Let
$$
O(1,3)=\{A\in GL_4(\mathbb R):A^TgA=g\}
$$
where $g$ is the diagonal matrix with $1$ on the first diagonal entry, and $-1$ on the other diagonal entries. I want to show that the Lie algebra consists of matrices $X$ such that $gXg=-X^T$. As I've already shown the spin homomorphisms $SL_2(\mathbb C)\to SO(1,3)_e$, I know that the Lie algebra of $O(1,3)$ is isomorphic to that of $SL_2(\mathbb C)$, which are traceless $2$-by-$2$ complex matrices. However, I don't think this is going to help particularly (except for a dimensional check). I was thinking of using properties of the exponential map, as we are looking for matrices $X$ such that
$$
(e^{tX})^T g e^{tX}=g.
$$
Now, I believe $(e^{tX})^T=e^{tX^T}$, which means that we have
$$
ge^{tX}g=e^{-tX^T},
$$
which almost seems to be $gXg=-X^T$. I'm not sure how to proceed, though.
Lie algebra of Lorentz Group $O(1,3)$
lie-algebras
Best Answer
Let's calculate the tangent space at the identity matrix $I$. Matrix $V\in \mathrm{End}(\mathbb R^4)$ is in the tangent space if and only if $$ \lim_{t\to 0} \frac{(I + tV)^T g (I + tV) - g}{t} = 0. $$ This is exactly $V^Tg + gV = 0$. Multiplying by $g$ from the right and using $g^2=I$, you get $$V^T + gVg= 0.$$