Suppose I have a regular polytope $P$ which I'm representing as a graph $G_P$ with vertices and edges. I can already put this data into a computer program to find a list of symmetries of $P$---they're just the graph isomorphisms from $G_P$ to itself.
But as I understand it, half(?) of these isomorphisms will be orientation-preserving and the other half will be orientation-flipping.
How can I make a list of the graph isomorphisms that are orientation-preserving? Does the adjacency data from the graph (which is abstract and not embedded in any geometric space) already determine which isomorphisms preserve/flip orientation, or is it necessary to supply some additional structure such as an embedding of the graph, a directed traversal of the edges, or a list of the vertices of a face in clockwise order? Given whatever additional data, what should the algorithm be?
Edit: I'm wondering if I can find the orientation-preserving symmetries using algebra: I believe the square of any symmetry is a rotation. So there's a subgroup generated by the squares, and they're all rotations. Is this all the rotations (I'd generally doubt it), or how might we get the rest?
Maybe there's only a few groups that satisfy the properties that (1) they contain exactly half the members of the symmetry group (2) they contain all the perfect squares. In that case, it might be that any one of those groups could be the rotation group given an appropriate embedding --- i.e. each one is isomorphic to the rotation group. In that case I'd wonder if we need any additional constraints to guarantee that all the subgroups that satisfy these constraints are (isomorphic to) the group of rotations.
Edit #2: I can prove that the sign of a group automorphism is independent of choice of embedding, but I am still searching for a constructive method to compute it from the graph without using coordinates as an intermediate.
The proof that the sign is independent of the choice of embedding is as follows:
Let $P$ be a regular polytope, fix $S\subset \mathbb{R}^n$ as points for $P$ in $\mathbb{R}^n$, let $G$ be the abstract graph of $P$ and let $V$ be its vertices.
A graph automorphism is any bijection $\rho:V\rightarrow V$ that preserves and reflects the adjacency relationships of the graph $G$.
A choice of coordinates for $G$ is a map $\chi: V\cong S\subset \mathbb{R}^n$. Let's fix one such coordinate choice as our canonical coordinates: $\chi :V\cong S$.
Every other choice of coordinates factors uniquely through $\chi$: if $\chi^\prime$ is any other choice of coordinates, then $\chi^\prime = \sigma\circ \chi$ for some unique isometry $\sigma:\mathbb{R}^n\rightarrow \mathbb{R}^n$.
Given a choice of coordinates $\chi$, every graph automorphism $\rho:V\rightarrow V$ corresponds to a unique isometry of Euclidean space $\sigma_\rho :\mathbb{R}^n\rightarrow \mathbb{R}^n$, such that $\sigma_\rho \circ \chi = \chi \circ \rho$ ("the isometry after embedding is equivalent to the automorphism before embedding")
Let $\rho:V\rightarrow V$ be any graph automorphism, and let $\chi, \chi^\prime : V\cong S \subset \mathbb{R}^n$ be two embeddings of the graph. We'll show that, under the two embeddings $\chi,\chi^\prime$, the automorphism $\rho$ induces two isometries $\sigma,\sigma^\prime$ with the same sign. This shows that the isometry induced by $\rho$ has the same sign under any choice of coordinates, which means that $\rho$'s sign is determined independent of any embedding.
By (5), we can find the unique isometries $\sigma$ and $\sigma^\prime$ such that $$\sigma\circ \chi = \chi \circ \rho$$ and $$\sigma^\prime \circ \chi^\prime = \chi^\prime \circ \rho.$$
By (4), we can factor $\chi^\prime = \tau\circ \chi$ for some unique isometry $\tau$. Factoring $\tau$ out of the second equation above gives: $$\sigma^\prime \circ \tau \circ \chi = \tau \circ \chi \circ \rho$$
Composing $\tau$ on the left of the first equation gives: $$\tau \circ \sigma \circ \chi = \tau \circ \chi \circ \rho $$
The right hand sides of these two equations are equal; hence their left hand sides must be equal as well:
$$\sigma^\prime \circ \tau \circ \chi = \tau \circ \sigma \circ \chi$$
The choice of coordinates is injective, so we can drop it, leaving just a commutative square: $\sigma^\prime \circ \tau = \tau \circ \sigma$.
Now every term in this equation is an isometry $\mathbb{R}^n\rightarrow \mathbb{R}^n$, meaning it has a determinant whose value is $\pm 1$.
$$\det(\sigma^\prime \circ \tau) = \det(\tau \circ \sigma)$$ $$\det(\sigma^\prime)\cdot \det(\tau) = \det(\tau) \cdot \det(\sigma)$$ $$\det(\sigma^\prime) = \det(\sigma)$$ Hence the two isometries induced by the two choices of coordinates for the graph automorphism $\rho$ have the same sign, QED.
Eg. for a $4$-simplex you have vertices labelled as $(1,2,3,4,5)$ An isomorphism will give you a permutation, eg $(2,1,4,5,3)$, and you check its parity.
For $n$-cubes and $n$-orthoplex, the embedding is the simplest way to go. The cube vertices will have coordinates $(\pm 1, \pm 1, \dots \pm 1)$ . The orthoplex vertices will have coordinates $(0,0,\dots, 0,\pm1, 0, \dots, 0)$.
You select $n$ vertices of non-zero determinant, and see if the isomorphism preserves det sign.
– user3257842 Nov 09 '22 at 08:01You could use those rules to associate (in a way unique up to isomorphism) embedding coordinates to your vertices, then do the determinant trick. The only question remaining is if this could be done without the intermediate step.
– user3257842 Nov 09 '22 at 09:49For the $n$-orthoplex, the $2n$ vertices should group into $n$ pairs of unconnected ones. Associate $(0,0,…,0,±1,0,…,0)$ for each of those pairs, and you have an embedding.
– user3257842 Nov 09 '22 at 09:59