# Set up

The Salmon's invarinat of degree 100 defines the so-called Eckardt hypersurface in $\mathbb{P}^{4}$. This hypersurface parametrizes the cubic surfaces with an Eckardt point. What are the cubic surfaces lying on the singular locus of this hypersurface?

# Solution

The singular locus of the Eckardt hypersurface determines two 2-dimensional rational family of cubic surfaces, intersecting in a rational curve.

In fact, the two rational families are given by Sylvester forms as follows:

$ax_0^3+bx_1^3+bx_2^3+cx_3^3+cx_4^3,\ \ x_0+x_1+x_2+x_3+x_4=0, \ \ \ a,b,c\in \mathbb{C}$

and

$ax_0^3+bx_1^3+cx_2^3+cx_3^3+cx_4^3,\ \ x_0+x_1+x_2+x_3+x_4=0, \ \ \ a,b,c\in \mathbb{C}$

Naturally, the two families intersect in a $\mathbb{P}^1$.

More precisely, the singular locus of the Eckardt hypersuarface has 30 linear irreducible components of dimension 2 and 3.

Five components correspond to the cubic surfaces with a degenerate Sylvester form, and each is a copy of $\mathbb{P}^3$.

Therefore, considering the natural dominant morphism $\mathbb{P}^4 \longrightarrow \mathbb{P}(1,2,3,4,5)$ the five components are contracted

to the point $(1:0:0:0:0)$.

The 25 residual components are copies of $\mathbb{P}^2$, and are divided in two groups of cardinality 10 and 15.

The components in each group are mapped to the same rational surfaces inside $\mathbb{P}(1,2,3,4,5)$. Therefore, the singular locus

is mapped to a reducible surface in $\mathbb{P}(1,2,3,4,5)$ with two rational components intersecting in a rational curve.