Lie Subgroups, Actions, and Homogeneous Spaces — Examples

By [A1], $i$ is an injective immersed homomorphism with dense image. Since its kernel is trivial, the canonical image structure in [A2] is transported from the one-dimensional source $\mathbb{R}$.

The image is proper. The point $(1,e^{\pi i\alpha })$ is not in it: equality of the first coordinate would force $t=n\in \mathbb{Z}$, while equality of the second would make $\alpha (n-\frac{1}{2})$ an integer, impossible because a nonzero rational multiple of irrational $\alpha$ is irrational. A proper dense subset is not closed.

For each $j\ge 1$, let $q_j$ be the least positive integer satisfying $\parallel q_j\alpha \parallel <1\mathrm{/}j$, whose existence is the finite-pigeonhole calculation in [A1]. Irrationality makes every fixed $\parallel q\alpha \parallel$ positive, so $q_j\to \mathrm{\infty }$. Nevertheless $i(q_j)=(1,e^{2\pi i\alpha q_j})\to (1,1)=i(0)$ in the ambient subspace. Hence the inverse of $i$ on its image is not continuous, so [F1] shows that the subgroup is not embedded. Leastness makes the sequence choice-free; ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited only through the general image supplier [A2]. The source dimension is exactly one, its tangent $(1,\alpha )$ is nonzero, and no endpoint is present.

The locus $\det \mathrm{\ne }0$ is open in the finite-dimensional real vector space underlying $M_n(\mathbb{F})$. Matrix multiplication is polynomial there, and the adjugate formula $A^{-1}=\mathrm{adj}(A)\mathrm{/}\det A$ makes inversion smooth, so this locus is the Lie group ${\mathrm{GL}}_n(\mathbb{F})$ in the sense of [F1]. The determinant is polynomial, hence smooth, and multiplicativity makes $\det :{\mathrm{GL}}_n(\mathbb{F})\to {\mathbb{F}}^{\times }$ a Lie-group homomorphism. Its identity fibre is exactly ${\mathrm{SL}}_n(\mathbb{F})$, so [A1] makes this fibre a closed embedded normal Lie subgroup.

In the Leibniz expansion of $\det (I+tX)$, the identity permutation contributes $1+t{\sum }_i{X}_{ii}+O(t^2)$, while every nonidentity permutation needs at least two off-diagonal factors and contributes $O(t^2)$. Thus $d(\det {)}_I(X)=\mathrm{tr}X$. This differential is onto $\mathbb{F}$: the matrix $\mathrm{diag}(z,0,\dots ,0)$ has trace $z$. Left multiplication transports surjectivity to every point of the identity fibre, so the fibre is regular and [F2] gives the same tangent kernel.

Combining steps 1.1 and 1.2 with [A1] yields $\mathrm{Lie}({\mathrm{SL}}_n(\mathbb{F}))=\ker d(\det {)}_I=\{X:\mathrm{tr}X=0\}$. For $n=1$ the subgroup and Lie algebra are both trivial; singular matrices $X$ are allowed as tangent vectors. The complex case is read as a real Lie group, and the complex-linear trace map is also onto as a real map. No endpoint or metric choice occurs. ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited exactly through [A1].

Multiplicativity in [F1] and polynomiality make determinant a smooth Lie-group homomorphism. By definition, its identity fibre is $\{A:\det A=1\}=S{L}_n(\mathbb{R})$.

For each $r\in {\mathbb{R}}^{\times }$, the diagonal matrix $\mathrm{diag}(r,1,\dots ,1)$ is invertible and has determinant $r$. Thus the image on $G{L}_n(\mathbb{R})$ is all of ${\mathbb{R}}^{\times }$. The same matrix lies in $G{L}_n^{+}(\mathbb{R})$ exactly when $r>0$, so the restricted image is ${\mathbb{R}}_{>0}$.

Apply [A1]. It gives surjective submersions $G{L}_n(\mathbb{R})\to {\mathbb{R}}^{\times },G{L}_n^{+}(\mathbb{R})\to {\mathbb{R}}_{>0}$ with fibres the left cosets of $S{L}_n(\mathbb{R})$, followed in each case by the evident inclusion of the image. Equivalently, the canonical intrinsic quotient by the kernel is isomorphic to the displayed image Lie group. For $n=1$ these maps are the identity on ${\mathbb{R}}^{\times }$ and its positive subgroup. The disconnected two-component image in the first case is intentional. Countable choice is used only through [A1].

The action is smooth and preserves $S^n$. Given $u,v\in S^n$, extend each to an oriented orthonormal basis whose last vector is respectively $u$ and $v$; when necessary, changing the sign of one of the first $n$ vectors corrects the orientation. The linear map carrying the first oriented basis to the second is in $SO(n+1)$ and sends $u$ to $v$. Thus the action is transitive.

A matrix in $SO(n+1)$ fixes $e_{n+1}$ exactly when it preserves $e_{n+1}^{\perp }$ and has block form $\mathrm{diag}(A,1)$. Orthogonality and determinant one then say precisely $A\in SO(n)$. Hence the stabilizer is the displayed copy of $SO(n)$.

Apply [A1] at $e_{n+1}$. The map $gSO(n)\mapsto g{e}_{n+1}$ is the asserted equivariant diffeomorphism. For $n=1$, $SO(1)=\{1\}$ and the quotient is $SO(2)\cong S^1$. The excluded value $n=0$ also has the analogous point quotient if one adopts $SO(0)=SO(1)=\{1\}$, but it is not needed for the stated family. Countable choice is used through [A1].

The actions on lines are smooth: in an affine projective chart where one coordinate is nonzero, the transformed line coordinates are ratios of linear functions with a nonvanishing denominator. Given two real lines, choose unit generators and extend them to oriented orthonormal bases; the resulting element of $SO(n+1)$ carries one line to the other. Given two complex lines, extend unit generators to unitary bases; the resulting element of $U(n+1)$ does the same. Hence both actions are transitive.

The stabilizer in $SO(n+1)$ of the line $\mathbb{R}{e}_0$ preserves its orthogonal complement and is therefore $S(O(1)\times O(n))=\{\mathrm{diag}(\epsilon ,A):\epsilon =\pm 1,A\in O(n),\epsilon \det A=1\}\mathrm{.}$ Conversely every such block matrix fixes the line. The stabilizer in $U(n+1)$ of $\mathbb{C}{e}_0$ is exactly the block subgroup $U(1)\times U(n)$: unitarity forces preservation of the orthogonal complement, and every such block matrix fixes the line.

Apply [A1] to the base lines. It yields the two displayed equivariant diffeomorphisms. The determinant-one condition in the real stabilizer is essential; replacing it by $O(1)\times O(n)$ would not be a subgroup of $SO(n+1)$. For $n=0$, both projective spaces are a point and the analogous quotient is trivial. Countable choice is used through [A1].

Given two $k$-planes, choose orthonormal bases for each and extend them to orthonormal bases of the ambient space. The orthogonal or unitary map between the adapted bases carries one plane to the other, proving transitivity. The stabilizer of the coordinate $k$-plane preserves it and its orthogonal complement, hence is exactly the block diagonal subgroup $O(k)\times O(n-k)$ or $U(k)\times U(n-k)$. Conversely every such block matrix stabilizes the plane.

For a flag with successive quotient dimensions $n_1,\dots ,n_s$, choose an orthonormal basis adapted to all members of the flag. Mapping one adapted basis to another proves transitivity. A unitary or orthogonal transformation fixes the coordinate flag exactly when it preserves each successive orthogonal block, which gives the stated product block subgroup. These actions are smooth in the usual graph-coordinate charts for subspaces.

Apply [A1] to steps 1.1 and 1.2. This gives all displayed equivariant diffeomorphisms. The cases $k=0$ or $k=n$ have stabilizer the whole group and quotient a point. Repeated or zero flag blocks are omitted because they do not change a flag; partial flags correspond to any positive composition of $n$. Countable choice is used through [A1].

Write $q=a+bi+cj+dk$. A coordinate check from [F1] gives $\overline{qr}=\bar{r}\bar{q}$ and hence $N(qr)=N(q)N(r)$. Thus the unit sphere $S^3\subset \mathbb{H}$ is a group, with inverse $q\mapsto \bar{q}$. It is a smooth three-manifold by [F2], because $N^{-1}(1)$ is regular: $d{N}_q(w)=2⟨q,w⟩$ is nonzero on every unit $q$. Multiplication and inversion are polynomial and linear respectively, so this is a Lie group.

The determinant-nonzero matrices form an open subset $G{L}_3(\mathbb{R})$ of the nine-dimensional matrix space. Matrix multiplication is polynomial and inversion is the smooth adjugate-over-determinant formula, so this open manifold is a Lie group. Inside it, the equations $R^{T}R=I$ and $\det R=1$ define a closed subgroup; [F2] therefore gives it the embedded Lie-group structure denoted $SO(3)$. This is the exact use of ${\mathrm{A}\mathrm{C}}_{\omega }$ in the construction.

For a unit $q=a+r$, with $r=bi+cj+dk$, and an imaginary quaternion $v$, direct multiplication gives the vector formula $qv{q}^{-1}=(a^2-\mathrm{\mid }r{\mathrm{\mid }}^2)v+2⟨r,v⟩r+2a(r\times v)$. It also gives zero real part. The identities $⟨r\times v,r⟩=⟨r\times v,v⟩=0$ and $\mathrm{\mid }r\times v{\mathrm{\mid }}^2=\mathrm{\mid }r{\mathrm{\mid }}^2\mathrm{\mid }v{\mathrm{\mid }}^2-⟨r,v{⟩}^2$ show from this formula that $\mathrm{\mid }qv{q}^{-1}\mathrm{\mid }=\mathrm{\mid }v\mathrm{\mid }$. Hence conjugation defines an orthogonal transformation of $\mathrm{Im}\mathbb{H}$.

The polynomial map $a+bi+cj+dk\mapsto \left(\begin{array}{cc}a+bi& c+di\\ -c+di& a-bi\end{array}\right)$ preserves products by the quaternion table. Its image consists exactly of the matrices in $SU(2)$, since their columns have the displayed form and the unitary and determinant-one equations reduce to $a^2+b^2+c^2+d^2=1$. It and its coordinate inverse are smooth, so it identifies the Lie group of step 1.1 with $SU(2)$.

The unit sphere is path connected: if $q\mathrm{\ne }-1$, normalize the nonzero segment $(1-t)q+t$, while $-1$ is joined to $1$ by $t\mapsto \cos (\pi t)+i\sin (\pi t)$. Consequently the determinant of the orthogonal map in step 1.3, a continuous function with values in $\{1,-1\}$, equals its value $1$ at $q=1$. Thus $\rho (q)\in SO(3)$. Associativity gives $\rho (q_1{q}_2)=\rho (q_1)\rho (q_2)$, and the coordinate formula in step 1.3 makes $\rho$ smooth.

Identify $T_{1}SU(2)$ with $\mathrm{Im}\mathbb{H}$. Differentiating conjugation along $q(t)=1+tr+O(t^2)$ gives $d{\rho }_1(r)(v)=rv-vr=2r\times v$. Differentiating $R^{T}R=I$ at $I$ shows that $T_{I}SO(3)$ is contained in the three-dimensional space of skew-symmetric endomorphisms. The displayed cross-product map takes values in that space and is injective: if $r\times v=0$ for every $v$, take a basis vector not parallel to nonzero $r$ to get a contradiction. Its three-dimensional image lies in $T_{I}SO(3)$, so the inclusions force $T_{I}SO(3)$ to be the full skew-symmetric space and $d{\rho }_1$ to be an isomorphism. By [F3], there are neighborhoods $U_0$ of $1$ and $W_0$ of $I$ such that $\rho {\mathrm{\mid }}_{U_0}:U_0\to W_0$ is a diffeomorphism. Choose a smaller open neighborhood $U\subseteq U_0$ with $U\cap (-U)=\mathrm{\varnothing }$ and put $W=\rho (U)$; the restriction $\rho {\mathrm{\mid }}_U:U\to W$ remains a diffeomorphism and $W$ is open.

The map $\rho$ is onto. For $R\in SO(3)$, $\det (R-I)=\det (R^T-I)=\det (R^{-1}-I)=\det (I-R)=-\det (R-I)$, so $\det (R-I)=0$. Choose a unit vector $u$ fixed by $R$. Its perpendicular plane is invariant, and the restriction of $R$ there is an orientation-preserving planar orthogonal map. By [F4], in a positively oriented orthonormal basis it is rotation through some angle $\theta$. Substituting $q=\cos (\theta \mathrm{/}2)+u\sin (\theta \mathrm{/}2)$ into the formula of step 1.3 gives Rodrigues' formula, so $\rho (q)=R$. Only this one finite-dimensional choice of an axis and basis is made.

If $\rho (q)$ is the identity, then $q$ commutes with $i,j,k$. Comparing $qi$ with $iq$ first gives $c=d=0$, and comparing $(a+bi)j$ with $j(a+bi)$ then gives $b=0$. Since $q$ is unit, $a=1$ or $a=-1$. Conversely both real unit quaternions act trivially. Therefore $\ker \rho =\{1,-1\}$, and every fibre is exactly $\{q,-q\}$.

Step 3.2 now gives ${\rho }^{-1}(W)=U\bigsqcup (-U)$, and both restrictions are diffeomorphisms onto $W$. For arbitrary $R_0\in SO(3)$ choose one $q_0$ above it using step 3.1; then $R_{0}W$ is evenly covered by the two translated sheets $q_{0}U$ and $-q_{0}U$. Hence $\rho$ is a covering homomorphism in the sense of Covering homomorphisms of Lie groups, with exactly two sheets. The groups are nonempty and three-dimensional; no zero-dimensional, endpoint, degenerate, or biconditional case is hidden. The proof itself makes only finitely many choices, while ${\mathrm{A}\mathrm{C}}_{\omega }$ is used through the closed-subgroup construction of $SO(3)$ in [F2].

Put $U_0=S^1\setminus \{-1\}$ and $U_1=S^1\setminus \{1\}$. For $-\pi <\theta <\pi$ define $s_0(e^{i\theta })=e^{i\theta \mathrm{/}2}$, and for $0<\theta <2\pi$ define $s_1(e^{i\theta })=e^{i\theta \mathrm{/}2}$. These are smooth and satisfy $s_j(w{)}^2=w$. The fibres of $p(z)=z^2$ are exactly $\{z,-z\}$, so ${\tau }_j:U_j\times H\to p^{-1}(U_j)$, ${\tau }_j(w,\epsilon )=s_j(w)\epsilon$, is an equivariant bijection with smooth inverse $z\mapsto (z^2,s_j(z^2{)}^{-1}z)$. Its second component takes values in the discrete zero-dimensional Lie group $H$, hence is locally constant and smooth. Thus the two ${\tau }_j$ are principal charts and $p$ is a smooth principal $H$-bundle in the sense of [F1].

The associated relation from [F2] is $(z,t)\sim (-z,-t)$, so $E=(S^1\times \mathbb{R})\mathrm{/}\sim$ is a smooth real line bundle over the base circle, with $[z,t]\mapsto z^2$. On the upper component of $U_0\cap U_1$, the two sections in step 1.1 agree. On the lower component, expressing the same angle in the second interval adds $2\pi$, so $s_1=-s_0$ and the associated fibre coordinate changes by $\rho (-1)=-1$. Thus its transition function is $+1$ on one overlap component and $-1$ on the other.

Writing $z=e^{\pi ix}$ identifies $E$ with $([0,1]\times \mathbb{R})\mathrm{/}((0,t)\sim (1,-t)),$ because the only two representatives in this half-circle fundamental domain are $(0,t)$ and $(1,-t)$. This is exactly the standard half-twisted-strip Möbius line bundle, and step 2.1 also records its nontrivial sign transition rather than merely the topology of the total space. The zero section and zero fibre vectors are fixed; the two-chart construction uses no choice principle.

Addition gives $0\cdot x=x$ and $(m+n)\cdot x=m\cdot (n\cdot x)$, so the formula is a left action. It is smooth because its restriction to every open component $\{n\}\times \mathbb{R}$ is the smooth translation $x\mapsto x+n$. If $n\cdot x=x$, then $n=0$, so the action is free.

The action is proper. Let $K\subseteq {\mathbb{R}}^2$ be compact and write $\mathrm{\Theta }(n,x)=(x+n,x)$. The continuous coordinate projections and difference map $(a,b)\mapsto a-b$ send $K$ to compact, hence bounded, subsets $P$ and $D$ of $\mathbb{R}$ by [F2]. Thus the first coordinate of every $(n,x)\in {\mathrm{\Theta }}^{-1}(K)$ lies in the finite set $E=\mathbb{Z}\cap D$, while $x\in P$. The set $E$ is finite and therefore compact; hence $E\times P$ is compact by [F2]. Because compact $K$ is closed and $\mathrm{\Theta }$ is continuous, ${\mathrm{\Theta }}^{-1}(K)$ is closed in $\mathbb{Z}\times \mathbb{R}$, and consequently is a closed subspace of the compact set $E\times P$. It is compact by [F2], proving properness.

The map $p(x)=e^{2\pi ix}$ is constant on translation orbits. Conversely, $p(x)=p(y)$ exactly when $x-y\in \mathbb{Z}$, so it induces a bijection $\bar{p}:\mathbb{R}\mathrm{/}\mathbb{Z}\to S^1$. For every $z_0=e^{2\pi i{x}_0}$, the restriction of $p$ to $(x_0-\frac{1}{2},x_0+\frac{1}{2})$ maps each sufficiently short subinterval diffeomorphically onto an open arc about $z_0$, with a smooth argument branch as inverse. These local inverse branches show that $p$ is a covering map and, using the quotient slice charts supplied by the free proper action, that $\bar{p}$ and ${\bar{p}}^{-1}$ are smooth. Thus $\bar{p}$ is a diffeomorphism.

By [F3], the orbit projection $\mathbb{R}\to \mathbb{R}\mathrm{/}\mathbb{Z}$ is a principal $\mathbb{Z}$-bundle for the right action $x\cdot n=x-n$. Transporting its base along the diffeomorphism $\bar{p}$ from step 3.1 gives precisely $p:\mathbb{R}\to S^1$. This verifies every claim, including both the quotient smooth structure and the principal-bundle assertion, without any choice principle.

Define the $\mathbb{R}$-action on ${\mathbb{T}}^2$ by $t\cdot (z,w)=(e^{2\pi it}z,e^{2\pi i\alpha t}w)\mathrm{.}$ By [F2], it is a smooth free left action.

Every orbit is dense by [F2].

The action is not proper. The map $u\mapsto e^{2\pi iu}$ identifies the wrap-metric circle in [F3] with the complex unit circle $S^1$: their chordal distance is $\mathrm{\mid }{e}^{2\pi iu}-e^{2\pi iv}\mathrm{\mid }=2\sin (\pi d(u,v))$, so the map is a homeomorphism. Thus [F3] makes $S^1$, then ${\mathbb{T}}^2\times {\mathbb{T}}^2$, compact. The full inverse image of this compact target under the action-graph map is $\mathbb{R}\times {\mathbb{T}}^2$. Were it compact, its continuous projection onto $\mathbb{R}$ would make $\mathbb{R}$ compact by [F4], contrary to the open cover $\{(-n,n):n\ge 1\}$, which has no finite subcover.

The quotient is not Hausdorff. Each orbit is a proper dense subset: it is dense by step 2.1. For a point $(z,w)$, choose $\theta \in \mathbb{R}$ with $z=e^{2\pi i\theta }$. Its orbit meets $\{1\}\times S^1$ only at the countable set $\{(1,e^{2\pi i\alpha (n-\theta )}w):n\in \mathbb{Z}\}$, so it cannot contain the whole circle $\{1\}\times S^1$ and is therefore proper. If the quotient were Hausdorff, a singleton orbit class would be closed and its inverse image under the quotient map would be a closed orbit, contradicting density and properness. This free, nonproper action therefore refutes both conclusions, without using any choice principle.

Let $SO(2)$ act on ${\mathbb{R}}^2$ by matrix multiplication. This is a smooth action, and $SO(2)$ is compact, so [F1] makes the action proper.

Every rotation fixes the origin. Thus the stabilizer of $0$ is all of $SO(2)$ rather than the trivial group, and the action is not free by [F2].

If the orbit projection ${\mathbb{R}}^2\to {\mathbb{R}}^2\mathrm{/}SO(2)$ were a principal $SO(2)$-bundle for this action (or for the equivalent right action $x\cdot g=g^{-1}x$), [F3] would make the action free on the fibre over the orbit of $0$. Step 2.1 contradicts this. Hence properness without freeness does not yield a principal bundle. No choice principle is used.

Define $\mathrm{\Theta }([g,X+\mathfrak{h}])=d(L_g^{G\mathrm{/}H}{)}_{eH}(\overline{d{q}_e}(X+\mathfrak{h}))\mathrm{.}$ This vector lies over $gH$. The formula is representative-independent. Indeed, $[gh,v]=[g,{\mathrm{Ad}}_{h}v]$ in the associated bundle, while [F1] gives $d(L_{gh}{)}_{eH}\overline{d{q}_e}(v)=d(L_g{)}_{eH}d(L_h{)}_{eH}\overline{d{q}_e}(v)=d(L_g{)}_{eH}\overline{d{q}_e}({\mathrm{Ad}}_{h}v)\mathrm{.}$

On the fibre over $gH$, $\mathrm{\Theta }$ is the composite of the linear isomorphisms $\overline{d{q}_e}$ and $d(L_g{)}_{eH}$, so it is a fibrewise-linear bijection. In a principal trivialization with smooth section $s:U\to G$, its coordinate expression is $(x,v)⟼d(L_{s(x)}{)}_{eH}\overline{d{q}_e}(v)\mathrm{.}$ This is smooth. Its inverse applies $d(L_{s(x{)}^{-1}}{)}_x$ and then ${\overline{d{q}_e}}^{-1}$, so it is smooth as well.

Hence $\mathrm{\Theta }$ is a smooth vector-bundle isomorphism over $G\mathrm{/}H$. If $H=G$, both sides are the zero bundle over a point; if $H=\{e\}$, this is the standard left trivialization $G\times \mathfrak{g}\cong TG$. Normality of $H$ is not needed; it is precisely the isotropy action, not an action assumed trivial, that makes step 1.1 work. Countable choice is inherited through [A1] and [F1].