John Baez, in his week 253, said:
"... The magic square gives vector space isomorphisms as follows:
f4 = so(R (+) O) (+) (R (x) O)^2
e6 = so(C (+) O) (+) (C (x) O)^2 (+) Im(C)
e7 = so(H (+) O) (+) (H (x) O)^2 (+) Im(H)
e8 = so(O (+) O) (+) (O (x) O)^2
Here f4, e6, e7 and e8 stand for the Lie algebras of the compact real forms of these exceptional Lie groups. R, C, H, and O are the usual suspects - the real numbers, complex numbers, quaternions and octonions. For any real inner product space V, so(V) stands for the Lie algebra of the rotation group of V. And, for each of the isomorphisms above, we must equip the vector space on the right side with a cleverly (but not perversely!) defined Lie bracket to get the Lie algebra on the left side. ...".
From an 8-Periodicity table for real Clifford algebras
(in the Table the Real Numbers, Complex Numbers, and Quaternions are denoted by R, C, H instead of R, C, Q) it can be seen that
so(R (+) O) = Spin(9) is the bivector Lie algebra of the Clifford algebra Cl(9) = M(16,C) the 16x16 Complex Matrix Algebra whose even subalgebra is M(16,R) the 16x16 Real Matrix Algebra, so that its spinors Spnr9 are 16-real dimensional (since 9 is odd-dimensional the spinors do not reduce to half-spinors), and can be identified with 1x8 + 1x8 = 16-real dimensional (R (x) O)^2, so that
so(C (+) O) = Spin(10) is the bivector Lie algebra of the Clifford algebra Cl(10) = M(16,Q) the 16x16 Quaternion Matrix Algebra whose even subalgebra is M(16,C) the 16x16 Complex Matrix Algebra, so that its full spinors Spnr10 are 16-complex-dimensional columns, and therefore 32-real dimensional (with its half-spinors being the real and imaginary parts of the full spinors), and its full spinors can be identified with 2x8 + 2x8 = 32-real dimensional (C (x) O)^2, so that
so(H (+) O) = Spin(12) is the bivector Lie algebra of the Clifford algebra Cl(12) = M(32,Q) the 32x32 Quaternion Matrix Algebra whose even subalgebra is M(16,Q) (+) M(16,Q) two copies of the 16x16 Quaternionic Matrix Algebra, so that its spinors Spnr12 are two 16-quaternion-dimensional columns, and therefore 2x16x4 = 128-real dimensional (with its half-spinors Spnr12+ and Spnr12- each being one copy of M(16,Q) and so 16x4 = 64-real dimensional), and its half-spinors Spnr12+ can be identified with 4x8 + 4x8 = 64-real dimensional (Q (x) O)^2, so that
so(O (+) O) = Spin(16) is the bivector Lie algebra of the Clifford algebra Cl(16) = M(256,R) the 256x256 Real Matrix Algebra whose even subalgebra is M(128,R) (+) M(128,R) two copies of the 128x128 Real Matrix Algebra, so that its spinors Spnr16 are two 128-real dimensional columns, and therefore 2x128 = 256-real dimensional (with its half-spinors Spnr16+ and Spnr16- each being one copy of M(128,R) and so 128-real dimensional), and its half-spinors Spnr16+ can be identified with 8x8 + 8x8 = 64-real dimensional (O (x) O)^2, so that
J. F. Adams in his book "Lectures on Exceptional Lie Groups" (Chicago 1996) said:
"... D4 = Spin(4)
D+
Ad /\1 ... we call the symmetry triality
D-
...
Under the inclusions Spin(2n) -> Spin(2n+1) -> Spin(2n+2) , we have
D+
/
D+ + D- <- D <
\
D-
...
Under Spin(2r) x Spin(2s) -> Spin(2r+2s)
D+ (x) D+ + D- (x) D- <- D+ D+ (x) D- + D- (x) D+ <- D-
Under Spin(2r) x Spin(2s+1) -> Spin(2r+2s+1)
D+ (x) D + D- (x) D <- D
Under Spin(2r+1) x Spin(2s+1) -> Spin(2r+2s+2)
D (x) D <- D+ D (x) D <- D-
... write I for the identity representation of S1 = U(1) ...
local type of H dim L(G)/L(H) as C rep. dim G dim Spin(9) 36 D 16 F4 52 Spin(10)xU(1)/Z4 46 D+(x)I^3 + D-(x)I^(-3) 32 E6 78 Spin(12)xSp(1)/Z2 69 D+(x)/\1 64 E7 133 Spin(16)/Z2 120 D+ 128 E8 248
..."
With respect to E6:
As to whether or not (CxO)P2 is isotropic, Joseph A. Wolf says in his book "Spaces of Constant Curvature" (Fifth Edition, Publish or Perish 1984) at pages 293-294:
"... M is called isotropic at x if I(M)x is transitive on the unit sphere in Mx; it is isotropic if it is isotropic at every point. ...M is isotropic if and only if it is two point homogeneous. ...
Let M be a riemmannian symmetric space. Then the following conditions are equivalent.
(i) M is two point homogeneous.
(ii) Either M is a euclidean space or M is irreducible and of rank 1. ...".
Since (CxO)P2 = E6 / SO(10) x SO(2) is rank 2, it is not isotropic.
With respect to E7:
By using triality, maybe Adams could also have used, for E7, either D- (x) /\1 or D+ (x) D-.
By varying signature, and instead of using Spin(12) of Cl(12) = M(32,Q), using Spin(2,10) of Cl(2,10) = M(64,R) for which full spinors would be 64-real dimensional (reducible to two half-spinors each of which being 32-real dimensional, Adams's table for E7 would have been
local type of H dim L(G)/L(H) as C rep. dim G dim Spin(12)xSp(1)/Z2 69 D+ + D- 64 E7 133
With respect to the geometry of the 128-real dimensional Spnr16+ part of E8:
Look at the Type EVIII rank=8 space E8 / Spin(16), This 248 -120 = 128 dimensional symmetric space that corresponds to a half-spinor representation Spnr16+ of Spin(16) is the octooctonionic projective plane (OxO)P2. In his book "Geometry of Lie Groups" (Kluwer 1997), Boris Rosenfeld says (slightly edited):
"... The group of motions ... in the octooctonionic Hermitian elliptic plane (OxO)P2 is the compact real simple Lie group E8 ...The Hermitian elliptic plane (OxO)P2 admits interpretation as the compact symmetric Riemannian space V128 ...
the stabilizers of points in V128 represent transformations in the families of 7-planar generators of the absolute hyperquadrics of S15 under spinor representations of the group SO(16) ...".
Rosenfeld goes on to describe much more geometry, such as the connection of V128 with focal varieties of congruences of 63-planes in S127.
With respect to the 120-real dimensional Spin(16) = so(O (+) O) part of E8:
Spin(16) is the 120-real dimensional bivector Lie algebra of the Clifford algebra Cl(16).
By 8-fold real Clifford periodicity Cl(16) = Cl(8) (x) Cl(8)
Each Cl(8) has a 28-real dimensional bivetor Lie algebra Spin(8) and 8-real dimensional vectors V8, so the tensor product Cl(8) (x) Cl(8) should produce bivectors that are
1 (x) Spin(8) i.e., scalar (x) bivector = bivector Spin(8) (x) 1 i.e., bivector (x) scalar = bivector V8 (x) V8 i.e., vector (x) vector = bivector
so that naturally gives you, for the Spin(16) part of E8,
The structure can be visualized as
oxxxxxxxxxxxxxxx ooxxxxxxxxxxxxxx oooxxxxxxxxxxxxx ooooxxxxxxxxxxxx oooooxxxxxxxxxxx ooooooxxxxxxxxxx oooooooxxxxxxxxx ooooooooxxxxxxxx ( 120 Spin(16)generators denoted by x ) oooooooooxxxxxxx ooooooooooxxxxxx oooooooooooxxxxx ooooooooooooxxxx oooooooooooooxxx ooooooooooooooxx ooooooooooooooox oooooooooooooooo o-------xxxxxxxx oo------xxxxxxxx ooo-----xxxxxxxx oooo----xxxxxxxx ooooo---xxxxxxxx oooooo--xxxxxxxx ooooooo-xxxxxxxx ooooooooxxxxxxxx ( 28 Spin(8) generators denoted by - ooooooooo+++++++ 64 V8xV8 generators denoted by x oooooooooo++++++ 28 Spin(8) generators denoted by + ) ooooooooooo+++++ oooooooooooo++++ ooooooooooooo+++ oooooooooooooo++ ooooooooooooooo+ oooooooooooooooo
With respect to the overall structure of E8:
I think the key to understanding E8 is the E8 lattice.
First, note that the 240 E8 root vectors form the Witting polytope of 8-dim close-packing (so E8 is the maximum size for rank-8 Lie groups) corresponding to 240 of the 248 generators of E8, with the other 8 forming a Cartan subalgebra of E8.
Then, if you look at the D8 root vectors of so(16) (nearest neighbors to the origin in a D8 lattice with a vertex at the origin) you see that there are 112 of them. They correspond to 112 of the 120 generators of so(16), with the other 8 forming a Cartan subalgebra of so(16).
In their Springer book "Sphere Packings, Lattices, and Groups", Conway and Sloane say (slightly edited):
"... The covering radius of Dn increases with n, and when n = 8 it is equal to the minimal distance between the lattice points.So when n = 8 [or greater] we can slide another copy of Dn in between the points of the original Dn ...
[the union of the two Dn lattices] is a lattice packing if and only if n is even. ... When n = 8 ... the lattice being known as E8 ...".
Since the first D8 lattice has a vertex at the origin, the second D8 lattice should have a hole at the origin.
Conway and Sloane also say in that book (slightly edited):
"... Zn is the n-dimensional cubic or integer lattice. ... Zn is self-dual. ...Dn is obtained by coloring the points of Zn alternately red and white with a checkerboard coloring, and taking the red points. Dn is sometimes called the checkerboard lattice. ...".
A cubic Z8 lattice with a hole at the origin has 2^8 = 256 vertices (the vertices of an 8-dim cube) surrounding the origin hole, so the second D8 lattice, a checkerboard with half of the vertices of Z8, has 128 vertices surrounding the origin hole.
Therefore,
and
the second copy of D8 lattice provides the other 240 - 112 = 128 root vectors of E8, and therefore the 128+ half-spinor generators of E8. Since 128 belong to the same second copy of D8 lattice, they should all have the same half-spinor helicity.
Since the 8-dim close-packing has 112 + 128 = 240 vertices that are nearest neighbors of the origin, there is no room for any of the mirror image 128- so(16) half-spinors to be in any E8 lattice.
For example, in the E8 lattice
±1, ±i, ±j, ±k, ±e, ±ie, ±je, ±ke,
(±1 ±ke ±e ±k)/2 (±i ±j ±ie ±je)/2
(±1 ±je ±j ±e)/2 (±ie ±ke ±k ±i)/2
(±1 ±e ±ie ±i)/2 (±ke ±k ±je ±j)/2
(±1 ±ie ±je ±ke)/2 (±e ±i ±j ±k)/2
(±1 ±k ±i ±je)/2 (±j ±ie ±ke ±e)/2
(±1 ±i ±ke ±j)/2 (±k ±je ±e ±ie)/2
(±1 ±j ±k ±ie)/2 (±je ±e ±i ±ke)/2
the 112 vertices
±1, ±i, ±j, ±k, ±e, ±ie, ±je, ±ke,
(±1 ±ke ±e ±k)/2 (±i ±j ±ie ±je)/2
(±1 ±je ±j ±e)/2 (±ie ±ke ±k ±i)/2
(±1 ±e ±ie ±i)/2 (±ke ±k ±je ±j)/2
correspond to the first D8 that has the 112 root vectors of 120-dim so(16). Note that the 4-component ones each of them has 2 componets with an e factor.
and the 128 vertices
(±1 ±ie ±je ±ke)/2 (±e ±i ±j ±k)/2
(±1 ±k ±i ±je)/2 (±j ±ie ±ke ±e)/2
(±1 ±i ±ke ±j)/2 (±k ±je ±e ±ie)/2
(±1 ±j ±k ±ie)/2 (±je ±e ±i ±ke)/2
correspond to the second D8 that has the 128 +half-spinors of so(16). Note that they are all 4-component and each of them has either 1 or 3 componets with an e factor.
There are 7 independent E8 lattices, all of which have the same general two D8 structure described above.
Taking all 7 E8 lattices together,
and
there are 7x8x16 = 896 that appear in the second type of D8 lattice representing half-spinors of so(16) with all 896 of those vertices being 4-component.
Since there are ( 8 choose 4 ) = 8x7x6x5 / 1x2x3x4 = 70 sets of 4 components and 2^4 = 16 sets of signs, there are 70 x 16 = 1,120 possible 4-component vertices. Since 1,120 = 224 + 896 the 7 E8 lattices taken together contain all possible 4-component vertices, and each of them contains all of the 16 1-component vertices.
None of the 7 E8 lattices contain any half-spinor vertices in common with any other E8 lattice.
Each of the 14x16 = 224 4-component so(16) vertices of the first D8 lattice appears in 3 of the E8 lattices:
More details can be found at http://www.valdostamuseum.org/hamsmith/E8.html