Mapping Tori and Monodromy

336 Views Asked by At

I have a question regarding the right setup for mapping tori. Let me give the definitions that I use first.

Let $I$ denote the interval $[0,2\pi]$ and let $I^*$ denote the quotient of $I$ by the relation $0 \sim 2\pi$. Further, let $F$ be a smooth manifold, $f \colon F \rightarrow F$ a diffeomorphism, and define an equivalence relation $\sim$ on $I \times F$ by $(0,x) \sim (2\pi,f(x))$. The Mapping Torus of $f$ is the manifold $E_f = (I \times F)/\sim$. It is not difficult to see that this is a locally trivial fibration over $I^*$. A map trivializing over $I^* \setminus \{\pi\}$ is, for instance, given by $$\psi([t,x]) = \begin{equation} \begin{cases} ([t],x), &\text{if } 0 \leq t < \pi \\ ([t],f^{-1}(x)), &\text{if } \pi < t \leq 2\pi. \end{cases} \end{equation}$$ A lot of references now claim that this fibration has monodromy $[f]$. First of all, to speak of monodromy we need to fix a loop. A natural choice would be $\gamma \colon S^1 \rightarrow I^*, ~\gamma(t) = [t]$. Let $q$ denote the obvious map $[0,2\pi] \rightarrow S^1$. Now the pullback bundle $E = (\gamma \circ q)^*E_f$ has a global trivialization given by $$h \colon F \times [0,2\pi] \rightarrow E, ~h(x,t) = (t,[t,x]).$$ We have that $h(x,0) = (0,[0,x])$ and, hence, viewed as a map $F \rightarrow E_0 \simeq F$, $h(\cdot,0)$ is exactly the identity $id_F$. Therefore, the monodromy associated to $\gamma$ is $h(\cdot,2\pi)$. However, this map $$h(x,2\pi) = (2\pi,[2\pi,x]) = (2\pi,[0,f^{-1}])$$ is actually $f^{-1}$, not $f$. Doing a similar calculation, we find that the monodromy associated to $\delta(t) = [2\pi-t]$ is actually $f$.

In conclusion, we would have assumed that the most natural setup (i.e. using $\gamma$) would lead to monodromy $f$. If my calculations are correct, then this does not hold, though. So, the question is: should we define the mapping torus via the relation $(0,x) \sim (2\pi,f^{-1}(x))$ instead, or do the textbooks simply not communicate that the "natural" monodromy of $E_f$ actually is $f^{-1}$? (Of course, the implicit question "Are my calculations corret?" also stands...)