Obtain the explicit form of Lorentz transformation matrix using Lie algebra

Question

Consider the Minkowski space $\mathbb R^4$ with the Minkowski metric tensor
\begin{align}
\langle,\rangle:\ \mathbb R^4\times\mathbb R^4&\longrightarrow\mathbb R
\\ (u,v)&\longmapsto\langle u,v\rangle=-u_0v_0+u_1v_1+u_2v_2+u_3v_3=[u]^\top\eta\,[v]
\end{align}
where $\eta=$ diag$(-1,I_3)$.

We know that the Lorentz transformation is represented by a matrix $L\in GL_4(\mathbb R)$ such that
$$L^\top\eta\,L=\eta. $$
Each Lorentz transformation is a transformation of space-time coordinates between a stationary inertial frame $(t,x)$ and a moving inertial frame $(t',y)$ in the direction of the velocity $v=(v_1,v_2,v_3)$ with respect to the frame $(t,x)$.

I want to derive the explicit form of $L$. One common approach is decomposing vector $v$ into two terms $v=v_{\perp}+v_{\parallel}$, do some classical algebra and geometry, and then we will get the concrete matrix
\begin{align}
L=\begin{bmatrix}
\gamma & -\gamma v
\\ -\gamma[v] & I_3+(\gamma-1)[\widetilde v][\widetilde v]^\top
\end{bmatrix}
\end{align}
This way is indeed quite simple, but it doesn't satisfy my desire. I interest in the approach using Lie algebra, which may involve more calculate, but it will be a good chance, a first step to get used to dealing with Lie algebra and from there do it proficiently in the next parts of General Relativity. At this time, I don't know much about Lie algebra, I don't know what it's about, I don't know what tool it uses. So I really hope that anyone will guide me.

Answer 1

Lie algebra is the linear approximation of the Lie group at the identity $$L = 1_4 + \epsilon \ l_{ik}$$ where $1_4$ is the 4d identity matrix and $$l_{ik}$$ a linear transformation in the 2d-plane with indices $i,k$
e.g. $l_{0,2}$

$$L=\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) \ + \ \epsilon \ \left( \begin{array}{cccc} a & 0 & b & 0 \\ 0 & 0 & 0 & 0 \\ c & 0 & d & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) $$

Now you can simply use

$$\partial_\epsilon\ \left( \left( 1_4 + \epsilon \ l_{ik}\right)^T \cdot \eta \cdot \left( 1_4 + \epsilon \ l_{ik}\right) - \eta \right) |_{\epsilon\to 0} =0 $$

and

$$\det\left( 1_4 + \epsilon \ l_{ik}\right) =1$$

This last equation, indeed, is the defining equation of the special orthogonal group $(SO(3,1))$, the Lorentz group.

As a Lie group, you can go to the limit of infinite powers

$$\left( 1_4 + \epsilon \ l_{ik} \right)\to \left( 1_4 + \frac{\epsilon}{n} \ l_{ik}\right)^n \to e^{\epsilon \ l_{ik}} $$ to retain the three 1-parameter subgroups of boosts and the three rotation groups in the six 2-planes of $\mathbb R^4$

It needs some analysis of matrix functions, to show that the product limit of powers of $$\lim_{n\to \infty }\ (1 + \frac{x}{n})^n = e^n\ = \ \sum{\frac{x^n}{n!}}$$ is working within a frame of convergencies in norms for matrix operators, and for operators in Hilbert spaces, generally.

The explicit forms are $$l_{0,1}=\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) $$ etc with exponential,

$$\exp\left(u\ \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array}\right)\right)$$

$$ 1_4 + \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) * \sum_0^\infty \ \frac{u^{2n+1}} {(2n+1)!} + \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) *\sum_1^\infty \frac{u^{2n}} {2n!} =\cosh u \ 1_4 + \sinh u \ l_{0,3}$$

and the same with alternating signs and trigonometric functions for the rotation sungroups