structural:expansion
The Lie group SO (3,1)

The set of transformations { B (ζ), R (θ), Λ (ζ, θ) } {\displaystyle \{B({\boldsymbol {\zeta }}),R({\boldsymbol {\theta }}),\Lambda ({\boldsymbol {\zeta }},{\boldsymbol {\theta }})\}} with matrix multiplication as the operation of composition forms a group, called the "restricted Lorentz group", and is the special indefinite orthogonal group SO (3,1). (The plus sign indicates that it preserves the orientation of the temporal dimension).

For simplicity, look at the infinitesimal Lorentz boost in the x direction (examining a boost in any other direction, or rotation about any axis, follows an identical procedure). The infinitesimal boost is a small boost away from the identity, obtained by the Taylor expansion of the boost matrix to first order about ζ = 0, B x = I + ζ ∂ B x ∂ ζ | ζ = 0 + ⋯ {\displaystyle B_{x}=I+\zeta \left.{\frac {\partial B_{x}}{\partial \zeta }}\right|_{\zeta =0}+\cdots } where the higher order terms not shown are negligible because ζ is small, and B is simply the boost matrix in the x direction. The derivative of the matrix is the matrix of derivatives (of the entries, with respect to the same variable), and it is understood the derivatives are found first then evaluated at ζ = 0, ∂ B x ∂ ζ | ζ = 0 = − K x. {\displaystyle \left.{\frac {\partial B_{x}}{\partial \zeta }}\right|_{\zeta =0}=-K_{x}\,.}

For now, K is defined by this result (its significance will be explained shortly). In the limit of an infinite number of infinitely small steps, the finite boost transformation in the form of a matrix exponential is obtained B x = lim N → ∞ (I − ζ N K x) N = e − ζ K x {\displaystyle B_{x}=\lim _{N\to \infty }\left(I-{\frac {\zeta }{N}}K_{x}\right)^{N}=e^{-\zeta K_{x}}} where the limit definition of the exponential has been used (see also characterizations of the exponential function). More generally

B (ζ) = e − ζ ⋅ K, R (θ) = e θ ⋅ J. {\displaystyle B({\boldsymbol {\zeta }})=e^{-{\boldsymbol {\zeta }}\cdot \mathbf {K} }\,,\quad R({\boldsymbol {\theta }})=e^{{\boldsymbol {\theta }}\cdot \mathbf {J} }\,.}

The axis-angle vector θ and rapidity vector ζ are altogether six continuous variables which make up the group parameters (in this particular representation), and the generators of the group are K = (K, K, K) and J = (J, J, J), each vectors of matrices with the explicit forms

K x =, K y =, K z = J x =, J y =, J z = {\displaystyle {\begin{alignedat}{3}K_{x}&={\begin{bmatrix}0&1&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0\\\end{bmatrix}}\,,\quad &K_{y}&={\begin{bmatrix}0&0&1&0\\0&0&0&0\\1&0&0&0\\0&0&0&0\end{bmatrix}}\,,\quad &K_{z}&={\begin{bmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&0\end{bmatrix}}\\J_{x}&={\begin{bmatrix}0&0&0&0\\0&0&0&0\\0&0&0&-1\\0&0&1&0\\\end{bmatrix}}\,,\quad &J_{y}&={\begin{bmatrix}0&0&0&0\\0&0&0&1\\0&0&0&0\\0&-1&0&0\end{bmatrix}}\,,\quad &J_{z}&={\begin{bmatrix}0&0&0&0\\0&0&-1&0\\0&1&0&0\\0&0&0&0\end{bmatrix}}\end{alignedat}}}

These are all defined in an analogous way to K above, although the minus signs in the boost generators are conventional. Physically, the generators of the Lorentz group correspond to important symmetries in spacetime: J are the rotation generators which correspond to angular momentum, and K are the boost generators which correspond to the motion of the system in spacetime. The derivative of any smooth curve C (t) with C (0) = I in the group depending on some group parameter t with respect to that group parameter, evaluated at t = 0, serves as a definition of a corresponding group generator G, and this reflects an infinitesimal transformation away from the identity. The smooth curve can always be taken as an exponential as the exponential will always map G smoothly back into the group via t → exp(tG) for all t; this curve will yield G again when differentiated at t = 0.

Expanding the exponentials in their Taylor series obtains B (ζ) = I − sinh ζ (n ⋅ K) + (cosh ζ − 1) (n ⋅ K) 2 {\displaystyle B({\boldsymbol {\zeta }})=I-\sinh \zeta (\mathbf {n} \cdot \mathbf {K})+(\cosh \zeta -1)(\mathbf {n} \cdot \mathbf {K})^{2}} R (θ) = I + sin θ (e ⋅ J) + (1 − cos θ) (e ⋅ J) 2. {\displaystyle R({\boldsymbol {\theta }})=I+\sin \theta (\mathbf {e} \cdot \mathbf {J})+(1-\cos \theta)(\mathbf {e} \cdot \mathbf {J})^{2}\,.} which compactly reproduce the boost and rotation matrices as given in the previous section.

It has been stated that the general proper Lorentz transformation is a product of a boost and rotation. At the infinitesimal level the product Λ = (I − ζ ⋅ K + ⋯) (I + θ ⋅ J + ⋯) = (I + θ ⋅ J + ⋯) (I − ζ ⋅ K + ⋯) = I − ζ ⋅ K + θ ⋅ J + ⋯ {\displaystyle {\begin{aligned}\Lambda &=(I-{\boldsymbol {\zeta }}\cdot \mathbf {K} +\cdots)(I+{\boldsymbol {\theta }}\cdot \mathbf {J} +\cdots)\\&=(I+{\boldsymbol {\theta }}\cdot \mathbf {J} +\cdots)(I-{\boldsymbol {\zeta }}\cdot \mathbf {K} +\cdots)\\&=I-{\boldsymbol {\zeta }}\cdot \mathbf {K} +{\boldsymbol {\theta }}\cdot \mathbf {J} +\cdots \end{aligned}}} is commutative because only linear terms are required (products like (θ · J)(ζ · K) and (ζ · K)(θ · J) count as higher order terms and are negligible). Taking the limit as before leads to the finite transformation in the form of an exponential Λ (ζ, θ) = e − ζ ⋅ K + θ ⋅ J. {\displaystyle \Lambda ({\boldsymbol {\zeta }},{\boldsymbol {\theta }})=e^{-{\boldsymbol {\zeta }}\cdot \mathbf {K} +{\boldsymbol {\theta }}\cdot \mathbf {J} }.}

The converse is also true, but the decomposition of a finite general Lorentz transformation into such factors is nontrivial. In particular, e − ζ ⋅ K + θ ⋅ J ≠ e − ζ ⋅ K e θ ⋅ J, {\displaystyle e^{-{\boldsymbol {\zeta }}\cdot \mathbf {K} +{\boldsymbol {\theta }}\cdot \mathbf {J} }\neq e^{-{\boldsymbol {\zeta }}\cdot \mathbf {K} }e^{{\boldsymbol {\theta }}\cdot \mathbf {J} },} because the generators do not commute. For a description of how to find the factors of a general Lorentz transformation in terms of a boost and a rotation in principle (this usually does not yield an intelligible expression in terms of generators J and K), see Wigner rotation. If, on the other hand, the decomposition is given in terms of the generators, and one wants to find the product in terms of the generators, then the Baker–Campbell–Hausdorff formula applies.