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In Frederic Schuller's lecture series Lectures on The Geometrical Anatomy of Theoretical Physics, he gives an example of a bundle $E\overset{\pi}{\rightarrow}M$ where different points of the base manifold have different fibres:

the bundle example

Or, as spelled out in Simon Rea's transcription of the lectures, found here (p. 39):

$$F_p:=\mathrm{preim}_\pi(\{p\}) \cong_\mathrm{top} \begin{cases} S^{1} & p < 0\\ \{ p \} & p = 0\\ [0,1] & p > 0\\ \end{cases}$$

The example is great for intuition. But I don't see how E can be a manifold with such an odd structure (as a reminder, the definition of $E\overset{\pi}{\rightarrow}M$ requires E to be a topological manifold).

Davide Radaelli
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    I don't believe such a set is a bundle, since it does not have consistent dimension. – J.V.Gaiter Jun 12 '21 at 20:48
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    It depends on your definition of bundle. In at least one fundamental text, the defintion of bundle is just a continuous surjection. – Randall Jun 12 '21 at 20:50
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    I also had this confusion when watching these lectures. Btw a simple example of $\pi:E\to M$ where $\pi$ is even a smooth surjective map between smooth manifolds, but yet the fibers are not all homeomorphic is the following. Let $M=\Bbb{R}$ and $E=\Bbb{R}^2\setminus {(0,1)}$, and let $\pi:E\to M$ be the mapping $\pi(x,y)=x$. So in this case, everything is nice and smooth, but for any $x\neq 0$, the fiber $E_x=\pi^{-1}({x})={x}\times \Bbb{R}$ but $E_0={0}\times ((-\infty,1)\cup (1,\infty))$. So $E_x$ for $x\neq 0$ are connected, but $E_0$ is not (draw a picture, this becomes obvious). – peek-a-boo Jun 12 '21 at 22:56

2 Answers2

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Indeed, the total space here fails to be a manifold -- more precisely, it's impossible to have any continuous surjection $E \to \mathbb{R}$ from a topological manifold $E$ with the prescribed preimages. To see this, note that removing the unique point in the preimage of $0$ disconnects $E$, so if $E$ were a manifold, it would have to be $1$-dimensional. However, $E$ cannot be a $1$-dimensional manifold, as follows.

It is a standard result that every non-compact, connected topological $1$-manifold is homeomorphic to $[0,\infty)$ or $\mathbb{R}$. There cannot be a continuous surjection from one of these spaces to $\mathbb{R}$ with the prescribed fibers, since $S^1$ is not homeomorphic to a subspace of $\mathbb{R}$ (every connected compact subspace of $\mathbb{R}$ is of the form $[a,b]$ with $a \leq b \in \mathbb{R}$, and these subspaces are simply connected while $S^1$ is not).

From a brief watch of a few minutes, it seems like these lectures have a few other problems with mathematical precision.

For example, the definition of "fiber bundle" given is incorrect, or at the very least extremely nonstandard. Please compare to the definition on Wikipedia, and see this post which may be referencing these same lectures.

Also, the lecturer claims that the Möbius strip is homeomorphic to the cylinder, which is simply not true (the Möbius strip is not homeomorphic to a subspace of $\mathbb{R}^2$, while the cylinder is).

So, I would take the mathematical details in these videos with a grain of salt (perhaps consider studying alongside a reputable math textbook on topological manifolds).

diracdeltafunk
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  • If a lecturer seriously claim that a Möbius strip and a cylinder are homeomorphic, then the lecturer's mathematical credentials are indeed very much in question. – Ted Shifrin Jun 12 '21 at 23:28
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    Here's a link to that timestamp (working as of June 12, 2021). Since I am not a physicist, I can only imagine (and hope) that these lectures are more helpful than harmful to a physicist's training. – diracdeltafunk Jun 12 '21 at 23:55
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    "Yeah, they must be." Oy. There is an interesting mathematics question (which has been addressed elsewhere on here): What's the simplest example of non-isomorphic bundles whose total spaces are homeomorphic? – Ted Shifrin Jun 13 '21 at 00:01
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It is not a bundle of topological manifolds, but it is a bundle of topological spaces.

If you want, you can remove the problematic point both in $E$ and in $M$, giving you something which is a (surjective) bundle of topological manifolds.

I say surjective because there are people for whom "bundles" need not be surjective. With this broader definition, it suffices to remove the point in $E$, and in that way $E$ is a topological manifold, the only thing is that the fibre over $q$ is now empty.

If you want a simple example of a surjective bundle over $\mathbb R$ which is not a fibre bundle, you can consider the map $\mathbb R\to\mathbb R$ given like this:bundle

Jackozee Hakkiuz
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    Note that these lectures explicitly define the word "bundle" to mean "a surjective continuous map between topological manifolds". – diracdeltafunk Jun 12 '21 at 22:45
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    @diracdeltafunk Yeah, I just wanted to convey the idea that it's not the only possible definition. I thought it was clear but I guess my phrasing was not the best. If you have an idea of how to improve the wording to convey this, please feel free to edit my post. :) – Jackozee Hakkiuz Jun 12 '21 at 22:47
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    No, your answer is excellent! I just wanted to clarify this point to anyone who might be reading this post (now or in the future) without having watched the video. – diracdeltafunk Jun 12 '21 at 22:48