From: jamesrheckman@gateway.netnospam (Jim Heckman) Subject: Re: The Monster and the Leech Date: 27 Sep 2000 10:04:45 GMT Newsgroups: sci.physics.research Summary: [missing] John Baez wrote in message <8qc20v$t3l$1@mortar.ucr.edu> >In article <8qa6ht$7gn$1@newsflash.concordia.ca>, >MCKAY john wrote: >>Dynkin diagrams give a description of the roots. The Leech >>lattice is characterized by having no roots. I was *really* hoping someone would keep this thread going awhile longer, but since no one else has, let me try to revive it. :-) > [...] > >But I don't know if that's what you mean by "root". If we're >trying to cook up a Dynkin diagram, what really matters we can >find a basis of lattice vectors s.t. reflections through these >vectors act as symmetries of the lattice. I guess these vectors >are really what deserve the name "roots". In the context of orthogonal groups in N-d Euclidean(?) space E^N, "roots" are indeed vectors that describe reflection symmetries across hyperplanes. In (finite?) reflection groups, i.e., those that can be generated by a set of their own reflections, the set of all roots contains subsets that form systems of "simple" roots which satisfy certain mutual obtuseness conditions. (Also, it turns out every root is conjugate to a simple root.) If the reflection group is "effective" or "essential", i.e., it leaves no non-trivial subspace of E^N invariant, then a system of simple roots is indeed a basis for E^N. >Is it true that the Leech lattice has no roots in this sense? My question exactly! If so, does this mean the orthogonal symmetry group of the Leech lattice has *no* det=1 operations, i.e., that it's a pure rotation group? For that matter, what *is* the full symmetry group of the Leech lattice? (I think) I know it's a group of twice the order of the sporadic simple group Co_1, which I can look up as having Out(Co_1) = 1, and Schur multiplier Z_2. So..., the Leech's symmetry group, which I believe is called Co_0 (right?) must be either Co_1 x Z_2 or the covering group of Co_1 (right?). If the latter, then Co_0 must indeed be a rotation group, since it's a finite subgroup of O(24,R) with no subgroup of index 2 (right?). > [...] > >Btw, there's also a cool sense in which the Leech lattice *is* >a Dynkin diagram! Do tell! -- Jim Heckman