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.
Best Answer
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 $X$.