Indeed, since the Monster is simple and non-Abelian, these unitary matrices all necessarily have determinant 1 (so are not reflections). Proof: Consider the composition of the representation rho : G --> U(n) with det : U(n) --> U(1). This yields a homomorphism from G to U(1). By simplicity, either the kernel or image must be trivial. The former case is impossible, as the Monster is non-Abelian, so every element must have determinant 1. -- APG.
Sent: Thursday, April 16, 2020 at 9:25 PM From: "Victor Miller" <victorsmiller@gmail.com> To: "math-fun" <math-fun@mailman.xmission.com> Subject: Re: [math-fun] More Ignorance about Monster Rep. Theory
If by rotations you mean orthogonal matrices, then, yes. It's a result (I think of Weyl) that, in the right basis, the image of the group elements by a representation is unitary. The wikipedia article on Monstrous Moonshine gives a good summary: https://en.wikipedia.org/wiki/Monstrous_moonshine
Victor
On Thu, Apr 16, 2020 at 3:44 PM Brad Klee <bradklee@gmail.com> wrote:
Here are a few more questions, possibly interesting, but currently beyond my depth:
Are all irreducible representations rotations?
If no, are any of the monster irreps rotational?
Does "Monstrous moonshine" imply that the J-invariant is a Hilbert Series for some rep. of monster group?
Or more specifically, can J(q) be computed by some sort of Molien equation or correlation table?
Do these last two questions make any sort of sense, or do they sound like drunken ravings?
In memoriam John Conway, right now could be a good time to throw out some speculation, even if it is over the moon, so to speak.
Cheers,
Brad _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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