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?

  • $\begingroup$ 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..."? $\endgroup$ – Andrew D. Hwang Jan 14 '17 at 18:47
  • $\begingroup$ 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. $\endgroup$ – Irregular User Jan 14 '17 at 19:02
  • $\begingroup$ 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, $\endgroup$ – Xaver Jan 14 '17 at 19:38
  • $\begingroup$ @Xaver I'm familiar with barycentric coordinates, but I don't see how this helps? $\endgroup$ – Irregular User Jan 14 '17 at 19:47
  • $\begingroup$ 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.... $\endgroup$ – 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 $\Delta^2$: 2-Simplex (red triangle) 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.

  • $\begingroup$ This doesn't answer the question unless you generalise it to a $3$-simplex and optionally a $4$-simplex. $\endgroup$ – Irregular User Jan 14 '17 at 20:45
  • $\begingroup$ 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. $\endgroup$ – Xaver Jan 14 '17 at 21:02
  • $\begingroup$ 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... $\endgroup$ – Xaver Jan 14 '17 at 21:11
  • $\begingroup$ "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? $\endgroup$ – Irregular User Jan 14 '17 at 21:24
  • $\begingroup$ I reformulated my answer and inserted a picture. I hope the situation becomes clearer now. $\endgroup$ – 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 ... ($n \geq 4$).

The convex hull of the four-dimensional standard basis is a regular tetrahedron

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.