# Least Action

Nontrivializing triviality..and vice versa.

## A lemma on a rather frequently encountered set of Hermitian matrices

I’m documenting a small ‘lemma’ that I think is worth mentioning. Suppose $M^{i}$ are 4 Hermitian matrices ($i = 1, \ldots, 4$) satisfying

$M^{i}M^{j} + M^{j}M^{i} = 2\delta^{ij}I$

where $I$ denotes the identity matrix. Then these matrices have eigenvalues $\pm 1$, are traceless and are necessarily of even dimension.

For $i = j$, the anticommutator above gives $(M^{i})^2 = I$. So, for any eigenvector $X$ and eigenvalue $\lambda$, we have

$M^{i}X = \lambda X \implies (M^{i})^2 X = \lambda^2 X$

or equivalently $\lambda = \pm 1$.

Traceless-ness has a neat proof:

Suppose $j \neq i$. Then

$tr(M^{i}) = tr(M^{i}(M^{j})^2) = tr(M^{i}M^{j}M^{j}) = tr(M^{j}M^{i}M^{j})$

where the last equality follows from $tr(ABC) = tr(CAB)$. But $M^{j}M^{i} = -M^{i}M^{j}$, so

$tr(M^{i}) = -tr(M^{i}M^{j}M^{j}) = -tr(M^{i}) \implies tr(M^{i}) = 0$.

Finally, suppose the numbers of +1 eigenvalues and -1 eigenvalues are $a$ and $b$ respectively. The dimension of the matrix is then $a + b$. Since the trace equals the sum of eigenvalues, we have

$a - b = 0$

So, the dimension is $a + b = 2a = 2b$, which is clearly always an even number.