Invariant tensors and the word structure

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

This lesson and the next two prove the free corner estimate, the central technical result of [OpenAI-288]. A II$_1$ factor $D$ generated by increasing matrix algebras sits freely next to a finite algebra $A_0$. An operator $G$ on the Hilbert space of the algebra they generate commutes with left multiplication by $D$, and its commutators with the elements of a small corner $e\mathcal Le$ are small. The estimate says that $G$ then almost commutes with every trace-zero symmetry $s\in A_0$ on $L^2(A_0)$, with a constant independent of the size of the corner.

The first step, carried out here, determines the shape of such an operator after averaging over the unitary groups of the matrix algebras. Invariant theory for the unitary groups, in the form of Schur–Weyl duality, writes the averaged operator on each word level in terms of permutations of tensor factors. Two size estimates remove all permutations except those that match source letters with test letters. The outcome is the word structure: on a word with a repeated centred letter from $D$, the operator acts only on the letters from $A_0$, by an operator $B_k$ that does not depend on the repeated letter.

We use: Words in free subalgebras (Theorem 1.3, Lemmas 2.1 and 2.2, Proposition 3.1); Schur–Weyl duality, Theorem 3.1 of its lesson, which says that on $V^{\otimes p}$ the algebra generated by $GL(V)$ and the algebra spanned by the permutations are each other's commutants; the commutation theorem $\{L_x\}'=\{R_x\}$ for a finite trace, Theorem 3.1 of Integration for a trace, and the duality $|\tau(ab)|\le\|b\|\,\|a\|_1$ of Section 1 there; Haar unitaries in II$_1$ factors and comparison of projections, Lemma 2.3 and Corollary 2.4 of Haar unitaries, freeness and tracial ultraproducts, with Proposition 14.2 of Projections and types (equivalent projections of a finite algebra are unitarily equivalent); and a left-invariant Haar probability measure on a compact group, Theorem 8.3 of Haar measure on locally compact groups.

1. The setting and the free corner estimate

Standing hypotheses (for this lesson and the next two). $D$ is a II$_1$ factor generated, as a von Neumann algebra, by an increasing sequence $D_1\subseteq D_2\subseteq\cdots$ of unital subalgebras $D_j\cong M_{m_j}(\mathbb C)$ with $m_j\to\infty$. $\mathcal L$ is a von Neumann algebra with a faithful normal tracial state $\tau$, generated by $D$ and a subalgebra $A_0$, and $D$ and $A_0$ are free. We write $E=L^2(D)$, $E^\circ=E\ominus\mathbb C1$, $B=L^2(A_0)$, $B^\circ=B\ominus\mathbb C1$, and $H=L^2(\mathcal L)$.

The restriction of $\tau$ to each $D_j$ is its normalized trace, the only tracial state of a full matrix algebra.

Theorem 1.1 (free corner estimate). Let $t=1/N$ for an integer $N\ge2$, let $G\in B(H)$ commute with $L_D=\{L_d:d\in D\}$, and let $\varepsilon\ge0$ be such that $\|[G,L_x]\|\le\varepsilon\|x\|\tag{1.1}$ for every projection $e\in D$ with $\tau(e)=t$ and every $x\in e\mathcal Le$. If $s=s^*=s^{-1}\in A_0$ and $\tau(s)=0$, then $\begin{gathered} \big\|P_B[G,L_s]|_B\big\|\le c_1\varepsilon,\\ c_1=3\pi^2(1+4c_0), \end{gathered}\tag{1.2}$ with $c_0=4\sqrt2$ the constant of Theorem 4.2 of Row, column and cyclic estimates. (The preprint [OpenAI-288] states the estimate with the larger constant $6\pi^2(1+4c_0)$ and a larger $c_0$; the size of the constant plays no role.)

The proof occupies this lesson and the next two and ends in Section 4 of The Hankel test. Two projections of $D$ with trace $t$ are unitarily conjugate in $D$ (Lemma 2.3 of the Haar unitary lesson and Proposition 14.2 of the projections lesson), and $G$ commutes with $L_v$ for unitaries $v\in D$; so (1.1) for one projection of trace $t$ implies it for all. For $x\in e\mathcal Le$, the commutator $[G,L_x]$ vanishes on $(1-e)H$ and takes values in $eH$, because $G$ commutes with $L_e$.

2. Factoriality

Lemma 2.1. $D'\cap\mathcal L=\mathbb C1$. In particular $\mathcal L$ is a II$_1$ factor.

Proof. By Theorem 1.3 of the free words lesson, $H\ominus E$ is spanned by the words containing at least one letter from $A_0^\circ$. Group the letters of such a word as $d\,m\,d'$, where $d$ is the part before the first $A_0$-letter and $d'$ the part after the last one (each a centred $D$-letter or absent), and $m$ starts and ends with an $A_0$-letter. Writing an arbitrary element of $D$ as scalar plus centred part, the map $d\otimes m\otimes d'\mapsto dmd'$ gives a unitary $\begin{gathered} H\ominus E\cong E\otimes\mathcal V\otimes E,\\ \mathcal V=\bigoplus_{n\ge1}B^\circ\otimes(E^\circ\otimes B^\circ)^{\otimes(n-1)}, \end{gathered}$ by the inner product formula of Lemma 1.2 there; left multiplication by $D$ acts on the first factor and right multiplication on the last.

Let $v\in D$ be a Haar unitary (Corollary 2.4 of the Haar unitary lesson) and $u_n=v^n$. The vectors $v^n$, $n\in\mathbb Z$, are orthonormal in $E$, so for bounded $\xi,\eta\in D$, $\langle u_n\xi,\eta\rangle=\langle v^n,\eta\xi^*\rangle\to0$ and $\langle\xi u_n^*,\eta\rangle=\langle v^{-n},\xi^*\eta\rangle\to0$. By density and uniform boundedness, $L_{u_n}$ and $R_{u_n^*}$ tend weakly to $0$ on $E$, and $L_{u_n}R_{u_n^*}=L_{u_n}\otimes I\otimes R_{u_n^*}$ tends weakly to $0$ on $E\otimes\mathcal V\otimes E$: its matrix coefficients between elementary tensors tend to $0$, and the operators are uniformly bounded. A vector $\zeta\in H\ominus E$ with $d\zeta=\zeta d$ for all $d\in D$ satisfies $u_n\zeta u_n^*=\zeta$, so $\|\zeta\|^2=\langle L_{u_n}R_{u_n^*}\zeta,\zeta\rangle\to0$, and $\zeta=0$.

Let $z\in D'\cap\mathcal L$ and let $E_D$ be the trace-preserving conditional expectation onto $D$. By bimodularity $dE_D(z)=E_D(dz)=E_D(zd)=E_D(z)d$ for $d\in D$, so $E_D(z)\in D'\cap D=\mathbb C1$. The vector $z-E_D(z)\in H\ominus E$ commutes with $D$ in the above sense, so it is $0$, and $z\in\mathbb C1$. The centre of $\mathcal L$ lies in $D'\cap\mathcal L$, so $\mathcal L$ is a factor; it is finite, with a faithful tracial state, and infinite-dimensional, hence of type II$_1$. $\square$

3. Symmetrization

For a unitary $u\in D$, Proposition 3.1 of the free words lesson, with $\alpha_1=\operatorname{Ad}u$ on $D$ and $\alpha_2=\mathrm{id}$ on $A_0$, gives a trace-preserving automorphism $\theta_u$ of $\mathcal L$ and a unitary $W_u$ on $H$ satisfying (3.1): $\begin{aligned} &W_uL_xW_u^*=L_{\theta_u(x)},\\ &W_u\xi=\xi\quad(\xi\in B),\\ &W_u(x_1\otimes\cdots\otimes x_n)\\ &\quad=\theta_u(x_1)\otimes\cdots\otimes\theta_u(x_n). \end{aligned}$ Here $\theta_u=\operatorname{Ad}u$ on $D$ and $\theta_u=\mathrm{id}$ on $A_0$; also $W_uW_{u'}=W_{uu'}$.

Lemma 3.1 (symmetrization). In proving Theorem 1.1 we may assume, in addition, that $G$ commutes with $W_u$ for every $u\in\bigcup_j\mathcal U(D_j)$. The hypothesis (1.1) then also holds for $G^*$ in place of $G$.

Proof. For each $j$, the map $u\mapsto W_u$ is strongly continuous on the compact group $\mathcal U(D_j)$: on a bounded word, $u\mapsto W_uw$ is a product of norm-continuous functions of $u$, and the $W_u$ are unitary. Let $\mu_j$ be the Haar probability measure of $\mathcal U(D_j)$ and $G_j=\int W_uGW_u^*\,d\mu_j(u)$, a weak operator integral.

Let $\mathcal S$ be the set of operators $Y$ with $\|Y\|\le\|G\|$ that commute with $L_D$ and satisfy (1.1). It is convex and closed in the weak operator topology, since each condition is a family of inequalities $|\langle Y\xi,\eta\rangle|\le\cdots$ or equalities between matrix coefficients. Each $W_uGW_u^*$ lies in $\mathcal S$: it commutes with $W_uL_dW_u^*=L_{udu^*}$, $d\in D$; and for $x\in e\mathcal Le$, $[W_uGW_u^*,L_x]=W_u[G,L_{\theta_u^{-1}(x)}]W_u^*$ with $\theta_u^{-1}(x)\in e'\mathcal Le'$, $e'=u^*eu$ of trace $t$. Hence every $G_j$ lies in $\mathcal S$, and so does a weak operator cluster point $\widetilde G$ of $(G_j)$. For $u\in\mathcal U(D_i)$ and $j\ge i$, invariance of $\mu_j$ under left translation by $u\in\mathcal U(D_i)\subseteq\mathcal U(D_j)$ gives $W_uG_jW_u^*=G_j$; so $\widetilde G$ commutes with $W_u$.

The operators $W_u$ fix $B$ pointwise and commute with $L_s$ (as $\theta_u(s)=s$). So for $\xi,\eta\in B$, $\begin{aligned} &\langle[W_uGW_u^*,L_s]\xi,\eta\rangle\\ &\quad =\langle W_u[G,L_s]W_u^*\xi,\eta\rangle=\langle[G,L_s]\xi,\eta\rangle, \end{aligned}$ and this identity passes to the averages and to $\widetilde G$: the compression $P_B[\,\cdot\,,L_s]|_B$ takes the same value on $G$ and on $\widetilde G$. Finally $[G^*,L_x]=-[G,L_{x^*}]^*$, and $x^*\in e\mathcal Le$ when $x\in e\mathcal Le$. $\square$

From now on $G$ denotes an operator satisfying the hypotheses of Theorem 1.1 and commuting with all $W_u$, $u\in\bigcup_j\mathcal U(D_j)$.

4. Word levels

Put $J_k=(B^\circ)^{\otimes k}\otimes B$ for $k\ge0$, and $\mathcal H_k=E\otimes(E^\circ)^{\otimes k}\otimes J_k . \tag{4.1}$ The elementary tensor $d_0\otimes d_1\otimes\cdots\otimes d_k\otimes(a_1\otimes\cdots\otimes a_k\otimes z)$ represents the vector $d_0a_1d_1a_2\cdots a_kd_kz$, for bounded letters $d_0\in D$, $d_i\in D^\circ$, $a_i\in A_0^\circ$ and $z\in A_0$.

Lemma 4.1. $H=\bigoplus_{k\ge0}\mathcal H_k$, the identification being isometric. Left multiplication by $d\in D$ acts on $\mathcal H_k$ as $\lambda(d)\otimes I$ on the first factor, and $W_u$ acts as $C_u^{\otimes(k+1)}\otimes I_{J_k}$, where $C_u\xi=u\xi u^*$ on $E$ and on $E^\circ$.

Proof. Expand $d_0=\tau(d_0)1+d_0^\circ$ and $z=\tau(z)1+z^\circ$. The four resulting products are centred alternating words of four different types, and every centred alternating word arises exactly once in this way: $k$ is the number of its $D$-letters preceded by an $A_0$-letter. By Theorem 1.3 and Lemma 1.2 of the free words lesson the identification is isometric and exhausts $H$. The statement about $L_d$ is Lemma 2.1 there: $\bigoplus_k(E^\circ)^{\otimes k}\otimes J_k$ is the closed span of $1$ and the words beginning with an $A_0$-letter. The statement about $W_u$ is (3.1): $\theta_u$ conjugates the $D$-letters, including the scalar part of $d_0$, and fixes the $A_0$-letters. $\square$

5. Invariant multilinear forms on matrices

For a permutation $\pi$ of $\{1,\dots,p\}$ let $P_\pi$ be the operator on $(\mathbb C^m)^{\otimes p}$ that sends $w_1\otimes\cdots\otimes w_p$ to $w_{\pi^{-1}(1)}\otimes\cdots\otimes w_{\pi^{-1}(p)}$.

Lemma 5.1. The linear span of $\{u^{\otimes p}:u\in U(m)\}$ in $\operatorname{End}((\mathbb C^m)^{\otimes p})$ equals the linear span of $\{g^{\otimes p}:g\in GL_m(\mathbb C)\}$.

Proof. Let $\mathcal W$ be the first span; it is a finite-dimensional algebra, since $u^{\otimes p}v^{\otimes p}=(uv)^{\otimes p}$. For a Hermitian matrix $X$, the entire function $z\mapsto(e^{izX})^{\otimes p}$ takes values in $\mathcal W$ for real $z$. Every linear functional vanishing on $\mathcal W$ therefore vanishes on this entire function along the real line, hence everywhere; so $(e^{izX})^{\otimes p}\in\mathcal W$ for all complex $z$, and in particular $Q^{\otimes p}\in\mathcal W$ for every positive definite $Q=e^{X}$. An invertible $g$ is $uQ$ with $u$ unitary and $Q=|g|$ positive definite, so $g^{\otimes p}=u^{\otimes p}Q^{\otimes p}\in\mathcal W$. The other inclusion is clear. $\square$

Corollary 5.2. The operators on $(\mathbb C^m)^{\otimes p}$ commuting with every $u^{\otimes p}$, $u\in U(m)$, are the linear combinations of the $P_\pi$. If $m\ge p$, the $P_\pi$ are linearly independent.

Proof. By Lemma 5.1 these are the operators commuting with every $g^{\otimes p}$, $g\in GL_m(\mathbb C)$, which by Schur–Weyl duality form the algebra spanned by the $P_\pi$. If $m\ge p$, the vectors $P_\pi(e_1\otimes\cdots\otimes e_p)$, which equal $e_{\pi^{-1}(1)}\otimes\cdots\otimes e_{\pi^{-1}(p)}$, are distinct basis vectors. $\square$

Let $\operatorname{tr}$ be the normalized trace on $M_m(\mathbb C)$. For $\pi\in S_p$ define the cycle-trace form $\Phi_\pi$ by (5.1): $\begin{aligned} &\Phi_\pi(v_1,\dots,v_p)\\ &\quad=\prod_{(i_1\,i_2\,\cdots\,i_r)}\operatorname{tr}(v_{i_1}v_{i_2}\cdots v_{i_r}), \end{aligned}$ the product over the cycles of $\pi$, each written with $\pi(i_s)=i_{s+1}$ and $\pi(i_r)=i_1$. By the cyclicity of the trace the factor does not depend on where a cycle starts. Let $c(\pi)$ be the number of cycles.

Proposition 5.3. Let $m\ge p$ and let $F:M_m(\mathbb C)^p\to\mathbb C$ be multilinear with $F(uv_1u^*,\dots,uv_pu^*)=F(v_1,\dots,v_p)$ for all unitaries $u$. Then $F=\sum_{\pi\in S_p}\lambda_\pi\Phi_\pi$ with unique coefficients $\lambda_\pi$.

Proof. Identify $M_m(\mathbb C)^{\otimes p}$ with $\operatorname{End}((\mathbb C^m)^{\otimes p})$. The trace pairing is nondegenerate, so $F(v_1,\dots,v_p)=\operatorname{Tr}(Y\,v_1\otimes\cdots\otimes v_p)$ for a unique $Y$. Invariance says $\operatorname{Tr}(Yu^{\otimes p}Xu^{*\otimes p})=\operatorname{Tr}(YX)$ for all $X$, that is, $Y$ commutes with every $u^{\otimes p}$; by Corollary 5.2, $Y=\sum_\pi c_\pi P_\pi$. A direct computation in the tensor basis shows that $\operatorname{Tr}(P_\pi\,v_1\otimes\cdots\otimes v_p)$ equals $m^{c(\pi)}\Phi_{\pi^{-1}}(v_1,\dots,v_p)$: the trace is a sum over indices $k_1,\dots,k_p$ of $\prod_l(v_l)_{k_l,k_{\pi^{-1}(l)}}$, which factors along the cycles of $\pi^{-1}$. So $F$ is a combination of cycle-trace forms. Uniqueness follows from Lemma 5.4. $\square$

Lemma 5.4 (evaluation on matrix units). Let $m\ge p$, $\pi\in S_p$ and $v_i=e_{i,\pi(i)}$ ($1\le i\le p$), matrix units of $M_m(\mathbb C)$. Then $\Phi_\rho(v_1,\dots,v_p)=m^{-c(\pi)}$ if $\rho=\pi$, and $0$ if $\rho\ne\pi$.

Proof. Let $(i_1\cdots i_r)$ be a cycle of $\rho$. The product $e_{i_1,\pi(i_1)}e_{i_2,\pi(i_2)}\cdots e_{i_r,\pi(i_r)}$ is nonzero exactly when $\pi(i_s)=i_{s+1}=\rho(i_s)$ for $s<r$, and then equals $e_{i_1,\pi(i_r)}$, whose normalized trace is $1/m$ if $\pi(i_r)=i_1=\rho(i_r)$ and $0$ otherwise. So the factor of this cycle is $1/m$ when $\pi$ and $\rho$ agree on the cycle, and $0$ otherwise. $\square$

6. Invariant intertwiners

Theorem 6.1. Let $k,l\ge0$ and let $Y:E\otimes(E^\circ)^{\otimes k}\to E\otimes(E^\circ)^{\otimes l}$ be bounded, with $\begin{aligned} &Y(\lambda(d)\otimes I)=(\lambda(d)\otimes I)Y\quad(d\in D),\\ &YC_u^{\otimes(k+1)}\\ &\quad =C_u^{\otimes(l+1)}Y\quad\Big(u\in\bigcup_j\mathcal U(D_j)\Big). \end{aligned}$ If $k\ne l$, then $Y=0$. If $k=l$, there are scalars $a_\sigma$ ($\sigma\in S_k$) with $\begin{aligned} &Y=I_E\otimes\sum_{\sigma\in S_k}a_\sigma U_\sigma,\\ &U_\sigma(\xi_1\otimes\cdots\otimes\xi_k)\\ &\quad =\xi_{\sigma^{-1}(1)}\otimes\cdots\otimes\xi_{\sigma^{-1}(k)}. \end{aligned}$

Proof. For $x\in D$ put $x^\circ=x-\tau(x)1$. For $a,x_1,\dots,x_k,z_1,\dots,z_l\in D$ define $\begin{aligned} &F(a,x,z)=\big\langle Y(1\otimes x_1^\circ\otimes\cdots\otimes x_k^\circ),\\ &\qquad a^*\otimes(z_1^*)^\circ\otimes\cdots\otimes(z_l^*)^\circ\big\rangle . \end{aligned}$ It is complex multilinear, since $a\mapsto a^*$ and $z\mapsto(z^*)^\circ$ are conjugate linear and enter the conjugate linear slot. Note $F(a,x,z)=F(a,x^\circ,z^\circ)$.

Step 1: an endpoint bound. Fix the tail letters. The operator $T$ on $E$ with $\langle T\xi,\eta\rangle=\langle Y(\xi\otimes x^\circ),\eta\otimes(z^*)^\circ\rangle$ commutes with left multiplication by $D$, so by the commutation theorem $T=R_b$ for some $b\in D$ with $\|b\|=\|T\|\le\|Y\|\prod_i\|x_i\|_2\prod_j\|z_j\|_2$. Hence $F(a,x,z)=\langle b,a^*\rangle=\tau(ab)$, and the duality between $D$ and $L^1(D)$ gives $\begin{aligned} &|F(a,x,z)|\le\|Y\|\,\|a\|_1\\ &\qquad\cdot\prod_{i=1}^k\|x_i\|_2\prod_{j=1}^l\|z_j\|_2 . \end{aligned} \tag{6.1}$ The same slice gives, for bounded $c,d\in D$, $\langle Y(d\otimes x^\circ),c\otimes(z^*)^\circ\rangle=\langle db,c\rangle$, which is $\tau(c^*db)=F(c^*d,x,z)$.

Step 2: finite stages. Put $p=1+k+l$ and restrict all variables to a stage $D_j\cong M_m(\mathbb C)$ with $m\ge p$. Since $C_u$ is unitary, fixes $1$ and commutes with centring, the intertwining hypothesis gives $F(uau^*,ux_iu^*,uz_ju^*)=F(a,x,z)$ for $u\in\mathcal U(D_j)$. Proposition 5.3 gives $F|_{D_j^p}=\sum_\pi\lambda_{\pi,j}\Phi_\pi$, where we list the variables as $(a,x_1,\dots,x_k,z_1,\dots,z_l)$ and the traces are those of $\tau$. The forms $\Phi_\pi$ on $D_{j+1}$ restrict to the same forms on $D_j$, because $\tau$ restricts to normalized traces; uniqueness in Proposition 5.3 shows that $\lambda_\pi=\lambda_{\pi,j}$ does not depend on $j$ once $m_j\ge p$.

Step 3: long cycles vanish. Take matrix units $v$ as in Lemma 5.4 in a stage of size $m\ge p$. Then $F(v)=\lambda_\pi m^{-c(\pi)}$. Every matrix unit has $\|e_{il}\|_1=1/m$ and $\|e_{il}\|_2=m^{-1/2}$, so (6.1) gives $|\lambda_\pi|m^{-c(\pi)}\le\|Y\|m^{-1-(k+l)/2}$. Letting $m\to\infty$, $\begin{gathered} \lambda_\pi=0\\ \text{unless}\qquad c(\pi)\ge1+\tfrac{k+l}2 . \end{gathered}\tag{6.2}$

Step 4: centred tails. Let the $k+l$ tail variables be centred. A form $\Phi_\pi$ in which some tail variable forms a cycle by itself vanishes. If no tail variable is alone and the cycle of the first variable $a$ has length $r$, the other cycles contain the remaining $k+l-(r-1)$ tail variables in groups of at least two, so $c(\pi)\le1+\frac{k+l-r+1}2$. By (6.2), $r=1$ and all tail cycles have length two. Hence, for all $a$ and centred tails in $\bigcup_jD_j$, the identity (6.3) holds: $\begin{aligned} &F(a,x,z)\\ &\quad=\tau(a)\sum_{\mathcal P}\lambda_{\mathcal P}\prod_{\{i,i'\}\in\mathcal P}\tau(y_iy_{i'}), \end{aligned}$ where $\mathcal P$ runs over the pairings of the tail slots and $(y_1,\dots,y_{k+l})=(x_1,\dots,x_k,z_1,\dots,z_l)$; there are no pairings when $k+l$ is odd. With $a=c^*d$ and the conjugate linear involution $\mathcal J\xi=\xi^*$ of $E$, the inner product of $Y(d\otimes\xi_1\otimes\cdots\otimes\xi_k)$ and $c\otimes w_1\otimes\cdots\otimes w_l$, with $\xi_i=x_i^\circ$ and $w_j=(z_j^*)^\circ$, equals $\langle d,c\rangle\sum_{\mathcal P}\lambda_{\mathcal P}\prod_{\mathcal P}\beta$, where $\beta=\langle\xi_i,\mathcal J\xi_{i'}\rangle$ for a pair of source slots, $\beta=\langle\xi_i,w_j\rangle$ for a source slot $i$ and a test slot $j$, and $\beta=\langle\mathcal Jw_j,w_{j'}\rangle$ for a pair of test slots. Both sides are continuous in each $L^2$-variable, so the formula holds for all $d,c\in E$ and all $\xi_i,w_j\in E^\circ$.

Step 5: no pairs within one side. Let $v\in D$ be a Haar unitary and $h_n=(v^n+v^{-n})/\sqrt2$ for $n\ge1$: these are self-adjoint, centred, and orthonormal, with $\mathcal Jh_n=h_n$ and $\tau(h_nh_{n'})=\delta_{nn'}$. Fix a pairing $\mathcal P$ with $r_s$ pairs of source slots and $r_t$ pairs of test slots, and give each pair its own infinite set of indices $n$, disjoint from the others. For a pair of a source and a test slot, put one fixed $h_n$ of its set in both slots. For a pair of source slots, sum the source tensor over $M$ members of its set, with the same $h_n$ in both slots; do this independently for all such pairs, and build the test tensor in the same way. Take $d=c=1$. In the formula of Step 4, a pairing $\mathcal P'\ne\mathcal P$ has a pair whose two slots carry members of different sets, so its product vanishes; the pairing $\mathcal P$ contributes $\lambda_{\mathcal P}$ for each of the $M^{r_s+r_t}$ terms. The source summands are orthonormal, and so are the test summands, so the two tensors have norms $M^{r_s/2}$ and $M^{r_t/2}$. Boundedness of $Y$ gives $|\lambda_{\mathcal P}|M^{r_s+r_t}\le\|Y\|M^{(r_s+r_t)/2}$, and $M\to\infty$ gives $\lambda_{\mathcal P}=0$ whenever $r_s+r_t>0$.

Step 6: conclusion. The surviving pairings join every source slot with a test slot, which requires $k=l$; if $k\ne l$, all coefficients vanish and $Y=0$. If $k=l$, these pairings are $\{(i,\sigma(i))\}$ for $\sigma\in S_k$, and with $a_\sigma=\lambda_{\mathcal P}$ the coefficient is $\begin{aligned} &\langle d,c\rangle\sum_\sigma a_\sigma\prod_{i=1}^k\langle\xi_i,w_{\sigma(i)}\rangle\\ &\quad =\Big\langle\Big(I_E\otimes\sum_\sigma a_\sigma U_\sigma\Big)(d\otimes\xi),\,c\otimes w\Big\rangle . \end{aligned}$ Elementary tensors are total, which proves the identity. $\square$

7. The word structure

Theorem 7.1 (word structure). The operator $G$ maps every level $\mathcal H_k$ into itself. There are operators $B_k\in B(J_k)$ with $\|B_k\|\le\|G\|$ such that, for every $d_0\in E$, every $x\in E^\circ$ and every $\alpha\in J_k$, $\begin{aligned} &G(d_0\otimes x^{\otimes k}\otimes\alpha)\\ &\quad =d_0\otimes x^{\otimes k}\otimes B_k\alpha,\\ &G^*(d_0\otimes x^{\otimes k}\otimes\alpha)\\ &\quad =d_0\otimes x^{\otimes k}\otimes B_k^*\alpha . \end{aligned}\tag{7.1}$ In particular $G|_{\mathcal H_0}=I_E\otimes B_0$, $G$ maps $B=1\otimes B\subseteq\mathcal H_0$ into itself, and $G|_B=B_0$.

Proof. Let $P_k$ be the projection onto $\mathcal H_k$; it commutes with $L_D$ and with every $W_u$ (Lemma 4.1). For $\alpha\in J_k$ and $\beta\in J_l$ define $Y_{\alpha,\beta}:E\otimes(E^\circ)^{\otimes k}\to E\otimes(E^\circ)^{\otimes l}$ by $\begin{gathered} \langle Y_{\alpha,\beta}\xi,\eta\rangle=\langle P_lGP_k(\xi\otimes\alpha),\eta\otimes\beta\rangle,\\ \|Y_{\alpha,\beta}\|\le\|G\|\,\|\alpha\|\,\|\beta\| . \end{gathered}$ By Lemma 4.1 it satisfies the hypotheses of Theorem 6.1. For $k\ne l$ all these slices vanish, so $P_lGP_k=0$ and the levels reduce $G$. For $k=l$, $Y_{\alpha,\beta}=I_E\otimes\sum_\sigma a_\sigma(\alpha,\beta)U_\sigma$.

Fix a unit vector $h\in E^\circ$, let $V_k\alpha=1\otimes h^{\otimes k}\otimes\alpha$, an isometry of $J_k$ into $\mathcal H_k$, and put $B_k=V_k^*GV_k$; then $\|B_k\|\le\|G\|$ and $\langle B_k\alpha,\beta\rangle=\langle Y_{\alpha,\beta}(1\otimes h^{\otimes k}),1\otimes h^{\otimes k}\rangle$, which is $\sum_\sigma a_\sigma(\alpha,\beta)$, because every $U_\sigma$ fixes $h^{\otimes k}$. For $d_0\in E$, $x\in E^\circ$ and $\eta\in E\otimes(E^\circ)^{\otimes k}$, $\begin{aligned} &\langle G(d_0\otimes x^{\otimes k}\otimes\alpha),\eta\otimes\beta\rangle\\ &\quad =\Big(\sum_\sigma a_\sigma(\alpha,\beta)\Big)\langle d_0\otimes x^{\otimes k},\eta\rangle\\ &\quad =\langle d_0\otimes x^{\otimes k}\otimes B_k\alpha,\eta\otimes\beta\rangle, \end{aligned}$ again because $U_\sigma x^{\otimes k}=x^{\otimes k}$. Such vectors $\eta\otimes\beta$ are total in $\mathcal H_k$, and $G$ preserves $\mathcal H_k$, which proves the first identity in (7.1). The operator $G^*$ also commutes with $L_D$ and every $W_u$, so the same argument applies to it, with $V_k^*G^*V_k=B_k^*$. At level zero, $J_0=B$ and (7.1) reads $G(d_0\otimes\alpha)=d_0\otimes B_0\alpha$. $\square$

Corollary 7.2. For every $s\in A_0$, $P_B[G,L_s]|_B=[B_0,L_s|_B]$.

Proof. $L_s$ preserves $B=L^2(A_0)$, and so does $G$, acting there as $B_0$. $\square$

8. Exercises

Exercise 8.1. For $p=2$ and $m\ge2$, write down the two cycle-trace forms and show directly that every bilinear form $F$ on $M_m(\mathbb C)$ invariant under simultaneous unitary conjugation is $\lambda_1\operatorname{tr}(v_1)\operatorname{tr}(v_2)+\lambda_2\operatorname{tr}(v_1v_2)$.

Exercise 8.2. Show that on $(\mathbb C^2)^{\otimes3}$ the operator $\sum_{\pi\in S_3}\operatorname{sgn}(\pi)P_\pi$ is zero, so that the permutation operators are linearly dependent when $m<p$.

Exercise 8.3. Let $v$ be a Haar unitary of $D$ and $h_n=(v^n+v^{-n})/\sqrt2$. Verify that the $h_n$, $n\ge1$, are self-adjoint, centred and orthonormal.

Exercise 8.4. Show that $Y=I_E\otimes U_\sigma$ on $E\otimes(E^\circ)^{\otimes k}$ satisfies the hypotheses of Theorem 6.1.

Exercise 8.5. Show that the operator $B_k$ of Theorem 7.1 does not depend on the unit vector $h\in E^\circ$ used in its definition.

9. Solutions

Solution 8.1. The permutations of $\{1,2\}$ are the identity, with two cycles, and the transposition, with one; so $\Phi_{\rm id}=\operatorname{tr}(v_1)\operatorname{tr}(v_2)$ and $\Phi_{(12)}=\operatorname{tr}(v_1v_2)$. Write $F(v_1,v_2)=\operatorname{Tr}(Y\,v_1\otimes v_2)$ with $Y$ commuting with all $u\otimes u$. For $m\ge2$ this commutant is spanned by $1$ and the flip $P_{(12)}$ (Corollary 5.2), and $\operatorname{Tr}(v_1\otimes v_2)=m^2\operatorname{tr}(v_1)\operatorname{tr}(v_2)$, $\operatorname{Tr}(P_{(12)}\,v_1\otimes v_2)=\operatorname{Tr}(v_1v_2)$, which is $m\operatorname{tr}(v_1v_2)$; the last identity is the computation $\sum_{k,l}(v_1)_{kl}(v_2)_{lk}$.

Solution 8.2. The operator is $3!$ times the projection onto the antisymmetric tensors. An antisymmetric tensor in $(\mathbb C^2)^{\otimes3}$ is zero: its coefficient on $e_{i_1}\otimes e_{i_2}\otimes e_{i_3}$ changes sign under exchanging two positions, and two of the three indices coincide.

Solution 8.3. $h_n^*=h_n$. Since $\tau(v^k)=0$ for $k\ne0$, $\tau(h_n)=0$ and, for $n,n'\ge1$, $\tau(h_nh_{n'})$ is half the trace of $v^{n+n'}+v^{n-n'}+v^{n'-n}+v^{-n-n'}$, which is $\delta_{nn'}$; and $\langle h_n,h_{n'}\rangle=\tau(h_{n'}h_n)$.

Solution 8.4. $\lambda(d)\otimes I$ acts on the first factor, which $I_E\otimes U_\sigma$ leaves alone; and $C_u^{\otimes(k+1)}=C_u\otimes C_u^{\otimes k}$ commutes with $I_E\otimes U_\sigma$ because a permutation of tensor factors commutes with applying the same operator to every factor.

Solution 8.5. By the proof, $\langle B_k\alpha,\beta\rangle=\sum_\sigma a_\sigma(\alpha,\beta)$, and the coefficients $a_\sigma(\alpha,\beta)$ are defined by $Y_{\alpha,\beta}$, which does not involve $h$.

References

[OpenAI-288] OpenAI, Kadison's similarity theorem through uniform derivation estimates, preprint, 23 September 2026, Section 3. https://github.com/openai/math/tree/main/preprints/Kadisons-similarity-theorem-through-uniform-derivation-estimates-September-23-2026