Let $K^2$ be the Klein bottle and $M = K^2\widetilde{\times}I$ be the twisted, orientable $I = [0,1]$-bundle over $K^2$. So, $M$ is geometrically atoroidal (there is only one $\mathbb{Z}\times\mathbb{Z}$ subgroup in $\pi_1(M)$, and it comes from $\partial M \sim T^2$). In particular, the JSJ decomposition of $M$ is trivial, hence the geometrization theorem says that $M$ is geometric.
My question is: what is its underlying geometry (since it is not closed, can it admit more than one?)? I looked around and couldn't find any explicit mention to that, if someone can help me out, I'd really appreciate it.