# How to find two inequivalent ,but weakly equivalent bundles?

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).

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

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.

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

• Do you mean to ask whether there is a pair of bundles $E, E'$ and some weak equivalence $(\bar{f}, f)$ that is a weak equivalence but not an equivalence, or do you mean to ask for a pair of bundles for which every weak equivalence is not an equivalence? I know you asked for the latter, but the former might be easier to write down. – John Hughes Apr 12 '16 at 13:18
• Well I want a pair of bundles that are inequivalent but weakly equivalent :)I want to know if it is true for my $B$ and if it is which ones are explicitly – user162343 Apr 12 '16 at 13:20
• @JohnHughes Does that answer your question ? – user162343 Apr 12 '16 at 13:51
• Yes...but unfortunately, that leads naturally to Qiaochu's answer, which doesn't seem to please you. :) S-T Hu wrote an article about this in rather old-fashioned language, back when it was a relatively new kind of question...but even then, he was talking about classifying spaces and characteristic classes. One problem is that things like cohomology were developed to address questions like these, so you're kinda saying "Can you build me a house..but not use any hammers?" :) – John Hughes Apr 13 '16 at 3:12
• Ok so let me see what else do I have as tools right ? And I tell so please review this often is it ok for you obviously :) – user162343 Apr 13 '16 at 3:15

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.

• Ok, let me check it, I have edited my post in fact :) – user162343 Apr 13 '16 at 23:51
• Can you provide a geometric vision of $E, E'$? – user162343 Apr 13 '16 at 23:54
• It's the Mobius bundle, on either the longitudinal or meridianal circle of the torus, pulled back to the whole thing. So each is twisted in one direction (meridian or longitude) and untwisted in the other. – user98602 Apr 13 '16 at 23:55
• Ok, let me continue checking :) – user162343 Apr 13 '16 at 23:57
• The "Mobius bundle" is a bundle over $S^1$. You can restrict vector bundles, and the Mobius bundle is obtained by the restriction of $E \to S^1 \times S^1 = T^2$ to $S^1 \times \{pt\}$. The Mobius bundle is also obtained by restricting $E$ to $\{pt\} \times S^1$. – user98602 Apr 14 '16 at 0:03

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.

• Thanks @QiaochuYuan but I don't know cohomology :) Can you provide a not too fancy answer ? – user162343 Apr 12 '16 at 16:42
• @user: what methods do you know to show that two bundles aren't isomorphic? In practice the easiest thing to do is compare characteristic classes; here I'm comparing the first Stiefel-Whitney class. – Qiaochu Yuan Apr 12 '16 at 16:50
• Well just the definition in fact no more tools :( sorry for the delay :) – user162343 Apr 12 '16 at 18:51
• @user162343, well, just with your bare hands it is usually extremely difficult to show that two bundles are not isomorphic — to solve precisely that sort of problem is that cohomology, characteristic classes and so on were invented in the first place! It is not an unreasonable answer to your question, given what Qiaochu wrote, to say: well, wait a bit until you have to tools to understand this and, for now, keep just the end result that the answer to your question is yes. – Mariano Suárez-Álvarez Apr 13 '16 at 23:27
• This should be motivation to learn the tools needed for this (which are not very many nor very difficult, in fact) – Mariano Suárez-Álvarez Apr 13 '16 at 23:27