Equations of the secant variety

Question

Let $X\subset\mathbb{P}^N$ be an irreducible nondegenerate (i.e. not contained in a hyperplane) projective complex algebraic variety, and let $\mathrm{Sec}(X)$ be the secant variety of $X$ (i.e. the union of all secant lines of $X$). Further suppose that the homogeneous ideal of $X$ is generated by forms $F_1,\ldots,F_m$ of degree two.

How do I write the equations of $\mathrm{Sec}(X)$ as a function of $F_0,\ldots,F_m$ ?

Let us focus attention on the simplest case, i.e. when $\mathrm{Sec}(X)$ is a cubic hypersurface. Then we have $\mathrm{Sec}(X)=V(G(x_0,\ldots,x_N))$, where $G(x_0,\ldots,x_N)=\sum_{i=1}^m L_i(x_0,\ldots,x_N) F_i(x_0,\ldots,x_N)$. The linear forms $L_0,\ldots,L_N$ depend only by $F_0,\ldots,F_N$, but how?

Thanks.

Answer 1

The computation of explicit equations of secant varieties is hard in general. However, things are simpler when one works with determinantal varieties, i.e. varieties whose equations are given by some minors of a matrix of homogeneous forms.

In fact, there is the following result, that one can find in [Harris, Algebraic Geometry, p. 145]:

Proposition. Let $M$ be the projective space of $m \times n$ matrices and let $M_k \subset M$ be the subvariety of matrices of rank at most $k$. Assume that $2k < \min\{m, n\}$. Then the secant variety $S(M_k)$ is equal to the subariety $M_{2k} \subset M$ of matrices of rank at most $2k$.

As an example, let us consider the Veronese surface $X \subset \mathbb{P}^5$, whose equations are the $2 \times 2$ minors of the matrix $$ M:=\pmatrix{z_0 & z_3 & z_4 \cr z_3 & z_1 & z_5 \cr z_4 & z_5 & z_2}.$$ In this case $m=n=3$ and $k=1$, that is $X=M_1$. Then the secant variety $S(X)$ coincides with the determinantal variety $M_2$, i.e. with the cubic hypersurface defined by $$\det M=0.$$ This is essentially due to the fact that the linear combination of two rank $1$ matrices can have rank at most $2$.

Developing the determinant along any line or column, you can find an expression of the equation of $S(X)$ as a linear combination of generators of the homogeneous ideal $I(X)$.

Moreover, in the same way one can find equations for the secant varieties of rational normal curves, see [Harris, Algebraic geometry, p. 103].

For a deeper treatment of this problem, see the paper by Ottaviani and Landsberg Equations for secant varieties of Veronese and other varieties, arXiv:1111.4567.

Answer 2

If $I(X)$, the ideal of $X$ is empty in degrees less than $d$, then there can be no equations of the secant variety until degree $d+1$, and the ideal in degree $d+1$ consists of all polynomials $P$ such that all partials of $P$ are in $I_d(X)$. There is a similar description for the ideal of the secant variety in higher degrees which I call "multi-prolongation", but one does not know when one has generators for the ideal by this method, and it becomes very difficult to compute. In your case, since you have a hypersurface the termination problem does not arise - your cubic is the unique cubic all of whose partial derivatives are in the ideal of $X$.