Spatial rotations of an inertial frame

Let $(o;e_0,e_1,e_2,e_3)$ be an inertial frame of the model (Inertial frames and inertial charts of the affine Minkowski model) and let $S:=\mathrm{span}(e_1,e_2,e_3)$ be its spatial three-space with $\delta$ the restriction of $\eta$ to $S$ (The Euclidean spatial structure and cross product of an inertial frame). Then $\delta$ is a positive-definite symmetric bilinear form, $(e_1,e_2,e_3)$ is $\delta$-orthonormal, and $S=\{w\in V:\eta (e_0,w)=0\}$.

A spatial rotation of the frame is a linear map $R:V\to V$ (Linear map between vector spaces over the same field) such that

1. $R(e_0)=e_0$;
2. $R(S)\subseteq S$, so that the restriction $R{\mathrm{\mid }}_S$ is an endomorphism of $S$;
3. $\delta (Rw,Rz)=\delta (w,z)$ for all $w,z\in S$; and
4. the restriction $R{\mathrm{\mid }}_S$ has determinant $+1$ computed in the ordered basis $(e_1,e_2,e_3)$ (The determinant of an endomorphism of a finite-dimensional vector space: its matrix determinant in an ordered basis in positive dimension, and $1$ on the zero space).

Condition 3 makes $R{\mathrm{\mid }}_S$ injective, since $Rw=0$ with $w\in S$ forces $\delta (w,w)=\delta (Rw,Rw)=0$ and hence $w=0$ by positive definiteness; $S$ being finite-dimensional, $R{\mathrm{\mid }}_S$ is then an automorphism of $S$, so in fact $R(S)=S$. The extension of such an $R{\mathrm{\mid }}_S$ by $R(e_0)=e_0$ is also written $R$ and called a spatial rotation of the frame.

Matrix form and the group $SO(3)$. Via the $\delta$-orthonormal basis $(e_1,e_2,e_3)$, a linear map of $V$ fixing $e_0$ and preserving $S$ corresponds to the block matrix $\mathrm{diag}(1,A)$ with $A\in M_3(\mathbb{R})$; under this correspondence the maps of conditions 1-4 are exactly those with

$A^{T}A=I_3\text{and}\det A=+1.$

Indeed, the matrix of the adjoint of an endomorphism of the inner product space $(S,\delta )$ in a $\delta$-orthonormal basis is the transpose of its matrix (In orthonormal bases, the matrix of the adjoint is the conjugate transpose of the matrix), and an endomorphism of a finite-dimensional inner product space preserves the inner product exactly when $T^{\ast }T=I$ (For an endomorphism in finite dimension, preserving lengths, preserving inner products, carrying orthonormal bases to orthonormal bases, and $T^{\ast }T=I$ are equivalent, Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces). Writing $SO(3):=\{A\in M_3(\mathbb{R}):A^{T}A=I_3,\det A=+1\}$, the orthogonal operators on $S$ form a group and each of them has determinant $\pm 1$ (Orthogonal and unitary operators form groups, and their determinants have modulus one), and $\det (S_1{S}_2)=\det (S_1)\det (S_2)$ for composable endomorphisms (For endomorphisms $S$ and $T$ of one finite-dimensional vector space, $\det (ST)=\det (S)\det (T)$); hence the determinant-one part is a subgroup, the spatial rotations of the frame form a subgroup of the general linear group of $V$, that is, of the group of invertible linear maps $V\to V$ under composition (Invertible linear maps, linear isomorphisms, and inverse linear maps), and the map $R\mapsto A(R)$ sending a rotation to the matrix of $R{\mathrm{\mid }}_S$ in the basis $(e_1,e_2,e_3)$ is a group isomorphism onto $SO(3)$. The determinant in condition 4 is independent of the ordered basis of $S$ (The determinant of an endomorphism of a finite-dimensional vector space: its matrix determinant in an ordered basis in positive dimension, and $1$ on the zero space). Thus, for fixed $\eta$, the rotation subgroup depends only on the time axis $e_0$: it fixes $e_0$ and acts on $S=e_0^{\perp }$ by determinant-one $\delta$-isometries. Changing the origin or the spatial orthonormal basis while retaining $e_0$ leaves this subgroup unchanged; changing $e_0$ can change which maps are spatial rotations. No claim is made here that spatial rotations are Lorentz maps; that is proved in Unique boost-rotation factorization of proper orthochronous Lorentz maps.