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\title{Partial permutation information implies Fourier bounds\\\large a complete, self-contained proof}
\author{}
\date{}
\begin{document}
\maketitle

\section*{Overview and the two theorems we prove}

A permutation can look perfectly uniform on almost every label and still be
far from uniform. The clearest example is parity. Let $\mu$ be uniform on the
alternating group $A_n$. The images of any $n-2$ labels are exactly uniform,
because fixing the images of $n-2$ labels leaves a transposition free, and a
transposition reverses parity. Yet $\mu$ is as far from the uniform law on
$S_n$ as it can be: $\|\mu-U_{S_n}\|_{\TV}=\tfrac12$.

So partial information, by itself, never forces full mixing. This note proves
the precise sense in which it almost does. There are two ingredients. The
first turns information about partial permutations into a bound on the
\emph{Fourier matrix} of $\mu$. The second uses the Fourier bound to force
mixing, provided the sign (parity) of $\mu$ is unbiased. The parity example
sits exactly at the boundary: it has perfect partial information but a
sign mean of $\pm1$.

Throughout, $G=S_n$, $U=U_{S_n}$ is uniform on $G$, and for a probability
$\mu$ we write
\[
 \|\mu-U\|_{\TV}=\tfrac12\sum_{g\in G}|\mu(g)-U(g)|.
\]
For an irreducible unitary representation $\rho$ of $G$ we set
\[
 \widehat\mu(\rho)=\sum_{g\in G}\mu(g)\,\rho(g),
 \qquad D_\rho=\dim\rho,
 \qquad \|A\|_{\HS}^2=\tr(A^*A),
\]
using the \emph{unnormalized} trace. For a partition $\lambda\vdash n$ we write
$D_\lambda$ for the dimension of the corresponding irreducible, and
\[
 C_u=\sup_{m\ge1}\sum_{\lambda\vdash m}D_\lambda^{-u},
 \qquad
 Z_u(n)=\sum_{\lambda\vdash n,\ \lambda\notin\{(n),(1^n)\}}D_\lambda^{-u}.
\]
The two partitions $(n)$ and $(1^n)$ are exactly the one-dimensional types
(trivial and sign); Lemma~\ref{lem:degrees} shows $C_u<\infty$ and
$Z_u(n)\to0$ for every $u>0$.

\begin{theorem}[Coset caps control the whole Fourier matrix]\label{thm:main}
Partition $\{1,\dots,n\}$ into $b$ nonempty blocks $M_1,\dots,M_b$ and let
$H_i=\Sym(M_i)$ act on its block and fix everything else. Let $f\ge0$ be a
measure of total mass at most $1$ obeying the coset cap
\begin{equation}\label{eq:cap}
 f(gH_i)\le \frac{B}{[G:H_i]}\qquad\text{for all }g\in G,\ 1\le i\le b.
\end{equation}
Then for every $u>0$ and every irreducible $\rho$ of degree $D$,
\begin{equation}\label{eq:main}
 \|\widehat f(\rho)\|_{\HS}^2\le b\,B\,C_u\,D^{-1+(u+2)/b}.
\end{equation}
\end{theorem}

The headline consequence needs the averaged information about partial
permutations, plus unbiased sign.

\begin{corollary}[Two convolutions suffice]\label{cor:headline}
Let $n\to\infty$ through multiples of $8$. Let $\mu_n$ be probabilities on
$S_n$ with sign mean $o(1)$, and suppose the images of one omitted block of
size $n/8$ are on average nearly uniform:
\[
 \E_{|M|=n/8}\,\|\mu_{n,H_M}-U_{G/H_M}\|_{\TV}=o(1),
\]
where $\mu_{n,H_M}$ is the pushforward of $\mu_n$ to the cosets $S_n/H_M$.
Then
\[
 \|\mu_n^{*2}-U_{S_n}\|_{\TV}\longrightarrow0.
\]
\end{corollary}

The proof has three independent parts, and we give each in full. Section
\ref{sec:degrees} proves the degree sums $C_u<\infty$, $Z_u(n)\to0$ together
with an exponential dimension lower bound. Section \ref{sec:trim} passes from
the averaged hypothesis to a single subprobability with unit coset cap.
Section \ref{sec:fourier} proves Theorem~\ref{thm:main} and, with Plancherel,
deduces Corollary~\ref{cor:headline}.

\section{Symmetric-group degrees}\label{sec:degrees}

Write $\lambda'$ for the transpose of $\lambda$ and let
\[
 L(\lambda)=\max(\lambda_1,\lambda'_1),\qquad k(\lambda)=n-L(\lambda)
\]
be the length of a longest row or column and the number of boxes below it.
The hook-length formula says
$D_\lambda=n!/\prod_{x\in\lambda}h(x)$, where $h(x)$ counts $x$ and the boxes
strictly to its right or below it.

\begin{lemma}[Tableau lower bound]\label{lem:tableau}
Let $\lambda\vdash n$ have first-row length $a$ (so the lower diagram has
$k=n-a$ boxes). For every integer $b$ with $0\le b\le\min(a,k)$,
\[
 D_\lambda\ \ge\ \binom{a}{b}.
\]
\end{lemma}
\begin{proof}
A Young subdiagram is one closed under moving a box up or left. Starting from
the lower diagram (a genuine Young diagram of size $k$) and repeatedly
deleting a lower corner yields a Young subdiagram $\zeta$ of every size
$0\le b\le k$; fix one of size $b$. Its width is at most $b$, because the
first row of any partition of $b$ has length at most $b$.

Count standard tableaux of the subdiagram made of the full first row (length
$a$) and $\zeta$. Put the labels $1,\dots,b$ in the first $b$ cells of the top
row. Choose any $b$ of the labels $\{b+1,\dots,a+b\}$ for the cells of $\zeta$,
and fill $\zeta$ in one fixed standard relative order (this exists since
$\zeta$ is a Young diagram). Fill the remaining $a-b$ top-row cells
increasingly with the leftover labels, of which there are exactly $(a)-(b)$.
This is a valid standard tableau: the top row is increasing, and every cell of
$\zeta$ lies in a column $1,\dots,b$ whose top cell already carries a label
$\le b$, smaller than any chosen label. Distinct choices of the $b$-subset
give distinct tableaux, so there are $\binom{a}{b}$ of them.

Every standard tableau of a Young subdiagram extends to a standard tableau of
$\lambda$, by adjoining the missing boxes in an order that adds outer corners
with increasing labels. Hence $D_\lambda\ge\binom ab$.
\end{proof}

We also use the standard partition bound: with $p(k)$ the number of partitions
of $k$ and $p(0)=1$,
\begin{equation}\label{eq:partition}
 p(k)\le e^{3\sqrt k}\qquad(k\ge1).
\end{equation}
For a diagram with first row (after orienting a longest line) of length $n-k$
there are at most $p(k)$ lower diagrams, hence at most $2p(k)$ shapes with
$L(\lambda)=n-k$ (the factor $2$ records the two possible orientations).

\begin{theorem}[Degree sums and an exponential deficit bound]\label{lem:degrees}
For every $u>0$: $C_u<\infty$ and $Z_u(n)\to0$. Moreover there are absolute
constants $c_1>0,c_2>0$ such that for all sufficiently large $n$ and every
$\lambda\vdash n$,
\begin{equation}\label{eq:deficit}
 D_\lambda\ \ge\ c_1\,e^{\,c_2\,k(\lambda)}.
\end{equation}
For each fixed positive $k$, $D_\lambda\to\infty$ uniformly over shapes with
$k(\lambda)=k$. One may take $c_1=\tfrac13$ and $c_2=\tfrac{\log 3}{48}$.
\end{theorem}
\begin{proof}
Orient a longest line as the first row; let its length be $L=n-k$, and if it
was originally the first column transpose everything (dimensions unchanged).
If $L\le n/16$, then every hook has length at most $2L\le n/8$, so
$\prod_x h(x)\le(n/8)^n$ and the hook formula gives
$D_\lambda\ge (n/e)^n/(n/8)^n=(8/e)^n\ge2^n$.

So assume $L>n/16$, hence $k=n-L$. Put $b=\min(k,\lfloor L/3\rfloor)$, so
$1\le b\le L/3$ and $b\le k$, and Lemma~\ref{lem:tableau} gives
\[
 D_\lambda\ \ge\ \binom{L}{b}\ \ge\ \Big(\frac{L}{b}\Big)^{b}\ \ge\ 3^{b}.
\]
If $b=k$ then $b=k\ge k/48$, while if $b=\lfloor L/3\rfloor$ then
$\lfloor L/3\rfloor>n/48-1\ge k/48-1$ (using $n=k+L$ and $L>n/16$). In both
cases $b\ge k/48-1$, whence
$D_\lambda\ge 3^{k/48-1}=\tfrac13\,3^{k/48}$. This is \eqref{eq:deficit} with
$c_1=\tfrac13,\ c_2=\tfrac{\log3}{48}$. When $L\le n/16$ the bound $2^n$
certainly dominates $\tfrac13 e^{c_2k}$ for large $n$ since $k\le n$. Finally,
if $b=k$ then $D_\lambda\ge\binom{n-k}{k}$, which tends to infinity for each
fixed $k$ as $n\to\infty$.

Here is the sum. Group shapes by deficit $k\ge1$: there are at most $2p(k)$ of
them, and each has $D_\lambda^{-u}\le 3^{u}e^{-uc_2k}$ by \eqref{eq:deficit}.
For fixed $k$ every term tends to $0$ as $n\to\infty$. The skeleton bound
$2p(k)\,3^u e^{-uc_2k}\le 2\cdot3^u e^{3\sqrt k-uc_2k}$ is summable in $k$ and
independent of $n$. Dominated convergence gives $Z_u(n)\to0$. Adding the two
linear types shows $\sum_{\lambda\vdash n}D_\lambda^{-u}\to2$, and the same
skeleton bound shows the full sum is uniformly bounded, so $C_u<\infty$.
\end{proof}

\section{One capped measure from averaged information}\label{sec:trim}

The averaged hypothesis mentions one omitted block at a time. We first select
a single favourable partition, then trim one measure that obeys every coset
cap at once.

\begin{lemma}[Selecting one partition]\label{lem:select}
Let $b$ divide $n$ and suppose a law $\mu$ on $S_n$ satisfies
$\E_{|M|=n/b}\|\mu_{H_M}-U_{G/H_M}\|_{\TV}\le\delta$. Then some partition into
$b$ equal blocks $M_1,\dots,M_b$ has
$\sum_{i=1}^b\|\mu_{H_{M_i}}-U_{G/H_{M_i}}\|_{\TV}\le b\delta$.
\end{lemma}
\begin{proof}
Choose a uniformly random ordered equal partition. Each block is a uniform
$n/b$-subset, so the expected sum of the $b$ errors is at most $b\delta$; some
partition attains at most the expectation.
\end{proof}

\begin{lemma}[Trimming to unit cap]\label{lem:trim}
Let $H_1,\dots,H_b\le G$, let $\mu$ be a probability, and put
$\delta_i=\|\mu_{H_i}-U_{G/H_i}\|_{\TV}$, $\delta=\sum_i\delta_i$. There is a
subprobability $0\le f\le\mu$ with
\[
 f(G)\ge 1-\delta,\qquad f(gH_i)\le \frac{1}{[G:H_i]}\quad(g\in G,\ 1\le i\le b).
\]
\end{lemma}
\begin{proof}
Trim the systems one at a time. For system $i$, in each coset $Q$ whose current
mass exceeds the uniform level $1/[G:H_i]$, scale all current point masses in
$Q$ down to that level; leave other cosets untouched. Trimming only lowers
masses, so it never breaks a cap established earlier, and it never raises a
coset above the level, so it establishes cap $i$. The mass removed while
trimming system $i$ is the total positive excess of the current coset masses
over uniform for system $i$. Since the current measure is pointwise
$\le\mu$, that positive excess is pointwise no larger than the positive excess
of $\mu$, so the removed mass is at most
$\sum_{Q:\,\mu(Q)>U(Q)}(\mu(Q)-U(Q))=\delta_i$. Summing over $i$ removes at
most $\delta$ in total.
\end{proof}

For our application we choose $b=8$, take the partition from
Lemma~\ref{lem:select} (so $\delta=o(1)$), and let $f$ be the subprobability
from Lemma~\ref{lem:trim}. Then $f\le\mu_n$, $f(S_n)=1-o(1)$, and
\eqref{eq:cap} holds with $B=1$ for all eight blocks.

\section{Plancherel, amplification, and the Fourier bound}\label{sec:fourier}

\subsection{Plancherel and the deficit inequality}

\begin{lemma}[Plancherel with a mass deficit]\label{lem:plancherel}
For every complex function $a$ on $G$,
\begin{equation}\label{eq:plancherel}
 |G|\sum_{g\in G}|a(g)|^2=\sum_{\rho\in\Irr(G)}D_\rho\,\|\widehat a(\rho)\|_{\HS}^2.
\end{equation}
Consequently, if $f\ge0$ has mass $z\le1$ and $U$ is uniform, then
\begin{equation}\label{eq:deficit-plancherel}
 \Big(\sum_{g\in G}|f(g)-U(g)|\Big)^2
 \le 2\,(1-z)^2
   +2\!\!\sum_{\rho\in\Irr(G),\,\rho\ne\mathrm{triv}}D_\rho\,\|\widehat f(\rho)\|_{\HS}^2 .
\end{equation}
\end{lemma}
\begin{proof}
The identity \eqref{eq:plancherel} is the Fourier inversion identity for the
regular representation, obtained by expanding both sides, or from the
orthogonality of matrix coefficients. For the second part write
$f-U=(f-zU)+(z-1)U$, whose trivial Fourier coefficients are $0$ and $z-1$.
By \eqref{eq:plancherel} applied to the complex function $f-zU$,
\[
 |G|\sum_g|f(g)-zU(g)|^2=\sum_{\rho\ne\mathrm{triv}}D_\rho\|\widehat f(\rho)\|_{\HS}^2 ,
\]
since the trivial coefficient of $f-zU$ vanishes. Cauchy--Schwarz and
$\sum_g U(g)=1$ give
\[
 \sum_g|f(g)-U(g)|\le\Big(|G|\sum_g|f(g)-zU(g)|^2\Big)^{1/2}+|z-1|.
\]
Squaring and using $(x+y)^2\le2x^2+2y^2$ gives \eqref{eq:deficit-plancherel}.
\end{proof}

\subsection{From Fourier bounds to mixing}

\begin{proposition}[Amplification]\label{prop:amplify}
Let $\mu_n$ be probabilities on $G$ with $|\widehat\mu_n(\sgn)|=o(1)$. Let
$f_n$ be either a subprobability with $f_n\le\mu_n$ and $f_n(G)\to1$, or a
probability with $\|f_n-\mu_n\|_{\TV}\to0$. Suppose there are constants
$A,\gamma>0$ with
\[
 \|\widehat f_n(\rho)\|_{\HS}^2\le A\,D_\rho^{-\gamma}
 \qquad\text{for every }\rho\ne\mathrm{triv}.
\]
Then for every fixed integer $r$ with $r\gamma>1$,
\[
 \|\mu_n^{*r}-U\|_{\TV}\longrightarrow0.
\]
The same conclusion holds if instead $\|\widehat f_n(\rho)\|_{\op}\le A D_\rho^{-\beta}$
for all $\rho\ne\mathrm{triv}$ and $r\beta>1$.
\end{proposition}
\begin{proof}
Write $z_n=f_n(G)$, so $z_n\to1$ in either alternative. The trivial
coefficient of $f_n^{*r}$ is $z_n^r\to1$, and the sign coefficient satisfies
$|\widehat f_n(\sgn)|\le|\widehat\mu_n(\sgn)|+(1-z_n)=o(1)$ (in the
subprobability case) or the same with $\|f_n-\mu_n\|_{\TV}$ replacing $1-z_n$.
Hence $|\widehat f_n(\sgn)|^{r}=o(1)$.

By Hilbert--Schmidt submultiplicativity
$\|\widehat f_n(\rho)^r\|_{\HS}\le\|\widehat f_n(\rho)\|_{\HS}^{r}\le A^{r/2}D_\rho^{-\gamma r/2}$.
Apply \eqref{eq:deficit-plancherel} to $f_n^{*r}$, whose mass is $z_n^r$ and
whose Fourier matrix is $\widehat f_n(\rho)^r$:
\[
 \Big(\sum_g|f_n^{*r}(g)-U(g)|\Big)^2
 \le 2(1-z_n^r)^2+2A^{r}\!\!\sum_{\rho\ne\mathrm{triv}}D_\rho^{1-\gamma r}
 =2(1-z_n^r)^2+2A^{r}Z_{\gamma r-1}(n).
\]
Both terms tend to $0$ because $\gamma r-1>0$ and $Z_{\gamma r-1}(n)\to0$ by
Theorem~\ref{lem:degrees}. Thus $\|f_n^{*r}-U\|_{\TV}\to0$. If $f_n\le\mu_n$
then $f_n^{*r}\le\mu_n^{*r}$, so
$\|\mu_n^{*r}-f_n^{*r}\|_{\TV}\le(1-z_n^r)/2\to0$; if $\|f_n-\mu_n\|_{\TV}\to0$
then $\|\mu_n^{*r}-f_n^{*r}\|_{\TV}\le r\|f_n-\mu_n\|_{\TV}\to0$ by the
triangle inequality for convolution. In either case $\|\mu_n^{*r}-U\|_{\TV}\to0$.

For the operator alternative, use $\|\widehat f^r\|_{\HS}\le\sqrt D\|\widehat f\|_{\op}^{r}\le A^{r}D^{1/2-\beta r}$
and the identical summation with $2\beta r-2>0$.
\end{proof}

\subsection{The coset-cap estimate}

\begin{lemma}[Cosetwise isotypic estimate]\label{lem:central}
Let $H\le G$, let $f\ge0$ have mass $z\le1$ and obey
$f(gH)\le B/[G:H]$ for all $g$, and let $\tau\in\Irr(H)$ have degree $a$.
Let $P_{\rho,H,\tau}$ be the orthogonal projection of the $G$-module $\rho$
onto the full $\tau$-isotypic subspace of $\rho|_H$. Then
\begin{equation}\label{eq:central}
 \sum_{\rho\in\Irr(G)}D_\rho\,\|\widehat f(\rho)P_{\rho,H,\tau}\|_{\HS}^2\le Bz\,a^2 .
\end{equation}
Every multiplicity copy of $\tau$ is included.
\end{lemma}
\begin{proof}
Define the signed mass $e(h)=\tfrac{a}{|H|}\overline{\chi_\tau(h)}$ for $h\in H$
and $e=0$ off $H$; character orthogonality shows $\widehat e(\rho)=P_{\rho,H,\tau}$.
Since $|\chi_\tau|\le a$ we have
$|e(h)|\le a^2 U_H(h)$ where $U_H(h)=1/|H|$ on $H$. Positivity of $f$ gives,
for every $g$,
\[
 |(f*e)(g)|\le\sum_h f(gh^{-1})|e(h)|\le a^2(f*U_H)(g)
 =a^2\,\frac{f(gH)}{|H|}\le\frac{a^2B}{|G|}.
\]
Also, using the counting inner product and $e*e=e$ (so right convolution by $e$
is an orthogonal projection),
\[
 |G|\sum_g|(f*e)(g)|^2
 =|G|\langle f,f*e\rangle
 \le a^2|G|\langle f,f*U_H\rangle .
\]
Now $(f*U_H)(g)=f(gH)/|H|$, so
$|G|\langle f,f*U_H\rangle=\frac{|G|}{|H|}\sum_g f(g)f(gH)
=\sum_{Q\in G/H}f(Q)^2\cdot[G:H]\le[G:H]\cdot\Big(\frac{B}{[G:H]}\cdot z\Big)$,
where the last step uses $\sum_Q f(Q)=z$ and $f(Q)\le B/[G:H]$. Multiplying,
$|G|\sum_g|(f*e)(g)|^2\le a^2 Bz$. By \eqref{eq:plancherel} the left side
equals $\sum_\rho D_\rho\|\widehat f(\rho)P_{\rho,H,\tau}\|_{\HS}^2$, giving
\eqref{eq:central}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
Fix an irreducible $\rho$ of degree $D$. Restrict to the commuting product
$H=H_1\times\cdots\times H_b$; every joint isotypic space is
$(V_{\tau_1}\otimes\cdots\otimes V_{\tau_b})\otimes\mathcal M$ and embeds into
$\rho|_H$, so $\prod_i\dim\tau_i\le D$. Hence some factor satisfies
$\dim\tau_i\le D^{1/b}$, and on that joint space the projection selecting all
$H_i$-types of degree $\le D^{1/b}$ acts as the identity. Summing over $i$ and
all such types, and using orthogonality of the joint spaces,
\begin{equation}\label{eq:cover}
 I\ \preceq\ \sum_{i=1}^b\ \sum_{\tau\in\Irr(H_i):\ \dim\tau\le D^{1/b}}
 P_{\rho,H_i,\tau}.
\end{equation}
Trace \eqref{eq:cover} against the positive operator $\widehat f(\rho)^*\widehat f(\rho)$.
Since $\tr\big(P\,\widehat f(\rho)^*\widehat f(\rho)\big)=\|\widehat f(\rho)P\|_{\HS}^2$,
\begin{equation}\label{eq:trace}
 D\,\|\widehat f(\rho)\|_{\HS}^2
 =\tr\big(\widehat f(\rho)^*\widehat f(\rho)\big)
 \le\sum_{i=1}^b\ \sum_{\dim\tau\le D^{1/b}}
        \|\widehat f(\rho)P_{\rho,H_i,\tau}\|_{\HS}^2 .
\end{equation}
By Lemma~\ref{lem:central} with $z\le1$, each term obeys
$D\|\widehat f(\rho)P_{\rho,H_i,\tau}\|_{\HS}^2\le B\,(\dim\tau)^2$
when summed over $\rho$; restricting to the single $\rho$ keeps the bound, so
$\|\widehat f(\rho)P_{\rho,H_i,\tau}\|_{\HS}^2\le B(\dim\tau)^2/D$. Summing this
over the selected types and using $(\dim\tau)^2\le D^{(u+2)/b}(\dim\tau)^{-u}$,
\begin{equation}\label{eq:typesum}
 \sum_{\dim\tau\le D^{1/b}}(\dim\tau)^2
 \le D^{(u+2)/b}\!\!\sum_{\tau\in\Irr(H_i)}(\dim\tau)^{-u}
 \le C_u\,D^{(u+2)/b}.
\end{equation}
Substituting into \eqref{eq:trace} gives
$D\|\widehat f(\rho)\|_{\HS}^2\le b\,B\,C_u\,D^{(u+2)/b}$, which is \eqref{eq:main}.
\end{proof}

\begin{proof}[Proof of Corollary~\ref{cor:headline}]
Apply Lemma~\ref{lem:select} with $b=8$ to the hypothesis, obtaining a partition
$M_1,\dots,M_8$ into blocks of size $n/8$ with
$\sum_{i}\|\mu_{n,H_{M_i}}-U_{G/H_{M_i}}\|_{\TV}=o(1)$. Apply Lemma~\ref{lem:trim}
to get a subprobability $f_n\le\mu_n$ of mass $1-o(1)$ obeying \eqref{eq:cap}
with $B=1$ for all eight blocks. By Theorem~\ref{thm:main} with $u=1$, for every
irreducible $\rho$ of degree $D>1$,
\[
 \|\widehat f_n(\rho)\|_{\HS}^2
 \le 8\cdot1\cdot C_1\cdot D^{-1+3/8}=8C_1\,D^{-5/8}.
\]
Now $f_n\le\mu_n$, $f_n(S_n)\to1$, and the sign mean condition gives
$|\widehat\mu_n(\sgn)|=o(1)$. Proposition~\ref{prop:amplify} applies with
$A=8C_1$, $\gamma=5/8$ and $r=2$, since $r\gamma=5/4>1$. Hence
$\|\mu_n^{*2}-U\|_{\TV}\to0$.
\end{proof}

\section*{Scope of this note}

We have proved the paper's main theorem (coset caps bound every nontrivial
Fourier matrix, Theorem~\ref{thm:main}) and its principal consequence (averaged
partial information on $7n/8$ labels plus unbiased sign forces mixing in two
convolutions, Corollary~\ref{cor:headline}). The remaining results of the
original document are \emph{alternative routes} to these same conclusions or to
variants that impose different hypotheses (a random omitted set, entropy
comparisons valid under every tilt, signed palettes, kernel/equivariant forms,
or a conditional-product form). Each is an independent strengthening or
reformulation under different input; none is needed for the theorem proved
here, and all rest on the same three mechanisms displayed above: the degree
sums of Section~\ref{sec:degrees}, the trimming lemma of Section~\ref{sec:trim},
and the central isotypic estimate combined with Plancherel.

\end{document}