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I think there must be a standard answer to this, for people in the know.

Let $X\subseteq\mathbb{A}^{n}$ be an affine (closed) variety of dimension $\geq d$, and fix some set $\{f_{i}\}_{i}$ of polynomials defining $X$. Let $\mathcal{G}=\mathrm{Gr}(n-d,\mathbb{A}^{n})$ be the Grassmannian of $(n-d)$-dimensional linear subspaces in $\mathbb{A}^{n}$. Since a generic $L\in\mathcal{G}$ is supposed to have non-empty intersection with $X$, my question is: is there a way to describe the closure of \begin{equation*} W=\{L\in\mathcal{G}|L\cap X=\emptyset\} \end{equation*} as a subvariety of $\mathcal{G}$ solely in terms of the $f_{i}$?

I would even settle for a description of any (closed) subvariety $W'\supseteq W$ inside $\mathcal{G}$, rather than $\overline{W}$ itself, as long as $\dim(W')<\dim(\mathcal{G})$.

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    $\begingroup$ Let $\overline{X}$ be the closure of $X$ in $\mathbb{P}^n$ and $Y$ be the intersection of $\overline{X}$ with the hyperplane $\mathbb{P}^{n-1}\cong\mathbb{P}^n-\mathbb{A}^n$. Further, let $G$ be the Grassmannian of $\mathbb{P}^{n-d}$'s in $\mathbb{P}^n$; $\mathcal{G}$ is an open subset of $G$. We have a closed subvariety $\tilde{W}$ of $G$ consisting of $M$ such that $M\cap\overline{X}\subset M\cap Y$. The variety $W$ should be its intersection with $\mathcal{G}$. $\endgroup$ – Kapil Apr 30 at 17:37
  • $\begingroup$ Thanks, I was also suggested in real life to think projectively: here we are saying that $M$ would intersect $X$ only at infinity, if I understand your idea correctly. But then I have the same standard question: why/how is $\tilde{W}$ itself a closed subvariety describable in terms of the $f_{i}$? $\endgroup$ – D. Dona May 2 at 12:47
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$L \cap X = \emptyset$ is the same as saying that $\{f_i\} \cup \{\ell_j\}$ has no solution, where the $\ell_j$ are the linear functions defining $L$. By the Nullstellensatz, that's the same as saying there exist polynomials $h_i,g_j$ such that $\sum_i h_i f_i + \sum_j g_j \ell_j = 1$. By an Effective Nullstellensatz there is a bound (doubly-exponential, but still finite) on the degrees of the $h_i,g_j$ that can possibly be needed here. So you can treat the coefficients of the $h_i,g_j$ as variables, and then the preceding equation is a (really big) linear equation $Ax = b$, where the $x$ vector consists of the coefficients of $h_i,g_j$, there is one equation for each monomial up to the relevant degree bound, and the $b$ vector is $e_1$ (because of the 1 on the RHS of the above equation). The values appearing in $A$ are the coefficients of the $f_i,\ell_j$.

So $L \cap X = \emptyset$ precisely when this big linear equation has solutions, which is to say when $e_1$ is in the column span of $A$. The latter can be written in terms of various minors vanishing, where those minors depend only on the coefficients of the $f_i, \ell_j$. (1)

Now, if you treat the $\ell_j$ themselves as variables, each such minor gives you an equation in the $\ell_j$.

Next, one has to convert back from the description of $L$ via its equations $\ell_j$ to a point on the Grassmannian. Local coordinates on the Grassmannian $G(k,n)$ are essentially given by the entries of a $k \times n$ matrix (with each chart determined by requiring that a fixed $k \times k$ submatrix is full rank), where the corresponding subspace is the rowspan of this matrix. Let $\mathcal{L}$ be the matrix with columns being the $\ell_j$; then the matrix $M$ corresponding to the space $L$ is just the left kernel of $\mathcal{L}$. So we get the equations $M \mathcal{L} = 0$. (2)

Finally, since we want the $\ell_j$ themselves to be variables, we have the equations (1) and (2) with the $\ell_j$ as variable, and then project onto the coordinates of $M$, and this should be the variety you want. Note that the only part here that requires anything like Gröbner bases is this projection step - the rest is pretty straightforward from the coefficients of the $f_i$.

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