# $\Bbb{S}^1$-bundle over $\Bbb{S}^1$ + reference request

I have the following questions:

Let $$p:\Bbb{T}^2\to \Bbb{S}^1$$ be a circle bundle. Is $$p$$ trivial ?

Before I show you what I attempted, I just want to let you know that I don't have any knowledge of fiber bundles. I just base my intuition on the little I know about vector bundles, and from what I quiclky read on Wikipedia I think that I'm allowed to do what follows.

Take the quotient map $$q:[0,1]\to S^1$$. We can pull back the bundle $$p$$ via $$q$$. Because $$[0,1]$$ is contractible, this pulled back bundle is trivial so we have $$\begin{array} {ccc} [0,1]\times \Bbb{S}^1 & \stackrel{h}{\longrightarrow} & E=\Bbb{T}^2\\ p' \Big\downarrow & & \Big\downarrow p\\ [0,1] & \stackrel q {\longrightarrow} & S^1 \end{array}$$

Then I guess I should glue the sides of the cylinder $$[0,1]\times \Bbb{S}^1$$ by factorizing $$h$$ but I don't know how to do it formally (I don't see how $$f$$ factors). Also should I have taken the universal cover $$\mathbb{R}\to \Bbb{S}^1$$ instead of $$q$$?

Second question: I am looking for a (short) introductory book to learn about the theory of fiber bundles. I don't know much about it so I can't really be specific in my request, but I'd like something which is "differentiable manifolds related". From what I have read I think that I'm looking for something like the 3rd chapter of Dale Husemoller, Fiber Bundles. I also know about Norman Steenford The topology of fibre bundles, but I'm not sure if this one is exactly what I'm looking for. Do you have any advices? (I've had a course of differential topology already but we didn't study fiber bundles)

• Basically $T^2$ is homeomorphic/diffeomorphic to $S^1 \times S^1$, which is what a "trivial $S^1$-bundle over $S^1$" should mean. You didn't really describe it, but I suppose your circle bundle is given by one of the projections? But I'm a fibre bundle newb too, only guessing – Jane Doé May 3 at 21:16
• @JaneDoé Yes but $p$ isn't necessarily the projection onto one factor right? I mean it's not direct from the definition? – Adam Chalumeau May 3 at 21:17
• I see, so you are asking whether an arbitrary $S^1$-bundle over $S^1$ is trivial. This is less clear. – Jane Doé May 3 at 21:18
• @JaneDoé yes and I am asking that the total space is diffeomorphic to $\Bbb{T}^2$. – Adam Chalumeau May 3 at 21:20
• Dear Adam, I realised that my answer was pretty similar to what you wrote, I added a few details. I hope this is helpful. – Nicolas Hemelsoet May 3 at 22:00

This is true.

Sketch : Let $$x \in S^1$$, $$I = S^1 \backslash \{x\} \cong (0,1)$$ and $$U = p^{-1}(I)$$. The map $$p_{|U} : U \to I$$ is a trivial fiber bundle because $$I$$ is contractible, i.e it's the projection $$S^1 \times I \to I$$.

Moreover, let us extend $$p_{|U}$$ to a map $$\overline p : \overline{I} := [0,1] \times S^1 \to [0,1]$$. By construction one can reconstruct $$p$$ from $$\overline p$$ by identifying $$0 \times S^1$$ and $$1 \times S^1$$ together.

The map should be a diffeomorphism, so it should either reverse or preserve the orientation. It's easy to see that you can only get the Klein bottle in the first case or the torus in the second case.

EDIT : here are more details. With OP's notation, it's clear that $$h$$ has the same values on $$0 \times S^1$$ and $$1 \times S^1$$ since the diagram is commutative. Thus $$h$$ is a surjection and since $$[0,1] \times S^1$$ is compact we get an homeomorphism $$[0,1] \times S^1 / \sim \cong T^2$$. This homeomorphism is determined by a continuous bijection $$S^1 \to S^1$$. Up to homotopy there is only two such maps : $$z \mapsto z$$ and $$z \mapsto \overline{z}$$. In the second case we would obtain the Klein bottle which is not possible, so the gluing map is homotopic to $$z \mapsto z$$ which means that the trivialisation extends to the whole space i.e the fiber bundle was trivial.

• So here is a distinction which confused me a little. Should we call the product $E \times F$ a trivial bundle, or any bundle isomorphic to $E \times F$? According to ncatlab.org/nlab/show/trivial+fiber+bundle , a fibre bundle that is isomorphic is called trivializable and is distinguished from the product $E \times F$, the trivial bundle. Your argument shows that $T^2$ is a trivializable $S^1$-bundle over $S^1$, but it may be homeomorphic with $S^1 \times S^1$ by a topologically interesting map given by a Dehn twist: en.wikipedia.org/wiki/Dehn_twist – Jane Doé May 3 at 21:46
• @JaneDoé : Indeed you are right, $p$ is trivialisable but not necessarily trivial since $p$ could be something else that a projection. However it is very usual to look a trivialisable fiber bundle as a trivial one since they are the same up to a diffeomorphism. – Nicolas Hemelsoet May 3 at 21:51
• @NicolasHemelsoet thanks for your help! I don't understand one thing: how do you know that there would be a continuous bijection to glue the sides of the cylinder? If I have such a map I know pretty much how to finish as you did with factorization etc – Adam Chalumeau May 4 at 14:55