A torus (equipped with a Riemannian or Lorentzian metric) which has constant curvature must be flat because of Gauss-Bonnet theorem.
Is it true that a Klein bottle (equipped with a Riemannian or Lorentzian metric) which has constant curvature must also be flat?
My first idea was to argue with Gauss-Bonnet, but then I realized that this only works for oriented surfaces. Can we lift the metric to the torus and then argue that this lifted metric is flat?
The Klein bottle, like the torus, has $\Bbb R^2$ as universal covering and the group of deck transformations is made of translations.
If we endow the Klein bottle with a metric of constant curvature and pull it back to $\Bbb R^2$ we get a metric of constant curvature which admits a group of translations acting as isometries.
The flat metric is the only such.
Alternatively, the Klein bottle admits the torus as twofold cover, so you can apply the Gauss-Bonnet argument to the constant curvature metric on the torus pullback of the metric on the bottle.