This is a reference request/nomenclature question. Let $A \subseteq \mathbb{P}^n$ be a finite set of points not contained in a hyperplane (over some field), and let $\sigma_r(A)$ be the $r$-th secant variety to $A$. This secant variety forms a subspace arrangement, i.e., a finite union of linear subspaces of $\mathbb{P}^n$. Is there a specific name for subspace arrangements of this form? Surely such arrangements have been studied, and I would be very grateful for a reference.
Secant Variety to a Zero-Dimensional Projective Variety
ag.algebraic-geometryhyperplane-arrangementsprojective-varietiesreference-request
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It is true that $Z$ spans $L$ — even if $X$ isn't ACM. You can also allow $X$ to be singular (but you do need $X$ irreducible and non-degenerate, of course). To illustrate one of the main ideas it is useful to first look at the case when $X$ is a curve.
If $X$ is a curve. Let $M$ be the span of $Z$ and suppose that $M\neq L$. (In the curve case, $L$ will be a hyperplane). Let $p$ be any point of $X$ outside of $Z$ and let $H$ be any hyperplane containing $M$ and $p$. Then $H\cap X$ contains at least $d+1$ points, so by Bezout's theorem the intersection cannot be zero dimensional. Since $X$ is irreducible and one dimensional, this means that the intersection must be all of $X$, so $X$ is contained in $H$, contrary to hypothesis.
The general case. The idea when $k\geqslant 2$ is to show that if $H$ is a general hyperplane containing $L$ then $H \cap X$ is irreducible and non-degenerate (i.e, the intersection $H\cap X$ is not contained in a smaller linear space of $H$). But now all dimensions have been reduced by $1$, and so iterating this procedure reduces us to the curve case, which we've already solved.
To set this up, note that hyperplanes in $\mathbb{P}^n$ containing $L$ are parameterized by a $\mathbb{P}^{k-1}$ (If $V$ is the underlying vector space of $\mathbb{P}^{n}$, $W$ the underlying vector space of $L$, then the hyperplanes are parameterized by the projectivization of $(V/W)^{*}$). We'll use $H$ to refer both to a point of $\mathbb{P}^{k-1}$ and the corresponding hyperplane in $\mathbb{P}^n$ containing $L$. Define $\Gamma\subset \mathbb{P}^{k-1}\times (X\setminus Z)$ to be the set
$$\Gamma = \left\{(H,p) \mid p\in H\right\}$$
i.e, the pairs $(H,p)$ so that $H$ is a hyperplane containing $L$, and $p$ a point of $H\cap X$ not on $Z$.
If we fix $p$, then the set of possible $H$'s satisfying this condition are simply the hyperplanes $H$ containing the span of $L$ and $p$, and this is parameterized by a $\mathbb{P}^{k-2}$. In other words, $\Gamma$ is a $\mathbb{P}^{k-2}$ bundle over $X\setminus Z$. (This fibration is where we use $k\geqslant 2$.) Since $X\setminus Z$ is irreducible this implies that $\Gamma$ is irreducible.
Let $\overline{\Gamma}$ be the Zariski-closure of $\Gamma$ in $\mathbb{P}^{k-1}\times X$. Then $\overline{\Gamma}$ is irreducible since $\Gamma$ is. For a fixed $H\in \mathbb{P}^{k-1}$ the fibre of the projection $\overline{\Gamma}\longrightarrow \mathbb{P}^{k-1}$ over $H$ is simply the intersection $X\cap H$, of dimension $k-1$.
Now let $q$ be any point of $Z$. Then $q\in X\cap H$ for every $H\in \mathbb{P}^{k-1}$ so $q$ gives a section of $\overline{\Gamma}\longrightarrow\mathbb{P}^{k-1}$. Since $Z$ consists of $d$ distinct points where $d$ is the degree of $X$ we conclude that $q$ is a smooth point of $X$. Finally, since $Z$ is the intersection of all $X\cap H$ for $H\in \mathbb{P}^{k-1}$ this implies that the general intersection $X\cap H$ is smooth at $q$. Summarizing, we have a section of the map which generically lies in the smooth locus of the fibres. Since $\overline{\Gamma}$ is irreducible, this implies that the generic fibre is irreducible, i.e, if $H$ is a generic hyperplane containing $L$, then $H\cap X$ is irreducible.
(The intuitive reason for this implication is that, generically over $\mathbb{P}^{k-1}$ the section lets us pick out precisely one irreducible component of the fibre. The union of these components gives us a subset of $\overline{\Gamma}$ which has the same dimension as $\overline{\Gamma}$, and hence whose closure must be all of $\overline{\Gamma}$ by irreducibility. But if there is more than one component in a general fibre, this is a contradiction, thus the general fibre must be irreducible. To make this intuitive construction rigorous requires passing to the normalization of $\overline{\Gamma}$ and then looking at the Stein factorization of the map from the normalization to $\mathbb{P}^{k-1}$. The section gives a generic section of the finite part of the Stein factorization, and that allows one to construct the ``union of the components containing the section''.)
Finally, the same trick as in the curve case also shows us that for any hyperplane $H$, $H\cap X$ must be non-degenerate. Let $Y=H\cap X$, so that $Y$ is a variety of degree $d$ and dimension $k-1$. Let $M$ be the span of $Y$. If $M\neq H$ then pick any point $p\in X\setminus Y$ and let $H'$ be any hyperplane containing $M$ and $p$. Then $H'\cap X$ can't be all of $X$ (since this would contradict the non-degeneracy of $X$), so $Y'=H'\cap X$ must be a subvariety of dimension $k-1$ (more precisely, all components of $Y'$ have dimension $k-1$) and degree $d$. But $Y$ is therefore a component of $Y'$, and the equality of degrees tells us that $Y'$ can't have any other components so we must have $Y'=Y$. This contradicts the fact that $p\in Y'$ and $p\not\in Y$.
Together this shows the required inductive step: If $H$ is a general hyperplane containing $L$ then $H\cap X$ is irreducible and non-degenerate.
Other remarks. I'm guessing from the setup of the question that you want to apply the result for a particular $L$ that you have chosen. If, in the application, you're allowed to pick a general $L$ then you can say something stronger. The classical uniform position principle (where ''classical'' in this case means ''established by Joe Harris in the 80's'') states that for a general subspace $L$ of dimension $n-k$ the finite set of $d$-points in $Z=L\cap X$ have the property that any subset of $r+1$ of the points (with $r\leqslant n-k$) span a $\mathbb{P}^{r}$. Picking $r=n-k$, this means that any subset of $n-k+1$ points of $Z$ spans all of $L$, and so in particular $Z$ spans $L$. (Note that $d\geqslant n-k+1$; for instance, as a consequence of the argument above: if $d < n-k+1$ then the $d$ points of $Z$ would never span $L$.)
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 $X$.
Best Answer
I believe this would be a dual arrangement of a star arrangement.
A star arrangement is a union of subspaces defined as follows. Let $H_1,\dotsc,H_d$ be a collection of hyperplanes and fix an integer $c$. The codimension $c$ star arrangement $X_c$ is the union of intersections of $c$ of the $H_i$, over all size $c$ subsets of $H_1,\dotsc,H_d$. Usually there is some hypothesis of linear generality so that any $c$ of the $H_i$'s are independent. See for example https://arxiv.org/abs/1203.5685, https://arxiv.org/abs/1801.04579.
Given a subspace arrangement $\mathcal{A} = \{W_1,\dotsc,W_s\}$ of subspaces in $\mathbb{P}V$, the dual arrangement is $\mathcal{A}^* = \{W_1^\perp,\dotsc,W_s^\perp\}$ in $\mathbb{P}V^*$.
Well, the subspace $W = H_{i_1} \cap \dotsb \cap H_{i_c}$ is dual to $W^\perp = \operatorname{span}\{H_{i_1}^\perp,\dotsc,H_{i_c}^\perp\}$. So the secant varieties you're asking about are dual arrangements of star arrangements. I don't know a better or alternative name and I'm not aware of any work on these arrangements specifically, but the authors of the papers I linked might be able to give more information if you write to them.
By the way, these dual arrangements of star arrangements are not necessarily themselves star arrangements. Let $A$ be a set of $5$ general points in $\mathbb{P}^3$ so $\sigma_2(A)$ is a set of $10$ lines. This arrangement of lines has $4$ through each node (point where lines meet) and $3$ in each plane spanned by two meeting lines. This arrangement of lines isn't a star arrangement. A star arrangement given by pairwise intersections of $5$ general hyperplanes would again consist of $10$ lines, but this time with $4$ in each plane spanned by two meeting lines, and $3$ lines through each node.