A quartic surface in $\mathbb{P}^3$ is said to be a "symmetroid" if its equation is obtained as the determinant of a 4x4 symmetric matrix of linear forms. It is well known that the general symmetroid has 10 nodes and the family of such surfaces seem to have the same dimension as the moduli space of 10 points in $\mathbb{P}^3$.

QUESTION: is any general set of 10 points in $\mathbb{P}^3$ the singular locus of a symmetroid? Or better, how can one detect whether such a point set comes from a symmetroid?


The answer is no.

In fact, let $\Gamma \subset \mathbb{P}^3$ be the set of $10$ nodes of a general quartic symmetroid. Then the Gale transform $\Gamma' \subset \mathbb{P}^5$ of $\Gamma$ must be a zero dimensional scheme of length $10$ given by the simple intersection points of two Veronese surfaces, and this condition is not satisfied by a collection of $10$ general points.

Another characterization is the following: the set $\Gamma$ arises as a linear section of the secant variety $\textrm{Sec}(C_6)$, where $C_6 \subset \mathbb{P}^6$ is a rational normal curve. Note that $\textrm{Sec}(C_6)$ has precisely degree $10$, since $10$ is the number of nodes of a general projection of $C_6$ to $\mathbb{P}^2$.

For (many) more details, see

(1) D. Eisenbud, S. Popescu: The projective geometry of the Gale transform, Journal of Algebra 230 (2000), 127-173,

in particular Example 6.3 p. 153.

(2) D. Eisenbud, K. Hulek and S. Popescu: A note on the intersection of Veronese surfaces, Commutative Algebra, Singularities and Computer Algebra, Volume 115 of the series NATO Science Series (2003), pp 127-139,

in particular Section 3.


No, a general set of 10 points in $\mathbb P^3$ do not form the singular locus of a symmetroid. This goes back to the papers of Cayley on quartic surfaces. If I remember correctly, to construct a symmetroid set, one can fix 7 of the points in $\mathbb P^3$ freely, then the 8th can be chosen in a 2-dimensional family $S$ (a "Cayley surface"), and the ninth along a curve on $S$ (and then the 10th point is uniquely determined for it to be a symmetroid). This is explained in Jessop's book 'Quartic surfaces with singular points'.

One can see that the nodes are special position as follows. Given a quartic surface $X\subset\mathbb P^3$ with nodes at $p_0,\ldots,p_9$ one can project from one of the nodes, say $p_0$ to get a double cover of the plane branched along a degree 6 curve $C$. If $X$ is a symmetroid, one can check that this curve is reducible and is generically a union of two cubic curves. It was showed by Cayley (and more recently in this paper) that also the converse holds - a quartic is a symmetroid if and only if the degree 6 curve splits into two cubics. Note that the images of the points $p_1,\ldots,p_9$ are the intersection of these cubics. So these 9 points are special in the sense that there is a pencil of cubics passing through them.


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