Geometrical interpretation of simplices

Question

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?

Answer 1

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.

Answer 2

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).