### One of the most outrageous open problems in operator/matrix theory is solved!

I want to report on a very exciting development in operator/matrix theory: the von Neumann inequality for $3 \times 3$ matrices has been shown to hold true. I learned this from a recent paper (with the irresistible title) “The von Neumann inequality for $3 \times 3$ matrices“, posted on the arxiv by Greg Knese. In this paper, Knese explains how the solution of this outstanding open problem follows from results in a paper by Lukasz Kosinski, “The three point Nevanlinna-Pick problem in the polydisc” that appeared on the arxiv about a half a year ago. Beautifully, and not surprisingly, the solution of this operator/matrix theoretic problem follows from deep new facts in complex function theory in several variables.

To recall the problem, let us denote $\|A\|$ the operator norm of a matrix $A$, and for every polynomial $p$ in $d$ variables we denote by $\|p\|_\infty$ the supremum norm $\|p\|_\infty = \sup_{|z_i|\leq 1} |p(z_1, \ldots, z_d)|$.

A matrix $A$ is said to be a contraction (or contractive) if $\|A\| \leq 1$.

We say that $d$ commuting contractions $A_1, \ldots, A_d$ satisfy von Neumann’s inequality if

(*) $\|p(A_1,\ldots, A_d)\| \leq \|p\|_\infty$

for every polynomial $p$ in $d$ variables.

It was known since the 1960s that (*) holds when $d \leq 2$. Moreover, it was known that for $d \geq 3$, there are counter examples, consisting of $d$ contractive $4 \times 4$ matrices that do not satisfy von Neumann’s inequality. On the other hand, it was known that (*) holds for any $d$ if the matrices $A_1, \ldots, A_d$ are of size $2 \times 2$. Thus, the only missing piece of information was whether or not von Neumann’s inequality holds or not for three or more contractive $3 \times 3$ matrices. To stress the point: it was not known whether or not von Neumann’s inequality holds for three three-by-three matrices. The problem in this form has been open for 15 years  – but the problem is much older: in 1974 Kaiser and Varopoulos came up with a $5 \times 5$ counter-example, and since then both the $3 \times 3$  and the $4 \times 4$ cases were open until Holbrook in 2001 found a $4 \times 4$ counter example. You have to agree that this is outrageous, perhaps even ridiculous, I mean, three $3 \times 3$ matrices, come on!

In Knese’s paper this story and the positive solution to the problem is explained very clearly and succinctly, and is recommended reading for any operator theorist. One has to take on faith the paper of Kosinski which, as Knese stresses, is where the major new technical advance has been made (though one should not over-stress this fact, because tying things together, the way Knese has done, requires a deep understanding of this problem and of the various ingredients). To understand Kosinki’s paper would require a greater investment of time, but it appears that the paper has already been accepted for publication, so I am quite confident and happy to see this problem go down.