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 $X$?

1$\begingroup$ Is $X$ smooth? If so, then you can check whether or not the projection from $p$ is singular. A computer algebra package such as Macaulay2 can compute defining equations of the projection. Then you take the Jacobian matrix of those equations, take the appropriate minors, put those together with your original equations (in characteristic $0$, sometimes unnecessary thanks to the Euler identity), and then compute the Hilbert polynomial of this singular locus  zero means that $p$ is not in the secant variety, anything else means that $p$ is in the secant variety. $\endgroup$– Jason StarrDec 1 '13 at 16:32

$\begingroup$ I'm looking for situation where $X$ might not be smooth, (the singularities are at most rational though). We can compute the secant variety as the elimination ideal of certain ruled join of variety as used by several authors. Since I'm just asking for one point, I'm wondering if there is something better to check. $\endgroup$– Thanh VuDec 1 '13 at 20:43

$\begingroup$ Can you give one example of an $X$ that you are considering? $\endgroup$– Jason StarrDec 1 '13 at 20:47

$\begingroup$ I'm interested in toric case. So for an example, that I computed with Macaulay2, X is a surface of codimension 3 given by $\endgroup$– Thanh VuDec 1 '13 at 20:55

$\begingroup$ $x_2^3  x_0*x_1*x_5,x_0*x_1*x_2x_3*x_4*x_5,x_1^3x_0*x_2*x_4,x_0^3x_1*x_2*x_3,x_1^2*x_2^2x_0^2*x_4*x_5,x_0^2*x_2^2x_1^2*x_3*x_5,x_0^2*x_1^2x_2^2*x_3*x_4$. $\endgroup$– Thanh VuDec 1 '13 at 21:03
This would be really easy to check if you had the generators of the ideal of the secant variety, or even settheoretic 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?

$\begingroup$ Thanks for your information. In some sense, what I have is a large collection of lattice points giving, (i.e., dimension is relatively small compared to codimension), I'd like to know if you can find a point corresponding to natural coordinates outside of the secant variety so I can project from that point. I'll see your reference. $\endgroup$– Thanh VuDec 7 '13 at 0:11
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 $e_{i,0} = ee_i$, 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

$\begingroup$ Thank you very much for your careful explanation. I'll study it more carefully. $\endgroup$– Thanh VuDec 3 '13 at 16:52

$\begingroup$ Unfortunately that polynomial has over 100 terms, so it is difficult to extrapolate any general features. $\endgroup$ Dec 3 '13 at 17:18

$\begingroup$ From the equation, the point $[1:0:...:0]$ is not on the tangent hypersurface. Also, can we check if $X$ is tangent defective or not? $\endgroup$– Thanh VuDec 3 '13 at 17:24

$\begingroup$ I should have said, $X$ will be tangent defective if and only if the kernel of $\widetilde{g}^*$ is not principal, i.e., it is tangent nondefective if and only if $\text{Ker}(\widetilde{g}^*)$ is principal. For your surface, since the kernel is principal, then it is tangent nondefective. $\endgroup$ Dec 3 '13 at 17:28

$\begingroup$ Sorry, I didn't pay much attention. In the case $n = 2r+1$, it is tangent nondefective iff it is of codimension $1$. $\endgroup$– Thanh VuDec 3 '13 at 18:00