identifying a specific 3d manifold: A torus fibered over a circle with a $Z_2$ twist.

Question

I am looking for literature on a specific 3-manifold, can someone help me with its name?

The three manifold in question is a torus fibered over a circle $S_1$. As I move along the circle the longitude and the meridian of the torus gets interchanged such that when I am back at my starting point, the meridian is glued to the longitude and vice versa. In a sense it is a 3d variant of the moebius strip.

Does anyone know what 3d manifold I am talking about? Thanks a lot in advance!

Answer 1

The torus' mapping class group allows classification of these surface bundles and since the mapping class group has only three involution elements which are determined by maps $\Bbb Z^2\to\Bbb Z^2$ and specified by the three matrices: $$f=\left(\begin{array}{cc} -1&0\\ 0&-1\end{array} \right)\quad,\quad g=\left(\begin{array}{cc} 1&0\\ 0&-1\end{array} \right)\quad,\quad h=\left(\begin{array}{cc} 1&1\\ 0&-1\end{array} \right),$$ then there are only three of such bundles. These three matrices are involutions (both these three squared are the identity matrix and why the structural group is $\Bbb Z_2$) and these three matrices give three 3-manifolds well known in terms of Seifert invariants. The three examples are Seifert fiber spaces.

The three bundles $E_f=T\times_f S^1$, $E_g=T\times_g S^1$ and $E_h=T\times_h S^1$ are respectively:

$E_f=(Oo,0|(1,2),(1,2),(1,-2),(1,-2))$,

$E_g=(NnI,2|(1,0))$ and

$E_h=(NnI,2|(1,1))$.

The second, $E_g$, is known to be the cartesian product $K\times S^1$, where $K$ is the Kleinbottle.

Answer 2

I'd say "It's the manifold you just described." But the description can be made a little simpler. Under the diffeomorphism $$ (\theta, \phi) \mapsto (\theta + \phi, \theta - \phi) $$ of the torus (or its inverse), the meridian maps to a meridian plus longitude, and a longitude maps to a meridian minus a longitude. Taking these new coordinates, your clutching function looks like $m' \to m'$, $\ell' \to -\ell'$, where $m'$ and $\ell'$ are the new meridian and longitude.

I'm not certain whether this is any clearer a description. Actually, it seems as if it might be...you've got an $S^1 \times S^1$ bundle over $S^1$. The first factor is a trivial bundle, so we can just say we've got a product of $S^1$ with $F$, an $S^1$ bundle over $S^1$. And the clutching map for $F$ is just $\theta \mapsto -\theta$. That means that $F$ is the Klein bottle.

Hmm. You've got $S^1 \times K$, where $K$ is the Klein bottle. I guess that is a simpler description.