Past exam of the geometry course of the University of Oxford / 2009 / B / 5

An inflection point of an algebraic plane curve is a smooth point where the tangent has intersection multiplicity at least three. For a homogeneous defining polynomial $P$ of degree $d$, its Hessian curve is defined by the determinant of the matrix of second partial derivatives, of degree $3(d-2)$. In characteristic zero, the smooth inflection points are exactly the smooth points of $C\cap \mathrm{Hess}(C)$.

For a local verification, move a smooth point to $[0:0:1]$ and its tangent to $y=0$, and scale so $P(x,y,1)=y+a{x}^2+bxy+c{y}^2+\cdots$. The tangent contact is at least three exactly when $a=0$. The homogeneous Euler identities give the Hessian matrix at that point as $\left(\begin{array}{ccc}2a& b& 0\\ b& 2c& d-1\\ 0& d-1& 0\end{array}\right)$, whose determinant is $-2a(d-1{)}^2$. This proves the criterion for $d>1$.

For the given cubic, the singularity equations are $x^2+\lambda yz=y^2+\lambda xz=z^2+\lambda xy=0$. No coordinate can vanish at a projective solution, since that would force all three to vanish. Multiplying the equations therefore gives $1=(-\lambda {)}^3$, so singularity requires ${\lambda }^3=-1$. Conversely, under that condition the point $[1:1:-1/\lambda ]$ satisfies all three equations, and the Euler identity places it on the cubic. Thus

The second-derivative matrix isIts determinant is $54[(4+{\lambda }^3)xyz-{\lambda }^2(x^3+y^3+z^3)]$. Hence the Hessian curve has equationIf $\lambda =0$, this is the union $xyz=0$ of three lines and is singular. If $\lambda \ne 0$, divide by ${\lambda }^2$ and write it as a Hesse cubic with parameter $\mu =-({\lambda }^3+4)/(3{\lambda }^2)$. The preceding singularity criterion gives ${\mu }^3=-1$, equivalentlyThus the complete condition isThe other values beyond the source's first family are zero and the three cube roots of eight.

On the nonsingular cubic, $x^3+y^3+z^3=-3\lambda xyz$. Substituting this into the Hessian equation leaves $-4({\lambda }^3+1)xyz=0$. Since ${\lambda }^3+1\ne 0$, a flex has a zero coordinate. The remaining cubic equation then gives the nine pointswhere $\omega =e^{2\pi i/3}$. All are distinct, smooth, and lie on both the cubic and its Hessian curve, so they are exactly the inflection points. In particular the nine flexes are independent of $\lambda$ within the nonsingular family. This is the flex configuration of the Hesse pencil.