Numerical index one characterises commutative JB*-triples

We establish that a JB${}^{\ast }$-triple $E$ is commutative if and only if its numerical index $n(E)$ is one. As a consequence, $n(E)=1$ implies $n(E^{\ast })=n(E^{\ast \ast })=1$. This settles a question left open in the literature since 2008.