Chapter 3 Lie Algebra

Definition 3.1 (Lie Algebra) Let \(\mathbb{F}\) be a field. A Lie algebra over \(\mathbb{F}\) is a vector space \(\mathfrak{g}\) over \(\mathbb{F}\) together with a map \([\cdot,\cdot] : \mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}\), called the Lie bracket, satisfying the following properties:

  1. Bi-linearity: The bracket \([\cdot,\cdot]\) is bilinear.

  2. Skew-symmetry: For all \(X,Y \in \mathfrak{g}\), \[[X,Y] = -[Y,X].\]

  3. Jacobi identity: For all \(X,Y,Z \in \mathfrak{g}\), \[[X,[Y,Z]] + [Y,[Z,X]] + [Z,[X,Y]] = 0.\]

Two elements \(X,Y \in \mathfrak{g}\) are said to commute if \([X,Y] = 0\). The Lie algebra \(\mathfrak{g}\) is called commutative (or abelian) if \([X,Y] = 0\) for all \(X,Y \in \mathfrak{g}\).

Example 3.1 Let \(\mathfrak{g} = \mathbb{R}^3\) and define the bracket \[[x,y] = x \times y, \quad x,y \in \mathbb{R}^3,\] where \(x \times y\) denotes the cross product. Then \(\mathfrak{g}\) is a Lie algebra.

Proof. Bilinearity and skew-symmetry are standard properties of the cross product. To verify the Jacobi identity, it suffices (by bilinearity) to check it when \[x = e_j, \quad y = e_k, \quad z = e_\ell,\] where \(e_1, e_2, e_3\) are the standard basis vectors of \(\mathbb{R}^3\).

  • If \(j,k,\ell\) are all equal, each term in the Jacobi identity is zero.
  • If \(j,k,\ell\) are all distinct, the cross product of any two of \(e_j, e_k, e_\ell\) is a multiple of the third, so again each term in the Jacobi identity is zero.
  • It remains to consider the case in which two of \(j,k,\ell\) are equal and the third is different. By reordering the terms in the Jacobi identity as necessary, it suffices to verify \[[e_j,[e_j,e_k]] + [e_j,[e_k,e_j]] + [e_k,[e_j,e_j]] = 0. \qquad (3.1)\]

The first two terms in (3.1) are negatives of each other, and the third is zero.

Thus the Jacobi identity holds, and \(\mathfrak{g}\) is a Lie algebra.

Example 3.2 Let \(A\) be an associative algebra and let \(\mathfrak{g}\) be a subspace of \(A\) such that \[XY - YX \in \mathfrak{g} \quad \text{for all } X,Y \in \mathfrak{g}.\]

Then \(\mathfrak{g}\) is a Lie algebra with bracket operation defined by \[[X,Y] = XY - YX.\]

Proof. It is very easy to show the bilinear and skew symmetry properties. So, just go for Jacobi Identity.

Let \(X,Y,Z \in \mathfrak{g}\), consider

\[\begin{align*} [X,[Y,Z]] &= X(YZ - ZY) - (YZ - ZY)X = XYZ - XZY - YZX + ZYX,\\ [Y,[Z,X]] &= Y(ZX - XZ) - (ZX - XZ)Y = YZX - YXZ - ZXY + XZY,\\ [Z,[X,Y]] &= Z(XY - YX) - (XY - YX)Z = ZXY - ZYX - XYZ + YXZ. \end{align*}\]

Adding these three expressions together, every monomial in \(X,Y,Z\) appears exactly twice, once with a plus sign and once with a minus sign:

\[\begin{align*} (XYZ - XYZ) + (XZY - XZY) + (YZX - YZX) + (YXZ - YXZ) \\+ (ZXY - ZXY) + (ZYX - ZYX) = 0. \end{align*}\]