differential-geometrynon-orientable-surfacesriemannian-geometrysemi-riemannian-geometrysurfaces
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?
Consider the embedding $\psi:\Bbb R\times\Bbb R\big/{\Bbb Z\times\Bbb Z}\hookrightarrow\Bbb R^3$
$$\psi([\theta,\tau])=
\begin{pmatrix}
\cos(2\pi\theta) &-\sin(2\pi\theta) & 0\\
\sin(2\pi\theta) &-\sin(2\pi\theta) & 0\\
0&0&1
\end{pmatrix}
\begin{pmatrix}
2+\cos(2\pi\tau)\\
\sin(2\pi\tau)\\
0
\end{pmatrix}$$
Then one verifies
(on the picture and then by doing the math) that $-\psi([\theta,\tau])=\psi([\theta+\frac12,-\tau])$
Thus the Klein bottle is the quotient of $\Bbb R\times \Bbb R$ by the (non-commutative) group $G=\langle v,t\rangle$ of homeomorphisms generated by vertical displacement by one $$v(\theta,\tau)=(\theta,\tau+1)$$ and a twist $$t(\theta,\tau)=(\theta+\frac12,-\tau)$$ (Notice that $t^2=h$ the horizontal displacement by one $h(\theta,\tau)=(\theta+1,\tau)$) You'll verify that $G(v(\theta,\tau))=G(\theta,\tau)=G(t(\theta,\tau))$ so $G$ descends to a map $\tilde{G}:K\to \Bbb R^4$. $\tilde{G}$ is an injective (easy verification) immersion ($G$ already was). By compactness of $K$, it is an embedding.
When you join up the edges of the flat square to make a projective plane, what happens to its corners? Each corner gets identified with the opposite corner. So at two points in the projective plane you have two corner bits of squares identified. They may be flat, but round the corner, you have only $\pi$ worth of angle instead of $2\pi$. This means you can't extend the flat metric to these corners, so the quotient isn't naturally a Riemannian manifold.
An easier illustration: consider the surface of a standard cube in $\Bbb R^3$. Each face is flat. What about the edges? They aren't trouble, one can think of each pair of adjacent faces as a $2\times1$ rectangle folded over, and that has a flat metric. So if you remove the vertices, the surface of the cube has a flat Riemannian metric all over it.
But those pesky vertices! If you make a little circuit about one of these, in effect you're going through an angle of $3\pi/2$. If you parallel-transport a vector round the path, it come back turned through a right angle. This means there's no way of extending this flat Riemannian metric to the vertex.
This works for every compact surface; you can express it as a simplicial complex, and if you delete the vertices then you can put a flat metric on it. Alas one can rarely extend it to the whole surface (certainly if it has nonzero Euler characteristic).
Best Answer
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.