Common crossed products

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

After the previous lesson, two close algebras $P$ and $Q$ on one Hilbert space are normalized by a single strongly continuous unitary group $U$. This lesson forms the two crossed products by the action $\operatorname{Ad}U$ on the same Hilbert space $L^2(\mathbb R,H)$. It describes them as fixed-point algebras (Lemma 1.1), shows that they are as close as $P$ and $Q$ at every matrix level (Lemma 2.1), and proves that if the crossed product of $Q$ is a factor of type II$_\infty$, then $P$ and $Q$ are conjugate by a unitary near the identity (Proposition 3.2). The proof conjugates finite corners of the crossed products with the theorem for finite factors, inflates the conjugating unitary, and descends to $H$ through the completely positive criterion. No factoriality of the crossed product of $P$ is assumed [OpenAI-289, Section 8].

We use: Close operator algebras (Lemmas 1.4 and 1.5); Averaging and the completely positive criterion (Lemma 2.2, Theorem 3.1, Proposition 4.1 and Lemma 5.1); Irreducible hyperfinite subfactors (Lemma 1.1); Stability of finite factors (Theorem 2.1, with $t_{\rm fin}$ and $\eta_{\rm fin}$); Frequency nests and analytic multipliers (Lemma 4.1); Corollary 5.4 of Traces on von Neumann algebras (equal trace gives equivalence in a finite factor); Lemma 4.3 of Measurable operators for a trace (subprojections of every trace); Proposition 7.4 of the tensor product lesson (a matrix unit with sum $1$ generates a copy of $B(\ell^2)$); and Corollary 5.2 of Injective von Neumann algebras (type I algebras are injective).

1. Crossed products as fixed points

Let $P\subseteq B(H)$ be a unital von Neumann algebra on a separable Hilbert space and $(U_s)_{s\in\mathbb R}$ a strongly continuous unitary group with $U_sPU_s^*=P$. On $\mathcal K=L^2(\mathbb R,H)=H\otimes L^2(\mathbb R)$ put $\begin{gathered} (\lambda_s\xi)(t)=\xi(t-s),\\ (\rho_s\xi)(t)=\xi(t+s),\\ (\pi(x)\xi)(t)=U_t^*xU_t\,\xi(t), \end{gathered}$ for $x\in B(H)$. Then $\pi$ is a faithful normal representation of all of $B(H)$, and completely isometric: at each matrix level its norm is the essential supremum of the norms of unitary conjugates. Let $C(P,U)=\bigl(\pi(P)\cup\{\lambda_s:s\in\mathbb R\}\bigr)'' ,$ the crossed product of $P$ by $\operatorname{Ad}U$. For a second algebra normalized by the same group, its crossed product lives on the same space. Let $Q_a$ be multiplication by $e^{iat}$ and $\beta_a=\operatorname{Ad}Q_a$.

Lemma 1.1 (fixed points). With $\alpha_s=\operatorname{Ad}(U_s\otimes\rho_s)$, $\begin{gathered} C(P,U)=\bigl(P\bar\otimes B(L^2(\mathbb R))\bigr)^\alpha,\\ C(P,U)^\beta=\pi(P). \end{gathered}$

Proof. Inclusion. $\pi(x)$ is multiplication by a field with values in $P$, and $\lambda_r\in1\otimes B(L^2(\mathbb R))$, so both lie in $P\bar\otimes B(L^2(\mathbb R))$. A direct computation gives $\alpha_s(\pi(x))=\pi(x)$, using $U_sU_{t+s}^*xU_{t+s}U_s^*=U_t^*xU_t$, and $\alpha_s(\lambda_r)=\lambda_r$.

Smoothing. Let $X$ be $\alpha$-fixed in $P\bar\otimes B(L^2(\mathbb R))$. For $f,g\in C_c(\mathbb R)$ let $\lambda(f)=\int f(s)\lambda_s\,ds$ and $Y=\lambda(f)X\lambda(g)^*$, again $\alpha$-fixed, since translations commute with $U_s\otimes\rho_s$. The operators $R_{f,t}:\mathcal K\to H$, $R_{f,t}\xi=\int f(t-s)\xi(s)\,ds$, have norm at most $\|f\|_2$, depend norm continuously on $t$, and $(\lambda(f)\xi)(t)=R_{f,t}\xi$. For compactly supported $\xi$, Fubini's theorem gives $\begin{gathered} (Y\xi)(t)=\int k(t,r)\xi(r)\,dr,\\ k(t,r)=R_{f,t}XR_{g,r}^* . \end{gathered}$ The kernel is bounded and jointly norm continuous. Its values lie in $P$: $R_{f,t}$ intertwines the constant action of $P'$, with which $X$ commutes. Invariance of $Y$ under $\alpha_s$ gives $k(t+s,r+s)=U_s^*k(t,r)U_s$, first almost everywhere and then everywhere by continuity; hence $k(t,r)=U_t^*k(0,r-t)U_t . \tag{1.1}$

Reconstruction. Let $\chi_j(v)=(1-|v|/j)_+$. It is the Fourier transform of the probability measure $\mu_j$ with density $\frac j{2\pi}\bigl(\frac{\sin(ja/2)}{ja/2}\bigr)^2$, so multiplying a kernel by $\chi_j(t-r)$ is the contraction $Y\mapsto\int Q_aYQ_a^*\,d\mu_j(a)$. The resulting operator is, by (1.1) and $(\rho_v\xi)(t)=\xi(t+v)$, $Y_j=\int_{\mathbb R}\chi_j(v)\,\pi(k(0,v))\,\rho_v\,dv ,$ a strong integral of a compactly supported strongly continuous function with values in $C(P,U)$, since $\rho_v=\lambda_{-v}$; so $Y_j\in C(P,U)$. On compactly supported vectors the kernels are bounded on sets of finite measure and $\chi_j\to1$, so $Y_j\to Y$ weakly by dominated convergence, and $Y\in C(P,U)$. Finally, with nonnegative $f_n$ of integral one and supports shrinking to $0$, $\lambda(f_n)\to1$ strongly, so $\lambda(f_n)X\lambda(f_n)^*\to X$ weakly and $X\in C(P,U)$.

Characters. $\beta_a(\pi(x))=\pi(x)$ and $\beta_a(\lambda_s)=e^{ias}\lambda_s$, so $\beta$ preserves $C(P,U)$. An element fixed by all $\beta_a$ commutes with all $Q_a$, so by Lemma 4.1 of Frequency nests and analytic multipliers it is multiplication by a bounded field $F$, with values in $P$ almost everywhere because it commutes with the constant action of $P'$. Invariance under $\alpha_s$ gives $F(t+s)=U_s^*F(t)U_s$ almost everywhere, for each $s$. Then $G(t)=U_tF(t)U_t^*$ is invariant under translations; testing on a countable dense set of pairs of vectors and convolving with approximate identities shows that each coefficient is almost everywhere constant, so $G=x$ almost everywhere, with $x\in P$ because $U$ normalizes $P$. So $F(t)=U_t^*xU_t$ and the element is $\pi(x)$. $\square$

With the invariant mean of the averaging lesson, every unitary representation $V$ of $\mathbb R_d$ gives the UCP map $E_V(T)=m_sV_sTV_s^*$, which fixes the operators commuting with all $V_s$ and preserves every von Neumann algebra invariant under the conjugations (Lemma 2.2 there).

2. Close crossed products

Lemma 2.1. Let $P,Q\subseteq B(H)$ be unital von Neumann algebras on a separable space, normalized by the same strongly continuous unitary group $U$. If $d_{\rm cb}(P,Q)<\gamma$, then $d_{\rm cb}(C(P,U),C(Q,U))<\gamma$.

Proof. Choose $\gamma'<\gamma$ with $d_{\rm cb}(P,Q)\le\gamma'$. By Lemma 5.1(1) of the averaging lesson, $d_{\rm cb}(P\bar\otimes B(L^2(\mathbb R)),Q\bar\otimes B(L^2(\mathbb R)))\le\gamma'$. Let $V_s=U_s\otimes\rho_s$. For a contraction $X$ in $M_n(C(P,U))$, choose a contraction $Y$ in $M_n(Q\bar\otimes B(L^2(\mathbb R)))$ within $\gamma'$ plus a small slack. The amplification of $E_V$ fixes $X$, by Lemma 1.1; it maps $Y$ to an $\alpha$-fixed element of $M_n(Q\bar\otimes B(L^2(\mathbb R)))$, that is, into $M_n(C(Q,U))$; and it is completely contractive. So $\operatorname{dist}(X,M_n(C(Q,U))_1)\le\gamma'$. Exchange $P$ and $Q$. $\square$

3. Conjugacy through a crossed product

Lemma 3.1 (matrix units). A type II$_\infty$ factor $B$ on a separable Hilbert space contains matrix units $(e_{ij})_{i,j\ge1}$ with $\sum_ie_{ii}=1$ strongly and $e_{11}$ finite and nonzero. The von Neumann algebra they generate is a copy of $B(\ell^2)$, hence injective.

Proof. Let $\tau$ be a faithful normal semifinite trace on $B$. A maximal family of mutually orthogonal nonzero projections of finite trace is countable, since $B$ is countably decomposable, and has sum $1$, since $\tau$ is semifinite; call them $q_1,q_2,\dots$. Then $\sum_k\tau(q_k)=\tau(1)=\infty$. Fix $c>0$ and fill consecutive bins of trace $c$: go through $q_1,q_2,\dots$ and, by Lemma 4.3 of the measurable-operators lesson (a type II factor has no minimal projections), split off from the current $q_k$ a subprojection of trace equal to the smaller of the trace still needed in the current bin and the trace remaining in $q_k$. Since $\sum_k\tau(q_k)=\infty$, each bin is filled after finitely many steps, and every $q_k$ is used up. This gives mutually orthogonal projections $f_i$, $i\ge1$, with $\tau(f_i)=c$ and $\sum_if_i=1$. For $i\ne j$, the corner $(f_i+f_j)B(f_i+f_j)$ is a finite factor in which $f_i$ and $f_j$ have equal trace, so they are equivalent by Corollary 5.4 of the traces lesson. Partial isometries $w_i$ with $w_i^*w_i=f_1$ and $w_iw_i^*=f_i$ give $e_{ij}=w_iw_j^*$. By Proposition 7.4 of the tensor product lesson, the algebra they generate is spatially a copy of $B(\ell^2)\otimes1$. $\square$

Proposition 3.2 (conjugacy through a crossed product). There are an absolute $\gamma_0>0$ and a nondecreasing function $\zeta:(0,\gamma_0)\to(0,\infty)$ with $\zeta(\gamma)\to0$ as $\gamma\to0$, with the following property. Let $P,Q\subseteq B(H)$ be unital von Neumann algebras on a separable Hilbert space with $d_{\rm cb}(P,Q)<\gamma<\gamma_0$, normalized by the same strongly continuous unitary group $U$. If $C(Q,U)$ is a factor of type II$_\infty$, there is a unitary $a\in B(H)$ with $aPa^*=Q,\qquad\|a-1\|\le\zeta(\gamma).$

Proof. Choose $\gamma_0$ so that, for $\gamma<\gamma_0$, $301\gamma<\min\{1/10,t_{\rm fin}\}$ and $e(\gamma):=2(150\gamma+\eta_{\rm fin}(301\gamma))<10^{-3}$; and put $\zeta(\gamma)=32e(\gamma)$. Write $\mathcal A=C(P,U)$ and $\mathcal B=C(Q,U)$; by Lemma 2.1, $d_{\rm cb}(\mathcal A,\mathcal B)<\gamma$.

A common copy of $B(\ell^2)$. Lemma 3.1 gives matrix units $(e_{ij})$ in $\mathcal B$ with $e=e_{11}$ finite and nonzero, generating an injective algebra $S$. Since $S\subset_\gamma\mathcal A$, Christensen's theorem gives a unitary $w$ with $wSw^*\subseteq\mathcal A$ and $\|w-1\|\le150\gamma$. Put $\mathcal A_1=w^*\mathcal Aw$. Then $S\subseteq\mathcal A_1\cap\mathcal B$ and $d(\mathcal A_1,\mathcal B)<301\gamma=:\kappa$.

Finite corners. Compressing by the common projection $e$ gives von Neumann algebras $A_0=e\mathcal A_1e$ and $B_0=e\mathcal Be$ on $e\mathcal K$, with identity $e$ and $d(A_0,B_0)<\kappa$. The corner $B_0$ of the II$_\infty$ factor by a finite projection is a II$_1$ factor. By Lemma 1.1 of the hyperfinite lesson, $A_0$ is a finite factor, and Theorem 2.1 of Stability of finite factors, applied on the separable space $e\mathcal K$ with the finite factor $B_0$, gives a unitary $v_0$ on $e\mathcal K$ with $v_0A_0v_0^*=B_0,\qquad\|v_0-e\|\le\eta_{\rm fin}(\kappa).$

Inflation. Put $V=\sum_{i\ge1}e_{i1}v_0e_{1i}$. It is the direct sum of copies of $v_0$ on the spaces $e_{ii}\mathcal K$, so it is a unitary, $\|V-1\|=\|v_0-e\|$, and $Ve_{ij}=e_{i1}v_0e_{1j}=e_{ij}V$. For $x\in\mathcal A_1$, the entries $e_{1i}xe_{j1}$ lie in $A_0$, so $e_{1i}VxV^*e_{j1}=v_0e_{1i}xe_{j1}v_0^*\in B_0$, and $VxV^*=\sum_{i,j}e_{i1}\bigl(e_{1i}VxV^*e_{j1}\bigr)e_{1j}$ is a strong limit of finite sums in $\mathcal B$. So $V\mathcal A_1V^*\subseteq\mathcal B$, and the same argument with $v_0^*$ gives equality. With $Z=Vw^*$, $\begin{gathered} Z\mathcal AZ^*=\mathcal B,\\ \|Z-1\|\le150\gamma+\eta_{\rm fin}(301\gamma). \end{gathered}$

Descent to $H$. Average the characters: $E_\beta(T)=m_aQ_aTQ_a^*$ on $B(\mathcal K)$. It fixes every $\pi(x)$, $x\in B(H)$, and maps $\mathcal B$ into $\mathcal B^\beta=\pi(Q)$ by Lemma 1.1. So $F(x)=\pi^{-1}\bigl(E_\beta(Z\pi(x)Z^*)\bigr)\qquad(x\in P)$ is a UCP map from $P$ to $Q$. Since $E_\beta$ fixes $\pi(x)$ and $\pi$ is completely isometric, $\|F-\mathrm{id}_P\|_{\rm cb}\le2\|Z-1\|\le e(\gamma)$, where $\mathrm{id}_P$ is the inclusion of $P$ in $B(H)$. The reverse ordinary gap from $Q$ to $P$ is less than $\gamma$, and $\gamma+e(\gamma)<1$. Proposition 4.1 of the averaging lesson gives a unitary $a$ with $aPa^*=Q$ and $\|a-1\|\le32e(\gamma)=\zeta(\gamma)$. $\square$

4. Exercises

Exercise 4.1. Check $\alpha_s(\pi(x))=\pi(x)$ and $\beta_a(\lambda_s)=e^{ias}\lambda_s$.

Exercise 4.2. Verify that $\chi_j(v)=(1-|v|/j)_+$ is the Fourier transform of the stated probability density.

Exercise 4.3. For $P=B(H)$ and $U$ trivial, identify $C(P,U)$ and check Lemma 1.1 directly.

Exercise 4.4. Show that $V=\sum_ie_{i1}v_0e_{1i}$ commutes with every $e_{ij}$.

Exercise 4.5. In Proposition 3.2, why is the factoriality of $C(P,U)$ not needed?

5. Solutions

Solution 4.1. $\bigl((U_s\otimes\rho_s)\pi(x)(U_s\otimes\rho_s)^*\xi\bigr)(t)$$=U_sU_{t+s}^*xU_{t+s}U_s^*\xi(t)$$=U_t^*xU_t\xi(t)$, since $U_{t+s}U_s^*=U_t$. For the second, $(Q_a\lambda_sQ_a^*\xi)(t)=e^{iat}e^{-ia(t-s)}\xi(t-s)$.

Solution 4.2. $\chi_j=j^{-1}\,1_{[-j/2,j/2]}*1_{[-j/2,j/2]}$, and $\int_{-j/2}^{j/2}e^{-iav}\,dv=2\sin(ja/2)/a$. Since the Fourier transform of a convolution is the product of the transforms, $\frac1{2\pi}\int e^{-iav}\chi_j(v)\,dv=\frac j{2\pi}\bigl(\frac{\sin(ja/2)}{ja/2}\bigr)^2$, and Fourier inversion gives $\chi_j(v)=\int e^{iav}\,d\mu_j(a)$. The total mass of $\mu_j$ is $\chi_j(0)=1$.

Solution 4.3. $\pi(x)=x\otimes1$, and $C(P,U)$ is generated by $B(H)\otimes1$ and the translations, so it is $B(H)\bar\otimes\lambda(\mathbb R)''$. The $\alpha$-fixed points of $B(H)\bar\otimes B(L^2(\mathbb R))$ are the operators commuting with $1\otimes\rho_s$, and the commutant of the right translations is the algebra generated by the left translations.

Solution 4.4. $Ve_{ij}=\sum_ke_{k1}v_0e_{1k}e_{ij}=e_{i1}v_0e_{1j}$ and $e_{ij}V=\sum_ke_{ij}e_{k1}v_0e_{1k}=e_{i1}v_0e_{1j}$.

Solution 4.5. Only the finite corner $A_0$ needs to be a finite factor, and Lemma 1.1 of the hyperfinite lesson derives this from its closeness to the II$_1$ factor $B_0$.

References

[OpenAI-289] OpenAI, Universal strong Kadison–Kastler stability, preprint, 23 September 2026, Section 8. https://github.com/openai/math/tree/main/preprints/Universal-strong-Kadison-Kastler-stability-September-23-2026