TITLE: Why is positivity of first entry sufficient for a matrix in $\mathrm{SO}(1,3)$ to be in $\mathrm{SO}^{+}(1,3)$?
QUESTION [7 upvotes]: This is a result physics books tell all the time, that the branch of proper Lorentz transformations with positive first entry forms the identity component of Lorentz group.
In mathematical language, for all $\Lambda\in\mathrm{SO}(1,3)$ and $(\Lambda)_{00} \geq 0$ (indices: 0,1,2,3), we have $\Lambda\in\mathrm{SO}^{+}(1,3)$. The converse is also true. But why? $(\Lambda)_{00} \geq 0$ is a very weak condition, and I have tried for hours but cannot even prove such $\Lambda$'s is closed under multiplication. Could anyone give me a hint?
In addition, is this result generalizable to arbitrary $\mathrm{SO}(1,n)$?

REPLY [2 votes]: The action of $SO(1,3)$ leaves invariant the quadratic form $t^2-x^2-y^2-z^2$. It is easily seen that the orbits of $SO(1,3)$ are precisely the sets of vectors $(t,x,y,z)$ for which this form takes a fixed value $q\in\mathbf R$. If $q=0$ this is a cone, otherwise it is a hyperboloid. But there is a qualitative difference between the case $q<0$, where the hyperboloid is connected, and the case $q>0$ where the hyperboloid has $2$ connected components, one on each side of the hyperplane defined by $t=0$. Since the action of $SO(1,3)$ is continuous, it follows that $SO(1,3)$ can be separated into two open subsets according to whether it sends a particular vector $v_0=(1,0,0,0)$ (where the form is positive) to a vector in one or the other of the connected components of its orbit, i.e., a vector with $t>0$ or with $t<0$ (in fact with $t\geq 1$ or with $t\leq-1$). Since the stabilizer subgroup of $v_0$ is isomorphic to $SO(3)$, and therefore connected, it follows that these two open subsets of $SO(1,3)$ are in fact the connected components of $SO(1,3)$.
And as a consequence there is nothing special about the choice of $v_0$: doing the same for any other vector where the quadratic form is positive (and whose orbit therefore has two components) would lead to the same separation of $SO(1,3)$ into two parts.