The uniform commutator estimate

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

The previous lesson bounded matrix commutators for a II$_1$ factor acting on finitely many copies of its standard space. This lesson removes every restriction on the von Neumann algebra and its representation [OpenAI-288, Section 7]. The result is the uniform commutator estimate: there is one numerical constant $C$ such that, for every von Neumann algebra $P\subseteq B(K)$, every $Y\in B(K)$ and every matrix $X\in M_h(P)$, $\|[Y^{(h)},X]\|\le C\,g_P(Y)\,\|X\| .$ In other words, the inner derivation $x\mapsto Yx-xY$ of $P$ is completely bounded with $\|\cdot\|_{\rm cb}\le C\|\cdot\|$. The proof treats factors on separable spaces by type: II$_1$ factors embed into countably many copies of their standard space, properly infinite factors absorb matrices through isometries with orthogonal ranges, and finite type I factors are handled by averaging over their compact unitary groups. The central decomposition then assembles factors, and a separable reduction removes the separability assumption.

We use: Proposition 4.1 of Matrix commutators in finite factors and its notation $g_P$; Proposition 2.2 of Row, column and cyclic estimates, with $c_{\rm row}=4\sqrt2$; the commutation theorem and the normality of the standard representation, Theorem 3.1 of Integration for a trace; from Projections and types of von Neumann algebras, Proposition 18.2 (if a von Neumann algebra has a separating vector, every positive normal functional on it is a vector functional), Corollary 7.3 (the types of factors), Corollary 10.4 (type I factors are $B(L)$) and Corollary 13.5 (a properly infinite projection is a sum of countably many mutually orthogonal projections equivalent to it); the central decomposition, Theorem 4.1, Lemma 3.1 and Fact 1.1 of The central decomposition and the types of the fibres; and Kaplansky's density theorem, Theorem 7.1, which applies to $*$-algebras that are not norm closed; and a left-invariant Haar probability measure on a compact group, Theorem 8.3 of Haar measure on locally compact groups. Set $\begin{aligned} &C_{\rm fac}=3c_1+2,\\ &C=3C_{\rm fac}+2,\\ &c_1=3\pi^2(1+4c_0),\ c_0\\ &\quad =4\sqrt2 . \end{aligned}\tag{0.1}$ Numerically $c_1<700$, $C_{\rm fac}<2102$ and $C<6308$. Note $C_{\rm fac}\ge1+2c_{\rm row}$ and $C_{\rm fac}\ge2$.

1. Factors on separable Hilbert spaces

Proposition 1.1. Let $P\subseteq B(K)$ be a factor on a separable Hilbert space. For every $Y\in B(K)$, $h\ge1$ and $X\in M_h(P)$, $\|[Y^{(h)},X]\|\le C_{\rm fac}\,g_P(Y)\,\|X\| .$

Proof. Put $\delta=g_P(Y)$. A nonzero factor is finite or properly infinite, and a finite factor is of type II$_1$ or of type I; a finite type I factor is $B(L)$ with $L$ finite-dimensional (Corollaries 7.3 and 10.4 of the projections lesson; $B(L)$ contains a non-unitary isometry when $L$ is infinite-dimensional).

Type II$_1$. $P$ has separable predual. Let $\lambda:P\to B(L^2(P))$ be the standard representation of its trace; it is a normal $*$-isomorphism onto a von Neumann algebra with the separating vector $1$ (commutation theorem). For $\xi\in K$, the functional $\lambda(x)\mapsto\langle x\xi,\xi\rangle$ is positive and normal on $\lambda(P)$, so by Proposition 18.2 of the projections lesson it is $\langle\lambda(x)\zeta,\zeta\rangle$ for some $\zeta\in L^2(P)$. The map $x\xi\mapsto\lambda(x)\zeta$ is then isometric, since both squared norms are $\langle x^*x\xi,\xi\rangle$, and it intertwines the two actions of $P$. By Zorn's lemma $K$ is an orthogonal sum of cyclic subspaces $\overline{P\xi_n}$, countably many because $K$ is separable. Hence there is an isometry of $K$ into $\widetilde K=L^2(P)^{\oplus\mathbb N}$ intertwining the action of $P$ on $K$ with its common left action on $\widetilde K$, and its range reduces this action. Transport $Y$ and extend it by $0$ on the orthogonal complement of the range; $g_P$ does not change, because the range projection commutes with $P$. Let $p_n$ be the projection onto the first $n$ copies. It commutes with $P$, so $[p_nYp_n,x]=p_n[Y,x]p_n$ and $g_P(p_nYp_n|_{p_n\widetilde K})\le\delta$. Proposition 4.1 of the previous lesson gives $\|[(p_nYp_n)^{(h)},X]\|\le(3c_1+2)\delta\|X\|$ on $L^2(P)^n$. For fixed $X$, these commutators $p_n^{(h)}[Y^{(h)},X]p_n^{(h)}$ converge strongly to $[Y^{(h)},X]$, and a norm bound passes to strong limits. Restricting to the original space gives the claim with $3c_1+2=C_{\rm fac}$.

Properly infinite. By Corollary 13.5 of the projections lesson, $1=\sum_ne_n$ with mutually orthogonal $e_n\sim1$; this gives isometries $v_1,\dots,v_h\in P$ with orthogonal ranges. The row $V=(v_1\ \cdots\ v_h):K^h\to K$ satisfies $V^*V=I$. The map $\Delta(x)=Yx-xY$ is a rectangular derivation of $P$ for the identity representation on both sides, with $\|\Delta\|=\delta$. Proposition 2.2 of the cyclic estimates lesson, applied to the row $V$ and the column $V^*$, gives $\begin{gathered} \|YV-VY^{(h)}\|\le c_{\rm row}\delta,\\ \|Y^{(h)}V^*-V^*Y\|\le c_{\rm row}\delta . \end{gathered}$ For $X\in M_h(P)$, $VXV^*\in P$ has norm $\|X\|$, and $\begin{aligned} [Y,VXV^*]&=(YV-VY^{(h)})XV^*\\ &\quad +V[Y^{(h)},X]V^*\\ &\quad +VX(Y^{(h)}V^*-V^*Y). \end{aligned}$ Since $V^*V=I$, $\|V[Y^{(h)},X]V^*\|=\|[Y^{(h)},X]\|$. Therefore $\|[Y^{(h)},X]\|\le(1+2c_{\rm row})\delta\|X\|$, which is at most $C_{\rm fac}\delta\|X\|$.

Finite type I. $P\cong M_n(\mathbb C)$ has a compact unitary group. Let $Y_0=\int_{\mathcal U(P)}uYu^*\,du$, with Haar probability measure. Then $Y_0\in P'$ and $\|Y-Y_0\|\le\sup_u\|uYu^*-Y\|$, which is $\sup_u\|[u,Y]\|\le\delta$. Since $[Y^{(h)},X]=[(Y-Y_0)^{(h)},X]$, the commutator has norm at most $2\delta\|X\|\le C_{\rm fac}\delta\|X\|$. $\square$

2. The central decomposition

Proposition 2.1. Let $P\subseteq B(K)$ be a von Neumann algebra on a separable Hilbert space. For every $Y\in B(K)$, $h\ge1$ and $X\in M_h(P)$, $\|[Y^{(h)},X]\|\le C\,g_P(Y)\,\|X\| .$

Proof. Put $\delta=g_P(Y)$. For a finite partition of unity $z_1,\dots,z_n$ by central projections, the unitaries $\sum_j\zeta_jz_j$, $|\zeta_j|=1$, form a compact group in $P$, and averaging $Y$ over it gives $\sum_jz_jYz_j$. Each conjugate $uYu^*$ of the average lies within $\|[u,Y]\|\le\delta$ of $Y$, so the averages lie within $\delta$ of $Y$. Direct the partitions by refinement and let $Y_1$ be a weak operator cluster point of the averages. Every central projection $z$ commutes with all averages over partitions refining $\{z,1-z\}$, so $Y_1$ commutes with the centre $Z(P)$. Moreover $\|Y_1-Y\|\le\delta$ and $g_P(Y_1)\le g_P(Y)+2\|Y_1-Y\|\le3\delta$.

Take the central decomposition of Theorem 4.1 of the central decomposition lesson: $K=\int^\oplus K(t)\,d\mu(t)$ with separable fibres, and $P=\int^\oplus M(t)$ with every $M(t)$ a factor; the centre is the diagonal algebra, so $Y_1$, which commutes with it, is decomposable, $Y_1=\int^\oplus Y_1(t)$ (Fact 1.1(f) there). We show $\begin{gathered} g_{M(t)}(Y_1(t))\le3\delta\\ \text{for almost every }t. \end{gathered}\tag{2.1}$ Lemma 3.1 of the central decomposition lesson gives a countable unital $*$-algebra $\mathcal Q\subseteq P$ over $\mathbb Q(i)$, fields $\beta_b$ representing $b\in\mathcal Q$, and a null set $N$ such that for $t\notin N$ the map $b\mapsto\beta_b(t)$ is a unital $*$-homomorphism of $\mathbb Q(i)$-algebras whose image has complex span strongly dense in $M(t)$. Enlarging $N$, we may assume $\|\beta_b(t)\|\le\|b\|$ for $t\notin N$ and $b\in\mathcal Q$ (Fact 1.2(b) there).

Let $\Theta(y)=y(1+y^*y)^{-1/2}$ on a C*-algebra. It is norm continuous and maps the algebra onto its open unit ball, with inverse $c\mapsto c(1-c^*c)^{-1/2}$. For $b\in\mathcal Q$ put $c_b=\Theta(b)\in P$, a contraction. Choosing polynomials $q_k$ with rational coefficients converging uniformly to $(1+x)^{-1/2}$ on $[0,\|b\|^2]$, the elements $bq_k(b^*b)\in\mathcal Q$ converge to $c_b$ in norm, and their fields converge off $N$ to $\Theta(\beta_b(t))$; since the norm of a decomposable operator is the essential supremum of the norms of its fibres (Fact 1.1(e) there), $c_b$ is represented by $t\mapsto\Theta(\beta_b(t))$. As $\|[Y_1,c_b]\|\le3\delta$, the same essential supremum formula gives $\|[Y_1(t),\Theta(\beta_b(t))]\|\le3\delta$ outside a null set $N_b$. Let $t\notin N\cup\bigcup_bN_b$, a countable union of null sets. Let $A(t)$ be the norm closure of the complex span of $\{\beta_b(t):b\in\mathcal Q\}$; the set $\{\beta_b(t)\}$ is norm dense in $A(t)$, because $\mathbb Q(i)$ is dense in $\mathbb C$. By continuity of $\Theta$, the contractions $\Theta(\beta_b(t))$ are norm dense in the open, hence in the closed, unit ball of $A(t)$, and by Kaplansky's density theorem strongly dense in the unit ball of $M(t)$. A norm bound on commutators passes to strong limits, which proves (2.1).

Each $M(t)$ is a factor on a separable space, so Proposition 1.1 gives $\|[Y_1(t)^{(h)},X(t)]\|\le3C_{\rm fac}\delta\|X(t)\|$ for almost every $t$, where $X(t)$ is the fibre of $X$ on $K(t)^h$, with $\|X(t)\|\le\|X\|$ almost everywhere. The operator $[Y_1^{(h)},X]$ is decomposable on $K^h=\int^\oplus K(t)^h$ with these fibres, so its norm is at most $3C_{\rm fac}\delta\|X\|$. Finally $[Y^{(h)},X]=[Y_1^{(h)},X]+[(Y-Y_1)^{(h)},X]$, and the last term has norm at most $2\delta\|X\|$. $\square$

3. Removing separability

Theorem 3.1 (uniform commutator estimate). For every Hilbert space $K$, every von Neumann algebra $P\subseteq B(K)$ with $1_K\in P$, every $Y\in B(K)$, every $h\ge1$ and every $X\in M_h(P)$, $\begin{aligned} &\|[Y^{(h)},X]\|\\ &\quad \le C\,g_P(Y)\,\|X\| . \end{aligned}\tag{3.1}$ Equivalently, the derivation $x\mapsto Yx-xY$ of $P$ into $B(K)$ has completely bounded norm at most $C$ times its norm.

Proof. Fix $h$, $X\in M_h(P)$ and a unit vector $\xi=(\xi_1,\dots,\xi_h)\in K^h$. Let $A_X\subseteq P$ be the unital C*-algebra generated by the entries of $X$. The C*-algebra generated by $A_X$ and $Y$ is norm separable, so the closed span $K_0$ of its images of $\xi_1,\dots,\xi_h$ is a separable subspace reducing $A_X$ and $Y$. Let $\sigma(a)=a|_{K_0}$ for $a\in A_X$, $Y_0=Y|_{K_0}$ and $P_0=\sigma(A_X)''\subseteq B(K_0)$. For $a\in A_X$, $\|[Y_0,\sigma(a)]\|\le g_P(Y)\|a\|$. Every $b\in\sigma(A_X)$ is $\sigma(a)$ with $\|a\|=\|b\|$ (Proposition 17.1 of C*-algebras, continuous functional calculus, automatic continuity and positive cones), so $\|[Y_0,b]\|\le g_P(Y)\|b\|$, and by Kaplansky's density theorem $g_{P_0}(Y_0)\le g_P(Y)$. Proposition 2.1 applied to $P_0$ and $\sigma_h(X)$ gives $\|[Y_0^{(h)},\sigma_h(X)]\|\le Cg_P(Y)\|\sigma_h(X)\|$, which is at most $Cg_P(Y)\|X\|$. Since $K_0$ reduces all operators involved and contains every $\xi_i$, $[Y^{(h)},X]\xi=[Y_0^{(h)},\sigma_h(X)]\xi$. Take the supremum over $\xi$. $\square$

Combined with Arveson's distance formula, Corollary 4.5 of Completely bounded homomorphisms and similarity, Theorem 3.1 says: for every von Neumann algebra $P\subseteq B(K)$ and every $Y\in B(K)$, $\operatorname{dist}(Y,P')\le\tfrac C2\,g_P(Y). \tag{3.2}$ A C*-algebra $A\subseteq B(K)$ has the same commutant and, by Kaplansky's density theorem, the same $g$ as its weak closure, so (3.2) holds for all C*-algebras containing $1_K$ as well.

4. Exercises

Exercise 4.1. Show that $g_P(Y)\le2\operatorname{dist}(Y,P')$, and that $g_P(Y)=0$ if and only if $Y\in P'$.

Exercise 4.2. Let $P$ be properly infinite. Construct isometries $v_1,\dots,v_h\in P$ with mutually orthogonal ranges from Corollary 13.5 of the projections lesson.

Exercise 4.3. Let $P=M_n(\mathbb C)\otimes1$ on $\mathbb C^n\otimes\mathbb C^k$. Compute $P'$, and show directly that $Y_0=\int uYu^*\,du$ is the conditional expectation $Y\mapsto(\operatorname{tr}\otimes\mathrm{id})(Y)$ onto $P'$ after the identification $P'=1\otimes M_k(\mathbb C)$.

Exercise 4.4. Show that the map $\Theta(y)=y(1+y^*y)^{-1/2}$ sends every element of a C*-algebra to its open unit ball, and that $c\mapsto c(1-c^*c)^{-1/2}$ is its inverse there.

5. Solutions

Solution 4.1. If $Z\in P'$, then $[Y,a]=[Y-Z,a]$, of norm at most $2\|Y-Z\|\|a\|$. If $g_P(Y)=0$, $Y$ commutes with the unit ball of $P$, hence with $P$.

Solution 4.2. Write $1=\sum_ne_n$ with $e_n\sim1$ mutually orthogonal, and let $v_n$ implement $1\sim e_n$: $v_n^*v_n=1$, $v_nv_n^*=e_n$. For $n\ne n'$, $v_n^*v_{n'}=v_n^*e_ne_{n'}v_{n'}=0$.

Solution 4.3. $P'=1\otimes M_k(\mathbb C)$. For $Y=a\otimes b$, the Haar average of $uau^*$ over $U(n)$ commutes with all unitaries, so it is a scalar, which equals $\operatorname{tr}(a)$ by taking traces; hence $Y_0=\operatorname{tr}(a)1\otimes b$. Extend linearly.

Solution 4.4. $\Theta(y)^*\Theta(y)$ equals $(1+y^*y)^{-1/2}y^*y(1+y^*y)^{-1/2}$, that is, $f(y^*y)$ with $f(x)=x/(1+x)$, whose values on the spectrum of $y^*y$ are at most $\|y\|^2/(1+\|y\|^2)<1$. For $\|c\|<1$, put $z=c(1-c^*c)^{-1/2}$. All functions of $c^*c$ commute, so $z^*z=c^*c(1-c^*c)^{-1}$ and $1+z^*z=(1-c^*c)^{-1}$, hence $\Theta(z)=z(1-c^*c)^{1/2}=c$; the other composition is checked in the same way with $y^*y$.

References

[OpenAI-288] OpenAI, Kadison's similarity theorem through uniform derivation estimates, preprint, 23 September 2026, Section 7. https://github.com/openai/math/tree/main/preprints/Kadisons-similarity-theorem-through-uniform-derivation-estimates-September-23-2026