# Geometrical interpretation of simplices

I've just started doing simplices, where the $n$-simplex has been defined to be $$\Delta^n = \{x \in \mathbb{R}^{n+1}\mid x_i \geq 0, \sum x_i=1\}.$$

It's easy to see that the $0$-simplex is the point $1$ in $\mathbb{R}^1$, the $1$-simplex is the line from $(1,0)$ to $(0,1)$ in $\mathbb{R}^2$, and the $2$-simplex is the triangle, including the interior, with vertices $(1,0,0), (0,1,0), (0, 0, 1)$ in $\mathbb{R}^3$.

But how do we justify that the $3$-simplex is a tetrahedron, including the interior, with points $(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)$ in $\mathbb{R}^4$?

Even worse, why is a $4$-simplex a pentachoron?

• As defined, the three-simplex is the convex hull of four mutually-equidistant points in the three-dimensional affine space $x_{1} + x_{2} + x_{3} + x_{4} = 1$. Can you articulate more explicitly why your intuition jumps from "it's easy to see..." to "how do we justify..."? – Andrew D. Hwang Jan 14 '17 at 18:47
• I'm finding it difficult to visualise why it must be a tetrahedron. Okay, the first three points $(1,0,0,0), (0,1,0,0)$ and $(0,0,1,0)$ are fine and can be thought of in 3 dimensions, but the fourth point $(0,0,0,1)$ is 4 dimensional and I can't see why these four points must join up to give a tetrahedron. – Irregular User Jan 14 '17 at 19:02
• To understand the geomety behind simplices, you have to understand the coordinate system. A good point to start would be to read something about the barycentric coordinate system, – Xaver Jan 14 '17 at 19:38
• @Xaver I'm familiar with barycentric coordinates, but I don't see how this helps? – Irregular User Jan 14 '17 at 19:47
• The solution set of $x_{1} + x_{2} + x_{3} + x_{4} = 1$ is a three-dimensional affine space, just as the solution set of $x_{1} + x_{2} + x_{3} = 1$ in $\mathbf{R}^{3}$ is a plane. Looking at how many coordinates each point has is a red herring; what matters is how many free parameters are needed to describe a solution space. Of course, visualizing subsets of $\mathbf{R}^{n}$ for $n \geq 4$ involves a certain amount of reasoning by analogy, and no amount of explanation can convey direct geometric apprehension.... – Andrew D. Hwang Jan 14 '17 at 20:08

Let's take a look at the definition of a simplex: $$\Delta^n = \{x \in \mathbb{R}^{n+1}\mid x_i \geq 0, \sum x_i=1\}.$$ Two aspects of this definition are important:

1. The first important thing to notice is that $x_i \geq 0$. This means: A point $P$ can only be part of $\Delta^n$, if $P\in [0,\infty)^{n+1}$ holds. Because the formula $\sum x_i = 1$ has to hold too, you can even say that $P\in [0,1]^{n+1}$ has to hold for every point $P$ of the $n$-simplex. Formally: $$P\in\Delta^n\ \Rightarrow\ P\in [0,1]^{n+1}.$$ Geometrically, the set $[0,1]^{n+1}$ is a hypercube.
2. The second important thing to notice is that $\sum x_i=1$ defines an affine hyperplane, let's call it $h$. Every point $P$ of $\Delta^n$ has to be on this affine hyperplane, i.e. $P\in h$ has to hold. Formally: $$P\in\Delta^n\ \Rightarrow\ P\in h$$

There are no other restrictions on the point $P$. So it holds that $P\in\Delta^n$ if and only if $$P\in[0,1]^{n+1}\cap h,$$ or - in other notation - that $$\Delta^n =[0,1]^{n+1}\cap h.$$ So to visualize a simplex, you can visualize the hypercube, visualize the affine hyperplane and visualize their cut-set, which is the simplex. Here is a visualization of $\Delta^2$: For $\Delta^2$, you are in a three-dimensional vector space. The hypercube is a cube and the hyperplane is a plane. In the above picture, the cube is visualized in green color and the cut-set of the cube with the plane (i.e. the simplex) is visualized in red.

Visualization in higher dimensions is difficult, but the concept that the simplex is the cut-set of a hypercube with an affine hyperplane also holds, so the geometrical situation is basically the same.

Please note: In the above geometrical description I have implicitly used the canonical basis vectors. You can visualize the simplex using another basis, too. When using another basis, replace hypercube with parallelepiped in the above description.

• This doesn't answer the question unless you generalise it to a $3$-simplex and optionally a $4$-simplex. – Irregular User Jan 14 '17 at 20:45
• The main part of the answer was: "In order to give the coordinates a geometrical meaning, you have to define a basis." You can do this in any dimension, so this answer also includes the 3-simplex and the 4-simplex. Please note: I only gave an example of a basis. With this basis, your 2-simplex will look like a regular triangle. But you can use any other basis as well. The same simplex will then look different. – Xaver Jan 14 '17 at 21:02
• What you see in this answer: Each of the basis vectors represents a point, so you have three points for the 2-simplex. Each simplex can be viewed as the convex hull of these points, so you get a triangle (including the interior). If you increase the dimension by one, your basis represents four points (3-simplex), and their convex hull is a tetrahedron. If you further increase the dimension by one, your basis represents five points (4-simplex), and their convex hull is a pentachoron. And so on... – Xaver Jan 14 '17 at 21:11
• "If you increase the dimension by one, your basis represents four points (3-simplex), and their convex hull is a tetrahedron. If you further increase the dimension by one, your basis represents five points (4-simplex), and their convex hull is a pentachoron." Why? – Irregular User Jan 14 '17 at 21:24
• I reformulated my answer and inserted a picture. I hope the situation becomes clearer now. – Xaver Jan 15 '17 at 9:25

Again, as I see it, the correct geometric intuition is to note that the locus $\sum_{i=1}^{n+1} x_{i} = 1$ in $\mathbf{R}^{n+1}$ is an $n$-dimensional (affine) Euclidean space, in which the standard basis vectors are mutually-equidistant, and therefore constitute the vertices of an equilateral triangle ($n = 2$) or a regular tetrahedron ($n = 3$) or ... ($n \geq 4$).

To elaborate this point a bit more:

Theorem: If $E^{N}$ denotes the Euclidean space of dimension $N$, and if $(p_{j})_{j=0}^{n}$ and $(q_{j})_{j=0}^{n}$ are two sets of $n + 1 \leq N + 1$ points of $E^{N}$ such that $$\|p_{i} - p_{j}\| = 1 = \|q_{i} - q_{j}\|\quad\text{for all i \neq j,}$$ then there exists a Euclidean isometry $T:E^{N} \to E^{N}$ such that $T(p_{i}) = q_{i}$ for all $i$.

In words, "there is a unique unit $n$-simplex up to isometry".

(To prove this, one might use a translation to move $p_{1}$ to $q_{1}$, then argue inductively, using the fact that the orthogonal group $O(k)$ acts transitively on the unit sphere in $E^{k}$ and the (isotropy) subgroup fixing one point is $O(k-1)$.)

Now, a unit regular tetrahedron is (the convex hull of) a set of four points in $E^{3}$ whose mutual distance is unity. Setting $\ell = 1/\sqrt{2}$, the four points $$(\ell, 0, 0, 0),\quad (0, \ell, 0, 0),\quad (0, 0, \ell, 0),\quad (0, 0, 0, \ell)$$ have mutual separation equal to unity. Consequently, their convex hull is isometric to a "standard" regular tetrahedron in $E^{3}$.

This argument generalizes in an obvious way to arbitrary finite dimension. Particularly, a four-simplex of unit side length (the convex hull of $\ell$ times the set of standard basis vectors in $E^{5}$) is isometric to whatever definition of a pentachoron is acceptable (e.g., the convex hull of five points in $E^{4}$ whose mutual separation is unity).