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.

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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 $\omega$.