In QM one would have CCR: [Q, P] = i I, (h-bar=1) with [P, I] = [Q, I] = 0, expressing the nilpotency of the Heisenberg Lie algebra; with FCCR we have equation (2.8) with: [G(n), Q(n)] := i 2 n^2 (n-1) X_2(n) (4.1) = i alpha^(-2)(n) X_2(n) [P(n), G(n)] := i 2 n^2 (n-1) X_1(n) (4.2) = i alpha^(-2)(n) X_1(n) [X_1(n), X_2(n)] := i alpha(n) X_3(n) (4.3) with the definition alpha(n) := (n sqrt(2(n-1)))^(-1) (4.4) (so alpha(n) goes to zero as n^(-3/2) when n->infinity) and with Einstein summation convention, permutation symbol, and Kronecker delta, 1 [X_i(n), X_j(n)] = i -------------- epsilon_(ijk) X_k(n) (4.5) n sqrt(2(n-1)) = i alpha(n) epsilon_(ijk) X_k(n) {X_i(n), X_j(n)} = (1/4) n^(-2) (n-1)^(-1) delta_i_j (4.6) = (1/2) alpha^2(n) delta_i_j 3 X_1^2(n) + X_2^2(n) + X_3^2(n) = ----------- (O(n-2) + I(2)) (4.7) 8n^(2)(n-1) = (3/4) alpha^(2)(n) (O(n-2) + I(2)) where i, j, k = 1, 2, 3. If on this two dimensional subspace we define the normalized Pauli spin matrices and the unit matrix: |0 1| sigma_1(n) = (1/2) | | |1 0| |0 -i| sigma_2(n) = (1/2) | | |i 0| |1 0| sigma_3(n) = (1/2) | | |0 -1| |1 0| sigma_0(n) = | | |0 1| (4.8) which obey [sigma_i(n), sigma_j(n)] = i epsilon_(ijk) sigma_k(n) (4.9) for i,j,k = 1,2,3 where epsilon_(ijk) is the standard completely antisymmetric epsilon-density, giving the 2x2 IRREP of the algebra su(2), which acts on the subspace of Hilb(n) spanned by |n, n-1> and |n, n-2>, then X_k(n) = alpha(n) sigma_k(n) (4.10) for k = 1,2,3 and Q(n) = Q(n-1) + (n alpha(n))^(-1) sigma_1(n) = Q(n-1) + (n alpha(n))^(-2) X_1(n) P(n) = P(n-1) + (n alpha(n))^(-1) sigma_2(n) = P(n-1) + (n alpha(n))^(-2) X_2(n) G(n) = G(n-1) + (n alpha(n))^(-2) sigma_3(n) = G(n-1) + (n alpha(n))^(-3) X_3(n) (4.11)The three X_k(n) obviously close on themselves under commutation forming an su(2) algebra, equation (4.5). The scale factor alpha(n) of the structure constants goes to zero as n -> infinity, as alpha(n) is of the order of n^(-3/2). If they are further commuted with Q(n), P(n) and G(n), the commutators close with the X_k(n) to span an su(3) algebra in its defing representation.

For the su(3) algebra, explicitly take algebra generators: |0 1 0| |0 0 1| lambda_1(n) = |1 0 0| lambda_4(n) = |0 0 0| |0 0 0| |1 0 0| |0 -i 0| |0 0 -i| lambda_2(n) = |i 0 0| lambda_5(n) = |0 0 0| |0 0 0| |i 0 0| |1 0 0| |0 0 0| lambda_3(n) = |0 -1 0| lambda_6(n) = |0 0 1| |0 0 0| |0 1 0| |0 0 0| |0 0 0| lambda_7(n) = |0 0 -i| lambda_8(n) = |0 1 0| |0 i 0| |0 0 -1| then sqrt(n-2) [Q(n), X_1(n)] = +i ------------ lambda_5(n) 2n sqrt(n-1) 1 sqrt(n-2) [Q(n), X_2(n)] = +i --- lambda_8(n) - i ------------ lambda_4(n) n 2n sqrt(n-1) 1 sqrt(n-2) [Q(n), X_3(n)] = -i --- lambda_7(n) + i ------------ lambda_2(n) n 2n sqrt(n-1) 1 sqrt(n-2) [P(n), X_1(n)] = -i --- lambda_8(n) - i ------------ lambda_4(n) n 2n sqrt(n-1) sqrt(n-2) [P(n), X_2(n)] = -i ------------ lambda_5(n) 2n sqrt(n-1) 1 sqrt(n-2) [P(n), X_3(n)] = +i --- lambda_6(n) - i ------------ lambda_1(n) n 2n sqrt(n-1) 1 [G(n), X_1(n)] = -i ------------- lambda_7(n) = +i n X_2(n) sqrt(2 (n-1)) 1 [G(n), X_2(n)] = -i ------------- lambda_6(n) = -i n X_1(n) sqrt(2 (n-1)) [G(n), X_3(n)] = 0

Repeating the procedure generates su(4). An extrapolation suggests that the ultimate closure is on the algebra su(n) in its defining representation. This "Aufbau" of su(m) terminating in su(n) by successive commutation produces a subalgebra chain, with containment:

su(2) < su(3) < ... < su(n-1) < su(n)

It can be seen that FCCR (2.8) can be represented in terms of the generators of the su(n) algebra. The commutator of either Q(n) or P(n) with any of these basis generators does not not vanish. For n > 1, su(n) is simple, that is, does not contain any invariant Lie subalgebra.

<Theorem 4.1>:

Q(n) and P(n) generate by commutation, the defining
representation of the Lie algebra su(n), and the operators

for k = 0, 1, 2, ..., n-1, h_k(n) = -(1/n) C!^(k+1)(n) G(n) C^(k+1)(n) so that Tr( h_k)(n) ) = 0form a Cartan subalgebra Cf. [Appendix B]. The real polynomials in Q(n) and P(n) are a nonfaithful representation of the universal enveloping algebra of su(n).

Proof:The h_k(n) are all diagonal, and since their traces vanish, there are n-1 linearly independent h_k(n). For the cyclic operator C(n)

C^n(n) = I(n) = C^0(n) C^(n-k)(n) = I(n) = C^k(n) Define, B_k(n) = (1//n) C!^(k+1)(n) B(n) C^(k+1)(n) so B_k!(n) = (1//n) C!^(k+1)(n) B(n) C^(k+1)(n) then since, [B!(n), B(n)] = -G(n) it follows that [B!_k(n), B_k(n)] = h_k(n) and of course [h_k(n), h_j(n)] = 0 and n n-1 SIGMA h_k(n) = SIGMA h_k(n) = 0 k=1 k=0Q(n) and P(n) are both Hermitean and of trace zero, so their commutator, -i [Q(n), P(n)] and subsequent commutators of commutators formed in this fashion will also be Hermitean and of trace zero. From the expansions of Q(n) and P(n) as linear combinations of the su(n) generators given in [Appendix B] it can be seen that the only matrices that can commute with both Q(n) and P(n) are multiples of the identity. By Schur's lemma, the algebra that Q(n) and P(n) generate is irreducible. It is then the full complement of nxn Hermitean matrices of trace zero, that is, the Lie algebra su(n).

That the h_k(n) span the Cartan subalgebra of su(n) can be seen by inspection.

QED

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Created: August 1997

Last Updated: May 28, 2000

Last Updated: August 7, 2002