# How to obtain the explicit form of Lorentz transformation matrix using Lie algebra?

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.

• It's all in the book by Wu-ki Tung. Due diligence. Commented Jan 29 at 22:28
• Actually it is much less calculation. Let $\psi$ be the rapidity i.e. $|\vec v|=c\tanh\psi$. Then prove $(\hat{\vec v}\cdot K)^3=\hat{\vec v}\cdot K$, because this immediately produces $e^{\psi\hat{\vec v}\cdot K}=\mathbb I+(\cosh\psi-1)(\text{stuff})+\hat{\vec v}\cdot K\sinh\psi$, and that is very easy to compare with the final answer. Commented Jan 30 at 2:37
• See Wikipedia for more info on what naturallyInconsistent is talking about. Commented Jan 30 at 6:35
• I don't know what tool it uses. Matrix exponentiation. Commented Jan 30 at 6:37
• @naturallyInconsistent what is $K$, and what operation is $\cdot$ ? Commented Jan 30 at 8:03

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

• so Lie algebra is the tangent space of a Lie group at the identity element ? Commented Jan 30 at 8:05
• By the Lie group identities the tangential space is translation invariant $$T_x(G) =g(x)^T\ T_1 g(x)$$ Commented Jan 30 at 8:15
• I am not sure that this answer really touches on the notion of Lie algebras as OP asks about in their question. The alternative that I have in mind is to start with the Lorentz algebra and use the exponential map to generate the restricted Lorentz group. Commented Jan 30 at 8:16
• @RolandF so here is my solution : I deduced that every $X\in\mathfrak{so}(1,3)$ takes the form $X=\theta\cdot J+v\cdot K$, where $K$ is sum of the matrices $l_{ik}$ as you mentioned, and $\theta,\,v\in\mathbb R^3$. Since the exponential map is bijective from $\mathfrak{so}(1,3)$ to $SO^+(1,3)$, each Lorentz transformation $L\in SO^+(1,3)$ has the form $L=e^X$ for some $X\in\mathfrak{so}(1,3)$. The general Lorentz boost is obtained when $\theta=0$, and hence $L=e^{v\cdot K}$. However I have no idea how to deduce that $L=e^{\psi (\widehat v\cdot K) }$ where $\tanh\psi=|v|$. Commented Feb 8 at 1:52

With the six generators in $$l_{ik}\in so(3,1)$$ of the one-parameter subgroups, it remains to show that the total of $$SO(3,1)$$ is the exponential of any linear combination of 3 boost parameters and 3 rotation angles $$L\left((u_{0,1},u_{0,2},u_{0,3}),\ (\phi_{1,2},\phi_{1,3},\phi_{2,3})\right)$$ such that $$L_{\vec u,\vec \phi} = e^{\vec u\cdot \vec l_t + \vec \phi \cdot \vec l_x}$$ and to show that $$L_{\vec u,\vec \phi}^T \ \eta \ L_{\vec u,\vec \phi} = \eta$$ $$\eta L^T \eta = L^{-1}$$

We know from the analoguous analysis of the rotation group SO(3), that its not a simply connected manifold by the fact that rotations of $$\pm \pi$$ in any 2-plane have to be identified. So a path crossing any greatest circle cannot be shrinked to a point.

SO(3,1) has this feature, too.

$$SO(3,1)$$ has 3 more components around the matrices of reflection of time and/or on any space 2-plane.

So its no wonder that there exists a single component, simply connected representation of the $$\mathit{so}(3,1)$$ algebra.

Namely $$\mathit{sl}(2,C)$$, the 2d special linear group of matrices with complex combinations of $$2\times2$$ hermitean, trace 0, Pauli matrices representing the basis.

The real combinations represent generators of boosts, the imaginary combinations generate rotations.

Like $$SU(2)$$ for $$SO(3,1)$$, the group $$SL(2,C)$$ represents the most elementary and topological simple representation of the Lorentz group.

At the price, that angles run over $$(0,4\pi)$$, rotations by $$2\pi$$ yield a $$-\text{identity}$$ on the space directions.

This fact eliminates the possibility to find a parity operator in the elementary representation.

The group homomorphism map $$SL(2,C) \to SO(3,1)$$ acts by identifying all matrices $$L,-L$$, rotations by $$\phi$$ and $$\phi +\pi$$, as a double cover of of the $$SO(3,1)$$-manifold.

An intuitive correct image of the group manifold of the Lorentz group $$SO(3,1)$$:

Four identical copies of the component around the identity, generated by the two inversions of time and space.

The copy of identity is a ball of radius $$\pi$$ with antipodes on the surface identified is a boost hyperbola, eg in the $$(x_1,t)$$ . At any point attached as a fiber is sitting a $$SU(2)$$ . Thats similar to the representation of the unit sphere as a half circle with a circle attached at each point with the radius equal to the distance from its axis.