The holonomy map associated to a mapping torus

Question

So I have a rather embarrassing problem, which is not really a "problem", so much as a mental block I seem to be unable to overcome. I am trying to understand the "holonomy map" of a mapping torus. To keep things concrete I would like to understand the following basic example: Take a $\mathbb{T}^2$ bundle over $\mathbb{S}^1$. One way to get an example of such a fiber bundle is by a mapping torus construction: take the product an interval $I$ with the fiber $\mathbb{T}^2$ and "glue" the ends using an homeomorphism of $\mathbb{T}^2$, $f$, to get the mapping torus $M_f$.

$$M_f =\frac{(I \times \mathbb{T}^2)}{(1,x)\sim (0,f(x))}$$

$f$ is then called the holonomy map of the mapping torus. Now consider the following bundles: the trivial bundle $\mathbb{T}^2\times \mathbb{S}^1$ and the bundle given by "rotating" an $\mathbb{S}^1$ factor of the $\mathbb{T}^2$ $n$ times as we travel around the $\mathbb{S}^1$ base of the fibre bundle once i.e. the torus bundle with monodromy

$\left( \begin{array}{ccc} 1 & n \\ 0 & 1 \\ \end{array} \right)$

Question: what is the holonomy map for these basic examples? There should be a particular $f$ for each $n$ so that the mapping torus $M_f$ is associated to $f$ is the torus bundle by twisting $n$ times. But I don't know how a single homeomorphism of the torus can distinguish between the cases. How can a single homeomorphism capture that the torus has turned $n$ times? I mean, everytime you rotate $2\pi$ the map at the "end point" is back to being the identity. I hope I have made my problem clear, I am aware that it is a bit silly but I have been turning it around in my head for a while and still can't fully see what is happening.

(again, this was asked on stack exchange and I switched to here due to lack of response)

Answer 1

If you represent the torus as $\mathbb R^2/\mathbb Z^2$, given a monodromy matrix $A \in GL_2(\mathbb Z)$, we can construct one example of the associated diffeomorphism. The multiplication-by-$A$ map $\mathbb R^2 \to \mathbb R^2$ sends $\mathbb Z^2$ to $\mathbb Z^2$, hence it defines a map $\mathbb R^2/\mathbb Z^2 \to \mathbb R^2 / \mathbb Z^2$.

As for why this is nontrivial, the simple answer is that the circles intermediate between the beginning and the endpoint rotate amounts intermediate between $0$ and $2\pi n$, and this distinguishes the map from the identity. As far as I know the simplest rigorous way to distinguish these maps up to isotopy is by either the fundamental group or first homology, just reducing to the monodromy matrix you have already written down.

Of course the same thing works for higher $n$. The only difference is that, for $n=2$, any map is isotopic to one of these (the one given by its monodromy matrix), but for $n \geq 5$ this is no longer true.