sets = simplices

Okay, clickbaity and reductive title aside, the reason I write this blog post is that there have been many times that sets with $n + 1$ elements have looked really gosh darn like rank $n$ simplices. For those who don't know, a "simplex" is a generalization of the triangle and tetrahedron you might be familiar with. The "rank" of a simplex is like the dimension of the space it lives in, although the technical details are a bit more nuanced. But, for example, a triangle is a rank $2$ simplex, a tetrahedron is a rank $3$ simplex, and this concept can be extended into further dimensions. Meanwhile, a line segment is a rank $1$ simplex, and a point is a rank $0$ simplex. Again, this is all very handwavy. So actually, let me define what a simplex is, in a way that will highlight why these things might be connected to sets.
Quick disclaimer: this blog post has nothing to do with simplicial sets. If you're into those, sorry.

but first, we need to talk about abstract polytopes

Think for a bit how you might define a polygon or polyhedron. At least for me, it's not obvious at all. Intuitively, we have this concept of what a polygon or polyhedron should be, but it's hard to write down the rules they must follow, isn't it?

Here's one perspective.

In a sense, a polyhedron is made up of many polygons stitched together along their line segment edges. A polygon, similarly, is made up of many line segments stitched together along the vertices at their endpoints. And a line segment is just two vertices stitched together along... well, it seems like nothing at all, so this seems to be where the pattern ends. What it seems is that a rank $n$ polytope is made up of rank $n - 1$ polytopes stitched together along rank $n - 2$ seams. (A "polytope" is the generalization of a polygon or polyhedron to any number of dimensions.) Indeed, this is how "abstract polytopes" are defined. In particular, an abstract polytope is a partially ordered set of subpolytopes ordered by "incidence", or whether one subpolytope is contained within the other. For example, think of a cube. Every face of the cube is incident to the cube itself in the sense that they are contained within the cube, if we think of this cube as including its boundary. Additionally, each edge is incident to two different faces of the cube, but no more than that. If an edge isn't part of a face, then it's not incident to it. This applies similarly to vertices as well: every vertex of the cube is incident to three edges and three faces, if you can visualize that. For technical reasons, we also consider there to be a bottom element of this poset of rank $-1$. It just makes the math work out nicer, but you can think of it as what the vertices are stitched together along, or empty space that is incident to everything. There are other details I'm leaving out, but for the purposes of this blog post, you can think of abstract polytopes as posets representing incidence relations.

the abstract polytope definition of simplices

Consider the powerset of a set with $3$ elements. It has a natural ordering by subset inclusion. Now label the vertices of a triangle with $A, B, C$ and label each subpolytope of the triangle by which vertices the subpolytope contains within it.

If we include empty space as being the empty string, then there is a precise correspondence between incidence of subpolytopes of a triangle and subset inclusion of the set $\{A, B, C\}$! This is no coincidence and in fact is one way to define what a simplex is, at least in the abstract polytope sense. We define the rank $n$ simplex as the abstract polytope whose incidence poset is the subset inclusion poset of a set with $n + 1$ elements.

now let's talk about automorphism groups

Random question: what is the automorphism group of a set with $n + 1$ elements? Recall that an automorphism of a set is just a permutation, that is, an invertible function from itself to itself. We often denote the group of permutations on a set with $n + 1$ elements as $S_{n + 1}$.

Random question: what is the automorphism group of a rank $n$ simplex? Well, we could define this entirely abstractly in terms of poset automorphisms, which are order-preserving permutations. However, I think it's more interesting to now embed our abstract rank $n$ simplex into the Euclidean space that we are all familiar with in real life. We could place the vertices all over space willy-nilly, but let's take a so-called "symmetric realization" and place them in a way that preserves the symmetries of the abstract structure. For example, a rank $2$ simplex becomes an equilateral triangle, a rank $3$ simplex becomes a regular tetrahedron, etc. Now, let's say that an automorphism is an isometry of the Euclidean space that maps the simplex to itself. In simpler terms, this is typically what we think of as a symmetry of a geometric figure. For example, an equilateral triangle has a symmetry group often denoted $D_3$, corresponding to the $3$ rotations and $3$ reflections that send the triangle to itself. It turns out that we can achieve any permutation of the vertices of a simplex via such isometries!

Here's a quick proof: select two vertices $A, B$ and define a rank $n - 1$ hyperplane by $\{C \mid d(A,C) = d(C,B)\}$, where $d$ represents the Euclidean distance function. (A hyperplane is just a generalization of a plane, which is a rank $2$ hyperplane. A rank $1$ hyperplane is a line, a rank $3$ hyperplane is hard to visualize but you can think of it as separating rank $4$ Euclidean space into two halves, etc.) It is evident by the symmetry of the simplex that every other vertex will be in this hyperplane, so we may take the isometry that consists of reflecting across it. This is purely a transposition of $A$ and $B$. Transpositions generate every permutation and we can apply this construction to any two points to obtain any transposition, QED.

In fact, because rank $n$ simplices have $n + 1$ vertices, we have that its isometry group is precisely $S_{n + 1}$! Further, rotations correspond to even permutations, and reflections correspond to odd permutations. Thus, the alternating group $A_{n + 1}$ is precisely the rotation group of a rank $n$ simplex!
Pretty cool, huh?

my final demonstration: projective geometry, $q$-analog combinatorics, and the field with $1$ element
(i swear im not a crank)

Finally, I am going to demonstrate one last thing, and in fact, this is the reason why I made this blog post to begin with. The other stuff before this was just some supporting facts, but this is the real kicker for me.

Firstly, recall that rank $n$ projective space is $(V - \{0\}) / \sim$ where $V$ is a rank $n + 1$ $k$-vector space and $v \sim w$ iff $cv = w$ for some non-zero $c \in k$. (I'm using the word "rank" here instead of dimension for consistency and brevity even though this word is usually reserved for when $k$ is not a field.) One natural question to ask is this: given a field with $q$ elements, how many distinct rank $k$ (projective) hyperplanes are there in rank $n$ projective space? Here's one way to find out. Firstly, note that the number of points (rank $0$ hyperplanes) in rank $n - 1$ projective space is

$[n]_q = \frac{q^n - 1}{q - 1} = 1 + q + q^2 + \ldots + q^{n - 1}$.

This is just because we have $q^n - 1$ elements in $V - \{0\}$ and we have $q - 1$ unique non-zero scalars in $k$. The bracket notation is stolen directly from $q$-analog combinatorics and represents the "$q$-analog" of $n$. Now, to define a rank $k$ hyperplane, we need $k + 1$ points. Each time we choose a point, we need to make sure that we don't pick from the smaller rank hyperplane defined by the points we've already chosen. In total, we end up with

$\prod_{i = 0}^{k} ([n + 1]_q - [i]_q)$

different possible choices. However, we need to account for how we can pick different points or the same points in a different order and still end up with the same hyperplane. To this end, we can divide by the number of unique bases of our hyperplane. Our hyperplane is rank $k$ and so this is just our formula applied to $n = k$; we pick points "in general position" until we can't anymore. In total, we end up with the formula

$\frac{ \prod_{i = 0}^{k} ([n + 1]_q - [i]_q) }{ \prod_{i = 0}^{k} ([k + 1]_q - [i]_q) }$.

Indeed, plug in $q, n, k$ and this will spit out the number of distinct rank $k$ hyperplanes in rank $n$ projective $\mathbb{F}_q$-space. Now, let $q = 1$. If we use the right hand side of the expression defining $[n]_q$, we get $[n]_1 = n$. (This is the core idea behind $q$-analog combinatorics; the $q$-analogs just become typical everyday numbers when we take the "limit" as $q \rightarrow 1$.) Our formula becomes

$\frac{ \prod_{i = 0}^{k} (n + 1 - i) }{ \prod_{i = 0}^{k} (k + 1 - i) } = \frac{ (n+1)!/(n-k)! }{(k+1)!} = {n+1 \choose k+1}$.

Huh. Does this have a sensible combinatorical interpretation?

yes it does, but we need to define projective $\mathbb{F}_1$-space
(wait please dont leave me)

Okay, you've read this far, so, please, hear me out.

For the sake of visualization, let $n = 2$. On the left, we have projective $\mathbb{F}_3$-space. Recall that rank $n$ projective space can be thought of as affine spaces of rank $n, n-1, \ldots, 1, 0$ stitched together. This is because we can usually normalize the first coordinate of a point in projective space to $1$ via scaling. However, if that coordinate is $0$, then we can't do this, and the subspace where this coordinate is $0$ is simply a rank $n - 1$ hyperplane. Applying this procedure recursively yields the desired decomposition. In the picture, this manifests as $3$ affine spaces of rank $2, 1, 0$ and $9, 3, 1$ points, respectively. Then, when $q = 2$, we have $4, 2, 1$ points. As $q$ tends to 1, we have a single point representing each affine space. This makes sense, because taken literally, affine $\mathbb{F}_1$-space is just the $0$ space, i.e. a $1$ point space. This yields a... you guessed it, rank $2$ simplex! Indeed, rank $n$ projective $\mathbb{F}_1$-space is a rank $n$ simplex, which, as we know, is a set with $n + 1$ elements. Further, a rank $k$ hyperplane is just a subset with $k + 1$ elements, so counting subspaces of rank $k$ is just choosing $k + 1$ elements from an $n + 1$ element set. This is precisely what our formula describes! (Hopefully you don't think I'm that crazy anymore.)

conclusion

You may be asking how you can actually apply this to actual math. My opinion is that whenever we deal with sets without any structure, we should visualize them as simplices to respect the inherent symmetries a structureless set has. Often we write out the elements in a line, one after another, but this breaks our symmetries when $\|S\| > 2$. Inherently, $3$ element sets want to live on the plane, and $4$ element sets want to live in 3D space. Regardless though, these are relatively obscure facts and I wanted to organize them all in one place for ease of sharing. So yeah, hope you enjoyed.