What is the geometric shape of the Monster sporadic group?

Question

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

Answer 1

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:

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.

Answer 2

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).

Answer 3

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 $M$.

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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.

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

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