Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 26I / c

The identity matrix belongs to $O(1,3)$. If $A,B\in O(1,3)$, thenso the set is closed under multiplication. Taking determinants in $A^{T}MA=M$ gives $(\det A{)}^2=1$, so $A$ is invertible. Rearranging the defining identity givesIts inverse also preserves $M$, either directly or becauseThus the Lorentz group $O(1,3)$ is a group under matrix multiplication.

Consider the smooth mapIts derivative isAt $A\in O(1,3)$ write $H=AY$. ThenThis derivative is onto the ten-dimensional space of symmetric matrices: for a symmetric $S$, taking $Y=\frac{1}{2}MS$ gives $Y^{T}M+MY=S$. Hence $M$ is a regular value. Since $M_4(\mathbb{R})$ has dimension 16, the regular level set theorem gives

The tangent vectors are exactly those for which the derivative vanishes:Writing $Y$ in one-plus-three block form shows explicitly thatThe three entries of $v$ and three independent entries of the skew-symmetric matrix $\Omega$ again display dimension six. This is the Lie algebra of the Lorentz group.

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