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Assume $M$ is a topological space and $f\in \operatorname{Homeo}(M)$, then $f$ obviously plays a significant role in the torus bundle

$$M_f = M\times I/\{(x,0)\sim (f(x),1)\mid x\in M\}.$$

Hence there should be some general result of the following type: $M_f$ and $M_g$ are bundle isomorphic (resp. diffeomorphic) if and only if "W", where W is a relation between $f$ and $g$.

Can someone help give W and explain? Thank you!

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  • $\begingroup$ I'm not sure "torus bundle" is right, I've heard "mapping torus" used for this construction. Do you want your maps (between mapping tori or whatever) to necessarily preserve the I-coordinate, or at least take each M x {t_0} to some M x {t_1}? $\endgroup$
    – Aaron Mazel-Gee
    Commented Apr 22, 2010 at 8:07
  • $\begingroup$ @ Aaron Mazel-Gee.sorry,i made a mistake.as you point out,what i mean is mapping torus. $\endgroup$
    – sara
    Commented Apr 22, 2010 at 9:00
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    $\begingroup$ @Sara - It would be nice if you edited your question to fix the title. $\endgroup$
    – Sam Nead
    Commented Apr 22, 2010 at 10:25
  • $\begingroup$ At least in 3-manifold topology, the term 'torus bundle' is usually used in the special case of this construction when M is a 2-torus. $\endgroup$
    – HJRW
    Commented Apr 22, 2010 at 18:25

2 Answers 2

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Your $M_f$ is usually called the "mapping torus" of $f$, as Aaron points out. It comes with a map to the circle $S^1 = I/(0 \sim 1)$ and this map is a fibre bundle with fibres isomorphic to $M$.

First of all, the bundle isomorphism type of the bundle $M_f \to S^1$ only depends on the the isotopy class of $f$; i.e. if $f_0$ and $f_1$ are homotopic through homeomorphisms $f_t$ then $M_{f_0}$ and $M_{f_1}$ will be isomorphic bundles. This reduces you to working with the mapping class group, which you probably already knew since I see the mapping-class-groups tag included.

Now, it's an easy exercise to check that $M_f$ and $M_g$ are isomorphic bundles if and only if $f$ and $g$ are conjugate in the mapping class group of $M$.

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  • $\begingroup$ I wonder if there is a subtle point lurking here... if $f_0$ and $f_1$ are isotopic (homotopic through homeomorphisms) then the two bundles will be homeomorphic. However, the question asked about diffeomorphism. This is not my area, so I don't see that a diffeotopy (homotopic through diffeomorphisms?) gives a diffeomorphism of bundles... It feels like the "quality" of the derivative of the time direction will be important? $\endgroup$
    – Sam Nead
    Commented Apr 22, 2010 at 10:32
  • $\begingroup$ For a diffeotopy it is exactly the same short argument; topologically isotopic maps give homeomorphic bundles, and smoothly isotopic maps give diffeomorphic bundles. And in fact, reducing to the mapping class group is unnecessary. If f and g are isotopic then they are conjugate. $\endgroup$
    – Jeffrey Giansiracusa
    Commented Apr 22, 2010 at 11:31
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    $\begingroup$ For the second question (in brackets): if the bundles are just diffeomorphic and not bundle-isomorphic, then the mapping classes need not be conjugate. See mathoverflow.net/questions/241822/… $\endgroup$
    – ThiKu
    Commented Oct 10, 2016 at 5:00
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Mapping tori come equipped with a projection to the circle: $p\colon M_f\to S^1$. Bundle isomorphism is isomorphism of the total space that commutes with the projection. A weaker notion of isomorphism for mapping tori is an isomorphism that commutes with the projection only up to homotopy. Pseudoisotopy is the name for the relation on isomorphisms of $M$ that replaces isotopy in this coarser notion of equivalence of mapping tori.

If $M$ is a simply connected manifold of high dimension, then pseudoisotopy implies isotopy and preserving the homotopy class of the map to the circle isn't a big deal, so isomorphism of mapping tori is pretty much isotopy of mapping classes. But if $M$ is not simply connected, there are two ways that mapping tori may be isomorphic without being bundle isomorphic: there may be more pseudoisotopies than isotopies and there may be isomorphisms that do not preserve the homotopy class of the map to the circle.

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  • $\begingroup$ I'm not sure about the high-dimensional situation, but in two dimensions it's worth noting that pseudoisotopy and isotopy are the same (by a non-obvious theorem). $\endgroup$
    – Dylan Thurston
    Commented Oct 10, 2016 at 5:31
  • $\begingroup$ @DylanThurston Right, in 2d pseudoisotopy implies homotopy implies isotopy. In both low and high dimensions (but not smooth 4), the answer is that the difference between isotopy and pseudoisotopy is a higher Whitehead group, which should vanish for aspherical spaces (Borel conjecture). Waldhausen proved the pseudoisotopy=isotopy for Haken 3-manifolds, suggesting that the K-theory gadget vanishes. Hence his seemingly sharp turn into K-theory. (I imagine that for some 3d lens space pseudoisotopy is coarser than isotopy.) $\endgroup$
    – Ben Wieland
    Commented Oct 13, 2016 at 2:43
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    $\begingroup$ @DylanThurston Actually, I was wrong. In high dimensions even if $Wh_2$ vanishes, as for free abelian groups, there is an additional obstruction for pretty much any space that is not simply connected. For the nicest spaces around, $S^n\times S^1$ and $(S^1)^n$, the group of homeomorphisms pseudo-isotopic to the identity but not isotopic is the $\mathbb Z/2$ vector space on the set of nontrivial conjugacy classes of elements in the fundamental group (or maybe the subspace invariant under inversion?). Same for diffeomorphisms. $\endgroup$
    – Ben Wieland
    Commented Feb 27, 2017 at 18:43

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