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 homogeneousequation $F\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{,}z\mathrm{)}\mathrm{=}\mathrm{0}$ ofdegree of homogeneity (http://planetmath.org/HomogeneousFunction) $n$.If $F$ has continuous first derivatives in a neighborhood ofa 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 $H$ isthe Hessian determinant:

Proof.

We may choose a system $x,y,z$ of homogenous coordinatessuch that the point $P$ lies at $(0,0,1)$ and the equationof the tangent to $C$ at $P$ is $y=0$. Using theimplicit function theorem, we may conclude that thereexists an interval $(-ϵ,ϵ)$ and a function$f:(-ϵ,ϵ)\to ℝ$ such that$F(t,f(t),1)=0$ when . Inother words, the portion of curve near $P$ may be describedin non-homogenous coordinates by $y=f(x)$. By the waythe 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 beexplained later, we do not yet want to make use of thefact 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 theequations 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 wehave $H(0,0,1)=0$.∎

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 coordinatesystem, 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 doesnot mean that the Hessian vanishes if and only if $P$ isan inflection point since the definition of inflectionpoint not only requires that $f^{\prime \prime }(0)=0$ but that thesign 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 purelyalgebraic and avoid introducing analysis where possible,so a variant of this result is preferred. The theoremmay 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 pointof $C$, 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 ofthe implicit function theorem to obtain $f$ with an expansionin 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 regularpoint 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 wedefine derivatives algebraically (http://planetmath.org/DerivativeOfPolynomial),we may proceed with the rest of the proof exactly as above.