Let us see where He(X)intersectsX. We assume that He(X)is a hypersurface of degree(n+ 1)(d−2) >0. A glance at the expression (1.21) reveals the following fact.
Proposition 1.1.19 Each singular point ofX belongs toHe(X).
Let us see now when a nonsingular pointa∈ X lies in its Hessian hyper- surface He(X).
By Corollary1.1.12, the inflection tangents inTa(X)sweep the intersection
of Ta(X)with the polar quadric Pad−2(X). Ifa ∈ He(X), then the polar
quadric is singular at some pointb.
Ifn= 2, a singular quadric is the union of two lines, so this means that one of the lines is an inflection tangent. A pointaof a plane curveX such that there exists an inflection tangent atais called aninflection pointofX.
Ifn >2, the inflection tangent lines at a pointa∈X∩He(X)sweep a cone over a singular quadric inPn−2 (or the wholePn−2if the point is singular).
Such a point is called aparabolic pointofX. The closure of the set of parabolic points is theparabolic hypersurfaceinX(it could be the wholeX).
Theorem 1.1.20 LetX be a hypersurface of degreed >2inPn. Ifn= 2,
thenHe(X)∩Xconsists of inflection points ofX. In particular, each nonsin- gular curve of degree≥3has an inflection point, and the number of inflections points is either infinite or less than or equal to3d(d−2). Ifn >2, then the setX∩He(X)consists of parabolic points. The parabolic hypersurface inX
is either the wholeXor a subvariety of degree(n+ 1)d(d−2)inPn.
Example 1.1.21 Let X be a surface of degreedinP3. If ais a parabolic
multiplicity higher than 3 or it has only one branch. In fact, otherwiseX has at least two distinct inflection tangent lines which cannot sweep a cone over a singular quadric inP1. The converse is also true. For example, a nonsingular
quadric has no parabolic points, and all nonsingular points of a singular quadric are parabolic.
A generalization of a quadratic cone is adevelopable surface. It is a special kind of aruled surfacewhich characterized by the condition that the tangent plane does not change along a ruling. We will discuss these surfaces later in Chapter 10. The Hessian surface of a developable surface contains this surface. The residual surface of degree2d−8 is called thepro-Hessian surface. An example of a developable surface is the quartic surface
(t0t3−t1t2)2−4(t21−t0t2)(t22−t1t3) =−6t0t1t2t3+4t31t3+4t0t32+t 2 0t 2 3−3t 2 1t 2 2 = 0. It is the surface swept out by the tangent lines of a rational normal curve of degree 3. It is also thediscriminant surfaceof a binary cubic, i.e. the surface parameterizing binary cubicsa0u3+ 3a1u2v+ 3a2uv2+a3v3with a multiple root. The pro-Hessian of any quartic developable surface is the surface itself [84].
Assume now thatX is a curve. Let us see when it has infinitely many in- flection points. Certainly, this happens whenX contains a line component; each of its points is an inflection point. It must be also an irreducible compo- nent of He(X). The set of inflection points is a closed subset ofX. So, ifX
has infinitely many inflection points, it must have an irreducible component consisting of inflection points. Each such component is contained in He(X). Conversely, each common irreducible component ofXand He(X)consists of inflection points.
We will prove the converse in a little more general form taking care of not necessarily reduced curves.
Proposition 1.1.22 A polynomialf(t0, t1, t2)divides its Hessian polynomial He(f)if and only if each of its multiple factors is a linear polynomial. Proof Since each point on a non-reduced component ofXred⊂V(f)is a sin-
gular point (i.e. all the first partials vanish), and each point on a line component is an inflection point, we see that the condition is sufficient forX ⊂He(f). Suppose this happens and letR be a reduced irreducible component of the curveX which is contained in the Hessian. Take a nonsingular point ofRand consider an affine equation ofRwith coordinates(x, y). We may assume that OR,xis included inOˆR,x ∼=C[[t]]such thatx=t, y =tr,where(0) = 1.
Thus the equation ofRlooks like
f(x, y) =y−xr+g(x, y), (1.22) whereg(x, y)does not contain termscy, c ∈C. It is easy to see that(0,0)is an inflection point if and only ifr >2with the inflection tangenty= 0.
We use the affine equation of the Hessian (1.21), and obtain that the image of h(x, y) = det d d−1f f1 f2 f1 f11 f12 f2 f21 f22 inC[[t]]is equal to det 0 −rtr−1+g 1 1 +g2 −rtr−1+g 1 −r(r−1)tr−2+g11 g12 1 +g2 g12 g22 .
Since every monomial entering ingis divisible byy2, xyorxi, i > r, we see that ∂y∂g is divisible bytand ∂g∂x is divisible bytr−1. Alsog11is divisible bytr−1. This shows that
h(x, y) = det 0 atr−1+· · · 1 +· · · atr−1+· · · −r(r−1)tr−2+· · · g12 1 +· · · g12 g22 ,
where· · · denotes terms of higher degree int. We compute the determinant and see that it is equal tor(r−1)tr−2+· · ·. This means that its image in
C[[t]]is not equal to zero, unless the equation of the curve is equal toy = 0,
i.e. the curve is a line.
In fact, we have proved more. We say that a nonsingular point ofXis an in- flection point oforderr−2and denote the order by ordflxXif one can choose an equation of the curve as in (1.22) withr≥3. It follows from the previous proof thatr−2is equal to the multiplicityi(X,He)xof the intersection of the curve and its Hessian at the pointx. It is clear that ordflxX =i(`, X)x−2, where`is the inflection tangent line ofX atx. IfXis nonsingular, we have
X
x∈X
i(X,He)x= X x∈X
ordflxX = 3d(d−2). (1.23)