Subalgebras versus ideals
A subspace M⊆L is a subalgebra if [M,M]⊆M, and an ideal if the stronger [L,M]⊆M holds. The distinction is the exact analogue of subgroup versus normal subgroup: an ideal is closed under bracketing with *everything*, just as a normal subgroup is closed under conjugation by everything. And just as with groups, you can form the quotient L/M precisely when M is an ideal, with bracket [x+M,y+M]=[x,y]+M.
Three ideals you always have: the center Z(L)={x:[x,y]=0 for all y} (the kernel of ad); the derived algebra [L,L], spanned by all brackets, the analogue of the commutator subgroup; and L itself together with 0. A homomorphism φ:L→L′ has kernel an ideal and image a subalgebra, and the isomorphism theorems hold verbatim: L/ker φ ≅ im φ, and so on. If you know the group versions, you already know these.
Example: the upper-triangular and strictly-upper-triangular subalgebras of gl(3).
b = { upper triangular } = span of E_11, E_22, E_33, E_12, E_13, E_23
n = { strictly upper triangular } = span of E_12, E_13, E_23
Is n an ideal of b? Check [b, n] subset of n.
Recall [E_ij, E_kl] = delta_jk E_il - delta_li E_kj.
[E_11, E_12] = E_12 - 0 = E_12 (in n)
[E_22, E_12] = 0 - E_12 = -E_12 (in n)
[E_12, E_23] = E_13 , [E_13, anything in n] = 0 here
Every bracket [diagonal or upper, strictly upper] lands in n. So n is an ideal of b.
Quotient b/n: brackets of diagonal parts vanish mod n, so
b/n is abelian, dim 3 (the diagonal h).
Meanwhile [b,b] = n exactly: the derived algebra of b is n.The bridge to Lie groups
Here is the picture that names the whole subject. A Lie group G is a group that is also a smooth manifold, with smooth multiplication and inversion — think GL(n,ℝ), SO(3), the unitary group U(n). Its Lie algebra is the tangent space at the identity, L=T_eG, equipped with a bracket. The bracket is not arbitrary: it is the *infinitesimal commutator*. If you take two near-identity elements, conjugate, and expand to second order, the leading nonabelian term is exactly [X,Y].
- Each X in L gives a one-parameter subgroup t↦exp(tX) in G; the matrix exponential exp(X)=I+X+X²/2+… is the prototype.
- Conjugation in G differentiates to the adjoint representation Ad of G on L; differentiating Ad again gives ad, and ad(X)(Y)=[X,Y].
- The Baker–Campbell–Hausdorff formula exp(X)exp(Y)=exp(X+Y+½[X,Y]+…) reassembles the group product from bracket data alone.