Partitioning.tex

\section{Partitioning $ab$-insensitive sets}

{\tiny \textbf{[[Ryan: In what follows, I use some notation ``$n_{\ref{thm:subsp}}(k,d,\eta/2,\pi)$ to mean a number large enough that, when $n \geq n_{\ref{thm:subsp}}()$, every $ab$-insensitive set $L \subseteq [k]^{n}$ with $\pi^{\otimes}(L) \geq \eta/2$ contains a nondegenerate $d$-dimensional subspace.]]}}\\

\begin{theorem} \label{thm:partition} Let $\pi$ be a distribution on $[k]$ and assume $\alpha = \min(\pi) > 0$. Let $L \subseteq [k]^n$ be $ab$-insensitive on $I$, and assume \[ |I| \geq n_{\ref{thm:partition}}(k,d,\eta,\pi) := 2m \left(\frac{k+d}{\alpha}\right)^m \ln(1/\eta), \] where \[ m = n_{\ref{thm:subsp}}(k,d,\eta/2,\pi). \] Then $L$ can be partitioned into a collection $\calS$ of disjoint nondegenerate $d$-dimensional subspaces, along with an error set $E$ satisfying $\pi^{\otimes n}(E) \leq \eta$. \end{theorem} \begin{proof} The proof proceeds in ``rounds, $t = 1, 2, 3, \dots$. In each round, we remove some disjoint subspaces from $L$ and put them into $\calS$; and, the set $I$ shrinks by $m$ coordinates. We will show that after the $t$th round, \begin{equation} \label{eqn:L-shrinks} \pi^{\otimes n}(L) \leq \left(1 - \frac{\alpha^m}{2(k+d)^m}\right)^t. \end{equation} We continue the process until $\pi^{\otimes n}(L) \leq \eta$, at which point we may stop and set $E = L$. Because of~\eqref{eqn:L-shrinks}, the process stops after at most $T = 2(\frac{k+d}{\alpha})^m \ln(1/\eta)$ rounds. By our choice of $n_{\ref{thm:partition}}(k,d,\eta,\pi) = mT$, the set $I$ never ``runs out of coordinates.\\

Suppose we are about to begin the $t$th round; hence, writing $\delta = \pi^{\otimes n}(L)$ we have \[ \eta < \delta \leq \left(1 - \frac{\alpha^m}{2(k+d)^m}\right)^{t-1}. \] The round begins by choosing an arbitrary $J \subseteq I$ with $|J| = m$. We have \[ \delta = \Pr_{x \sim \pi^{\otimes n}}[x \in L] = \Ex_{y \sim \pi^{\otimes \barJ}}[\pi^{\otimes J}(L_y)], \] where we have written $\barJ = [n] \setminus J$ and $L_y = \{z \in [k]^J : (y,z) \in L\}$. Hence $\pi^{\otimes J}(L_y) \geq \delta/2$ for at least a $\delta/2$ probability mass of $y$'s (under $\pi^{\otimes \barJ}$); call these $y$'s ``good. Since $J \subseteq I$ and $L$ is $ab$-insensitive on $I$, it follows that each $L_y$ is $ab$-insensitive on $J$. Since $|J| = m = n_{\ref{thm:subsp}}(k,d,\eta/2,\pi) \geq n_{\ref{thm:subsp}}(k,d,\delta/2,\pi)$, it follows from Theorem~\ref{thm:subsp} that for each good $y$ there must exist a nondegenerate $d$-dimensional subspace $\rho \subseteq L_y$. Since the number of $d$-dimensional subspaces in $[k]^m$ is at most $(k+d)^m$, there must exist a \emph{fixed} nondegenerate $d$-dimensional subspace $\rho_0 \subseteq [k]^J$ such that \begin{equation} \label{eqn:L-mass-decr} \Pr_{y \sim \pi^{\otimes \barJ}}[\rho_0 \subseteq L_y] \geq \frac{\delta}{2(k+d)^m}. \end{equation} Let $R \subseteq [k]^{\barJ}$ be the set of $y$'s with $\rho_0 \subseteq L_y$. Since $L$ is $ab$-insensitive on $I$, it is easy to see\noteryan{Honest.} that $R$ is $ab$-insensitive on $I \setminus J$. Thus $R \times \rho_0$ is $ab$-insensitive on $I \setminus J$; hence so too is $L \setminus (R \times \rho_0)$. We therefore complete the round by setting $I = I \setminus J$ and transferring $R \times \rho_0$ (a disjoint union of subspaces $\{y\} \times \rho_0$) from $L$ into $\calS$. This shrinks the number of coordinates in $I$ by $m$, as promised. And since we can crudely bound $\pi^{\otimes J}(\rho_0) \geq \alpha^m$, we conclude \[ \pi^{\otimes n}(R \times \rho_0) \geq \frac{\delta}{2(k+d)^m} \cdot \alpha^m, \] as required to establish~\eqref{eqn:L-shrinks} inductively. \end{proof}