# Variety of subspaces not intersecting $X$

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 $$\dim(W')<\dim(\mathcal{G})$$.

• 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 $\mathcal{G}$. – Kapil Apr 30 at 17:37
• 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 $f_{i}$? – D. Dona May 2 at 12:47

## 1 Answer

$$L \cap X = \emptyset$$ 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 $$f_i$$.