The Jacobi Identity

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Section 3.3 The Jacobi Identity

The Jacobi identity, introduced in Section 3.1, states that

\begin{equation} \bigl[X,[Y,Z]\bigr] + \bigl[Y,[Z,X]\bigr] + \bigl[Z,[X,Y]\bigr] = 0\tag{3.3.1} \end{equation}

for elements \(X\text{,}\) \(Y\text{,}\) \(Z\) of some Lie algebra \(\gg\text{.}\) If \(\gg\) is a matrix Lie algebra, we can verify (3.3.1) by direct computation. Working out the first term, we have

\begin{align} \bigl[X,[Y,Z]\bigr] \amp= X(YZ-ZY) - (YZ-ZY)X\notag\\ \amp= (XYZ-YZX) + (ZYX-XZY)\tag{3.3.2} \end{align}

where we have deliberately regrouped the terms in the last expression according to whether they are an even or odd permutation of the original order. It now follows from the cyclic nature of (3.3.1) that the even and odd terms cancel in the sum.

If \(\gg\) is not a matrix Lie algebra, then the Jacobi identify must be checked separately. An example is given in Section 3.4.

Using the antisymmetry of the commutator, the Jacobi identity can be rearranged into two equivalent forms, which lie at the heart of the properties of Lie algebras, namely

\begin{gather} \bigl[X,[Y,Z]\bigr] = \bigl[[X,Y],Z\bigr] + \bigl[Y,[X,Z]\bigr] ,\tag{3.3.3}\\ \bigl[[X,Y],Z\bigr] = \bigl[X,[Y,Z]\bigr] - \bigl[Y,[X,Z]\bigr] .\tag{3.3.4} \end{gather}

The first form, (3.3.3), states that the Lie bracket (commutator) is a derivation, that is, it satisfies the Leibniz product rule:

\(X\) acting on the product \(YZ\) is the sum of the product of \(X\) acting on \(Y\) with \(Z\) and the product of \(Y\) with \(X\) acting on \(Z\text{.}\)

where both "acting on" and "product" are to be interpreted as commutators.

The second form, (3.3.4), states that the action of the Lie algebra on itself is a Lie algebra homomorphism, that is, the action of the commutator (left-hand side) is the same as the commutator of the actions (right-hand side).