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Denote the matrĩ $\eta=$ diag$(-1,1,1,1)$. The group $O(1,3)$, called Lorentz group, is the group of all matrix $L\in M_4(\mathbb R)$ such that \begin{align} L^\top\eta\,L\,=\,\eta.\tag1 \end{align} The subgroup of all Lorentz matrix in $O(1,3)$ of determinant 1 is denoted as $SO(1,3)$. My aim is to find the explicit description of each matrix $L\in SO(1,3)$.

Since $O(1,3)$ and $SO(1,3)$ is a Lie group, they admits a Lie algebra. Indeed, at the present I have not so much understand of the formal definition of the Lie algebra of a Lie group. However, in this concrete context, I know that the Lie algebra of $SO(1,3)$ is given by \begin{align} \mathfrak{so}(1,3)\,=\,\Big\{X\in M_4(\mathbb R)\,\big|\,e^{tX}\in SO(1,3)\ \, \forall t\in\mathbb R\Big\}.\tag2 \end{align} By the definition of exponential matrix, for any $X\in \mathfrak{so}(1,3)$, $\exists A\in M_4(\mathbb R)$ such that \begin{align} e^{tX}\,=\,I_4+A.\tag3 \end{align} Since $e^{tX}\in SO(1,3)$, we have \begin{align} \eta\,&=\,(I_4+A)^\top\eta\, (I_4+A) \\ &=\,\eta+\eta A+A^\top\eta+A^\top\eta\,A\tag4 \end{align} Omitting the term $A^\top\eta\,A$, we obtain \begin{align} (\eta A)^\top\,=\,-\eta A.\tag5 \end{align} We can deduce that, then, $A$ is given by \begin{align} A\,=\,\begin{bmatrix} 0&a&b&c \\ a&0&d&e \\ b&-d&0&f \\ c&-e&-f&0 \end{bmatrix},\ a,b,c,d,e,f\in\mathbb R.\tag6 \end{align} This implies \begin{align} A\,&=\,\theta_1L_1+\theta_2L_2+\theta_3L_3+\lambda_1K_1+\lambda_2K_2+\lambda_3K_3 \\ &=\,\theta\cdot L+\lambda\cdot K\tag7 \end{align} where $\theta=\big(\theta_1,\theta_2,\theta_3\big),\,\lambda=\big(\lambda_1,\lambda_2,\lambda_3\big)\in\mathbb R^3$ and \begin{align} L_1=\begin{pmatrix} 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&-1 \\ 0&0&1&0 \end{pmatrix}\ \ \ L_2=\begin{pmatrix} 0&0&0&0 \\ 0&0&0&1 \\ 0&0&0&0 \\ 0&-1&0&0 \end{pmatrix}\ \ \ L_3=\begin{pmatrix} 0&0&0&0 \\ 0&0&-1&0 \\ 0&1&0&0 \\ 0&0&0&0 \end{pmatrix}\tag8 \end{align} and \begin{align} K_1=\begin{pmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{pmatrix}\ \ \ K_2=\begin{pmatrix} 0&0&1&0 \\ 0&0&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \end{pmatrix}\ \ \ K_3=\begin{pmatrix} 0&0&0&1 \\ 0&0&0&0 \\ 0&0&0&0 \\ 1&0&0&0 \end{pmatrix}.\tag8 \end{align} These matrices satisfy (5).

By direct calculating, we can show that \begin{align} \big[L_i,L_j\big]\,&=\,\varepsilon_{ijk}L_k \\ \big[J_i,K_j\big]\,&=\,\varepsilon_{ijk}K_k \\ \big[K_i,K_j\big]\,&=\,-\varepsilon_{ijk}L_k\tag{10} \end{align} where the coefficients $\varepsilon_{ijk}$ are the Levi-Cevita symbol and the commutator $\big[L_i,K_j\big]=L_iK_j-K_jL_i$.

Substituting (7) into (3), we get \begin{align} e^{tX}\,=\,I_4+\sum_1^3\theta_i L_i+\sum_1^3\lambda_i K_i\,=\,\begin{bmatrix} 1&\lambda_1&\lambda_2&\lambda_3 \\ \lambda_1&1&-\theta_3&\theta_2 \\ \lambda_2&\theta_3&1&-\theta_1 \\ \lambda_3&-\theta_2&\theta_1&1 \end{bmatrix}.\tag{11} \end{align} Finally, any Lorentz matrix of $SO(1,3)$ is given by \begin{align} L\,=\,e^{\theta\cdot L+\lambda\cdot K}\tag{12} \end{align} for arbitrary $\theta,\lambda\in\mathbb R^3 $.

In this solution, there're some move that I still yet to understand.

1/ Why could we neglect the term $A^\top\eta\,A$ in (5) ?

2/ Why do we need to consider the commutator in (10), what is its role in the proof ?

3/ How can one get the result at (12) from the preceding moves ?

4/ How is the formula (12) related to the formula of generel Lorentz transformation \begin{align} \begin{bmatrix} t'\\ y_1\\ y_2\\ y_3 \end{bmatrix} = \begin{bmatrix} \gamma & -\gamma v_1 & -\gamma v_2 & -\gamma v_3 \\ -\gamma v_1 & 1+(\gamma-1)\displaystyle\frac{v_1^2}{v^2} & (\gamma-1)\displaystyle\frac{v_1v_2}{v^2} & (\gamma-1)\displaystyle\frac{v_1v_3}{v^2} \\ -\gamma v_2 & (\gamma-1)\displaystyle\frac{v_1v_2}{v^2} & 1+(\gamma-1)\displaystyle\frac{v_2^2}{v^2} & (\gamma-1)\displaystyle\frac{v_2v_3}{v^2} \\ -\gamma v_3 & (\gamma-1)\displaystyle\frac{v_1v_3}{v^2} & (\gamma-1)\displaystyle\frac{v_2v_3}{v^2} & 1+(\gamma-1)\displaystyle\frac{v_3^2}{v^2} \end{bmatrix} \begin{bmatrix} t\\ x_1\\ x_2\\ x_3 \end{bmatrix} \end{align} where \begin{align*} \gamma=\frac{1}{\sqrt{1-v^2}},\ \ \ v^2=\sqrt{v_1^2+v_2^2+v_3^2} \end{align*} ?

I hope someone would help me to clarify those impedents. Thanks.

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In (3), the left-hand side is meant to be an "infinitesimal" matrix, that is, $t$ is infinitesimally small. Therefore, $A$ is also infinitesimal. It follows that in (4), where we obtain $\eta A + A^T \eta + A^T \eta A = 0$, the terms linear in $A$ must cancel, because a term that is quadratic in an infinitesimal cannot cancel a nonzero term that is linear in an infinitesimal.

We can also arrive at the same result by considering the function $f : t \mapsto e^{tX}$. If $f$ maps the real line to a one-parameter subgroup of the Lorentz group, it must be that $f(t)^T \eta f(t) = \eta$ for all $t$. By differentiating both sides and setting $t = 0$, we arrive at $X^T \eta + \eta X = 0$.

The commutators of the matrices $K$ and $L$ are essential to the definition of the Lorentz Lie algebra. In general, a Lie algebra is completely specified by a basis $e_1, \ldots, e_D$ and the Lie brackets $[e_i, e_j]$ for each pair $1 \le i, j \le D$. This, in turn, is crucial for understanding the representation theory of the Lorentz group more deeply, for example, classifying all irreducible finite-dimensional representations of the Lorentz group. However, we will not need the commutators if our goal is purely to derive the representation of the boosts in the vector representation.

Your third question is discussed in the link provided by mike stone in the comments. However, if we limit our consideration to only those elements of the Lorentz group that represent pure boosts (and only in the vector representation), then it's a lot easier to prove that the exponential map is surjective. On purely physical grounds, a pure boost $B$ is always related by a rotation of axes to a boost of the same speed along the positive x-axis

$$ B = R^{-1} B_x(v) R $$

for some rotation matrix $R$. Now (again on physical grounds) $B_x(v)$ has the pair of eigenvectors $(v, \pm v, 0, 0)$ corresponding to the 4-momentum of light travelling parallel or antiparallel to the x-axis, and it also leaves the vectors $(0, 0, 1, 0)$ and $(0, 0, 0, 1)$ invariant. From this we derive that it takes the form

$$ B_x(v) = \begin{bmatrix} a & b & 0 & 0 \\ b & a & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$

and by further imposing the condition (1) we obtain that $a^2 - b^2 = 1$. Therefore, there exists $w$ such that $a = \cosh w, b = \sinh w$, and it is not hard to show that $B_x(v) = \exp \omega$ where

$$ \omega = \begin{bmatrix} 0 & w & 0 & 0 \\ w & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$

for some real number $w$. Since $B_x(v) = \exp \omega$ it follows that $B = \exp(R^{-1} \omega R)$, and it turns out that $R^{-1} \omega R$ is also a linear combination of the $K$ matrices, so it's also an element of $\mathfrak{so}(3, 1)$.

For a boost in an arbitrary direction, it's easiest to derive the general form of the Lorentz transformation matrix by performing a rotation of axes as we've done above. However, it's also possible to calculate the matrix exponential directly. In particular, if $A = \lambda_1 K_1 + \lambda_2 K_2 + \lambda_3 K_3$ then $A^3$ is a scalar multiple of $A$, which allows us to derive a closed form for the power series for $\exp A$ using the $\sinh$ and $\cosh$ functions.

Lastly, once the general form of a boost is derived (whether by this procedure or otherwise) we need to relate the entries of the matrix with the velocity of the boost. To do so we use the fact that a boost with velocity $\vec{v}$ must map $(1, \vec 0)$ (the four-velocity of a particle that is stationary in the original frame, as measured by an observer in that frame) to $(\gamma, -\gamma \vec{v})$ (the four-velocity of the same particle, now viewed from the boosted frame).

Note: A derivation for the general form of a Lorentz transformation (which is the composition of a rotation and a boost) in the vector representation is given here.

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  • $\begingroup$ okay, so here is my updated solution : first we admit that the exponential map $exp:\ \mathfrak{so}(1,3)\longrightarrow SO^+(1,3), \ X\longmapsto e^{X}$ is surjective. Hence, for any matrix $L\in SO^+(1,3)$, there exists $X$ such that $L=e^X$. We have \begin{align} \big(e^{X}\big)^\top\,\eta\,e^X=\eta. \end{align} By properties of exponential matrix, we imply \begin{align} I_4=e^{X^\top}\eta\,e^X\,\eta^{-1}=e^{X^\top}e^{\eta\,X\,\eta^{-1} } \end{align} hence $X^\top+\eta\,X\,\eta=0$, or $(\eta X)^\top=-\eta X $. $\endgroup$
    – PermQi
    Feb 6 at 15:49
  • $\begingroup$ Therefore, we have \begin{align} X&=\theta\cdot L+\lambda\cdot K \newline &=\theta\cdot J+ |\lambda|\left(\frac{1}{|\lambda|}\lambda\right)\cdot K \newline &=\theta\cdot J+ \psi(v\cdot K) \end{align} This imply \begin{align} L=e^X&=e^{\theta\cdot J}e^{\psi(v\cdot K)} \end{align} $\endgroup$
    – PermQi
    Feb 6 at 16:10
  • $\begingroup$ The general Lorentz boost is corresponding to $\theta=0$, which will get us \begin{align} L&=e^{\psi(v\cdot K)}. \end{align} Using the fact that $(v\cdot K)^3=v\cdot K$, we have \begin{align} L &=I_4+\psi(v\cdot K)+\frac{\psi^2}{2!}(v\cdot K)^2+\frac{\psi^3}{3!}(v\cdot K)^3+\cdots \newline &=I_4+\left(\psi+\frac{\psi^3}{3!}+\frac{\psi^5}{5!}+\cdots \right)(v\cdot K)+\left(\frac{\psi^2}{2!}+\frac{\psi^4}{4!} +\frac{\psi^6}{6!}+\cdots\right)(v\cdot K)^2 \newline &=I_4+(\sinh\psi)v\cdot K+(\cosh\psi-1)(v\cdot K)^2. \end{align} $\endgroup$
    – PermQi
    Feb 6 at 16:34
  • $\begingroup$ Let $\cosh\psi=\gamma$, then $\sinh\psi=\sqrt{\gamma^2-1} $. We have \begin{align} L &=I_4+\sqrt{\gamma^2-1} v\cdot K+(\gamma-1)(v\cdot K)^2 \end{align} But here, I feel like it seems not the same to the formula of generel Lorentz boost in the question 4/. I don't know why. $\endgroup$
    – PermQi
    Feb 6 at 16:36

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