# Continuous outer actions of locally compact groups *Written by GPT-6.1 Sol (OpenAI), Ultra, September 2026. Self-checked by the writing AI. New original text: public domain (CC0).* Every locally compact group with a countable base acts faithfully and continuously on the hyperfinite \(\mathrm{II}_1\) factor, with every nonidentity element acting outerly. The construction repeats the regular representation countably many times and then applies the tracial CAR functor. Repetition makes the Hilbert–Schmidt innerness criterion fail for every nonidentity group element. We use left Haar measure, its Radon regularity and the density of compactly supported continuous functions in \(L^2\). The action and innerness inputs are proved in [Bogoliubov automorphisms and the Hilbert–Schmidt criterion](bogoliubov-automorphisms-and-innerness.md). The identification of a separable AFD \(\mathrm{II}_1\) factor with the tracial CAR factor is the previously proved [finite AFD uniqueness theorem](hyperfinite-finite-factors.md#theorem-5-1). The group hypothesis is **locally compact Hausdorff with a countable base**. The freely accessible [Blattner paper](https://msp.org/pjm/1958/8-4/pjm-v8-n4-p03-s.pdf), Section 6, explicitly uses the second countability axiom for its outer-action corollary. We state that hypothesis directly. A countable dense subset alone is not silently used to infer metrizability or separability of the regular Hilbert space. The target factor is separable AFD of type \(\mathrm{II}_1\). The concrete CAR construction has these properties, and the local uniqueness theorem transports the action to every factor satisfying those hypotheses. ## 1. A faithful continuous regular representation Let \(G\) have the above hypotheses, let \(m\) be a left Haar measure, and put \(H_0=L^2(G,m)\). Define \[ (\lambda(g)f)(t)=f(g^{-1}t),\qquad g,t\in G. \tag{1} \] Left invariance of \(m\) makes these operators unitary. Substitution gives \(\lambda(g)\lambda(h)=\lambda(gh)\). **Lemma 1.1.** The Hilbert space \(H_0\) is separable, and \(\lambda\) is strongly continuous and faithful. **Proof.** Local compactness and second countability give a countable cover by relatively compact open sets. To see this, refine a countable base to those elements whose closures lie in some compact neighborhood; that subfamily still covers \(G\). Their finite unions have finite Haar measure. Thus \(m\) is \(\sigma\)-finite. Take the countable ring generated by this base, using finite unions, intersections and differences. On every finite-measure member of the covering sequence, its simple functions with rational complex coefficients form a countable family with dense span in \(L^2\). Here is the density justification: the class of Borel subsets whose indicators belong to the closed span is closed under increasing and decreasing limits inside that finite-measure set, by \(L^2\) dominated convergence. The ring generates the Borel sigma algebra, so the monotone-class theorem includes all Borel subsets. Simple functions then approximate all \(L^2\) functions, and truncating to the covering sequence handles the whole space. Completion of Haar measure does not change the equivalence classes of \(L^2\) functions. This proves separability. For \(f\in C_c(G)\), the function \((g,t)\mapsto f(g^{-1}t)\) is jointly continuous. Restrict \(g\) to a compact neighborhood of the identity. All translated supports then lie in one compact set, and the difference from \(f\) converges uniformly to zero there as \(g\to e\). Its \(L^2\) norm tends to zero, since that compact set has finite Haar measure. The density of \(C_c(G)\), and the norm-one bound for every \(\lambda(g)\), prove strong continuity on \(H_0\). Translation of this conclusion gives continuity at every \(g\). Fix \(g\ne e\). Hausdorffness and continuity of multiplication give a relatively compact nonempty open neighborhood \(V\) of \(e\) with \[ V\cap gV=\varnothing. \tag{2} \] For example choose disjoint neighborhoods of \(e\) and \(g\), and shrink \(V\) so it lies in the first and \(gV\) in the second. Haar measure is positive on nonempty open sets and finite on \(V\). Consequently \[ \xi=\frac{1_V}{m(V)^{1/2}},\qquad \|\lambda(g)\xi-\xi\|^2=2. \tag{3} \] Thus \(\lambda(g)\ne1\), proving faithfulness. \(\square\) For completeness, positivity of a nonempty open set follows from nonzero Haar measure and regularity. If such a set had measure zero, translates would cover any compact set by finitely many measure-zero sets. Every compact set would have measure zero, contradicting nonzero Radon measure. ## 2. Repetition removes Hilbert–Schmidt perturbations Let \[ K=H_0\otimes\ell^2(\mathbb N),\qquad U(g)=\lambda(g)\otimes1. \tag{4} \] This is a separable infinite-dimensional Hilbert space, even when \(G\) is finite or trivial. A nonzero vector in \(H_0\) tensored with the standard basis of \(\ell^2\) supplies infinitely many orthogonal vectors. The operators \(U(g)\) form a unitary representation. Strong continuity follows on elementary tensors from that of \(\lambda\), then on finite sums, then on all of \(K\) by density and the common norm-one bound. **Lemma 2.1.** For every \(g\ne e\), \[ U(g)-1\notin\mathcal L^2(K). \tag{5} \] In fact \(U(g)-1\) is not compact. **Proof.** Choose the unit vector \(\xi\) from (3), and write \(\eta_n=\xi\otimes e_n\). They are orthonormal, and \[ \|(U(g)-1)\eta_n\|^2=2 \qquad(n\ge1). \tag{6} \] Extending this orthonormal sequence to a basis, its contribution to the defining Hilbert–Schmidt sum is already infinite. This proves (5). The output vectors are mutually orthogonal and have norm \(\sqrt2\). They have no convergent subsequence, so the operator is not compact. \(\square\) The same argument works with any faithful strongly continuous unitary representation on a separable Hilbert space: repeat a single nonzero displacement infinitely often. The regular representation gives one concrete choice for every group under consideration. ## 3. The outer action Let \(R_K\) be the von Neumann algebra of the trace representation of \(\operatorname{CAR}(K)\), and define \[ \beta_g(a(f))=a(U(g)f). \tag{7} \] This is the normally extended Bogoliubov automorphism, rather than merely an algebraic prescription. **Theorem 3.1.** The map \[ \beta:G\longrightarrow\operatorname{Aut}(R_K) \tag{8} \] is a continuous injective homomorphism, preserves the normalized trace, and satisfies \[ \beta_g\text{ is inner}\quad\Longleftrightarrow\quad g=e. \tag{9} \] For any separable AFD factor \(R_0\) of type \(\mathrm{II}_1\), transporting (8) along a normal isomorphism gives such an action on \(R_0\). **Proof.** The Bogoliubov extension theorem gives normality and trace preservation, and its homomorphism law gives \(\beta_g\beta_h=\beta_{gh}\). Its continuity theorem, composed with the strong continuity of \(U\), gives point-predual norm continuity of (8). The Hilbert–Schmidt innerness theorem says that \(\beta_g\) is inner exactly when \(U(g)-1\) is Hilbert–Schmidt. Lemma 2.1 excludes every \(g\ne e\); \(g=e\) gives the identity. If \(\beta_g=\beta_h\), then \(\beta_{h^{-1}g}\) is the identity and hence inner, so \(h^{-1}g=e\). This proves injectivity. The CAR factor \(R_K\) is separable AFD and of type \(\mathrm{II}_1\). The uniqueness theorem supplies a normal isomorphism \(\theta:R_K\to R_0\). Set \[ \gamma_g=\theta\beta_g\theta^{-1}. \tag{10} \] Conjugation by \(\theta\) is a homeomorphism of the automorphism groups in point-predual norm: a normal functional on \(R_0\) pulls back isometrically to one on \(R_K\). It also preserves innerness, because an implementing unitary is transported by \(\theta\). Thus all conclusions hold on \(R_0\). \(\square\) This construction requires neither amenability nor connectedness. It provides an action on one finite factor; it does not identify an arbitrary group von Neumann algebra as injective. ## 4. Why countably many copies matter A faithful representation on a finite-dimensional space can have every group element at Hilbert–Schmidt distance from the identity. Its CAR action then gives inner automorphisms. Infinite repetition removes this possibility while keeping the Hilbert space separable. For the two-element group, its regular representation on \(\mathbb C^2\) sends the nonidentity element to the flip. One copy has a finite-rank difference from the identity. Countably many copies contain infinitely many copies of its \(-1\) eigenvector, so the Hilbert–Schmidt sum diverges. The corresponding automorphism of the infinite CAR trace factor is outer. The countable-base hypothesis also has a topological role. If a compact group admits a continuous injective map into the Hausdorff Polish group \(\operatorname{Aut}(R_K)\), that map is a homeomorphism onto its image: a continuous bijection from a compact space onto a Hausdorff space is closed. Its domain must therefore be metrizable. Thus a theorem for compact nonmetrizable groups cannot follow by simply replacing second countability with the existence of a countable dense set. ## 5. Exercises with complete solutions **Exercise 1.** Verify the representation law in (1). Explain why no modular-function factor occurs. *Solution.* For \(f\in L^2(G,m)\), \[ (\lambda(g)\lambda(h)f)(t) =f(h^{-1}g^{-1}t)=f((gh)^{-1}t). \] Left translation preserves the chosen left Haar measure, so the change of variables in its \(L^2\) norm has Jacobian one. A right-translation formula against left Haar measure would require the modular correction; the left-translation formula here does not. **Exercise 2.** For a finite group \(G\) with counting measure, calculate the squared Hilbert–Schmidt norm of \(\lambda(g)-1\), \(g\ne e\). Then calculate it on \(N\) repeated copies. *Solution.* Left multiplication by \(g\ne e\) fixes no basis vector \(\delta_h\). Hence \(\lambda(g)\delta_h\) and \(\delta_h\) are orthogonal, and each squared difference norm is \(2\). Summing gives \[ \|\lambda(g)-1\|_{\mathrm{HS}}^2=2|G|. \] On \(N\) orthogonal copies the sum is \(2N|G|\). The countably infinite sum diverges, which is the mechanism in Lemma 2.1. **Exercise 3.** Prove that any nonzero bounded operator \(T\) has a non-Hilbert–Schmidt infinite repetition \(T\otimes1_{\ell^2}\). *Solution.* Choose a unit vector \(\xi\) with \(\|T\xi\|=c>0\). The orthonormal vectors \(\xi\otimes e_n\) have images of norm \(c\), so their contribution to the squared Hilbert–Schmidt norm is \(\sum_n c^2=\infty\). The images are mutually orthogonal, also proving noncompactness. Apply this with \(T=\lambda(g)-1\). **Exercise 4.** Show that repetition preserves strong continuity by an explicit tail estimate. *Solution.* Identify \(K\) with \(\bigoplus_n H_0\), and write \(\eta=(\eta_n)\). Then \[ \|(U(g)-U(g_0))\eta\|^2 =\sum_n\|(\lambda(g)-\lambda(g_0))\eta_n\|^2. \] Given \(\varepsilon>0\), choose \(N\) with \(4\sum_{n>N}\|\eta_n\|^2<\varepsilon^2/2\). Strong continuity makes the finite prefix sum less than \(\varepsilon^2/2\) for \(g\) close enough to \(g_0\). The tail is bounded by the chosen quantity, so the whole norm is less than \(\varepsilon\). **Exercise 5.** For \(G=\mathbb R\), choose an explicit displacement vector for each fixed \(s\ne0\). *Solution.* Use Lebesgue measure and left translation \((\lambda(s)f)(t)=f(t-s)\). Choose an interval \(V=(0,r)\) with \(0