Proof. Fix \(j\) and omit its superscript. By (18), revealing \(v_{t+1}\) does not update the old hidden-card law \(p_t\). Given \(\mathcal F_t\) and this direction, the matching edges incident to observed cards are known. Their next positions reveal exactly the fresh coins on those edges. Those coins are independent of the old hidden position, so their revelation also leaves \(p_t\) unchanged; all other edge coins remain independent and fair.

An edge containing two positions of \(D_t\) averages their two masses. An edge containing exactly one position of \(D_t\) transports its mass to the unique position on that edge unoccupied by observed cards after the switch. Edges with no available position carry no hidden mass. These rules give the conditional vector at the exact field \(\mathcal F_{t+1}\) in (16). They also transport the uniform vector on \(D_t\) to the uniform vector on \(D_{t+1}\). Transport preserves squared error, while averaging two masses reduces it by \(\frac12(p_t(x)-p_t(y))^2\). This proves the exact identity.

For the expectation, split \(D_t=D^0\cup D^1\) across \(J_t\) and put \(z_x=p_t(x)-1/r_j\), \(u=|D^0|\), and \(w=|D^1|\). Since \(\sum z_x=0\), \[ \sum_{x\in D^0,y\in D^1}(z_x-z_y)^2 =w\sum_{D^0}z_x^2+u\sum_{D^1}z_y^2 +2\left(\sum_{D^0}z_x\right)^2 \ge\min(u,w)\sum_{D_t}z_x^2. \tag{21}\] Every cross pair has matching probability \(2/n\) in the hyperplane case. Combining this with the factor \(1/2\) in the exact decrement proves (20). In the proper-subspace case, every cross difference lies outside \(W_t\) and has probability \(1/(n-|W_t|)\ge1/n\); discard the other eligible pairs to obtain the denominator \(2n\). The calculation covers an empty side. The algebraic identity (21) holds for any real centered vector; a later use for another vector must also identify its averaging rule. ◻

Proof. For fixed \(\Lambda,\Xi\), apply (23). At index \(j\) use the coarser field \[\mathcal H_j =\sigma(\Lambda,\Xi,Z_T(c_1),\ldots,Z_T(c_{j-1})).\] The tower property and (24) give \(\Pr\{Z_T(c_j)=x\mid\mathcal H_j\} =\mathbb E[p_T^j(x)\mid\mathcal H_j]\). The set \(D_T^j\) is \(\mathcal H_j\)-measurable, so Jensen’s inequality for the \(\ell^1\) norm compares both vectors to the same uniform law \(U_{D_T^j}\). Cauchy–Schwarz gives \[\|p_T^j-U_{D_T^j}\|_{\mathrm{TV}} \le\frac12\sqrt{(n-j+1)\Delta_T^j}.\] Consequently, almost surely in the terminal data and initial data, \[ \|\mathcal L(Z_T(C)\mid\Lambda,\Xi)-\pi_q\|_{\mathrm{TV}} \le\frac12\sum_{j=1}^q \mathbb E\!\left[\sqrt{(n-j+1)\Delta_T^j}\,\middle|\,\Lambda,\Xi\right]. \tag{26}\] Averaging and applying Cauchy–Schwarz once more proves the lemma. ◻

Proof. Use \(v_i=e_i\) for \(1\le i\le d\). For \(d\le t<T\) put \[ J_t=\mathop{\mathrm{span}}\{v_{t-d+2},\ldots,v_t\} \tag{30}\] and choose \(v_{t+1}\) uniformly in \(V\setminus J_t\), with fresh direction randomness. Every \(d\) successive directions form a basis. Conditional on the complete schedule, use independent fair switches to form \(Z\). Independently choose \(\mathcal M\) uniformly among labeled equal-size partitions. For a fixed \(a\), let \(\Xi\) be this partition and let \(C\) be any deterministic ordering of \(V\setminus M_a\), with \(q=n-m\). This is the online setup above.

For each hidden index \(j\), the available set contains the positions of all omitted labels: \[ Z_t(M_a)\subseteq D_t^j. \tag{31}\] Write \(D_t^{j,0}=D_t^j\cap J_t\) and \(D_t^{j,1}=D_t^j\setminus J_t\). If \(J_t^0=J_t\) and \(J_t^1=V\setminus J_t\), then \[\left\{\min_i|D_t^j\cap J_t^i|<m/4\right\} \subseteq \left\{\min_i|Z_t(M_a)\cap J_t^i|<m/4\right\}.\] Fix the direction and switch history through time \(t\), but not \(\mathcal M\). This fixes \(Z_t\) and \(J_t\), while \(Z_t(M_a)\) is a uniform \(m\)-subset of \(V\).

For a uniform \(m\)-subset and a fixed half of \(V\), let \(X\) be its count in that half. View the subset as an ordered sample without replacement. A specified set of \(\ell\) sampled indices all lie in the half with probability at most \(2^{-\ell}\). If \(I_i\) are the membership indicators, expansion gives \[\mathbb E3^X=\mathbb E\prod_{i=1}^m(1+2I_i) \le\sum_{A\subseteq\{1,\ldots,m\}}2^{|A|}2^{-|A|}=2^m.\] If one half contains fewer than \(m/4\) sampled points, the other contains more than \(3m/4\). Markov’s inequality and a union bound therefore yield, for \(d\le t<T\), \[ \Pr\{\min(|D_t^{j,0}|,|D_t^{j,1}|)<m/4\} \le2\frac{2^m}{3^{3m/4}}=b_n. \tag{32}\] This bound averages the independent partition and the shuffle; it is not a concentration assertion conditional on the observed paths.

Off the event in (32), the conditional loss in (20) is at least \(\alpha\Delta_t^j\). Its loss is always nonnegative. Applying (22) gives \[ \mathbb E\Delta_{t+1}^j \le(1-\alpha)\mathbb E\Delta_t^j+\alpha b_n. \tag{33}\] Starting with \(\Delta_d^j\le1\) and iterating for \(T-d=16kd\) steps, we obtain, uniformly in \(a,j\), \[ \mathbb E\Delta_T^j\le(1-\alpha)^{T-d}+b_n \le e^{-4d}+b_n. \tag{34}\]

Let \(\widetilde\delta_a(\mathbf v,\mathcal M)\) be the auxiliary tuple distance conditional on the complete schedule and partition. Use Lemma 8 with \(\Lambda=\mathbf v\). Its terminal identification is automatic, since \(\mathcal F_T^j\) already contains the complete schedule and predecessor paths. Equations (25) and (34) give \[ \mathbb E_{\mathbf v,\mathcal M} \widetilde\delta_a(\mathbf v,\mathcal M) \le\frac{n^{3/2}}2\sqrt{e^{-4d}+b_n} =\varepsilon_n. \tag{35}\] The schedule remains inside the distance on the left.

For every fixed schedule and partition, choose \(\boldsymbol\iota=(1,\ldots,d)\) in Lemma 4. The first directions are \(v_j=e_j\), so \(L_0=I\). Corollary 5, followed by the terminal rotation, therefore gives the same-list identity \[ \widetilde\delta_a(\mathbf v,\mathcal M)=\delta_a(\mathcal M). \tag{36}\] The frame maps enter only this fixed-schedule endpoint comparison. Summing (35) over \(a\) now gives \(\mathbb E_{\mathcal M}\sum_a\delta_a(\mathcal M)\le k\varepsilon_n\). At least one deterministic partition attains this average bound. Finally, \(n^{3/2}e^{-2d}=\exp((\tfrac32\log2-2)d)\to0\), while \(2/3^{3/4}<1\) makes \(b_n\) exponentially small in \(n\). Thus \(\varepsilon_n\to0\). ◻

Proof. For a variable \(Z\) in an interval of length \(b\), put \(\psi(s)=\log\mathbb Ee^{sZ}\). Its second derivative is the variance under the exponentially tilted law, at most \(b^2/4\). Integrating from \(\psi(0)=0\) and \(\psi'(0)=\mathbb EZ\) gives \(\mathbb Ee^{s(Z-\mathbb EZ)}\le e^{s^2b^2/8}\). Multiplication and minimization in \(s\) prove the independent-variable bound. The conditional version, with interval length \(2c_j\), gives \(\mathbb Ee^{s(M_r-M_0)}\le e^{s^2\sum_jc_j^2/2}\) and the martingale bound. These are the exponential-moment arguments underlying the inequalities of Hoeffding and Azuma [10, 3].

Expose a uniform subset in a uniform random order and form the Doob martingale for its final half-count. The expected final count changes by at most one at each draw: completions following two possible next draws can be coupled by exchanging those points. The initial expectation is \(m/2\). Use \(r=m\), \(c_j=1\), and \(a=m/4\). ◻

Proof. Fix the complete direction schedule. The direction \(v_{s-d}\) crosses the two cosets of \(W_s\), because the \(d\) directions starting at \(s-d\) form a basis. The intervening directions lie in \(W_s\) and preserve each coset.

Condition on the observed configuration immediately before step \(s-d\). Let \(a\) be the number of matching pairs with two available positions and \(b\) the number with one, so \(m=2a+b\). Immediately after this crossing switch, the available count on either chosen side is \(a+\operatorname{Bin}(b,1/2)\), using \(b\) independent fair coins. Its mean is \(m/2\). If \(b>0\), Lemma 16 bounds a deviation of at least \(m/4\) by \(2\exp(-2(m/4)^2/b)\le2e^{-m/8}\); if \(b=0\) the count is exactly \(m/2\). The intervening switches preserve the count. Remove the conditioning to obtain the claim. ◻

Proof. For a successive hidden-card problem, its number \(m\) of available positions is at least \(N/L\). Off the event in (61), Lemma [shear:lem:energy] gives conditional expected loss at least \(\gamma\Delta_t\), where \(\gamma=1/(4L)\). The loss is always nonnegative. Equation (22) therefore gives, for \(t\ge d\), \[ \mathbb E\Delta_{t+1}\le(1-\gamma)\mathbb E\Delta_t +2\gamma e^{-N/(8L)}. \tag{66}\] Starting from \(\Delta_d\le1\) yields \(\mathbb E\Delta_T\le(1-\gamma)^{T-d}+2e^{-N/(8L)}\). Apply Lemma 8 with terminal data \(\Lambda=\mathbf v\). For every complete schedule, the first basis is \((e_{\iota_1},\ldots,e_{\iota_d})\), so \(L_0=I\) in Lemma 4. The same-list equality (14) transfers the conditional tuple bound to \(X^{\boldsymbol\iota}\), giving (62).

Each switch is an involution. Reversing the independent factors therefore identifies \((X^{\boldsymbol\iota}_T)^{-1}\) with a coordinate chain whose first axis is \(i(T)\) and whose periodic order is reversed. This is another permitted ordering, so the same estimate applies. The relation \(Y_T=R^TX_T\) transfers the forward physical law by an output bijection. Its inverse is \(X_T^{-1}R^{-T}\); the input bijection only changes the fixed list, and the bound is uniform in that list.

For (63), weaken the additive term in (66) to \(2e^{-N/64}\) and sum the geometric series, giving the factor \(64\). At \(257d\) keep the sharper recurrence and use \((31/32)^{256d}\le e^{-8d}\). More generally, at \(T=(1+64L)d\), the weakened additive term \(2e^{-N/(8L)}\) gives \(e^{-16d}+8Le^{-N/(8L)}\), yielding (65). These rates dominate the factor \(N^3\) in the squared tuple bound, and \(e^{-16d}=o(N^{-4})\). ◻

Proof. The map \(S_s\) is an involution in \(C_{e_{i(s)}}\), because \(b_se_{i(s)}=0\). The increment of \(Z\) is \[A_{s-1}(S_sh_s)A_{s-1}^{-1}.\] For fixed \(\Lambda\), the factors \(S_sh_s\) are independent and uniform in their coordinate matching groups. Conjugation gives the asserted product-switch law, including the joint prefix identity (67).

Extend \(i(s)\) periodically beyond \(T\) and define the deterministic next-read time of a column by \[\tau(s,j)=\min\{t>s:i(t)=j\}\qquad(j\ne i(s)).\] Right multiplication by \(S_s\) leaves the selected column unchanged and adds \(b_{sj}v_s\) to column \(j\). Thus the direction at a column’s first read, for \(t\le d\), is \[v_t=e_{i(t)}+\sum_{1\le u<t}b_{u,i(t)}v_u.\] These first \(d\) directions form a triangular basis. At later reads, \[ v_t=v_{t-d}+\sum_{t-d<u<t}b_{u,i(t)}v_u\qquad(t>d). \tag{68}\] In either display the coefficients used at time \(t\) are exactly the entries whose next-read time is \(t\). Inductively, all earlier directions are functions only of entries with next-read time less than \(t\). The batch read at \(t\) is therefore independent and fair conditional on those directions. The recurrence also shows inductively that every consecutive \(d\) directions form a basis. For \(t>d\), its fresh \(d-1\) coefficients make \(v_t\) uniform on \(v_{t-d}+W_t=V\setminus W_t\).

Finally, (67) says that the entire virtual path prefix and \(\Lambda\) are conditionally independent given the direction prefix. Hence observing any portion of that path prefix cannot reveal the fresh batch that determines the next direction. ◻

Proof. Use the direct direction process with standard first basis, sampled independently of \(B\), and take \(\Xi=B\) in the online construction. At every time the available set for a hidden card contains \(Z_t(B)\). Fix the full direction schedule and the full permutation \(Z_t\), but not \(B\). The set \(Z_t(B)\) is then a uniform \(b\)-subset, and each coset of the next-direction hyperplane has size \(N/2\).

Its count in a specified half has the distribution of the number of ones among the first \(b\) of \(N\) independent fair bits, conditional on their total being \(N/2\). This conditioning event has probability at least \(1/(N+1)\), because the central binomial coefficient is maximal. For the unconditioned first-\(b\) count \(J\), \[\Pr\{J\le b/4\} \le2^{b/4}\mathbb E2^{-J} =\bigl(2^{1/4}3/4\bigr)^b.\] The inclusion of \(Z_t(B)\) in the available set and a union bound give \(p_{\mathrm{bad}}=2(N+1)\theta^b\) as an upper bound for the probability that either available half has fewer than \(b/4\) positions. This bound is unconditional in the observed paths.

Off this event the conditional expected loss is at least \(\Delta_t/32\). Equations (22) and (20) give \[\mathbb E\Delta_{t+1}\le(31/32)\mathbb E\Delta_t+p_{\mathrm{bad}}/32, \qquad \mathbb E\Delta_T\le e^{-8d}+p_{\mathrm{bad}}.\] Use Lemma 8 with \(\Lambda=\mathbf v\). For each fixed schedule and \(B\), the standard first basis gives \(L_0=I\), so the conditional auxiliary distance equals the distance for \(X_T\) on that same complement. Averaging over \(B\) proves the bound. ◻

Proof. Use the direct process with the standard first basis and take \(\Xi=C\). For a hidden label at index \(j\), let \(h=N-j+1\ge m\) be the number of available positions. Fix the complete direction and switch history, but not \(C\). At each fixed time, the preceding positions form a uniform ordered sample, so the available set is a uniform \(h\)-subset. The next-direction hyperplane is fixed under this conditioning.

If \(X\) is the count of a uniform \(h\)-subset in a fixed half, expand the product of \(1+I_i\) over an ordered sample without replacement. Any specified \(\ell\) sample indices lie in that half with probability at most \(2^{-\ell}\). Thus \[ \mathbb E2^X\le(3/2)^h,\qquad \Pr\{\min(X,h-X)<h/4\}\le2\vartheta^h. \tag{74}\] The second inequality follows by applying Markov’s inequality to the count in each half above \(3h/4\). It is used only after averaging out the input list, not conditional on observed paths.

On the balanced event, the conditional expected loss is at least \(h\Delta_s/(4N)\). With \(\gamma_j=h/(4N)\), the unconditional recurrence and its iteration give \[\mathbb E\Delta_{s+1}\le(1-\gamma_j)\mathbb E\Delta_s +2\gamma_j\vartheta^h,\qquad \mathbb E\Delta_t\le(1-1/(4b))^{t-d}+2\vartheta^m.\] Lemma 8, with \(\Lambda=\mathbf v\), bounds the average distance conditional on the full schedule and list. For every fixed schedule the first basis is standard, so the same-list frame equality transfers that distance to \(X_t\); the terminal rotation transfers it to the physical shuffle. This proves (73). A uniform starting deck sends any fixed ordered label list to a uniform input tuple, which proves the last assertion. ◻

Proof. The respective contracting times are \(512d\), \(2048d\), and \(32bd\), with energy factors at most \(e^{-8d}\), \(e^{-32d}\), and \(e^{-8d}\). Use \(e^{-8d}\le N^{-8}\) for the first and \(N^{3/2}e^{-16d}=o(N^{-10})\) for the second. Exponential decay in \(N\) handles the other term in each case. ◻

Proof. For the hidden label at index \(j\), let \(h=N-j+1\ge N/8\). Conditional on the full direction and switch history but not on \(C\), its available set is a uniform \(h\)-subset. For its count in the fixed half \(J_t\), choose each point independently with probability \(h/N\) and condition on selecting exactly \(h\) points. That size has probability at least \(1/(N+1)\): it is a mode of the binomial size distribution, as comparison of consecutive probabilities shows. Before conditioning, the half-count has mean \(h/2\) and is a sum of \(N/2\) independent bounded variables. Lemma 16 gives \[ \Pr\{\min(|D_t^j\cap J_t|,|D_t^j\setminus J_t|)<h/4\} \le2(N+1)e^{-h^2/(4N)} \le\beta_N,\qquad \beta_N=2(N+1)e^{-N/256}. \tag{78}\] For \(h=N\) the count is deterministic and the bound is automatic.

The proper-subspace part of Lemma [shear:lem:energy] gives conditional expected loss at least \(h\Delta_t/(8N)\ge\Delta_t/64\) off this event. Hence, from time zero, \[\mathbb E\Delta_{t+1}\le(63/64)\mathbb E\Delta_t+\beta_N/64,\qquad \mathbb E\Delta_T\le e^{-T/64}+\beta_N =e^{-16d}+\beta_N=O(N^{-20}).\] Apply the terminal tuple lemma with \(\Lambda=\mathbf v\) and then the per-schedule averaged equality (76). This proves (77). ◻

Proof. Use the random initial basis process above. At a hidden index, the number \(h\) of available positions is at least \(N/b\). Conditional on the complete direction and switch history, and before sampling the list, this set is a uniform \(h\)-subset. Thus (74) bounds its imbalance across the chosen \(J_t\) by \(2\vartheta^h\).

On the balanced event the proper-subspace contraction gives loss at least \(\Delta_t/(8b)\). Using the weaker additive term \(2\vartheta^{N/b}\) in the unconditional recurrence yields \[\mathbb E\Delta_T\le(1-1/(8b))^T+2T\vartheta^{N/b}.\] The terminal tuple comparison and then (76) give \[\begin{align*} \mathbb E_{\mathbf v,C} \|\mathcal L(Z_T(C)\mid\mathbf v,C)-\pi_q\|_{\mathrm{TV}} &=\mathbb E_C\|\mathcal L(X_T(C)\mid C)-\pi_q\|_{\mathrm{TV}}\\ &\le\frac12N^{3/2} \left((1-1/(8b))^T+2T\vartheta^{N/b}\right)^{1/2}. \tag{79}\end{align*}\] Here \((1-1/(8b))^T\le e^{-4d}\), so the right side tends to zero. Since \(T\) is a multiple of \(d\), \(X_T=Y_T\) in the rotating-frame coupling, which proves the physical statement. ◻

Proof. The degree assertions follow from [22]. The same companion gives separate proofs at powers \(1,2,10\) in Lemmas , , and ; these also justify the fixed-power sums used below. The two modifications, including their common-measure and mass claims, follow from [22]. The normalization assertion for \(f\) follows from Lemma 29 below; dividing the exact cap by \(z\) gives the normalized cap. ◻

Proof of Proposition 28. Write \(A=\widehat f(\rho)\) and abbreviate \(S_i(D)\) to \(S_i\). If \(z=0\), then \(f=0\) and every assertion is immediate. Assume \(z>0\).

Proof. The regular-character expansion in Lemma 14, with coefficients \(a(g)\overline{a(h)}\), proves (91). Apply it to \(\xi-mU_G\) and \(\xi-U_G\). Their trivial coefficients are respectively \(0\) and \(m-1\), and their nontrivial coefficients are those of \(\xi\). Cauchy–Schwarz gives (93). If \(\xi\le\mu\), the nonnegative difference has mass \(1-m\). Also \(\sum_g|\xi(g)/m-\xi(g)|=1-m\); the triangle inequality and the factor \(1/2\) in TV prove the normalization claim. ◻

Proof. Let \(T=256d\) and \(\mu=q_d^{*T}\). For \(d\ge3\), partition the labels into eight sets \(M_i\) of size \(N/8\). The fixed-list estimate (63) bounds each complementary image-tuple error by \[\delta_N=\frac{N^{3/2}}2 \left((31/32)^{255d}+64e^{-N/64}\right)^{1/2}=o(1).\] Exact trimming by Proposition 27 gives \(0\le f\le\mu\) with mass \(z\), where \(1-z\le8\delta_N=o(1)\), and cap one on all eight coset systems.

For this covering partition, the companion isotypic estimate [22], with inverse-degree parameter \(1\), gives \(\|\widehat f(\rho)\|_{\mathrm{HS}}^2\le8zC_1D^{-5/8}\le8C_1D^{-5/8}\). Consequently \[\sum_{\substack{\lambda\vdash N\\ \lambda\notin\{(N),(1^N)\}}} D_\lambda\|\widehat f(\rho_\lambda)^2\|_{\mathrm{HS}}^2 \le(8C_1)^2Z_{1/4}(N)=o(1),\] since \(1-2(5/8)=-1/4\) and [22] gives the inverse-quarter limit. The sign coefficient of \(f\) is at most \(1-z\) in magnitude, and the trivial coefficient of \(f^{*2}\) is \(z^2\). Lemma 29 therefore gives \[\left(\sum_g|f^{*2}(g)-U_G(g)|\right)^2 \le(1-z^2)^2+(1-z)^4+(8C_1)^2Z_{1/4}(N)=o(1).\] Adding back the missing mass \(1-z^2=o(1)\) proves \(\|\mu^{*2}-U_G\|_{\mathrm{TV}}=o(1)\). The two physical blocks take \(2T=512d\) shuffles, and (2) gives uniformity over starts. ◻

Proof. For \(T=256d\), let \(\mu=q_d^{*T}\) and use eight equal label parts. The error \(\delta_N\) in the preceding proof bounds each complementary marginal. Deleting heavy cosets loses \(\eta\le16\delta_N=o(1)\) and produces a probability \(\nu\) with cap \(K\le4\) for large \(d\). Equation (84) with \(r=8\) gives \[\|\widehat\nu(\rho)\|_{\mathrm{op}} \le C D^{-1/4},\qquad C=\sqrt{32C_2}.\] For eight factors the nonsign, nontrivial Plancherel contribution is at most \[C^{16}\sum_{\substack{\lambda\vdash N\\ \lambda\notin\{(N),(1^N)\}}} D_\lambda^{\,2-16/4}=C^{16}Z_2(N)=o(1).\] The sign contribution is at most \((2\eta)^{16}\), and the trivial coefficient is one. Thus \(\nu^{*8}\) tends to uniform in TV. Restoring the true factors costs at most \(8\eta=o(1)\), and \(8T=2048d\) physical shuffles.

For the fixed-list subprobability route take \(T=257d\). Equation (64) makes the summed error on any partition into eight equal parts \(o(1)\). Exact trimming gives \(f\le\mu=q_d^{*T}\) of mass \(z=1-o(1)\) and cap one. The orbit bound is \(\|\widehat f(\rho)\|_{\mathrm{op}}\le\sqrt{8C_2}D^{-1/4}\). In (94), the trivial term is \((z^8-1)^2\), the sign term is at most \((1-z)^{16}\), and the remaining sum is at most \((8C_2)^8Z_2(N)\). These terms and the restored mass \(1-z^8\) all vanish. The eight time blocks take \(2056d\) physical shuffles.

For the random omitted set route, Proposition 21 and the partition averaging above give a partition into eight equal parts with summed error \(o(1)\) at \(T=257d\). Heavy-coset deletion loses \(\eta=o(1)\) and gives cap \(K=2/(1-\eta)\). The orbit bound is \(\sqrt{16C_2/(1-\eta)}D^{-1/4}\). The eight-factor nonsign sum is at most \([16C_2/(1-\eta)]^8Z_2(N)=o(1)\), the sign term is at most \((2\eta)^{16}\), and restoring the true factors costs \(8\eta=o(1)\). Again \(8T=2056d\). Equation (2) handles every start. ◻

Proof. The ordinary marginal assertion is Lemma 25, specifically (79). The partition averaging above selects 64 equal parts whose complementary errors sum to \(o(1)\). Exact trimming by Proposition 27 gives \(0\le f\le\mu_T\), of mass \(z=1-o(1)\), with cap one for all 64 coset systems. This is the stated domination of the complementary marginals.

Use the direct inverse-tenth input cited in Proposition 27 and the orbit estimate (82) with \(r=64\), \(u=10\), and \(z\le1\). It gives \[\|\widehat f(\rho)\|_{\mathrm{op}} \le\sqrt{64C_{10}}D^{-13/32}, \qquad -\frac12+\frac{12}{128}=-\frac{13}{32}.\] At sixteen factors the nonsign, nontrivial Plancherel sum is at most \[(64C_{10})^{16} \sum_{\substack{\lambda\vdash N\\ \lambda\notin\{(N),(1^N)\}}} D_\lambda^{\,2-32(13/32)} \le(64C_{10})^{16}Z_{10}(N)=o(1),\] because the displayed exponent is \(-11\). The sign contribution is at most \((1-z)^{32}\), and the trivial term is \((z^{16}-1)^2\). Equation (94) and restoration of the missing mass \(1-z^{16}=o(1)\) prove convergence for \(\mu_T^{*16}\). The block length \(T=2048d\) is a multiple of \(d\), so these are \(16T=32768d\) physical shuffles. Equation (2) gives the assertion from every deterministic start. ◻

Proof. Use 32 equal label parts. Equation (65) makes all complementary errors \(o(1)\). Exact trimming gives \(f\le\mu=q_d^{*T}\) of mass \(z=1-o(1)\). Its normalization \(\nu=f/z\) is within \(1-z\) of \(\mu\) in TV and has cap at most two for large \(d\). The coefficient estimate (85), with \(r=32\) and \(K=2\), gives \[\|\widehat\nu(\rho)\|_{\mathrm{HS}}^2 \le C'D^{-5/8},\qquad C'=128C_{10}.\] For 32 factors the nonsign, nontrivial Plancherel sum is at most \[(C')^{32}\sum_{\substack{\lambda\vdash N\\ \lambda\notin\{(N),(1^N)\}}}D_\lambda^{-19} \le(C')^{32}Z_{10}(N)=o(1).\] The sign term is at most \([2(1-z)]^{64}\), and the trivial coefficient is one. Restoring the true factors costs \(32(1-z)=o(1)\). The physical time is \(32T=32(2049d)=65568d\); translation invariance gives the same conclusion from every deterministic start. ◻

Proof. At \(T=1024d\), choose eight equal label parts whose summed complementary error is \(o(1)\). Heavy-coset deletion produces a probability \(\nu\) within \(\eta=o(1)\) of \(\mu=q_d^{*T}\), with cap at most four. For every \(\lambda\notin\{(N),(1^N)\}\), the branching estimate and (97), with \(r=8\), give \[\|\widehat\nu(\rho_\lambda)\|_{\mathrm{HS}}^2 \le4rM(k_\lambda)D_\lambda^{-1+2/r} \le D_\lambda^{-1/2}.\] Four factors therefore leave the inverse-first Plancherel sum: \[4\|\nu^{*4}-U_G\|_{\mathrm{TV}}^2 \le(2\eta)^8+ \sum_{\substack{\lambda\vdash N\\k_\lambda\ge1}}D_\lambda^{-1} =(2\eta)^8+Z_1(N)=o(1).\] The displayed sign term uses (95); the trivial coefficient is one. Restoring the four true factors costs \(4\eta=o(1)\). Their physical time is \(4T=4096d\), and (2) gives the same conclusion from every start. ◻

Proof. Lemma 2 gives distance at least \(1-(e/\sqrt N)^N>1/4\) for every integer \(t\le d\) when \(d\) is large. Lemma 3 gives the stated one-card bound for every \(d\). The upper bound follows from Corollary 30; (5) gives the sharper lower asymptotic \(2d-O(1)\). ◻

Proof. Lemma [shear:lem:general-frame] gives exactly the asserted path-law identity, with \(L_0e_i=v_i\). Its construction applies to every fixed schedule with consecutive basis windows and permits an arbitrary initial basis. It remains to check what happens to trajectory labels.

If labels \(a,b\) are paired in the cyclic process just before layer \(t\), then \(g_{t-1}(b)=g_{t-1}(a)+w_t\). The column identity in that frame construction is \(L_{t-1}w_t=v_t\). Hence their transformed positions, with initial labels changed by \(L_0\), differ by \(v_t\) and are paired in the transformed process. The implication reverses because \(L_{t-1}\) is invertible. When \(L_0=I\), the labels themselves do not change. For general \(L_0\), its bijection preserves the uniform law of the input list. The terminal bijection \(L_T\) preserves uniform injections and total variation. This proves both assertions. ◻

Proof. The online averaging and transport rules are those of Lemma [shear:lem:energy]; here we must also justify conditioning on complete observed paths. A feasible specification of those paths prescribes precisely the coins on the visited time-edge pairs. Necessity is immediate. Conversely, those prescribed coins force the paths, by induction over the layers. The path event is therefore a cylinder in the independent coin array. All other coins remain independent and fair, including unvisited coins at earlier times.

On an edge used by an observed label the coin is fixed; on an edge with two unobserved positions it remains fair. Future path constraints use later time-edge pairs and do not alter this earlier recursion. Thus the conditional vector equals the same forward averaging and transport flow obtained from the observed prefix. Applying the exact decrement in Lemma [shear:lem:energy], with its available set equal to \(F_{t-1}\), proves the displayed identity and transports the uniform comparison vector as stated. ◻

Proof. First fix the list, the complete schedule, and feasible complete trajectories of the preceding labels. Lemma 38 identifies the conditional mass on the left of (101) with the forward mass computed from the list, the direction prefix, and the observed path prefix through \(t\). It is therefore already \(\mathcal F_t^{C,j}\)-measurable. The tower property, using \(\mathcal F_t^{C,j}\subseteq\mathcal H^{C,j}\), proves (101).

Conditional on the complete schedule, the observed path prefix depends only on the direction prefix, by the product-switch hypothesis. Since the list is independent, its observation and that of the preceding paths do not change the next-direction law \(K_t\). Also, the field generated by \(\mathcal F_t^{C,j}\) and \(v_{t+1}\) is contained in \(\mathcal H^{C,j}\). Taking the tower of (101) to this intermediate field shows that the hidden mass remains \(p_t\) after \(v_{t+1}\) is revealed. Combining these two statements proves (102). ◻

Proof. Apply the energy calculation in Lemma [shear:lem:energy] to the available set \(F_0\cup F_1\) and the centered vector \(z\). That proof uses only \(\sum_xz_x=0\), not positivity of a probability vector. In its notation, take \(D^0=F_0\) and \(D^1=F_1\); the exact cross-pair loss is the displayed sum divided by \(n\). When the forbidden subspace is smaller than \(J\), every cross-coset difference remains allowed and has probability at least \(1/n\), instead of \(2/n\). Its averaging loss is still half the squared difference, giving the denominator \(2n\). ◻

Proof. The transverse-layer proof of Lemma 17 is uniform over the entire fixed direction schedule. It therefore applies here even though the first basis is prescribed. Explicitly, fix all directions and the observed-label configuration just before layer \(t-d\). The direction at that layer crosses the cosets of \(J_t\), because \(v_{t-d},\ldots,v_{t-1}\) form a basis. The following \(d-1\) directions lie in \(J_t\) and preserve the two coset counts. If \(a\) crossing pairs contain two unobserved positions, one count after the crossing is \(a+\operatorname{Binomial}(m-2a,1/2)\). The conditional fair-coin law is unchanged by fixing the directions. The binomial tail from that lemma is at most \(2e^{-m/8}\); if \(m=2a\) the count is deterministic. Removing this conditioning proves the stated unconditional bound.

Equation (101) makes the energy vector measurable in the chronological field, and (102) permits averaging the next direction there. Lemma 40 then contracts energy by at least \(\alpha=1/(4r)\) whenever the two counts are at least \(m/4\). We use the balance estimate only after taking expectations. Since \(0\le E\le1\), \[\mathbb E E_t\le(1-\alpha)\mathbb E E_{t-1} +2\alpha e^{-n/(8r)}.\] Iterate from \(E_d\le1\) and sum the geometric series. This proves (105) without asserting concentration conditional on the subsequently observed paths. ◻

Proof. Markov’s inequality and a union bound give the lost mass. For each complete direction schedule the law of \(L_T^{-1}Z_T\) is \(\mu\), so restricting to \(\mathcal G_{\mathcal P}\) gives \(0\le\nu\le\mu\). For an order, its \(j\)-th energy event is measurable from all directions and the preceding paths. On it, every endpoint has conditional probability at most \[\frac1{n-j+1}+n^{-3}\le\frac{1+n^{-2}}{n-j+1}.\] To bound a specified list of distinct outputs intersected with \(\mathcal G_{\mathcal P}\), first discard requirements belonging to other orders. Condition on all directions and the first \(q-1\) paths, bound the final target on its own energy event, and then discard that event. Continue downwards through the list. Each remaining target event and energy constraint is measurable in the conditioning data at its turn. The target in variable coordinates is known because \(L_T\) is determined by all directions. We obtain \[\Pr(\mathcal G_{\mathcal P},\text{specified outputs}) \le\prod_{j=1}^q\frac{1+n^{-2}}{n-j+1} \le2\frac{(n-q)!}{n!}.\] These output specifications are precisely the left cosets \(gK_i\). Normalization costs \(1/a\), at most two eventually. The total-variation bound for \(\bar\nu\) follows from Lemma 29. ◻

Proof. Take \(d\ge5\), which suffices for all block sizes in this proposition. For the first assertion take \(r=8\) and \(T=512d\). Equation (105) is at most \(n^{-12}\) for large \(d\), since \((31/32)^{511d}\le e^{-511d/32}\). The exceptional mass is at most \(8n\,n^6n^{-12}=o(1)\). Use the normalized law \(\bar\nu\) in Lemma 42 and the direct inverse-square estimate of Lemma . Proposition , with \(p=2\), gives \(\|\widehat{\bar\nu}(\rho)\|_{\rm op}\le\sqrt{32C_2}D^{-1/4}\). Eight convolutions have Plancherel exponent \(2-16/4=-2\). The original block has unbiased sign because one independent switch already has fair parity; the normalized modification has sign mean \(o(1)\). Proposition  applies. Since \(T=512d\) is a whole number of coordinate sweeps, eight physical blocks give \(8T=4096d\).

For the second construction take \(r=32\) and \(T=(1+80r)d\). The first term in (105) is at most \(e^{-20d}\le n^{-20}\), so the exceptional mass is at most \(32n^7(n^{-20}+2e^{-n/256})=o(1)\). Keep the unnormalized subprobability, whose coset factor is \(K=2\). The paragraph “Central-binomial powers eight and twenty” following Lemma  in Subsection  of [22] gives \[C_8=\sup_{s\ge1}\sum_{\lambda\vdash s}(f^\lambda)^{-8}<\infty, \qquad \sum_{\substack{\lambda\vdash s\\ \lambda\notin\{(s),(1^s)\}}}(f^\lambda)^{-8} \longrightarrow0\qquad(s\longrightarrow\infty).\] Applying Proposition  of the same companion with \(p=8\), \(K=2\), and \(r=32\) gives exponent \(-1/2+10/64=-11/32\) and prefactor \(\sqrt{64C_8}\). Sixteen convolutions leave exponent \(2-32(11/32)=-9\). The retained measure is dominated by the physical block law, whose sign mean is zero. Proposition , including its trivial and sign terms, therefore applies. Here \(T=2561d\) is again a whole number of sweeps, so \(16T=40976d\) physical shuffles suffice. ◻

Proof. Start with the standard basis and generate directions by independent fair coefficients \[v_t=v_{t-d}+\sum_{t-d<s<t}c_{s,t}v_s\quad(t>d).\] Write \(\mathbf c=(c_{s,t}:d<t\le T,\ t-d<s<t)\) for the complete coefficient array. Conditional on this array, let \(Z_0=I\) and \(Z_t=\zeta_tZ_{t-1}\), where the \(\zeta_t\) are independent uniform switches in directions \(v_t\). This is the same conditional uniform-complement schedule as before. For each fixed \(\mathbf c\), the frame comparison starts at \(L_0=I\). Thus the distance for the fixed endpoint list is unchanged by the output relabeling, and equals the average over \(\mathbf c\) of the conditional auxiliary distance.

For later use, let \[ \mathcal A_h=\sigma\bigl( c_{s,u}:d<u\le h,\ u-d<s<u;\quad \zeta_u:1\le u\le h\bigr). \tag{107}\] This field fixes the directions and the complete switch history through layer \(h\). The conditional law of these switches uses only the directions through \(h\), hence only coefficients with target time at most \(h\). Coefficients with later target times therefore remain independent fair bits given \(\mathcal A_h\).

Fix the input list \(a_1,\ldots,a_q\), write \(Y_t(j)=Z_t(a_j)\), and let \(F_{j,t}=V\setminus\{Y_t(i):i<j\}\), of size \(n_j=n-j+1\ge m\). Define the chronological field \[\mathcal F_{j,t} =\sigma\bigl(v_1,\ldots,v_t,\, Y_s(i):0\le s\le t,\ i<j\bigr).\] The conditional path law given \(\mathbf c\) depends only on its direction sequence. Equations (101) and (103), with \(\Lambda=\mathbf c\), therefore identify the two versions of the hidden mass: \[p_{j,t}(x) :=\Pr\{Y_t(j)=x\mid\mathbf c,\, Y_s(i):0\le s\le T,\ i<j\} =\Pr\{Y_t(j)=x\mid\mathcal F_{j,t}\}.\] Put \[E_{j,t}=\sum_{x\in F_{j,t}} \left(p_{j,t}(x)-\frac1{n_j}\right)^2.\] Thus \(E_{j,t}\) is measurable in \(\mathcal F_{j,t}\), even though its first definition uses the complete coefficient array and complete preceding paths. For \(t>d\) set \(J_t=\operatorname{span}(v_{t-d+1},\ldots,v_{t-1})\). Say that step \((j,t)\) is balanced if both its cosets contain at least \(m/4\) members of \(F_{j,t-1}\). Let \(\mathcal B_j(t)\) require this for all steps \(2d,\ldots,t\), with no requirement when \(t=2d-1\). The event \(\mathcal B_j(t)\) is \(\mathcal F_{j,t-1}\)-measurable, before the direction at \(t\) is revealed. Equation (102) and Lemma 40, followed by \(\mathcal B_j(t)\subseteq\mathcal B_j(t-1)\), give \[\mathbb E[\mathbf1_{\mathcal B_j(t)}E_{j,t}] \le(1-1/(4b))\, \mathbb E[\mathbf1_{\mathcal B_j(t-1)}E_{j,t-1}].\] Thus, for \(\mathcal B_j=\mathcal B_j(T)\), \[ \mathbb E[\mathbf1_{\mathcal B_j}E_{j,T}] \le(1-1/(4b))^{T-2d+1}. \tag{108}\] The events decrease as \(j\) increases because the hole sets shrink and the common threshold \(m/4\) stays fixed.

The energy on the nested balance events is now controlled. It remains to estimate the probability that the smallest reservoir fails balance; this will be the only exceptional-event cost in the tuple comparison. Fix \(t\ge2d\) and write \(h=t-d\). Let \(\ell_t\) be the unique linear form vanishing on \(J_t\) and taking value one on \(v_h\). Given \(\mathcal A_h\), this form is uniform among the forms with \(\ell_t(v_h)=1\). Indeed use the past basis \(v_{h-d+1},\ldots,v_h\). The equations \(\ell_t(v_{h+i})=0\), \(1\le i<d\), give \[\ell_t(v_{h-d+i}) =\sum_{s=h-d+i+1}^{h}c_{s,h+i}\ell_t(v_s).\] Solve backwards from \(i=d-1\) to \(i=1\). At each stage the term \(c_{h,h+i}\ell_t(v_h)=c_{h,h+i}\) is a fresh fair bit, independent of the coefficients at later stages, of the other coefficients in the current equation, and of \(\mathcal A_h\). This proves the uniformity claim.

For \(F=F_{q,h}\) put \(Z=\sum_{x\in F}(-1)^{\ell_t(x)}\). The set \(F\) is \(\mathcal A_h\)-measurable, and \(\mathbb E[Z^2\mid\mathcal A_h]\le|F|=n_q\): diagonal terms are one, off-diagonal terms with difference \(v_h\) are minus one, and every other off-diagonal term has mean zero. The directions at times \(h+1,\ldots,t-1\) lie in \(J_t\), so the coset counts from the post-step-\(h\) configuration to time \(t-1\) are preserved. A count below \(m/4\) forces \(|Z|>n_q-m/2\ge n_q/2\). Chebyshev and a union bound give \[ \Pr(\mathcal B_q^c)\le4T/n_q\le4T/m. \tag{109}\]

We finish by coupling the endpoint lists, conditional on \(\mathbf c\). Write \(\pi_q\) for the uniform injection law on \(q\) positions, and put \[\delta_{\rm aux}(\mathbf c) =\left\|\mathcal L((Y_T(1),\ldots,Y_T(q))\mid\mathbf c)-\pi_q \right\|_{\rm TV}.\] At stage \(j\), prescribe the real endpoint kernel to be \(p_{j,T}\), evaluated on the preceding real paths. This prescription is used for every joint coupling history; it is not replaced by a law conditioned on earlier coupling success. Prescribe the comparison kernel to be uniform on the positions not yet used by the comparison list.

On previous success the two lists have the same preceding endpoints, so their available endpoint sets agree. If \(\mathcal B_j\) holds, maximally couple the prescribed kernels, declaring failure if their endpoints differ. If balance fails, declare failure and use any coupling of them. Once a failure has occurred, continue using any coupling of the prescribed kernels, with the comparison list’s own remaining set. After choosing the real endpoint, sample its full real path from the conditional path law given that endpoint, \(\mathbf c\), and the preceding real paths. These prescriptions preserve the real path law and the uniform comparison-list law at every stage. The probability of a balance failure before any mismatch is at most \(\Pr(\mathcal B_q^c\mid\mathbf c)\) by nesting. The mismatch probability on a balanced, previously successful stage is bounded by \(\tfrac12\sqrt{n_jE_{j,T}}\). Averaging over \(\mathbf c\) and dropping the preceding-success indicator gives the total bound \[\begin{align*} \mathbb E_{\mathbf c}\delta_{\rm aux}(\mathbf c) &\le \Pr(\mathcal B_q^c) +\frac12\sum_{j=1}^q \mathbb E\!\left[\mathbf1_{\mathcal B_j} \sqrt{n_jE_{j,T}}\right]\\ &\le\frac{4T}{m} +\frac12n^{3/2}(1-1/(4b))^{(T-2d+1)/2}, \end{align*}\] by (108), (109), and Cauchy–Schwarz. The fixed-list frame comparison identifies the left side with the physical marginal distance, proving (106). Finally the exponential factor is at most \(e^{-5d/2}\), so \(n^{3/2}e^{-5d/2}\to0\), and \(T/m\to0\). ◻

Proof. Use any fixed partition into sixteen equal blocks and apply Proposition 44 to its complements. Theorem  gives exponent \(a=1/8\) with constant one. Sixteen convolutions have exponent \(2-2(16)/8=-2\). The original block has zero sign mean, and Proposition  applies to its pruned subprobability. The total number of physical steps is \(16(322d)=5152d\). ◻

Proof. The conditioned paths are feasible, because they come from \(g\). Their complement at time \(t\) is exactly the set of available positions \(g_t(F)\). Lemma 38 gives the conditional mass flow on these positions. We express its three edge cases in the reference indices.

If a matching edge has one position in \(g_{t-1}(F)\), its coin is fixed by the path of the observed card on that edge. It is therefore the coin used by the reference history, and it carries the available position \(g_{t-1}(v)\) to \(g_t(v)\). Its mass keeps index \(v\). If both endpoints belong to \(g_{t-1}(F)\), say with reference indices \(v,v'\), the two conditional output masses are the average of the preceding two masses. They are equal, so the reference history’s choice to swap or fix that edge does not change the two indexed output values. An edge with no available position has no coordinate in the row. These cases give \(r_t^{u,F}=r_{t-1}^{u,F}B_t^F\). Initially card \(u\) is at \(u\), so the initial row is \(e_u^{\mathsf T}\). Iteration proves the row formula, and the index of the realized endpoint \(g_T(u)\) is \(u\). ◻

Proof. We estimate the distribution of these matching functionals using the directly generated variable-direction process. Start its directions with the standard basis, and choose each later direction uniformly outside the span of its preceding \(d-1\) directions. At each layer use fresh independent fair switch coins. Denote its cumulative permutation by \(\widetilde g_t\). Lemma 37 says that, for each complete direction schedule, its matching sequence on trajectory labels has the same law as the cyclic matching sequence. Thus every functional \(Z_T^F\) of those matchings has the same law in the two models. For the following calculation, \(B_t^F\), \(Q_t^F\), and \(Z_t^F\) denote these same matrix functions evaluated on the auxiliary matching sequence.

Take \(F\) uniform of a fixed size \(m\ge n/L\), independently of this auxiliary process. Condition on \(F\), the directions through time \(t-1\), and the complete switch history through that time. For a row of \(Q_{t-1}^F-J_F\), write its entries as \(z_u\), \(u\in F\). This row is known under the conditioning and has sum zero. Transfer it to the current positions by \[\widetilde z(\widetilde g_{t-1}(u))=z_u\quad(u\in F).\] The label pairs averaged by \(B_t^F\) correspond exactly to the position pairs contained in \(\widetilde g_{t-1}(F)\) in the new matching. Thus the squared row loss is the position-space loss in Lemma 40.

For \(t>d\), let \(J_t\) be the span of the preceding \(d-1\) directions, and let \(h,h'\) be the sizes of \(\widetilde g_{t-1}(F)\cap J_t\) and \(\widetilde g_{t-1}(F)\setminus J_t\). The next direction is fresh under the conditioning above. Applying the lemma to each row and summing gives \[\mathbb E[Z_{t-1}^F-Z_t^F\mid F,\text{ complete generated past through }t-1] \ge\frac{\min(h,h')}{n}Z_{t-1}^F.\]

Now fix the generated past but leave \(F\) unobserved. The map \(\widetilde g_{t-1}\) and the half-space \(J_t\) are fixed, and \(\widetilde g_{t-1}(F)\) is a uniform \(m\)-subset. The elementary half-count estimate gives \[\Pr\{\min(h,h')<m/4\mid\text{ generated past}\} \le 2\theta^m, \qquad \theta=(3/2)2^{-3/4}<1.\] For example, expansion of the product for a uniform sample without replacement gives \(\mathbb E2^X\le(3/2)^m\) for either half-count \(X\); Markov’s inequality applied to the larger half yields this bound. Set \(c=-\log\theta>0\). With \(Z_{t-1}^F\le n\) and \(m\ge n/L\), the unconditional recurrence is \[\mathbb E Z_t^F\le(1-1/(4L))\mathbb E Z_{t-1}^F +O(ne^{-cn/L})\qquad(t>d).\] There are \(T-d=100Ld\) useful layers. Starting with \(Z_d^F\le n\), the initial contribution is at most \(ne^{-25d}\) and the accumulated error is exponentially small in \(n\). Hence \(\mathbb E Z_T^F\le n^{-20}\) for large \(d\), uniformly in \(m\).

The equality of matching laws proved at the start transfers this expectation to the cyclic histories on which \(\mathcal G_{\mathrm{ent}}\) was defined. For each \(m\), Markov’s inequality applied to the subset average at threshold \(n^{-8}\) gives failure probability at most \(n^{-12}\). There are at most \(n\) sizes, so \(\mathbb P(\mathcal G_{\mathrm{ent}}^c)\le n^{-11}=o(1)\). ◻

Proof. Fix an ordered list \(i_1,\ldots,i_q\). Let \(\tau_j\) and \(y_j\) be the complete trajectory and endpoint of \(i_j\). For the current history let \(p_j\) be the conditional probability under \(\mathbb P\) of its realized endpoint \(y_j\), given \(\tau_1,\ldots,\tau_{j-1}\). Relative-entropy chain rule and conditional data processing give \[\begin{align*} D(\nu\|\mathbb P) &\ge\sum_{j=1}^q\bigl[-\operatorname{Ent}_\nu (y_j\mid\tau_1,\ldots,\tau_{j-1})-\mathbb E_\nu\log p_j\bigr]\\ &\ge-\operatorname{Ent}_\nu(y_1,\ldots,y_q) -\sum_{j=1}^q\mathbb E_\nu\log p_j. \end{align*}\] The second inequality uses that the preceding endpoints are functions of the preceding trajectories. Put \(F_j=V\setminus\{i_1,\ldots,i_{j-1}\}\) and \(m_j=|F_j|=n-j+1\). The uniform injection assigns each endpoint list mass \(\prod_jm_j^{-1}\), so the last inequality is \[D(\nu\|\mathbb P)\ge D(\nu_X\|U_X) -\sum_{j=1}^q\mathbb E_\nu\log(m_jp_j), \qquad X=\{i_1,\ldots,i_q\}.\]

Lemma 46 applies with observed complement \(V\setminus F_j\) and gives \(p_j=Q_T^{F_j}(i_j,i_j)\) on each realized history. For a uniform ordered input list, \(F_j\) is a uniform subset of size \(m_j\). The bound (112) is independent of which \(i_j\in F_j\) is chosen. On every history in \(\mathcal G_{\mathrm{ent}}\), Jensen’s inequality therefore gives \[\mathbb E_{\text{ordered list}}\log(m_jp_j) \le m_j\mathbb E_{F:\,|F|=m_j}\sqrt{Z_T^F} \le n\sqrt{n^{-8}}=n^{-3}.\] All these sizes are at least \(n/L\). Sum over \(q\le n\) indices and average the preceding entropy inequality over lists and over \(\nu\). The induced set \(X\) is uniform of size \(q\), and ordering its coordinates does not change its relative entropy from a uniform injection. This proves the claimed error \(n^{-2}\). ◻

Proof. If \(a=\mathbb P(\mathcal G_{\mathrm{ent}})=0\), the probability on the left side of (115) is zero. Otherwise apply the all-tilt coefficient theorem, [22]. Its reference probability is the law \(\mathbb P\) of cyclic switch histories, its endpoint map is \(g_T\), and its common event is \(\mathcal G_{\mathrm{ent}}\) from Lemma 47. Lemma 48 verifies the theorem’s hypothesis for every probability \(\nu\) supported on that same event, with \(q=n-n/128\) and additive error \(n^{-2}\). The theorem therefore gives the tail exponent \(1/256=1/(2L)\) at threshold \(D^{-1/8}\). For \(a>0\), splitting each absolute coefficient at \(D^{-1/8}\) under \(\mathbb P(\,\cdot\mid\mathcal G_{\mathrm{ent}})\) gives the stated operator bound with factor \(C/a\). By Lemma 47, \(a=1-o(1)\), so \(a\ge1/2\) eventually. Since \(L=128\) and \(D\ge1\), \(D^{-1/8}\le D^{-1/(2L)}\), which gives \(C'=1+2C\). ◻

Proof. For all sufficiently large \(d\), put \(a=\mathbb P(\mathcal G_{\mathrm{ent}})>0\) and \(\eta=\mathcal L(g_T\mid\mathcal G_{\mathrm{ent}})\). The original block law \(\mu=\mathcal L(g_T)\) has zero sign mean, so \[|\widehat\eta(\operatorname{sgn})| =\frac{|\mathbb E[\operatorname{sgn}(g_T)\mathbf1_{\mathcal G_{\mathrm{ent}}^c}]|}{a} \le\frac{1-a}{a}=o(1).\] For the nonlinear irreducibles, Proposition 49 gives the operator bound with the absolute constant \(C'\). The trivial coefficient of \(\eta^{*b}\) is one. Plancherel therefore gives \[|G|\sum_g|\eta^{*b}(g)-U_G(g)|^2 \le |\widehat\eta(\operatorname{sgn})|^{2b} +(C')^{2b}\sum_{\rho\notin\{\mathbf1,\operatorname{sgn}\}} D_\rho^{\,2-b/L}=o(1).\] Indeed \(2-b/L=-30\), and the nonlinear sum vanishes by [22]. Cauchy–Schwarz gives total-variation convergence. The \(b\) independent histories all belong to \(\mathcal G_{\mathrm{ent}}\) with probability \(a^b\). Removing their conditioning costs at most \(1-a^b\le b(1-a)=o(1)\), because \(b\) is fixed. The physical block length is a multiple of \(d\), and its \(b\)-fold product is the actual shuffle at time \(32L(1+100L)d\). ◻

Proof. Set \(A_i=B_i^{-1}\). Since \(S_i\) is an involution, \(A_i=A_{i-1}S_i\) and \(w_i=A_{i-1}e_{\iota_i}\), which are the forward frame and selected directions of Lemma 19. Remark 20 gives the inverse coupled identity \(X_i=B_iY_i\) and the conditional coordinate-switch law. The prefix product-switch law and consecutive-basis property are conclusions of that lemma for the same periodic axis order. The prefix law makes the old virtual path and the next direction conditionally independent given the direction prefix; Corollary 39 records the resulting hidden-position factorization after specified virtual path prefixes through layer \(i-1\) are observed. The uniform-complement law for \(i>d\) is the lemma’s fresh-direction conclusion.

In particular, the conditional law of the full virtual trajectory given \(\Lambda\) depends on \(\Lambda\) only through \(W\). Thus the trajectory and \(\Lambda\) are conditionally independent given \(W\). Conditioning also on a fixed observed part of the trajectory preserves this factorization for the remaining trajectories. ◻

Proof. Lemma 52 gives the conditional product-switch law and the fresh direction in the field \(\mathcal F_{i-1}^j\). Apply Lemma [shear:lem:energy] with initial datum \(\mathcal O\), \(t=i-1\), available set \(D_{i-1}\), and hidden mass \(q_{i-1}\). Its averaging and transport rules preserve the uniform comparison law on the holes and give \[ E_{i-1}-E_i =\frac12\sum_{\substack{\{x,y\}\text{ an edge of layer }i\\ x,y\in D_{i-1}}} \bigl(z_{i-1}(x)-z_{i-1}(y)\bigr)^2. \tag{119}\] In particular \(0\le E_i\le E_{i-1}\le1\) for every realization.

Suppose now that \(i>d\). Before its direction is drawn, put \(J_i=\operatorname{span}(w_{i-d+1},\ldots,w_{i-1})\) and split \(D_{i-1}=D^0\cup D^1\) across its two cosets, with sizes \(h_0,h_1\). Lemma 52 makes the next direction uniform outside \(J_i\), so each cross pair has conditional matching probability \(2/n\). The cross-pair calculation in Lemma [shear:lem:energy], with centered vector \(z_{i-1}\), gives \[\begin{align*} \mathbb E(E_{i-1}-E_i\mid\mathcal F_{i-1}^j) &=\frac1n\sum_{x\in D^0,\ y\in D^1} \bigl(z_{i-1}(x)-z_{i-1}(y)\bigr)^2\\ &\ge\frac{\min(h_0,h_1)}n E_{i-1}. \tag{120}\end{align*}\] The inequality is (21); it also covers an empty part.

We use a simultaneous balance estimate for every affine half-space. At a fixed time, condition on \(\Lambda\) and the complete virtual switch history but not on \(\mathcal O\). The permutation \(Y_{i-1}\) is then fixed, and the independent ordering makes \(D_{i-1}\) a uniform \(m\)-subset of \(V\). The simultaneous estimate below is valid even for a half-space chosen after that subset is seen.

For \(m<n\), a uniform \(m\)-subset has the law of independent Bernoulli choices of parameter \(m/n\), conditional on making exactly \(m\) choices. This conditioning event has probability at least \(1/(n+1)\): \(m\) is a mode of the binomial distribution and that distribution has \(n+1\) values. In a specified affine half-space let \(X\) be the unconditioned number of choices. It has mean \(m/2\), and \[\begin{align*} \Pr\{X<m/4\} &\le 2^{m/4}\mathbb E2^{-X} =2^{m/4}\left(1-\frac{m}{2n}\right)^{n/2}\\ &\le e^{-(1-\log2)m/4}\le e^{-m/16}, \end{align*}\] where the last inequality uses \(\log2<3/4\). There are fewer than \(2n\) affine half-spaces arising as a linear hyperplane or its other coset. Since \(m\ge n/r\), a union bound after conditioning gives \[ \mathbb P\bigl(\text{some such half-space contains fewer than }m/4 \text{ holes}\bigr)\le\delta_n. \tag{121}\] For \(m=n\) the balance assertion is deterministic.

On the complement of the event in (121), the conditional loss in (120) is at least \(mE_{i-1}/(4n)\ge E_{i-1}/(4r)\). On the exceptional event we retain the deterministic energy nonincrease and the bound \(E_{i-1}\le1\). Taking expectations consequently gives \[\mathbb E E_i\le \left(1-\frac1{4r}\right)\mathbb E E_{i-1}+\delta_n \qquad(i>d).\] Only the unconditional estimate (121) has been used. Iterating over \(T-d=100rd\) layers, starting from \(E_d\le1\), yields \[\mathbb E E_T\le \left(1-\frac1{4r}\right)^{100rd}+T\delta_n \le e^{-25d}+T\delta_n.\] ◻

Proof. For each \(j<k\), the terminal factorization in Lemma 52 and (103) give \[\Pr\{Y_T(a_{j+1})=x\mid \Lambda,\mathcal O,\, Y_s(a_h):0\le s\le T,\ h\le j\}=q_T(x).\] Thus the terminal-data hypothesis of Lemma 8 holds with initial datum \(\mathcal O\) and terminal datum \(\Lambda\). Write \[\delta_{\rm virt}(\Lambda,\mathcal O) =\left\|\mathcal L\bigl((Y_T(a_1),\ldots,Y_T(a_k)) \mid\Lambda,\mathcal O\bigr)-\pi_k\right\|_{\rm TV}\] for the conditional virtual tuple distance. The lemma’s coarsening to preceding endpoints and sequential comparison, together with Lemma 53, give \[\mathbb E_{\Lambda,\mathcal O}\delta_{\rm virt}(\Lambda,\mathcal O) \le\frac{k}{2}\sqrt{n\bigl(e^{-25d}+T\delta_n\bigr)}.\] For each fixed \(\Lambda\) and \(\mathcal O\), applying \(B_T\) to every output preserves \(\pi_k\) and total variation. The conditional law of \(X_T\) given \(\Lambda\) is \(\mu\), and the independent ordering contributes no further information about that law. Hence \(\delta_{\rm virt}(\Lambda,\mathcal O)\) equals the distance under \(\mu\) for the specified list \(C\). Averaging proves (122). Finally, \(n=2^d\) gives \(n^{3/2}e^{-25d/2}\to0\), while \(\delta_n\) is exponentially small in \(n\) up to a polynomial factor. Hence \(\varepsilon_n=o(1)\). ◻

Proof. The proof of Lemma  bounds the complete inverse-twentieth sum by \(K_0\). Apply Proposition  with \(p=20\), \(K=2\), and \(r=128\). That proposition assigns a component \(v_i\) of a given vector to the small \(H_i\)-types and sets \(W_i=\operatorname{span}\{\rho(h)v_i:h\in H_i\}\). Its orbit bound is \[ \dim W_i \le\sum_{\substack{\tau\in\operatorname{Irr}(H_i)\\ \dim\tau\le D^{1/r}}}(\dim\tau)^2 \le K_0D^{22/r}. \tag{125}\] The sum counts each irreducible subgroup type once: the orbit of a fixed vector in all copies of a type of degree \(f\) has dimension at most \(f^2\). Combining this estimate with the coset cap and Schur averaging in that proposition gives \(\sqrt{2rK_0}D^{-1/2+11/r}\). ◻

Proof. For \(d\ge4\), put \(m=n/16\), \(T=513d\), and \(\mu=q_d^{*T}\). Corollary 23 gives \(\mathbb E_{|A|=n-m}\|\mu_A-U_A\|_{\mathrm{TV}}\le\varepsilon_n=o(1)\). In a uniformly chosen ordered partition \((M_1,\ldots,M_{16})\) into sets of size \(m\), each complement \(V\setminus M_i\) is uniform. Hence the expected sum of the sixteen complementary errors is at most \(16\varepsilon_n\), and some partition has sum at most this value.

For that partition, the restriction to \(V\setminus M_i\) identifies the left coset \(gH_i\). Delete the union of all cosets whose original \(\mu\)-mass exceeds twice uniform. The calculation in the proof of Lemma 10 gives a common subprobability \(f\le\mu\) of mass \(z=1-\eta\), with \(\eta\le32\varepsilon_n=o(1)\), such that \(f(gH_i)\le2/[G:H_i]\) for every \(g,i\). Equation (128), with \(z\le1\), therefore gives \(\|\widehat f(\rho)\|_{\mathrm{HS}}^2\le32C_2D^{-3/4}\). Hilbert–Schmidt submultiplicativity yields \[D\|\widehat f(\rho)^4\|_{\mathrm{HS}}^2 \le D\|\widehat f(\rho)\|_{\mathrm{HS}}^8 \le(32C_2)^4D^{-2}.\] The trivial coefficient of \(f^{*4}\) is \(z^4\). Lemma 15 gives \(|\widehat f(\mathop{\mathrm{sgn}})|\le\eta\), since the original block has sign mean zero. Plancherel and (129) now imply \[|G|\|f^{*4}-z^4U_n\|_2^2 \le \eta^8+(32C_2)^4Z_2(n)=o(1).\] Cauchy–Schwarz gives \(\|f^{*4}-z^4U_n\|_1=o(1)\). Positivity gives \(f^{*4}\le\mu^{*4}\), and their mass difference is \(1-z^4=o(1)\). Consequently \[\|\mu^{*4}-U_n\|_{\mathrm{TV}} \le 1-z^4+\tfrac12\|f^{*4}-z^4U_n\|_1=o(1).\] The four independent physical blocks have total length \(4T=2052d\) and law \(\mu^{*4}=q_d^{*(2052d)}\). Equation (2) gives the stated maximum over starts. ◻

Proof. For sufficiently large \(d\), in particular \(d\ge4\), set \(T=2049d\), \(\mu=q_d^{*T}\), and \(K=P_d^T\). The identity \(K(g,z)=\mu(zg^{-1})\) shows that \(K\) is doubly stochastic and that it commutes with the right action of every label permutation. A uniform source state in \(X_i\) places the specified labels outside \(M_i\) at a uniform injection. Their output law is the quotient row \(p_i(x,\cdot)\), so Corollary 23 gives \[\frac1L\sum_{x\in X_i}\frac12\sum_{y\in X_i} |p_i(x,y)-L^{-1}|\le\delta_n,\qquad \delta_n=n^{-10}.\] Choose \(\epsilon=n^{-1}\) in (130). A removed quotient atom has positive excess over uniform at least \(\epsilon/(1+\epsilon)\) times its mass. Thus the mean mass removed is \[\gamma_i:=\frac1{|G|}\sum_{g,z}(K(g,z)-T_i(g,z)) \le\frac{1+\epsilon}{\epsilon}\delta_n=o(1).\] Each \(T_i\) is row- and column-substochastic because \(0\le T_i\le K\). Moreover \(H_i\) is normal in \(H\), and right equivariance gives \(p_i(xh,yh)=p_i(x,y)\) for \(h\in H\). The cutoff therefore preserves the action of the whole product \(H\).

We now verify the sign condition with its full multiplicity. Fix \(i\) and choose a deck \(g_x\) in each fiber \(x\in X_i\). For \(a,b,c\in H_i\), put \(w_{xy}(c)=T_i(g_x,g_yc)\). Right equivariance gives \[T_i(g_xa,g_yb)=w_{xy}(ba^{-1}),\qquad \sum_cw_{xy}(c)=t_i(x,y) :=p_i(x,y)\mathbf1\{p_i(x,y)\le(1+\epsilon)/L\}.\] On fiber \(x\), define the normalized sign vector by \(\phi_x(g_xa)=\mathop{\mathrm{sgn}}(a)/\sqrt{m!}\), and set it to zero off that fiber. Each fiber is a regular \(H_i\)-set, so these \(L\) orthonormal vectors span the full sign isotypic space of \(\ell^2(G)\). In this basis the multiplicity operator has entries \[ T_i^{\mathop{\mathrm{sgn}}}(x,y)=\langle\phi_x,T_i\phi_y\rangle =\sum_{c\in H_i}w_{xy}(c)\mathop{\mathrm{sgn}}(c), \qquad |T_i^{\mathop{\mathrm{sgn}}}(x,y)|\le t_i(x,y). \tag{131}\]

Suppose the ignored positions in \(x\) contain a matched pair of the first physical shuffle, and let \(s\) be its position transposition. Then \(\tau=g_x^{-1}sg_x\in H_i\) is an odd transposition and \(g_x\tau=sg_x\). In the first physical step \(RS\), the switch \(S\) is uniform in \(C_{e_1}\). Since \(s\in C_{e_1}\), one has \(Ss\stackrel{\mathrm{law}}=S\). The first-step laws from \(g_x\tau\) and \(g_x\) are therefore identical, and so \(K(g_x\tau,z)=K(g_x,z)\) for every output deck \(z\). The cutoff depends only on the source and target quotient states, so the same identity holds for \(T_i\). Right equivariance now gives \[w_{xy}(c) =T_i(g_x\tau,g_yc) =T_i(g_x,g_yc\tau^{-1}) =w_{xy}(c\tau^{-1}).\] Since \(\mathop{\mathrm{sgn}}(\tau)=-1\), the signed sum in (131) is its own negative. Every outgoing sign coefficient from this fiber is zero.

For a uniform fiber \(x\), its ignored positions form a uniform \(m\)-subset of \(V\). The fraction of these sets containing no full first-layer pair is \[a_{n,m}=\prod_{j=0}^{m-1}\frac{n-2j}{n-j} \le\exp\left(-\frac{m(m-1)}{2n}\right).\] After \(j\) positions have been sampled without a partner, exactly \(j\) partners are forbidden among the \(n-j\) remaining positions. The product follows, and \(\log(1-u)\le-u\) gives the exponential bound. Only these \(La_{n,m}\) fiber rows can contribute to the sign matrix. For each such row the quotient cap gives \[\sum_y|T_i^{\mathop{\mathrm{sgn}}}(x,y)|^2 \le\sum_y t_i(x,y)^2 \le\frac{1+\epsilon}{L}\sum_y t_i(x,y) \le\frac{1+\epsilon}{L}.\] Consequently the squared Hilbert–Schmidt norm on the entire sign multiplicity space is \[\|T_i^{\mathop{\mathrm{sgn}}}\|_{\mathrm{HS}}^2 =\sum_{x,y}|T_i^{\mathop{\mathrm{sgn}}}(x,y)|^2 \le(1+\epsilon)a_{n,m} \le(1+\epsilon)e^{-m(m-1)/(2n)}=o(1).\]

Here \(m=n/16\to\infty\) and \(\delta_n/\epsilon=n^{-9}\to0\). The companion theorem therefore gives mean row error \(o(1)\) for \(K^{16}\). The right action of all label permutations is transitive on decks and commutes with \(K\), so every row has the same error. Finally \(K^{16}=P_d^{16T}\) and \(16T=32784d\), proving the assertion. ◻

Proof. For sufficiently large \(d\), in particular \(d\ge10\), set \(b=1024\), \(T=(1+32b)d\), and \(\mu=q_d^{*T}\). Corollary 23, in the set form established at the start of this section, gives the displayed hypothesis with \(\delta_n=o(1)\). Let \(\nu\) be the resulting nearby probability. Lemma 15 gives \(|\widehat\nu(\mathop{\mathrm{sgn}})|\le2\|\nu-\mu\|_{\mathrm{TV}}=o(1)\). For eight factors, Plancherel and \(\|A\|_{\mathrm{HS}}\le\sqrt D\,\|A\|_{\mathrm{op}}\) give \[4\|\nu^{*8}-U_n\|_{\mathrm{TV}}^2 \le|\widehat\nu(\mathop{\mathrm{sgn}})|^{16} +C^{16}Z_2(n)=o(1),\] using (129); the nonlinear exponent is \(2-16(1/4)=-2\). Replacing the eight factors one at a time costs at most \(8\|\nu-\mu\|_{\mathrm{TV}}=o(1)\). Thus \(\mu^{*8}\) converges to uniform. These physical blocks have total length \(8T=8(1+32\cdot1024)d=262152d\), so \(\mu^{*8}=q_d^{*(262152d)}\). Equation (2) gives the same error from every start. ◻

Proof. Proposition 18 covers every periodic ordering of the standard axes, including every starting phase, and the inverse product obtained by reversing independent involutive layers. With its omitted-block parameter equal to \(2\) and time \(T=100d\), it gives, for \(\eta=\pi\) or \(\eta=\pi^{-1}\), \[\|\mathcal L(\eta(c_1),\ldots,\eta(c_q))-\pi_q\|_{\mathrm{TV}} \le\frac{n^{3/2}}2 \left(e^{-99d/8}+2e^{-n/16}\right)^{1/2}=o(1),\] where \(\pi_q\) is the uniform ordered injection law. The fixed-list estimate uses the hidden-card vector from the chronological field (16). The terminal comparison uses (24) with \(\Lambda=\mathbf v\), disclosing the complete schedule and only the complete predecessor paths.

For the independent shear realization of the coordinate chain, (71) gives the local block identity \[ B_0=I,\qquad M_t=B_t^{-1}Q_tB_{t-1},\qquad \pi=Q_L\cdots Q_1=B_LM_L\cdots M_1. \tag{132}\] Here \(Q_t\) are the original coordinate layers and \(M_t\) the virtual layers coupled to them. Conditional on the independent shears, the \(M_t\) are independent uniform matching switches. The known terminal \(B_L\) relabels only final positions, so the specified input list is unchanged. Equation (69) identifies its virtual terminal posterior with its chronological vector after all shears are disclosed. The independent shear realization starts with \(B_0=I\); the general first-basis realization retains the input map \(L_0^{-1}\) from (14). ◻

Proof. Let \(Y'\) be an independent copy and put \(u=(Y')^{-1}Y\). Independence and unitarity give \[\frac{\operatorname{tr}(T_\alpha^*T_\alpha)}{D_\alpha} =\mathbb E\frac{\chi_\alpha(u)}{D_\alpha} \le\mathbb E\frac{|\chi_\alpha(u)|}{D_\alpha}.\] If \(f\) counts the fixed points of \(u\), then, for \(1\le k\le b\), \[ \Pr(f\ge k)\le\binom bk b^{-k/2}. \tag{136}\] Indeed, condition on \(Y'\) and prescribe \(Y(x)=Y'(x)\) at a chosen set of \(k\) inputs. Each input-output pair determines one route because each coordinate is updated once. A consistent family of routes visits at least \(hk/2\) distinct switch bits: a switch serves at most two routes. It therefore has probability at most \(2^{-hk/2}=b^{-k/2}\). A union bound over the input set proves (136).

Write \(D=D_\alpha\). The assertion is immediate for \(D=1\). For \(D>1\) set \[\kappa=\frac{\log D}{2\log b},\qquad k=\left\lfloor\max\{b^{9/10},\kappa\}\right\rfloor+1.\] Since \(D^2\le b!\), one has \(\kappa\le b/4\), so \(k\le b\) for large \(b\). For \(k\ge b^{9/10}\) and sufficiently large \(b\), \[\binom bk b^{-k/2} \le\left(\frac{e b^{1/2}}k\right)^k \le b^{-k/4}\le D^{-1/8}.\] Outside an event of probability at most \(D^{-1/8}\), therefore, \(f\le b^{9/10}\) or \(f\le\kappa\). The first case is covered by (134). In the second, restrict \(\rho_\alpha\) to the symmetric group on the \(b-f\) moving points. Every irreducible constituent of degree \(E\) satisfies \[E\ge\frac{D}{[S_b:S_{b-f}]}\ge b^{-f}D\ge D^{1/2}.\] For the first inequality, translates of one constituent subspace by coset representatives span a nonzero \(S_b\)-invariant subspace and hence the whole irreducible representation. Since \(f\le\kappa\le b/4\), the moving-point group has size tending uniformly to infinity. The permutation has no fixed points there. Applying (134) to each constituent and averaging with its dimension weight, including multiplicity, bounds the normalized character by \(D^{-1/100}\). On the exceptional event use the bound one. Thus the normalized trace is at most \(D^{-1/100}+D^{-1/8}\le2D^{-1/100}\). ◻

Proof. Restrict to the elementary abelian subgroup generated by a fixed matching of \(n/2\) transpositions. If every constituent assigns the same sign to all these generators, their representing matrices are equal. Conjugation then makes any two disjoint transpositions have equal matrices. A third transposition disjoint from two overlapping ones exists when \(n\ge6\), so all transpositions have the same matrix. The image is generated by one involution, and irreducibility gives the trivial or sign representation. Otherwise a constituent assigns negative sign to exactly \(r\) generators, with \(1\le r\le n/2-1\). The normalizer permutes the matching pairs, so all \(\binom{n/2}{r}\ge n/2\) reordered characters occur. Their eigenspaces are linearly independent. ◻

Proof. Let \(G_j\) be the symmetric group supported on \(J_j\) and put \(H=G_0\times G_1\). If a positive operator \(Q\) commutes with \(\rho(G_{1-j})\), then \(\rho(\pi)^*Q\rho(\pi)\) is determined by \(Q_j(\pi)\). Indeed, equal ordered preimages of \(J_j\) mean \(\pi'\pi^{-1}\in G_{1-j}\). The separate marginal cap and positivity therefore give, for every unit vector \(v\), \[ \mathbb E\langle\rho(\pi)v,Q\rho(\pi)v\rangle \le4\,\mathbb E_{g\sim U_{S_n}} \langle\rho(g)v,Q\rho(g)v\rangle =4\,\frac{\operatorname{tr}Q}{D}. \tag{138}\] The final equality is Schur averaging. This comparison uses one capped preimage marginal at a time; it imposes no joint density bound.

Set \(A_j=\mathbb E\rho(\sigma_j)\) and \(A=A_0A_1\). These factors and their adjoints commute, and they are contractions. Independence and convexity give \[ \|A\,\mathbb E\rho(\pi)v\|^2 \le\mathbb E\langle\rho(\pi)v,A^*A\rho(\pi)v\rangle. \tag{139}\] For each \(j\), let \(P_{j,\le}\) select the full \(G_j\)-isotypic spaces of degree at most \(D^{1/100}\), and put \(P_{j,>}=I-P_{j,\le}\). These projections commute with \(H\) and with each other. On a joint half-type, write \(X_j=A_j^*A_j\). The \(X_j\) are positive contractions on separate tensor factors. If type \(j\) is large, \(X_0X_1\preceq X_j\); if both types are small, \(X_0X_1\preceq I\). Hence \[ A^*A\preceq A_0^*A_0P_{0,>}+A_1^*A_1P_{1,>}+P_{0,\le}P_{1,\le}. \tag{140}\] Every term is positive. The first two commute with the complementary half group, and the last commutes with both groups, so (138) applies to each.

On each \(G_j\)-type selected by \(P_{j,>}\), Lemma 61 bounds the normalized trace by \(2D_\alpha^{-1/100}\le2D^{-1/10000}\). Averaging with the restriction dimension weights, including all multiplicities, gives \[ \frac1D\operatorname{tr}(A_j^*A_jP_{j,>}) \le2D^{-1/10000}. \tag{141}\]

It remains to bound the fraction of the space on which both half-types are small. If \(m_{\alpha\beta}\) is the multiplicity of \(\alpha\otimes\beta\) in \(\rho|_H\), then \[ m_{\alpha\beta}\le\sqrt{b+1}\,D_\alpha D_\beta. \tag{142}\] By Frobenius reciprocity [8], \(m_{\alpha\beta}\) is the multiplicity of \(\rho\) in \(W=\operatorname{Ind}_H^{S_n}(\alpha\otimes\beta)\), so \(m_{\alpha\beta}^2\le\dim\operatorname{End}_{S_n}(W)\). Realize \(W\) in fibers of dimension \(D_\alpha D_\beta\) over \(S_n/H\). An equivariant block map at one representative ordered pair of cosets determines its transported maps on that pair orbit. The stabilizer can impose linear constraints on the representative map and can only reduce the number of choices. Cosets correspond to \(b\)-subsets, and intersection size classifies their ordered-pair orbits, of which there are \(b+1\). Each representative map has at most \((D_\alpha D_\beta)^2\) parameters. This proves (142).

The complete inverse-first sum gives \[\sum_{D_\alpha\le D^{1/100}}D_\alpha^2 \le D^{3/100}\sum_{\alpha\in\widehat{S_b}}D_\alpha^{-1} \le C_\zeta D^{3/100}.\] Lemma 63 gives \(b\le D\). Therefore \[ \frac{\operatorname{tr}(P_{0,\le}P_{1,\le})}{D} \le\frac{\sqrt{b+1}}D \left(\sum_{D_\alpha\le D^{1/100}}D_\alpha^2\right)^2 =O(D^{-1/2+6/100}). \tag{143}\] Equations (138)–(143) bound the squared operator norm by \(16D^{-1/10000}+O(D^{-1/2+6/100})\). Taking square roots proves the stated exponent. Finally, the pass \(\sigma\) contains an independent fair switch for large \(d\). Toggling it changes sign, and its independence from the modified \(\pi\) gives zero sign coefficient for the block. ◻

Proof of Theorem 59. Take \(r=60000\) successive blocks, modifying their long segments independently. The product law \(\mu\) differs from the unmodified product by at most \(r\,o(1)=o(1)\). The block phases may differ, but the estimates are uniform. Their Fourier matrices multiply in chronological order, and Proposition 64 gives \[\|\mathbb E_\mu\rho\|_{\mathrm{op}} \le C_1D_\rho^{-3}, \qquad C_1=C_0^{60000}, \qquad \rho\notin\{\mathbf1,\operatorname{sign}\}.\] The sign coefficient is zero. Plancherel, the Hilbert–Schmidt inequality, and Lemma 63 yield \[4\|\mu-U_{S_n}\|_{\mathrm{TV}}^2 \le C_1^2\sum_{\rho\notin\{\mathbf1,\operatorname{sign}\}}D_\rho^{-4} \le C_1^2(2/n)^3\sum_{\rho\in\widehat{S_n}}D_\rho^{-1}=o(1).\] Restoring the modified segments still costs \(o(1)\).

The unmodified coordinate product has \(t=60000(101d-1)\) layers and equals \(R^{-t}Y_t\) in the notation of Section 3. Restoring the one terminal rotation preserves uniform measure and total variation even when \(t\) is not divisible by \(d\). A deterministic initial deck is a right translate, so it has the same distance. Since \(t\le6060000d\), the upper bound follows. The lower bound is the earlier one-card or support lower bound. ◻

Proof. Write \(S=(S_1,S_2)\) for an output partition and \(p,u\) for its actual and uniform laws. For fixed \(S\), a permutation is a pair of bijections \(J_i\to S_i\). Let \(\mu_S\) be their conditional joint law, with marginals \(\mu_{i,S}\), and let \(U_{i,S}\) be uniform on the \(i\)th bijection space. Either injection determines \(S\) completely. The uniform injection law has partition marginal \(u\) and conditional bijection law \(U_{i,S}\). Replacing \(u\) by \(p\) in this target changes it by exactly \(\epsilon_{\mathrm{part}}\) in total variation. The triangle inequality therefore gives \[\sum_Sp(S)\|\mu_{i,S}-U_{i,S}\|_{\mathrm{TV}} \le\epsilon_i+\epsilon_{\mathrm{part}}.\]

For every \(S\) with \(p(S)>0\), take a maximal coupling of \(\mu_{i,S}\) with \(U_{i,S}\) for each \(i\). Given the original pair, apply the two coupling kernels using separate auxiliary randomness. Each resulting coordinate has its prescribed uniform marginal. The probability that the pair changes is at most the sum of the two marginal coupling errors. Thus replacing the conditional joint laws while retaining \(p\) costs at most \(\epsilon_1+\epsilon_2+2\epsilon_{\mathrm{part}}\). On partitions of zero \(p\)-mass choose any joint law with uniform marginals. Finally replace \(p\) by \(u\), retaining these conditional joint laws. This costs exactly \(\epsilon_{\mathrm{part}}\) and proves (144). ◻

Proof of Theorem 65. Choose \(C\) from the shared half-list lemma and a fixed positive integer \(K\) with \(K\delta>3\). Take \(K\) successive blocks, repairing each \(A_0\) independently. The phases and omitted coordinates may differ, but the estimates above are uniform over these choices. Let \(\mu\) be the product law of the repaired blocks. Its Fourier matrix is a product of \(K\) matrices satisfying (149). Using submultiplicativity of the operator norm and \(\|Q\|_{\mathrm{HS}}\le\sqrt D\|Q\|_{\mathrm{op}}\), Plancherel gives \[4\|\mu-U_n\|_{\mathrm{TV}}^2 \le C_2^K\sum_{\lambda\ne(n),(1^n)}D_\lambda^{2-K\delta} =o(1).\] Indeed \(2-K\delta<-1\), and the inverse-first-degree sum outside the trivial and sign representations tends to zero. The sign term vanishes exactly. This estimate uses no normality of the Fourier matrices.

Undoing all \(K\) repairs costs \(o(1)\) by coupling the independent blocks, since \(K\) is fixed. Their original total length is \(K((d-1)+Cd)\) coordinate layers. At that single terminal time, the relation \(Y_t=R^tX_t\) from Section 3 restores the deterministic coordinate rotation. It preserves uniform measure and total variation, whether or not the block length is divisible by \(d\). Each coordinate layer represents one physical shuffle. Finally, applying the position permutation to any fixed deck is a bijection of the state spaces, so the same convergence holds with the stated worst-start quantifier. ◻