Global Lorentz stabilizers and their connected-cover preimages

A Lorentz transformation fixing $p_m=m{e}_0$ fixes $e_0$ and preserves its orthogonal complement, so it is $\mathrm{diag}(1,A)$ with $A\in SO(D-1)$. For a future null vector use the frame in the statement. Any $T(A,v)$ defined by $T(A,v)e_{+}=e_{+},T(A,v)x=Ax+(v\cdot Ax)e_{+},T(A,v)e_{-}=e_{-}+v+\frac{1}{2}\mathrm{\mid }v{\mathrm{\mid }}^2{e}_{+}\mathrm{.}$ with $A\in SO(n)$ and $v\in {\mathbb{R}}^n$ preserves all pairings of the frame: $e_{+}$ stays null, $\eta (e_{+},T{e}_{-})=-1$, transverse pairings are $\eta (Ax,Ay)=x\cdot y$, and the only remaining checks are $\eta (Tx,T{e}_{-})=v\cdot Ax-v\cdot Ax=0$ and $\eta (T{e}_{-},T{e}_{-})=\mathrm{\mid }v{\mathrm{\mid }}^2-\mathrm{\mid }v{\mathrm{\mid }}^2=0$. Its matrix is triangular in the ordered frame with diagonal blocks $1,A,1$, so its determinant is $1$. Since it fixes the future null $e_{+}$, $\eta (T{e}_0,e_{+})=\eta (e_0,e_{+})<0$; hence $T{e}_0$ is future timelike and $T(A,v)\in L$.

For $x\in {\mathbb{R}}^{D-1}$ put $r=\sqrt{1+\mathrm{\mid }x{\mathrm{\mid }}^2}$ and $B(x)=\left(\begin{array}{cc}r& x^T\\ x& I+\frac{x{x}^T}{1+r}\end{array}\right)\mathrm{.}$ On the plane spanned by $e_0$ and $x$ (with $x=0$ giving $B(0)=I$), this is the boost with entries $r,\mathrm{\mid }x\mathrm{\mid },\mathrm{\mid }x\mathrm{\mid },r$ and determinant $r^2-\mathrm{\mid }x{\mathrm{\mid }}^2=1$; on its orthogonal complement it is the identity. Hence $B(x)\in L$ and $B(x)e_0=(r,x)$. Every $g\in L$ has $g{e}_0=(r,x)$ for a unique $x$, and $B(x{)}^{-1}g$ fixes $e_0$, so it is $\mathrm{diag}(1,A)$. Conversely every such pair gives an element of $L$. These mutually inverse maps are smooth, giving $L\cong {\mathbb{R}}^{D-1}\times SO(D-1)$ and ${\pi }_1(L)\cong {\pi }_1(SO(D-1))$.

Conversely, if $g\in L$ fixes $e_{+}$, write $g{e}_{-}=\alpha e_{+}+e_{-}+v$; the coefficient of $e_{-}$ is $1$ from $\eta (g{e}_{+},g{e}_{-})=-1$, and $\alpha =\mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}2$ from $\eta (g{e}_{-},g{e}_{-})=0$. For $x\in {\mathbb{R}}^n$, $\eta (gx,e_{+})=0$ gives $gx=Ax+\lambda (x)e_{+}$; preservation of transverse pairings makes $A$ orthogonal, and $\eta (gx,g{e}_{-})=0$ gives $\lambda (x)=v\cdot Ax$. The determinant of $g$ is $\det A=1$, so $A\in SO(n)$. On $x\in {\mathbb{R}}^n$, composition gives $ABx+(v\cdot ABx+w\cdot Bx)e_{+}=ABx+((v+Aw)\cdot ABx)e_{+}$; on $e_{-}$ it gives $e_{-}+(v+Aw)+\frac{1}{2}\mathrm{\mid }v+Aw{\mathrm{\mid }}^2{e}_{+}$. Thus $T(A,v)T(B,w)=T(AB,v+Aw)$, so these matrices form exactly ${\mathbb{R}}^n⋊SO(n)$, including $n=1$ and $n=0$. The zero vector is fixed by all of $L$. Continuity of the orbit map makes each stabilizer closed, and $K_p=c^{-1}(L_p)$ because $c$ is onto.

$SO(1)$ is trivial, $SO(2)\cong S^1$ has ${\pi }_1=\mathbb{Z}$ by [F1], and $SU(2)\cong S^3$ is simply connected and covers $SO(3)$ twice by [F1], so its deck group gives ${\pi }_1(SO(3))\cong \mathbb{Z}\mathrm{/}2$ by [L2]. For $r\ge 4$, the map $SO(r)\to S^{r-1}$ sending a matrix to its last column has fiber $SO(r-1)$. Near $e_r$, with $u=x-(x\cdot e_r)e_r$ and $J=u{e}_r^T-e_r{u}^T$, the smooth matrix $I+J+J^2\mathrm{/}(1+x\cdot e_r)$ is in $SO(r)$ and sends $e_r$ to $x$; translating this local section gives local sections near every point. These local sections give bundle charts. Compactness supplies a finite trivializing subcover, and [F1] supplies a subordinate smooth partition, so this bundle is numerable and hence a Serre fibration. Its homotopy sequence and ${\pi }_1(S^{r-1})={\pi }_2(S^{r-1})=0$ from [F1] show ${\pi }_1(SO(r-1))\to {\pi }_1(SO(r))$ is an isomorphism. Thus ${\pi }_1(SO(r))=\mathbb{Z}\mathrm{/}2$ for every $r\ge 3$.

Let $q_i:E_i\to B$ ($i=1,2$) be covering homomorphisms of Lie groups with $E_1$ connected, and let $f:E_1\to E_2$ be a based covering isomorphism over $B$. The maps $(x,y)\mapsto f(xy)$ and $(x,y)\mapsto f(x)f(y)$ lift the same map $(x,y)\mapsto q_1(x)q_1(y)$ through $q_2$, and agree at $(e,e)$. Since $E_1\times E_1$ is connected by [F3], uniqueness of lifts [F2] makes them equal; the covering-chart description of $f$ also makes it smooth, so $f$ is a Lie-group isomorphism. Now take an arbitrary connected covering homomorphism $c:K\to L$ and apply the connected-group quotient theorem [L3] to this $K$. Its simply connected covering group $q_K:\tilde{K}\to K$ has discrete central kernel ${\mathrm{\Gamma }}_K$ and identifies $K$ with $\tilde{K}\mathrm{/}{\mathrm{\Gamma }}_K$ over $K$. The composite $f=c\circ q_K:\tilde{K}\to L$ is a surjective Lie-group homomorphism with invertible differential, hence a local diffeomorphism. Any surjective homomorphism $q$ that is a local diffeomorphism is a covering: choose a symmetric identity neighborhood $U$ on which $q$ is a diffeomorphism onto an open set $V$ and with $U^{-1}U\cap \ker q=\{e\}$; then $q^{-1}(V)={\bigcup }_{\gamma \in \ker q}\gamma U$ is a disjoint union of sheets, each mapped homeomorphically onto $V$. Thus $f$ is a universal cover of $L$. By uniqueness of based universal covers, $f$ is isomorphic over $L$ to $\tilde{L}\to L$, and the lift argument just given makes that isomorphism a Lie-group isomorphism. Consequently $c:K\to L$ is isomorphic over $L$ to $\tilde{L}\mathrm{/}\mathrm{\Gamma }\to L$, where $\mathrm{\Gamma }$ is the corresponding subgroup of the universal deck group $\mathrm{\Delta }=\ker (\tilde{L}\to L)\cong {\pi }_1(L)$. The group $\mathrm{\Delta }$ is discrete central by [L3] applied to $L$. Conversely, every subgroup $\mathrm{\Gamma }\le \mathrm{\Delta }$ is discrete central, and $\tilde{L}\mathrm{/}\mathrm{\Gamma }\to L$ is a surjective local-diffeomorphism homomorphism, hence a connected covering by the same neighborhood argument. Under $\mathrm{\Delta }\cong {\pi }_1(L)$, its induced fundamental-group subgroup is exactly $\mathrm{\Gamma }$: a loop lifts to a path ending at its deck element $\delta$, and its image in $\tilde{L}\mathrm{/}\mathrm{\Gamma }$ closes exactly when $\delta \in \mathrm{\Gamma }$. The corollary [L2] now distinguishes these already-constructed covers by their induced subgroups. By step 1.2, ${\pi }_1(L)$ is trivial for $D=2$, is $\mathbb{Z}$ generated by the positive $2\pi$ rotation for $D=3$, and is $\mathbb{Z}\mathrm{/}2$ for $D\ge 4$. Its subgroups give only $L$ in $D=2$; $K_N=\tilde{L}\mathrm{/}⟨z^N⟩$ ($N\ge 1$) and $K_{\mathrm{\infty }}=\tilde{L}$ in $D=3$; and $L,\tilde{L}$ in $D\ge 4$.

In $D=3$ the massive stabilizer is the rotation circle, parametrized by angle $t$ modulo $2\pi$; its lifted one-parameter group has period $2\pi N$ in $K_N$ and no period in $K_{\mathrm{\infty }}$, giving $\mathbb{R}\mathrm{/}(2\pi N\mathbb{Z})$ or $\mathbb{R}$. The null stabilizer in $L$ is the unipotent $\mathbb{R}$ subgroup from step 2.1; its lift is simply connected, and every other component of its full preimage is a unique deck translate. The deck element $z$ is central by [L1], so the preimage is $\mathbb{R}\times \mathbb{Z}$ and quotienting by $⟨z^N⟩$ gives $\mathbb{R}\times (\mathbb{Z}\mathrm{/}N\mathbb{Z})$.

For $D\ge 4$, the massive stabilizer $M=SO(D-1)$ includes in $L\cong {\mathbb{R}}^{D-1}\times SO(D-1)$ as the second factor, so ${\pi }_1(M)\to {\pi }_1(L)$ is an isomorphism. Its full preimage in $\tilde{L}$ is connected: loop monodromy from $M$ reaches every deck element because this inclusion is onto, and lifting paths in $M$ then connects every point of the preimage to the identity. As a cover of $M$, its induced subgroup is the kernel of this isomorphism, namely $0$; it is the universal cover $\mathrm{S}\mathrm{p}\mathrm{i}\mathrm{n}(D-1)\to SO(D-1)$ by [L2]. The based cover isomorphism is a Lie-group isomorphism by step 3.1. Put $n=D-2$ and $H={\mathbb{R}}^n⋊SO(n)$, which is connected. Its fundamental group is that of $SO(n)$ because ${\mathbb{R}}^n$ is contractible. Choose the standard null frame with $e_{+},e_{-}$ in the $e_0,e_{D-1}$-plane; then the rotation factor $SO(n)$ is the standard subgroup of the spatial $SO(n+1)$. For $n\ge 3$ the same sphere-fibration argument as in step 2.2 makes ${\pi }_1(H)\to {\pi }_1(L)$ an isomorphism. For $n=2$, the generator loop in $SO(2)$ lifts under $SU(2)\to SO(3)$ to $\cos (t\mathrm{/}2)+k\sin (t\mathrm{/}2)$ and ends at $-I$ when $t=2\pi$, so its image is the nontrivial element of ${\pi }_1(L)=\mathbb{Z}\mathrm{/}2$. To check connectedness of the full null preimage, join the projection of any point in $c^{-1}(H)$ to the identity by a path in $H$ and lift that path; the endpoint then differs from a point over the identity by a deck transformation. Two points over the identity are joined within the preimage exactly when their deck difference is the monodromy of a loop in $H$. Thus components correspond to cosets of the image of ${\pi }_1(H)$ in the deck group, and the surjectivity just proved makes $c^{-1}(H)$ connected. As a cover of $H$, its induced subgroup is $\ker ({\pi }_1(H)\to {\pi }_1(L))$, since a loop in $H$ lifts closed exactly when its image loop in $L$ does. The next paragraph identifies its group. In $D=2$ both nonzero stabilizers are trivial.

For $n\ge 3$, the map ${\pi }_1(H)\to {\pi }_1(L)$ in step 4.2 is an isomorphism, so the connected full preimage has induced subgroup $0$ in ${\pi }_1(H)$. The connected covering ${\mathbb{R}}^n⋊\mathrm{S}\mathrm{p}\mathrm{i}\mathrm{n}(n)\to H$ also has induced subgroup $0$, since $\mathrm{S}\mathrm{p}\mathrm{i}\mathrm{n}(n)\to SO(n)$ is universal. For $n=2$, ${\pi }_1(H)=\mathbb{Z}$ and the inclusion maps a generator onto the nontrivial element of $\mathbb{Z}\mathrm{/}2$, so the full preimage has induced subgroup $2\mathbb{Z}$; the connected double cover ${\mathbb{R}}^2⋊(\mathbb{R}\mathrm{/}(4\pi \mathbb{Z}))\to H$ has that same subgroup. In each case [L2] gives a unique based covering isomorphism over $H$ between the displayed semidirect product and the full preimage. Both maps to $H$ are homomorphisms of Lie groups, so step 3.1 proves that this based isomorphism preserves multiplication (and is smooth). Hence the full null preimage is exactly ${\mathbb{R}}^n⋊\mathrm{S}\mathrm{p}\mathrm{i}\mathrm{n}(n)$ with the stated global group structure, including its sheets and not merely its Lie algebra.

The computed fundamental groups are countable, so every connected cover here has at most countably many sheets. A countable basis of evenly covered neighborhoods in $L$, lifted sheet by sheet, gives a countable basis upstairs; each cover is a locally compact Hausdorff Lie group by [L1]. Therefore each $K$ is second-countable locally compact Hausdorff, every inverse-image stabilizer is closed by step 2.1, and all stabilizers and preimages in the statement have been computed.