Examples of non trivial vector bundles

Question

Once you see the notion of vector bundle, next thing you want to see are examples of non trivial vector bundles.

Here, I want to collect such examples with justification of one or two lines saying why this vector bundle is non trivial.

Please add one per answer.

Paul Orland · Answer 1 · 2020-12-07T06:17:33.240

7

The usual example is the Möbius bundle. Let $X = [0,1]\times\mathbb{R}$, and define an equivalence relation $\sim$ where $(0,y)\sim(1,-y)$. Let $E = X/\sim$ be our total space and let $M = [0,1]$. Then a vector bundle $\pi:E \rightarrow M$ is given by $[(x,y)] \mapsto x$.

Edit to give a justification: if $E$ were trivial, then it would admit a smooth global frame, or a non-vanishing, smooth, global section. This would amount to a nonvanishing function $f : [0,1] \mapsto \mathbb{R}$ with $f(0) = -f(1)$, which can't exist according to the intermediate value theorem.

edited Dec 07 '20 at 06:17

answered Mar 06 '19 at 18:33

Paul Orland

7,138

I am asking for an example with an explanation of why it is non trivial :) not just example of vector bundles :) – Mar 06 '19 at 18:45

score 7 · Answer 2 · answered Mar 06 '19 at 23:22

You get an example for every non-orientable smooth manifold $M$:

A smooth $n$-dimensional manifold $M$ is orientable iff there exists a nowhere vanishing $n$-form i.e. a nowhere vanishing section of the bundle $\Lambda^n(T^*M)$ whose fiber at $p$ is the vectorspace of all multlinear alternating maps from $(T_pM)^n$ to $\mathbb R$. Since this bundle is 1-dimensional the existence of a nowhere vanishing section is equivalent to the trviality of the bundle.

As an example if $n$ is even then $\mathbb{RP^n}$ is not orientable. The reason :

Any $n$-form $\omega$ on $\mathbb{RP^n}$ can be pulledback to a $n$-form $\bar\omega$ on $S^n$ via the quotient map $q:S^n\rightarrow\mathbb{RP^n}$. Then after identifying $TS^n\subset S^n\times\mathbb{R^{n+1}}$ there is $f\in C^\infty(S^n)$ such that $\bar\omega_p(X_1,...,X_n)=f(p)\det(X_1,...,X_n,p)$. As $q$ cannot distinguish antipodal points $\bar\omega_{p}(X_1,...,X_n)=\bar\omega_{-p}(-X_1,...,-X_n)$. In particular setting $X_i=e_i$, $p=e_{n+1}$ yields $f(-p)=-f(p)$ so as $S^n$ is connected $f$ has a zero so $\bar\omega$ vanishes somewhere and hence so does $\omega$.

score 5 · Answer 3 · answered Mar 06 '19 at 18:46

5

Given a map $f:S^{k-1} \to GL(n)$, construct a rank-$n$ bundle $\xi_f \to S^k$ by attaching the trivial bundles $\theta^n_+, \theta^n_-$ over the two hemispheres $S^k_+, S^k_-$ by $f$. This map (the "clutching construction") actually gives a bijection between $\pi_{k-1}(GL(n))$ and vector bundles (up to isomorphism) over $S^k$. Proving the latter is fact is mostly a matter of unwinding the definition, noting that any bundle over a contractible space (e.g., each hemisphere) is trivial. Like most of algebraic topology, proving that a particular bundle usually involves showing that certain invariants of it (e.g., the Euler class) are nontrivial, rather than trying to prove anything directly.

answered Mar 06 '19 at 18:46

anomaly

26,475

So, you are saying $\xi_f\rightarrow S^k$ is non trivial bundle.. can you give one or two lines explanation saying why it is non trivial? please keep in mind these has to be examples just after introducing the notion of vector bundles... – Mar 06 '19 at 18:50
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This is an exceptionally important class of explicit examples, but as the answer points out, proving that any particular vector bundle constructed this way is non-trivial is typically done indirectly using tools from algebraic topology. Indeed, it's kind of a mathematical fact of life that non-triviality of any given vector bundle is often most readily detectable indirectly via algebraic topology. – Branimir Ćaćić Mar 06 '19 at 18:52
@PraphullaKoushik If you want to impose such a condition on answers, you should include in the question to begin with---otherwise people will spend time writing answers that don't meet you unspoken criteria. (But like Branimir points out, for this particular question that restriction might be too much, as the relevant tools genearlly aren't available "just after introducting the notion of vector bundles".) – Travis Willse Mar 06 '19 at 19:00
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@PraphullaKoushik: As BranimirĆaćić said, it's generally very difficult to prove whether a particular bundle is trivial from first principles, just as it's hard to prove that two spaces aren't homeomorphic just from the definition. Characteristic classes, for example, work very well for this sort of thing, but it's just not possible to do much from scratch. At the very least, it's not a one- or two-sentence thing. – anomaly Mar 06 '19 at 19:01
@BranimirĆaćić that may be true but I am looking for examples which you can do prove non triviality without using algebraic topology... I do not doubt if some thing can be proved to be non trivial using algebraic topology I am asking for basic examples where one can prove easily from definitions, – Mar 06 '19 at 19:02
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@PraphullaKoushik: Right, but what we're saying is that there are no easy examples. – anomaly Mar 06 '19 at 19:09
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@PraphullaKoushik If you really want to get your hands dirty, I think the most elementary conceivable non-trivial vector bundles (from the standpoint of proving they're non-trivial) are the non-trivial complex line bundles on $S^2 \cong \mathbb{C}P^1$. In this case, your clutching map basically amounts to a function $f : S^1 \cong \text{equator of $S^2$} \to \mathbb{C} \setminus {0}$, and once you fix a point on the equator, you find that the resulting line bundle is trivial if and only if the class $[f] \in \pi_1(\mathbb{C} \setminus {0}) \cong \pi_1(S^1) \cong \mathbb{Z}$ is zero. – Branimir Ćaćić Mar 06 '19 at 23:03
@PraphullaKoushik Indeed, I think you'll find it a very useful exercise to convince yourself that giving a global trivialisation of the line bundle defined by $f$ is tantamount to giving a nullhomotopy of the class $[f]$. – Branimir Ćaćić Mar 06 '19 at 23:05

score 3 · Answer 4 · 2019-03-06T19:07:36.797

3

A very important case of vector bundle is the tangent bundle $TM$, the disjoint union of all of the tangent spaces of a manifold. It being trivial is equivalent to $n(=\dim(M))$ linearly indipendent vector fields on the manifold (that is, $n$ sections of the tangent bundle).

The question if a manifold is parallel is not easy (for references, go here) :

For $S^2$ this is not true, and it's a consequence of the famous hairy ball theorem. For the n-sphere in general, the only parallelizable $S^n$ are for $n=0,1,3,7$, that is to say:

$TS^n$ is non trivial for every $n\neq 0,1,3,7$

For very short and self contained proof of the Hairy ball theorem (and of a little more actually) by Milnor, click here

For the general result on parallelizable sphere, the original work can be found here

For references, look up this Wikipedia page

edited Mar 06 '19 at 19:07

answered Mar 06 '19 at 18:38

Please consider giving an example and an explanation of a bundle being non trivial... – Mar 06 '19 at 18:46
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As the answer explains, non-triviality of the vector bundles discussed here is actually a well-known and rather non-trivial theorem, the so-called "hairy ball theorem," whose proof would be a bit difficult to sketch in one or two sentences. – Branimir Ćaćić Mar 06 '19 at 18:53
@PraphullaKoushik a full explanation, as Branimir pointed out, would be quite long, as it relies on some really non trivial theorems (for which, by the way, I'll add some references). If, on the other and, is the concept of parallelizable manifold the problem, I'll add the definition. – Mar 06 '19 at 18:59
Thanks for the links. +1 – Mar 06 '19 at 19:04

Examples of non trivial vector bundles

4 Answers4