be realized as the quotient G/A of the plane by a discrete lattice A Z2c (3, embedded in p2 by the map associated to the complete linear system of a divisor H r + r2 + r3 -t(Zl)+ t(z2)+ t(z3) of degree 3. Inasmuch as C is cubic and projectively normal, a point Po t(Zo) will be a flex of C if and only if the divisor 3po is linearly equivalent to H; by Abel’s theorem this will be the case if and only if

3zo =-- Z -]" 22 "l- z3(A). It follows that if Po t(Zo) is a flex, then a point p t(z) on C will be a flex if and only if Z-ZoA so that for any flex P0 of C the set F of flexes on C has naturally the structure of a 2-dimensional vector space 1/2A V-- --X--Z Z A 3 3 over Z/3 with P0 the origin; thus F has the structure of a 2-dimensional over Z/3. Note that for any three points p, q, r of F, p + q + r will be zero or not independently of which point of F is chosen as the origin, and conversely specifying when three points of F add up to 0 determines the Z/3-affine space structure" for any two points q and r in F, -q- r will always be the point s such that s + q + r 0; and for any q F if we take p as the origin q will be the point r such that p + q + r -0. The affine space structure on F is thus geometrically visible, inasmuch as three flexes qi t(wi) will add up to zero---i.e., w + wE + w3 will be congruent to zl + zE + z3 mod A--if and only if q, qE and q3 are collinear. Indeed, F has more structure: under the identification of the lattice A with the integral first homology H(C,Z), intersection of cycles and the natural orientation on C give a skew-symmetric unimodular form A x A-Z; the form 9Q. 1/2 A 1/2 A- Z then descends to give a pairing v v z/3. so that F has the structure of an affine space with volume element. 694 JOE HARRIS

Note that all this structure must be preserved under the monodromy action on F: for example, if Pl, P2 and P3 are flexes of a smooth cubic C l, 7(0 C(t) a loop in the space U c W of smooth cubics with base point C, and qi the image of Pi under the induced monodromy action on F, then we must have q +q2+q3=0 if i(t)=(C(t);_pi(t),li(t)) is a lifting of ), to I3 with i(0)=(C; pi,li) and i(1)=(C;qi,mi), we may define a lifting a of / with a (0) (C; P3,/3) by setting a(t) (C(t); p(t),l(t)) where p(t) is the third point of intersection of the line p(t), p2(t) with the cubic C(t) and l(t)= Tp(oC(t). By uniqueness of lifting, then, qa a so that qa =p(1) and q l, q2 and q3 are collinear. The monodromy group M of F is thus a priori a subgroup of the affine special linear group ASLz(Z/3) (that is, the group generated by the special linear transformations and translations on a 2-dimensional vector space over Z/3). We will prove now that in fact, PROPOSITION. M ASL2(Z/3). Proof. We will show first that for any flex p of a general cubic C, under the identification of the set F of flexes on C with V, the stabilizer of P0 in M is all of SLE(Z/3). To see this, write C as the G/A, where A (mw + no2}n,m z with Po corresponding to the origin in C (indeed, with a suitable choice of homogeneous coordinates in p2 we may take z [1, ], where 62(z) is the Weierstrass function on G periodic with respect to A; we will then have P0 [1, 0, 0] with tangent line [0, 1,0]). The flexes of C will then be the images of the third lattice points

1/2A tl A ( mT+n-- /A. Now let Y=( )be any element of SL2(Z/3), and let A=( ) be an element of SL2(Z) reducing to A mod 3. Choose an arc A (t)= ( ) in SL2(FI) with A (0) I, A (1) A, and set (01(t), w2(t))= A ( t)( 2 and GALOIS GROUPS OF ENUMERATIVE PROBLEMS 695

Let 62 be the Weierstrass function on G periodic with respect to the lattice A t, and let C c p2 be the image of G/A under the map ,,.z--,[

We have then Co C C, with Po a flex of C with tangent lo for all t. The arc is thus a loop in U with base point C; and for any flex point q=(v), v mobil3 + mo2/3, on C with flex line l we may lift 3’ to an arc q in 13 with "7(0) (C; q, l) by setting q(t)=(C,;q,,l,) where ,,,,(t) qt + n tt m---- 3 ) and Tq,(Ct)._.The monodromy action on F effected by 3’ is thus the linear transformation A. It follows easily now that M ASL2(7/3): since the stabilizer in ASL2(Z/3) of any point is SL2(Z/3), it follows that M contains any element of ASL2(Z/3) fixing a point. Moreover, we see that M is transitive, either by the argument in the following section in case d > 4 or by noting that the stabilizer in M of any point p F is transitive on F -p. If q is any element of ASL2(Z/3 ), then, and p is in F, we can find 3’ M with @(p)= q(p); the automorphism @-lo # then fixes p, hence lies in M, so # M. Q.E.D. Now, the group ASL2(Z/3) is solvable; and this in turn yields an explicit formula for the flexes of a general cubic curve. First, to see that ASL2(Z/3) is solvable, we note that it acts on the associated P( V) P17/3, (realized as the set of pencils of parallel lines on V), the translations fixing P(V). Since Pit/3 consists of four points, we have a map ASL2(Z/3) ]4, with kernel G the subgroup of ASL2(Z/3) generated by translations and the involution -I. (Note that the image of ql is the alternating group A4). The group G is clearly solvable; one way to give a composition series is to fix one point 1 of P(V), corresponding to three parallel lines q in V, and let G act on (); we have then a surjection IP2" G --) Y3 with kernel G 2 Z/3. - 696 JOE HARRIS

To apply this to the problem of determining the flexes of the cubic C, given on the locus F(X ) Eai,XXX2,

2 we note that a line in the plane V- (7/3) consists of three points D1,D2,t)3 adding up to zero--that is, three collinear flexes. A point in P(V), then, corresponds to three lines (L) in the plane, each passing through three flexes of C and together containing all nine flexes. Now, there is a pencil of cubics in the plane passing through the flexes of C, of which the cubics Cg= EL are elements; indeed, they are exactly the singular elements of the pencil. To find the equations of the C i, let H(X) Y. bijkX(1X2’ ’ be the Hessian of F(X); the coefficients b/j, of H are readily given as rational functions in the aij, as are likewise the coefficients c of the discriminant A(h) disc(F(X) + H(X))

AQ,) will be of degree 12, with four triple roots o l, so that the coefficients of the reduced polynomial 4

i=1 are expressible as rational functions of the c (e.g., d3 11/3). Solving A, we find equations for o and hence for the cubic Gi(X ) F(X) + o)iH(X ) defining the curves C i. Having determined GI(X ), we need only solve the cubic resultant of (OGI(X)/OXj) to find the singular points L2.LJ of C 1, and hence the lines (LI); similarly we can find the lines (L2) of C 2 and hence the nine flexes (L1N L). There is an amusing historical sidelight to this discussion. The problem of the Galois group G of the nine flexes to a cubic was first studied by Jordan [8] in 1870, who asserted simply that G was a subgroup of the affine general linear group AGLE(Z/3) and hence solvable. The problem was studied further by Dickson and Weber, who stated that G was the full affine general linear group AGLE(Z/3), not the special group as stated above. The reason for the disparity is this: Weber and Dickson were taking as their base field the rational numbers O. Now, if C is a real cubic curve, then C will have exactly 3 real flexes; of the GALOIS GROUPS OF ENUMERATIVE PROBLEMS 697

12 lines, exactly four will be real (one of which, of course, contains the three real flexes of C; the other three each contain one real flex and two complex conjugate flexes); and of the four cubics consisting of three of these lines, two will be real. The whole is as shown in Fig. 1, with the points (P0,i) real and .Pli complex conjugate to P2i, the lines (Poi)i and (Pji}j real and the lines (P0i, Pl, i+I’ P2i+2) and (P0i’ Pl, i+2’ P2, i+l) complex conjugate; and the two cubics and 2:,{ ff )i real, the remaining two complex conjugate. The upshot of all this is that complex conjugation acts as a linear transformation of determinant -1 on the vector space of flexes, with a real flex chosen as origin. This is reflected in the fact that the discriminant of the quartic polynomial A in , is minus a square in the function field q(aijk) i.e., over Q the Galois group maps via ql to the full symmetric group 2]4.

FIGURE 1.

The flexes of a cubic is the only case among those considered in this paper in which this disparity between the Galois group over G and over F1 occurs; in all other examples the action of complex conjugation on the complex solutions of an enumerative problem defined over FI is already in the monodromy group. 3. Flexes of a plane curve of degree d > 4. We turn now to the problem of determining the monodromy group of the map r’Ia Wa in case d > 4. The 698 JOE HARRIS answer in this case is that The monodromy M of rr, for d > 4, is the full symmetric group E3d(d_2) The argument for this assertion is in two steps: we show first that (a) M is twice transitive; and then that (b) M contains a simple transposition. It will then follow that M contains all simple transpositions, and hence that M 3d(d-2) The argument for part (a) is straightforward. To begin with, we note that since the fibers of the projection map / of I c Wd Io on the second factor are simply linear spaces of dimensiom dim Wd -3 and I0 is irreducible, I is irreducible. It follows then that the Zariski open V rr-I(U)c I is connected, so that for C a general plane curve of degree d and (p,l), (p’,l’) any two flexes of C, we can draw an arc on V with (0)= (C; p,l) and (1)= (C; p’; l’); the monodromy action associated to the arc 2, r will then carry (p, l) to (p’, l’). Similarly, to see that M is twice transitive, let C be. a general d- ic, (Po, 10) a flex of C, and let

C gl/" J/V’ "r/’(l-l(flO,/0)) (C" mpo(lo.C ) 3} d I’= ((C; po,l):mp(l. C) > 3,p vp0, l v 10) c W’X I. r/ then maps I’ onto the Zariski open (p v P0, va lo)c Io, with fibers linear spaces of codimension 3 in W’; so as before I’ is irreducible and hence any Zariski open subset is connected. Since C, being general, does not have any other flexes with tangent line o (or, for that matter, any other flexes at Po), all points of the fiber F rr-1(C) except (C; p, l) lie in I’. If (p’, l’), (p", l") are any two other flexes of C, then, we can find an arc q in I’ connecting them; and the monodromy associated to the arc 3’ rr. q in U will send (p,l) to (p,l), (p’,l’) to (p",l"). The stabilizer in M of a point (C; p,l) E rr-(C) thus acts transitively on the remaining points; and so M is doubly transitive. The procedure for showing that M contains a simple transposition is expressed in the following LEMMA. Let rr" Y---> X be a holomorphic map of degree n. If there exists a point p E X such that the fiber of Y over p consists of exactly n distinct points--i.e., n- 2 simple points q1,..., qn-2 and one double point qn_ 1--and if Y is locally irreducible at q_, then the monodromy group M of rr contains a simple transposition. Proof. We may take A C X a small disc around p, so that the points q, q.-2 have disjoint neighborhoods A mapping isomorphically via rr to A, and such that GALOIS GROUPS OF ENUMERATIVE PROBLEMS 699

the connected component An_ of r-lA containing qn-1 is disjoint from A An_ 2 and irreducible. Let U c X be a Zariski open set such that r" V=rr-l(U)o U is unramified, and let /6AU U; we may write r-l(/) =1+’’" +n with iAi for i= 1,...,n-2 and n-l,n A A f) An-l" Since n_ is irreducible V is connected, so that we may draw an arc in An_ f)V with (0)= n-1 and (1)= n; the monodromy action associated to the loop 7 r in U will then fix l, n-2 and interchange q’-l and n. Q.E.D. Returning to the monodromy on the flexes of a plane curve, to apply the above we must locate a plane curve C of degree d with exactly 3d(d-2)- flexes. The general Pliicker formulas tell us how to do this: if C is a plane d- ic with one simple hyperflex (Po,/o)-that is, a line 0 having a point P0 of contact of order 4 with C and meeting C transversely elsewhere--and if C has no other 700 JOE HARRIS lines l for which m(1) e(l) > 2, then by formula (2) above, C will have exactly 3d(3- 2)- 2 flexes other than 0. We have to show then, that such a curve C exists, and that I is irreducible at each point of r--I(C). To see this, we choose (Po, lo) lo and let W" ( C’mpo(lo. C) > 4) c W be the linear system of d ics with a hyperflex at (Po, 10). The general C W" is smooth at P0, and since for d > 4 W" has no base points away from Po, the general C Wf is smooth with 1o a simple hyperflex of C. To see that C has no other exceptional behavior, we consider the incidence correspondence I" cA W" I0 defined by I"= {(C’p,l)’mp(C. l) > 3,pvapo, lva/o}.

We note that since the fibers of /:I" Io are all linear spaces of codimension 3 in W", I" is irreducible of dimension dim/" dim W"; and all the points of the fiber of I over C except (Po, lo) lie in I". Now let K" c I be given by K"= ((C; p, 1):mc(l ) -ec(l ) > 2).

K" is a closed subvariety of I", and from the irreducibility of I" it follows that either or (ii) dim K" < dim I" dim W", i.e., either every flex line va o of every C W" has mc(1) -ec(l ) > 2, or else for general C W" we have mc(1 ) ec(l ) < fl =/= 1o. But now for any l 4:1o and p l0, the system of d- ics C with mpo(lo C) > 4 and mp,(l. C) > 3 cuts out the complete series [Ip3,(d)[ on l, so that l will be a simple flex line of a general such C, and possibility (i) is eliminated. By the Pliicker formulas, then, for d >.> 4 the gene.ric C W" has exactly 3d(d- 2)- distinct flexes. Since the fibers of 7:1 I0 are all linear spaces, Ia is smooth and hence irreducible at every point; so that by our lemma the monodromy group M of the flexes on a general plane curve of degree d > 4 is the full symmetric group. 4. Bitangents to a smooth plane quartic A note about quadratic forms over 7/2. Over a field k of any other than 2, there is a one-to-one correspondence between symmetric bilinear forms ,: V V k on a vector space V over k, and homogeneous quadratic polynomials q:V- k, obtained by q(v) X(v,v) GALOIS GROUPS OF ENUMERATIVE PROBLEMS 701 and 2X(v, w)= q(v + w)- q(v)- q(w). If k has characteristic 2, however, this correspondence breaks down: in the first equation, we note that q( + w) a( + w, + w) a(v, v) + X(w, w) q(v) + q(w), i.e., any quadratic polynomial q obtained by restricting a symmetric bilinear form to the diagonal is in fact the square of a linear form; likewise, the second equation above makes no sense unless q(v + w) q(v) q(w) = 0 for all v and w. We make, accordingly, the following definitions: first, since there is no a priori distinction between symmetric and skew-symmetric bilinear forms on a vector space in characteristic 2, we will say that (v, w) is strictly skew-symmetric if (v,v)= 0 /v V. Next, if (v,w) is any strictly skew-symmetric bilinear form on V, we define a quadratic refinement q" V k of to be a function q satisfying w)= q(v + w)- q(v)- q(w) for all v and w in V; a quadratic form on V will be a bilinear form with quadratic refinement. We note from [11] (although we shall not have logical need of this fact) that there is, up to isomorphism, only one non-degenerate strictly skew-symmetric bilinear form , on a vector space V of dimension 2k over Z/2 (this may be represented by the form k y)= + i’-I and there are, up to isomorphism, exactly two quadratic refinements of 2 (represented by the functions

q+ (x) ] XiXi+ k i=1 and k (X)-- XiXi+k "q- X "q- 2k)’ i----1E these refinements have, respectively, 2k-1(2k + 1) and 2k-1(2k- 1) zeros. We shall denote by OEk(Z/2) the group of linear of V preserving and by 02+k(Z/2) and O(Z/2) the subgroups of OEk(Z/2) preserving q+ and q We turn now to the problem of locating the bitangents to a general plane curve. To set up in general, we let W Wd be as before the linear system of plane curves of degree d, A c p2 p2 the diagonal and J0 p2 p2_ A, and 702 joe HARRIS for d > 4 let J Jd C Wd X Jo be the incidence correspondence

Jd { (C;/91,/02)" me, (C" oi/2 ) > 2 }, r and /the projection maps J- W and J J0; we ask for the Galois group of the map r" J W. We will look first at the case of plane quartics, where we find a situation at first glance similar to that of the flexes of a cubic" if l --PlP2 is any bitangent to C, then the divisor 2pl + 2/02 is the complete intersection of C with the line l; i.e., since C is canonical, 2(/Ol + P2)" Kc. Conversely, since C is projectively normal, any divisor in the canonical series IKcl of the form 2pl + 2p2 is cut out in C by a line, so that the bitangent lines l iP2 to C correspond exactly to the effective semicanonical divisors i.e., divisors D p + 1)2 on C such that 2D Kc. Now, choosing any semicanonical divisor x on C (effective or not), for any bitangent line iP to C we have 2(x Pl /02) 0’ and sincepl + P2 q + q2 forpl, P2 v ql,._q2 (C is not hyperelliptic), we see that the bitangents form a subset of the group F of points of order 2 on the Jacobian (C) of the curve C. The Jacobian (C) is the quotient of the three-dimensional complex vector space H(C,2) *--- 133 by the discrete lattice A H(C,Z) Z6; the points of order two on (C) thus form a six-dimensional vector space

V= A (7’/2)6 over Z/2. Note that, as in the case of the flexes on a cubic, the additive structure of F c V may be realized geometrically: irrespective of the choice of origin x in V, four bitangents PP2, qq2, rr2 and will add up to zero if the divisor Pl + 192 + ql + q2 + rl + r2 + s1 d- Sz,2Kc; since C is projectively normal, this will be the case if and only if the points Pi, qi, r and s all lie on a conic. Now, since (as we shall see) the set F of bitangents spans the affine space F of semicanonical divisors, the monodromy group M of F is a priori a subgroup of the group AGL6(Z/2) of affine linear transformations of V. There is moreover, additional structure" under the natural identification A H,(C), Z)- H,(C, Z) GALOIS GROUPS OF ENUMERATIVE PROBLEMS 703 we have the skew-symmetric, nondegenerate pairing Q "A A--+Z given by intersection of cycles; 4Q then induces a strictly skew-symmetric bilinear form Q" v v z/2 on V. Thus the monodromy group is further constrained to be a subgroup of the affine orthogonal group A 06(Z/2). Finally, we see that the action of the monodromy group M on V has to preserve the subset F c V of effective semicanonical divisors. To characterize this subgro.up of A 06(Z/2), choose a normalized basis X X6 for A, i.e., one such that Q(,i, Xj)-- 6j, i+3- i, j+3; the form Q on the vector space Xi Z6 } /A is then given by 3 Q(v,w)-- (DiWi+ Wi’Di+3). i=1 "- Let be the (even) quadratic refinement of Q given by 3 q (1) ) 2 l) l) + 3 i--1 Now, we can write 3 g+j 2 Zij’ki i---1 for Z (Z,) a symmetric matrix; taking Xl,X2,X3 as a complex basis for 133 Riemann’s theta function for C is given by

ril’z’q+2ril’’z 3 O(z) e for Z (Zl,Z2,Z3) C IZ and Riemann’s theorem tells us that there exists a semicanonical divisor on C such that for any divisor class E (Zl,Zz,Z3)/A (C) of degree 0 on C, h( + E) mult,0. (Note that since the curve C is canonically embedded, it cannot be hyperelliptic; thus for any such E we must have h(x + E)= 0 or 1). Now let 3 k 3 kg+i l) 4;- 2 i V i=120i- i=1 T 704 JoB be a point of order 2 on (); we have 0() E eril" Z" ’t+ril" ’a+ril" Z. /Z Taking two copies of this series and pairing the term corresponding to each l Z3 with the term corresponding to -/3- l, we have 20 (v) E (e ri,. Z .t, + ril. tot + ril. Z. tfl Z

q_ e ri( B- t) z ’(-B-l) + ri( B- t) ’a + ri( B-t)" Z

,it. z. ’t + ,iz. + ,it. z. zt). , e ’ ( + e’i( ) Z ’ (1 -b e -ri’’a - since l. ta Z. Thus in case e -riB" ’a lmi.e. if (v) 1--we have O(v) O. Similarly, playing the same game with the partials OO/zi, we have )0 2 (V)= E (lJ"eril’Z’tl+ril’ta+ril’z’t Z -[- (-- j. lj )e i( fl- l) Z" t( fl- l) +i( fl- l) tt +rri( fl- l) Z" tfl ) 0 --.O(V) -I- (1 e -ri#’’a ) (v) and since 0 cannot vanish together with all its first partials at v, we see that e -’iE" ’ 1--i.e., (v) 0--implies O(v) va O. Finally, if xo x + vo is any effective semicanonical divisor on C, set q(v) (v + Vo)- (Vo); note that since is even and (Vo) 1, q is odd. We then have, for any v, w V, q(v + w) + q(v) + q(w) ((v + w + Vo) + (v + Vo) + (w + Vo) + (Vo) (v + w + Vo) + ((v + Vo) + q(w) + (w + Vo) + (Vo) + (w)

Q(v + vo, w) + O(vo, w) O(v, w) so that q(v) is again a quadratic refinement of Q, and we see that For any element v V, the divisor xo + v is effective if and only if q(v) O. GALOIS GROUPS OF ENUMERATIVE PROBLEMS 705

The monodromy group M of the 28 bitangents must therefore be contained in the Steiner group H c AO6(Z/2) (Cf. [8] p. 229) consisting of affine orthogonal transformations of V preserving the zeros of q; that is, the group of affine transformations VAv + b" Q(Av, Aw)= Q(v,w) /v,w H { q(Av + b) q(v) /v V V) We claim now that The monodromy group of the 28 bitangents equals H. As in the case of the flexes, this will follow once we prove that The stabilizer in M of the bitangent xo is the odd orthogonal group 0-(7/2). The proof of this assertion will have to be deferred until we discuss the monodromy action on the lines of a cubic surface. In the meantime, we note a curious coincidence: we have a homomorphism :H--> 06(Z/2) given simply by sending the affine transformation A v + b to the orthogonal transformation A; as it turns out, q is an isomorphism. To see this, let A 06(Z/2) be any orthogonal map. The condition that Av + b be in H is then that q(Av + b)= q(v)

q(6)=0 and Q(Av, b) q(Av)- q(v). (2) But now q(A (v + w))- q(v + w)= Q(Av, Aw)- q(Av)- q(Aw) + Q(v,w)- q(v)- q(w) (q(Av) q(v)) + (q(Aw)- q(w)) i.e., q(Av)- q(v) is a linear functional in v. Since Q is nondegenerate, then, there is a unique b V satisfying equation (2). We have, then for every v V, q(Av + b) + q(Av) + q(b)= Q(Av, b) q(Av) + q(v) i.e., q(b) q(Av + b) + q(v). But now since v Av + b is one-to-one on V and q(v)= for 36 of the 64 706 JOE HARRIS vectors in V, we must have for some v V, q(Av + b)= q(v)= 1; it follows that q(b) 0 and hence H O6(Z/2). What this has to say about the problems of locating the bitangents of a general plane quartic will be discussed in section 4 of Chapter 3. 5. Bitangents to a plane curve, degree d > 5. We consider now the case d > 5; in this case we claim that the monodromy M on the bitangents of a general curve C of degree d > 5 is the full symmetric group. The argument will follow exactly the same lines as in the case of flexes. First, to see that M is transitive, we note that since the fibers of /:J J0 are all linear spaces of codimension 4 in W, Jo is irreducible; so the inverse image r-(U) in J of any Zariski open U c X is connected. Likewise, to see that M is twice transitive, let C be a general d- ic, Pl, P2 a bitangent of C and set W’= r(r/- l(p 1,/02)) (C" mp,(C. /Ol/O2) >/2) c W J’= ((C;q,q2)’mq,(C" q-]) > 2, qi=/:Pj, qlq2 . plp ) Again, since the fibers of J’ over Jo are linear subspaces of codimension 4 in W’, J’ is irreducible; and since J’ contains all the points of the fiber of r over C except (C; p, p2)i.e., C will be smooth at p and P2, and have no other points of tangency with pp2it follows that the subgroup of M fixing (Pl, P2) acts transitively on F (p, P2). To locate a simple transposition in M, we look again to the Pliicker formulas; they tell us that a smooth plane curve C will have exactly one fewer than the generic number of bitangents if it possesses a simple flex bitangentthat is, a bitangent line PiP2 with meeting C transversely away from P and P2; and no other lines with mc(l ) >/3. To show that such a curve exists, choose ql q2 p2 and set W’= (C’mq,(C.--) > 2, mq2(C’-) > 3). Since for d > 5, W" is a linear system without base points except q and q2, we see that in general C W" is smooth, with q lq2 a simple flex bitangent of C. Let I" c W" Io be given by I"= {(C; p,l)’mp(C, l) >/3,pva ql,q2; the fibers of r/:I"o I0 being linear spaces of codimension 3 in W", I" is irreducible of dimension equal to dim IV". The subvariety K" C I" defined by K"= {(C; p,l):mc(l ) > 3) GALOIS GROUPS OF ENUMERATIVE PROBLEMS 707 is thus either equal to I"--in which case every flex line of every C W" has mc(l ) > 3 or of dimension < dim W", in which case every flex line other than qlq2 of a general C W" has mc(l ) 2. Since we can construct as before a curve C W" with a simple flex (pl,ll) meeting l transversely away from Pl, the latter must hold. Similarly, the subvariety J" c W" x J0 given by /191, p2)’mpi(C" /01102 ) > 2, pi =2(= qj, /Ol102 ::/(= q--- ) is irreducible of dimension dim W". Again, we can find a curve C W" with a simple bitangent line PiP2 meeting C transversely away from Pl and P2, so that the subvariety L" C W" given by L"= {(C; p,l):mc(l ) > 3) has dimension strictly less than dim W"; and it follows that for general C W" every bitangent line of C other than has mc(ff) 2. Since finally any line with mc(l ) > 3 must be either a flex or a bitangent or both, we conclude that a general C W" has no lines 1 with mc(l ) > 3. C thus has exactly one fewer bitangent line than a general d- ic; and since J is, as before, smooth, we conclude that the monodromy M on the bitangents of a general plane curve of degree d is the full symmetric group.

III. Lines on hypersudaces 1. The Fano scheme. The next enumerative problem we shall consider is this: how many lines lie on a general hypersurface V c pn of degree d 2n 3? As in the case of flexes and bitangents to plane curves, this problem breaks up into a famous casemthe 27 lines on a smooth cubic surfacemand the general case n > 4; we will prove that [n The Galois group of the lines on a general hypersurface V2n_ 3 C for n > 4 is the full symmetric group. In order to do this, we have, as in the previous chapter, to determine not only the answer to the enumerative problem but also when the general answer fails to hold, i.e., when we have solutions with multiplicity; to. this end we will introduce the Fano scheme Z(V) of lines on V. The Fano scheme may be realized as a subscheme of the G-- G(2,n + 1) of lines in P" as follows ([1]) Let S G(2,n + 1) be the universal subbundle over G, that is, the bundle whose fiber over any point u G is the two-dimensional subspace A c In+l corresponding to u. If F(X) is any polynomial of degree d in the homogeneous coordinates X1,..., X of a, then, we may define a section oe of the dth symmetric power SymdS * of the dual of S simply by setting o (u) 708 JOE HARRIS Of course, OF(U) 0--that is, F(X) is identically zero on Aumif and only if the line c pn lies on the hypersurface V defined by F; the zero-locus Z(V) of oF is the Fano scheme of the hypersurface V. Note that since SymdS * is a vector bundle of rank d + 1, we would expect the dimension of Z(V) to be dimZ dimG (d+ 1) 2n 3 d so that in case d 2n- 3 we expect Z(V) to consist of a finite collection of points. In this case, the answer to the question "how many lines are there on V" is, under the assumption that Z(X) is zero-dimensional and reduced,--i.e., that all zeros of oe are simple--the class of Z(X) in nan-4(G) 7. This is just the top Chern class c2,,_2(Sym2n-3s*), and may be computed as follows: factoring the total Chern class c(S*)= + c(S*)+ c2(S* ) formally, we may write c(S*) + + where 1 and 2 satisfy 1 + 2 O1 and 1’2 011" We then have c(Sym2-3S*) (1 + a" + b2) a+b=2n-3II a,b )O so in particular C2n_2(Sym2n-3s,) rl (al + b’2)" a+ b=2n-3 a,b )O Since (al + b[2)(bl + a’2)= ab( + )+ (a 2 + b2)12 ab( + ’2)2 + (a2- 2ab + b2)[12 abo + (a b)2011, this yields n-2 c2n_2(Sym2"-3S*) (2n 3)2. Ol1" (a(2n 3 a)o + (2n 3 2a)O’ll ) a=lH and since t, o2k 2k! (ok.Ol-l-k) G(2, n+ 1)-- , ]a(2, k+2)-- k! (k + 1)! (cf. [10]), we have the answer to our enumerative problem: (2k)! (d- 2i)2 degZ(V)= f(d)= d. d!. II k k! (k + 1)’ Ic(1 n-2} iI i(d- i) #1=n-2-k where, of course d deg V 2n 3. In fact, we shall see in the following section that for generic V, Z(V) is reduced so that a general V will have exactly f(d) lines. GALOIS GROUPS OF ENUMERATIVE PROBLEMS 709

We now ask" when may a point of the Fano scheme Z(V) of a hypersurface V c pn (of any degree) be singular? To answer this, we write out the Jacobian of the defining equations for Z(V) at a point l0 G and interpret the condition that it have less than maximal rank. First, we let V c pn be the locus of the polynomial F(X) .alXI and let o be the line (X2 X. 0); 0 may be represented as the row space of the matrix

0 0 0 and any line in a neighborhood of 0 may similarly be given as the row space of a unique matrix of the form 0 U02 U03 UOn (10 Ul2 ul3 ul, ) The functions uj are standard Pliicker coordinates on the Grassmannian. We can also write

[kO,l,U020 "- u12 UonkO "1- Ulnl] }[Xo,)kllpl. Now, )to and X are sections of the bundle S*, nonzero and independent in a neighborhood of 10; the sections Xg, xg-lXl,..., Xl correspondingly give a trivialization of the dth symmetric power of S* near o. In terms of this trivialization, the section oF may be represented by the (d+ 1)-vector (fo,..., fd), where we write Fit fo(u) Xg + + fa(u) Xal the Fano scheme Z will be given locally by the ideal generated by the polynomials f/. Explicitly, we have

.(U)-- ad_i,i, 0 0 "l-ad-i-l,i,l,O 0U02 + aa-i- 1, i, 0, 1, 0 0U03

r ad- 1, i, 0 o, non + ad_i,i_l, 1, 0 0U12 + ad-i,i- 1, 0, 1, 0 0it13

@ad-i,i-l,O 0, lUln

The Jacobian matrix Jf(O)= (Of/Ou(O)) of the functions fo,..., fa evaluated at 710 JOE HARRIS

0 may thus be written in block form as

n2

where B is the 2 by d + matrix

0 aa, o l, o ao, a l, o whose rows we may recognize as the coefficients of the polynomials

OF and OF XO O X )1 O X The matrix Jf(O) will thus fail to have rank d + exactly when the polynomials ().OF/OXi} fail to span the vector space of all polynomials of degree d in . Note in particular that this will be the case whenever the polynomials OF/OX all have a common root, that is, when V is singular at a point of 10; thus if V is singular along o then Z is necessarily singular at the point u ---O. In case V is smooth, we can interpret this condition in terms of the normal bundle N to 0 in V as follows: any section of the normal bundle to l0 in pn may be written as

o(X) m2(X) +’’" +mn(X) ua 2 OXn where the m are homogeneous linear functionals in . Such a section owill take its values in N exactly when F F m222 + +mn =0 so h(N) is just the number 2n- 2 of (n- 1)-tuples (m2,..., m,) of linear functionals in , less the dimension of the span of the polynomials (Xj F/3Xi}; in other words, Z will be singular at u -0 if and only if h(N) > 2n 3- d. Finally, since el(N ) n- d- 1, by Riemann-Roch h(N)= n- d- l+n-2+hl(N) 2n 3 d+ hi(N) GALOIS GROUPS OF ENUMERATIVE PROBLEMS 711 so this condition may be restated as: Z will be singular at a point u if and only if the normal bundle to in V is special. We ask now what the normal bundle N to a line l in a smooth hypersurface V in [::n may be. Specifically, since N is a vector bundle of rank n- 2 on pl we may write N ((e,) +... + and ask what collections of n- 2 integers el,..., en_ 2 actually occur as the degrees of the direct-sum factors of the normal bundles of lines in hypersurfaces. Two things we know right away: since N is a subbundle of the normal bundle to in pn--which is just the direct sum of n- copies of <9(1)--it cannot have a sub line bundle of degree two or more; thus e < for all i; (1) and from before, cl(N ) ,e n d- 1. (2)

On the other hand, it turns out that any collection of integers e2,... en_ 2 satisfying conditions (1) and (2) may occur" we may exhibit the lines and hypersurfaces which do it. The line will always be as before (X2 X,, -0); for the hypersurface, if we write

el,...,es <0

es+ en_ 2 then we take F so that F )X2 OF xg+el-2Sl el +1

F xg+e,+ +ei-i-lx{-e ei+i

F OXs+2 OF =0 for j> s+2 712 JOE HARRIS i.e., setting e0 0, we take

F(X) + i=0 where G(X) is a general polynomial of degree d vanishing to order 2 along 1. The normal bundle N to in V then has sections

-Xl-e’+l l’X)l(X-e’+lO-’---Ox3 )OX2

02 -e2+ --Xl-e2+l---O X2(Xd -’--Ox4 )OX3

os__Xs(xes+l OXx+20 Xl_e,+ OXs+lO

Os+ X 00Xx+ 3

X On_2-" 0 )Xn Each section o defines a sub line bundle Li O(ei) of N; since the wedge product o / / On_ 2 is a section of the line bundle A"-EN, we have c,(An-2N) c,(N) > with equality holding if and only if the line bundles L are everywhere linearly independent, and it follows that N (9 L @lg(ei) as desired. Note that in case d 3, N cannot have a factor of degree -2 or less, and so must be nonspecial; while if d > 4, h(N) may be non-zero. In other words, the Fano scheme of lines on a cubic hypersurface V will be singular at a point u if and only if V is singular somewhere along lu; but if d deg V > 4, then Z(V) may be singular even if V is smooth. 2. Lines on a hypersurface in pn, n > 4. Let W denote the linear system of hypersurfaces of degree d 2n- 3 in pn, and let I C W G(2,n + 1) be given by I= ((V,l):l C V) GALOIS GROUPS OF ENUMERATIVE PROBLEMS 713

We can now prove without much difficulty that in case n > 4 the monodromy group M of the map r: I W is the full symmetric group. First of all, to see that M is transitive, we simply note that since the fibers of the map ,r I-- G G(2, n + 1) are linear spaces of codimension d + in W, I is irreducible. Next, to show that M is twice transitive, choose a line l0 c pn and set

I’= {(X,l)’l c X, l=/= lo) C W’ G Since two disjoint lines in pn impose independent conditions on [tqp, (2n- 3)1, the fibers of I’ G over the Zariski open U { l" l f3 l0 O) will be linear subspaces of "codimension d + in W’; the fibers of over the Schubert cycle on_2(10) {1" f)l0 vaO) will have codimension d in W’. Since dim on_2(lo) dim G(2,n + 1)- (n 2)= n, we see that in case n > 4 dim (o,, 2(lo)) dim W’ n + 3 < dimW’ so that I’ contains only one irreducible component dominating W’. Thus all the points other than (V, 10) of the fiber of ,r over a general V W’ lie in one irreducible component of I’, and hence the stabilizer of l0 in M is transitive on Z(V)- o. (Note that in case n 3, 1-1(o(lo)) does in fact form a second irreducible component of I’; we have I’ r/-I(U) + /-1(oi(/o)) with the two components mapping to W’ with degrees 16 and 10 respectively). Next, to show that M contains a simple transposition, we construct a hypersurface V whose Fano scheme has a point of multiplicity 2, as follows: we start with a V containing the line 10 (X0 Xn 0) such that the normal bundle N of 10 in V has hi(N)= 1, i.e., N (9(- 2) @ (9 (9(- 1)... (9(- 1). By the computations above, this may be achieved by taking V to be the locus of a polynomial F of the form G + H, where

a(g) gg-ls2 + gg-4g31g + 0 l,X4 -[-- gg-7SS5 .-[.-

+ + and H(X) vanishes to order 2 along l0. The Fano scheme of lines on V will then 714 JOE HARRIS be cut out in a neighborhood of u--0 by the functions fo(u) u02 + u2 k(u)- u,

f(u) Uo u f4(u)- Uo * u, .[u]

f6(u)- u05 +[u2 fv(u) u,5 +[u 2]

l,-,u Uo, +lull f,(u) u,, + [u’].

Now, the locus of the polynomials {f,.)i4=2 is a curve smooth at 0 with tangent line (uj 0, j 4 3, 4; u03 u14 0; u04 + ula 0); to make 0 a double point of Z we thus have to make sure that f2 vanishes to order exactly 2 at u 0, and that this line does not lie in its tangent cone at u 0. Writing out f2(u) more fully as

f2(u) ’ao, d-2 li lj o" UoiUoj

+ -’ao, d- 2 2i 0 Ui

+ al,a-3 1, lj o" (UoiUlj + UliUoj)

+ ’al,a-3 2, 0 2UoiUli

+ ’a2,a-4 1, 1 o" UliUj

+ a2, d- 4 2 0 Ui we see that this is in fact the case for a general choice of coefficients, i.e., for F(X) G(X)+ H(X) as above with H generic. Finally, we wish to show that for a generic hypersurface V of the linear system W"= {G(X)+ H(X)}, all the remaining points of the Fano scheme are simple. We consider accordingly the incidence correspondence I" c W"x G GALOIS GROUPS OF ENUMERATIVE PROBLEMS 715 given by "= ((V,): c V, / 0). The fiber of ,/:I" G over a general point of G is of codimension d + in W"; over a general point of the Schubert cycle of lines meeting lo, the fiber has codimension d; and over a point of the subvariety c of lines I having contact of order > 2 with- G(X) at (q [0 it has codimension d-1. Since has dimension strictly smaller than o, we see as before- that I" can have only irreducible component dominating W"; so that either for a general V W"- all lines c V other than 0 are simple points of the Fano scheme Z(V), or for all V W" they are all multiple. The latter possibility we can eliminate again by direct examination: in the expression F G + H, take n--2 H(X) Xa- ’-2ix2i-- 1Xi + H’(X) i=0E where H’(X) vanishes to order 3 along 0 and to order 2 along the line m0 (X0 X,,_ 2 0); the line m0 will then be a simple point of the Fano scheme of lines on the hypersurface given by F. Thus, the Fano scheme of lines in a generic V W contains only the one double point 10, M contains a simple transposition, and we are done. 3. Lines on a cubic surface. We have already seen that the Galois group of the 27 lines on a cubic surface S c p3 cannot be the full symmetric group, inasmuch as it must take pairs of lines which intersect to pairs that intersect and vice versa. To describe the incidence relations among the lines, recall that a smooth cubic surface may be realized as the blow-up 152 of pz at six points P,..., P6; so that the second H2(S,Z) is seven-dimensional. Let V be the primitive second cohomology p2(S, 7/2) of S mod 2, that is, the subspace of H2(S,Z/2) having intersection number 0 with the hyperplane class H -Ks mod 2. V is a six-dimensional vector space over Z/2, and we note that the intersection pairing Q: V V7/2 is strictly skew-symmetric: for any integral homology class E HZ(s,z) we have E.E-E.H Z ,r(E) 2 + so that E. H 0(2) implies E. E _= 0(2). Moreover, Q possesses a natural quadratic refinement q: V Z/2, defined by setting

E.E mod2 q()= 2 for any class E H2(S, 7) lifting E. 716 JOE HARRIS

Now, if L is any line on S, the class L-H has intersection number (L- H). H 3 0(2) with H, and so represents a primitive cohomology class L V. We note that 0 if Q(L,L’)--(H- L).(H- L’)_= if LOL’=O (1) and q(/)=(H-L)2=3-2- 1=0. (2) In particular, (1) tells us that the set of 27 lines on S injects onto V--for any L,L’ we can find L" meeting L but not L’--and that L is never zero. Indeed, if we choose 6 disjoint lines L1,..., L6, the classes e L form a basis for V, in terms of which we have Q(ei,ej) aj and q(’xiei) _a xixj; i

Now, there are 72 different sets of six disjoint lines on Sthere are ai-- (Ei )

Bij: Ei, Ej, Ek,tmn)m, nvi,j, k % (Ei, Ci,Zjk}kv

Oijk ( Ci, Cj., Ck,tmn)m, nvi,j, k E-- {Ci) corresponding to the 72 unordcrcd bases { ei} for V P 2(S, Z/2) in terms of which the quadratic form q is given by q(Y ,ei) i

# # # #G’ f’ Z6 ’. 720 must divide #O-(Z/2), we see that #fl’ divides 72; but since the elementary automorphisms permuting the lines (El,... ) act transitively on each of the subsets (A }, (Bijk ), (CO.}, (Dijk} and (E }, 2’ must be a union of these subsets. These subsets having orders l, 20, 30, 20 and l, respectively, we conclude that 2’ f, i.e., O6- (7/2) is generated by elementary automorphisms. Our result will now follow from the PROPOSITION. The monodromy group M contains all elementary automor- phisms.

Proof. Suppose o 06-(7/2) sends the six disjoint lines MI,..., M6 on S to Mo(1),..., Mo(6). Blow down the lines M on S to obtain p2; let/o be the image point of Mi. Draw real arcs _pl(t) p6(t) in the plane, such that p (o) =p,; Since the set U c (p)6 of sextuples of points in the plane in general position is the complement of a proper subvariety of (p2)6 and hence connected, we may 718 JOE HARRIS assume that the points pl(t),... ,_p6(t) are in general position for all t; let 5t2 be the plane blown up at_PE(t),...,/06( 0. Let ql, q4 H(15, (9(-K)) be the pullback to 15-- S of the homogeneous coordinates X on p3; since the group GL4(C ) is connected, we may choose ql(t), q4(t) H 0(Pt,~2 (9(-K)) varying continuously with such that qi(0)= qi(1) Vi and ql(t) A A q,4(t)0; let S be the image of [Stz under the map to[a3 given by [ql(t),..., q4(t)]. Then the monodromy action associated to the loop t- S in the space U c F19 of smooth cubics effects the permutation on the lines Mi. Q.E.D. Now, the group O6-(Z/2) is not solvable--indeed, it has a subgroup of index 2 (order 25920) which is simple (cf. [8])Bso that there does not exist a formula for the 27 lines of a general cubic surface. We may still ask, if we are given a cubic surface S (EalXI= 0) and the coordinates of some of its lines, can we solve for the remaining ones? The answer may be worked out as follows: 1. Suppose we are given the coordinates of two skew lines L 1,L2 of S. The Galois group G’ of the field K(I3) over the subfield generated by K(W3) and the coordinates of the two lines is, of course, the stabilizer in 06-(Z/2) of the two classes e,e2 represented by H-L and H-L2 in p2(S,Z/2). Since the intersection form is nondegenerate on the subspace of p2(S,Z/2) spanned by e and e2, this is just the orthogonal group of the orthogonal complement { e e2)l, i.e., G’ 04-(Z/2). Since 06- (Z/2) acts transitively on the 27.16 pairs of skew lines of S, we have

51840 120; #O4-(Z/2) 27.16 but as one can readily check there are exactly five normal bases for (e,e2} and any automorphism fixing them all is the identity. Thus G’= 04- (Z/2) 5 is the permutation group on 5 letters, and is not solvable. 2. Suppose we are given the coordinates of three skew lines on S. The subgroup G’ of O6-(Z/2) fixing their three classes is now of order 12 (indeed, it’s not hard to see that G’ is a dihedral group) and is correspondingly solvable.

Explicitly, if we call the three lines given E l, E2, and E3, then the quadric surface Q c p3 containing them is rationally determined, as are the two rulings { Mx), {N) of Q by lines. Now S, intersecting Q in three lines E of one ruling {Mx), must contain also three lines of the other ruling {N}; writing the restriction to Q of the equation of S out in terms of the coordinates , and/, factoring out the equations , ,i of the lines E and solving the resulting cubic polynomial in/z, we find the coordinates of three more lines of S. Since each of the new lines meets all three of the Ei, these must be the lines C4, C5 and C6. Now, for any pair Ei, C. of lines, one from the original three and one from the new, the plane H/j spanned by the pair intersects S residually in the line L/j; restricting F(X) to H/j and factoring out the linear functions corresponding to E GALOIS GROUPS OF ENUMERATIVE PROBLEMS 719 and Cj, we have the coordinates of L/j. We now have Ei, Cj and L/j for < < 3, 4 < j < 6; to obtain another linemsay L12--we have to solve a quadratic equation: the condition that L12 meet each of E E2, L34 and L35 are four linear relations on the PliJcker coordinates {//j} of the line L (cf. [6], p. 221); restricting the standard Pliicker relation E/j/t 0 to the line they define, we have a quadratic polynomial whose two roots are the coordinates of the lines L12 and C This the are 3 solved, finally, remaining lines all rationally determined: E is residual to C3 and L3j in the plane they span, and similarly C is residual to L12 and Ei, L3i residual to E and C3, and Ljk residual to Ej and Ck. 3. Suppose finally that we are given the coordinates of two lines on S which meet. The group G’ is again the stabilizer in O-(Z/2) of the corresponding classes e and e:; and since the quadratic form q is identically zero on the subspace they span, G’ must preserve the flag 0c (ei) c {ei, e:) c {el,e:)’c {ei)lc V= p2(S,Z/2).

2 In particular, G’ acts on {el,e:}l/{el,e2} (Z/2) as GL2(Z/2)- 2;3, the kernel acts on (el,e2) as Hom((el,e2)l/(el,e2}, {el,e2} ) (Z/2)4, and the kernel of that is order 2; so G’ is solvable. In practice, this works out as follows: given two lines of S which meetsay E and C6--we consider the pencil (Ht) planes in 3 through E There will be five such planes which intersect S residually in two lines, one of which (the plane spanned by E and C6) we already know about. Take F(X) restricted to H and factor out the equation of the line E H to obtain a quadric polynomial Gt(X ). The discriminant disc(Gt(X)) will be a quintic in t; factoring out the one known root we obtain a quartic which we can then solve for the equations of the planes (El, Cj,Llj}j=3 6" For each of these planes, solving a quadratic polynomial will give us the equations of the two individual lines in the plane; indeed, having found Cj and Llj for j 3, 4 and 5 we note that the rest of the lines are determined: C is the unique line other than E meeting each of LIE L13 Ll4 and L15 so that we can solve as in the previous case for the Pliicker coordinates of C1; E6 is the unique line other than E meeting C2, Ca, C4 and C5; L16 is to residual C and E6 and C6 residual to E and LI6 E is residual to Ll and Ci, and LO. is residual to E and Cj. In conclusion, then, we may state: given a cubic surface S and either two incident or three skew lines of S, we may solve explicitly for the remaining lines; but no smaller configuration suffices. 4. Bitangents to a plane quartic II. We may now complete the computation of the monodromy group of the 28 bitangents to a smooth plane quartic. To begin with, we recall from [6] the following: Let S be a smooth cubic surface in p3, p S a point of S not lying on any line of S, and re P the projection map from the blow-up of S at P to a plane P2c p3. Then the branch locus of the 2-sheeted map re is a smooth quartic curve C; and the bitangents of C are the images L of the 27 lines L C S, 720 JOE HARRIS plus the image D of the plane section D Te(S ) q S of S. The points of tangency of L with C are the images of the points of intersection of__L with the conic C residual to L in the plane P, Li; the points of tangency of D with C are the images of the tangent lines to D at P. Now, we have seen that the 27 lines of S are identified with elements of the vector space V P 2(S, Z/2) (Z/2)6, and likewise the 28 bitangents to C may be identified with a subset of the vector space W of points of order 2 on (C) with D corresponding to the origin. We claim now, that under these identifications, the map re extends to a linear isomorphism V W. To see this, we have to prove the

LEMMA. If L1,Lz,L are any__thre_..e lines_ on S andpl,_ 132, ql,q2, rl,r2 and sl,s2 the points of tangency of D, LI, L2 and L3 with C, then the class L + L2 4- L3 3 H is divisible by 2 in H2(S, 7)=0 { Pi, qi, ri,si) all lie on a conic.

Proof. We first note that the class L + L2 4- L3 3H is divisible in H2(S, 7) if and only if L L2 and L3 are coplanar" certainly if they are coplanar, L + L + L 3H -2H is divisible; while if L + L2 4- L3 3H is divisible by 2, we have Li.(L + L+ L3-3H)-0 (2) Vi Li.().iL))-O (2) Vi and

(L 4- L2 4- L3 -3H)2--0 (4) Li.Lj-2 (4) ivj from which it follows that L Lj for 4 j, i.e., L L2, and L3 lie in a plane H cP 3. Now, recall that the inverse image ( c S of the branch locus of re is the intersection of S with a quadric T c p3 called the polar quadric of S at P. Note that since T is tangent to S at P, the two lines/l,/ on T through P, lie in Te(S); and since the map re is unramified on the curve D Te(S ) fq S away from Pso that/l,/ cannot meet D except at Pit follows that/l,/_ are the tangent lines to D at P. We see then that the conic T re(T q H) contains Pi re( ti H), as well as the points { qi, ri,si}. Q.E.D. Inasmuch as the lines on S and bitangents to C generate V and W, it follows that re extends uniquely to a linear isomorphism V--> W. Note that this is an isomorphism of vector spaces with quadratic forms, since re sends a zero of the quadratic refinement q on I/to one in W. Now, to prove the assertion made in 4 of Chapter 2 about the monodromy group of the 28 bitangents to a quartic, we return to the proof of !!w proposition in the last section. Choose a point P on the cubic surface S not lying on. any line of Si.e., the image of a point p p2 not collinear with any two of the points Pi, GALOIS GROUPS OF ENUMERATIVE PROBLEMS 721 and not on any of the conics passing through five; we may assume P= [1,0,0,0,] and Tp(S)=(X3=O). Let CO be the branch locus of the projection rre. We may then choose our arcs p(t),..., p6(t) so that for each no three of the points p, pl(t), p6(t) are collinear and no six are a conic. We can finally choose our sections (t), 4(t)Ht’[52,,t,(-K)) so that 2(t), 3(t), 4(t) all vanish at p for each t, and so that ta(t) vanishes to order 2 at p for each t. Let S be as before: we have PES

Te(S ) (X3 0) so that if C is the branch locus of S under projection from P to the plane X0 0, (Ct) is an arc in the space U c [214 of smooth quartic curves, and the monodromy action of (Ct) on the 28 bitangents to CO corresponds, via re, to the action of (St) on the 27 lines of S. Since any quartic plane curve C c [::2 may be realized as the branch locus of the projection re of a cubic surface S C 3 from a point P S not on any line of S, with any bitangent to C the image of T.(S), it follows that the stabilizer, in the Galois group J of the 28 bitangents to a plane quartic, of any given bitangent is the odd orthogonal group O-(Z/2) and finally, by virtue of the computation made in 4 of Chapter 2, that the Galois group of the 28 bitangents to a plane quartic is the ordinary orthogonal group O6(Z/2). The answer to the problem of finding the bitangents of a quartic C, then, is that given any three bitangents whose points of contact lie on a conic, or an); four whose points of contact do not, we can solve for the remaining ones; but no smaller set will suffice.

After choosing one of the given bitangents 0 as the origin, the procedure for solving may be carried out either on the cubic surface--if C is the locus of the quartic f(x, y), the cubic surface S is the image of the surface (z 2- f(x, y)= O) under the map given by (x, y,z) (. 0(, y), y. lo(x, y),z q(x, y)) where f(x,y)= q(x,y)2 on /o--Or directly in the plane: for example, the bitangent "residual" to l and to a of incident bitangents- l2 corresponding pair lines on S will be the line joining the two residual points of intersection with C of the conic containing the points of contact of lo, l and l2.

IV. The problem of five conics. The last problem we take up is to determine the Galois group associated to a classical enumerative problem: given five conics in general position, how many will be tangent to all five? (cf. [5], [6] and 722 JOE HARRIS

[9]). The Galois group of the problem has been studied recently by Higman [8], who concluded that it was twice transitive; we will prove here that it is in fact the full symmetric group. The proof of this assertion will follow the same lines as used above, but each of the steps in the proofshowing double transitivity, exhibiting a specialization of the problem having exactly one double solution, and showing local irreducibility around that double solutionwill be somewhat harder; in particular, finding a double solution to the problem was only recently made possible by the formulas in Fulton’s and MacPherson’s paper [5]. To set up, let W p5 be the linear system of conics in the plane, W c W the locus of conics of rank 1, and I/P W- W Set X W 5, and let Y c X ffz be given by Y ((C,..., C5; C): C is tangent to C ’i); let F be a general fiber of rrl: Y--> X and G the Galois group of r acting on F. C conics to X For any conic D , let HD W be the locus of tangent D, D the locus of conics bitangent to D and YD the locus of conics having a point of contact of order > 3 with D. We first establish some facts about these subvarieties of W:

For D smooth, HD, XD and YD are irreducible; and for E not tangent to D, Ho f’) He and HD f’) Ye are irreducible. Proof. For the first part, letIDC WD,IC WD Dandlzc W D be given by ID= {(C;p)’rn(C" D ) > 2} / ((C; p,q)" mp(C. D ),mq(C. D) >/2} and I’ {(C, p)" m,(C D) > 3). Since the fibers of the projection maps %’1oD, r’I-D D and r’ I --> D are surjective with fibers p3, la and p2 respectively, we see that Io, I and I are all irreducible; so their images HD, XD and YD in W are irreducible. Note one additional point: since the map r :ID HD is one-to-one away from XD, we see that Ho is irreducible at any point of HD XD. Similarly, if E meets D transversely, let ID, C W D E and I, C ID, e be given by -- ID, F ((C; p,q)" mp(C. D ),mq(C. E) >/2} and D,e {(C; p,q) mp(C. D) > 2, mq(C. E) >/3}

The fibers of the projection map 1D, e D E are P’s, except over the eight points {(iv, p)’p D Cl E} and {(p,q)" T,(D)= Tq(E)}, where the fiber is GALOIS GROUPS OF ENUMERATIVE PROBLEMS 723

The image Hz N He of lz,e in W is thus irreducible in the complement of eight planes P2c W; since HD He has pure dimension 3 it follows that it is irreducible. Likewise, I’D,E maps 1- onto D E except over those eight p l. points, where the fiber is so that the image Hz f3 Ye of I’D, E is irreducible outside of eight lines and hence irreducible. Q.E.D. Now, to show that the Galois group of the map r l" YX is the full symmetric group, we first show that it is twice transitive. To see that G is transitive, consider the map r2 from the incidence correspondence Y to 1. The fibers (Hz) of Y over the Zariski open set U c W of smooth conics are irreducible of dimension 20; and the fibers over a conic D of rank 2, while reducible, are again of dimension 20. Thus Y can have only one irreducible component Y of dimension dim X 25. The points of F thus all lie in one irreducible component of Y, and so by the standard argument G is transitive. Similarly, to see that the stabilizer of a given point C F in G is transitive, let Xc= H C X be the quintuples of conics tangent to C, and define YcCXc l-(C)by Yc ((C1, C5; C’): C tangent to C’ Vi) Over the Zariski open set of smooth conics C’ transverse to C, the fiber (HC 0 /-/C,) of ’772 is, by the above, irreducible of dimension 5; over the locus of conics C’ tangent to C and/or singular, the fiber of ,’/72 is reducible but still only 15-dimensional. Thus Yc can have only one irreducible component dominating Xc, and as before it follows that the monodromy in r Yc Xc is transitive. Next, to show that G contains a simple transposition, we have to locate a fiber of r" Y X having one double point and the rest simple. To do this, we make use of the following PROPOSITION [5]. If no two of the five conics C1, C5 are tangent, no three are concurrent and no three are tangent to an), line, then the intersection Hc, fq Hcs fq W consists of isolated points with mc(Hc,... Hcs) 3264; C Hc fq fq Hc fq and the intersection multiplicity of Hc,, Hc at a point C W is given by

pcnci

To find a double point of a fiber of ,771 Y o X, then, choose a conic C W, and take C1,..., C4 generic conics tangent to C, and C5 a generic conic having a point of contact of order 3 with Ci.e., choose (C C5) a generic point of X H Yc Clearly C1,..., C5 will then satisfy the conditions of the proposition, and we have mc(Hc,...Hc)= 2; to show that the remaining points of the fiber of ,27" over (C1,..., C5) are all simple--i.e., that mc,(Hc, Hc) /C’ 4: C Hc, (q fq Hc (q IYV, we argue as follows: 724 JOE HARRIS let K c X if/be given by Kc {(C1,..., C5;C’)’C tangent to C’ V i, C’ 4: C } and let K’ C K be the closed subvariety K ((Cl,... C5;C’):mc,(Hcl. Hc,) 2}; we want to show that K’ cannot dominate But now the fiber of the X+. 4 projection r2 Kc X: over a point C’ is just (Hc N Hc,) (Hc Yc’), which we have seen is irreducible of dimension 14 if C’ is smooth and meets C transversely; it is reducible but still of dimension 14 for C’ Yc; and is of dimension 15 for C’ Yc. We see then that Kc can have only one irreducible component Kc dominating the 19-dimensional X+, and this component cannot be contained in K" for C’ smooth and Hc, and (D1, Ds;C’) generic in the fiber of r over C’, D,..., D will all be simply tangent to C’. Thus Kb cannot dominate Xc. Finally, it remains to see that for (C1,..., C5) generic in X, our original incidence correspondence Y is irreducible at the point (CI,..., Cs;C). But the fiber of r2 Y-- W over a point D W is irreducible at points (DI, Ds;D) such that D XD V i; and since no conic in a small neighborhood of C will be bitangent to a conic in a small neighborhood of C for any i, we see that I is irreducible at (C, C5;C ). Thus the Galois group G of r Yo X contains a simple transposition, and hence is the full symmetric group.

REFERENCES l. A. ALTMAN AND S. L. KLEIMAN, Foundations of the theory of Fano schemes, Compositio Math. 34 (1977), 3-47. 2. A. CAVLEV, A memoir on cubic surfaces, Phil. Trans. Royal Soc. London CLIX (1869), 231-326. 3. , Uber die Anwedung der Abelschen Functionen in Der Geometrie, Jour. de M. Borchardt, LXIII. 4. L. DICKSON, H. F. BLICHFELDT, AND G. A. MILLER, Theory and applications of finite groups, John Wiley, New York, 1916. 5. W. FULTON AND R. MACPHERSON, "Defining algebraic intersection," in Algebraic Geometry, Proceedings of the Tromso Symposium 1977, Springer Verlag. 6. P. GRIFFITHS AND J. HARRIS, Principles of Algebraic Geometry, Wiley-Interscience, New York, 1978. 7. G. HIGMAN, unpublished notes. 8. C. JORDAN, Trait des Substitutions, Gauthier-Villars, Paris, 1870. 9. S.L. KLEIMAN, Chasle’s enumerative theory of conics: A historical introduction, Aarhus preprint, 1976. O. Problem 15, Rigorous foundation of Schubert’s enumerative calculus, Proc. of Symposia in Pure Mathematics, Amer. Math. Soc. 28 (1976), 445-482. 11. J. MILNOR AND D. HUSEMOLLER, Symmetric Bilinear Forms, Springer-Verlag, New York, 1973. 12o STEINER JAKOa, Ueber die Glachen Dritten Grades, Crelle’s Journal. 13. H. WEaER, Zhrbuch der Algebra, Chelsea Publishing Co., New York, 1941.

DEPARTMENT OF MATHEMATICS, MASSACHUSETTS INSTITUTE OF TECHNOLOGY, CAMBRIDGE, MASSACHUSETTS 02139