Sum-of-Squares Lower Bounds for Sherrington-Kirkpatrick via Planted Affine Planes

TL;DR

This paper establishes high-degree Sum-of-Squares lower bounds for the Planted Affine Planes problem, which implies similar bounds for certifying energy bounds in the Sherrington-Kirkpatrick model, highlighting computational hardness.

Contribution

It introduces the Planted Affine Planes problem and proves degree-$n^{ ext{Omega}( ext{epsilon})}$ SoS lower bounds, connecting it to the SK model's certification hardness.

Findings

High-degree SoS fails to refute the PAP problem for $m \,\leq\, n^{3/2-\epsilon}$

Lower bounds imply hardness for certifying SK Hamiltonian energy bounds

Connects average-case hardness of PAP to SK model via SoS lower bounds

Abstract

The Sum-of-Squares (SoS) hierarchy is a semi-definite programming meta-algorithm that captures state-of-the-art polynomial time guarantees for many optimization problems such as Max-$k$-CSPs and Tensor PCA. On the flip side, a SoS lower bound provides evidence of hardness, which is particularly relevant to average-case problems for which NP-hardness may not be available. In this paper, we consider the following average case problem, which we call the \emph{Planted Affine Planes} (PAP) problem: Given $m$ random vectors $d_1,\ldots,d_m$ in $\mathbb{R}^n$, can we prove that there is no vector $v \in \mathbb{R}^n$ such that for all $u \in [m]$, $\langle v, d_u\rangle^2 = 1$? In other words, can we prove that $m$ random vectors are not all contained in two parallel hyperplanes at equal distance from the origin? We prove that for $m \leq n^{3/2-\epsilon}$, with high probability, degree-$n^{\Omega(\epsilon)}$ SoS fails to refute the existence of such a vector $v$. When the vectors $d_1,\ldots,d_m$ are chosen from the multivariate normal distribution, the PAP problem is equivalent to the problem of proving that a random $n$-dimensional subspace of $\mathbb{R}^m$ does not contain a boolean vector. As shown by Mohanty--Raghavendra--Xu [STOC 2020], a lower bound for this problem implies a lower bound for the problem of certifying energy upper bounds on the Sherrington-Kirkpatrick Hamiltonian, and so our lower bound implies a degree-$n^{\Omega(\epsilon)}$ SoS lower bound for the certification version of the Sherrington-Kirkpatrick problem.