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.
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1Hint: The fundamental group of the fiber is not central in the fundamental group of the unit tangent bundle – Moishe Kohan Nov 14 '19 at 23:40
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1this boundary is the 3 manifold obtained from $T^2\times I$ and glueing two boundary component(torus) by the matrix diagonal(1,-1) [monodromy map]. You can see this by Glutching function construction as well. And its double cover is $T^2\times S^1$ as you said. I do not know if this manifold has some specific name or not, maybe Thurston's classification could help you – Anubhav Mukherjee Nov 15 '19 at 04:30
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1This manifold is called the dicosm in https://arxiv.org/abs/math/0311476. It can also be realized as a quotient of the special Euclidean group $ SE_2 $ by the cocompact discrete subgroup $$ { \begin{bmatrix} 1 & 0 & n \ 0 & 1 & m \ 0 & 0 & 1 \end{bmatrix}: n,m \in \mathbb{Z} } \cup { \begin{bmatrix} -1 & 0 & n \ 0 & -1 & m \ 0 & 0 & 1 \end{bmatrix}: n,m \in \mathbb{Z} } $$ This discrete subgroup is the semidirect product $ \mathbb{Z}^2 \rtimes {\pm1} $ of a planar lattice $ \mathbb{Z}^2 $ by a two element group of transformations $ \pm I $ preserving the lattice. – Ian Gershon Teixeira Feb 19 '22 at 13:17
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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 $T^3$.
