How Do Inflection Points Intersect with the Hessian on Algebraic Curves?

[evinda]

Hi! (Smile)

Let the algebraic curve $f(x_0, x_1, x_2) \in K[x_0, x_1, x_2]$. The inflection points are the non-singular points of the curve that are the intersection points with the hessian.

If we have the curve $x^3+y^3+z^3=0$ the hessian is equal to $216 \cdot x \cdot y \cdot z$. How can we find the non-singular points of the curve that are the intersection points with the hessian? (Thinking)

 

[jvicens]

Hello! It's great to see your interest in algebraic curves and their inflection points. To answer your question, we can use the concept of partial derivatives to find the non-singular points of the curve that intersect with the hessian.

First, let's rewrite the given curve as $f(x_0, x_1, x_2) = x_0^3 + x_1^3 + x_2^3 = 0$. Now, we can take partial derivatives of this function with respect to each variable, giving us:

$\frac{\partial f}{\partial x_0} = 3x_0^2$
$\frac{\partial f}{\partial x_1} = 3x_1^2$
$\frac{\partial f}{\partial x_2} = 3x_2^2$

Next, we can set these partial derivatives equal to 0 and solve for $x_0, x_1, x_2$ to find the critical points of the curve. In this case, we get $x_0 = x_1 = x_2 = 0$.

Now, we can use the hessian matrix, which is the matrix of second partial derivatives of our function, to determine the type of critical point at $(x_0, x_1, x_2) = (0,0,0)$. The hessian matrix for our curve is:

$H = \begin{bmatrix} 6x_0 & 0 & 0 \\ 0 & 6x_1 & 0 \\ 0 & 0 & 6x_2 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

This matrix has a determinant of 0, which means that it is a singular point on the curve. In other words, the curve does not have any non-singular points that intersect with the hessian.

I hope this helps answer your question. Keep exploring and asking questions about algebraic curves – they are fascinating objects to study!