How to find two inequivalent ,but weakly equivalent bundles?

Question

I want to know if I can have an inequivalent but weakly equivalent bundles over the Torus, i.e. if we can find explicitly an example of a bundle map $(\overline{f},f)$ where we require that $\overline{f}$ is an isomorphism from one fibre to another and $f$ an homeomorphism from the torus to the torus but such that $f$ is not the identity map (if this happens we will have an equivalence).

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Notes

• This restlessness is taken from Spivak's A comprehensive introduction to differential geometry,third edition,chapter 3.

• I'm a beginner here so I don't know anything about cohomology, fancy differential geometry or that stuff thanks for your support.

Definition

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A weak equivalence between two bundles over the same base space $B$ is a bundle map $(\overline{f},f)$ where $\overline{f}$ is an isomorphism on each fibre, and $f$ is a homeomorphism of $B$ into itself.

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Example

When this curiosity came up to me it came up to me with two examples that are as follows. Imagine the Möbius strip and the infinite torus $\times \mathbb{R}$ then we mapped the Möbius strip to infinite jail cell and we project infinite torus $\times \mathbb{R}$ to only the infinite torus, this is for $E_1$ and for $E_2$ we only consider the maps the other way around.

Now, the other example was this. We pick the disjoint union of two circles and for $E_1$ we map the Möbius strip to one circle and $S^{1} \times \mathbb{R}$ to the other one, and again, for $E_2$ we only consider the maps the other way around. I cannot explain more because I am not very acquainted in this area but I can provide an image

Answer 1

The real line bundles over a torus $T^2$ are classified by $H^1(T^2, \mathbb{F}_2) \cong \mathbb{F}_2^2$. Weak equivalence allows you to act on this cohomology group by $\text{Aut}(T^2)$, which contains $GL_2(\mathbb{Z})$. The $GL_2(\mathbb{Z})$ action acts transitively on the nonzero elements of $H^1(T^2, \mathbb{F}_2)$, so any two distinct such elements come from two line bundles which are "weakly equivalent" (incidentally, I think this is a very bad name) but not isomorphic.

Answer 2

Here is a way of rephrasing what Qiaochu said in more-or-less elementary language. Go much further in thinking about vector bundles and you won't be able to do this anymore.

Let $E = \Bbb R^3/(x,y,z) \sim (x+1,y,-z) \sim (x,y+1,z)$. This has a map $E \to T^2 = \Bbb R^2/(x,y)\sim (x+1,y)\sim(x,y+1)$. It has the structure of a vector bundle over $T^2$ (do you see why?)

You can also define $E'= \Bbb R^3/(x,y,z) \sim (x+1,y,z) \sim (x,y+1,-z)$. This is weakly equivalent to $E$ (write down the weak equivalence) but not equivalent.

If it was equivalent, then their restrictions to the circle in $T^2$ corresponding to the circle given by the $x$-coordinate would be equivalent, as well. But for $E$ this restriction is the Mobius bundle, and for $E'$ it's trivial. You can prove that these bundles are not isomorphic by showing that the Mobius bundle has no nonvanishing section, while the trivial bundle does.