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Finding Lie-Algebra Invariants -- an Explicit Procedure?

  1. Aug 2, 2014 #1
    I've found this paper: [physics/9712033] Closed Expressions for Lie Algebra Invariants and Finite Transformations and I've attempted to interpret it.

    In some representation, a general Lie-algebra element A = aiLi for generators L (repeated dummy indices summed over, like i here).
    det(I - x*A) = xii(A)

    where the functions φi(A) are combinations of traces of powers of A: Tr(Ak). They can be expanded
    φi(A) = aj1aj2...ajiCD,j1,j2,...,ji

    giving a generalized Casimir invariant
    CD(i) = CD,j1,j2,...,jiLj1Lj2...Lji

    The D is for determinant. I don't get their jumping between lowered-index L's and raised-index L's. Something unstated somewhere, I think. I also don't get their jumping from φ's to C's.

    But if one grants that, one can find an alternate set of invariants from the traces of products:
    CT,j1,j2,...,ji = Tr(Lj1Lj2...Lji)

    CT(i) = CT,j1,j2,...,jiLj1Lj2...Lji

    with the CM(i)'s in terms of the CT(i)'s and vice versa. Each CT(i) contains CM(i) and vice versa.


    I've seen other forms of the invariants, like in terms of the commutator and the Killing form:
    [Li,Lj] = fijkLk, gij = fiklfjlk, gij = inverse of gij

    This makes it work only for semisimple Lie algebras, because the Killing form must be nonsingular here.

    Construct matrix Mij = fikjgklLl

    The Casimir invariants are CC(i) = Tr(Mi) taking power i of the matrix M with its indices as defined above, then the trace with those indices.

    So if that paper's results are valid, there must be some relationship between the CC's and the CT's and CM's.


    One can make conjectures in some special cases as to the form of the Casimir invariants, notably for orthogonal and pseudo-orthogonal groups SO(n) and SO(n1,n2), n=n1+n2. Let the commutation go
    [LijLkl] = hikLjl - hilLjk - hjkLil + hjlLik

    where h is some real symmetric nonsingular n*n matrix. The Killing form is
    gi1i2,j1j2 = (something) * (hi1,j2hi2,j1 - hi1,j1hi2,j2)

    Define a variant of the previous matrix M:
    Mij = Likhkj where hij is the inverse of hij.

    Then CO(i) = Tr(Mi), trace of power i using this M.

    One can show that the CC's contain the CO's, with CC(i) containing CO(i).

    For even n, there is a possible one that uses the antisymmetric symbol:
    CX = εijkl...LijLkl...

    It's easy to show that CO(2n) contains the square of CX.


    There are further questions about these operators. For semisimple Lie algebras, is there any halfway simple way of calculating them for generators in the Cartan-Weyl basis? That would give the operators' values in terms of the representations' highest weights. I've found A. M. Perelomov, V. S. Popov, “Casimir operators for semisimple Lie groups”, Izv. Akad. Nauk SSSR Ser. Mat., 32:6 (1968), 1368–1390, which does more-or-less that, but it's hard for me to see how Perelomov's and Popov's versions relate to the others.

    I tried calculating CD for some of the algebras that the papers' authors mentioned. It worked OK for the Lorentz group, SO(3,1), but not for the Poincaré group, SEuc(3,1) (Special Euclidean). For the Poincaré group, it got the same invariants as for the Lorentz group, though it got the right ones for the latter group. For the Galilean group, it only got the angular-momentum invariant.

    I'll note a list of sizes of irreducible invariants. For rank n, an algebra has n of them.
    • U(n): 1, 2, 3, ..., n
    • A(n), SU(n+1): 2, 3, ..., n, n+1
    • B(n), SO(2n+1): 2, 4, ..., 2n
    • C(n), Sp(2n): 2, 4, ..., 2n
    • D(n), SO(2n): 2, 4, ... 2n-2, n (instead of 2n)
    • G2: 2, 6
    • F4: 2, 6, 8, 12
    • E6: 2, 5, 6, 8, 9, 12
    • E7: 2, 6, 8, 10, 12, 14, 18
    • E8: 2, 8, 12, 14, 18, 20, 24, 30
    For a real or pseudoreal irrep, an odd-sized invariant will vanish. It's not surprising that irreducible odd-sized invariants only occur in the algebras that have complex irreps: U(n), SU(n) for n >= 3, SO(4n+2), E6.
  2. jcsd
  3. Aug 14, 2014 #2
    I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?
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