# What is the symmetry group of this compound of two polytopes?

The geometric shape in question is a compound of two polytopes: an 11-hypercube with edge length $$2$$ and an 11-simplex with edge length $$\sqrt6$$ whose vertices are a subset of the hypercube’s. What is the structure of this shape’s symmetry group?

Edit: proof that it’s possible: https://www.math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Markov.pdf

And explicit coordinates:

(–1,–1,–1,–1,–1,–1,–1,–1,–1,–1,–1) (1,–1,1,–1,–1,–1,1,1,1,–1,1) (1,1,–1,1,–1,–1,–1,1,1,1,–1) (–1,1,1,–1,1,–1,–1,–1,1,1,1) (1,–1,1,1,–1,1,–1,–1,–1,1,1) (1,1,–1,1,1,–1,1,–1,–1,–1,1) (1,1,1,–1,1,1,–1,1,–1,–1,–1) (–1,1,1,1,–1,1,1,–1,1,–1,–1) (–1,–1,1,1,1,–1,1,1,–1,1,–1) (–1,–1,–1,1,1,1,–1,1,1,–1,1) (1,–1,–1,–1,1,1,1,–1,1,1,–1) (–1,1,–1,–1,–1,1,1,1,–1,1,1)

• There is no such simplex. Feb 3, 2021 at 1:27
• Are you sure about that? Feb 3, 2021 at 2:19
• Wouldn't that hypercube have edges of length $2$, not $1$? Feb 3, 2021 at 2:39
• Oops, you’re right. Feb 3, 2021 at 2:53
• In fact, a regular simplex of dimension $n$ can be inscribed in a hypercube of dimension $n$ if and only if $n+1$ is the order of a Hadamard matrix. If we also want every automorphism of the simplex to extend to an automorphism of the hypercube, then the $n=2^m-1$ result is correct. Feb 3, 2021 at 13:55

The automorphism group of this configuration $$C'$$ is the Mathieu group $$M_{11}$$.

Firstly, we construct a larger configuration $$C$$ consisting of a 12-dimensional orthoplex inscribed in a 12-dimensional hypercube.

In particular, the orthoplex in $$C$$ consists of the vectors $$\{ \pm v : v \textrm{ is a column of } H \}$$, where $$H$$ is a 12-by-12 Hadamard matrix whose first row is $$(1, 1, \dots, 1)$$. The hypercube in $$C$$ simply consists of all $$2^{12}$$ vectors whose entries are $$\pm 1$$.

The automorphism group of $$C$$ contains the order-2 centre $$\{\pm I\}$$, modulo which it is isomorphic to the Mathieu group $$M_{12}$$. This is described here in terms of the [unique up to isomorphism] order-12 Hadamard matrix:

The two different 12-element permutation representations of $$M_{12}$$ correspond to permuting:

1. The 12 'coordinate axes' of the hypercube;
2. The 12 pairs of opposite vertices of the inscribed orthoplex;

The subgroup $$K$$ that fixes a particular one of the 12 coordinate axes also has the order-2 centre $$\{ \pm I \}$$, modulo which it is isomorphic to $$M_{11}$$. This follows from $$M_{11}$$ being describable as the stabiliser of a point in the permutation representation of $$M_{12}$$.

Indeed, $$K$$ factors as a direct product $$C_2 \times M_{11}$$, where the first factor indicates whether the first coordinate axis is flipped or not, and the second factor indicates the permutation of the remaining 11 coordinate axes.

Consequently, $$M_{11}$$ is precisely the subgroup of the symmetry group of $$C$$ which fixes the unit vector $$v_1 = (1, 0, \dots, 0)$$.

Finally, note that the OP's configuration $$C'$$ can be described as the restriction of $$C$$ to the 11-dimensional hyperplane $$\{ x . v_1 = 1 \}$$. (The 11-simplex is obtained as a facet of the 12-orthoplex, and the 11-hypercube is a facet of the 12-hypercube.) Another construction of $$C'$$ is to take the vertices of the simplex to be the columns of the rectangular matrix $$H'$$ obtained by deleting the first row from $$H$$; the vertices of the hypercube are again the $$\pm 1$$-vectors.

It follows, therefore, that the symmetry group of $$C'$$ is $$M_{11}$$. This has an 11-element permutation representation (permuting the coordinate axes of the hypercube) and a 12-element permutation representation (permuting the vertices of the simplex).

• Doesn't the orthoplex have dimension 11, not 12? Feb 3, 2021 at 14:00
• No, $C$ is a 12-dimensional orthoplex (24 vertices) inscribed in a 12-dimensional hypercube (4096 vertices). The codimension-1 configuration $C'$ is an 11-dimensional simplex (12 vertices) inscribed in an 11-dimensional hypercube (2048 vertices). Feb 3, 2021 at 15:20