Hessian and inflexion points

Theorem 1.

Suppose that $C$ is a curve in the real projective plane $\mathrm{R}\mathit{}{\mathrm{P}}^{\mathrm{2}}$ given by a homogeneous equation $F\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{,}z\mathrm{)}\mathrm{=}\mathrm{0}$ of degree of homogeneity (http://planetmath.org/HomogeneousFunction) $n$. If $F$ has continuous first derivatives in a neighborhood of a point $P$ and the gradient of $F$ is non-zero at $P$ and $P$ is an inflection point of $C$, then $H\mathit{}\mathrm{(}P\mathrm{)}\mathrm{=}\mathrm{0}$, where $H$ is the Hessian determinant:

Proof.

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 tangent to $C$ at $P$ is $y=0$. Using the implicit function theorem, we may conclude that there exists an interval $(-ϵ,ϵ)$ and a function $f:(-ϵ,ϵ)\to ℝ$ such that $F(t,f(t),1)=0$ when . 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 $f^{\prime }(0)=0$. Because $P$ is an inflection point, we also have $f^{\prime \prime }(0)=0$.

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

$0={\displaystyle \frac{d}{dt}}F(t,f(t),1)$ $={\displaystyle \frac{\partial F}{\partial x}}(t,f(t),1)+f^{\prime }(t){\displaystyle \frac{\partial F}{\partial y}}(t,f(t),1)$

$0={\displaystyle \frac{d^2}{d{t}^2}}F(t,f(t),1)$ $={\displaystyle \frac{{\partial }^{2}F}{\partial x^2}}(t,f(t),1)+f^{\prime }(t){\displaystyle \frac{{\partial }^{2}F}{\partial x\partial y}}(t,f(t),1)$

$\mathrm{\hspace{1em}}+{\left(f^{\prime }(t)\right)}^2{\displaystyle \frac{{\partial }^{2}F}{\partial y^2}}(t,f(t),1)+f^{\prime \prime }(t){\displaystyle \frac{\partial F}{\partial y}}(t,f(t),1)$

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 $f^{\prime \prime }(0)=0$:

${\displaystyle \frac{\partial F}{\partial x}}(0,0,1)$ $=0$

${\displaystyle \frac{{\partial }^{2}F}{\partial x^2}}(0,0,1)$ $=-f^{\prime \prime }(0){\displaystyle \frac{\partial F}{\partial y}}(0,0,1)$

Since $F$ is homogenous, Euler’s formula holds:

$$x\frac{\partial F}{\partial x}+y\frac{\partial F}{\partial y}+z\frac{\partial F}{\partial z}=nF$$

Taking partial derivatives, we obtain the following:

$x{\displaystyle \frac{{\partial }^{2}F}{\partial x^2}}+y{\displaystyle \frac{{\partial }^{2}F}{\partial x\partial y}}+z{\displaystyle \frac{{\partial }^{2}F}{\partial x\partial z}}=(n-1){\displaystyle \frac{\partial F}{\partial x}}$

$x{\displaystyle \frac{{\partial }^{2}F}{\partial x\partial y}}+y{\displaystyle \frac{{\partial }^{2}F}{\partial y^2}}+z{\displaystyle \frac{{\partial }^{2}F}{\partial y\partial z}}=(n-1){\displaystyle \frac{\partial F}{\partial y}}$

$x{\displaystyle \frac{{\partial }^{2}F}{\partial x\partial z}}+y{\displaystyle \frac{{\partial }^{2}F}{\partial y\partial z}}+z{\displaystyle \frac{{\partial }^{2}F}{\partial z^2}}=(n-1){\displaystyle \frac{\partial F}{\partial z}}$

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

${\displaystyle \frac{\partial F}{\partial z}}(0,0,1)$ $=0$

${\displaystyle \frac{{\partial }^{2}F}{\partial x\partial z}}(0,0,1)$ $=0$

${\displaystyle \frac{{\partial }^{2}F}{\partial y\partial z}}(0,0,1)$ $=(n-1){\displaystyle \frac{\partial F}{\partial y}}(0,0,1)$

${\displaystyle \frac{{\partial }^{2}F}{\partial z^2}}(0,0,1)$ $=0$

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

$H(0,0,1)$

$={(n-1)}^2{\left({\displaystyle \frac{\partial F}{\partial y}}(0,0,1)\right)}^2{f}^{\prime \prime }(0)$

Since $P$ is an inflection point, $f^{\prime \prime }(0)=0$, 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 $\partial F/\partial x=0$ and $\partial F/\partial z=0$ by the way we set up our coordinate system, we must have $\partial F/\partial y\ne 0$. Thus, we see that, if $n\ne 1$, then $H(0,0,1)=0$ if and only if $f^{\prime \prime }(0)$. 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 $f^{\prime \prime }(0)=0$ but that the sign of $f^{\prime \prime }(t)$ change as $t$ passes through $0$.

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

Theorem 2.

Suppose that $C$ is a curve in the real projective plane $\mathrm{R}\mathit{}{\mathrm{P}}^{\mathrm{2}}$ given by an equation $F\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{,}z\mathrm{)}\mathrm{=}\mathrm{0}$ where $F$ is a homogenous polynomial of degree $n$. If $C$ is regular at a point $P$ and $P$ is an inflection point of $C$, then $H\mathit{}\mathrm{(}P\mathrm{)}\mathrm{=}\mathrm{0}$, 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 $f(t)={\sum }_{k=0}^{\mathrm{\infty }}{c}_k{t}^k$ such that $F(t,f(t),1)=0$. Then, if we define derivatives 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