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The orientable Seifert fibered manifold $M=(1|(1,b))$ that fibers over the torus with no exceptional fibers but non-zero obstruction term $b$ has an embedded non-separating fibered torus (lift a nontrivial loop from the torus base space of the fibration). Cutting along this torus would leave a trivially fibered annulus cross $S^{1}$: $A\times S^{1}$. Therefore, $M=(1|(1,b))$ can be reobtained by gluing together the two torus boundary components of $A\times S^{1}$ in a particular fiber preserving way. What is this gluing map?

My guess was that the map would look like $d(u,v)=(u^{-1},u^{b}v)$ (assuming that the tori are positively oriented), but I can't compute the fundamental group.

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  • $\begingroup$ I think Mayer-Vietoris would give the first homology group as $\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}_{b} $ which would be correct, but I'm unsure of this. $\endgroup$
    – B. Peet
    Jul 11, 2017 at 22:08

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You guess at the gluing map seems close to correct, although perhaps it might be $(u,v) \mapsto (u,u^bv)$ (I'm not quite sure what your coordinates $u,v$ represent, they might be $T^2$ coordinates in my displayed equation below).

To compute the fundamental group, use the fact that gluing the two boundary components of the space $$A \times S^1 \approx (S^1 \times [0,1]) \approx (S^1 \times S^1) \times [0,1] \approx T^2 \times [0,1] $$ gives an example of a mapping torus. By doing exercise 11 in Section 1.2 of Hatcher's book "Algebraic Topology", you will learn by example a general technique for using Van Kampen's theorem to compute the fundamental group of a mapping torus.

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  • $\begingroup$ Thank you Dr. Mosher! $\endgroup$
    – B. Peet
    Jul 17, 2017 at 21:37

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