Proving not all generators of the Lorentz algebra are (anti)-hermitian

Question

Suppose we have the Lorentz algebra $\mathfrak{so}(1,3)$ with the generators $M_{\mu\nu} = - M_{\nu\mu}$ and commutators between them are $$[M_{\mu\nu}, M_{\sigma\rho}] = g_{\nu\sigma}M_{\mu\rho} - g_{\mu\sigma}M_{\nu\rho} - g_{\nu\rho}M_{\mu\sigma} + g_{\mu\rho}M_{\nu\sigma}.$$ The metric is given in the $(-,+,+,+)$ signature.

I know that there are 6 independent generators, and that they are not all hermitian or anti-hermitian. But how do we prove the latter? Given that they are all hermitian, the commutator must be anti-hermitian, but I don't seem to get any further than this.

Any help is appreciated!

Answer 1

Personally, I never find a list of commutator relations very enlightening. Instead lets derive the algebra from its definition. $\mathfrak{so}(3,1)$ is the set of linear maps which are skew for an $(3,1)$-signature bilinear form $(\cdot, \cdot)$. That is $(Xv,w) + (v,Xw) = 0$. Choosing an orthonormal basis for this form we can write it as the matrix $$B = \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&-1\\\end{pmatrix}$$

Here $(v,w) = v^TBw$. Now, our skew condition implies $X^TB+BX = 0$. But multiplying by $B$ on the left simply changes the sign of the bottom row and similarly multiplying by it on the right flips the sign of the right hand column. So writing $X$ as a block matrix in our basis:

$$ X = \begin{pmatrix}A&B\\C&D\end{pmatrix},$$ where $A$ is $3\times 3$ and $D$ is $1\times 1$, we get

$$ X^TB+BX = \begin{pmatrix}A^T&-C^T\\B^T&-D^T\end{pmatrix} +\begin{pmatrix}A&B\\-C&-D\end{pmatrix} = 0 $$

And thus $A^T = -A$, $B^T = -C$ and $D^T = -D$ (which forces $D=0$). So we have a block diagonal skew-symmetric part and a block off-diagonal symmetric part both of dimension $3$, so we can clearly choose a basis of $3$ symmetric ones and $3$ anti-symmetric ones.

Of course you can choose a basis of elements that don't fall in either category if you wanted.

Answer 2

$ \newcommand\R{\mathbb R} \newcommand\Cl{\mathrm{Cl}} \newcommand\trans[1]{#1^{\mathrm T}} $This is actually a geometric fact. We will show that the fact that $\mathfrak{so}(1,3)$ has three generators which are symmetric and three which are anti-symmetric is exactly the fact that $\R^4$ under the Minkowski metric contains three planes which are invariant under time reversal and three planes which are invariant under spatial point-reflections such that all six of these are mutually (partially) orthogonal.

We will use the space-time algebra, which is the Clifford algebra $\Cl_{1,3}(\R)$. This is an algebra generated by vectors in $\R^4$. The subscripts mean that there is an orthonormal basis of vectors $\gamma_0,\gamma_1,\gamma_2,\gamma_3$ such that $$ \gamma_0^2 = 1,\quad \gamma_1^2 = \gamma_2^2 = \gamma_3^2 = -1. $$ We are interested in the space of bivectors $\Cl^2_{1,3}(\R)$, which are sums of exterior products $v\wedge w$ of vectors $v, w$. Such an exterior product is naturally identified with the plane spanned by $v, w$. The elements $$ \gamma_0\gamma_1,\quad \gamma_0\gamma_2,\quad \gamma_0\gamma_3,\quad \gamma_1\gamma_2,\quad \gamma_2\gamma_3,\quad \gamma_3\gamma_1 $$ form a basis for $\Cl^2_{1,3}(\R)$. Notice that there are six of these. Together with the commutator product $$ X\times Y = \frac12(XY - YX), $$ these bivectors are isomorphic to $\mathfrak{so}(1,3)$.

We will use square brackets to denote the action of a matrix on a vector. For every $R \in \mathrm{SO}(1,3)$, there is a bivector $B$ such that for any vector $v$ $$ e^{-B}ve^B = R[v]. $$ If $\eta$ is the Minkowski metric tensor, then $\trans R\eta R = \eta$ so that $\trans R = \eta R^{-1}\eta$. We can also find that $\eta[v] = \gamma_0v\gamma_0$, which is exactly a spatial point-reflection $\gamma_0 \mapsto \gamma_0$ and $\gamma_i \mapsto -\gamma_i$ for $i = 1,2,3$. If $M \in \mathfrak{so}(1,3)$ which is symmetric or anti-symmetric and $B$ is a bivector corresponding to $e^M$, then we find $$ e^{-B}ve^B = e^M[v] = e^{\pm\trans M}[v] = \trans{(e^{\pm M})}[v] = (\eta e^{\mp M}\eta)[v] = \gamma_0e^{\pm B}\gamma_0v\gamma_0e^{\mp B}\gamma_0 $$$$ \implies X^{-1}vX = v,\quad X = e^B\gamma_0e^{\pm B}\gamma_0 = e^Be^{\pm\gamma_0 B\gamma_0}. $$ This, in fact, implies that $X$ must be a scalar. Because $e^B$ must have unit magnitude, $X = \pm1$ and $$ e^B = \pm e^{\mp\gamma_0B\gamma_0} \impliedby B = \mp\gamma_0B\gamma_0. $$ The $\pm$ in front of $e$ is unrelated to that in the exponent. Thus it is sufficient that we find bivectors which are invariant under spatial point-reflections $x \mapsto \gamma_0x\gamma_0$ and under time-reversal $x \mapsto -\gamma_0x\gamma_0$. But $$ \gamma_0\gamma_1,\quad \gamma_0\gamma_2,\quad \gamma_0\gamma_3 $$ are bivectors invariant under time reversal, and $$ \gamma_1\gamma_2,\quad \gamma_2\gamma_3,\quad \gamma_3\gamma_1 $$ are bivectors invariant under point-reflection, and these span the entire space of bivectors (which I remind are isomorphic to $\mathfrak{so}(1,3)$).