From: greg@math.ucdavis.edu (Greg Kuperberg) Subject: Re: F4 and the exceptional Jordan algebra Date: 5 Sep 2000 21:00:03 -0500 Newsgroups: sci.math.research Summary: [missing] In article <8p1b9a$o9j$1@Urvile.MSUS.EDU>, John Baez wrote: >The exceptional Jordan algebra is 27-dimensional, and its subspace >of traceless elements is 26-dimensional. F4 is the group of >automorphisms of the exceptional Jordan algebra, and it is >52-dimensional. 26 + 26 = 52. Is there any natural way to write >the Lie algebra of F4 as the direct sum of two copies of the space >of traceless elements in the exceptional Jordan algebra? Perhaps >it would help that this Lie algebra has a representation on this >26-dimensional space. But of course the adjoint representation of >F4 is irreducible, so it does not decompose as the sum of 2 copies >of this 26-dimensional representation. That puts a lower bound on >how sneaky we'd need to be to get this idea to work. You can at least come close. I'll use the vulgar notation V_26 and V_52 for the two representations. The weight diagram of V_26 is the vertices of a 24-cell plus two at the origin. (The 24-cell is a regular polytope that arises as the Voronoi region of conjectured densest sphere packing in R^4. The vertices are at (+-1,+-1,+-1,+-1) and (+-2,0,0,0) and permuted coordinates.) The weight diagram of V_52 is a super-multiset of that of V_26: It has multiplicity 4 at the origin instead of 2, and it also has the vertices a second, dual 24-cell that is larger by a factor of sqrt(2). So you would have your splitting if you could decide how to divide the 4-dim weight space at 0 into two 2-dim weight spaces. Also F_4 contains D_4 and the restriction of V_52 and V_26 is edifying but not quite enoguh. V_26 splits as 8+8+8+2, where we use the three triality representations of D_4 (the defining representation of SO(8) and the two semi-spinorial representations.) V_52 splits as 28+8+8+8, using the adjoint representation of D_4 and the triality representations. So the splitting into long and short roots again arises, but no particular splitting of the weight space at 0 does. Actually you see the same thing with G_2: V_14 looks like a short V_7 and a long V_7, but restriction to A_2 does not give you a splitting of the 0 weight space. Hmm... -- /\ Greg Kuperberg (UC Davis) / \ \ / Visit the Math ArXiv Front at http://front.math.ucdavis.edu/ \/ * All the math that's fit to e-print * ============================================================================== From: "Noam D. Elkies" Subject: Re: F4 and the exceptional Jordan algebra Date: 6 Sep 2000 14:30:04 -0500 Newsgroups: sci.math.research In article <8p3rl9$pa7$1@manifold.math.ucdavis.edu>, Greg Kuperberg wrote: >In article <8p1b9a$o9j$1@Urvile.MSUS.EDU>, >John Baez wrote: >>The exceptional Jordan algebra is 27-dimensional, and its subspace >>of traceless elements is 26-dimensional. F4 is the group of >>automorphisms of the exceptional Jordan algebra, and it is >>52-dimensional. 26 + 26 = 52. Is there any natural way to write >>the Lie algebra of F4 as the direct sum of two copies of the space >>of traceless elements in the exceptional Jordan algebra? Perhaps >>it would help that this Lie algebra has a representation on this >>26-dimensional space. But of course the adjoint representation of >>F4 is irreducible, so it does not decompose as the sum of 2 copies >>of this 26-dimensional representation. That puts a lower bound on >>how sneaky we'd need to be to get this idea to work. >You can at least come close. I'll use the vulgar notation V_26 and >V_52 for the two representations. The weight diagram of V_26 is the >vertices of a 24-cell plus two at the origin. (The 24-cell is a regular >polytope that arises as the Voronoi region of conjectured densest sphere >packing in R^4. The vertices are at (+-1,+-1,+-1,+-1) and (+-2,0,0,0) >and permuted coordinates.) The weight diagram of V_52 is a super-multiset >of that of V_26: It has multiplicity 4 at the origin instead of 2, and it >also has the vertices a second, dual 24-cell that is larger by a factor >of sqrt(2). So you would have your splitting if you could decide how >to divide the 4-dim weight space at 0 into two 2-dim weight spaces. >Also F_4 contains D_4 and the restriction of V_52 and V_26 is edifying >but not quite enoguh. V_26 splits as 8+8+8+2, where we use the three >triality representations of D_4 (the defining representation of SO(8) >and the two semi-spinorial representations.) V_52 splits as 28+8+8+8, >using the adjoint representation of D_4 and the triality representations. >So the splitting into long and short roots again arises, but no particular >splitting of the weight space at 0 does. >Actually you see the same thing with G_2: V_14 looks like a short V_7 >and a long V_7, but restriction to A_2 does not give you a splitting of >the 0 weight space. Hmm... Actually, there are not two but three cases of this. Here they are, together with their Dynkin diagrams and relevant dimensions etc.: G |rank(G)|dim(V)|dim(G)| p |Dynkin diagram ---+-------+------+------+---+--------------- B_2| 2 | 5 | 10 | 2 | @=>=@ G_2| 2 | 7 | 14 | 3 | @=>=@ [sorry, no triple edge in ASCII] F_4| 4 | 26 | 52 | 2 | @--@=>=@--@ In each case there is a representation V of dimension half of dim(G); the Dynkin diagram is symmetrical about a directed edge of multiplicity p; and there are dim(V)-rank(G) short roots, and as many long roots in a homothetic configuration expanded by a factor sqrt(p). There does not seem to be a way to exploit this in characteristic zero to construct G as per John Baez's query. However, some remarkable things happen in characteristic p. Morally speaking, in characteristic p you can't distinguish short from long roots. So, for instance, the Lie groups B_n and C_n (which have the same diagram except that the double edge points the other way) become isogenous mod 2. In more down-to-earth terms, this means that the symplectic group Sp_{2n} and the orthogonal group O_{2n+1} look almost the same in characteristic 2. This is because in this characteristic, the pairing =Q(x+y)-Q(x)-Q(y) obtained from an orthogonal form Q(.) is alternating, and if Q is nondegenerate on a space V of odd dimension 2n+1 then <.,.> has 1-dimensional kernel and the quotient of V by this kernel is a symplectic space of dimension 2n. For n=2, we get an isogeny between B_2 and C_2 -- but B_2 and C_2 are already isomorphic, so there is a new isogeny from B_2 to itself! It turns out that this isogeny reverses the direction of the edge, and its square is the Frobenius map that takes each element of B_2 to its coordinatewise square. This can be explained as follows: B_2 is the Lie group of both Sp_4 and O_5. Those groups are always isogenous: from a symplectic space W of dimension 4, we get an orthogonal space V of dimension 5 as the orthogonal complement in wedge^2(W) of the symplectic form; conversely from V we recover W as a spin representation. But in characteristic 2, we have another way to get a 4-dimensional representation from V: take n=2 in the description in the previous paragraph. Exercise: check that this yields a map from Sp_4 to itself whose square is the Frobenius map. Remark: if we look at Sp_4(F) where F is a finite field of order 2^k, and k is odd, there is another such map, namely the (k+1)/2 power of Frobenius. In Sp_4(F), these maps are automorphisms, and composing one with the inverse of the other yields an unexpected involution of Sp_4(F). The subgroup of Sp_4(F) fixed under this involution is isomorphic with the Suzuki group Sz(2^k), which is simple once k>1; these are the only sporadic simple groups whose order is not a multiple of 3. It turns out that for G=G_2 and G=F_4 there are similar maps from G to G in characteristic p which reverse the arrow and whose square is the Frobenius map. Again these yield unexpected involutions of G(F) for a finite field F of order p^k if k is odd, and the subgroup of G(F) fixed under an involution is a simple group once k>1; these are the two Ree families of finite simple groups. Now to return to John Baez's question. For the following information I thank Benedict Gross, who gives Jacob Lurie's thesis as a reference. [This is an *undergraduate* thesis! Jacob graduated from Harvard last year.] In each case, the adjoint representation of G becomes reducible in characteristic p: it contains a copy of V, and the quotient of the adjoint representation by this copy of V is a new irreducible representation V' almost isomorphic with V. "Almost" because V' is obtained from V via our arrow-reversing isogeny from V to itself. In particular, V' has the same dimension as V, explaining the numerology dim(G)=2*dim(V) in the table above. Thus the answer to John's question "Is there any natural way to write the Lie algebra of G as the direct sum of two copies of V?" is: not quite, but in characteristic p the Lie algebra can be obtained as an extension of V by the almost-isomorphic representation V'! --Noam D. Elkies Department of Mathematics, Harvard University ============================================================================== From: "Noam D. Elkies" Subject: Re: F4 and the exceptional Jordan algebra Date: 6 Sep 2000 17:30:01 -0500 Newsgroups: sci.math.research In article <8p63li$gc5$1@news.fas.harvard.edu>, I wrote: >[...] > > G |rank(G)|dim(V)|dim(G)| p |Dynkin diagram >---+-------+------+------+---+--------------- >B_2| 2 | 5 | 10 | 2 | @=>=@ >G_2| 2 | 7 | 14 | 3 | @=>=@ [sorry, no triple edge in ASCII] >F_4| 4 | 26 | 52 | 2 | @--@=>=@--@ > >In each case there is a representation V of dimension half of dim(G); >the Dynkin diagram is symmetrical about a directed edge of multiplicity p; >and there are dim(V)-rank(G) short roots, and as many long roots in a >homothetic configuration expanded by a factor sqrt(p). Oops -- change "dim(V)-rank(G)" to "dim(V)-(rank(G)/2)". --Noam D. Elkies Department of Mathematics, Harvard University