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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.

Sha Vuklia
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  • Not sure whether this will be helpful as it's been a while since i've looked at lie algebras, but I don't think $(e^{tX})^T = e^{-tX}$. I think it could be helpful if u write $g$ as $e^M$ for some matrix $M$. ( not sure if thats possible but it could be using pauli matrices ). Then, maybe the commutator relations might help in combining the exponentials into something useful. – Rohan Nuckchady Jun 21 '20 at 19:00
  • @RohanNuckchady Ah thanks, that was a stupid typo. I've corrected it. – Sha Vuklia Jun 21 '20 at 20:15

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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 $$V^T + gVg= 0.$$

Paweł Czyż
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  • Ah, of course! I had completely forgotten to think of the tangent space at the identity as consisting of velocity vectors of smooth curves passing through the identity. Right, so basically you take some smooth curve $t\mapsto X(t)$ such that $X(0)=1$, and by considering the Taylor expansion, we can write the derivative as the limit that you wrote down. Right thanks:) That was a lot easier than I expected. – Sha Vuklia Jun 21 '20 at 20:36