Subspaces of all real matrices invariant under Lorentz group

Question

This question concerns Jeevangee book "An introduction to tensors and group theory" 2nd edition page 258, example 5.39. Let $\eta$ be the Minkowski metric represented as the diagonal matrix $(1,1,1,-1)$. Every real $4\times 4$ matrix $M$ can be decomposed into the sum of three matrices $A+B+C$, where (1) $A$ is real scalar multiple of $\eta$. (2) $B$ is an anti-symmetric matrix. (3) $C$ is a symmetric $\eta$-"traceless" matrix, namely, $C_{ij}=C_{ji}$ and $\sum_{i,j=1}^4{\eta_{ij}C_{ij}}=0$. Equivalently, the linear space of all $4\times 4$ real matrices is the direct sum of $3$ subspaces. The book claims that these $3$ subspaces are $O(3,1)$-invariant. I don't know what he means by that. He probably means that these $3$ subspaces are invariant relative to some representation of $O(3,1)$. But which representation? If $L\in O(3,1)$, and $A$ belong to one of the three subspaces, I tried $LAL^T$ and $LAL^{-1}$ but none works for all three subspaces. Can any one help?

Answer 1

Let $g$ be the metric matrix of the defining 4d matrix representation of $SO(3,1)$. $S$ means 'special' and denotes the set of group components with $\det = +1$.

Since there are two different reflections with

$$\det R_1 =-1\ : \ x_1\to -x_1 , \det R_4 =-1 : \ x_4\to -x_4$$

the $SO(3,1)$ has two components, one around the identity matrix and one copy around the double reflection $ R_1 R_4$

Any $4\times 4$ matrix $L$ with $\det L^t \ L \ = \ \det L^2 \ =\ 1 $ respects the invariance of the Lorentz dot product

$$ x^t \ g \ y = (L \ x)^t \ g \ L \ y = x^t \ (L^t\ g \ L ) \ y $$

This implies

$$ \det (LL^t \ g ) = \det g \ \text{ and } \ \text{Tr} \ (LL^t \ g ) = \text{Tr} \ g = 2 $$

and

$$ L^{-1} = g L^t g $$

Determinant and trace as product and sum of the eigenvalues in the characteristic polynomial are two trivial abelian repersentations. Since the determinant is fixed, the other classifing group homomorphism, the abelian trace group is partioning the matrices into character classes. The rest of the proof is index permutation check in the transformation equation.

Answer 2

The action of $X \in O(3,1)$ is $M \mapsto XMX^T$. In index notation, we're just applying $X$ to each index of $M^{ij}$: $$ M^{ij} \mapsto X^i_kM^{kl}X^j_l. $$

Evidently the $A$ subspace is preserved by $X$ in this way since it is a multiple of $\eta$ and by definition $X\eta X^T = X^T\eta X = \eta$.

The $B$ subspace is also preserved since $$ (XBX^T)^T = XB^TX^T = -XBX^T $$ so $XBX^T$ is antisymmetric.

Finally, the $C$ subspace is preserved: just like above $XCX^T$ is symmetric, and then we compute $$ \eta_{ij}X^i_kC^{kl}X^j_l = (X^i_k\eta_{ij}X^j_l)C^{kl} = \eta_{kl}C^{kl} = 0. $$

Answer 3

A more conceptual (or at least basis-free) answer: An element $g \in O(3,1)$ is exactly one for which $\eta(gv,gw) = \eta(v,w)$. A linear transformation $X$ is symmetric if $\eta(Xv,w) = \eta(v,Xw)$ and anti-symmetric if $\eta(Xv,w) = - \eta(v,Xw)$. So consider the action of $O(3,1)$ on the set of matrices: $g\cdot X = g^{-1} X g$. Then: $$\eta((g^{-1} X g) v,w) = \eta(Xgv,gw) = \pm \eta(gv,Xgw) = \pm \eta(v,g^{-1}Xgw)$$

So $g\cdot X$ is (anti-) symmetric when $X$ is.

Changing perspective, $g$ preserves $\eta$ by definition so whatever linear transformation $\eta$ represents we have $g\cdot \eta = \eta$ and equally $\eta$ must itself be symmetric. So we only need to find a complement preserved by each $g$. And this is given by the elements $X$ of the form $\mathrm{tr}(\eta X) = 0$. Again note $\mathrm{tr}(\eta g^{-1}Xg) = \mathrm{tr}(g\eta g^{-1}X) = \mathrm{tr}((g^{-1}\cdot\eta)X) = \mathrm{tr}(\eta X) = 0$ so this is indeed an invariant subspace and $\mathrm{tr}(\eta^2) \neq 0$ so it is complementary (the trace form is a nondegenerate bilinear form).