Wikipedia says:

a regular polytope is a polytope whose symmetry is transitive on its flags,

How is "symmetry" defined?

Is transitive for a relation on the set of flags?

Can someone explain the meaning of "whose symmetry is transitive on its flags"?

Thanks in advance!

  • 1
    $\begingroup$ I think it means that the symmetry group of the polytope acts transitively on the set of flags. $\endgroup$
    – user39280
    Dec 15, 2012 at 22:30

2 Answers 2


A symmetry of a geometric object is a function $f$ that takes each point of the object to some other point (possibly the same point) of the object, which is a "rigid motion", which means that it must not change the distance between any two points.

For example, one symmetry of a square is a vertical reflection, which is a rigid motion that takes each point on the square to the point that is opposite to it across the horizontal midline. If the square has its center at $(0,0)$ and sides of length 2, its vertices are at $(-1,-1), (-1,1), (1,-1), $ and $(1,1)$, and the function $f$ is the one that takes each point $(x,y)$ of the square to the point $(x, -y)$.

Another symmetry is a clockwise quarter-turn rotation around the center of the square; here $f(x,y) \mapsto (y, -x)$. Yet another is the "motion" that does nothing at all and which has $f(x,y)=(x,y)$ for every point of the square.

The requirement that $f$ map every point of the square to some other point rules out operations like rotating the square a quarter-turn around a point other than the center, since then some points that were in the square end up outside, and vice-versa. The requirement that $f$ preserve distances rules out all sorts of crazy things, such as squashing the entire square down onto a small part, or reflecting only the left half of the square, or permuting the points of the square in a completely arbitrary way.

(In more general contexts, a symmetry of an object requires only that $f$ be "bijective", which means that for each point $p$ there is exactly one $q$ for which $f(q) = p$, but in geometric contexts this is implied by the rigidity requirement.)

The Wikipedia statement is a little bit imprecise. When it says "a polytope whose symmetry is transitive on its flags" they really mean "a polytope whose symmetry group is transitive on its flags". The symmetry group is the set of all symmetries (together with the observation that if you compose two symmetries by applying one and then the other, the result is again a symmetry), and it is "transitive" on some set $S$ if, for any two elements of $S$, say $s_1$ and $s_2$, there is some symmetry $f$ in the group for which $f(s_1) = s_2$.

So for example, an equilateral triangle is a regular polytope, because if you take any two flags, each of which is a sequence consisting of some vertex, some edge which contains it, and then the entire triangle, there is some symmetry of the triangle that maps the first flag to the second:

Equilateral triangle

Say for example you take $s_1$ to be the flag $(A, \overline{AB}, \triangle ABC)$ and take $s_2$ to be the flag $(B, \overline{AB}, \triangle ABC)$. Then the symmetry that takes $s_1$ to $s_2$ is a reflection along the line through $C$ that bisects $\overline{AB}$. The equilateral triangle has six flags, and it has six symmetries, and for any two flags there is some symmetry that maps the first flag to the second.

In contrast, consider a non-equilateral isosceles triangle with sides $DE=EF$.

isosceles triangle

This triangle has a symmetry (two, in fact), but it is not regular: $s_1$ is a flag that includes edge $DE$ and $s_2$ is a flag that includes edge $DF$, but there is no symmetry that can possibly map $s_1$ to $s_2$. If there were, it would map edge $DE$ to edge $DF$, and that means that the distance between points $E$ and $F$ would not be the same after the mapping, and therefore the mapping is not a symmetry.

The situation is more interesting in three dimensions and higher, where you can consider (for example) an equilateral pentagonal dipyramid in which the symmetries are transitive on the faces, but not on the edges or vertices, and hence not on the flags.


It specifically means that the automorhpism group of the polytope is transitive. If you denote your polytope by $\mathcal{P}$. then the automorphism group is generally denoted $\Gamma(\mathcal{P})$. $\Gamma$ is said to be transitive on flags if for every pair of flags $\Phi$ and $\Psi$ there is an element $\sigma\in\Gamma$ such that $\sigma(\Phi)=\Psi$.

Essentially, you need to be able to map every flag to every other flag via an automorphism. There are standard ways for determining the generators of the automorphism group, and the number of these generators is solely determined by the rank of the polytope.


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