The Kadison-Singer problem

In quantum mechanics, unlike in classical mechanics, one cannot make precise predictions about how a system will behave. Instead, one is concerned with mere probabilities. Consequently, it is a very important task to determine the basic probabilities associated with a given system. In this snapshot we will present a recent uniqueness result concerning these probabilities

Is the Algorithmic Kadison-Singer Problem Hard?

We study the following $\mathsf{KS}_2(c)$ problem: let $c \in\mathbb{R}^+$ be some constant, and $v_1,\ldots, v_m\in\mathbb{R}^d$ be vectors such that $\|v_i\|^2\leq \alpha$ for any $i\in[m]$ and $\sum_{i=1}^m \langle v_i, x\rangle^2 =1$ for any $x\in\mathbb{R}^d$ with $\|x\|=1$. The $\mathsf{KS}_2(c)$ problem asks to find some $S\subset [m]$, such that it holds for all $x \in \mathbb{R}^d$ with $\|x\| = 1$ that $$\left|\sum_{i \in S} \langle v_i, x\rangle^2 - \frac{1}{2}\right| \leq c\cdot\sqrt{\alpha},$$ or report no if such $S$ doesn't exist. Based on the work of Marcus et al. and Weaver, the $\mathsf{KS}_2(c)$ problem can be seen as the algorithmic Kadison-Singer problem with parameter $c\in\mathbb{R}^+$. Our first result is a randomised algorithm with one-sided error for the $\mathsf{KS}_2(c)$ problem such that (1) our algorithm finds a valid set $S \subset [m]$ with probability at least $1-2/d$, if such $S$ exists, or (2) reports no with probability $1$, if no valid sets exist. The algorithm has running time O\left(\binom{m}{n}\cdot \mathrm{poly}(m, d)\right)~\mbox{ for }~n = O\left(\frac{d}{\epsilon^2} \log(d) \log\left(\frac{1}{c\sqrt{\alpha}}\right)\right), where $\epsilon$ is a parameter which controls the error of the algorithm. This presents the first algorithm for the Kadison-Singer problem whose running time is quasi-polynomial in $m$, although having exponential dependency on $d$. Moreover, it shows that the algorithmic Kadison-Singer problem is easier to solve in low dimensions. Our second result is on the computational complexity of the $\mathsf{KS}_2(c)$ problem. We show that the $\mathsf{KS}_2(1/(4\sqrt{2}))$ problem is $\mathsf{FNP}$-hard for general values of $d$, and solving the $\mathsf{KS}_2(1/(4\sqrt{2}))$ problem is as hard as solving the \mathsf{NAE\mbox{-}3SAT} problem

Twice-Ramanujan Sparsifiers

We prove that every graph has a spectral sparsifier with a number of edges linear in its number of vertices. As linear-sized spectral sparsifiers of complete graphs are expanders, our sparsifiers of arbitrary graphs can be viewed as generalizations of expander graphs. In particular, we prove that for every $d>1$ and every undirected, weighted graph $G=(V,E,w)$ on $n$ vertices, there exists a weighted graph $H=(V,F,\tilde{w})$ with at most \ceil{d(n-1)} edges such that for every $x \in \R^{V}$, $$x^{T}L_{G}x \leq x^{T}L_{H}x \leq (\frac{d+1+2\sqrt{d}}{d+1-2\sqrt{d}})\cdot x^{T}L_{G}x$$ where $L_{G}$ and $L_{H}$ are the Laplacian matrices of $G$ and $H$, respectively. Thus, $H$ approximates $G$ spectrally at least as well as a Ramanujan expander with $dn/2$ edges approximates the complete graph. We give an elementary deterministic polynomial time algorithm for constructing $H$