Denote the matrix $\eta=$ 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_1J_1+\theta_2J_2+\theta_3J_3+\lambda_1K_1+\lambda_2K_2+\lambda_3K_3
\\ &=\,\theta\cdot J+\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}
J_1=\begin{pmatrix}
0&0&0&0 \\ 0&0&0&0
\\ 0&0&0&-1
\\ 0&0&1&0
\end{pmatrix}\ \ \ J_2=\begin{pmatrix}
0&0&0&0
\\ 0&0&0&1 \\ 0&0&0&0
\\ 0&-1&0&0
\end{pmatrix}\ \ \ J_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[J_i,J_j\big]\,&=\,\varepsilon_{ijk}J_k
\\ \big[J_i,K_j\big]\,&=\,\varepsilon_{ijk}K_k
\\ \big[K_i,K_j\big]\,&=\,-\varepsilon_{ijk}J_k\tag{10}
\end{align}
where the coefficients $\varepsilon_{ijk}$ are the Levi-Cevita symbol and the commutator $\big[J_i,K_j\big]=J_iK_j-K_jJ_i$.
Substituting (7) into (3), we get
\begin{align}
e^{tX}\,=\,I_4+\sum_1^3\theta_i J_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 J+\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 ?
I hope someone would help me to clarify those impedents. Thanks.
Best Answer
That argument is a little handwavy and I suspect the answer is: $A$ is "small enough". A better proof is to just differentiate $\exp(tX)^T\eta\exp(tX) = \eta$ by product rule which yields: $$ \frac{d}{dt}\exp(tX)^T\eta\exp(tX) = X^T\eta\exp(tX) + \exp(tX)^T\eta X = 0$$ and thus $$ \left.\frac{d}{dt}\right|_{t=0}\exp(tX)^T\eta\exp(tX) = X^T\eta + \eta X = 0$$ I'm not the hugest fan of using the transpose here though if I'm honest. For orthogonal transformations, transpose is the same as inverse but that is not true for the Lorentz group and I think that leads to misunderstandings down the line. The transpose really assumes you've picked a basis or more directly a definite inner product. I'd much rather just say that the elements of the group preserve the indefinite symmetric bilinear form and that differentiating gives you elements of the Lie algebra as skew for the same bilinear form by the same product rule style argument.
Lie algebras have a Lie bracket. That's what makes them Lie algebras. The commutator is the standard example of a Lie bracket. And indeed by Ado's Theorem any Lie bracket (on a finite dimensional Lie algebra) can be described as the commutator by taking some faithful representation of the Lie algebra. So in your case this is just realising $\mathfrak{so}(3,1)$ in its representation on $\mathbb{R}^{3,1}$
This is just not true as Moishe pointed out in the comments. $SO(3,1)$ is not connected so the exponential map cannot be surjective here. It is, however, surjective onto the connected component of the identity in this case. In physics that subgroup is often denoted $SO^+(3,1)$ and called the set of orthochronous Lorentz transformations.
I don't really know what is going into the matrix if I'm perfectly honest. It does seem to have only 3 parameters in it which wouldn't really make sense as the group is 6-dimensional but maybe those are complex numbers or I'm missing something else.