Cubic surfaces on the singular locus of Eckardt hypersurafce

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.

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