The centre of the associative algebra A.
The centralizer of the subalgebra S of the associative algebra A, that is, the subalgebra of A commuting elementwise with S.
Side: MonStgElt Default: "Both"
Given an associative algebra A and a subalgebra B of A, compute the idealizer of B in A, that is, the largest subalgebra of A in which B is an ideal. By default the two-sided idealizer, that is, the largest subalgebra in which B is a two-sided ideal, is found; the left- or right-idealizer can be found by setting the parameter Side to "Left" or "Right" respectively.
For an associative structure constant algebra A, return the structure constant algebra L with product given by the Lie bracket (a, b) |-> a * b - b * a. As a second value the map identifying the elements of A and L is returned.
Let A and B be subalgebras of an associative algebra with underlying module M. This function returns the submodule of M which is spanned by the elements [a, b] = a * b - b * a, a in A, b in B.
For two subalgebras A and B of an associative algebra, return the ideal generated by all [a, b] = a * b - b * a, a in A, b in B.
For two subalgebras A and B of an associative algebra, return the left annihilator of B in A; that is, the subalgebra of A consisting of all elements a such that a * b = 0 for all b in B.
For two subalgebras A and B of an associative algebra, return the right annihilator of B in A; that is, the subalgebra of A consisting of all elements a such that b * a = 0 for all b in B.
We create the Lie algebra sl_3(Q) as a structure constant algebra. First, we construct gl_3(Q) from the full matrix algebra M_3(Q) and get sl_3(Q) as the derived algebra of gl_3(Q).
> gl3 := LieAlgebra(Algebra(MatrixRing(Rationals(), 3))); > sl3 := gl3 * gl3; > sl3; Lie Algebra of dimension 8 with base ring Rational FieldLet's see how the first basis element acts.
> for i in [1..8] do > print sl3.i * sl3.1; > end for; (0 0 0 0 0 0 0 0) ( 0 -1 0 0 0 0 0 0) ( 0 0 -2 0 0 0 0 0) (0 0 0 1 0 0 0 0) (0 0 0 0 0 0 0 0) ( 0 0 0 0 0 -1 0 0) (0 0 0 0 0 0 2 0) (0 0 0 0 0 0 0 1)Since it acts diagonally, this element lies in a Cartan subalgebra. The next candidate seems to be the fifth basis element.
> for i in [1..8] do > print sl3.i * sl3.5; > end for; (0 0 0 0 0 0 0 0) (0 1 0 0 0 0 0 0) ( 0 0 -1 0 0 0 0 0) ( 0 0 0 -1 0 0 0 0) (0 0 0 0 0 0 0 0) ( 0 0 0 0 0 -2 0 0) (0 0 0 0 0 0 1 0) (0 0 0 0 0 0 0 2)This also acts diagonally and commutes with sl3.1, hence we have luckily found a full Cartan algebra in sl_3(Q). We can now easily work out the root system. Obviously the root spaces correspond to the pairs (sl3.2, sl3.4), (sl3.3, sl3.7) and (sl3.6, sl3.8). The product of a positive root with its negative should lie in the Cartan algebra.
> sl3.2*sl3.4; ( 1 0 0 0 -1 0 0 0) > sl3.3*sl3.7; (1 0 0 0 0 0 0 0) > sl3.6*sl3.8; (0 0 0 0 1 0 0 0)Clearly some choices have to be made and we fix sl3.3 as the element e_alpha corresponding to the first fundamental root alpha, sl3.7 as e_(-alpha) and get sl3.1 as h_alpha = e_alpha * e_(-alpha). For the other fundamental root beta we have to find an element e_beta such that e_alpha * e_beta is non-zero.
> sl3.3*sl3.2; (0 0 0 0 0 0 0 0) > sl3.3*sl3.4; ( 0 0 0 0 0 -1 0 0) > sl3.3*sl3.6; (0 0 0 0 0 0 0 0) > sl3.3*sl3.8; (0 1 0 0 0 0 0 0)We choose sl3.8 as e_beta, sl3.6 as e_(-beta) and consequently -sl3.5 as h_beta. This now determines e_(alpha + beta) to be sl3.2 and e_(-alpha - beta) to be sl3.4.
The centralizer of the element s of the associative algebra A, that is, the subalgebra of A commuting with s.
The Lie bracket a * b - b * a of a and b, where a and b are elements of an associative algebra A.
For an associative algebra A of dimension n return an isomorphic matrix algebra. If A contains the identity-element, the matrix algebra will be of degree n, otherwise it will be of degree n + 1.
Side: MonStgElt Default: "Right"
For an associative algebra A of dimension n over R return its regular representation. If B = (e_1, e_2, ..., e_n) is the stored basis for A, an element a in A is mapped to the matrix in R^(n x n) which has as its i-th row the coordinates of e_i * a with respect to B. As a second map, the homomorphism of A onto the regular representation is returned.[Next] [Prev] [_____] [Left] [Up] [Index] [Root]
By default, the right-regular representation is computed. This can be changed to the left-regular representation (in which the i-th row of the image of a contains the coordinates of a * e_i) by setting the parameter Side to "Left".