Elements Outside the Identity Component $SO^+(1,\,3)$ of the Lorentz Group $O(1,\,3)$

Question

I have been answering a question on Physics Stack exchange to do with the difference between the "proper orthochronous" (i.e. identity component of) the Lorentz group $SO^+(1,\,3)$ and the Lorentz group $O(1,\,3)$ itself, i.e.

$$SO^+(1,\,3) \cong O(1,\,3) / V_4$$

where $V_4$ is the Klein "fourgroup", wontedly written in physics as $V_4=\{\mathrm{id},\,P,\,T,\,P\,T\}$ with:

$$P=\text{"parity flipper"} = \mathrm{diag}[1,\,-1,\,-1,\,-1];\\T=\text{''time flipper''} = \mathrm{diag}[-1,\,1,\,1,\,1]$$

Naturally, as a topological group, the identity component $SO^+(1,\,3)$ is a normal subgroup of $O(1,\,3)$, so we can think of $V_4 \cong O(1,3)/SO^+(1,3)$.

To "sniff out" an element $L\in O(1,3);\,L\not\in SO^+(1,3)$ that doesn't belong to the identity component I can work as follows:

1. Compute $\det L$; if $\det L = -1$, then I know that $L$ belongs either to the $P$ or $T$ coset of $SO^+(1,3)$.

2. So now we have to tell whether $L\in SO^+(1,3)$ or whether it belongs to the $P\,T$ coset $P\,T\,SO^+(1,3)$. All of these matrices have determinant +1.

This shouldn't be hard, but I'm not coming up with the goods today: can anyone tell me how to tell the difference between a member of $SO^+(1,3)$ and the $P\,T$ coset $P\,T\,SO^+(1,3)$?

Afterword: I believe physicists are taught that a matrix $L\in O(1,3)$ is in the $T$ or $P\,T$ coset if the first leading diagonal element $L_{1,1}$ (written as $L_{0,\,0}$ in physics) is negative. I'm skeptical: is it this simple? If so, then this answers my question, so if this is true is there a simple way to prove it?

Answer 1

Your criterion for $SO^+(1,3)$ is correct. $O(1,3)$ is the group of invertible matrices that preserve the quadratic form $q=t^2-x^2-y^2-z^2$. It thus leaves invariant the set of vectors $v$ with $q(v)>0$, which has two connected components, distinguished by the sign of the $t$ coordinate. $SO^+(1,3)$ is the subgroup that leaves each of the two components invariant. So to test if an element $L\in SO(1,3)$ is in $SO^+(1,3)$, it is enough to take one element $v_0$ from one of the components, apply $L$ to it, and check if $Lv_0$ is in the same component of $v_0$. If you take the element $v_0=(1,0,0,0)^t$ (column vector) and apply $L$ to it, the result is the 1st column of the matrix representing $L$. It is in the same component of $v_0$ if its 1st coordinate (the $t$ coordinate) is positive, which is the test you indicate.