Checking if a point is in the secant variety

Question

Let $X \subset {\mathbb P}^n$ be a projective variety with an explicitly given defining ideal $I$. How can we check if the point $(1:0:\cdots:0)$ belong to the secant variety of $X$?

Answer 1

This would be really easy to check if you had the generators of the ideal of the secant variety, or even set-theoretic defining equations.

Perhaps you can do something like we did here: http://arxiv.org/abs/1212.1515

In particular we showed that the secant to the Segre (a nice toric variety) is covered by open normal toric varieties, so locally it is defined by nice binomial equations (in cumulants). So to test if a point is on that secant variety, you can just check the local defining equations.

Maybe you can express the secant of your toric variety in secant cumulants and then the defining equations will be easy to recognize and use?

Answer 2

This is the answer to the question that arose in the comments -- how to compute the defining equation of the tangent hypersurface $Y$ to an $r$-dimensional toric variety $X$ in $\mathbb{P}^n$, where $n=2r+1$. This is not an answer to the original question, how to determine whether a point is secant, although the two questions are related: if $X$ is not secant defective, then every point of $\mathbb{P}^n\setminus Y$ is a second point (of course many points of $Y$ will also be secant points, typically).

Just to clarify, the dense torus in the toric variety $X$ will be identified as $\mathbb{G}_m^r$ with coordinates $(t_1,\dots,t_r)$. Let $\mathbb{P}^n$ have homogeneous coordinates $[x_0,x_1,\dots,x_n]$.
For a monomial morphism, $$f:\mathbb{G}_m^r \to \mathbb{P}^n, \ f^*x_i = t_1^{e_{i,1}}\cdots t_r^{e_{i,r}}, \ e_{i,j}\in \mathbb{Z},$$ the morphism does not change if we multiply every $f^*x_i$ by an invertible monomial. Thus, after multiplying, assume that every $e_{i,j}$ is nonnegative, and for every $j=1,\dots,r$, there exists $i$ with $e_{i,j}=0$. In other words, the matrix $E=(e_{i,j})$ is a matrix of nonnegative integers, and every column contains at least one zero entry. Denote by $e_i$ the sum of entries in the $i^{\text{th}}$ row, $$e_i = e_{i,1}+\dots+e_{i,r}.$$ Also denote by $e$ the maximum of $e_1,\dots,e_n$.
Finally, introduce a dummy variable $t_0$ and define nonnegative $e_{i,0} = e-e_i$, so that all of the terms $$g_i = t_0^{e_{i,0}}t_1^{e_{i,1}}\cdots t_r^{e_{i,r}}, \ g_i \in k[t_0,\dots,t_r]$$ are homogeneous of degree $e$, and for every $j=0,1,\dots,r$, there exists $i$ with $e_{i,j}=0$. If we think of $\mathbb{G}_m^r$ with its standard embedding in $\mathbb{A}^r$, and if we think of $\mathbb{A}^r$ as $D_+(t_0)$ inside $\mathbb{P}^r$, then the rational transformation $$ [g_0,g_1,\dots,g_n]:\mathbb{P}^r \dashrightarrow \mathbb{P}^n,$$ is the maximal extension of $f$ to a rational transformation. Denote by $\widetilde{E}$ the matrix $E$ augmented to include the column $(e_{i,0})$. This is a matrix of nonnegative integers, and every column contains at least one zero entry.

Now introduce variables $(u_0,u_1,\dots,u_r)$. The tangent bundle of $\mathbb{A}^r$ is canonically isomorphic to $\mathbb{A}^r\times \mathbb{A}^r$, and the coordinates are $((t_1,\dots,t_r),(u_1,\dots,u_r)$. Of course $\mathbb{A}^r\times \mathbb{A}^r$ sits as $D_+(t_0)\times D_+(u_0)$ in $\mathbb{P}^r\times \mathbb{P}^r$, which has homogeneous coordinates as follows, $$([t_0,t_1,\dots,t_r],[u_0,u_1,\dots,u_r])\in \mathbb{P}^r\times \mathbb{P}^r.$$
The morphism $f$ extends to a tangent morphism, $$ \widetilde{f}:\mathbb{G}_m^r \times \mathbb{A}^r \to \mathbb{P}^n, \ \widetilde{f}^*x_i = t_1^{e_{i,1}}\cdots t_r^{e_{i,r}}(1+e_{i,1}u_1+\dots+e_{i,r}u_r). $$ There is a maximal extension of this morphism to a rational transformation, $$ \widetilde{g}: \mathbb{P}^r\times \mathbb{P}^r \dashrightarrow \mathbb{P}^n, $$ where each $\widetilde{g}^*x_i$ is the bihomogeneous polynomial, $$ \widetilde{g}_i = t_0^{e_{i,0}}t_1^{e_{i,1}}\cdots t_r^{e_{i,r}}(u_0+e_{i,1}u_1+\dots+e_{i,r}u_r) \in k[t_0,t_1,\dots,t_r,u_0,u_1,\dots,u_r]_{(e,1)}. $$ Altogether, this defines a graded ring homomorphism, $$ \widetilde{g}^*:k[x_0,\dots,x_n] \to k[t_0,t_1,\dots,t_r,u_0,u_1,\dots,u_r]. $$

Now I will assume that $n$ equals $2r+1$ and that the closed subvariety $\overline{\text{Image}(f)}$ is not "tangent defective", i.e., the tangent variety is a hypersurface. Then the kernel of the ring homomorphism $\widetilde{g}^*$ is a principal ideal, and a generator is a defining equation of the tangent hypersurface.

In particular, for the example in the comments, we have $r=2$, $n=5$, and $$ g_0=t_0t_1^2t_2, \ g_1=t_0t_1t_2^2, \ g_2=t_0^2t_1t_2, \ g_3=t_1^4, \ g_4=t_2^4, \ g_5=t_0^4. $$ Thus we have, $$ \widetilde{g}_0=t_0t_1^2t_2(u_0+2u_1+u_2), \ \widetilde{g}_1=t_0t_1t_2^2(u_0+u_1+2u_2), $$ $$\widetilde{g}_2=t_0^2t_1t_2(u_0+u_1+u_2), \ \widetilde{g}_3=t_1^4(u_0+4u_1), \ \widetilde{g}_4=t_2^4(u_0+4u_2), \ \widetilde{g}_5=t_0^4u_0. $$ When I plug this into Macaulay2, after a few minutes it gives me a degree $24$ defining equation of the tangent hypersurface. The formula is too large to include here, but it is included (commented) in the Macaulay2 script available at the following URL.
http://www.math.sunysb.edu/~jstarr/toric.m2