Almost-connected groups and the solvable radical

Original course text, October 2026. Original expression is public domain (CC0).

Introduction

An amenable group can contain a noncompact solvable part. The positive affine group is an example. For an almost-connected locally compact group, the obstruction lies in the quotient by that part: amenability is equivalent to compactness of the quotient by the solvable radical. We prove this at the full locally compact scope. A closed discrete free subgroup supplies the obstruction; explicit projective matrices make that subgroup visible.

Read Haar averages and compact translation control and Closed subgroups and continuous averaging first. Their full locally compact mean, fixed-point, normal-extension and closed-subgroup proofs are used here. The reduced-word proof for the discrete free group is in Means, Følner sets, and regular representations, Example 4.4.

Throughout, groups are locally compact Hausdorff unless explicitly stated otherwise. No countable base, separability, metrizability or unimodularity is assumed. Section 1A proves the compact-normal quotient input S3 in full, at its stated compact-representation and finite-dimensional Lie prerequisites. Section 1B proves the compact case and the reduction of arbitrary locally compact Lie approximation to the metrizable case, including the quotient topology and metrization proofs. Section 1C constructs the Lie radical and proves the full locally compact radical assertion S1 assuming S2. Sections 1D–1E construct compact conjugation and prove the closed rank-one subgroup input S4 at the explicit root, Lie and geometry prerequisites. Section 1B also constructs the metrizable Lie quotient and lifts it to arbitrary locally compact groups; its Haar and Peter–Weyl prerequisites are explicit. The other arguments, examples and exercise solutions are given below.

1. Means, components, and the structural inputs

A left invariant mean is a positive unital functional $m$ on Haar $L^\infty(G)$ such that $$m(L_gf)=m(f),\qquad L_gf(x)=f(g^{-1}x). \tag{1.1}$$ The group is amenable when such a mean exists. Use the locally completed Haar convention of the Haar lesson for groups without a sigma-compact hypothesis. Theorem 1.2 there proves equivalence with the uniformly continuous function formulation; Theorem 2.1 proves the compact convex fixed-point criterion. Thus choosing one of these equivalent criteria makes the same definition. This is the complete terminology of Takesaki's Definition XIII.4.2.

Write $G^\circ$ for the identity component. It is closed and characteristic: translations identify all components, and every continuous automorphism fixes the component containing the identity. A group is almost connected when $G/G^\circ$ is compact. A subgroup is solvable when its algebraic derived series reaches the identity after finitely many steps.

Here are the four structure inputs used in the proof. They concern general locally compact groups or finite-dimensional Lie groups as specified.

In these assertions, semisimple means that the connected solvable radical is trivial. Section 1A proves S3 by establishing Iwasawa's compact-normal centralizer factorization and the compact-kernel component theorem. Sections 1D–1E construct compact conjugation, a real restricted-weight split triple, a closed adjoint matrix image and its closed inverse-image component in the original group. Section 1C proves S1 using S2 and the stated finite-dimensional prerequisites. Those finite-dimensional results are prerequisites for the local S3 proof.

We also use the elementary Lie correspondence between closed subgroups and Lie subalgebras, the exponential chart, and the fact that a connected group locally isomorphic to $\mathrm{SL}_2(\mathbb R)$ has adjoint quotient $\mathrm{PSL}_2(\mathbb R)$. The latter follows from integration of the adjoint Lie algebra: the adjoint image is the connected inner automorphism group of $\mathfrak{sl}_2(\mathbb R)$, and its kernel is the center. No finiteness assumption on that center is required.

Lemma 1.1 (radical bookkeeping). The radical of $G^\circ$ equals the radical of $G$. The quotient $G/R$ is semisimple. If $G$ is almost connected, so is $G/R$, and $$(G/R)^\circ=G^\circ/R,\qquad (G/R)/(G/R)^\circ\cong G/G^\circ. \tag{1.2}$$

Proof. Every connected normal solvable subgroup of $G$ lies in $G^\circ$, and hence in $\operatorname{rad}(G^\circ)$. Conversely the latter radical is characteristic in $G^\circ$, so is normal in $G$. This proves equality.

To check semisimplicity of $G/R$, let $P$ be a connected solvable normal subgroup of that quotient. Replacing it by its closure preserves these three properties: continuity of commutators preserves a finite derived-length bound, and closure preserves connectedness and normality. Its inverse image $E$ has connected kernel $R$ and connected quotient $P$, so is connected. Indeed, a separation of $E$ would separate its connected coset fibers; both pieces would be saturated and their images would separate $P$, since the quotient map is open. The extension of two solvable groups is solvable: after as many derived steps as needed for $P$, the derived subgroup lies in $R$, and then terminates. Thus $E$ is connected, solvable and normal in $G$, so $E\subset R$. Therefore $P$ is trivial.

The component quotient $G/G^\circ$ is totally disconnected, as proved in Section 1A.4 using Lemma 1A.5. The closed connected subgroup $G^\circ/R$ of $G/R$ is consequently its identity component: any larger connected component would have a nontrivial connected image in $G/G^\circ$. This gives (1.2) and the almost-connected conclusion. $\square$

The solvability convention is compatible with closed derived series. If $D$ is dense in a group, continuity of commutators puts each successive commutator of $\overline D$ in the closure of the corresponding derived subgroup of $D$, by induction. A finite termination bound therefore passes to the closure. This is the fact used above and in the Haar lesson's proof that solvable locally compact groups are amenable.

1A. Compact normal quotients and connected solvable lifts

A compact normal subgroup need not be solvable or connected. Lifting a solvable subgroup of the quotient through the whole kernel therefore does not immediately produce a solvable subgroup of the original group. The centralizer supplies the right lift: after replacing the kernel by its compact abelian centre, the finite derived series acquires at most one extra step. The component theorem then produces a connected normal lift.

For a group $A$, define its algebraic derived series by $D^0A=A$ and $D^{j+1}A=[D^jA,D^jA]$, where the brackets mean the subgroup generated by all commutators. “Solvable” means $D^dA=\{e\}$ for some finite integer $d$. A connected solvable normal subgroup need not initially be closed.

We first prove these two mechanisms. The conclusion is Theorem 1A.10, which supplies the compact-normal quotient input used in Proposition 4.1. Section 1C proves existence of the largest solvable radical assuming S2. General Lie approximation remains a separate structure input; Sections 1D–1E prove the closed rank-one subgroup theorem at their exact root, Lie and geometry prerequisites.

The representation prerequisites are finite-dimensional complete reducibility, finite-dimensionality of compact irreducibles, Schur orthogonality, and Peter–Weyl point separation. They apply to arbitrary compact Hausdorff groups. In the finite matrix models we use the closed Lie subgroup theorem, matrix exponentials, and smooth quotients and local sections. These inputs are applied inside finite unitary groups, where their finite-dimensional hypotheses hold.

The classical compact-normal centralizer theorem is due to Kenkichi Iwasawa, as credited in Rickert, Theorem 1.4, printed page 434. Rickert's Lemma 3.4 gives the compact-normal semisimple quotient conclusion. The proof expression below is original and imports no prose from those papers.

1A.1. Connected actions and irreducible classes

Lemma 1A.1. Suppose a connected topological group $A$ acts continuously by automorphisms on a compact Hausdorff group $K$. For every continuous irreducible unitary representation $\pi$ of $K$ and every $a\in A$, the representations $\pi\circ\alpha_a$ and $\pi$ are unitarily equivalent, where $\alpha_a$ is the action automorphism.

Proof. Write $\chi_\pi(k)=\operatorname{tr}\pi(k)$. Joint continuity of $(a,k)\mapsto\chi_\pi(\alpha_a(k))$, and compactness of $K$, give continuity of $$\begin{gathered}a\longmapsto \chi_\pi\circ\alpha_a\quad\text{in }C(K)\\\text{with the uniform norm}.\end{gathered} \tag{1A.1}$$ For clarity, at $a_0$, apply continuity at each $(a_0,k)$, then take a finite cover of $K$ and intersect the corresponding neighbourhoods of $a_0$. The triangle inequality gives the desired common bound on the entire fibre.

An automorphism preserves irreducibility. Schur orthogonality gives $$\|\chi_\sigma\|_2=1,\qquad \|\chi_\sigma-\chi_\pi\|_2=\sqrt2 \quad\text{if }\sigma\not\simeq\pi. \tag{1A.2}$$ Since normalized Haar measure has mass one, the $L^2$ norm is bounded by the uniform norm. Thus a uniform neighbourhood of radius less than $\sqrt2$ contains only characters belonging to the same irreducible class. The representation-class orbit in (1A.1) is locally constant. Connectedness makes it constant, and at the identity its class is $[\pi]$. Equivalent unitary representations have a unitary intertwiner: an invertible intertwiner can be replaced by its unitary polar factor. $\square$

In particular $\ker\pi$ is invariant under every $\alpha_a$. For every finite set $F$ of irreducible classes, the closed normal subgroup $$N_F=\bigcap_{\pi\in F}\ker\pi \tag{1A.3}$$ is $A$-invariant. The finite direct sum $\rho_F=\bigoplus_{\pi\in F}\pi$ is also equivalent to its transform by every $\alpha_a$. Its compact image $$L_F=\rho_F(K)\subset U\!\left(\bigoplus_{\pi\in F}V_\pi\right) \tag{1A.4}$$ is a closed matrix subgroup and therefore a compact Lie group, possibly disconnected. The map identifies $K/N_F$ topologically with $L_F$, since a continuous bijection from compact to Hausdorff is a homeomorphism. The action descends to a continuous action on $L_F$: the map $A\times K\to A\times K/N_F$ is an open quotient map, so joint continuity descends. Peter–Weyl point separation gives $$\bigcap_F N_F=\{e\}. \tag{1A.5}$$ Every index here is an arbitrary finite subset of the complete irreducible dual. No sequence exhausts the representations.

1A.2. Normalizers of disconnected compact matrix groups

The next argument is the crucial finite-dimensional innerness proof. It handles disconnected groups without reducing their automorphisms merely to automorphisms of the Lie algebra.

Lemma 1A.2. Let $L$ be a closed subgroup of $U(n)$, and put $$\begin{gathered}N=N_{U(n)}(L),\qquad C=C_{U(n)}(L),\\\mathfrak l=\operatorname{Lie}(L).\end{gathered}$$ Then $N,C$ are compact Lie groups, $C$ is normal in $N$, and $$\mathfrak n=\mathfrak l+\mathfrak c,\qquad N^\circ=L^\circ C^\circ. \tag{1A.6}$$

Proof. The centralizer is an intersection of closed commuting equations. The normalizer is closed: if a net $u_i\in N$ converges to $u$, then $u\ell u^{-1}\in L$ for every $\ell\in L$, and applying the same argument to $u_i^{-1}$ gives equality rather than just containment. Both are closed in the compact Lie group $U(n)$, hence compact embedded Lie subgroups by the exact closed-subgroup theorem. Conjugating a centralizer element by a normalizer element again centralizes $L$, so $C\lhd N$.

On $\mathfrak u(n)$ use the real positive definite invariant form $$\langle X,Y\rangle=-\operatorname{Re}\operatorname{tr}(XY). \tag{1A.7}$$ For skew-Hermitian $X$, $-\operatorname{Re}\operatorname{tr}(X^2)=\operatorname{tr}(X^*X)>0$ unless $X=0$. Cyclicity of trace makes conjugation by $U(n)$ orthogonal and gives the usual infinitesimal invariance identity. In particular $\mathfrak l^\perp$ is stable under $\operatorname{ad}\mathfrak l$, and under $\operatorname{Ad}(L)$.

Take $X\in\mathfrak n$, and decompose it orthogonally as $X=Y+Z$, with $Y\in\mathfrak l$ and $Z\perp\mathfrak l$. Since $\exp(tX)$ normalizes $L$, differentiation of its adjoint action gives $[X,\mathfrak l]\subset\mathfrak l$. Therefore $[Z,\mathfrak l]\subset\mathfrak l$. But $\mathfrak l^\perp$ is $\operatorname{ad}\mathfrak l$-stable, so this bracket also lies in $\mathfrak l^\perp$, and consequently $$[Z,\mathfrak l]=0. \tag{1A.8}$$ In particular $[Y,Z]=0$. Thus $$\exp(tZ)=\exp(tX)\exp(-tY)\in N,$$ so $Z\in\mathfrak n$ as well.

This is where the disconnected components must be checked. For any $\ell\in L$, the curve $$\exp(tZ)\ell\exp(-tZ)\ell^{-1}$$ lies in $L$, begins at the identity, and has derivative $$Z-\operatorname{Ad}_\ell Z\in\mathfrak l. \tag{1A.9}$$ Both terms on the left belong to $\mathfrak l^\perp$, since $L$ preserves that orthogonal complement. Hence (1A.9) is zero. It holds for every component and every $\ell\in L$, so $Z$ centralizes all of $L$, not just $L^\circ$. Therefore $Z\in\mathfrak c$. We have proved $\mathfrak n\subset\mathfrak l+\mathfrak c$; the opposite inclusion follows because $L,C\subset N$.

The groups $L^\circ$ and $C^\circ$ commute elementwise. Their product is a connected subgroup; it is compact, hence closed, and its Lie algebra is $\mathfrak l+\mathfrak c=\mathfrak n$. One can see the last assertion directly from the differential of $(\ell,c)\mapsto\ell c$. The submersion theorem makes its image contain an identity neighbourhood of $N^\circ$. It is therefore an open subgroup of $N^\circ$; an open subgroup is also closed, and connectedness of $N^\circ$ forces equality. $\square$

Lemma 1A.3 (connected automorphisms of a compact Lie group are inner). Let a connected topological group $A$ act continuously by automorphisms on a compact Lie group $L$, including a disconnected one. Each action automorphism is conjugation by an element of $L^\circ$.

Proof. Choose a faithful continuous unitary representation $\rho:L\hookrightarrow U(n)$. Such a representation exists by the no-small-subgroups lemma and the compact Lie embedding theorem; in the matrix-model application (1A.4), the inclusion itself is faithful. Finite-dimensional complete reducibility and Lemma 1A.1 imply $\rho\circ\alpha_a\simeq\rho$. A unitary intertwiner $u_a$ then satisfies $$\rho(\alpha_a(\ell))=u_a\rho(\ell)u_a^{-1}\qquad(\ell\in L). \tag{1A.10}$$ Thus $u_a\in N=N_{U(n)}(\rho(L))$. Two implementers differ by $C=C_{U(n)}(\rho(L))$. There is a uniquely determined homomorphism $\psi:A\to N/C$, taking $a$ to its implementer coset.

There is no assumption that an implementer can be chosen continuously. Instead the map $$\begin{aligned}N/C&\longrightarrow C(L,U(n)),\\uC&\longmapsto\bigl[\ell\mapsto u\rho(\ell)u^{-1}\bigr].\end{aligned} \tag{1A.11}$$ is a continuous injection into the space of continuous maps with its uniform topology. Its source is compact and its target Hausdorff, so it is a homeomorphism onto its image. Joint continuity of the action and compactness of $L$ make the right side of (1A.10), as a function of $a$, uniformly continuous at each parameter in the sense proved in Lemma 1A.1. Composing with the inverse of (1A.11) proves continuity of $\psi$.

For the compact Lie quotient $N/C$, the identity component is the image of $N^\circ$. Here this elementary Lie statement needs no general locally compact component theorem: the smooth quotient/local-section theorem makes $N\to N/C$ a submersion. Since $N^\circ$ is open in $N$, its image is open; it is connected, and its cosets are open and closed. Therefore its image is exactly $(N/C)^\circ$. Connectedness of $A$ puts $\psi(A)$ in this image. By Lemma 1A.2 an implementing element in $N^\circ$ factors as $\ell c$, with $\ell\in L^\circ$ and $c$ centralizing $L$. Equation (1A.10) is consequently conjugation by $\ell$. $\square$

Equivalently, in the compact-open topology, $$\operatorname{Aut}(L)^\circ=\{\operatorname{Ad}_\ell:\ell\in L^\circ\}. \tag{1A.12}$$ To justify this phrasing directly, apply the proof to the connected identity component of the automorphism group. Evaluation is continuous for the compact-open topology on a compact locally compact domain. Conversely the conjugation image of connected $L^\circ$ is a connected set of automorphisms containing the identity. Formula (1A.12) is not being imported as an unproved Lie-algebra assertion.

1A.3. Inner action and the compact central extension

Theorem 1A.4 (compact normal centralizer factorization). Let $G$ be a connected topological group and $K$ a compact Hausdorff normal subgroup, with continuous conjugation action. Then $$G=K\,C_G(K). \tag{1A.13}$$

Proof. Apply Section 1A.1 to the conjugation action. Every $N_F$ is $G$-invariant, and $G$ acts continuously on the compact Lie matrix group $K/N_F$. Lemma 1A.3 makes this action inner. Fix $g\in G$. For each finite $F$, put $$\begin{aligned} \mathcal I_F(g)=\{k\in K:\;&gxg^{-1}N_F\\ &=kxk^{-1}N_F\\&\text{for every }x\in K\}. \end{aligned} \tag{1A.14}$$ Innerness on $K/N_F$ and surjectivity of $K\to K/N_F$ make this set nonempty. It is closed in $K$: each equality in (1A.14) is a closed equalizer in the Hausdorff quotient, and one intersects over all $x\in K$.

For finitely many indices $F_1,\ldots,F_m$, set $F=\bigcup_jF_j$. Since $N_F\subset N_{F_j}$, every implementer in $\mathcal I_F(g)$ belongs to each $\mathcal I_{F_j}(g)$. These compact closed sets have the finite intersection property. Compactness of $K$ supplies $k$ lying in every one. For every $x\in K$, the two conjugation values now agree modulo every $N_F$; (1A.5) makes them equal in $K$. Thus $k^{-1}g\in C_G(K)$, proving (1A.13). $\square$

The proof needs no local compactness of $G$ at this stage. In particular it does not derive the centralizer factorization from Gleason–Yamabe. It also does not demand a compatible choice of implementers: the closed implementer sets, rather than the individual selected elements, supply compatibility.

If $G$ is locally compact Hausdorff, write $C=C_G(K)$. This is closed, since it is the intersection of closed commuting equations, and normal in $G$, since $K\lhd G$. Furthermore $$C\cap K=Z(K),\qquad C/Z(K)\simeq G/K \tag{1A.15}$$ as topological groups. The map in (1A.15) is the restricted quotient map. Here is its topology, not just its algebra. If $B$ is closed in $C$, it is closed in $G$. The product $BK$ is closed because $K$ is compact. Indeed, from a net $b_i k_i\to x$, a subnet of $k_i$ converges to $k\in K$, and then $b_i\to xk^{-1}\in B$. Equivalently the same argument proves that every point in the closure belongs to the product. Therefore $q(B)$ is closed in $G/K$, as its inverse image is $BK$. The induced continuous bijection in (1A.15) is closed and hence a homeomorphism. In particular $C\to G/K$ is an open quotient map with compact central, therefore abelian, kernel $Z(K)$.

1A.4. Components through compact kernels

To lift a connected quotient subgroup, we must know exactly what happens to identity components. We prove the necessary component theorem through compact kernels, including the nonabelian compact-open-subgroup argument. The compact-space lemma below works for every compact Hausdorff space.

Lemma 1A.5 (open maps with connected fibres). If $f:X\to Y$ is an open continuous surjection with connected fibres, then the inverse image of every connected subset of $Y$ is connected.

Proof. For connected $B\subset Y$, put $E=f^{-1}(B)$. The restricted map is open: if $U$ is open in $X$, then $f(U\cap E)=f(U)\cap B$. If $E$ had a separation into relatively open nonempty parts $E_1,E_2$, each connected fibre would be wholly in one part. Thus both parts would be saturated, their images would be disjoint nonempty relatively open sets covering $B$, and $B$ would be disconnected. $\square$

For any topological group $H$, its identity component $H^\circ$ is closed, since closure preserves connectedness; it is a subgroup, since the continuous image of connected $H^\circ\times H^\circ$ under $(x,y)\mapsto xy^{-1}$ lies in the identity component; and it is normal, since conjugation fixes the identity and preserves connectedness. Its cosets are all the components by translation.

The group quotient map $p:H\to H/H^\circ$ is open, because $p^{-1}(p(U))=UH^\circ$ is open for open $U$. Its fibres are connected. Lemma 1A.5 therefore says that any connected subset of $H/H^\circ$ has connected inverse image in $H$; that inverse image must lie in a single component, so the subset is a singleton. We have proved $$H/H^\circ\text{ is totally disconnected}. \tag{1A.16}$$ This component statement itself does not require local compactness.

We will also use the elementary quotient topology facts. The quotient by a closed subgroup is Hausdorff: the relation $x^{-1}y\in J$ is closed, and the open surjection $p\times p$ maps its open complement onto the complement of the quotient diagonal. Quotients by closed normal subgroups are topological groups, since the quotient maps and their products are open and hence quotient maps. An open quotient map sends a compact identity neighbourhood to a compact identity neighbourhood, so a quotient of a locally compact group is locally compact. A closed subgroup is locally compact by intersecting a compact neighbourhood with it.

Lemma 1A.6 (compact-space components and clopen neighbourhoods). In a compact Hausdorff space $X$, the component of $x$ is the intersection $Q$ of all clopen neighbourhoods of $x$. If $X$ is totally disconnected, clopen sets form a neighbourhood basis.

Proof. Every connected set through $x$ lies in every such clopen set. To prove the reverse inclusion, suppose the compact intersection $Q$ were separated into two nonempty compact parts. Normality of a compact Hausdorff space gives disjoint open neighbourhoods $U,V$ of those parts, with $x\in U$. The compact complement of $U\cup V$ misses $Q$. By the definition of the intersection, each of its points is omitted by some clopen neighbourhood of $x$. A finite subcover of the complements gives a finite intersection $D$ of such clopen neighbourhoods with $Q\subset D\subset U\cup V$. Then $D\cap U$ is clopen in $X$, contains $x$, and omits the nonempty part of $Q$ in $V$, a contradiction. Therefore $Q$ is connected and equals the component. Normality used here follows by twice taking finite subcovers of Hausdorff separating neighbourhoods for two disjoint compact closed sets.

If every component is a singleton and $O$ is a neighbourhood of $x$, every point of the compact set $X\setminus O$ is omitted by a clopen neighbourhood of $x$. Finitely intersecting those neighbourhoods gives a clopen neighbourhood contained in $O$. $\square$

Lemma 1A.7 (van Dantzig, with no commutativity assumption). In every totally disconnected locally compact Hausdorff group $T$, compact open subgroups form a neighbourhood basis at the identity.

Proof. Let $W$ be an identity neighbourhood, and choose a compact identity neighbourhood $B$. Take a symmetric open identity neighbourhood $V$ with $V^2\subset W\cap\operatorname{int}B$. The inclusion $\overline V\subset V^2$ follows because, for $z\in\overline V$, the open neighbourhood $zV$ meets $V$, giving $z\in VV^{-1}=V^2$. Thus $Q=\overline V$ is compact, lies in $W$, and contains the identity in its interior.

This compact space is totally disconnected. Lemma 1A.6 supplies a subset $D$ clopen in $Q$, with $e\in D\subset\operatorname{int}Q$. It is compact and also open in $T$: relative openness in $Q$ and containment in the interior turn it into an ambient open set.

Use its left stabilizer $$S=\{t\in T:tD=D\}. \tag{1A.17}$$ It is a subgroup. For each $d\in D$, continuity of multiplication gives an identity neighbourhood $A_d$ and a neighbourhood $O_d$ of $d$ such that $A_dO_d\subset D$. Finitely many $O_d$'s cover compact $D$. Intersect their identity neighbourhoods and shrink to a symmetric open $A$. Then $AD\subset D$. Both $aD\subset D$ and $a^{-1}D\subset D$ hold for $a\in A$, so $aD=D$. Hence $A\subset S$, and $S$ is open.

An open subgroup is closed, because its other cosets are open. Also $S\subset D$, since $e\in D$. Therefore $S$ is compact and $S\subset W$, proving the assertion. The left stabilizer argument uses no abelian law. $\square$

Corollary 1A.8. A quotient of a totally disconnected locally compact Hausdorff group by a closed normal subgroup is totally disconnected.

Proof. Let $r:T\to T/J$ be the open quotient map. In the inverse image of any quotient identity neighbourhood, Lemma 1A.7 gives a compact open subgroup $S$. The image $r(S)$ is a compact open subgroup in that neighbourhood, and is closed because the quotient is Hausdorff. These clopen neighbourhoods separate the quotient identity from every other point. Translations then exclude every connected set with two different points. $\square$

Theorem 1A.9 (components through a compact kernel). If $H$ is locally compact Hausdorff and $N\lhd H$ is compact, then the quotient map satisfies $$q(H^\circ)=(H/N)^\circ. \tag{1A.18}$$

Proof. Set $M=q(H^\circ)$. Products of a closed set with a compact subgroup are closed, by the argument following (1A.15). Hence $q$ is a closed map, so $M$ is closed; it is connected and normal as an image of $H^\circ$.

Put $T=H/H^\circ$, and let $p:H\to T$ be the quotient map. By (1A.16), $T$ is totally disconnected; it is locally compact Hausdorff by the quotient facts proved above. The subgroup $p(N)$ is compact and normal. Corollary 1A.8 makes $T/p(N)$ totally disconnected.

There is a topological group isomorphism $$(H/N)/M\ \simeq\ (H/H^\circ)/p(N). \tag{1A.19}$$ Both iterated quotient maps out of $H$ are open surjections with kernel $H^\circ N$, which proves the topology as well as the algebra of (1A.19). The connected image of $(H/N)^\circ$ in this totally disconnected quotient is trivial. Therefore $(H/N)^\circ\subset M$. The reverse inclusion follows because $M=q(H^\circ)$ is connected and contains the identity. $\square$

Compactness of the kernel has a precise role: it makes $q(H^\circ)$ closed. No step assumes local connectedness or path connectedness of $H$, $N$, or the solvable quotient subgroup.

1A.5. The compact-normal quotient theorem

Theorem 1A.10 (compact-normal quotients). Let $G$ be a connected locally compact Hausdorff group whose only connected algebraically solvable normal subgroup is $\{e\}$. For every compact normal subgroup $K$, the quotient $G/K$ has the same property. No countability, linearity or finite-centre hypothesis is required.

Proof. Put $C=C_G(K)$, and identify $G/K$ with $C/Z(K)$ using (1A.15). Suppose $P$ is a connected solvable normal subgroup of $G/K$, of derived length at most $d<\infty$. Replace $P$ by its closure. This remains connected and normal and has the same finite derived-length bound: continuity of commutators gives, by induction, $$D^j(\overline P)\subset\overline{D^jP}. \tag{1A.20}$$ Thus $\overline P$ terminates at step $d$, and it suffices to handle closed $P$.

Let $$E=C\cap q^{-1}(P).$$ It is a closed subgroup, hence locally compact Hausdorff, and is normal in $G$, since both $C$ and $q^{-1}(P)$ are normal in $G$. Its restricted quotient onto $P$ is open: restricting an open quotient map to the full preimage of a subgroup gives the subgroup quotient topology, because $q_C(O)\cap P=q_C(O\cap E)$ for $O$ open in $C$. Its kernel is precisely $Z(K)$. This is compact and abelian. Consequently $$D^dE\subset Z(K),\qquad D^{d+1}E=\{e\}. \tag{1A.21}$$ The identity component $E^\circ$ is characteristic in $E$ under topological automorphisms, and therefore normal in $G$; it is connected and algebraically solvable. The assumption on $G$ forces $E^\circ=\{e\}$. Theorem 1A.9, applied to $E\to P$, gives $q(E^\circ)=P^\circ=P$. Hence $P=\{e\}$, as required. $\square$

One also obtains the stronger centralizer form used by Rickert: for connected locally compact $G$, Theorem 1A.9 makes the map $C^\circ\to G/K$ onto, so $$G=K C^\circ. \tag{1A.22}$$ In that case $C^\circ\cap K$ is central in $G$: it commutes with $K$ by belonging to $C$, and with $C^\circ$ by belonging to $K$, while those two groups generate $G$.

Open diagram at full size

Figure 2. The compact-normal mechanism proving S3 (Theorem 1A.10). The domain is a connected locally compact Hausdorff group $G$, with arbitrary compact normal $K$, and no nontrivial connected algebraically solvable normal subgroup in $G$. Lemma 1A.1 fixes irreducible classes and gives all finite models $K/N_F$. The projection in Lemma 1A.2 checks every component of each matrix group; here $C_m=C_{U(n)}(L)$. Lemma 1A.3 makes the continuous implementer coset an inner action; no continuous choice of individual implementers is assumed. Theorem 1A.4 uses closed implementer sets and finite-union refinement to prove $G=K C_G(K)$, and proves the homeomorphism $C_G(K)/Z(K)\cong G/K$. Lemmas 1A.5–1A.7 and Corollary 1A.8 supply the component mechanism of Theorem 1A.9. Theorem 1A.10 then lifts a connected solvable normal subgroup $P$ to $E$, with $D^dE\subset Z(K)$, $D^{d+1}E=\{e\}$, and $q(E^\circ)=P$; the hypothesis gives $E^\circ=P=\{e\}$. The finite index sample and orthogonal axes are abstractions, not an exhaustion or dimensions of $K$. Human credit: Iwasawa’s compact-normal theorem, identified in Rickert, Theorem 1.4, p. 434, and Lemma 3.4, p. 440.

1B. Lie approximation: finite kernels and the countability reduction

The general Lie-approximation input S2 concerns a connected locally compact Hausdorff group, which may have no countable neighbourhood base. Two preliminary steps are sometimes hidden when that input is cited: the compact case needs only finitely many representations at each neighbourhood, and the general case can first pass through a compact kernel to a metrizable open subgroup. We prove those steps here. The metrizable proof follows the preliminary reductions: Theorem 1B.20 constructs the compact normal kernel and the Lie structure, and Corollary 1B.21 gives the unrestricted conclusion.

For classical credit, the finite-kernel argument is the compact Peter–Weyl route to Lie approximation. The countability reduction and the metrization argument are the Kakutani–Gleason part of that route; see Terence Tao, Hilbert's Fifth Problem and Related Topics, Sections 1.4–1.5, especially Theorem 1.4.14, Theorem 1.5.2 and Exercise 1.5.4. The author's freely accessible preliminary version supplies a reading source. We organize the argument around compact fibres and kernel lifting, and give a direct word-cost proof of metrization.

Proposition 1B.1 (the compact case, without a countable base). Let $C$ be a compact Hausdorff group and $U$ an open neighbourhood of its identity. There is a compact normal subgroup $N\subset U$ such that $C/N$ is a compact Lie group.

Proof. Peter–Weyl point separation says that, for each $x\ne e$, some finite-dimensional continuous unitary representation $\pi_x$ has $\pi_x(x)\ne I$. The open sets $$\{y\in C:\pi_x(y)\ne I\},\qquad x\in C\setminus U,$$ cover the compact set $C\setminus U$. Choose a finite subcover, with representations $\pi_1,\ldots,\pi_r$, and take their direct sum $\pi$. Its kernel $N$ is closed and normal in $C$, hence compact, and the covering property gives $N\subset U$. If $U=C$, the trivial representation gives the same conclusion with $N=C$.

The image $\pi(C)\subset U(n)$ is compact and therefore closed. The closed Lie subgroup theorem makes it a compact Lie group. The induced continuous bijection $C/N\to\pi(C)$ is a homeomorphism, because its source is compact and its target Hausdorff. This proves the assertion. Only finitely many representations were chosen for this one neighbourhood; no countable family separating all of $C$ was assumed. $\square$

We next record the topology needed when a compact kernel is lifted. These facts apply to arbitrary locally compact Hausdorff groups.

Lemma 1B.2 (compact kernels give proper quotient maps). Let $H$ be locally compact Hausdorff and $N\lhd H$ compact. The quotient map $q:H\to H/N$ is open and closed. The quotient is locally compact Hausdorff, and $q^{-1}(B)$ is compact for every compact subset $B\subset H/N$.

Proof. For open $O\subset H$, its saturation $ON$ is a union of translates of $O$, hence open; the quotient topology gives openness of $q$. For closed $F\subset H$, the product $FN$ is closed. To verify this without a sequence assumption, let a net $f_i n_i$ converge to $h$. Compactness of $N$ supplies a subnet with $n_i\to n\in N$. Then $f_i=(f_i n_i)n_i^{-1}\to hn^{-1}\in F$, so $h\in FN$. Thus the saturation of $F$ is closed and $q$ is closed. Closedness of $N$ makes the group quotient Hausdorff: for $x\notin N$, choose an identity neighbourhood $W$ with $W^{-1}xW\cap N=\varnothing$, and the open images of sufficiently small neighbourhoods of $e$ and $x$ are disjoint. Translation gives separation of any two distinct cosets.

Choose a relatively compact open identity neighbourhood $V\subset H$. The compact set $q(\overline V)$ contains the open neighbourhood $q(V)$; thus the quotient is locally compact. Now let $B$ be compact. Finitely many translates $q(h_jV)$ cover $B$. Therefore $$q^{-1}(B)\subset\bigcup_{j=1}^m h_j\overline V N.$$ The right side is compact, and the left side is closed because $B$ is a compact subset of a Hausdorff quotient. It follows that $q^{-1}(B)$ is compact. $\square$

Lemma 1B.3 (a compact kernel with a countable quotient base). Let $G$ be locally compact Hausdorff and $U$ an open identity neighbourhood. There are an open sigma-compact subgroup $H\subset G$ and a compact normal subgroup $N\lhd H$ such that $N\subset U$ and $H/N$ has a countable neighbourhood base at its identity. If $G$ is connected, $H=G$.

Proof. Choose a symmetric relatively compact open identity neighbourhood $V$ with $\overline V\subset U$. The subgroup $$H=\bigcup_{m\geq1} V^m$$ is open. It is sigma-compact, since it is covered by the compact sets $(\overline V)^m$, each of which lies in $H$. To check the last point, if $x\in\overline V$, the open set $xV$ meets $V$, and consequently $x\in V^2\subset H$.

Set $E=\overline V$, a compact symmetric generating set for $H$. We construct symmetric relatively compact open identity neighbourhoods $W_1,W_2,\ldots$ in $H$, with $\overline{W_1}\subset V$, and $$\begin{gathered} \overline{W_{j+1}}\subset W_j,\qquad W_{j+1}^3\subset W_j,\\ k^{-1}W_{j+1}k\subset W_j\quad(k\in E). \end{gathered} \tag{1B.1}$$ Here the conjugation containment is uniform over $E$. Indeed, continuity of $(k,w)\mapsto k^{-1}wk$ at each $(k,e)$ gives a neighbourhood of $k$ and an identity neighbourhood for $w$ whose images lie in $W_j$. A finite cover of $E$, followed by intersection of the identity neighbourhoods, makes the latter independent of $k$. Continuity of multiplication also gives a symmetric identity neighbourhood whose cube lies in $W_j$. Intersect the two choices and then choose a symmetric relatively compact open neighbourhood with its closure inside that intersection. Local compactness and regularity justify this final shrinking.

Define $$N=\bigcap_{j\geq1}\overline{W_j}=\bigcap_{j\geq1}W_j. \tag{1B.2}$$ The equalities follow from the first containment in (1B.1). The intersection is a nonempty compact set, since its closed sets are nested in the compact set $\overline{W_1}$ and contain $e$. It lies in $U$. If $a,b\in N$, then $a,b^{-1}\in W_{j+1}$ for every $j$, so $ab^{-1}\in W_j$. Thus $N$ is a subgroup. The conjugation containment shows $k^{-1}Nk\subset N$ for every $k\in E$; applying it to $k^{-1}\in E$ gives equality. Since $E$ generates $H$, $N\lhd H$.

By Lemma 1B.2 the quotient $L=H/N$ is locally compact Hausdorff and $q(W_j)$ are open identity neighbourhoods. They form a countable base. In fact, if $O\subset L$ is an open identity neighbourhood, its inverse image contains $N$. Some $\overline{W_j}$ must lie in $q^{-1}(O)$: otherwise the nested compact sets $\overline{W_j}\setminus q^{-1}(O)$ would have a common point in $N\setminus q^{-1}(O)$, which is impossible. For that $j$, $q(W_j)\subset O$.

Every open subgroup is closed, since its other cosets form its open complement. If $G$ is connected, this nonempty open and closed subgroup $H$ equals $G$. $\square$

For completeness, a countable identity base really does supply a metric compatible with the topology; no metrizability hypothesis has been smuggled into Lemma 1B.3.

Lemma 1B.4 (word-cost metrization). A Hausdorff topological group $L$ with a countable identity neighbourhood base has a compatible left invariant metric. If $L$ is also sigma-compact, it is second countable.

Proof. Refine the given base to symmetric open sets $A_1\supset A_2\supset\cdots$ such that $A_{j+1}^3\subset A_j$, and put $A_0=L$. Hausdorffness gives $\bigcap_j A_j=\{e\}$. Define the cost of a finite word $x_1\cdots x_r$, with $x_i\in A_{j_i}$ and integers $j_i\geq0$, as $\sum_i2^{-j_i}$. Let $p(x)$ be the infimum of all such costs for words with product $x$; allow the empty word, of cost zero, for $e$. Every element has a one-letter word of cost one. Reversing a word and taking inverses preserves its cost, and concatenating words adds costs. Therefore $$p(x^{-1})=p(x),\qquad p(xy)\leq p(x)+p(y). \tag{1B.3}$$

The needed lower bound comes from this word estimate: $$\begin{gathered} \sum_{i=1}^r2^{-j_i}<2^{-n},\quad n\geq1\\ \Longrightarrow\quad x_1\cdots x_r\in A_{n-1}. \end{gathered} \tag{1B.4}$$ Prove it by induction on the length $r$, simultaneously for all $n$. For a single letter, its cost is less than $2^{-n}$, so $j_1\geq n+1$ and the assertion holds. For a longer word, write $s$ for its total cost. Choose the letter at which the running sum first reaches $s/2$. The words strictly before and strictly after that letter each have total cost at most $s/2<2^{-(n+1)}$. They have shorter length, so by induction their products lie in $A_n$; an empty side has product $e\in A_n$. The middle letter belongs to $A_n$ as well, because its cost is at most $s<2^{-n}$. The full product lies in $A_n^3\subset A_{n-1}$. This proves (1B.4); it does not rearrange the letters, which matters in a noncommutative group.

If $p(x)<2^{-n}$, an approximating word of cost less than $2^{-n}$ shows $x\in A_{n-1}$. If $x\in A_n$, its one-letter word gives $p(x)\leq2^{-n}$. Hence $p(x)=0$ forces $x=e$, and $$d(x,y)=p(x^{-1}y)$$ is a metric by (1B.3), invariant under common left multiplication. The two inclusions just proved show that its balls and the sets $A_j$ give the same neighbourhood system at $e$, and translation gives the same topology everywhere.

If $L$ is sigma-compact, write it as a countable union of compact sets. Each compact metric space has a countable dense subset: for each integer $m\geq1$, take the centres of a finite cover by balls of radius $1/m$, and unite those finite sets. Their union over the compact covering sets is countable and dense in $L$. A metric space with a countable dense subset has the countable base of balls with those centres and positive rational radii. Thus $L$ is second countable. $\square$

Proposition 1B.5 (the exact reduction for S2). Suppose the following metrizable theorem is available: every metrizable locally compact Hausdorff group $L$, and every open identity neighbourhood $O\subset L$, admit an open subgroup $L'\subset L$ and a compact normal subgroup $B\lhd L'$, with $B\subset O$ and $L'/B$ a Lie group. Then the same assertion holds for every locally compact Hausdorff group $G$. For connected $G$, the resulting open subgroup is all of $G$, giving exactly S2.

Proof. Fix an open identity neighbourhood $U\subset G$. Use Lemma 1B.3 to obtain $H$ and $N\subset U$, and let $q:H\to L=H/N$. Lemma 1B.4 makes $L$ metrizable. There is an open identity neighbourhood $W\subset H$ with $WN\subset U$. To verify this shrinking, for each $n\in N$, continuity gives an identity neighbourhood $W_n$ and a neighbourhood $T_n$ of $n$ with $W_nT_n\subset U$. Take a finite cover of the compact set $N$ by these $T_n$, and intersect the corresponding $W_n$.

The set $O=q(W)$ is an open identity neighbourhood in $L$, and $q^{-1}(O)=WN\subset U$. Apply the stated metrizable theorem to obtain $L'$ and $B$. Set $$G'=q^{-1}(L'),\qquad K=q^{-1}(B).$$ Then $G'$ is open in $H$, hence in $G$; $K$ is normal in $G'$, is contained in $U$, and is compact by Lemma 1B.2. The surjective composite $G'\to L'\to L'/B$ is continuous and open, with kernel $K$. It consequently induces a topological group isomorphism $G'/K\cong L'/B$, a Lie group. If $G$ is connected, $G'=G$ because an open subgroup is closed. $\square$

This reduction retains arbitrary locally compact scope. Theorem 1B.20 below proves its metrizable input, including the small-subgroup, exponential-neighbourhood and analytic-atlas arguments. Corollary 1B.21 then gives S2 without an extra countability or local-connectedness hypothesis.

Exercise 1B.1 (what a finite compact quotient can see). Level 2. Let $I$ be uncountable, let $C=\mathbb T^I$ with the product topology, and let $U$ be any open identity neighbourhood. Prove directly that there is a compact normal subgroup $N\subset U$ with $C/N$ a finite-dimensional torus. Prove also that $C$ has no countable identity neighbourhood base. Explain why these two facts are consistent with Proposition 1B.1.

Solution. Choose a basic product neighbourhood inside $U$. It restricts only finitely many coordinates, say those in $F\subset I$, and each restriction contains $1$. The subgroup $$N=\{z\in\mathbb T^I:z_i=1\text{ for every }i\in F\}$$ is a closed subgroup of the compact Hausdorff product, hence compact, and it is normal because the product is abelian. It lies in the chosen product neighbourhood. Projection onto $F$ is a continuous surjection with kernel $N$; compactness and Hausdorffness identify $C/N$ topologically with $\mathbb T^F$.

Suppose that $O_1,O_2,\ldots$ were a countable identity base. For each $n$, choose a basic product neighbourhood $B_n\subset O_n$, restricting a finite set $F_n$. Their union is countable, so choose $i\in I\setminus\bigcup_nF_n$. Let $W$ be the identity neighbourhood requiring $z_i$ to lie in a fixed proper open arc around $1$. A base would give $O_n\subset W$ for some $n$, and then $B_n\subset W$. But $B_n$ leaves coordinate $i$ unrestricted, a contradiction. Each neighbourhood permits a finite Lie quotient, while different neighbourhoods may require different finite coordinate sets; the proposition does not assert a countable family sufficient for every neighbourhood.

Exercise 1B.2 (why the open subgroup is part of the statement). Level 1. Let $D$ be an uncountable discrete group and $G=\mathbb R\times D$, with the product topology. Prove that $G$ is locally compact Hausdorff but is not sigma-compact. For $V=(-1,1)\times\{e\}$, identify the subgroup $H=\bigcup_{m\geq1}V^m$ in Lemma 1B.3. Prove that the connectedness qualification at the end of Proposition 1B.5 is exactly what removes this proper open subgroup.

Solution. Each point has a neighbourhood with compact closure, namely an interval with compact closure times a singleton of $D$; the product is Hausdorff. A compact subset of $G$ projects to a compact subset of the discrete space $D$, hence to a finite set. A countable union of compact subsets therefore projects to at most countably many elements of $D$, so it cannot cover $G$.

Addition in the real coordinate gives $V^m=(-m,m)\times\{e\}$ and $H=\mathbb R\times\{e\}$. This is a proper open and closed sigma-compact subgroup. Every connected subset containing the identity has constant projection to $D$, and $\mathbb R\times\{e\}$ itself is connected; thus it is $G^\circ$. The construction cannot force $H=G$ here. For a connected group, however, the complement of any proper open subgroup is a nonempty union of open cosets, giving a separation. This is the step that forces $H=G$, and subsequently $G'=G$, in the connected case.

The metrizable case: compact kernels, escape and coordinates

We now prove the metrizable theorem used in Proposition 1B.5. The route follows the classical convolution and escape mechanisms in [Tao], Sections 1.2.5, 1.3.2–1.3.3 and 1.5. We begin with the desired quotient, then construct its metric and coordinates. The proof includes regularization, the full product limit, finite-dimensionality, an exponential neighbourhood and a global analytic atlas.

For a metrizable locally compact Hausdorff $G$, take the compact-generated, sigma-compact open subgroup $P$ constructed in the first paragraph of Lemma 1B.3. Lemma 1B.4 makes $P$ second countable. Every open subgroup of $P$ is open in $G$. If $G$ is connected, $P=G$. Thus this reduction adds no hypothesis to the original group.

The proof uses two exact programme prerequisites. The Haar lesson, Theorem 8.3, Proposition 9.1 and Theorem 14.2(6)–(7), supplies the measure and convolution facts. Peter–Weyl, Theorem 4.1, supplies point separation for every compact Hausdorff group. Their precise uses are:

• Haar input: each locally compact Hausdorff group used below, including $P$ and the quotient $H=P'/N$, has a nonzero left-invariant Radon measure $\mu$, finite on compact sets and positive on nonempty open sets. Uniqueness is unused. Convolution of bounded Borel functions with precompact support is well-defined; the usual $L^1$/$L^2$ translation continuity gives a continuous convolution. Where continuity is not needed, we say so.
• Completed compact approximation input: for compact Hausdorff $C$ and an open identity neighbourhood $O\subset C$, there is a compact normal $K\subset O$ such that $C/K$ is topologically isomorphic to the compact image of a finite-dimensional continuous unitary representation. This is the existing finite Peter–Weyl argument in Proposition 1B.1, not a new countable-family assumption.

All convolutions below have bounded Borel factors with compact support. The first half of Haar Theorem 14.2(7) therefore makes them pointwise defined and continuous, without unimodularity; it applies because the first factor is in $L^1$ and the second is bounded. Left invariance gives the difference identities directly. The selected group $P$, its compact-generated open subgroups and their quotients are sigma-compact. On these groups the locally completed Haar convention agrees with the completion of the ordinary sigma-finite Borel measure. No global sigma-finiteness is imposed on the original $G$.

Only the finite representation and kernel part of the compact argument is consumed here. Its matrix image is NSS by the next direct calculation; the closed Lie-subgroup theorem appearing in the conclusion of Proposition 1B.1 is unnecessary for this construction. Thus the eventual NSS recognition is not being presumed through a compact-quotient recognition step.

Such a matrix image is NSS. Indeed, in its ambient matrix Lie group choose a norm ball $B_R(0)$ on which matrix exponential is injective and which supplies a logarithm neighbourhood. A subgroup contained in $\exp B_{R/3}(0)$ cannot contain $g=\exp X\ne e$: at the first integer $k$ with $k|X|\ge R/3$, one has $R/3\le k|X|<2R/3<R$, whereas $g^k$ would have a logarithm of norm less than $R/3$, contradicting injectivity on $B_R(0)$. Intersect the witness with the matrix image.

For bounded functions define $$\begin{gathered} \tau_g f(x)=f(g^{-1}x),\\ \partial_g=I-\tau_g,\\ p_f(g)=\|\partial_g f\|_\infty. \end{gathered}$$ Translations are isometries in the supremum norm. Therefore $$\begin{gathered} p_f(g^{-1})=p_f(g),\\ p_f(gh)\le p_f(g)+p_f(h). \tag{GY.1} \end{gathered}$$ For convolution $f*\eta(x)=\int f(y)\eta(y^{-1}x)\,d\mu(y)$, left invariance and direct expansion give the exact identities $$\begin{gathered} \partial_b(f*\eta)=(\partial_b f)*\eta,\\ \partial_a\partial_b(f*\eta)(x) \\ =\int(\partial_b f)(y)\, \partial_{y^{-1}ay}\eta(y^{-1}x)\,d\mu(y). \end{gathered} \tag{GY.2}$$ For a fixed $q$, powers of its translation commute and $$n\partial_q=\partial_{q^n} +\sum_{i=0}^{n-1}\partial_{q^i}\partial_q. \tag{GY.3}$$ This follows by subtracting $I-\tau_q^n=\sum_{i=0}^{n-1}\tau_q^i(I-\tau_q)$ from $n(I-\tau_q)$. These formulas specify the operator order and the conjugation twist used throughout.