Proof. Convolution with \(e=e_{H,\tau}\) acts separately in each left coset \(Q=xH\). Since \(\sum_{h\in H}|e(h)|^2=a^2/|H|\), the triangle inequality for a positive weighted sum of translates gives \[\sum_{h\in H}|(f*e)(xh)|^2\le f(Q)^2\frac{a^2}{|H|}.\] One may equivalently normalize the weights by \(f(Q)\) and use Jensen; the zero-mass coset contributes zero. Also \(\sum_Q f(Q)^2\le Bz/[G:H]\). Plancherel on \(G\) proves (7).

There is a second exact expression for the same calculation. Let \(f_Q(h)=f(xh)\) and \(\widehat f_Q(\tau)=\sum_hf_Q(h)\tau(h)\). Subgroup Plancherel yields \[\sum_\rho D_\rho\|\widehat f(\rho)P_{\rho,H,\tau}\|_{\mathrm{HS}}^2 =[G:H]\sum_Q a\|\widehat f_Q(\tau)\|_{\mathrm{HS}}^2.\] Unitarity gives \(\|\widehat f_Q(\tau)\|_{\mathrm{HS}}\le\sqrt a\,f(Q)\) and recovers the same bound. This identity makes the normalization of the Hilbert–Schmidt norm explicit.

There is also a positive-projection proof of the same bound. Right convolution by \(e\) is an orthogonal projection, since \(e*e=e\) and \(e^*(h)=\overline{e(h^{-1})}=e(h)\). With the counting inner product, \[\begin{align*} |G|\|f*e\|_2^2 &=|G|\langle f,f*e\rangle\\ &\le a^2|G|\langle f,f*U_H\rangle \le Bza^2. \tag{8}\end{align*}\] Indeed \(|e(h)|\le a^2U_H(h)\), so positivity of \(f\) bounds the absolute convolution pointwise; and \((f*U_H)(g)=f(gH)/|H|\le B/|G|\). This proves the estimate from idempotence and positivity, independently of the cosetwise norm calculation. ◻

Proof. For each \(i\), take all \(H_i\)-isotypic projections with degree at most \(D^{1/b}\). On every product type in the restriction to \(\prod_iH_i\), at least one such projection is identity. The projections commute and include all multiplicities, so \[ I\preceq\sum_{i=1}^b\sum_{\tau:\dim\tau\le D^{1/b}}P_{\rho,H_i,\tau}. \tag{9}\] Trace this inequality against the positive operator \(\widehat f(\rho)^*\widehat f(\rho)\), apply Lemma 10 after discarding other ambient irreducibles, and use (6). The resulting inequality is \(D\|\widehat f(\rho)\|_{\mathrm{HS}}^2\le Bf(G)bC_uD^{(u+2)/b}\). ◻

Proof. Use the same partition and subprobability as in that corollary. Proposition 11 with \(u=1\) gives HS squared at most \(8C_1D^{-5/8}\). Two powers have nonsign contribution at most \((8C_1)^2 Z_{1/4}(n)\), since \(1-2(5/8)=-1/4\). Proposition 6 handles sign and mass. ◻

Proof. Schur orthogonality makes the functions \(\sqrt{\dim\tau}\,\tau(h)_{jk}\) an orthonormal basis of this coefficient space. At every \(h\), the sum of their squared absolute values is \(\sum_{\tau\in\Lambda}(\dim\tau)^2\), because each \(\tau(h)\) is unitary. Cauchy–Schwarz for expansion in this basis proves the assertion. Multiple copies of a representation give the same coefficient functions and do not enlarge the space. ◻

Proof. Let \(P_i\) project onto all \(H_i\)-types of degree at most \(D^{1/b}\). The \(P_i\) commute, and \(\prod_i(I-P_i)=0\) by the product-degree bound. Thus \(Q_i=P_i\prod_{j<i}(I-P_j)\) are pairwise orthogonal projections summing to identity. Write \(v_i=Q_iv\).

For fixed \(g,w\), the function \(h\mapsto\langle w,\rho(gh)v_i\rangle\) on \(H_i\) uses only the selected types. Lemma 13 gives \[|\langle w,\rho(g)v_i\rangle|^2 \le C_uD^{(u+2)/b}\mathbb E_{h\sim U_{H_i}}|\langle w,\rho(gh)v_i\rangle|^2.\] Average over \(g\) with mass \(f(g)\). Right averaging replaces \(f\) by \(f*U_{H_i}\), whose density relative to \(U_G\) is at most \(B\). Schur orthogonality on \(G\) gives \[\sum_gf(g)|\langle w,\rho(g)v_i\rangle|^2 \le BC_uD^{-1+(u+2)/b}\|w\|^2\|v_i\|^2.\] Jensen for a measure of mass at most one and \(\sum_i\|v_i\|\le\sqrt b\|v\|\) finish the proof. This argument proves the same numerical inequality as Proposition 8, using evaluation of functions instead of dimensions of orbit subspaces. ◻

Proof. Put \(F=|G|f\). On \(xH\), project \(h\mapsto F(xh)\) onto the conjugates of the matrix coefficients of \(\tau\). Its uniform squared norm is \[a\left\|\mathbb E_{h\in H}F(xh)\tau(h)\right\|_{\mathrm{HS}}^2\le B^2a^2,\] because the matrix in the norm has operator norm at most the nonnegative function’s mean, which is at most \(B\). The entries of \(\rho(xh)P_{\rho,H,\tau}\) belong to the \(\tau\) coefficient space. Replacing \(F\) by this projection therefore preserves the tested Fourier matrix. Averaging the bound over cosets and applying full-group Plancherel proves (10).

For the second proof, the character bound \(|\chi_\tau(h)|\le a\) gives \[|(f*e_{H,\tau})(g)|\le\frac{a^2}{|H|}f(gH)\le\frac{Ba^2}{|G|}.\] The density of this convolution has squared \(L^2(U_G)\) norm at most \(B^2a^4\). Plancherel gives (11). No bound on the conditional law inside a coset was required in either proof. ◻

Proof. Trace the covering inequality (9) against \(\widehat f^*\widehat f\). Sum the corresponding estimate of Lemma 15 over all selected subgroup types. For \(j=2,4\), \[\sum_{\dim\tau\le D^{1/b}}(\dim\tau)^j \le C_uD^{(u+j)/b}.\] This proves both bounds, including every multiplicity. ◻

Proof. On \(H_i\) take the central projector density \(k_i(h)=\sum_{\dim\tau\le D^{1/b}}(\dim\tau)\overline{\chi_\tau(h)}\), relative to uniform subgroup measure. Its transform is \(P_i\). Pointwise \(|k_i|\le R_i\), so convolution of the density \(|G|\eta\) with \(k_i\) is bounded in absolute value by \(BR_i\). Plancherel yields the first displayed bound. Alternatively project onto all selected coefficient spaces on each coset. They are orthogonal; their squared mean sums to at most \(B^2R_i\) by the first proof of Lemma 15. This gives the second bound.

Use the orthogonal \(Q_i=P_i\prod_{j<i}(I-P_j)\), apply either bound to \(Q_iv\), and sum their norms. The inequalities \(\sum_i\|Q_iv\|\le\sqrt b\|v\|\) and \(R_i\le C_uD^{(u+2)/b}\) give the conclusions. ◻

Proof. Young branching says that each diagram \(\tau\) occurring under restriction to a block symmetric group is contained in \(\lambda\); conjugating a block to the standard subgroup does not change the set of occurring types [6]. Orient a longest row or column of \(\lambda\) horizontally, transposing the restricted diagrams too if necessary. Each \(\tau\) then has at most \(k_\lambda\) boxes below its first row. Given those lower boxes and the block size, its first row is determined. Thus at most \(M(k_\lambda)\) types occur, regardless of multiplicity. Apply (7) to each selected small type and bound its squared degree by \(D^{2/b}\) in (9). ◻

Proof. A Young subdiagram is closed under moving up and left. Add its remaining boxes in row-major order with successive larger labels; every predecessor is already present. To prove the first bound, retain the whole first row and a Young subdiagram of \(b\) lower boxes. Such a subdiagram is obtained by removing lower corners. Fill the first \(b\) top boxes with \(1,\ldots,b\); choose the \(b\) lower labels from the remaining \(a\) labels and fill them in the relative order of one fixed standard tableau. All lower boxes lie in the first \(b\) columns, so filling the rest of the top row increasingly is valid. For the second bound, reserve only the first \(c\) top labels and choose all \(k\) lower labels from the remaining \(a+k-c\) labels.

For the ballot bound, fix a standard order of the lower boxes and interleave their additions with those of the top row. Require each prefix to have at least as many top-row additions as lower additions. The \(j\)th lower box lies in a column at most \(j\), so its top predecessor has been inserted. Reflection at the first violating prefix counts the valid words by the displayed binomial difference. A valid balanced word with \(k\) additions of each kind, followed by \(a-k\) top additions, gives the last bound. For an actual two-row rectangle the word condition is also necessary, proving the exact count.

Finally, put successive batches of labels on increasing diagonals, in arbitrary order within each diagonal. Every row or column predecessor lies on an earlier diagonal. This gives the product of factorials. The first estimate uses \(t!\ge2^{t-1}\); the second uses \(t!\ge(t/e)^t\) and convexity of \(t\log t\) to obtain \(\sum_jt_j\log t_j\ge n\log(n/q)\). ◻

Proof. The lower hooks are unchanged, and the top hooks are \(a-j+1+b_j\), which proves the identity. The sequence \(b_j\) decreases while \((a-j+1)^{-1}\) increases. The opposite-order rearrangement inequality therefore gives \[\sum_{j=1}^a\frac{b_j}{a-j+1} \le\frac{k}{a}\sum_{j=1}^a\frac1j \le\frac{k(1+\log a)}a.\] For completeness, the rearrangement inequality follows by expanding \(\sum_{i,j}(b_i-b_j)(c_i-c_j)\le0\) for oppositely ordered sequences \(b_i,c_i\). Apply \(\log(1+x)\le x\) to the product. For the last estimate, \(b_j=0\) for \(j>k\), so every relevant denominator is at least \(a-k+1\), and \(\sum_jb_j=k\). ◻

Proof. Put \(L=L(\lambda)\) and \(k=n-L\). If \(L\le n/16\), every hook is at most \(2L\), so \(D_\lambda\ge(n/(2eL))^n\ge2^n\). Otherwise orient the diagram with first row \(L\), and for \(k\ge1\) retain \(b=\min(k,\lfloor L/3\rfloor)\) lower boxes. For sufficiently large \(n\), \(b\ge1\), and Lemma 20 gives \[D_\lambda\ge\binom Lb\ge(L/b)^b\ge3^b, \qquad b\ge k/48-1.\] Indeed \(L>n/16\) implies \(\lfloor L/3\rfloor>n/48-1\ge k/48-1\). Thus one may take \(c_1=1/3\) and \(c_2=(\log3)/48\) after enlarging the size threshold. For a fixed positive \(k\), eventually \(b=k\), and \(D_\lambda\ge\binom{n-k}{k}\to\infty\) uniformly.

There are at most \(2p(k)\) shapes at deficit \(k\). By (17) and (25), their total inverse \(u\)th powers are bounded by a constant depending on \(u\) times \(e^{3\sqrt k-uc_2k}\). This is summable in \(k\), and each fixed positive-\(k\) contribution tends to zero. Dominated convergence proves \(Z_u(n)\to0\). The full sum tends to two; its finitely many initial values are finite, including the value one at \(n=1\). ◻

A ballot proof of the summability assertions. For \(k\ge1\) and \(L=n-k\ge n/2\), (19) gives \[D_\lambda\ge\binom nk\frac{n-2k+1}{n-k+1} \ge\frac{2^k}{k+1}.\] For the second inequality, the ballot fraction is at least \(1/(k+1)\) and \(\binom nk\ge(n/k)^k\ge2^k\). The majorant \(2p(k)(k+1)^u2^{-uk}\) is summable, and the first binomial expression diverges for each fixed \(k\ge1\). If \(L<n/2\) but \(L\ge n/10\), retain a row of length \(L\) and \(L\) lower boxes. The ballot count \(\binom{2L}L/(L+1)\) is exponential in \(n\). If \(L<n/10\), the hook formula gives \((5/e)^n\). The subexponential bound for \(p(n)\) handles both remaining ranges. Adding the one-dimensional terms proves the uniform bound. ◻

Proof. Equation (23) has exponent \(k/(n-2k+1)\le1/2\), proving (26). If \(k>n/4\) and \(L\ge n/8\), keep the first row and \(b=\lfloor n/32\rfloor\) lower boxes. For large \(n\), \(1\le b\le L/4\). The same hook bound gives dimension at least \(\frac12\binom{L+b}b\ge4^b/2\). If \(L<n/8\), every hook is shorter than \(n/4\), giving \((4/e)^n\). Both exceed \(e^{n/100}\) for sufficiently large \(n\). The near-row inverse sum is bounded by \(4\sum_{1\le k\le n/4}p(k)/\binom nk\). Each term vanishes at fixed \(k\) and is bounded by \(4p(k)4^{-k}\); the remaining contribution is at most \(p(n)e^{-n/100}=o(1)\). This proves both assertions about inverse-first sums. ◻

Proof. The partition bounds give \(\log M(k)=O(\sqrt k\log(k+1))=o(k)\). For \(k>n/4\), Lemma 23 gives \(\log D_\lambda\ge n/100\), whereas \(\log M(k)=o(n)\) uniformly for \(k\le n\). For \(k\le n/4\), use \(D_\lambda\ge\frac12\binom nk\ge\frac12 4^k\) once \(k\) exceeds a fixed sufficiently large cutoff. The finitely many positive \(k\) below that cutoff are handled by the uniform divergence of \(\binom nk\). One size threshold works for all shapes. ◻

Proof. Define the complex mass function \(e_{J,\chi}\) to be \(\overline{\chi(j)}/|J|\) on \(J\) and zero elsewhere. Its Fourier transform is the orthogonal projection \(P\) onto vectors satisfying \(\rho(j)v=\chi(j)v\) for all \(j\in J\). With \(U_H,U_J\) denoting uniform probability masses on the subgroups, positivity gives \[|f*e_{J,\chi}|\le f*U_J\le I(f*U_H)\le\frac{IB}{|G|}, \qquad \sum_g|(f*e_{J,\chi})(g)|\le z.\] Plancherel therefore implies \(D\|\widehat f(\rho)P\|_{\mathrm{HS}}^2\le IBz\). Every conjugate of \((J,\chi)\) inside \(H\) satisfies the same bound.

Average the corresponding projections, writing \(\overline P=|H|^{-1}\sum_{h\in H}\rho(h)P\rho(h)^*\). On the full \(H\)-isotypic space \(V_\eta\otimes\mathcal A_\eta\), a group-algebra element acts on \(V_\eta\) tensored with the identity on the multiplicity space \(\mathcal A_\eta\). Schur’s lemma thus gives \[\overline P|_{V_\eta\otimes\mathcal A_\eta} =\frac{\operatorname{rank}(P|_{V_\eta})}{d_\eta}I.\] These scalars are nonnegative and the scalar for \(\eta=\tau\) is at least \(1/d_\tau\). Hence \(\overline P\ge d_\tau^{-1}Q_\tau\). Test this positive-operator inequality against \(\widehat f(\rho)^*\widehat f(\rho)\), and average the preceding Plancherel bounds. The result is (29). ◻

Proof. For each type \(\tau\) of \(H_i\), Lemma 26 and the assumed subcharacter give \[D\|\widehat f(\rho)Q_{i,\tau}\|_{\mathrm{HS}}^2 \le Bz d_\tau^{a+1}.\] Restrict \(\rho\) to the product of the \(H_i\). In any occurring joint type, the product of the factor dimensions is at most \(D\), including when that type has multiplicity greater than one. Some factor therefore has dimension at most \(R=D^{1/b}\). The sum of the commuting projections \(Q_{i,\tau}\) with \(d_\tau\le R\) dominates the identity. Taking its trace against \(\widehat f(\rho)^*\widehat f(\rho)\) gives \[D\|\widehat f(\rho)\|_{\mathrm{HS}}^2 \le Bz\sum_{i=1}^b\sum_{d_\tau\le R}d_\tau^{a+1} \le BzbC_1R^{a+2}.\] The last inequality uses \(d_\tau^{a+1}\le R^{a+2}d_\tau^{-1}\). Lemma 25 supplies \(a=1024\). ◻

Proof. First let \(v\) lie in a single \(H_i\)-isotypic space of type \(\tau_i\). The orbit span \(W=\mathop{\mathrm{span}}\{\rho_\lambda(h)v:h\in H_i\}\) has dimension at most \(d_{\tau_i}^2\). To see this in the presence of multiplicity, write the full \(H_i\)-isotypic space as \(V_{\tau_i}\otimes\mathcal A\); the orbit lies in the image of the linear map \(A\mapsto(A\otimes I)v\) from \(\operatorname{End}(V_{\tau_i})\).

For a test vector \(w\), \[|\langle w,\rho_\lambda(g)v\rangle| \le\|v\|\,\|P_{\rho_\lambda(g)W}w\|.\] The squared projected norm is constant on each \(gH_i\), and its uniform mean is \((\dim W/D_\lambda)\|w\|^2\): the average of the conjugate projections is scalar by Schur’s lemma, and its trace is \(\dim W\). Cauchy–Schwarz for the mass function \(f\), followed by its coset cap, gives \[|\langle w,\widehat f(\lambda)v\rangle|^2 \le zB\frac{\dim W}{D_\lambda}\|v\|^2\|w\|^2 \le\frac{zB d_{\tau_i}^2}{D_\lambda}\|v\|^2\|w\|^2.\] Thus this single-type bound holds before any joint decomposition. In a joint type of (31), the factor degrees satisfy \(\prod_i d_{\tau_i}\le D_\lambda\), since the entire tensor representation occurs at least once. Choose \(i\) with \(d_{\tau_i}\le D_\lambda^{1/b}\) and apply the preceding bound. For an arbitrary \(v\), decompose it into its \(K_\lambda\) orthogonal joint components and sum their bounds. Their norms have sum at most \(\sqrt{K_\lambda}\|v\|\), proving (34). ◻

Proof. Theorem 29 gives \(\|\widehat f(\lambda)\|_{\mathrm{op}} \le\sqrt{2K_\lambda}D_\lambda^{-1/4}\), and Lemma 28 gives \(\log(2K_\lambda)=O(\sqrt k)\). We verify directly that \(\log D_\lambda/\sqrt k\to\infty\) uniformly over \(k\ge1\).

Orient a longest line as the first row, of length \(a=n-k\). For \(1\le k\le n/3\), the tableau construction of Lemma 20 gives \[D_\lambda\ge\binom{n-k}k\ge((n-k)/k)^k.\] For \(k\) above a fixed cutoff, divide the logarithm by \(\sqrt k\) and use \((n-k)/k\ge2\); for the finitely many smaller positive \(k\), the same expression tends to infinity with \(n\). For \(k>n/3\) and \(a\le n/10\), the hook formula gives \(D_\lambda\ge(n/(2ea))^n\ge(5/e)^n\). If instead \(a>n/10\), retain the first row and \(r=\lfloor a/2\rfloor\) lower boxes; there are enough lower boxes because \(k>n/3>a/2\). Their tableaux extend to the full diagram, and the same construction gives \(D_\lambda\ge\binom ar\ge2^r\ge e^{cn}\) for an absolute \(c>0\) and all large \(n\). This proves the asserted uniform divergence. Consequently \(2K_\lambda\le D_\lambda^{1/4}\) eventually, which yields the displayed operator bound. ◻

Proof. Here the joint-type estimate is \(\sqrt{2K_\lambda}D_\lambda^{-3/8}\). For \(k\le n/3\), use \(K_\lambda\le M(k)^8\) and \(D_\lambda\ge\binom{n-k}k\) exactly as above to get \(\log(2K_\lambda)/\log D_\lambda\to0\) uniformly. For \(k>n/3\), the equal block sizes give the coarser count \(K_\lambda\le p(n/8)^8\), whose logarithm is \(O(\sqrt n)\). Let \(a\) again be the longest line. If \(a\ge n^{3/4}\), retain that line and \(r=\lfloor a/3\rfloor\) lower boxes, possible because \(k>n/3>a/3\). Tableau extension gives \(D_\lambda\ge\binom ar\ge3^r\), so \(\log D_\lambda\ge c n^{3/4}\). If \(a<n^{3/4}\), all hooks are at most \(2n^{3/4}\), and \[D_\lambda\ge\frac{n!}{(2n^{3/4})^n} \ge\left(\frac{n^{1/4}}{2e}\right)^n.\] Thus \(2K_\lambda\le D_\lambda^{1/4}\) uniformly for all large \(n\), and the result follows. ◻

Proof. Use densities \(A=|S_n|\alpha\) and \(B=|S_n|\beta\) with respect to uniform probability on \(G=S_n\). Write \(\int_G\) and \(\int_{H_i}\) for normalized counting averages. The two caps read \[ \int_{H_i}A(h^{-1}g)\,dh\le B_L, \qquad \int_{H_i}B(gh^{-1})\,dh\le B_R. \tag{37}\] Indeed these averages are respectively \([G:H_i]\alpha(H_i g)\) and \([G:H_i]\beta(gH_i)\). The density transform \(\int_G A(g)\rho_\lambda(g)\,dg\) equals the mass transform \(\widehat\alpha(\lambda)\), and similarly for \(B\). These conventions give \(\widehat{\beta*\alpha}=\widehat\beta\widehat\alpha\).

For a type \(\tau\) of \(H_i\) of degree \(d_\tau\) and character \(\chi_\tau\), its full isotypic projector is \[P_{i,\tau}=d_\tau\int_{H_i} \overline{\chi_\tau(h)}\rho_\lambda(h)\,dh.\] Centrality makes this scalar on each irreducible carrier, and character orthogonality identifies the scalar as one on type \(\tau\) and zero otherwise. It acts as the identity on all multiplicity copies of \(\tau\). Consider the density \[C(g)=d_\tau\int_{H_i}\overline{\chi_\tau(h)}A(h^{-1}g)\,dh.\] Weighted Cauchy–Schwarz and nonnegativity give \[|C(g)|^2\le d_\tau^2 \left(\int_{H_i}A(h^{-1}g)\,dh\right) \left(\int_{H_i}|\chi_\tau(h)|^2A(h^{-1}g)\,dh\right).\] The first factor is at most \(B_L\). The average over \(G\) of the second factor is \(a\int_{H_i}|\chi_\tau|^2=a\). The Fourier matrix of \(C\) is \(P_{i,\tau}\widehat\alpha(\lambda)\). Plancherel, retaining the \(\lambda\) term, therefore yields \[ D_\lambda\|P_{i,\tau}\widehat\alpha(\lambda)\|_{\mathrm{HS}}^2 \le B_La d_\tau^2, \qquad D_\lambda\|\widehat\beta(\lambda)P_{i,\tau}\|_{\mathrm{HS}}^2 \le B_Rc d_\tau^2. \tag{38}\] For the second inequality use \(d_\tau\int_{H_i}\overline{\chi_\tau(h)}B(gh^{-1})\,dh\), whose transform is \(\widehat\beta(\lambda)P_{i,\tau}\), and the second cap in (37).

The projectors for the different disjoint factors commute, and \(Q_{\boldsymbol\tau}=\prod_iP_{i,\tau_i}\) are the orthogonal joint projections in (31). On a nonzero joint space, \(\prod_i d_{\tau_i}\le D_\lambda\), so some \(d_{\tau_i}\le D_\lambda^{1/b}\). Since \(Q_{\boldsymbol\tau}\le P_{i,\tau_i}\), the two bounds in (38) imply \[\begin{split} \|\widehat\beta Q_{\boldsymbol\tau}\widehat\alpha\|_{\mathrm{HS}} &\le\|\widehat\beta Q_{\boldsymbol\tau}\|_{\mathrm{HS}} \|Q_{\boldsymbol\tau}\widehat\alpha\|_{\mathrm{HS}}\\ &\le\sqrt{B_LB_Rac}\,D_\lambda^{-1+2/b}. \end{split}\] Sum over the \(K_\lambda\) joint types before squaring, and then multiply by \(D_\lambda\). This gives (36). The factor \(K_\lambda^2\) is retained; no multiplicity-free restriction has been assumed. ◻

Proof. Use Lemma 28 with \(\eta=1-4/b>0\) in Theorem 32. It makes the sum of the nonlinear Plancherel terms for \(\beta_n*\alpha_n\) tend to zero. Its trivial coefficient is the product of the two masses, tending to one. Domination bounds the error in each sign coefficient by the corresponding lost mass, so its sign coefficient also tends to zero. Lemma 5 now gives convergence to uniform in \(\ell^1\). Finally, \(\beta_n*\alpha_n\le\nu_n*\mu_n\) and the difference has mass one minus the product of the retained masses. Adding that mass proves the conclusion. ◻

Proof. We first bound the deleted mass, then use each of the sixteen quotient caps on its corresponding factor. Write \[\gamma_i=\frac1{|\Omega|}\sum_{x,y}(K(x,y)-T_i(x,y)).\] On an atom with \(q_i(a,b)>(1+\varepsilon)/L_i\), its positive excess over uniform is at least \(\varepsilon q_i(a,b)/(1+\varepsilon)\). Summing over such atoms and then averaging over \(a\) gives \[ \gamma_i\le\frac{1+\varepsilon}{\varepsilon}\delta=o(1). \tag{43}\] Every \(T_i\) is row- and column-substochastic. A uniform starting distribution remains dominated by uniform after any product of these kernels. The average mass lost in \(T\) is consequently at most \(\Gamma=\sum_i\gamma_i\). Also \(0\le T\le K^{16}\) entrywise. The triangle inequality followed by Cauchy–Schwarz over all \(|\Omega|^2\) entries gives \[ \frac1{|\Omega|}\sum_x \|K^{16}(x,\cdot)-U_\Omega(x,\cdot)\|_{\mathrm{TV}} \le \frac\Gamma2+\frac12\|T-U_\Omega\|_{\mathrm{HS}}. \tag{44}\]

Fix \(i\) and first decompose only under \(H_i\). Each quotient fiber is a regular \(H_i\)-set. After choosing one representative per fiber, the block of \(T_i\) between fibers \(a,b\) is convolution by nonnegative weights of total mass \[t_i(a,b)=q_i(a,b) \mathbf1\{q_i(a,b)\le(1+\varepsilon)/L_i\}.\] On the multiplicity space of a regular-fiber type \(\alpha\), of degree \(f_\alpha\), this block is a weighted sum of unitary \(f_\alpha\) by \(f_\alpha\) matrices. Its Hilbert–Schmidt norm is at most \(\sqrt{f_\alpha}\,t_i(a,b)\). Since each row of \(t_i\) has mass at most one, the cap gives \[ \|T_i^\alpha\|_{\mathrm{HS}}^2 \le f_\alpha\sum_{a,b}t_i(a,b)^2 \le(1+\varepsilon)f_\alpha. \tag{45}\]

Now decompose under the whole product \(H\). Its \(\boldsymbol\alpha\)-isotype is \[\left(\bigotimes_{j=1}^{16}V_{\alpha_j}\right) \otimes\mathcal M_{\boldsymbol\alpha}.\] Viewed only as an \(H_i\)-space, it contributes \(D/f_i\) copies of \(\mathcal M_{\boldsymbol\alpha}\) to the multiplicity of \(\alpha_i\). Thus (45) implies \[\frac D{f_i}\|T_{i,\boldsymbol\alpha}\|_{\mathrm{HS}}^2 \le(1+\varepsilon)f_i.\] Multiplying these inequalities and using Hilbert–Schmidt submultiplicativity proves (42).

For \(u>0\), write \(\mathcal Z_u(m)=\sum_{\alpha\in\widehat S_m}(\dim\alpha)^{-u}\). The degree estimate of Theorem 22 gives \(\mathcal Z_{13}(m)\to2\). The total squared Hilbert–Schmidt contribution of all product types with \(D>1\) is therefore at most \[ (1+\varepsilon)^{16} \left(\prod_{i=1}^{16}\mathcal Z_{13}(m_i)-2^{16}\right)=o(1). \tag{46}\] The subtracted types are precisely the products of trivial and sign representations, since eventually every \(m_i\ge2\).

It remains to handle these \(2^{16}\) scalar carrier types; their multiplicity spaces have not been discarded. If a product type has a sign coordinate \(i\), its \(T_i\) block has squared Hilbert–Schmidt norm at most \(\beta_i\) by (41). All the other factors are operator contractions. Hence its product block has squared Hilbert–Schmidt norm at most \(\beta_i\). The sum over these finitely many types tends to zero.

For the all-trivial type, let \(T_{i,0}\) be its multiplicity operator and let \(u=|\Omega|^{-1/2}\mathbf1\) be the constant vector in that space. The restriction of \(U_\Omega\) is \(P_u=uu^*\). Equation (45) for the trivial \(H_i\) type gives \(\|T_{i,0}\|_{\mathrm{HS}}^2\le1+\varepsilon\), whereas \[\langle u,T_{i,0}u\rangle=1-\gamma_i.\] Consequently \[ \|T_{i,0}-P_u\|_{\mathrm{HS}}^2 =\|T_{i,0}\|_{\mathrm{HS}}^2+1 -2\langle u,T_{i,0}u\rangle \le\varepsilon+2\gamma_i. \tag{47}\] Telescoping the product against \(P_u^{16}=P_u\), with operator contractions on either side of each difference, gives \[\|T_{1,0}\cdots T_{16,0}-P_u\|_{\mathrm{HS}} \le\sum_{i=1}^{16}\sqrt{\varepsilon+2\gamma_i}=o(1).\] This is a bound on the entire all-trivial multiplicity space.

The three estimates prove \(\|T-U_\Omega\|_{\mathrm{HS}}\to0\), so (44) proves average mixing. Finally, equivariance under a transitive group makes all row distances of \(K^{16}\) from uniform equal. That additional symmetry gives the worst-start conclusion. ◻

Proof. For fixed \(s\), the entries of each \(q_i\) sum to one. The same is true for fixed \(s'\) by (49). Therefore the positive excess over uniform in either a quotient row or a quotient column is at most \(\delta\). On an atom exceeding twice uniform, its mass is at most twice its positive excess. Each cutoff loses at most \(2\delta\) in every row and column. Taking the union of the three deleted sets proves the loss bound \(6\delta\). The retained kernel is row- and column-substochastic, so its blocks are operator contractions. Also its \(i\)th quotient marginal is everywhere bounded by \(2/M_i\): an original atom above that value has been entirely deleted.

We check the Fourier orientation before estimating the blocks. The unitary regular decomposition of the right \(H\)-action has carrier \(V_\rho\) and multiplicity \(D\) on each fiber \(s\). In the matrix-coefficient basis \(\rho(g^{-1})_{ab}\), left multiplication \(g\mapsto hg\) replaces \[\rho(g^{-1})\quad\hbox{by}\quad \rho(g^{-1}h^{-1})=\rho(g^{-1})\rho(h^{-1}).\] For each fixed carrier index \(a\), the index \(b\), together with \(s\), is therefore acted on by (53). Each such matrix occurs \(D\) times. In particular composition of the deck kernels gives powers of \(B_\rho\), with no adjoint or reversal of the factors.

Fix \(s,s'\) and an index \(i\). First sum over its group coordinate: \[F(h_{-i})=\sum_{h_i}\widetilde p_{s,s'}(h)\rho_i(h_i^{-1}).\] The quotient cap and unitarity give \(\|F(h_{-i})\|_{\mathrm{HS}}\le2\sqrt{D_i}/M_i\). Applying Plancherel entry by entry on \(H_{-i}\) yields \[\begin{align*} &\sum_{\sigma\in\widehat H_{-i}}(\dim\sigma) \left\|\sum_{h_{-i}}\sigma(h_{-i}^{-1})\otimes F(h_{-i})\right\|_{\mathrm{HS}}^2\\ &\hspace{25mm}=|H_{-i}|\sum_{h_{-i}}\|F(h_{-i})\|_{\mathrm{HS}}^2 \le\frac{4D_i}{M^2}. \end{align*}\] Keep the term \(\sigma=\bigotimes_{j\ne i}\rho_j\) and sum over the \(M^2\) pairs \((s,s')\). A unitary permutation of tensor factors identifies the resulting matrix with \(B_\rho\), proving the first bound in (54). Choose an index with \(D_i\le D^{1/3}\). Then \[\|B_\rho\|_{\mathrm{HS}}^2\le4D^{-1/3},\qquad D\|B_\rho^{15}\|_{\mathrm{HS}}^2 \le D(4D^{-1/3})^{15}=2^{30}D^{-4}.\] Hilbert–Schmidt submultiplicativity suffices here; no normality is assumed. By Theorem 22, the sum over \(D>1\) is at most \[2^{30}\left(\prod_{i=1}^3 \mathcal Z_4(m_i)-8\right)=o(1).\]

We have controlled every carrier of dimension greater than one. The eight scalar carriers remain: one is all-trivial, six contain both a trivial and a sign factor, and one is all-sign. In every case the corresponding \(M\) by \(M\) matrix has Hilbert–Schmidt norm at most two, by (54).

Consider first one of the six mixed types. Choose an index \(i\) where \(\rho_i\) is trivial, so that its matrix entries are the signed sums of the retained marginal \(\widetilde q_i\). Replacing that marginal by \(1/M_i\) gives zero, since at least one of the other two factors is sign and its group sum vanishes. In each row and column, \[\sum|\widetilde q_i-M_i^{-1}| \le\sum|q_i-M_i^{-1}|+\sum(q_i-\widetilde q_i) \le8\delta.\] Thus \(B_\rho\) has absolute row and column sums at most \(8\delta\), and the same Cauchy–Schwarz argument used for nonnegative kernels, applied to absolute entries, gives \(\|B_\rho\|_{\mathrm{op}}\le8\delta\). Hence \[\|B_\rho^{15}\|_{\mathrm{HS}} \le2(8\delta)^{14}=o(1).\]

For the all-trivial type, uniform replacement instead gives the matrix \(J\) with every entry \(1/M\). Write \(B_0=J+E\). The preceding row and column estimate gives \(\|E\|_{\mathrm{op}}\le8\delta\), while \(\|E\|_{\mathrm{HS}}\le\|B_0\|_{\mathrm{HS}}+\|J\|_{\mathrm{HS}} \le3\). Expand \((J+E)^{15}\) into ordered words. The word containing only \(J\) equals \(J\), because \(J\) is an orthogonal projection. Every word containing both letters has a factor \(J\) of Hilbert–Schmidt norm one and at least one factor \(E\) of operator norm at most \(8\delta\), so its Hilbert–Schmidt norm tends to zero. The word \(E^{15}\) has norm at most \(3(8\delta)^{14}\). There are only \(2^{15}\) words. This proves \[\|B_0^{15}-J\|_{\mathrm{HS}}=o(1)\] without bounding the size \(M\) of the trivial multiplicity space.

Finally consider the all-sign type. Its untrimmed matrix is \(A_\chi\). Deleting at most \(6\delta\) mass from each row and column changes each corresponding absolute sum by at most \(6\delta\). Consequently \[\|B_\chi\|_{\mathrm{op}}\le a+6\delta, \qquad \|B_\chi^{15}\|_{\mathrm{HS}} \le2(a+6\delta)^{14}=o(1).\] This is precisely where the separate all-sign hypothesis is used.

The uniform kernel has only the all-trivial block \(J\). Summing all blocks with their carrier multiplicities \(D\) now gives \(\|\widetilde K^{15}-U_\Omega\|_{\mathrm{HS}}=o(1)\). For all sufficiently large members of the sequence \(6\delta<1\). Every row of \(\widetilde K^{15}\) has mass at least \((1-6\delta)^{15}\) and is dominated by the corresponding row of \(K^{15}\). Thus, as in (44), \[\frac1{|\Omega|}\sum_x \|K^{15}(x,\cdot)-U_\Omega(x,\cdot)\|_{\mathrm{TV}} \le45\delta+\frac12\|\widetilde K^{15}-U_\Omega\|_{\mathrm{HS}} =o(1).\] The final transitive-equivariance assertion follows because the untrimmed kernel then has the same row distance at every state. ◻

Proof. Put \(a=1/8\) and \(K_i=H_{R_i}\). Fix an irreducible \(\rho\) of degree \(D>1\). Every tensor type in its restriction to \(\prod_iK_i\) has product of factor degrees at most \(D\). Assign each entire isotypic space to one factor whose degree is at most \(D^{1/b}\). This yields an orthogonal decomposition of any unit vector \(v\) as \(v=\sum_i v_i\), where \(v_i\) contains only these small \(K_i\)-types. For \(W_i=\operatorname{span}\rho(K_i)v_i\), Lemma 7 counts each type once, including all its multiplicities. Together with Lemma 23, it gives \[\dim W_i\le C_1D^{3/b},\qquad C_1=\sup_{m\ge1}\sum_{\tau\in\widehat{S_m}}(\dim\tau)^{-1}.\] For fixed unit \(u,v\), the score \[H_{\rho,u,v,i}(g)=\|\operatorname{Proj}_{\rho(g)W_i}u\|\] depends only on \(g_{-R_i}\). Its uniform second moment is \(\dim W_i/D\) by Schur averaging [4]. Consequently \[ U_n\{H_{\rho,u,v,i}>D^{-a}/(2b)\} \le4b^2C_1D^{-1+2a+3/b}. \tag{55}\] If \(|\langle u,\rho(g)v\rangle|>D^{-a}/2\), at least one score exceeds \(D^{-a}/(2b)\), because \(\|v_i\|\le1\) and \(|\langle u,\rho(g)v_i\rangle|\le H_{\rho,u,v,i}(g)\).

Start with the measure \(\nu=\mu_n\). Whenever its claimed operator bound fails, choose \(\rho,u,v\) with \(|\langle u,\widehat\nu(\rho)v\rangle|>D^{-a}\), record this test, and delete all remaining mass on \[\{g:|\langle u,\rho(g)v\rangle|>D^{-a}/2\}.\] The mass deleted is greater than \(D^{-a}/2\): outside this event the coefficient has absolute value at most \(D^{-a}/2\), everywhere it has absolute value at most one, and the current measure has mass at most one. Since all deleted pieces are disjoint, any one representation of degree \(D\) can be selected at most \(2D^a\) times. There are finitely many irreducibles, so the procedure terminates with all the required bounds. The choices are made from deterministic current measures; thus the eventual finite list of tests is fixed, not selected anew for a sampled permutation.

For each \(i\), let \(E_i\) be the union of its score events in (55) over the complete list of recorded tests. It is a single event determined by \(g_{-R_i}\). Counting tests for each representation gives \[U_n(E_i)\le8b^2C_1\sum_{\rho:\dim\rho>1} D^{-1+3a+3/b} =8b^2C_1\sum_{\rho:\dim\rho>1}D^{-7/16}=o(1),\] by Theorem 22. Apply the marginal hypothesis to this entire union to get \(\mu_n(E_i)\le U_n(E_i)+\delta_n\). Every deleted point belongs to some \(E_i\), so the total mass loss is at most \(\sum_i\mu_n(E_i)=o(1)\). This proves the theorem. ◻

Proof. For any probability \(a\) and uniform probability \(b\) on a finite set, \[\sum_{x:a(x)>2b(x)}a(x) \le2\sum_{x:a(x)>b(x)}(a(x)-b(x)) =2\|a-b\|_{\mathrm{TV}}.\] Apply this to each restriction law. Under \(g\sim\mu\), the expected fraction of sets with \(e_R(g)=0\) is at most \(2\delta\). Markov’s inequality gives \(\mu(E^c)\le16\delta\). Conditioning on \(E\) changes the law in total variation by exactly \(\mu(E^c)\).

Conjugating \(H_R\) by \(a\in S_n\) replaces it by \(H_{aR}\) and conjugates \(P_R\) by \(\rho(a)\). Hence the average projection commutes with \(\rho(S_n)\) and is scalar. To estimate its trace, choose one partition \((R_1,\ldots,R_q)\) into equal blocks. The restriction of \(V\) to \(\prod_iH_{R_i}\) is an orthogonal sum of tensor types, with arbitrary multiplicities. If a type has factor degrees \(d_1,\ldots,d_q\), then \(\prod_i d_i\le D\), since at least one copy of that type occurs in \(V\). Consequently at most \(1/\xi\) factors have degree greater than \(D^\xi\). On this tensor type, \(I-P_{R_i}\) is either the identity or zero according to that inequality. Summing traces, including multiplicities, gives \[\sum_{i=1}^q\operatorname{rank}(I-P_{R_i})\le D/\xi.\] All these ranks agree by conjugacy. This proves (56). Commutation was used only for the disjoint subgroups in this rank calculation.

For the orbit bound, Lemma 7 gives, with all multiplicities included, \[\dim W_R(z)\le\sum_{\dim\tau\le D^\xi}(\dim\tau)^2 \le D^{4\xi}\sum_\tau(\dim\tau)^{-2},\] which is (57).

It remains to turn the orbit estimate into a coefficient estimate. The space \(W=W_R(z)\) is \(H_R\)-invariant, so \(\rho(g)W\) depends only on \(gH_R\). For unit \(u\), Schur averaging and the trace give \[\mathbb E_{g\sim U_n}\|\operatorname{Proj}_{\rho(g)W}u\|^2 =\frac{\dim W}{D}.\] On a coset with \(e_R=1\), the \(\nu\)-mass is at most four times its uniform mass: the original cap is two and normalization costs at most two. Since \(|\langle u,\rho(g)P_Rz\rangle|\le \|\operatorname{Proj}_{\rho(g)W}u\|\), Cauchy–Schwarz gives \[\mathbb E_\nu e_R(g)|\langle u,\rho(g)P_Rz\rangle| \le\left(4\frac{\dim W}{D}\right)^{1/2}.\] This proves (58). ◻

Proof. Suppress \(n\), fix \(\rho\), and use Lemma 37 with \(q=1024\) and \(\xi=1/16\). Thus \(c\ge63/64\), and the right side of (58) is \(A D^{-3/8}\), where \(A=2\sqrt{C_2}\). For \(g\in E\) define \[Q_g=\mathbb E_R e_R(g)P_R,\qquad F_g=I-Q_g/c=c^{-1}\mathbb E_R(1-e_R(g))P_R.\] The equality for \(F_g\) uses the scalar identity \(\mathbb E_RP_R=cI\); it does not require the individual projections to commute. These are positive operators, and \(\|F_g\|_{\mathrm{op}}\le1/(8c)\). Put \(s=\lceil\log D\rceil\), with natural logarithms. The finite identity \[ I=\sum_{j=0}^{s-1}(Q_g/c)F_g^j+F_g^s \tag{59}\] holds because \(Q_g/c=I-F_g\).

We now bound the expected coefficient of each term while preserving its projection order. For fixed unit vectors \(u,w\), the \(j\)th summand in (59) expands as an average over independently chosen \(R_0,\ldots,R_j\), with scalar factor \(c^{-(j+1)}\), indicator \[e_{R_0}(g)\prod_{i=1}^j(1-e_{R_i}(g)),\] and vector \(P_{R_0}P_{R_1}\cdots P_{R_j}w\). Fix the sets before integrating over \(g\). The vector \(z=P_{R_1}\cdots P_{R_j}w\) is then deterministic with norm at most one. Take absolute values and discard the failure indicators for \(i\ge1\). Equation (58) now bounds the expected coefficient by \(c^{-(j+1)}AD^{-3/8}\). In particular the argument never exchanges two projections.

The remainder has norm at most \((8c)^{-s}\). Summing gives \[\mathbb E_\nu|\langle u,\rho(g)w\rangle| \le As c^{-s}D^{-3/8}+(8c)^{-s} \le CD^{-1/4}.\] For the last inequality, \(c^{-s}\le c^{-1}D^{\log(1/c)}\) and \(\log(1/c)\le\log(64/63)<1/32\); the remaining logarithmic factor is absorbed by \(D^{3/32}\). Also \(\log(8c)>1/4\). Taking the supremum over the unit vectors proves the operator bound.

To obtain the convolution conclusion, the sign coefficient of \(\nu_n\) is \(o(1)\), since the original sign coefficient is \(o(1)\) and \(\|\nu_n-\mu_n\|_{\mathrm{TV}}=o(1)\). Plancherel and \(\|B\|_{\mathrm{HS}}\le\sqrt D\|B\|_{\mathrm{op}}\) yield \[4\|\nu_n^{*8}-U_n\|_{\mathrm{TV}}^2 \le o(1)+C^{16}\sum_{\rho:\dim\rho>1}D^{2-16/4} =o(1)+C^{16}\sum_{\rho:\dim\rho>1}D^{-2}\longrightarrow0\] by Lemma 58. Replacing the eight independent factors \(\nu_n\) by \(\mu_n\) costs at most \(8\|\nu_n-\mu_n\|_{\mathrm{TV}}=o(1)\). ◻

Proof. Use Lemma 37 with \(q=256\) and \(\xi=1/16\). In particular \(c\ge15/16\). For a retained permutation put \[M_g=\mathbb E_R e_R(g)P_R.\] Removing at most \(1/8\) of the projections from their scalar average gives \[ \tfrac12 I\le\tfrac{13}{16}I\le M_g\le I. \tag{60}\] For fixed unit vectors \(u,v\) and every integer \(p\ge1\), \[ \mathbb E_\nu|\langle u,\rho(g)M_g^p v\rangle| \le AD^{-3/8},\qquad A=2\sqrt{C_2}. \tag{61}\] Indeed, expand \(M_g^p\) over \(R_1,\ldots,R_p\) in that order. After taking absolute values, retain only the indicator \(e_{R_1}(g)\). For the fixed sets, apply (58) to the deterministic vector \(z=P_{R_2}\cdots P_{R_p}v\), of norm at most one. This proves (61), even when the omitted sets overlap.

The estimates for positive powers must now recover the original vector. Put \(L=\lfloor(\log_2D)/8\rfloor\). The identity \[v=\sum_{j=0}^{L-1}M_g(I-M_g)^jv+(I-M_g)^Lv\] holds also when \(L=0\), with an empty sum. Its remainder has norm at most \(2^{-L}\) by (60). Expand each \(M_g(I-M_g)^j\) as a polynomial in \(M_g\). The absolute values of its coefficients sum to \(2^j\), so (61) gives \[\mathbb E_\nu|\langle u,\rho(g)v\rangle| \le2^{-L}+A\sum_{j=0}^{L-1}2^jD^{-3/8} \le2^{-L}+A2^LD^{-3/8} \le CD^{-1/8}.\] Here \(2^{-L}\le2D^{-1/8}\) and \(2^L\le D^{1/8}\), including \(L=0\). This proves the asserted operator bound.

The sign coefficient is again \(o(1)\). For \(32\) factors the remaining Plancherel sum is at most \[C^{64}\sum_{\rho:\dim\rho>1}D^{2-64/8} =C^{64}\sum_{\rho:\dim\rho>1}D^{-6}\longrightarrow0,\] because \(D^{-6}\le D^{-2}\) and Lemma 58 applies. The total-variation cost of replacing the \(32\) factors is at most \(32\|\nu_n-\mu_n\|_{\mathrm{TV}}=o(1)\). Finally Pinsker’s inequality and Jensen’s inequality give \[\delta_n\le \left(\tfrac12\mathbb E_R D((\mu_n)_{-R}\|U_{-R})\right)^{1/2},\] which proves the entropy version. In this normalization Pinsker’s inequality reads \(\|p-u\|_{\mathrm{TV}}^2\le D(p\|u)/2\). To check the factor, take \(A=\{p\ge u\}\), set \(a=p(A)\) and \(b=u(A)\), and group the entropy sum over \(A\) and its complement. The log-sum inequality gives \(D(p\|u)\ge D(\operatorname{Ber}(a)\| \operatorname{Ber}(b))\). As a function of \(a\), the binary relative entropy on the right vanishes with zero derivative at \(a=b\) and has second derivative \(1/[a(1-a)]\ge4\). It is therefore at least \(2(a-b)^2=2\|p-u\|_{\mathrm{TV}}^2\), with boundary cases obtained by limits. ◻

Proof. The event \(\mathcal G\) and the retained endpoint law \(\mu\) are fixed before choosing the palette. Fix now a palette on \(B\) and a coset \(xH\), and put \(E=\mathcal G\cap\{g\in xH\}\). If \(P(E)=0\) the conclusion is immediate. Otherwise write \(Q=P(\,\cdot\mid E)\). The three laws are related by \[P(E)=z\mu(xH),\qquad D(Q\Vert P)=-\log P(E).\] Choose the reveal order independently of the setting, using the \(AB\) convention under both \(P\) and \(Q\). Since \(E\) does not involve the order, conditioning the setting on \(E\) leaves this order uniformly distributed. For a fixed order, data processing followed by the entropy chain rule gives \[-\log P(E)\ge \sum_{r=0}^{n-v-1}\mathbb E_Q D\bigl(\mathcal L_Q(Y_{r+1}^o\mid\mathcal F_r^o) \Vert\mathcal L_P(Y_{r+1}^o\mid\mathcal F_r^o)\bigr).\] For labels in \(A\), membership in \(xH\) prescribes a singleton target; for a label in \(B_i\) it prescribes the target set \(xB_i\). At a history that occurs under \(Q\), let \(a_r\ge1\) be the number of unoccupied sites in this target. The next endpoint lies in that target with \(Q\)-probability one. For any probability \(R\) supported on an event \(F\) of positive reference probability, the identity \(D(R\Vert P')=D(R\Vert P'(\,\cdot\mid F))-\log P'(F)\) shows that its entropy is at least \(-\log P'(F)\). Apply this observation to the next observation and then use (70). Since \(m=n-r\le n\), \[\begin{split} -\log\left(\frac{a_r}{m}+n\Delta_r^o\right) &=\log(m/a_r)-\log\left(1+\frac{nm\Delta_r^o}{a_r}\right)\\ &\ge\log(m/a_r)-n^2\Delta_r^o. \end{split}\] This lower bound is valid even when the displayed upper bound on the reference probability exceeds one. Averaging the chain rule over orders therefore yields \[ -\log P(E)\ge \mathbb E_{AB}\mathbb E_Q \sum_{r=0}^{n-v-1}\log((n-r)/a_r)-n^2a_n. \tag{73}\] The error bound uses (71) pointwise on \(E\). It does not use an average error estimate under \(P\) after the change of law to \(Q\).

It remains to count the information lost by omitting the last \(v\) labels. If all \(n\) labels were revealed, the numerator counts would multiply to \(n!\), while the target counts would multiply to \(\prod_i l_i!\). The full logarithmic sum is consequently \(\log[S_n:H]\), independently of the compatible observations. The last \(v\) labels are in \(B\). For \(v=0\) the missing contribution is zero. For \(v>0\), if their color counts are \(t_1,\ldots,t_u\), it is \[\log\frac{v!}{\prod_i t_i!} \le v\mathcal H((t_i/v)_i).\] The inequality follows because the multinomial probability of these counts under probabilities \(t_i/v\) is at most one. The last \(v\) labels form a uniform subset of \(B\) under the order average, so \(\mathbb E_{AB}(t_i/v)=p_i\). Concavity of entropy gives expected missing contribution at most \(v\mathcal H(p)\). Insert this bound in (73), exponentiate, and use \(\mu(xH)=P(E)/z\). The \(BA\) order gives the same result for palettes on \(A\). Every error term is independent of the chosen palette and coset, which proves simultaneous uniformity. ◻

Proof. Transpose the diagram if necessary, so its first row has length \(L\). Apply the first-row hook bound (22) with row length \(L\), \(k\) lower boxes, and total size \(b\). The dimension of the lower diagram is at least one, so \[\log D_\gamma\ge\log\binom bL-\frac{k(1+\log L)}{L}.\] If \(k\le b/2\), the correction is \(O(k\log b/b)\) and \(\binom bk\ge(b/k)^k\), proving the first case. If \(b/10\le L<b/2\), the binomial coefficient has logarithm at least \(L\log(b/L)\ge(b/10)\log2\), and the correction is \(O(\log b)\). If \(L<b/10\), every hook is at most \(2L<b/5\), so \(D_\gamma\ge(b/e)^b/(b/5)^b=(5/e)^b\). These bounds prove the second case and the stated consequence. ◻

Proof. Starting with \(\beta\), remove a first row or a first column of maximal length, and repeat on the remaining partition until it is empty. The successive lengths are the \(l_i\). Assign a trivial character to a row removal and a sign character to a column removal. If a first row is removed, the remaining partition interlaces the original one: putting back its size is a horizontal-strip addition in the Pieri rule. Transposing gives a vertical-strip addition for a first column. Iterated Pieri, transitivity of induction, and Frobenius reciprocity give the character occurrence. Conjugation supplies every placement.

Let \(L=l_1\) and \(k=h-L\). Every hook in a strip of length \(l\) at the time of its removal is at most \(2l\), because \(l\) is the longer of the current first row and first column. Its hook product is at most \((2l)^l\le(2e)^l l!\). For the first strip use the sharper bound \(L!e^k\), obtained from \(\log(1+x)\le x\) and the fact that the sum of the numbers of boxes below the strip is \(k\). Remaining hooks are unchanged by each removal. Thus, if \(M=h!/\prod_i l_i!\), the hook formula yields \[\log M\le\log D_\beta+(1+\log(2e))k.\] To compare this multinomial count with entropy, set \(f(j)=j\log j-\log j!\), with \(f(0)=0\). Then \(f(j)\ge0\) and \(0\le f(j)-f(j-1)\le1\). Consequently \[h\mathcal H((l_i/h)_i)-\log M =f(h)-\sum_i f(l_i)\le f(h)-f(L)\le k.\] If \(k>0\), Lemma 42 absorbs this total \(O(k)\) error into \(C_0\log D_\beta\). If \(k=0\), the representation is trivial or sign, and the one-color palette has entropy zero. ◻

Proof. Put \(J=\widehat\mu(\lambda)\) and \(D=D_\lambda\), using a unitary model \(\rho_\lambda\). We first bound \(J\) on each half-isotypic component, including its multiplicities. Combining the halves will bound each joint component; we then absorb the number of joint types into a small power of \(D\). For a type \(\beta\) of \(S_B\), choose its signed palette from Lemma 43. Equations (76) and (77) give, for large \(n\), \[ \mu(xH)\le\frac{2D_\beta^{1/10}}{[S_n:H]}. \tag{80}\] This holds for every conjugate of the chosen subgroup within \(S_B\).

Let \(P_{H,\theta}\) be the orthogonal projection onto the \(\theta\)-character space in the restriction to \(H\). Define the signed mass \(\theta_H\) to equal \(\overline{\theta(h)}/|H|\) on \(H\) and zero elsewhere. Right convolution by \(\theta_H\) has Fourier matrix \(JP_{H,\theta}\). On a coset \(xH\) each resulting mass has absolute value at most \(\mu(xH)/|H|\). Hence \[|S_n|\sum_{x\in S_n}|(\mu*\theta_H)(x)|^2 \le [S_n:H]\sum_{C\in S_n/H}\mu(C)^2 \le 2D_\beta^{1/10}.\] Plancherel, retaining the \(\lambda\) term, proves \[ D\|JP_{H,\theta}\|_{\mathrm{HS}}^2\le2D_\beta^{1/10}. \tag{81}\]

Let \(\Pi_{B,\beta}\) denote the entire \(\beta\)-isotypic projection. Average \(P_{H,\theta}\) over conjugation by \(S_B\). On the \(\beta\) carrier its rank is at least one, so Schur’s lemma makes this average at least \(D_\beta^{-1}I\) on that carrier. The group algebra acts as the identity on its multiplicity space. The average is therefore at least \(D_\beta^{-1}\Pi_{B,\beta}\) in positive semidefinite order on the full representation. Tracing against \(J^*J\) in (81) gives \[ \|J\Pi_{B,\beta}\|_{\mathrm{HS}}^2 \le \frac{2D_\beta^{11/10}}{D}. \tag{82}\] The identical estimate holds for a type \(\alpha\) of \(S_A\).

Every constituent of the product of the two halves has a small carrier on at least one side. Indeed, the commuting projections \(\Pi_{A,\alpha}\Pi_{B,\beta}\) decompose the identity orthogonally, and every nonzero such component has \(D_\alpha D_\beta\le D\). Use (82) for the smaller degree to obtain \[ \|J\Pi_{A,\alpha}\Pi_{B,\beta}\|_{\mathrm{HS}}^2 \le 2D^{-9/20}. \tag{83}\]

We verify that the number of type pairs costs only a vanishing power of \(D\). Let \(L=\max(\lambda_1,\lambda'_1)\) and \(k=n-L>0\). Branching forces every half-type diagram to be contained in \(\lambda\). If the longest line is a column, transpose \(\lambda\) and every contained half-type diagram for this count. Containment and dimensions are preserved, and each resulting tail beyond the first row has at most \(k\) boxes. The number of choices for either half-type is at most \(B_k=\sum_{j=0}^k p(j)\), where \(p(j)\) is the number of partitions of \(j\). The partition bound (17) gives \[B_k\le(k+1)e^{3\sqrt k}=e^{O(\sqrt k)}.\] Lemma 42 implies \(\log B_k/\log D\to0\) uniformly: if \(k\le\sqrt n\), its denominator is at least \((c_0/2)k\log n\), and if \(k>\sqrt n\), it is at least \(c_1k\). Thus \(B_k^2=D^{o(1)}\) uniformly over the nontrivial, nonsign diagrams. Orthogonality of the domain projections in (83) now yields \[\|J\|_{\mathrm{op}}^2\le\|J\|_{\mathrm{HS}}^2 \le 2B_k^2D^{-9/20}\le D^{-1/3}\] for sufficiently large \(n\). This proves both (78) and (79).

For four factors, Hilbert–Schmidt submultiplicativity, Plancherel and Cauchy–Schwarz give \[4\|\mu^{*4}-U_{S_n}\|_{\mathrm{TV}}^2 \le |\widehat\mu(\mathrm{sgn})|^8 +\sum_{\lambda\notin\{(n),(1^n)\}}D_\lambda^{1-4/3} =o(1),\] since the nonlinear sum is \(Z_{1/3}(n)\to0\) by Theorem 22. Replacing the four factors by \(\mu_0\) costs at most \(4\|\mu_0-\mu\|_{\mathrm{TV}}=o(1)\). Convolution contracts total variation and preserves uniform measure, so each four-factor conclusion implies its thirty-factor counterpart.

There is also a direct thirty-factor completion using only the operator bound (79). It gives \[4\|\mu^{*30}-U_{S_n}\|_{\mathrm{TV}}^2 \le |\widehat\mu(\mathrm{sgn})|^{60} +\sum_{\lambda\notin\{(n),(1^n)\}}D_\lambda^{-8}.\] The sum tends to zero. To see this directly, group the diagrams by \(k=n-\max(\lambda_1,\lambda'_1)\). There are at most \(2p(k)\) diagrams in each group. Lemma 42 gives the summable majorant \(2e^{3\sqrt k-8c_1k}\), while each fixed \(k>0\) has \(D_\lambda\to\infty\) as \(n\to\infty\). Dominated convergence applies. Replacing the thirty factors one at a time changes total variation by at most \(30\|\mu_0-\mu\|_{\mathrm{TV}}\). ◻

Proof. Apply the clipping construction of Theorem 45 with \(q=16\), \(w=\mu\), \(K=E_n\), and \(t=b_n\). It restricts \(\mu\) to the event \(\mathcal G\) on which all sixteen original quotient densities are at most \(e^{b_n}\). The retained subprobability has mass \(z\ge1-\eta_n\) and satisfies every coset cap with constant \(e^{b_n}\). Here \(\eta_n=o(1)\) by (92), so eventually \(z\ge1/2\). Normalizing this one retained measure increases all caps by the same factor \(z^{-1}\le2\), giving (93). Since the construction is an event restriction, its normalization is exactly \(\mu(\,\cdot\mid\mathcal G)\), at total-variation distance \(1-z\) from \(\mu\). ◻

Proof. If both the first row and the first column are shorter than \(n/8\), every hook has length less than \(n/4\). The hook formula then gives \(D_\lambda\ge(4/e)^n\), which suffices. Otherwise transpose if necessary so the first row has length \(m=\max(\lambda_1,\lambda'_1)\ge n/8\). The number \(n-m\) of remaining boxes is at least \(k/2\): it is \(k\) without transposition, and after transposition the assumed bound on the number of rows gives \(n-m\ge n/2\ge k/2\).

Retain this first row and a Young subdiagram of \(s=\min(n-m,\lfloor m/2\rfloor)\) boxes below it. For large \(n\), \(s\ge k/17\). To construct standard tableaux of the retained diagram, put \(1,\ldots,s\) in the first \(s\) boxes of its top row. Choose any \(s\) of the remaining \(m\) labels for the lower diagram, placed in a fixed standard order; fill the remaining top row in increasing order. Every column of the lower diagram is under one of the first \(s\) top boxes, so these are valid tableaux. There are \(\binom ms\ge(m/s)^s\ge2^s\) choices, and each extends to the full diagram by successively adding the omitted boxes in a Young order. Thus \(D_\lambda\ge2^s\), proving the lemma. ◻

Proof. Put \(D=D_\lambda\) and restrict \(V_\lambda\) to \(H_1\times\cdots\times H_{16}\). Its irreducible types are tensor products of sixteen carriers, with arbitrary multiplicities. For an occurring tensor type, the product of its carrier dimensions is at most \(D\), so at least one carrier has dimension at most \(D^{1/16}\). Assign the whole isotypic component of that type to one such index \(j\). This decomposes every unit vector \(v\) into an orthogonal sum \(v=\sum_{j=1}^{16}v_j\), where \(v_j\) is supported only on \(H_j\)-types of dimension at most \(D^{1/16}\).

Branching shows that each \(H_j\)-type diagram is a subdiagram of \(\lambda\); its tail below its first row has at most \(k\) boxes. Let \(p(i)\) count the partitions of \(i\), with \(p(0)=1\). There are at most \[B_k=\sum_{i=0}^k p(i),\qquad \log B_k=O(\sqrt k\log(k+1)),\] such types, by the elementary multiplicity-and-list count in (17). Its sharper exponential bound also suffices.

Let \(W_j\) be the span of the \(H_j\)-orbit of \(v_j\). For one type of degree \(a\), collect all equivalent copies as \(V_a\otimes\mathcal M_a\). The orbit of the component of \(v_j\) lies in the image of \[\operatorname{End}(V_a)\longrightarrow V_a\otimes\mathcal M_a, \qquad T\longmapsto(T\otimes I)v_j.\] Its dimension is at most \(a^2\), regardless of the multiplicity. Consequently \[ \dim W_j\le B_k D^{1/8}. \tag{96}\] This is the point at which grouping equivalent copies is essential: counting each copy separately would lose the degree saving.

We now use the caps to average these small orbit spaces over the full group. Since \(W_j\) is \(H_j\)-invariant, the space \(\rho_\lambda(x)W_j\) depends only on \(xH_j\). Its projection, averaged uniformly over the group, is \((\dim W_j/D)I\) by Schur’s lemma and the trace. For a unit vector \(w\), Cauchy–Schwarz and the coset cap therefore give \[\begin{split} |\langle w,\widehat\nu(\lambda)v_j\rangle| &\le\|v_j\| \left(\mathbb E_{x\sim\nu} \|P_{\rho_\lambda(x)W_j}w\|^2\right)^{1/2}\\ &\le\|v_j\|\sqrt{M_n B_kD^{-7/8}}. \end{split}\] Because \(\sum_j\|v_j\|\le4\), summing yields the quantitative bound \[ \|\widehat\nu(\lambda)\|_{\mathrm{op}} \le4\sqrt{M_n B_k}\,D^{-7/16}. \tag{97}\]

It remains to absorb the growing cap and the type count. Lemma 49 gives \(\log D\ge ck\), whereas \(\log M_n=O(n^{1/50})\) and \(\log B_k=O(\sqrt k\log(k+1))\). For \(k>n^{1/20}\), both of the latter quantities are \(o(k)\) uniformly. Thus, for large \(n\), \(\log(16M_nB_k)\le(5/24)\log D\). Squaring (97) and using this inequality gives \(16M_nB_kD^{-7/8}\le D^{-2/3}\), as required. ◻

Proof. Put \(\sigma=\nu^{*12}*K^{*40}\). The Fourier matrices multiply in this order. Each factor is a contraction, since it is an average of unitary matrices. For \(\lambda'_1\le n/2\) and \(k>n^{1/20}\), Proposition 50 gives \(\|\widehat\sigma(\lambda)\|_{\mathrm{op}}\le D_\lambda^{-4}\). For more than \(n/2\) rows, (99) gives zero. For \(1\le k\le n^{1/20}\), (98) gives \(\|\widehat\sigma(\lambda)\|_{\mathrm{op}}\le n^{-4k}\). These statements use submultiplicativity and do not require normality of any Fourier matrix.

Plancherel bounds four times the squared total-variation distance by \(\sum_{\lambda\ne(n)}D_\lambda^2 \|\widehat\sigma(\lambda)\|_{\mathrm{op}}^2\). For each fixed \(k\), there are at most \(p(k)\) diagrams with \(n-\lambda_1=k\), since the boxes below the first row form a partition of \(k\). By Lemma 49, the contribution from the large-degree range is at most \[\sum_{k>n^{1/20}}p(k)e^{-6ck}=o(1).\] For the small-degree range, branching shows that \(V_\lambda\) occurs in the permutation representation on ordered distinct \(k\)-tuples: its restriction to \(S_{n-k}\) contains the trivial representation because its first row has length \(n-k\). Thus \(D_\lambda\le(n)_k\le n^k\), and that contribution is at most \[\sum_{1\le k\le n^{1/20}}p(k)n^{2k}n^{-8k} \le\sum_{k\ge1}e^{3\sqrt k}n^{-6k}=o(1).\] This also explains why no separate sign hypothesis is needed here: the sign diagram has more than \(n/2\) rows and is annihilated by \(K\). The first assertion follows. Convolution contracts total variation, so replacing twelve factors changes the distance by at most \(12\|\mu-\nu\|_{\mathrm{TV}}=o(1)\), proving the second. ◻

Proof. The subgroup constructed in Lemma 25 is a product of symmetric groups on disjoint assigned row or column sets. Those sets partition the \(m\) labels, so after omitting empty sets it is conjugate to \(J=S_{s_1}\times\cdots\times S_{s_r}\). The occurring character \(\chi\) is trivial or sign on each factor. The full-group and trivial-subgroup choices in that lemma have the same form.

Frobenius reciprocity now places \(V_\lambda\) in \(\operatorname{Ind}_J^{S_m}\chi\). Its coset basis is indexed by the ordered set partitions of sizes \(s_1,\ldots,s_r\), and each group element permutes this basis with signs because \(\chi\) takes values in \(\{1,-1\}\). It is therefore a signed tabloid representation, with \[M=[S_m:J]\le D_\lambda^{1024}\le D_\lambda^{C_0}.\] ◻

Proof. Fix two tabloids of part sizes \(s_1,\ldots,s_r\), and let \(E_0\) be the event that a permutation takes the first to the second. For buffer \(B_z\), let \(s_i^z\) count the elements of the first tabloid’s part \(i\) lying in \(B_z\), and set \[M=\frac{m!}{\prod_i s_i!},\qquad M_z=\frac{|B_z|!}{\prod_i s_i^z!}.\] Relax \(E_0\) by requiring the correct target part only for positions outside \(B_z\). This relaxed event is determined by the corresponding image tuple. Conditional on any successful outside assignment, the unassigned target parts have sizes \(s_i^z\), so a uniform completion succeeds with probability \(1/M_z\). Since \(U_E(E_0)=1/M\), the relaxed event has uniform probability \(M_z/M\). Its probability under \(\nu\), and hence \(\nu(E_0)\), is at most \(2M_z/M\).

The product \(\prod_zM_z\) counts the ways to assign the part names inside the buffers with these fixed counts; all such assignments are among the \(M\) unrestricted assignments with totals \(s_i\). Thus \(\prod_zM_z\le M\), and some buffer has \(M_z\le M^{1/L}\). We have proved the uniform entry bound \[\nu(E_0)\le2M^{-1+1/L}.\] In the signed tabloid representation, each absolute entry of the averaged matrix is bounded by this probability. Each column has absolute sum at most \(\nu(S_E)\le1\), since every group element acts by a signed permutation matrix. Its squared Hilbert–Schmidt norm is consequently at most \(2M^{1/L}\).

Choose the signed tabloid containing \(\gamma\) from Lemma 53. Orthogonal decomposition into irreducibles shows that its squared Hilbert–Schmidt norm bounds the contribution of any one copy of \(\gamma\). Hence \[\frac{\|\widehat\nu(\gamma)\|_{\mathrm{HS}}^2}{D_\gamma} \le2D_\gamma^{C_0/L-1}\le2D_\gamma^{-1/2}.\] The bound by one follows separately because the average of unitaries against a subprobability is a contraction. ◻

Proof. The Littlewood–Richardson rule counts certain tableaux of skew shape \(\lambda/\alpha\) and content \(\beta\) whose reading words satisfy the lattice condition. The word determines the tableau. Every prefix of such a word has weakly decreasing counts of the successive letters. All words of content \(\beta\) with this prefix condition are in bijection with standard tableaux of shape \(\beta\): put the successive integers in the next available box of the row specified by the letter. The prefix condition makes each partial shape a partition. There are \(D_\beta\) such words, so the multiplicity is at most \(D_\beta\). Interchanging the two factors gives the bound by \(D_\alpha\). ◻

Proof of Theorem 52. Fix an irreducible \(\rho=\rho_\lambda\) of \(S_V\), and put \(D=D_\lambda>1\). For almost every \(z\), let \[B_z=\sum_{y\in H}(\nu_A^z\otimes\nu_C^z)(y)\rho(y).\] Then \[\widehat\theta(\lambda) =\mathbb E_Z\sum_x\xi(x)\rho(t_Z)B_Z\rho(x).\] Jensen’s inequality remains valid for a subprobability by adding the zero vector with its missing mass. Since \(\rho(t_Z)\) is unitary, for every vector \(u\) we obtain \[ \|\widehat\theta(\lambda)u\|^2 \le\mathbb E_Z\sum_x\xi(x)\|B_Z\rho(x)u\|^2. \tag{105}\] It is enough to bound the positive operator in the right side.

Fix \(z\) for the moment. The restriction to \(H\) decomposes as \[V_\lambda\big|_H =\bigoplus_{\alpha,\beta\vdash h} V_\alpha\otimes V_\beta\otimes \mathbb C^{c_{\alpha\beta}^\lambda}.\] On one occurring block write \(a=D_\alpha\), \(b=D_\beta\) and \(c=c_{\alpha\beta}^\lambda\). Because the conditional measure is a product, the block of \(B_z\) is exactly \[B_A(\alpha)\otimes B_C(\beta)\otimes I_c, \qquad B_J(\gamma)=\widehat{\nu_J^z}(\gamma).\] Both factor matrices are contractions. The contribution of this block to \(B_z^*B_z\) is therefore bounded in positive-semidefinite order by the operator \(Q_A\) whose block is \[B_A(\alpha)^*B_A(\alpha)\otimes I_b\otimes I_c\] and which is zero on the other blocks. In particular \(Q_A\) commutes with \(\rho(S_C)\).

If \(x\) and \(x'\) have the same ordered inverse images on \(A\), then \(x'=kx\) for some \(k\in S_C\). Thus \(\rho(x)^*Q_A\rho(x)\) depends only on \(x^{-1}|_A\). It is positive, so the pointwise cap (101) gives an operator inequality: \[\begin{align*} \sum_x\xi(x)\rho(x)^*Q_A\rho(x) &\preceq 2\mathbb E_{x\sim U_V}\rho(x)^*Q_A\rho(x)\\ &=2\frac{\operatorname{Tr}Q_A}{D}I =2\frac{cab}{D} \frac{\|B_A(\alpha)\|_{\mathrm{HS}}^2}{a}I. \end{align*}\] Here \(P\preceq Q\) means that \(Q-P\) is positive semidefinite; the uniform average is scalar by irreducibility, and its trace determines the scalar. Dropping the \(A\) factor instead gives the same estimate with \(\|B_C(\beta)\|_{\mathrm{HS}}^2/b\), using the cap on \(x^{-1}|_C\). Consequently the averaged contribution of this block is bounded by \[ 2\frac{cab}{D} \min\left\{1,2a^{-1/2},2b^{-1/2}\right\}I, \tag{106}\] where Lemma 54 supplies the two normalized norms.

The weights \(cab/D\) sum to one. For blocks with \(\max(a,b)\ge D^{1/10}\), summing (106) therefore costs at most \(4D^{-1/20}I\). On each remaining block, Lemma 55 gives \(c\le D^{1/10}\), so \(cab/D\le D^{-7/10}\). By Theorem 22, for all sufficiently large \(h\) the number of partitions of \(h\) with degree less than \(D^{1/10}\) is at most \[D^{1/10}\sum_{\alpha\vdash h}D_\alpha^{-1} \le3D^{1/10}.\] There are at most \(9D^{1/5}\) such pairs. Their total contribution to (106) is at most \(18D^{-1/2}I\). Thus, uniformly in \(z\), \[\sum_x\xi(x)\rho(x)^*B_z^*B_z\rho(x) \preceq(4D^{-1/20}+18D^{-1/2})I \preceq22D^{-1/20}I.\] Insert this into (105) and take square roots. This proves (103). ◻

Proof. Write \(q=\theta(S_{2h})\) and \(j=100\). Convolution preserves pointwise domination of nonnegative measures, so \(\theta^{*j}\le\mu^{*j}\), and the missing mass is \(1-q^j=o(1)\). Also \[|\widehat\theta(\mathrm{sgn})| \le |\widehat\mu(\mathrm{sgn})|+(1-q)=o(1).\] The trivial coefficient of \(\theta^{*j}-U\) is \(q^j-1=o(1)\), and its sign coefficient is \(\widehat\theta(\mathrm{sgn})^j=o(1)\). For all other irreducibles, operator-norm submultiplicativity and \(\|B\|_{\mathrm{HS}}\le\sqrt D\|B\|_{\mathrm{op}}\) give \[\begin{align*} \sum_{D_\lambda>1}D_\lambda \|\widehat\theta(\lambda)^j\|_{\mathrm{HS}}^2 &\le K_0^{2j}\sum_{D_\lambda>1}D_\lambda^{2-2j/40}\\ &=K_0^{200}\sum_{D_\lambda>1}D_\lambda^{-3}=o(1), \end{align*}\] by Theorem 22. No normality assumption on the Fourier matrices is used.

Finite-group Plancherel, including the trivial and sign terms, now yields \[(2h)!\sum_g\left|\theta^{*j}(g)-\frac1{(2h)!}\right|^2=o(1).\] Cauchy–Schwarz gives vanishing absolute sum of this difference. Adding the nonnegative missing measure of mass \(1-q^j\) proves the asserted total-variation convergence for \(\mu^{*j}\). ◻

Proof. Orient the first row to have length \(L=L(\lambda)\) and put \(k=n-L\). If \(L<n/8\), then \(D_\lambda\ge(4/e)^n\), so the contribution is at most \([2(e/4)^{20}]^n=o(1)\). For \(L\ge n/8\) and \(k\ge1\), (22) gives \[D_\lambda\ge\binom nk \exp\left(-\frac{k(1+\log L)}L\right).\] There are at most \(2\cdot2^k\) shapes at each \(k\). For \(1\le k\le\sqrt n\), the exponent is uniformly \(o(1)\) and \(\binom nk\ge(n/k)^k\ge n^{k/2}\). The contribution is at most \[2e^{o(1)}\sum_{k=1}^{\lfloor\sqrt n\rfloor}(2n^{-10})^k=o(1).\] For \(\sqrt n<k\le7n/8\), use \(\binom nk\ge(8/7)^k\) and \(k(1+\log L)/L\le7(1+\log n)\). This contribution is at most \[2e^{140}n^{140} \sum_{k>\sqrt n}[2(7/8)^{20}]^k=o(1),\] since \(2(7/8)^{20}<1\). The full inverse-twentieth sum is uniformly bounded after adjoining the row and column and the finitely many small sizes. Each type of dimension at most \(R\) contributes at least \(R^{-20}\) to that sum, proving (107). ◻

Initial-row interleavings and diagonals. Orient a longest line as the first row \(L=n-k\). For \(1\le k\le n/3\), the bound \(D_\lambda\ge\binom{n-k}{k}\ge2^k\) and the composition count \(2p(k)\le2\cdot2^k\) give the majorant \(2\cdot2^{-k}\), with fixed-\(k\) divergence. If \(k>n/3\) and \(L\ge n/8\), retain \(\lfloor L/2\rfloor\) lower boxes to obtain \[D_\lambda\ge\binom L{\lfloor L/2\rfloor} \ge\frac{2^L}{L+1}\ge\frac{2^{n/8}}{n+1}.\] If \(L<n/8\), diagonal filling gives \(D_\lambda\ge2^{3n/4}\). The shared summation step completes this hook-free proof.

The alternative diagonal cutoff \(L<n/4\) gives \(D_\lambda\ge2^{n/2}\). For \(L\ge n/4\), retain \(b=\min(k,\lfloor L/2\rfloor)\): \(k\le L/2\) gives the same \(2^k\) majorant, while \(k>L/2\) gives the exponential central-binomial bound. ◻

Initial-row interleavings with factorial or hook tails. At \(k\le n/4\), use \(\binom{n-k}k\ge3^k\), with inverse-square majorant \(2(2/9)^k\); at \(k\le n/3\), use \(2^k\) and \(2\cdot2^{-k}\). Either binomial diverges for fixed positive \(k\). The following rows give alternative completions of these near-row bounds. In each middle range retain the first row and \(b\) lower boxes: \[\begin{array}{c|c|c} \text{middle range}&b&\text{lower bound for }D_\lambda\\ \hline k>n/4,\ L\ge n/10&\lfloor L/4\rfloor&4^b\\ k>n/4,\ L\ge n/8&\lfloor L/4\rfloor&4^b\\ k>n/4,\ L\ge n/8&\lfloor n/32\rfloor&4^b\\ k>n/4,\ L\ge n/8&\lfloor n/16\rfloor&2^b\\ k>n/3,\ L\ge n/10&\lfloor L/3\rfloor&3^b\\ k>n/4,\ L\ge n^{0.6}&\lfloor L/4\rfloor&4^b \end{array}\] Every row follows from \(D_\lambda\ge\binom Lb\ge(L/b)^b\). There are enough lower boxes: \(k>n/4\) implies \(k>L/3\), and \(k>n/3\) implies \(k>L/2\); the stated cutoffs also ensure \(b\le L\). The fixed-cutoff rows are exponential in \(n\). Below \(L=n/10\) or \(n/8\), the factorial form of diagonal filling gives, respectively, \((5/e)^n\) or \((4/e)^n\). These are hook-free completions; using the hook formula gives the same two bounds independently. For the moving cutoff, the middle bound is \(e^{cn^{0.6}}\) and hooks below the cutoff give \(n!/(2n^{0.6})^n\). Thus every row completes the inverse-square sum. ◻

First-row hook ratios. For \(k\le n/4\), (23) gives \(D_\lambda\ge e^{-1}\binom nk\), so the inverse-square sum is at most \[2e^2\sum_{1\le k\le n/4}2^k(k/n)^{2k}=o(1).\] The summands inside the sum are at most \(8^{-k}\) and vanish at fixed \(k\). In the remainder \(L<3n/4\), use \((5/e)^n\) for \(L\le n/10\); otherwise retain \(b=\lfloor L/4\rfloor\) lower boxes. The hook ratio gives \(D_\lambda\ge e^{-1}\binom{L+b}b\ge e^{-1}4^b\), where \(b\ge n/50\) for large \(n\).

Two alternative hook cutoffs retain different quantitative losses. For \(L\ge n/2\), opposite rearrangement gives \[D_\lambda\ge\binom nk e^{-k(1+\log L)/L} \ge2^k e^{-k(1+\log L)/L}\ge(\sqrt2)^k\] for large \(n\). Here use \(2p(k)=e^{O(\sqrt k)}\) for domination; the binomial gives fixed-\(k\) divergence. For \(n/10\le L<n/2\), \(\binom nL\) is exponential in \(n\) and the logarithmic correction is \(O(\log n)\); below \(n/10\) use \((5/e)^n\). Alternatively, split at \(L=3n/4\) and \(n/8\): the first range has short-tail loss \(k/(L-k+1)\le1\), the middle uses the same \(b=\lfloor L/4\rfloor\) hook subdiagram, and the last uses \((4/e)^n\). Each choice proves the limit by the shared summation argument. ◻

Second-row width and two-row rectangles. Write \(c=\lambda_2\) after orientation. At \(1\le k\le n/4\), \[D_\lambda\ge\binom{n-c}k\ge\binom{n-k}k.\] Its ratio base is at least \(\sqrt n/2\) for \(k\le\sqrt n\) and at least three for \(\sqrt n<k\le n/4\), so the composition count gives vanishing geometric sums. In the remainder, hooks handle \(L<n/10\). For \(L\ge n/10\) and \(c\le L/2\), \[D_\lambda\ge\binom{k+L-c}k \ge2^{\min(k,L-c)}\ge2^{n/20}.\] The middle inequality follows by retaining \(\min(k,L-c)\) factors in the binomial product. If \(c>L/2\), the diagram contains a two-row rectangle of width \(\lfloor n/20\rfloor\), whose Catalan number is exponential in \(n\). The subexponential total count finishes. The alternative near-row cutoff \(k\le n/3\) gives majorant \(2\cdot2^{-k}\); in its remainder \(n/10\le L<2n/3\), the same minimum is at least \(n/20\), and the rectangle may have width \(c\). ◻

Direct reciprocal bounds by interleaving and ballot words. Interleaving gives \(D_\lambda\ge\binom{n-k}k\ge3^k\) on \(1\le k\le n/4\). Use the sharper deficit count \(2p(k)\le2e^{3\sqrt k}\) to sum \(2p(k)3^{-k}\). For \(k>n/4\), hooks give \(2^n\) if \(L<n/(4e)\); otherwise \(b=\min(k,\lfloor L/2\rfloor)\) is a positive linear fraction of \(n\) and \(\binom Lb\ge2^b\). Equivalently, split at \(L=n/8\), using \(b=\lfloor n/16\rfloor\) above and \((4/e)^n\) below. Both give \(C_1<\infty\) and \(Z_1(n)\to0\) without a near-row hook ratio.

Ballot words permit the coarser composition count near a long row. For \(L\ge n/2\), the binomial difference in (19) diverges at fixed \(k\), while its Catalan bound is at least \(4^k/((k+1)(2k+1))\). The reciprocal contribution is therefore dominated by a constant times \((k+1)(2k+1)2^{-k}\). For \(n/10\le L<n/2\), retain \(L\) lower boxes and use their Catalan bound; below \(n/10\), hooks give \((5/e)^n\).

A quarter-tail ballot completion instead splits at \(L=n/8\) and \(3n/4\). The first range has hook bound \((4/e)^n\); in the middle, \(b=\lfloor L/4\rfloor\) lower boxes give \[D_\lambda\ge\binom{L+b}b\frac{L-b+1}{L+1},\] exponential in \(n\). For \(L>3n/4\), the ballot fraction is at least \(1/2\), so \(D_\lambda\ge\frac12\binom nk\ge\frac12 4^k\). The deficit count and fixed-\(k\) divergence finish this version as well. ◻

A reciprocal hook proof. The hook estimates in Lemma 25 give \(D_\lambda\ge e^{n/64}\) for \(n\ge64\) and \(L\le3n/4\), which beats the subexponential total count. For \(L>3n/4\) and \(k\ge1\), \[D_\lambda\ge e^{-1}(n/k)^k\ge e^{-1}4^k.\] The composition count gives the reciprocal majorant \(2e\,2^{-k}\), and the first bound diverges at fixed \(k\). Thus \(Z_1(n)\to0\) and \(C_1<\infty\) follow independently from the hook comparisons used to control the subgroup index. ◻

Proof. If \(1\le k\le n/3\), the initial-row construction gives \(D_\lambda\ge\binom{n-k}k\ge2^k\), with at most \(2\cdot2^k\) shapes. The contribution is bounded by \(2\cdot2^{-(u-1)k}\), and each fixed-\(k\) contribution tends to zero. For \(k>n/3\) and \(L\ge n/8\), retain \(\lfloor L/2\rfloor\) lower boxes to get \(D_\lambda\ge2^{n/8}/(n+1)\). The remaining contribution in this range is at most \[(n+1)^u2^{(1-u/8)n}=o(1).\] For \(L<n/8\), diagonal filling gives \(D_\lambda\ge2^{3n/4}\), and the contribution is at most \(2^{(1-3u/4)n}=o(1)\). At \(u=16\) these two bounds are exactly \((n+1)^{16}2^{-n}\) and \(2^{-11n}\). At \(u=13\) they vanish because \(13/8>1\) and \(39/4>1\). The finitely many small-size sums give the uniform constants. ◻

Proof. Let \(a\) be the largest value of \(i+j-1\) over cells \((i,j)\). Diagonal filling gives \(D_\lambda\ge2^{n-a}\). Every cell satisfies \(ij\le n\), so its smaller index is at most \(\sqrt n\) and \(a\le L+\sqrt n\). If \(k<n/2\), orient the first row as \(L\). A lower cell satisfies \((i-1)j\le k\), whence \(i+j-1\le k+1\le L\); thus \(a=L\). If \(k\ge n/2\), then \(n-a\ge k-\sqrt n\ge k/2\) for large \(n\). This proves the first bound in both cases. A non-row, non-column shape contains \((L,1)\) after orientation; that hook has \(L\) tableaux, proving the second bound. For each \(k\), at most \(2p(k)\le2\cdot2^k\) shapes occur, so their inverse fourth powers are bounded by \(2\cdot2^{-k}\). At fixed positive \(k\) they tend to zero by \(D_\lambda\ge\sqrt n\). Dominated convergence and the finite-small-size argument finish. ◻

Proof. Orient the longest line as the first row \(L\ge\sqrt n\). For \(1\le k<L/2\), use \(D_\lambda\ge\binom Lk\ge2^k\); the contribution at \(k\) is at most \(2\cdot2^{-9k}\) and tends to zero for fixed \(k\). For \(k\ge L/2\), retain \(\lfloor L/2\rfloor\) lower boxes. The bound \(D_\lambda\ge2^L/(L+1)\) gives total contribution at most \[e^{3\sqrt n}(n+1)^{10}2^{-10\sqrt n}=o(1).\] This proves the limit and the uniform bound. ◻

A direct fourth-power proof. For power \(4\), one may also use the near-row majorant \(2\cdot2^{-3k}\) on \(k\le n/3\), the middle bound \(2^L/(L+1)\) on \(n/8\le L<2n/3\), and the hook bound \((4/e)^n\) on \(L<n/8\). This is a different use of the fourth power from the tail-adapted diagonal argument in Lemma 60. It uses the subexponential total partition count in its last two ranges. ◻

Proof. Let \(P_i\) now select factor degrees at most \(D^{1/8}\). At most eight factors of any product type can exceed that threshold, so \(\bar P=32^{-1}\sum_iP_i\succeq3I/4\). Lemma 13, the coset cap, and full-group Schur orthogonality give \[|\langle w,\widehat f(\rho)P_iv\rangle|^2 \le4C_1D^{-5/8}\|w\|^2\|P_iv\|^2.\] Here the selected coefficient space has dimension at most \(C_1D^{3/8}\). Thus \(\|\widehat fP_i\|_{\mathrm{op}}\le2\sqrt{C_1}D^{-5/16}\). Average over \(i\) and multiply by \(\bar P^{-1}\), whose norm is at most \(4/3\). ◻

Proof. Let \(P_i\) select degrees at most \(D^{2/b}\). At least half the factors of every product type meet this condition, so \(\bar P=b^{-1}\sum_iP_i\succeq I/2\). A vector in the range of \(P_i\) has orbit span of dimension at most \(C_{32}D^{68/b}\) by Lemma 7. The coset cap and Schur averaging, exactly as in Proposition 8, give \[\|\widehat fP_i\|_{\mathrm{op}}\le2\sqrt{C_{32}}D^{-1/2+34/b} \le2\sqrt{C_{32}}D^{-1/4},\] since \(68/256\le1/2\). Average and use \(\|\bar P^{-1}\|_{\mathrm{op}}\le2\). This proof retains the geometric averaging mechanism and only uses the reciprocal power \(32\). ◻

Proof. Decompose a vector orthogonally into full product types \(v=\sum_{\boldsymbol\tau}v_{\boldsymbol\tau}\). If \(d_i\) are the four factor degrees, their product is at most \(D\). The sum of \((\prod_i d_i)^{-1/4}\) over all possible product types is at most \(C_{1/4}^4\). Thus the number \(T\) of occurring types is at most \(C_{1/4}^4D^{1/4}\).

For one type choose \(i\) with \(d_i\le D^{1/4}\). The coefficient function on \(gH_i\) belongs to a single type’s coefficient space, even when that type occurs repeatedly. Lemma 13 bounds its supremum by \(d_i\) times its uniform root mean square. Sum over cosets with cap two, then apply Cauchy–Schwarz over the uniform cosets. Full-group Schur orthogonality yields \[\sum_gf(g)|\langle w,\rho(g)v_{\boldsymbol\tau}\rangle| \le2d_iD^{-1/2}\|w\|\|v_{\boldsymbol\tau}\| \le2D^{-1/4}\|w\|\|v_{\boldsymbol\tau}\|.\] Finally \(\sum_{\boldsymbol\tau}\|v_{\boldsymbol\tau}\|\le\sqrt T\|v\|\), giving the claimed exponent and constant. The count here is of entire product types, unlike the blockwise assignment in Proposition 14. ◻

Proof. Remove the longer of the first row and first column of the current diagram. In a row removal, the realigned lower diagram \(\mu\) is contained in the original diagram \(\lambda\), and \(\lambda/\mu\) is a horizontal strip: every column loses exactly its bottom box. The Pieri rule and Frobenius reciprocity show that restriction to \(S_s\times S_r\) contains \(V_\mu\) tensored with the trivial representation of \(S_r\). For a column removal use the sign representation and the transposed argument. Iteration supplies a product subgroup and a product of trivial and sign characters occurring in the original representation.

To prove the index bound, it remains to compare dimensions at each step. Transpose so that a row is removed; then the diagram has height at most \(r\). If \(c_j\) counts its lower boxes in column \(j\), the hook formula gives \[\frac D{D'}=\frac{\binom{r+s}r}{R},\qquad R=\prod_{j=1}^r\left(1+\frac{c_j}{r-j+1}\right), \qquad \sum_jc_j=s.\] The case \(s=0\) is immediate. For \(r\ge512\) and \(s\ge16r\), the bound \(c_j\le r\) gives \[R\le\binom{2r}r\le4^r, \qquad \binom{r+s}r\ge(1+s/r)^r\ge17^r.\] Thus \(R\le\binom{r+s}r^{1/2}\). For \(1\le s\le16r\), the opposite ordering of \(c_j\) and \((r-j+1)^{-1}\) instead gives \[\log R\le\frac sr\sum_{j=1}^r\frac1j \le\frac{s(1+\log r)}r, \qquad \log\binom{r+s}r\ge s\log(1+r/s).\] Since \((1+\log r)/r\) decreases for \(r\ge512\) and is at most \(\frac12\log(17/16)\) there, this also gives \(D/D'\ge\binom{r+s}r^{1/2}\).

If \(r<512\) and \(s>0\), then \(r\ge2\) and \(r+s\le r^2\). Each standard ordering of the lower diagram yields two distinct standard tableaux: fill the first row before that ordering, or fill its first \(r-1\) boxes, the first lower box, the last top-row box, and then the remaining lower boxes in that ordering. The first lower box is in column one, so the second filling is also valid. These two constructions are injective and have disjoint images. Hence \(D/D'\ge2\), while \(\binom{r+s}r\le2^{r^2}\le2^{C_{\rm peel}}\). This proves (117) in every case. Multiplying it through the successive steps telescopes the dimension ratios. The product of the binomial coefficients is exactly the index of the product subgroup constructed above, proving the assertion. ◻

Proof. We use \(p(h)\le e^{3\sqrt h}\), which follows by evaluating the partition generating function at \(e^{-1/\sqrt h}\): the logarithm of the product is at most \(\sqrt h\sum_{j\ge1}j^{-2}\le2\sqrt h\), and the coefficient bound adds \(\sqrt h\).

The first removed length \(r\) is at least \(\sqrt m\) because both width and height are at most \(r\). If \(r<m/2\), (117) gives \[\log D_\lambda\ge \frac{r\log(m/r)}{C_{\rm peel}}.\] The concavity of \(r\log(m/r)\) bounds it below on \([\sqrt m,m/2]\) by the smaller of \(\frac12\sqrt m\log m\) and \(\frac12m\log2\). After division by \(\sqrt m\), this lower bound tends to infinity. It therefore dominates the logarithm \(3\sqrt m\) of the number of all diagrams, proving a vanishing inverse sum in this range.

If \(m/2\le r<m\), put \(h=m-r\ge1\). At most \(2p(h)\) diagrams have this value, counting both orientations, and (117) gives \(D_\lambda\ge(m/h)^{h/C_{\rm peel}}\). For fixed \(h\) their inverse-degree contribution tends to zero. It is bounded by \(2e^{3\sqrt h}2^{-h/C_{\rm peel}}\), a summable sequence. Dominated convergence handles this range. The remaining diagrams are the row and column, of degree one. Once the nonlinear reciprocal sum is at most one, at most \(Y\) nonlinear types have degree at most \(Y\); adding the two linear types gives \(Y+2\le3Y\). ◻

Proof. Lemma 26, using Lemma 67, bounds a full \(H_i\)-isotypic projection of degree \(d_\tau\) by \(2d_\tau^{C_{\rm peel}+1}/D\) in squared Hilbert–Schmidt norm. As in Theorem 27, the projections with \(d_\tau\le Y=D^{1/b}\) cover the identity. By Lemma 68 there are at most \(3Y\) such types for each factor. Summing gives \(D^{-1}\cdot2b\cdot3Y\cdot Y^{C_{\rm peel}+1}\), which is (118). ◻