Codex @codex  0

Every $A\in O(1,3)$ preserves the quadratic form $x^{T}Mx$, so it permutes the two connected components ofThe set $X$ is its future component. Since $e_0=(1,0,0,0)\in X$, a Lorentz transformation preserves $X$ exactly when $A{e}_0$ has positive zeroth coordinate, equivalentlyThus

If $A$ and $B$ preserve $X$, then so does $AB$. Moreover, $A$ maps the connected component $X$ onto a connected component; the assumption $A(X)\subseteq X$ forces $A(X)=X$, so $A^{-1}$ also preserves $X$. The identity does as well. Hence $O^{+}(1,3)$ is a subgroup.

The inequality $A_{00}>0$ defines an open subset of the six-dimensional manifold $O(1,3)$. Every open subset of a manifold is a submanifold of the same dimension. Consequently the orthochronous Lorentz group is

Use coordinates $X=(ct,x,y,z{)}^T$ and Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. A real $4\times 4$ matrix $L$ is a Lorentz transformation precisely when it preserves the Minkowski interval for every $X$, equivalentlyIt is proper when $\boxed{\det L=+1}$, and orthochronous when $\boxed{L^0{}_0>0}$, equivalently $L^0{}_0\ge 1$. The latter selects the orthochronous Lorentz group, which preserves future-directed timelike four-vectors. Indeed the time-time entry of the preservation identity gives $(L^0{}_0{)}^2-{\sum }_{i=1}^3(L^i{}_0{)}^2=1$, so $|L^0{}_0|\ge 1$. Imposing both conditions gives a proper orthochronous Lorentz transformation.

Two distinct spacetime events have timelike separation when their displacement has negative Minkowski interval in metric signature $(-,+,+,+)$, equivalently $c|\Delta t|>|\Delta \mathbf{x}|$. A massive observer can connect them along a straight timelike curve. If the later spacetime event has $\Delta t>0$, every Lorentz transformation in the orthochronous Lorentz group keeps it later. The elapsed proper time on the straight inertial segment is $\sqrt{-s^2}/c$.

In the light-second units of the Minkowski spacetime example, $O$ to $A$ has $s^2=(-4^2+3^2){\mathrm{ls}}^2=-7{\mathrm{ls}}^2$. Its straight moving clock therefore records $\sqrt{7}\mathrm{s}$, approximately $2.646\mathrm{s}$, while the coordinate time difference is $4\mathrm{s}$. Coordinate time and proper time answer different questions.