Can someone explain how to picture or construct the Monster Group?
What is the simplest way to fathom the Monster Group?
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4@user3163: With great difficulty? – 2010-11-05
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0It isn't the symmetry of any solid that you could make, I'm not even sure if it can be the symmetries of a higher dimensional object. – 2010-11-05
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1See http://en.wikipedia.org/wiki/Monster_group and the references therein. Of course every finite group is the symmetry group of some object in some $\mathbb{R}^n$. – 2010-11-05
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0@Robin Chapman, oh cool - that's not obvious to me though. – 2010-11-05
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8@Robin and muad: Any finite group is the group of hyperbolic isometries of some compact hyperbolic surface. It's apparently a theorem of Gromov's that such a surface isometrically embeds in $\mathbb R^5$: http://mathoverflow.net/questions/37708/nash-embedding-theorem-for-2d-manifolds – 2010-11-05
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2Every finite group is the automorphism group of some finite simple graph $G$. If $G$ has $n$ vertices start with a regular $n$-simplex $S$ in $\mathbb{R}^{n-1}$. Each edge of $G$ corresponds to an edge of $S$. Push the midpoint of that edge out slightly to create a new vertex. One gets a convex polytope with the same symmetry group as $G$. – 2010-11-05
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7Rumor has it that the simplest way to fathom the "Monster Group" is to behold it under the moonshine at John Conway's annual Halloween party. But, alas, you're a few days late for that. It was truly a surreal experience. – 2010-11-05
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0You can try reading "Symmetry and the Monster: The Story of One of the Greatest Quests of Mathematics" by Mark Ronan. It attempts to explain the math behind the monster at a layman (or maybe undergraduate math) accessible level. This book will only give you an overview of the classification of finite simple groups - to truly understand all the math would require years of dedicated study. – 2010-11-05
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1I was thinking it would be good to have something along the lines of this, if possible: http://www.scientificamerican.com/article.cfm?id=puzzles-simple-groups-at-play – 2010-11-05
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3I tried to read "Symmetry and the Monster" and found it very frustrating. Too much gee-whiz. – 2010-11-05
1 Answers
I think the answer really depends on what you mean by "fathom".
If you want a short construction of the Monster, there is a sketch by Conway in one of the later chapters of Sphere Packings, Lattices, and Groups (occasionally available on Google Books). The construction there goes through a nice progression of increasingly complicated exceptional objects, like the Golay code and the Leech lattice.
If you want to understand a "natural" object on which the Monster acts by symmetries, you should read up about vertex algebras. At our current state of knowledge (and depending on who you ask), the most natural object on which the monster acts is the monster vertex algebra $V^\natural$ (also known as the moonshine module, or the monster VOA), which is a graded vector space, together with some extra structure like a rather complicated multiplication operation $V^\natural \otimes V^\natural \to V^\natural((z))$. The construction of $V^\natural$ is given in the book Vertex operator algebras and the Monster by I. Frenkel, Lepowsky, and Meurman. A string theorist might say that it is given by orbifolding the Leech lattice CFT.
If you want to consider some basic facts about the Monster, you can have a look at the ATLAS of finite groups, or play around with the software GAP. Both have the character table, and orders of centralizers of elements, etc. Wilson showed that the monster is a Hurwitz group, so Ryan Budney's comment about acting on a Riemann surface holds for the minimum possible genus (about $10^{52}$).
If you want to understand finer points about the structure of the Monster group, you're pretty much out of luck. It's big enough that there are plenty of explicit questions whose answers we don't know. For example, the conjugacy classes of homomorphisms from $\mathbb{Z} \times \mathbb{Z}$ (i.e., pairs of commuting elements) are not classified, and $H^4$ with coefficients in $\mathbb{Z}$ is still unknown (very annoying).