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 highdimensional 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?

2$\begingroup$ 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 at 10:16

1$\begingroup$ Here is described a "Monster graph" whose automorphism group is the Monster group (described by Griess). Maybe this graph could be the 1skeleton of a polytope in 196883 dimensions? neverendingbooks.org/themonstergraphandmckaysobservation $\endgroup$ – Ian Agol Mar 24 at 5:08
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 1dimensional 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 vertextransitive convex polytope with symmetry group isomorphic to the Monster group.

7$\begingroup$ 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 at 12:05

$\begingroup$ Hi Adam, thanks for answering. You say, "sharply vertextransitive convex polytope"  I understand convex polytope as a higherdimensional hexagon, say. But what does "sharply vertextransitive" 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 highdimensional regular hexagon or similar? It's not a completely regular polytope, is it? $\endgroup$ – JamesEadon Mar 23 at 13:13

6$\begingroup$ @JamesEadon If you did the same thing with the alternating group A5 and its 3dimensional irreducible representation, you'd get a [not necessarily uniform] snub dodecahedron (see en.wikipedia.org/wiki/Snub_dodecahedron for a picture). 'Sharply vertextransitive' 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 at 13:37
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 vertexregular 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 onetoone 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 vertextransitive polytopes."

1$\begingroup$ Which kind of "set of points" is $X$ in the unitary case? In which sense it is "196883dimensional"? $\endgroup$ – Qfwfq Mar 23 at 11:13

1$\begingroup$ @Qfwfq It is a set of points in $\Bbb C^d$, so 196883 complex dimensions. $\endgroup$ – M. Winter Mar 23 at 11:13

1$\begingroup$ (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 at 11:18

$\begingroup$ Dear M. Winter. The jargon does give me a feel for the amazing and beautiful mathematics involved. It's mindblowing actually. If I may, please can you translate your answer into something understandable by a MSc Physics / Computer Science grad? Is this thing a highdimensional 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 at 13:20

4$\begingroup$ @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 196883dimensional 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 at 13:34