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The Bracket: Lie Algebras from the Ground Up

A Lie algebra is a vector space with a bilinear, antisymmetric product satisfying the Jacobi identity. We build the definition from the commutator [x,y]=xy−yx, read off structure constants, and meet the examples that anchor everything to come: gl, sl, so, and the tiny but decisive sl(2).

Where the bracket comes from

Take any associative algebra — say the n×n matrices over a field — and ask how badly two elements fail to commute. The measure is the commutator [x,y]=xy−yx. Throw away the original multiplication and keep only this bracket: what survives is a Lie algebra. The point is that the bracket forgets the order-dependent part of multiplication and retains exactly the infinitesimal symmetry that continuous groups carry.

Formally, a Lie algebra over a field k is a vector space L with a bilinear map [·,·]:L×L→L, the Lie bracket, that is antisymmetric ([x,x]=0, hence [x,y]=−[y,x]) and satisfies the Jacobi identity [x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0. Note what is *not* required: no associativity, no unit. The Jacobi identity is the shadow that associativity casts onto the bracket.

Structure constants and the worked example sl(2)

Once you fix a basis e_1,…,e_n, the bracket is determined by the structure constants c_{ij}^k defined by [e_i,e_j]=Σ_k c_{ij}^k e_k. Antisymmetry says c_{ij}^k=−c_{ji}^k; the Jacobi identity becomes a quadratic relation among the c's. Everything about a finite-dimensional Lie algebra is, in principle, encoded in this table of numbers — though, as we will see, the *useful* invariants are far more economical than the raw table.

sl(2,C): traceless 2x2 complex matrices, dim 3.
Basis:
  e = [0, 1; 0, 0]   h = [1, 0; 0, -1]   f = [0, 0; 1, 0]

Compute the bracket [x,y] = xy - yx directly:

  [h,e] = h e - e h
        = [0, 2; 0, 0] - [0, -2; 0, 0] = [0, 4; 0, 0]?  -- recompute
  h e = [1,0;0,-1][0,1;0,0] = [0, 1; 0, 0]
  e h = [0,1;0,0][1,0;0,-1] = [0, -1; 0, 0]
  [h,e] = [0,1;0,0] - [0,-1;0,0] = [0, 2; 0, 0] = 2e

  [h,f] = h f - f h = [0,0;-1,0] - [0,0;1,0] = -2f
  [e,f] = e f - f e = [1,0;0,0] - [0,0;0,1] = [1,0;0,-1] = h

Structure constants (the whole table):
  [h,e] = 2e ,  [h,f] = -2f ,  [e,f] = h
  (all others fixed by antisymmetry, e.g. [e,h] = -2e)

Check Jacobi on (e,f,h):
  [e,[f,h]] + [f,[h,e]] + [h,[e,f]]
  = [e, 2f] + [f, 2e] + [h, h]
  = 2[e,f] + 2[f,e] + 0 = 2h - 2h + 0 = 0.  OK.
The defining relations of sl(2,ℂ) — the single most important example in the whole theory. Memorize [h,e]=2e, [h,f]=−2f, [e,f]=h.

Three families will recur throughout: the general linear algebra gl(n,k) of all n×n matrices with the commutator; the special linear algebra sl(n,k) of traceless matrices (an ideal, since tr[x,y]=0 always); and the orthogonal so(n) of skew-symmetric matrices. These are the *classical* Lie algebras, and the classification at the end of this track is, in essence, the statement that over ℂ they (with the symplectic family and five exceptions) are all the simple ones.

The adjoint representation: a Lie algebra acting on itself

Fix x in L and consider the map ad(x):L→L, ad(x)(y)=[x,y]. This is the adjoint representation, and it is the engine of the entire subject. The Jacobi identity, read the right way, says exactly that ad(x) is a derivation of the bracket: ad(x)[y,z]=[ad(x)y,z]+[y,ad(x)z]. It also says x↦ad(x) is a homomorphism of Lie algebras into gl(L), i.e. ad([x,y])=ad(x)ad(y)−ad(y)ad(x).