In $\mathbb{R}^d$, the collection of hyperplanes $\mathcal{H}=\left\{x_{j}=x_{k}: 1 \leq j<k \leq d\right\}$ is called the $d$-dimensional real braid arrangement. I’m interested in the intersections of those hyperplanes, specifically with the added order by reverse inclusion ($x \preceq y \Leftrightarrow x \supseteq y$).
As an example, for $d = 3$, in the arrangement we have the three hyperplanes given by $x_1 = x_2, x_1 = x_3$, and $x_2 = x_3$. Rewriting them as tuples (i.e. $(x,x,x_3),(x,x_2,x),(x_1,x,x)$, respectively), it is easy to see that any two hyperplanes intersect at a line, which all coincide. The intersection lattice would looks like this:
(If we intersect just $1$ hyperplane, we get that hyperplane back. Intersecting $2$ hyperplanes gives us a line. By the same idea, the result of intersecting $0$ hyperplanes is defined to be $R^d$.)
This is a lot harder in dimension $4$ and up. I’ve computed that there are $d \choose 2$ hyperplanes for each dimension $d$, so the first line of thinking to consider all $2^{d \choose 2}$ sets of hyperplanes. But going this way there are some questions I’m still not able to answer:
- Do we have intersections of every dimension from $1$ to $d - 2$?
- If so, do some of them coincide, like how the three lines in $\mathbb{R}^3$ do?
In general I’m just looking for a systematic way to think about this problem. Any suggestion would be appreciated!
These intersections are just the equivalence relations on a set with $d$ elements. Namely, for any equivalence relation $\sim$ on $\{1,\dots,d\}$, there is an intersection consisting of all elements $x\in\mathbb{R}^d$ such that $i\sim j$ implies $x_i=x_j$, and every intersection has this form for a unique $\sim$. So you are just asking about the lattice of equivalence relations on $\{1,\dots,d\}$, ordered by inclusion.
So for your first question, an intersection of dimension $e$ is just an equivalence relation with $e$ equivalence classes. This can easily be achieved for any $e$ from $1$ to $d$: just partition the set into $e$ subsets. Two collections of hyperplanes have the same intersection iff the corresponding collections of ordered pairs of elements of $\{1,\dots,d\}$ generate the same equivalence relation. So, for any equivalence relation that has an equivalence class of size greater than 2, there is more than one way to write it as an intersection of hyperplanes: for instance, if $\{i,j,k\}$ is one of the equivalence classes, you could write use any two of the hyperplanes corresponding to two of $i,j,$ and $k$ to make them all equivalent.