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I don't understand how to solve the following problem. Can you help me?

Prove that there are only two vector bundles of rank $k$ with base $S^{1}$ $-$ trivial $1_{k}$ and non-trivial $\eta_{k}$.

Then let $S^{1}=\{z\in C, |z|=1\}$ and $f:S^{1} \to S^{1}$, $f(z)=z^{n}$. Find the bundle $f^{*}\eta_{k}$.

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  • $\begingroup$ Do you know clutching functions? $\endgroup$ Sep 18, 2013 at 15:05
  • $\begingroup$ @JasonDeVito No :c $\endgroup$
    – xxxxx
    Sep 18, 2013 at 15:14

2 Answers 2

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Here's the idea of clutching functions. (Almost all of what I'm going to say works, more generally, for any sphere.)

Start with a rank $k$ vector bundle $\xi$ over $S^1$. Let $U$ be the open northern hemisphere, extended ever so slightly beyond the equator. Likewise, let $V$ be the open southern hemisphere, slightly extended. So, $S^1 = U\cup V$ and $U\cap V$ consists of two small disjoint open intervals around $(\pm 1,0)$.

Since both $U$ and $V$ are contracible, $\xi|_U$ and $\xi|_V$ must be trivial. The full bundle $\xi$ can be recovered by gluing $\xi|_U \cong U\times \mathbb{R}^k$ and $\xi|_V \cong V\times\mathbb{R}^k$ together appropriately. Such a gluing is given by a map $f:U\cap V\rightarrow Gl(k,\mathbb{R})$, called the clutching function of $\xi$.

Fact: If $f$ and $g:U\cap V\rightarrow Gl(k,\mathbb{R})$ are homotopic, then the bundles formed by gluing via $f$ and $g$ are isomorphic.

Since in our case, $U\cap V$ is homotopy equivalent to $S^0$, a disjoint union of two points, the set of homotopy classes of maps from $U\cap V$ to $Gl(k,\mathbb{R})$ is in natural bijective correspondence with $\pi_0(Gl(k,\mathbb{R}))$. But this is well known to be a 2 pointed set. It follows that there are at most 2 bundles of any given rank over $S^1$.

To show there are at least 2 bundles of every rank over $S^1$, begin with the observation that the Moebius bundle $M$ is a nontrivial bundle over $S^1$ of rank $1$. Further, the Moebius bundle is non-orientable. Now, convince yourself that $M\oplus 1_{k-1}$ must also be nonorientable and rank $k$, so this provides a nontrivial bundle of rank $k$ over $S^1$.

Finally, for the last argument, we've now proven that $\eta_k$ is isomorphic to $M\oplus 1_{k-1}$. Now, to compute $f^\ast \eta_k$, it's clear that it's the same as $(f^\ast M) \oplus 1_{k-1}$ so it's enough to figure out that $f^\ast M$ is. This can be done with bare hands (and pictures!). Just to give a hint, the answer is that $f^\ast M$ is trivial iff $n$ is even.

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    $\begingroup$ A good reference for this is Milnor and Stasheff's Characteristic classes book, in case you need a reference. $\endgroup$ Sep 18, 2013 at 18:03
  • $\begingroup$ Oh, thank you for your answer! I'll read it a bit later, and will try to understand this. $\endgroup$
    – xxxxx
    Sep 18, 2013 at 18:05
  • $\begingroup$ elegant solution, thank you :) $\endgroup$
    – xxxxx
    Sep 20, 2013 at 14:13
  • $\begingroup$ math.stackexchange.com/questions/515169/…, could you tell me what's wrong with my argument? $\endgroup$
    – lee
    Oct 5, 2013 at 0:45
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    $\begingroup$ "Clutching functions" are elsewhere called "transition functions," right? $\endgroup$
    – Neal
    Oct 5, 2013 at 2:49
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You can also use the fact that $S^1$ is the 1-pt compactification of the unit interval $I=[0,1]$, together with the fact that $I$ is contractible, so that every bundle over $I$ is trivial. Now, if you glue $0$~$1$ in $I$ to get $S^1$ from I , then the $\mathbb R^k$- bundle over $S^1$ will also glue along this map; informally, the fiber over $0$ will glue with the fiber over $1$ according to this map (which is called the monodromy). Like Jason said above, homotopic gluing maps give rise to isomorphic bundles. We then only need to consider the homotopy classes of homeomorphisms from I to itself , up to isotopy( so that we can glue the copy of $\mathbb R^k$ over $1$ with the copy of $\mathbb R^k$ over $0$. There are , up to isotopy, just two self-homeomorphisms; $+I$ and $-I$

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