Hessian and inflexion points

Theorem 1.

Suppose that C is a curve in the real projective plane RP2 given by a homogeneous equation F(x,y,z)=0 of degree of homogeneity (http://planetmath.org/HomogeneousFunction) n. If F has continuousMathworldPlanetmathPlanetmath first derivativesMathworldPlanetmath in a neighborhood of a point P and the gradient of F is non-zero at P and P is an inflection pointMathworldPlanetmath of C, then H(P)=0, where H is the Hessian determinant:


We may choose a system x,y,z of homogenous coordinates such that the point P lies at (0,0,1) and the equation of the tangentPlanetmathPlanetmathPlanetmath to C at P is y=0. Using the implicit function theorem, we may conclude that there exists an intervalMathworldPlanetmathPlanetmath (-ϵ,ϵ) and a function f:(-ϵ,ϵ) such that F(t,f(t),1)=0 when -ϵ<t<ϵ. In other words, the portion of curve near P may be described in non-homogenous coordinates by y=f(x). By the way the coordinates were chosen, f(0)=0 and f(0)=0. Because P is an inflection point, we also have f′′(0)=0.

Differentiating the equation F(t,f(t),1)=0 twice, we obtain the following:

0=ddtF(t,f(t),1) =Fx(t,f(t),1)+f(t)Fy(t,f(t),1)
0=d2dt2F(t,f(t),1) =2Fx2(t,f(t),1)+f(t)2Fxy(t,f(t),1)

We will now put t=0 but, for reasons which will be explained later, we do not yet want to make use of the fact that f′′(0)=0:

Fx(0,0,1) =0
2Fx2(0,0,1) =-f′′(0)Fy(0,0,1)

Since F is homogenous, Euler’s formulaMathworldPlanetmathPlanetmath holds:


Taking partial derivativesMathworldPlanetmath, we obtain the following:


Evaluating at (0,0,1) and making use of the equations deduced above, we obtain the following:

Fz(0,0,1) =0
2Fxz(0,0,1) =0
2Fyz(0,0,1) =(n-1)Fy(0,0,1)
2Fz2(0,0,1) =0

Making use of these facts, we may now evaluate the determinant:

H(0,0,1) =|-f′′(0)Fy(0,0,1)2Fxy(0,0,1)02Fxy(0,0,1)2F2y(0,0,1)(n-1)Fy(0,0,1)0(n-1)Fy(0,0,1)0|

Since P is an inflection point, f′′(0)=0, so we have H(0,0,1)=0. ∎

Actually, we proved slightly more than what was stated. Because the gradient is assumed not to vanish at P, but F/x=0 and F/z=0 by the way we set up our coordinate systemMathworldPlanetmath, we must have F/y0. Thus, we see that, if n1, then H(0,0,1)=0 if and only if f′′(0). However, note that this does not mean that the Hessian vanishes if and only if P is an inflection point since the definition of inflection point not only requires that f′′(0)=0 but that the sign of f′′(t) change as t passes through 0.

This result is used quite often in algebraic geometryMathworldPlanetmathPlanetmath, where F is a homogenous polynomialPlanetmathPlanetmath. In such a context, it is desirable to keep demonstrations purely algebraic and avoid introducing analysisMathworldPlanetmath where possible, so a variant of this result is preferred. The theoremMathworldPlanetmath may be restated as follows:

Theorem 2.

Suppose that C is a curve in the real projective plane RP2 given by an equation F(x,y,z)=0 where F is a homogenous polynomial of degree n. If C is regularPlanetmathPlanetmathPlanetmath at a point P and P is an inflection point of C, then H(P)=0, where H is the Hessian determinant.

To make our proof purely algebraic, we replace the use of the implicit function theorem to obtain f with an expansion in a formal power series. As above, we choose our x,y,z coordinates so as to place P at (0,0,1) and make C tangent to the line y=0 at P. Then, since P is a regular point of C, we may parameterize C by a formal power series f(t)=k=0cktk such that F(t,f(t),1)=0. Then, if we define derivativesPlanetmathPlanetmath algebraically (http://planetmath.org/DerivativeOfPolynomial), we may proceed with the rest of the proof exactly as above.

Title Hessian and inflexion points
Canonical name HessianAndInflexionPoints
Date of creation 2013-03-22 18:22:26
Last modified on 2013-03-22 18:22:26
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 14
Author rspuzio (6075)
Entry type Theorem
Classification msc 53A04
Classification msc 26A51