Conway made the comment that the Monster group represents the symmetries of a shape in 196,883 dimensions, something like a "star you hang on a Christmas tree." My question is, What do we know (or conjecture) about the enigmatic shape whose symmetries are captured by the Monster? Is the shape analogous to a star (pointy) or convex like a polygon / polytope, or is it more likely to be something with a different topology, analogous to a high-dimensional torus? Conversely, what do we know it is not like, what have we ruled out? P.S. Also, if anyone knows Conway, please wish him well, and maybe they might ask him this question? Update May 2020 RIP Mr Conway

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    $\begingroup$ I don't know what Conway meant, but when you have a faithful representation, then you can built an orbit polytope in that dimension with the monster group as its symmetry group. If its a real representation, then its a convex polytope, otherwise its more like a set of point in a complex vector space. $\endgroup$ – M. Winter Mar 23 '20 at 10:16
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    $\begingroup$ Here is described a "Monster graph" whose automorphism group is the Monster group (described by Griess). Maybe this graph could be the 1-skeleton of a polytope in 196883 dimensions? neverendingbooks.org/the-monster-graph-and-mckays-observation $\endgroup$ – Ian Agol Mar 24 '20 at 5:08

It is possible that Conway was referring to the generic construction that works for all finite groups equipped with faithful representations, given in the other answers. However, I think it is more likely that Conway was referring to a construction that is specific to the monster, hinted at in Ian Agol's comment.

In section 14 of Conway's 1985 Inventiones paper, "A simple construction for the Fischer-Griess monster group", Conway points out that in the 196883 dimensional representation, there is a collection of distinguished lines fixed by centralizers of elements in conjugacy class 2A (in fact, 196883 decomposes as 1+4371+96255+96256 under the action of such a centralizer). Elements in this conjugacy class are known as Fischer involutions or transpositions, and the lines are called axes of transpositions. The centralizer of a transposition is a double cover of Fischer's Baby Monster sporadic group, so there are $\frac{|\mathbb{M}|}{2|\mathbb{B}|} \sim 9 \times 10^{19}$ of these axes. Choosing a nonzero point on an axis, and taking its orbit yields an arrangement of points (or we may consider the convex hull). I claim that this arrangement has Monster symmetry.

To prove this, we use the fact that for a pair $(x,y)$ of 2A elements, their product lies in 2A if and only if the angle between the axes is a particular value. In fact, the conjugacy class of the product is uniquely determined by the angle, except for classes 3C and 4B, as we can see from the "McKay E8 diagram" in Conway's paper:

Fig.3, Inner products of transposition vectors

We therefore have a natural construction of a graph from this polytope, whose vertices correspond to axes (or orbit elements on the axes), and whose edges correspond to those pairs whose product lies in class 2A. This graph is the "monster graph" mentioned in Ian Agol's comment, and Griess showed that its automorphism group is precisely the monster.

These axes appear in the theory of vertex algebras, in the following way. The 196883 dimensional representation is naturally embedded as the space of Virasoro primary vectors in the weight 2 subspace of the "Moonshine Module" vertex operator algebra $V^\natural$. On each axis, there is a distinguished "Ising vector" that generates a vertex algebra isomorphic to the $L(1/2,0)$ minimal model. Miyamoto showed that any Ising vector in a vertex algebra yields a "Miyamoto involution", and for $V^\natural$, these are precisely the 2A elements.

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    $\begingroup$ Ooh, I much prefer this construction! The minimal permutation representation of the Monster has the same number of elements (97239461142009186000) as there are vertices in your polytope, so your polytope is (in a certain sense) the 'simplest' polytope with Monster symmetry. Elegant! $\endgroup$ – Adam P. Goucher May 5 '20 at 9:03
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    $\begingroup$ @JamesEadon Fundamentally, it is a collection of about $10^{20}$ points in 196883-dimensional space. If you had five points symmetrically arranged in the plane, you could choose to connect them like a pentagon, or like a star (or perhaps try some more exotic thing). Similarly, there are options here, but I do not have the vocabulary to describe them. Most people would choose the analog of "pentagon", i.e., the convex hull, because it is conceptually simpler. $\endgroup$ – S. Carnahan May 7 '20 at 4:38
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    $\begingroup$ @JamesEadon The pentagon was an analogy that was useful because of the link to star shapes. The points are very regularly spaced, in the sense that the shape has lots of symmetry. $\endgroup$ – S. Carnahan May 11 '20 at 4:01
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    $\begingroup$ @Rudi_Birnbaum Assuming you are asking whether the boundary of the convex hull decomposes into flat 196882-simplices joined to their neighbors at nonzero angles, I do not have an answer to that question. $\endgroup$ – S. Carnahan May 13 '20 at 14:50
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    $\begingroup$ @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. $\endgroup$ – S. Carnahan May 24 '20 at 22:29

In the penultimate chapter of Sphere Packings, Lattices and Groups, the authors define a $196884$-dimensional real vector space and a faithful representation of the Monster group on that space.

Now, because we know the degrees of the irreducible representations of the Monster, this representation must necessarily decompose as the direct sum of a trivial 1-dimensional representation and a faithful real $196883$-dimensional representation.

Choose an vector $v$ in that $196883$-dimensional subspace in general position, and normalise it to have unit length. Let $X$ be the orbit of $v$ under the action of the Monster; it follows that $X$ has the same number of elements as the Monster. Moreover, $X$ is a subset of the unit sphere.

Let $P$ be the convex hull of $X$. Then $P$ is a $196883$-dimensional sharply vertex-transitive convex polytope with symmetry group isomorphic to the Monster group.

EDIT: S. Carnahan's answer provides a more elegant construction, taking $v$ to be a point on a line fixed by a Fischer involution instead of a point in general position. The resulting polytope has 97239461142009186000 vertices, which is minimal (as any permutation representation of the Monster has at least this many vertices).

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    $\begingroup$ I just want to point out that the claim $\mathrm{Aut}(P)\cong M$ (where $M$ is the monster group) is far from obvious. It can be proven though. $\endgroup$ – M. Winter Mar 23 '20 at 12:05
  • $\begingroup$ Hi Adam, thanks for answering. You say, "sharply vertex-transitive convex polytope" - I understand convex polytope as a higher-dimensional hexagon, say. But what does "sharply vertex-transitive" mean, in language a university MSc. physics / computer science grad can understand, in the context of the Monster? Basically, I'm trying to visualise the shape of the monster in terms of a 2D or 3D analogue. Is it like a high-dimensional regular hexagon or similar? It's not a completely regular polytope, is it? $\endgroup$ – JamesEadon Mar 23 '20 at 13:13
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    $\begingroup$ @JamesEadon If you did the same thing with the alternating group A5 and its 3-dimensional irreducible representation, you'd get a [not necessarily uniform] snub dodecahedron (see en.wikipedia.org/wiki/Snub_dodecahedron for a picture). 'Sharply vertex-transitive' means that, given any ordered pair $(u, v)$ of vertices, there's a unique element $g$ of the symmetry group such that $g(u) = v$. $\endgroup$ – Adam P. Goucher Mar 23 '20 at 13:37
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    $\begingroup$ @JamesEadon to me it seems it cant be a $196883$ dimensional simplex, because that would have the symmetry of the fully symmetric group of the $97239461142009186000$ vertices ($S_{97239461142009186000}$), I suppose. $\endgroup$ – Raphael J.F. Berger May 13 '20 at 7:59
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    $\begingroup$ @Rudi_Birnbaum I was thiking myself about the symmetry. The monster group has a fair bit of structure, I believe, implying (I'm guessing) that the shape might not be as simple as some kind of a high-D simplex. $\endgroup$ – JamesEadon May 15 '20 at 16:28

It is not too surprising that the Monster group $M$ is the symmetry group of something geometrical. E.g. every group is the group of symmetries of some convex polytope.

You can even make it a vertex-regular polytope $P$, which means that $\mathrm{Aut}(P)\cong M$ acts regularly on the vertices of $P$. So the vertices of $P$ can be set in one-to-one correspondence with the elements of $M$.

A convex polytope is not much more than its set of vertices, and if you are okay with that point of view than it is easy to construct from any faithful representation a discrete set of $>10^{20}$ points in $>$ 196883 (complex) dimensions for which the (unitary) symmetry group is isomorphic to $M$.

Here are the details

Let $\rho:M\to \mathrm O(\Bbb R^d)$ be some irreducible faithful orthogonal representation of the monster group $M$. Then $\Gamma:=\mathrm{im}(\rho)\subset\mathrm O(\smash{\Bbb R^d})$ is a matrix group isomorphic to $M$. For every point $p\in\smash{\Bbb R^d}\setminus\{0\}$ we obtain the so called orbit polytope

$$P:=\mathrm{Orb}(\Gamma,p):=\mathrm{conv}(\Gamma p)\subset\Bbb R^d,$$

that is, the convex hull of the orbit of $p$ under $\Gamma$. This is a $d$-dimensional convex polytope and clearly has $\Gamma\subseteq\mathrm{Aut}(P)$. One can show [1] that for an appropriate choice of $p$ this gives equality, that is $\mathrm{Aut}(P)\cong M$.

Now, your representation might not be real, but complex valued. But from that we can get a real representation of twice the dimension (and therefore a convex polytope of twice the dimension). Alternatively, you can let $\rho:M\to\smash{\mathrm U(d)}$ be a unitary representation, $\Gamma:=\mathrm{im}(\rho)\subset \mathrm U(n)$ as well as $p\in\smash{\Bbb C^d\setminus\{0\}}$. In that case the orbit $\Gamma p\subset\smash{\Bbb C^d}$ does not yield a convex polytope but just a discrete set of points in a complex vector space. Again, it is possible [2] to choose $p$ appropriately so that

$$M\cong \mathrm{Aut}_{\mathrm U(n)}(\Gamma p),$$

a set of points in (complex) dimension 196883 with the monster group $M$ as its (unitary) symmetry group.

[1] Babai, László. "Symmetry groups of vertex-transitive polytopes."

[2] Friese, Erik. "Unitary Groups as Stabilizers of Orbits"

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    $\begingroup$ Which kind of "set of points" is $X$ in the unitary case? In which sense it is "196883-dimensional"? $\endgroup$ – Qfwfq Mar 23 '20 at 11:13
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    $\begingroup$ @Qfwfq It is a set of points in $\Bbb C^d$, so 196883 complex dimensions. $\endgroup$ – M. Winter Mar 23 '20 at 11:13
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    $\begingroup$ (Ah, yes, it's just a discrete set. For some reason I didn't realize you don't make an analogous operation to the convex envelope in the unitary case.) $\endgroup$ – Qfwfq Mar 23 '20 at 11:18
  • $\begingroup$ Dear M. Winter. The jargon does give me a feel for the amazing and beautiful mathematics involved. It's mind-blowing actually. If I may, please can you translate your answer into something understandable by a MSc Physics / Computer Science grad? Is this thing a high-dimensional hexagon or something? I'm just trying to visualise it, in 3D or 2D analogue terms. A weirdly special shape for such symmetry to only exist at such a specific high dimension? $\endgroup$ – JamesEadon Mar 23 '20 at 13:20
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    $\begingroup$ @JamesEadon In essence: yes, there is a high dimensional shape whose symmetries are exactly described by the Monster group. In fact, this shape is a polytope (a 196883-dimensional version of a hexagon, cube, ...). But it is the nature of the Monster group that this polytope exists only in such a high dimension, and all smaller dimensional examples will miss some of the complexity of that group. This polytope is nothing special though, as something like this exists for almost all groups. It is also not regular, or otherwise special besides its relation to the Monster group (as far as I know). $\endgroup$ – M. Winter Mar 23 '20 at 13:34

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