 Solution

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