Lie, Jordan algebras and the associator identity

Before I continue the quantum mechanics algebraic series, I want to first state my happiness for the defeat of the far (alt)-right candidate in France despite Putin's financial and hacking support. Europe has much better antibodies against the scums like Trump than US. In US the diseases caused by the inoculation of hate perpetuated over many years by Fox News has to run its course before things will get better.

Back to physics, first I will show that the product \(\alpha\) is indeed a Lie algebra. This is utterly trivial because we need to show antisymmetry and the Jacobi identity:

\(a\alpha b = -b\alpha a\)

\(a\alpha (b\alpha c) + c\alpha (a\alpha b) + b\alpha (c\alpha a) = 0\)

We already know that  the product \(\alpha\) is antisymmetric and we know that the it obeys Leibniz identity:

\(a\alpha (b\circ c) =  (a\alpha b) \circ c + b\circ (a\alpha c) \)

where \(\circ\) can stand for either \(\alpha\) or \(\sigma\). When \(\circ = \alpha\) we get:

\(a\alpha (b\alpha c) =  (a\alpha b) \alpha c + b\alpha (a\alpha c) \)

which by antisymmetry becomes

\(a\alpha (b\alpha c) = - c \alpha (a\alpha b) - b\alpha (c\alpha a) \)

In other words, the Jacobi identity.

Therefore the product \(\alpha\) is in fact a Lie algebra. Now we want to prove that the product \(\sigma\) is a Jordan algebra.

This is not as simple as proving the Lie algebra, and we will do it with the help of a new concept: the associator. Let us first define it. The associator of an arbitrary product \(\circ\) is defined as follows:

\([a,b,c]_{\circ} = (a\circ b)\circ c - a\circ (b\circ c)\)

as such it measures the lack of associativity. 

It is helpful now to look at the concrete realizations of the products \(\alpha\) and \(\sigma\) in quantum mechanics to know where we want to arrive. In quantum mechanics the product alpha is the commutator, and the product sigma is the anticommutator:

\(A \alpha B = \frac{i}{\hbar}[A,B] = \frac{i}{\hbar}(AB - BA)\)

\(A\sigma B = \frac{1}{2}\{A, B\} = \frac{1}{2}(AB+BA)\)

Let's compute alpha and sigma associators:

\([A,B,C]_{\alpha} = \frac{-1}{\hbar^2}([AB-BA, C] - [A, BC-CB]) = \)

\(=\frac{-1}{\hbar^2}(ABC-BAC-CAB+CBA - ABC+ACB+BCA-CBA)\)

\(= \frac{-1}{\hbar^2}(-BAC-CAB +ACB+BCA)\)

\([A,B,C]_{\sigma} = \frac{1}{4}(\{AB+BA, C\} - \{A, BC+CB\}) = \)

\(=\frac{1}{4}(ABC+BAC+CAB+CBA - ABC-ACB-BCA-CBA) = \)

\(=\frac{1}{4}(BAC+CAB -ACB-BCA)  \)

and so we have the remarkable relationship:

\([A,B,C]_{\sigma} + \frac{i^2 \hbar^2}{4}[A,B,C]_{\alpha} = 0\)

What is remarkable about this is that the Jordan and Lie algebras lack associativity in precisely the same way and because of this they can be later combined into a single operation. The identity above also holds the key for proving the Jordan identity.

Next time I'll show how to derive the identity above using only the ingredients we proved so far and then I'll show how Jordan identity arises out of it. Please stay tuned.