Kadison–Singer problem

Let $\ell^2$ denote the Hilbert space of square-summable sequences of complex numbers, with standard orthonormal basis $(e_n)_{n\in\mathbb{N}}$. Let $B$ be the C*-algebra of all bounded (continuous) linear operators from $\ell^2$ to $\ell^2$, and let $D\subseteq B$ be the C*-subalgebra consisting of those operators that are diagonal with respect to $(e_n)_{n\in\mathbb{N}}$, i.e. of those $T\in B$ for which there is a bounded sequence $(\lambda_n)_{n\in\mathbb{N}}$ of complex numbers with $Te_n=\lambda_n e_n$ for all $n$. Both $B$ and $D$ contain the identity operator $I$.

For a C*-algebra $A$ with multiplicative identity $I$, a state on $A$ is a continuous linear functional $\varphi:A\to\mathbb{C}$ such that $\varphi(I)=1$ and $\varphi(T)\geq 0$ for every positive $T\in A$ (i.e. every $T$ of the form $T=S^*S$). The set of states on $A$ is convex, and a state $\varphi$ is called pure if it is an extreme point of this set, that is, if $\varphi=t\varphi_1+(1-t)\varphi_2$ with $0<t<1$ and $\varphi_1,\varphi_2$ states on $A$ forces $\varphi_1=\varphi_2=\varphi$. A state $\psi$ on $B$ is said to extend a state $\varphi$ on $D$ if $\psi(T)=\varphi(T)$ for all $T\in D$.

For every pure state $\varphi$ on $D$ there exists exactly one state $\psi$ on $B$ that extends $\varphi$.