PlanetMath: Hessian and inflexion points

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 $-\epsilon < t < \epsilon$ . 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 $f'(0) = 0$ . 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:

$\displaystyle 0 = {d \over dt} F(t, f(t), 1)$	$\displaystyle = {\partial F \over \partial x} (t, f(t), 1) + f'(t) {\partial F \over \partial y} (t, f(t), 1)$
$\displaystyle 0 = {d^2 \over dt^2} F(t, f(t), 1)$	$\displaystyle = {\partial^2 F \over \partial x^2} (t, f(t), 1) + f'(t) {\partial^2 F \over \partial x \partial y} (t, f(t), 1)$
	$\displaystyle \quad+ \left(f'(t)\right)^2 {\partial^2 F \over \partial y^2} (t, f(t), 1) + f''(t) {\partial F \over \partial y} (t, f(t), 1)$

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 $f''(0) = 0$ :

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

Since $F$ is homogenous, Euler's formula holds: $$ x {\partial F \over \partial x} + y {\partial F \over \partial y} + z {\partial F \over \partial z} = n F $$ Taking partial derivatives, we obtain the following:

$\displaystyle x {\partial^2 F \over \partial x^2} + y {\partial^2 F \over \part... ...artial^2 F \over \partial x \partial z} = (n - 1) {\partial F \over \partial x}$
$\displaystyle x {\partial^2 F \over \partial x \partial y} + y {\partial^2 F \o... ...artial^2 F \over \partial y \partial z} = (n - 1) {\partial F \over \partial y}$
$\displaystyle x {\partial^2 F \over \partial x \partial z} + y {\partial^2 F \o... ...} + z {\partial^2 F \over \partial z^2} = (n - 1) {\partial F \over \partial z}$

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:

$\displaystyle H (0,0,1)$	$\displaystyle = \left\vert \begin{matrix}- f''(0) {\partial F \over \partial y}... ... 0 & (n - 1) {\partial F \over \partial y} (0,0,1) & 0 \end{matrix} \right\vert$
	$\displaystyle = (n - 1)^2 \left( {\partial F \over \partial y} (0,0,1) \right)^2 f'' (0)$

Since $P$ is an inflection point, $f''(0) = 0$ , so we have $H(0,0,1) = 0$ . $\qedsymbol$

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 $f''(0)$ . 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 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:

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, we may proceed with the rest of the proof exactly as above.