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Explicit construction of isomorphisms of low-rank Lie algebras.

  1. Sep 23, 2014 #1
    I've worked that out in my Mathematica notebook "Lie-Algebra Matrices.nb" in directory "Extra Mma notebooks" in SemisimpleLieAlgebras.zip, but I must concede that some of my derivations seem rather inelegant, not using features of the algebras very well. The last comment in How should I show that the Lie algebra so(6) of SO(6) is isomorphic to the Lie algebra su(4) of SU(4)? - Mathematics Stack Exchange is the sort of thing that I'm looking for.

    I'm going into detail about this so you people can check on whether I'm on the right track if you are so inclined. I'm hoping that I did not make any major conceptual mistakes.

    Here is what I've been working with. First, real vs. complex Lie algebras. A real algebra has values given by (values).(generators) where the generators have real commutator structure functions, though they need not be real themselves. A complex one is like these but with complex numbers instead. A complex algebra may have several real algebras as subalgebras. So what I did was to check the isomorphism of several real algebras:
    Each (generator of algebra 2) = (real numbers) . (generators of algebra 1)
    I succeeded, even if rather inelegantly.

    I checked on the families of algebras associated with SO(2), SO(3), SO(4), SO(5), and SO(6), the ones with all the low-rank isomorphisms. The isomorphisms with the other algebras were with the spinor versions of these algebras, Spin(2) to Spin(6) and their relatives.

    SO(n) generators (a,b): ## L_{ab,ij} = \delta_{ai}\delta_{bj} - \delta_{aj}\delta_{bi} ##
    Spin(n) generators: usual Clifford-algebra version in parallel with the SO(n) generators above.
    SO(n1,n2) generators: diag(g) . SO(n) generators, where g is a vector with n1 +1's and n2 -1's and n=n1+n2.
    Its generators (a,b) can be expressed as ## diag(\sqrt{g}) \cdot (L_{ab} \sqrt{g_a} \sqrt{g_b}) \cdot diag(1/\sqrt{g}) ##
    Thus, the Spin(n1,n2) generators (a,b) are ##\sqrt{g_a} \sqrt{g_b}## times the SO(n) ones.
    SO(n) and SO(n1,n2) here are both SO(n,R) and SO(n1,n2,R), R = real numbers.
    SO(n,H) has generators with quaternion elements, where quaternions are {I(2), -i*(Pauli matrices)} (H = William Rowan Hamilton, discoverer of quaternions). Its generators are antisymmetric, and it has as many generators as SO(2n).
    Spin(n,H) I got by finding the SO(n,H) generators as multiples of SO(2n) ones, then multiplying the Spin(2n) generators by them.
    Note: SO(n,H) is sometimes called SO*(2n)

    Of the others, U(n1,n2) is constructed from U(n) and Sp(2n1,2n2) constructed from Sp(2n) both in the fashion of SO(n1,n2) from SO(n), by multiplying by diag(metric). SU(n1,n2) is all the traceless elements of U(n1,n2), in parallel with SU(n) vs. U(n).

    The general linear group's generators GL(n,X) are n*n matrices over field X. GL(n,C) can be constructed from GL(n,R) by making two copies of GL(n,R), and then multiplying each copy's elements by {{1,0},{0,1}} and {{0,1},{-1,0}} respectively. This makes its elements more distinct, since Mathematica works in complex numbers. GL(n,H) uses the quaternion basis I'd mentioned earlier, multiplying four copies of GL(n,R). SL(n,R) is traceless elements of GL(n,R) and likewise for SL(n,H). SL(n,C) is constructed like GL(n,C), with two copies of SL(n,R).
    Note: SL(n,H) is sometimes called SU*(2n).

    The symplectic algebra Sp(2n) is really USp(2n,C), both unitary and symplectic. It's possible to construct a real-valued version, Sp(2n,R), of it.

    My results:

    SO(2) family:
    SO(2) ~ Spin(2) ~ U(1)
    SO(1,1) ~ Spin(1,1) ~ GL(1,R+) (OK to use R+, the positive real numbers, here, since we only multiply)
    Only one element.
    GL(1,C) ~ GL(1,R+) * U(1) (the magnitude and the argument)
    Only two elements.

    SO(3) family:
    Spin(3) ~ SU(2) ~ SL(1,H) ~ Sp(2)
    Easy. Sort the canonical forms' generators and they match.
    Spin(2,1) ~ SU(1,1) ~ SL(2,R) ~ Sp(2,R)
    A bit more difficult. The first two must be transformed into real form with T.L.T-1 for some appropriate matrix T. I found which elements to match by finding the Killing forms of all the algebras and rearranging them to get the same structure functions.

    SO(4) family:
    Spin(4) ~ SU(2) * SU(2)
    In Spin(4), I took sums and differences of (1,2) and (3,4), (1,3) and (2,4), and (1,4) and (2,3). Each half of Spin(4) could then be identified with one of the SU(2)'s.
    Spin(3,1) ~ SL(2,C)
    Rather tricky, since I could not split SL(2,C). But I could identity both halves of Spin(3,1) with SL(2,C).
    Spin(2,2) ~ SU(1,1) * SU(1,1)
    Just like the Spin(4) case.
    Spin(2,H) ~ SU(2) * SU(1,1)
    Weird, with one half of Spin(2,H) matching SU(2) and the other half SU(1,1)

    SO(5) family:
    Spin(5) ~ Sp(4)
    Spin(4,1) ~ Sp(2,2)
    Spin(3,2) ~ Sp(4,R)
    Solving these cases was rather tricky. I found the Killing forms of all the metrics and then used them to find which elements of each side match. That still left a lot of undetermined coefficients, so I did a lot of experimentation and trial and error to find some solutions.

    SO(6) family:
    Spin(6) ~ SU(4)
    The two halves of Spin(6) match SU(4) and its complex conjugate.
    Spin(5,1) ~ SL(2,H)
    Of the two halves of Spin(5,1), one of them matches SL(2,H) and the other is unmatched.
    Spin(4,2) ~ SU(2,2)
    Like Spin(6)
    Spin(3,3) ~ SL(4,R)
    Like Spin(5,1)
    Spin(3,H) ~ SU(3,1)
    Like Spin(6)
    I attempted to solve using what I'd done for the SO(5) family, but it was much more tricky. I could do Spin(5,1) and Spin(3,3) together, and Spin(4,2) and Spin(3,H) together, however. For Spin(6), I used the trick that I found in that stackexchange article. I took the quaternion basis, then constructed 4*4 matrices with a Kronecker product, like what I'd used for the complex and quaternionic algebras earlier. This gave me 16 matrices, and I removed the identity matrix. The remaining 15 were good basis sets for both halves of Spin(6) and for SU(4), and vice versa. I then used this identification to find SU(4) from one half of Spin(6). With this identification, the other half of Spin(6) matched onto the complex conjugate of SU(4).
     
  2. jcsd
  3. Oct 2, 2014 #2
    OP1 and Lorentzian Geometry (John Baez's Octonions) contains something that I wanted to verify.

    Algebra isomorphisms:
    • SO(2,1) ~ SL(2,R) -- real
    • SO(3,1) ~ SL(2,C) -- complex
    • SO(5,1) ~ SL(2,H) -- quaternions / Pauli matrices
    • SO(9,1) ~ SL(2,O) -- octonions
    I could verify the first three but not the fourth. I couldn't figure out how to construct SL(2,X) with octonions, since octonions are not associative. JB's discussion of this issue seemed rather roundabout and arcane, so it was not much help either.

    I was, however, able to verify that the automorphism group of the octonions is G2.
     
  4. Oct 9, 2014 #3
    I think I found the solution to that conundrum. Construct a Hermitian matrix in values of a Cayley-Dickson algebra, something like {{a, x}, {conjg(x), b}}. Then take its determinant. What linear transformations on the variables will leave the determinant value unchanged?

    For 2*2 matrices, it's O(n+1,1) for a n-D algebra. Its continuous-with-identity part is SO+(n+1,1) or SO(n+1,1).
    For n*n real matrices, it's SL(n,R)
    For n*n complex matrices, it's SL(n,C)
    For n*n quaternionic matrices, it's SL(n,H) or SU*(2n)
    That would make it SL(n,O) for octonions.

    Thus, the odd result that I'd mentioned earlier.
     
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