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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?

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    $\begingroup$ take the double cover : it is orientable so you can use gauss bonnet; as it is flat the surface itself is (flatness is a local property) $\endgroup$
    – Thomas
    Feb 23, 2019 at 8:06
  • $\begingroup$ Why does user454042 say that Gauss-Bonnet only works for orientable manifolds? On Wikipedia Gauss-Bonnet is stated as $ \frac{1}{2\pi} \int_M K dA =\chi(M) $ for all closed two manifolds. $\endgroup$ Mar 31, 2020 at 16:36

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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.

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    $\begingroup$ Sorry but what would be the precise argument? Because there are also non-flat tori even though the torus is a quotient of the cylinder. $\endgroup$
    – user454042
    Feb 23, 2019 at 0:05
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    $\begingroup$ @user454042, are there non-flat tori with constant curvature? Once you lift the metric to the plane, the resulting metric in $\Bbb R^2$ is of constant curvature with a group of translations acting as isometries. $\endgroup$ Feb 23, 2019 at 0:32
  • $\begingroup$ Ok. Does this mean thay any mfd with constant curvature which has $\mathbb{R}^n$ as universal covering is flat? $\endgroup$
    – user454042
    Feb 23, 2019 at 7:54
  • $\begingroup$ And also I’m not getting all the details of that argument. Why does the existence of an action by translations imply flatness? $\endgroup$
    – user454042
    Feb 23, 2019 at 8:01
  • $\begingroup$ @user454042 Also a compact Riemann surface $X$ of genus $\geq2$ has $\Bbb R^2$ (or the disc) as universal covering, but the group of covering transformations is not made of translations, thus they're not isometries and the flat metric in $\Bbb R^2$ does not descend to $X$. $\endgroup$ Feb 23, 2019 at 15:25

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