# What are the “simplest” polytopes with an automorphism group of $\mathrm M_{11} \hspace {-1.25pt}$?

Do any polytopes have an automorphism group of the smallest of the sporadic groups, the Matthieu group $$\mathrm M_{11} \hspace {-1pt}$$? Indeed, they must exist. What are the simplest such polytopes we can construct, where "simplest" can be interpreted in a number of equally valid ways such as possessing:

• The lowest dimensionality,
• The fewest vertices and/or facets,
• And/or particularly elegant faces and/or topological structure (the abstract polytope it generates)?

This exercise was inspired by a comment made by @Rudi_Birnbaum:

I have found an interesting quote: "The Mathieu groups M11 and M22 are not automorphism groups of any polytopes.". Is this in contradiction to what has been written above?

@S.Carnahan replied:

@Rudi_Birnbaum I cannot read the article because of the paywall. However, given the name of the article, I suspect the author is restricting his view to regular polytopes, and didn't bother to include the word "regular" in that particular sentence.

@Rudi_Birnbaum I couldn't find that sentence in the linked article, but it does appear in this one. S. Carnahan is correct that the author is discussing regular polytopes, but they are also "abstract", which refers to a very different concept. Of course, many important lattices such as the Leech lattice include $$\mathrm M_{11}$$ as a subgroup of their automorphism group.

In the answer, I will explain what the modifiers "regular" and "abstract" mean in this context. I will then fully characterize a specific non-regular, geometric (as opposed to abstract) polytope with $$66$$ vertices in $$\mathbb Z^{11} \!$$ which $$\color {red} {\textbf{does not}}$$ have an automorphism group of precisely $$\mathrm M_{11} \hspace{-1pt}$$. @M.Winter has shown that my solution is, in fact, incorrect.

• I'm not sure that posting a "write-up" (as you called it yourself in the previous thread) in the form of you posting a question and answering it immediately is really what this site is for. Otherwise, every paper could be its own post, with the question being the motivation and then the answer being the result. – verret Dec 18 '20 at 3:28
• @verret I'm not certain that it is appropriate either; I'm new to MO. I wanted to reply to a comment, but needed way more than 500 chars, so I tried to make the Q&A as didactic and broadly interesting as possible. This page makes it seem like answering your own question is encouraged. Perhaps I should reframe it in more general terms, as I'm less interested in M₁₁ specifically, but more broadly want to discuss lattices, polytopes, codes, &c. with sporadic or exceptional symmetry. – OzoneNerd Dec 18 '20 at 7:18
• As I also noted in my answer below the linked question, every finite group is the symmetry group of a convex polytope. – M. Winter Dec 18 '20 at 12:06
• Thank you @OzoneNerd! that's a very interesting elaboration on the topic! – Raphael J.F. Berger Dec 18 '20 at 15:19
• @M.Winter I suppose I'm particularly interested in relatively simple structures. A polytope with $\sim 10^{19}$ vertices with an automorphism group of $\mathrm {Co}_0$ would be far less interesting than using the minimal vectors of the Leech lattice. – OzoneNerd Dec 18 '20 at 19:35

# $$\color {red} {\textbf {WARNING}}$$

I believe @M.Winter has shown in the comments that the following answer is, in fact, $$\color {red} {\textbf {incorrect}}$$. It would appear that I have merely constructed a rectified $$11$$-simplex. I was misled by Magma's AutomorphismGroup(P) : TorPol -> GrpMat function, which only finds subgroups of $$\mathrm {GL}_d(\mathbb Z)$$. However, they also provided references showing that the feat is indeed possible. I will begin working on providing a correct answer tomorrow.

In the meantime, I invite others to try their hand at the problem. In general, simple polytopes, should be preferred, where "simple" can be interpreted in a number of equally valid ways. I provided some examples in the question.

Unless there are objections, I will leave this answer up, as I think it still contains some valuable information and valid ideas for how to go about fixing it.

## Regular and/or Abstract Polytopes

Regular polytopes possess the greatest degree of symmetry a polytope can have. Their automorphism group acts transitively on its faces of any dimension. This includes their vertices, edges, $$\ldots$$, and facets. The actual definition is even stricter than this, requiring that the automorphism group act transitively on its flags, but I won't get into that level of detail.

In our geometric setting, we are only concerned with real, Euclidean, convex regular polytopes. This rules out self-intersecting star polytopes, complex polytopes, tesselations of the hyperbolic plane, and infinite apeirotopes. Alongside regularity, these requirements are highly restrictive:

1. In two dimensions, there are an infinite number of regular polygons.
2. In three dimensions, the regular polytopes are more commonly known as the five Platonic solids.
3. There are six four-dimensional regular polytopes, up to similarity.
• In all higher dimensions $$d > 4$$, there are only three regular polytopes, up to similarity:

There is no regular geometric polytope with an automorphism group of $$\mathrm M_{11} \hspace {-1pt}$$.

Abstract polytopes can be generated from the more familiar geometric polytopes by "forgetting" the geometry of the polytope and retaining only the topological information regarding how its faces of various dimensions are connected to each other. One can also construct valid abstract polytopes that can not be "realized" as geometric polytopes such as the $$11$$-cell and the $$57$$-cell.

By modifying the definition slightly, the concept of regularity can be extended to abstract polytopes, and there are far more regular abstract polytopes than there are regular geometric polytopes. Hence, one can construct regular abstract polytopes with automorphism groups which do not apply to any of the regular geometric polytopes.

It is important to note that the automorphism groups of a geometric polytope and the abstract polytope generated from it need not be isomorphic. By forgetting the geometric information, the latter can become larger than the former. The former should always be a subgroup of the latter.

If my tentative understanding of the papers linked in the question and some of their references is correct, there are abstract polytopes with an automorphism group of $$\mathrm M_{11} \hspace {-1pt}$$, but none of them are regular. I will proceed to construct a non-regular geometric polytope with an automorphism group of $$\mathrm M_{11} \hspace {-1pt}$$.

## Construction of an $$\mathrm M_{11} \!$$ Polytope

$$\mathrm M_{11} \!$$ has irreducible $$\mathbb Q$$-linear representations of degrees $$1$$, $$10$$, $$11$$, $$20$$, $$32$$, $$44$$, $$45$$, and $$55$$.

The degree-$$11$$ representation, corresponding to the character $$\chi_5 \hspace {-0.5pt}$$, is particularly appealing because it's absolutely irreducible over the rationals, and its invariant form is just the identity matrix $$I_{11} \hspace {-1pt}$$. Thus, the lattice it generates is simply $$\mathbb Z^{11} \!$$. For these reasons, I will use it below.

It is worth noting, however, that the degree-$$32$$ representation, which corresponds to $$\chi_6 + \chi_7 \hspace {-0.5pt}$$ and is not absolutely irreducible over the rationals, generates a lattice with an automorphism group of $$\mathbb Z_2 \! \times \mathrm M_{11} \hspace {-1pt}$$. This is, I think, as close to an automorphism group of $$\mathrm M_{11} \hspace {-1pt}$$ as can be achieved, as lattices, by definition, must have inversion symmetry.

I begin by enumerating the orbits of the short vectors of $$\mathbb Z^{11} \!$$ under the degree-$$11$$ representation of $$\mathrm M_{11} \hspace {-0.5pt}$$:

1. The $$22$$ norm-$$1$$ unit vectors fall into a single orbit. This is merely the $$11$$-orthoplex.
2. The $$220$$ vectors of (squared) norm $$2$$ fall into a single orbit. These can't have an automorphism group of just $$\mathrm M_{11} \hspace {-1pt}$$.
3. The $$1320$$ norm-$$3$$ vectors fall into two orbits, $$O_1 \! \hspace {1pt}$$ and $$O_{-1}$$, both of of size $$660$$. Each one is non-antipodal, in the sense that $$v \in O_i \! \implies \! {-v} \notin O_i$$. However, they form a pair, i.e. $$v \in O_i \! \implies \! {-v} \in O_{-i}$$. The representation doesn't act primitively on them, and it seems doubtful that such a large set of vectors could have an automorphism group of just $$\mathrm M_{11} \hspace {-1pt}$$.
4. The orbits of $$5302$$ vectors of norm $$4$$ include a scaled up orthoplex from norm $$1$$, two pairs of non-antipodal orbits of sizes $$330$$ and $$1320$$, and one antipodal orbit of size $$1980$$. The same doubtful comment from norm $$3$$ applies here as well.
5. There are $$15\,224$$ vectors of norm $$5$$. They include a pair of non-antipodal of orbits of size $$66$$. Notably, the representation acts primitively on these vectors, corresponding to the $$S_5$$ maximal subgroup of $$\mathrm M_{11} \hspace {-1pt}$$. We can choose one arbitrarily and proceed.

(At norms $$8$$ and $$11$$, one finds two more pairs of orbits with primitive actions of sizes $$165$$ and $$12$$, respectively. The latter is just a simplex, but perhaps the former could be used to construct a polytope which actually has an automorphism group of $$\mathrm M_{11} \hspace {-1pt}$$.)

At this point, I noticed something quite astonishing: The union of these two orbits are precisely the weight-$$5$$ code words of the unextended ternary Golay code, where the identity element of $$\mathbb F_3$$ is mapped to $$0$$, one non-identity element to $$+1$$, and the other to $$-1$$. So we can just take a non-antipodal coset of these code words, such that they are invariant under the action of our degree-$$11$$ representation of $$\mathrm M_{11} \hspace {-1pt}$$, as the vertices of our polytope. Explicitly, they can be given by the cyclic permutations of the following six vectors:

$$\pmb {v_1}$$ $$\pmb {v_2}$$ $$\pmb {v_3}$$ $$\pmb {v_4}$$ $$\pmb {v_5}$$ $$\pmb {v_6}$$ $$\pmb {v_7}$$ $$\pmb {v_8}$$ $$\pmb {v_9}$$ $$\pmb {v_{10}}$$ $$\pmb {v_{11}}$$ $$\pmb {\sum v_i}$$
$$+1$$ $$+1$$ $$+1$$ $$\hphantom + 0$$ $$+1$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$+1$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$+5$$
$$+1$$ $$\hphantom + 0$$ $$-1$$ $$+1$$ $$-1$$ $$-1$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$-1$$
$$-1$$ $$-1$$ $$+1$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$-1$$ $$+1$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$-1$$
$$+1$$ $$-1$$ $$\hphantom + 0$$ $$-1$$ $$\hphantom + 0$$ $$+1$$ $$\hphantom + 0$$ $$-1$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$-1$$
$$-1$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$-1$$ $$-1$$ $$\hphantom + 0$$ $$+1$$ $$+1$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$-1$$
$$+1$$ $$\hphantom + 0$$ $$+1$$ $$-1$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$-1$$ $$\hphantom + 0$$ $$-1$$ $$\hphantom + 0$$ $$\hphantom + 0$$ $$-1$$

Magma manages to construct the polytope without too much effort, and after a few days of computation, $$\color {red} {\textbf {fails to}}$$ confirm that its automorphism group is indeed $$\mathrm M_{11} \hspace {-1pt}$$.

## Polytope Details

The polytope has a volume of $$\frac {9 \cdot 509} {32 \cdot 25 \cdot 7 \cdot 11}$$ Remarkably, this fills only $$\frac {3^5 \cdot 509} {2^{11} \cdot 5^{13/2} \, \pi^5} \approx 5.649 \cdot 10^{-6}$$ of the volume of the circumscribed ball $$B^{11} \! \left ( \sqrt 5 \right )$$.

Its face structure is:

$$\pmb d~~$$ $$\textbf{Simplices}~~$$ $$~~~\textbf{Volume}\\~~\textbf{(per face)}$$ $$\hspace{0.5pt}\textbf{Rectified}~~\\\textbf{Simplices}~~$$ $$~~~\textbf{Volume}\\~~\textbf{(per face)}$$
$$0~~$$ $$~~~~~~~~\hspace{3pt}66~~\\\text {vertices}~~$$
$$1~~$$ $$~~\hspace{4pt}660~~\\\text {edges}~~$$ $$~~\,\text {(length)}\\~~~~~\,\sqrt 6$$
$$2~~$$ $$~~~~~~~2200~~\\\text {triangles}$$ $$~~\hspace{2pt}\text {(area)}\\~~\hspace{1pt}3 \sqrt 3/2$$
$$3~~$$ $$~~~~~~~~~\hspace{3pt}3960~~\\\text {tetrahedra}~~$$ $$~\sqrt 3$$ $$~~~~~~~~~~~495~~\\\text {octahedra}~~$$ $$~~~~~\,4 \sqrt 3$$
$$4~~$$ $$5544~~$$ $$~~~3 \sqrt 5/8$$ $$792~~$$ $$~~~~~~~~33 \sqrt 5/8$$
$$5~~$$ $$5544~~$$ $$~~9 \sqrt 2/40$$ $$924~~$$ $$~~~~~117 \sqrt 2/20$$
$$6~~$$ $$3960~~$$ $$~~3 \sqrt 7/80$$ $$792~~$$ $$~~~~~171 \sqrt 7/80$$
$$7~~$$ $$1980~~$$ $$~3 \sqrt 6/280$$ $$495~~$$ $$~~~~~~~~~~\,9 \sqrt 6/7$$
$$8~~$$ $$660~~$$ $$~~~27/4480$$ $$220~~$$ $$~~~~\,6669/4480$$
$$9~~$$ $$132~~$$ $$\sqrt {30}/4480$$ $$66~~$$ $$251 \sqrt {30}/2240$$
$$~~~~~~~~~10~~\\\text {(facets)}~~$$ $$12~~$$ $$~~\displaystyle \frac {3 \sqrt {11}} {44\,800}$$ $$12~~$$ $$~~~\displaystyle \frac {3039 \sqrt {11}} {44\,800}$$

$$d$$-simplices have $$d + 1$$ vertices, and rectified $$d$$-simplices (such as the rectified $$10$$-simplex) have $$d(d + 1)/2$$ vertices.

The automorphism group of the abstract polytope generated by this polytope is the symmetric group $$S_{12} > \mathrm M_{11} \hspace {-1pt}$$. It acts transitively on both of the types of facets listed in the bottom row of the table.

The inner products between any given vertex and the full set of vertices is the multiset $$\left \{5, 2^{20}, -1^{45} \right \}$$. By taking both orbits and rescaling the vertices down to the unit sphere, we can create an antipodal $$(11, 132, 2/5)$$ spherical code. This is superior to the $$(11, 78, 2/5)$$ code listed on Dr. Neil Sloane's webpage. I was optimistic that it might even be optimal, but Prof. Henry Cohn very kindly calculated an even better $$(11, 172, 2/5)$$ code at my prompting.

## Further Extensions

Is this construction known? What about the dual of this polytope? It has only $$24$$ vertices, although they no longer lie on a single sphere. Do polytopes exist with automorphism groups matching any given finite group?

What about lattices? Is the $$\mathbb Z_2 \! \times G$$ factor necessary? If so, when? I found that it also occurred when constructing two $$76$$- and one $$77$$-dimensional lattices from the invariant forms corresponding to the (absolutely irreducible over the rationals) $$\chi_4$$, $$\chi_5$$, and $$\chi_6$$ characters of the sporadic pariah Janko $$\mathrm J_1 \hspace {-1pt}$$ group.

• Every finite group is the automorphism group of a graph. – Mark Sapir Dec 18 '20 at 2:55
• A very interesting polytope indeed! Can you say something about the action of the symmetry group on the edges? How many edge orbits are there? Which vertices form an edge, and how long are they? – M. Winter Dec 18 '20 at 21:47
• I am getting the impression that the polytope you describe might be the "rectified 11-simplex": it has 66 vertices, 660 edges, all its faces are simplices or rectified simplices, it is vertex- and edge-transitive, and if its circumradius is $\sqrt 5$ then its edge length is $\sqrt 6$. But the symmetry group of the rectified simplex is $A_{11}$ if I am not missing something. If your $M_{11}$-polytope is truly different then this similarity is astonishing. – M. Winter Dec 19 '20 at 0:16
• I am especially suspicious because all facets of your polytope are (rectified) simplices. As far as I know, every vertex-transitive polytope with such types of facets should be classified. They are called (Elte's) semi-regular polytopes, but the list on wikipedia does not reach to dimension 11. – M. Winter Dec 19 '20 at 0:18
• @OzoneNerd A polytope with all vertices on the sphere and all edges of the same length has a unique realization of its combinatorial type (aka. abstract polytope) up to reorientation (this follows from Cauchy's rigidity theorem). Could it be that magma is not computing the full Euclidean symmetry group, but only transformations that are defined over, say, $\Bbb Q$. This would still be interesting. So, I don't think you should delete your answer, but extend it by a remark! – M. Winter Dec 19 '20 at 9:49