Unit (co)tangent bundle of Klein bottle

Question

The unit (co)tangent bundle of the 2-torus is trivial, whose total space is the 3-torus. Since the torus is a double cover of the Klein bottle, I would imagine this 3-torus double covers the unit (co)tangent bundle $X$ of the Klein bottle. But what is $X$ as a 3-manifold? Is it still the 3-torus? Note that the tangent bundle of the Klein bottle is not trivial, since the Klein bottle is non-orientable, although the Euler characteristic of the Klein bottle is zero.

Answer 1

The unit tangent bundle of the 2-torus is the 3-torus, where we use the standard parallelization of $\Bbb R^2$ to trivialize the tangent bundle of $T^2 = \Bbb R^2/\Bbb Z^2$. The involution on $\Bbb R^2/\Bbb Z^2$ whose quotient is the Klein bottle is $$\iota(x, y) = (x+1/2, -y).$$

Thus the induced action on $T^1(T^2) = T^3$ is $$\iota(x,y,\theta) = (x+1/2, -y, \bar \theta).$$ Here $\bar \theta$ is complex conjugation on the unit circle.

That is to say, this is $$S^1 \times_{\Bbb Z/2} T^2,$$ where $\Bbb Z/2$ acts by the antipodal map on the circle and by $-1$ on $\Bbb R^2/\Bbb Z^2$ (so with 4 fixed points), or equivalently the fiber product $K \times_{S^1} K$ where $K \to S^1$ is the obvious projection. I am not sure what other descrip you could want. Of course this is not $T^3$.