---
tag: BTCXG67H69
kind: definition
name: Wythoff construction
msc: [51M20, 20F55, 51F15]
defines: J4EDDYCXMR
uses: [G6FND7X69J, 06BRDF3F0E, 0YQPZB5Q1G, 0FQSDC2EQ2, 300K6CB1KH]
sources:
  - url: https://en.wikipedia.org/wiki/Wythoff_construction
---

## Definition

Let $W$ be a finite group of isometries of $\R^n$ generated by [[reflections|06BRDF3F0E]] in hyperplanes through the origin, a finite [[Coxeter group|G6FND7X69J]].
Assume that no nonzero vector is fixed by all of $W$.
Let $C$ be a fundamental chamber of $W$: the closure of one connected component of the complement of all the reflecting hyperplanes of $W$.
Then $C$ is a simplicial cone bounded by $n$ hyperplanes $H_1, \dots, H_n$, and the reflections $s_1, \dots, s_n$ in them generate $W$.
Choose a point $v \in C$ with $v \ne 0$.
The *Wythoff construction* produces the polytope
$$P(v) = \operatorname{conv}(W v),$$
the [[convex hull|0FQSDC2EQ2]] of the [[orbit|0YQPZB5Q1G]] $W v = \set{w(v) : w \in W}$.

## Remarks

All points of $W v$ lie on the sphere of radius $\norm{v}$, so each of them is a vertex of $P(v)$, and $W$ acts transitively on the vertices of $P(v)$.
The polytope depends, up to similarity, mainly on the set of mirrors $H_i$ that do not contain $v$; in a Coxeter diagram one marks the corresponding nodes with rings.
If $v$ does not lie on the mirror $H_i$, then $v$ and $s_i(v)$ are joined by an edge of length twice the distance from $v$ to $H_i$.
When $v$ has the same distance to every mirror that does not contain it, all edges of $P(v)$ have the same length, and $P(v)$ is a [[uniform polytope|BZXHGNGQAX]].
Let $W$ be the symmetry group of a regular polytope.
If $v$ lies on all mirrors except the one at an end of the Coxeter diagram, then $P(v)$ is that regular polytope or its dual, up to similarity.

Example: let $W$ be the symmetry group of the square, of order $8$, generated by the reflections in the lines $y = 0$ and $y = x$, with chamber $C = \set{(x, y) : 0 \le y \le x}$.
For $v = (1, 0)$, on the mirror $y = 0$, the orbit is $\set{(\pm1, 0), (0, \pm1)}$, and $P(v)$ is a square.
For $v = (1, 1)$, on the mirror $y = x$, the orbit is $\set{(\pm1, \pm1)}$, again a square.
For $v = (1, \tan(\pi/8))$, which is at equal distance from both mirrors, the orbit has $8$ points spaced by angles $\pi/4$, and $P(v)$ is a regular octagon.

Non-example: the choice $v = 0$ is excluded; its orbit is the single point $0$, and the construction gives no polytope of dimension $n$.