# Is $\text{Conf}_n(\mathbb{A}^2)$ an affine variety?

Given a variety $X$ (over a field $k$) we can define the configuration space of $X$ as a hyperplane arrangement.

$$\text{Conf}_n(X) = X^n \backslash \bigcup_{1 \leq i,j \leq n}\{ x_i = x_j \}$$

Is this a projective or an affine variety?

In the case $X = \mathbb{C}$ or $X = \mathbb{R}^2$ or $X = \mathbb{A}^2$, how can we cover this space $\text{Conf}_n(X)$ with affine varieties (making a chart or atlas of some kind). Do we need to resort to schemes just yet?

I am flipping through notes for some definitions. In any case, I think we have as sets:

$$\text{Conf}_n (\mathbb{A}^2) \subset \mathbb{A}^{2n}$$

so I think we are describing an affine variety, possibly with a single equation.

• This is almost never projective and seldom affine even if $X$ is. Did you check in your example with $n=2$? – Mohan Jul 28 '18 at 17:10
• @Mohan can you cover it with affine varieties? – cactus314 Jul 28 '18 at 19:34
• Any quasi-projective variety can be covered by affine open subsets. If $X$ is quasi-projective so is the configuration space. – Mohan Jul 28 '18 at 19:43
• @Mohan does this discussion get any easier if we use schemes? – cactus314 Jul 28 '18 at 19:54

Explicitly, with $X=\Bbb A^2$ and $n=2$ points, your configuration space is the complement of the variety given by $(x_1-w_1, x_2-w_2)$ inside $\Bbb A^4$. This is not affine, because it is the complement of a nonempty variety of codimension 2: if it were, this would contradicts the fact that the complement of an affine subset of a variety is codimension 1 if nonempty. In general, you can see that the set you need to throw away from $X^n$ will not be codimension 1 very often: consider the number of equations you need to define it, and that these equations are generally independent. It can also be seen fairly quickly that it is difficult to be projective as well: it is very, very rare for the complement of a projective subvariety of a projective variety to be again projective.
Proof of affine-codimension fact: Let $U\subset X$ be an affine open set where $X$ is separated scheme of finite type over a field. Then we may reduce to $X$ affine, noetherian, normal, integral by replacing $X$ and $U$ with their normalizations (doesn't change codimension), and intersect against a well-chosen affine open cover of $X$. Thus $U\subset X$ is an open embedding of affine schemes, which means it is totally determined by $k[X]\to k[U]$.
Since $X$ is integral, this map is injective. If $X\setminus U$ were of codimension 2 or more, then any function $f\in k[U]$ would be regular at all codimension 1 points of $X$ (because $U$ necessarily contains these points) and thus we may apply algebraic Hartog's to see that our map $k[X]\to k[U]$ is surjective and thus an isomorphism. So $U\to X$ is an isomorphism if $X\setminus U$ is codimension 2 or more, and therefore if $X\setminus U$ is nonempty it must be codimension 1.