# 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:

$$H=\left|\begin{array}{ccc}\hfill \frac{{\partial}^{2}F}{\partial {x}^{2}}\hfill & \hfill \frac{{\partial}^{2}F}{\partial x\partial y}\hfill & \hfill \frac{{\partial}^{2}F}{\partial x\partial z}\hfill \\ \hfill \frac{{\partial}^{2}F}{\partial y\partial x}\hfill & \hfill \frac{{\partial}^{2}F}{\partial {y}^{2}}\hfill & \hfill \frac{{\partial}^{2}F}{\partial y\partial z}\hfill \\ \hfill \frac{{\partial}^{2}F}{\partial z\partial x}\hfill & \hfill \frac{{\partial}^{2}F}{\partial z\partial y}\hfill & \hfill \frac{{\partial}^{2}F}{\partial {z}^{2}}\hfill \end{array}\right|$$ |

###### 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^{} $(-\u03f5,\u03f5)$ and a function
$f:(-\u03f5,\u03f5)\to \mathbb{R}$ 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 ${f}^{\prime \prime}(0)=0$:

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

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

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:

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

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

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

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

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

$H(0,0,1)$ | $=\left|\begin{array}{ccc}\hfill -{f}^{\prime \prime}(0)\frac{\partial F}{\partial y}(0,0,1)\hfill & \hfill \frac{{\partial}^{2}F}{\partial x\partial y}(0,0,1)\hfill & \hfill 0\hfill \\ \hfill \frac{{\partial}^{2}F}{\partial x\partial y}(0,0,1)\hfill & \hfill \frac{{\partial}^{2}F}{{\partial}^{2}y}(0,0,1)\hfill & \hfill (n-1)\frac{\partial F}{\partial y}(0,0,1)\hfill \\ \hfill 0\hfill & \hfill (n-1)\frac{\partial F}{\partial y}(0,0,1)\hfill & \hfill 0\hfill \end{array}\right|$ | ||

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

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 |