From: Lubos Motl Subject: Re: octonions; d=11 SUGRA; F-theory Date: Sun, 16 Sep 2001 16:31:10 GMT Newsgroups: sci.physics.research Summary: Global distinctions among similar Lie groups You disagreed about some global properties of the maximal compact subgroup of E8(8). Neither of you is correct, I think. > > J.Baez: I think Spin(16) is a subgroup of E8(8), not SO(16). Are you > > claiming otherwise... Therefore I think, zirkus, that your reference is not relevant: > Yes, and please also read my longer reply if it gets posted. By coincidence, I > just looked at a brand new paper [1] which shows in the first paragraph on page > 8 that SO(16) really *is* the maximal compact subgroup of E8(8).... > ... [1] http://arxiv.org/abs/hep-th/0109005 because it discusses the commutators only, i.e. the paper analyzes the groups locally - at the level of Lie algebras. If you open Green, Schwarz, Witten: Superstring Theory; appendix 6.A in the first volume, you will find a construction of E8. It is constructed as the 120-dimensional SO(16) adjoint plus spinorial generators transforming as 128 added to it. This "SO(16)" subgroup is not really SO(16) but rather Spin(16)/Z2. The John's answer Spin(16) is surprisingly not correct either, I think. The truth is just in between both of you: Spin(16)/Z2 (similarly like the heterotic string has gauge group Spin(32)/Z2, see the bottom of this letter). You know that Spin(16) has a center Z2 x Z2 (just like Spin(4) = SU(2) x SU(2), for example, or any SO(4k)). Clearly, the rotation by 2.pi is not contained in the center because the spinors in 128 are odd under that. Note that SO(16) is Spin(16) over another Z2 and SO(16) still has a center Z2 so you can construct SO(16)/Z2. The other independent Z2 in the center of SO(16) (that is seen at the end of the previous short paragraph, too) is generated by the element rotating all 8 independent planes by 1.pi: it acts as the minus unit matrix in the vector representation of SO(16); it is a spatial reflection. We must be careful whether this generator of Z2 involves the rotation by 2.pi; the difference is the sign when acting on the 128 spinor. The minus sign seen in the vector representation is cancelled for the (adjoint) tensors because it has two indices. However this 8-uple rotation by pi acts trivially also on the spinor 128 because they are eigenstates of this 8-uple pi-rotation and the eigenvalue is (-1) to the power of the sum of all the weight-coordinates. This sum is odd because of chirality (we have an odd number of -1/2 for the positively-chiral spinors) and also for the adjoint roots e_i+e_j and e_i-e_j. You might think that a rotation by "pi" (around x or y) changes all "up" into "down" and therefore the spinor is not an eigenstate of the multiple rotation. But it is just a misleading 3-dimensional intuition. The 8-uple rotation is more like a product of pi-rotation around the "z" axis under which the "up" and "down" states are eigenstates. I discussed the case of E8 because I believe that E8(8) behaves in the same way. You can still construct it as 120+128 of SO(16); the only difference is that you add a minus sign on the RHS of the {Q,Q}=J commutator; this step effectively multiplies Q by "i" and creates another (noncompact) version of the group while the Spin(16)/Z2 subgroup is not affected. These are the only two forms of E8 with an Spin(16)/Z2 subgroup, I believe. By the way, the heterotic-O theory has a gauge group Spin(32)/Z2 because it contains spinor states (there the 2.pi rotations is not trivial and we cannot start with SO(32)) but it contains the spinor states of one chirality only. The self-dual lattice of Spin(32)/Z2 is generated by the adjoint lattice plus a spinor root of one chirality (i.e. it has 2 shifted copies of the root lattics of SO(32)), but it still contains 50% of the full weight lattice of Spin(32) only (this one is 4 times bigger than the root lattice). The sum of all the coordinates of the weights is even for Spin(32)/Z2 - because the spinors have an odd number of -1/2. The element (-1)^{sum-of-the-coordinates} is an element of the maximal torus and acts trivially. Best wishes Lubos ______________________________________________________________________________ E-mail: lumo@matfyz.cz Web: http://www.matfyz.cz/lumo tel.+1-805/893-5025 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Superstring/M-theory is the language in which God wrote the world. ============================================================================== From: abergman@Princeton.EDU (Aaron J. Bergman) Subject: Re: octonions; d=11 SUGRA; F-theory Date: 22 Sep 2001 04:59:01 GMT Newsgroups: sci.physics.research In article , zirkus wrote: >In article , Brett McInnes >says... >>Certainly the relevant subgroup of E8 isn't SO(16)---but it isn't Spin(16) >>either! It's Spin(16)/Z_2. >IIRC (and I'm pretty sure that I do), John Baez already mentioned on page 50 >of his octonions paper [1] that Adams had already demonstrated this. However, >neither you nor Lubos have convinced me that the same should be true in the >non-compact case. By the Global Iwasawa Decomposition, we know that any Lie group is diffeomorphic to K x A x N where A and N are simply connected. In other words, K and G have the same fundamental group. Now, it might be possible to have finite quotients of the split form of E_8, but usually when one writes that down, one means the universal cover which is, by definition, simply connected. So, K must also be simply connected and, thus, it must be Spin(16). (Unless there's another simply connected form of so(16), but that doesn't seem right.) Aaron -- Aaron Bergman ============================================================================== From: baez@galaxy.ucr.edu (John Baez) Subject: Re: octonions; d=11 SUGRA; F-theory Date: Sat, 22 Sep 2001 22:25:35 +0000 (UTC) Newsgroups: sci.physics.research Summary: [missing] In article , zirkus wrote: >In article , Brett McInnes >says... >>Certainly the relevant subgroup of E8 isn't SO(16)---but it isn't Spin(16) >>either! It's Spin(16)/Z_2. >IIRC (and I'm pretty sure that I do), John Baez already mentioned on page 50 >of his octonions paper [1] that Adams had already demonstrated this. Hey, you're right! I'd forgotten this! Duh! I hereby publicly apologize for running around telling everyone that the compact real form of E8 contains Spin(16) as a subgroup. I should have said Spin(16)/Z_2, where Z_2 is the subgroup of Spin(16) that acts trivially on right-handed spinors. A proof is in John F. Adams, Lectures on Exceptional Lie Groups, eds. Zafer Mahmud and Mamoru Mimira, University of Chicago Press, Chicago, 1996. More precisely: you can build the compact real form of E8 as follows. First let the vector space e8 be the direct sum of so(16) and its right-handed spinor representation, S+. Define the Lie bracket on e8 using the obvious maps so(16) x so(16) -> so(16) (the Lie bracket on so(16)) so(16) x S+ -> S+ (the action of so(16) on S+) and S+ x S+ -> so(16) defined by dualizing the previous map, where we use the natural inner products on so(16) and S+ to identify these vector spaces with their duals. It's nontrivial to check that this Lie bracket satisfies the Jacobi identity, but it does. Next, form the simply connected Lie group E8 whose Lie algebra is e8. From the above it's obvious that e8 has so(16) as a Lie subalgebra, so E8 has a unique connected subgroup G whose Lie algebra is so(16). What's G? There aren't many choices for what G could be, since the only connected Lie groups with so(16) as Lie algebra are the simply-connected group Spin(16) and various groups of the form Spin(16)/H where H is a subgroup of the center of Spin(16). What's the center of Spin(16)? Well, clearly it has at least 2 elements, since there is always a copy of Z_2 in the center of Spin(n) such that when we mod out by this subgroup we get SO(n). However, when n is even, SO(n) itself has a nontrivial center, namely the matrices 1 and -1. The inverse image of these elements is the center of Spin(n). Since Spin(n) -> SO(n) is a double cover we get 4 elements in the center of Spin(n) when n is even. This implies that the center of Spin(16) is either Z_2 x Z_2 or Z_4. Now, I seem to recall that the center of Spin(2n) alternates between being Z_2 x Z_2 and being Z_4 depending on whether n is even or odd. The books where I store this information are in my office at UCR, and I'm at home now, but it seems plausible. Clearly for n = 2 we get Spin(4) = SU(2) x SU(2) with center Z_2 x Z_2. Unfortunately the case n = 1 doesn't quite fit the pattern, since Spin(2) isn't a simple Lie group; however, Spin(2) = U(1) contains Z_4 as a subgroup, not Z_2 x Z_2, so I think I'm on the right track here. So: if I had to make a guess or be ruthlessly executed by my interrogators, I would guess the center of Spin(16) is Z_2 x Z_2. Let me assume this from now on. Z_2 x Z_2 has five subgroups: 1) the whole group Z_2 x Z_2 2) the left copy of Z_2, 3) the right copy of Z_2, 4) the diagonal copy of Z_2, and 5) the trivial group. If we call one of these five guys be G, we get a connected Lie group Spin(16)/G with Lie algebra so(16). I think the five groups we get this way are: 1) SO(16)/{+-1} 2) Spin(16) mod the Z_2 subgroup that acts trivially on left-handed spinors 3) Spin(16) mod the Z_2 subgroup that acts trivially on right-handed spinors 4) Spin(16) mod the Z_2 subgroup that acts trivially on vectors, i.e. SO(16) and 5) Spin(16). What Adams showed is that group number 3) is the subgroup of E8 whose Lie algebra is so(16). This makes a lot of sense if you think about it. Lubos Motl seems to have sketched the argument - I need to reread his post more carefully. I hope everyone here is in agreement on these points. If I screwed up, I hope someone will correct me! By the way, if anyone is worried about the apparent left/right asymmetry of the above description of E8 and its Spin(16)/Z_2 subgroup, fear not: if you switch the words "left" and "right" in everything I've written above, I believe everything will still be true. (Except, of course, in phrases such as "Hey, you're right!") >However, neither you nor Lubos have convinced me that the same should >be true in the non-compact case. Right; I've just been talking about the compact real form of E8. Now we need to turn to the noncompact real form which goes by the name of E8(8). Aaron Bergman has given an argument that *this* group has Spin(16) as a subgroup. I should probably talk about this in a separate post.