[math-fun] More Ignorance about Monster Rep. Theory
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
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
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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Actually, I was thinking more in an analogy with groups for three-dimensional solids, where it is always possible, regardless of degree, to specify generators in terms of Euler angles + the so-called "Wigner D matrices". This should be more restrictive than asking for unitary or orthogonal matrices in general. So I think this is probably not the case for Monster that group operations tie back into SO(3), this sounds wrong since the dimensions are so high. Maybe there are super-rotations where the Euler angles and D matrices are generalized? What I could be looking for, rather blindly and uncertainly, is a minimal set of parameters with some extra addition rules that give a cohesive explanation of //any// representation. As far as I know, Conway was not satisfied with any of the proposed geometric interpretations. Unfortunately, I'm also ignorant as to what he would have considered a "good-enough" explanation. Any thoughts? --Brad On Thu, Apr 16, 2020 at 3:26 PM Victor Miller <victorsmiller@gmail.com> wrote:
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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see also: https://www.youtube.com/watch?v=lbN8EMcOH5o On Thu, Apr 16, 2020 at 4:13 PM Brad Klee <bradklee@gmail.com> wrote:
Actually, I was thinking more in an analogy with groups for three-dimensional solids, where it is always possible, regardless of degree, to specify generators in terms of Euler angles + the so-called "Wigner D matrices". This should be more restrictive than asking for unitary or orthogonal matrices in general.
So I think this is probably not the case for Monster that group operations tie back into SO(3), this sounds wrong since the dimensions are so high.
Maybe there are super-rotations where the Euler angles and D matrices are generalized?
What I could be looking for, rather blindly and uncertainly, is a minimal set of parameters with some extra addition rules that give a cohesive explanation of //any// representation.
As far as I know, Conway was not satisfied with any of the proposed geometric interpretations. Unfortunately, I'm also ignorant as to what he would have considered a "good-enough" explanation. Any thoughts?
--Brad
On Thu, Apr 16, 2020 at 3:26 PM Victor Miller <victorsmiller@gmail.com> wrote:
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I took a course entitle "Simple Groups" from Conway. The only students left after the first few weeks were me and David Moulton (another Math Funster). Near the end of the course Conway did all sorts of amazing calculations in Monster and Baby Monster aided by a ping pong paddle which had various formulae pasted to both sides. He would calculate, aided with the paddle, turning it back and forth at a fierce pace. On Thu, Apr 16, 2020 at 5:23 PM Brad Klee <bradklee@gmail.com> wrote:
see also: https://www.youtube.com/watch?v=lbN8EMcOH5o
On Thu, Apr 16, 2020 at 4:13 PM Brad Klee <bradklee@gmail.com> wrote:
Actually, I was thinking more in an analogy with groups for three-dimensional solids, where it is always possible, regardless of degree, to specify generators in terms of Euler angles + the so-called "Wigner D matrices". This should be more restrictive than asking for unitary or orthogonal matrices in general.
So I think this is probably not the case for Monster that group operations tie back into SO(3), this sounds wrong since the dimensions are so high.
Maybe there are super-rotations where the Euler angles and D matrices are generalized?
What I could be looking for, rather blindly and uncertainly, is a minimal set of parameters with some extra addition rules that give a cohesive explanation of //any// representation.
As far as I know, Conway was not satisfied with any of the proposed geometric interpretations. Unfortunately, I'm also ignorant as to what he would have considered a "good-enough" explanation. Any thoughts?
--Brad
On Thu, Apr 16, 2020 at 3:26 PM Victor Miller <victorsmiller@gmail.com> wrote:
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Hi Victor, Thanks, this is very helpful. Now in addition to the thing about Christmas ornaments, we can guess that Conway's preferred explanation would also have something to do with the game of ping pong. Perhaps the Christmas ornaments could be volleyed back-and-forth, plus and minus, until one player makes a sign error, at which time the number of generated elements would increase by one. It could go on for a while... I'm busy checking whatever literature I can find to see if the ping-pong angle is mentioned anywhere else, but so far have only found an intro: Christopher S. Simons, "An Elementary Approach to the Monster", American Mathematical Monthly, 2005. https://www.jstor.org/stable/30037469?seq=1#metadata_info_tab_contents If anyone is interested, it looks like this article gives the elegant construction that Adam was mentioning in the other thread. --Brad On Thu, Apr 16, 2020 at 4:34 PM Victor Miller <victorsmiller@gmail.com> wrote:
I took a course entitle "Simple Groups" from Conway. The only students left after the first few weeks were me and David Moulton (another Math Funster). Near the end of the course Conway did all sorts of amazing calculations in Monster and Baby Monster aided by a ping pong paddle which had various formulae pasted to both sides. He would calculate, aided with the paddle, turning it back and forth at a fierce pace.
On Thu, Apr 16, 2020 at 5:23 PM Brad Klee <bradklee@gmail.com> wrote:
see also: https://www.youtube.com/watch?v=lbN8EMcOH5o
On Thu, Apr 16, 2020 at 4:13 PM Brad Klee <bradklee@gmail.com> wrote:
Actually, I was thinking more in an analogy with groups for three-dimensional solids, where it is always possible, regardless of degree, to specify generators in terms of Euler angles + the so-called "Wigner D matrices". This should be more restrictive than asking for unitary or orthogonal matrices in general.
So I think this is probably not the case for Monster that group operations tie back into SO(3), this sounds wrong since the dimensions are so high.
Maybe there are super-rotations where the Euler angles and D matrices are generalized?
What I could be looking for, rather blindly and uncertainly, is a minimal set of parameters with some extra addition rules that give a cohesive explanation of //any// representation.
As far as I know, Conway was not satisfied with any of the proposed geometric interpretations. Unfortunately, I'm also ignorant as to what he would have considered a "good-enough" explanation. Any thoughts?
--Brad
On Thu, Apr 16, 2020 at 3:26 PM Victor Miller <victorsmiller@gmail.com
wrote:
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (3)
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Adam P. Goucher -
Brad Klee -
Victor Miller