# Hessian and inflexion points

###### Theorem 1.

Suppose that $C$ is a curve in the real projective plane $\mathbb{R}\mathbb{P}^{2}$ given by a homogeneous equation $F(x,y,z)=0$ 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 $H(P)=0$, where $H$ is the Hessian determinant:

 $H=\left|\begin{matrix}{\partial^{2}F\over\partial x^{2}}&{\partial^{2}F\over% \partial x\partial y}&{\partial^{2}F\over\partial x\partial z}\\ {\partial^{2}F\over\partial y\partial x}&{\partial^{2}F\over\partial y^{2}}&{% \partial^{2}F\over\partial y\partial z}\\ {\partial^{2}F\over\partial z\partial x}&{\partial^{2}F\over\partial z\partial y% }&{\partial^{2}F\over\partial z^{2}}\end{matrix}\right|$
###### 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   $(-\epsilon,\epsilon)$ and a function $f\colon(-\epsilon,\epsilon)\to\mathbb{R}$ such that $F(t,f(t),1)=0$ when $-\epsilon. 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:

 $\displaystyle 0={d\over dt}F(t,f(t),1)$ $\displaystyle={\partial F\over\partial x}(t,f(t),1)+f^{\prime}(t){\partial F% \over\partial y}(t,f(t),1)$ $\displaystyle 0={d^{2}\over dt^{2}}F(t,f(t),1)$ $\displaystyle={\partial^{2}F\over\partial x^{2}}(t,f(t),1)+f^{\prime}(t){% \partial^{2}F\over\partial x\partial y}(t,f(t),1)$ $\displaystyle\quad+\left(f^{\prime}(t)\right)^{2}{\partial^{2}F\over\partial y% ^{2}}(t,f(t),1)+f^{\prime\prime}(t){\partial F\over\partial y}(t,f(t),1)$

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

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

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

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

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

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

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

 $\displaystyle H(0,0,1)$ $\displaystyle=\left|\begin{matrix}-f^{\prime\prime}(0){\partial F\over\partial y% }(0,0,1)&{\partial^{2}F\over\partial x\partial y}(0,0,1)&0\\ {\partial^{2}F\over\partial x\partial y}(0,0,1)&{\partial^{2}F\over\partial^{2% }y}(0,0,1)&(n-1){\partial F\over\partial y}(0,0,1)\\ 0&(n-1){\partial F\over\partial y}(0,0,1)&0\end{matrix}\right|$ $\displaystyle=(n-1)^{2}\left({\partial F\over\partial y}(0,0,1)\right)^{2}f^{% \prime\prime}(0)$

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 $\partial F/\partial y\neq 0$. Thus, we see that, if $n\neq 1$, 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 $0$.

###### Theorem 2.

Suppose that $C$ is a curve in the real projective plane $\mathbb{R}\mathbb{P}^{2}$ given by an equation $F(x,y,z)=0$ 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 $H(P)=0$, 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 $f(t)=\sum_{k=0}^{\infty}c_{k}t^{k}$ 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 HessianAndInflexionPoints 2013-03-22 18:22:26 2013-03-22 18:22:26 rspuzio (6075) rspuzio (6075) 14 rspuzio (6075) Theorem msc 53A04 msc 26A51