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I can't see or understand if it is true or not if all vector bundles on over an open set of $\mathbb{R}^n$ are trivial or not. Is there an easy way to see it?

The problem comes from the fact that we define what a local trivialization is for a bundle and we define an atlas which trivializies the bundle. Since every chart is diffeomorphic to an open set of the euclidean space, if all the open sets have trivial bundles it should say that a trivialization could be global. The idea is that a vector bundle of rank $r$ over some manifold, with a trivialing atlas $\{(U_{\alpha},\phi_\alpha,\chi_\alpha)\}$ is equivalent to some limit of the sheaf $C^\infty(U_\alpha,\mathbb{R}^r)$ over all the atlas. The point of the question is if such atlas can always be taken to be a maximal atlas.

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    $\begingroup$ Not unless the set is contractible. Try an annulus - by treating it essentially as a circle you can build a Möbius strip-like line bundle. $\endgroup$ Jul 3 '16 at 11:43
  • $\begingroup$ @AnthonyCarapetis I knew homology would get the answer but I couldn't get why. Thank you. $\endgroup$
    – Lolman
    Jul 3 '16 at 13:39
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If $X$ is a paracompact topological space, $\operatorname{Vect}_k(X) \cong [X, BO(k)] = [X, \operatorname{Gr}_k(\mathbb{R}^{\infty})]$ where $\operatorname{Vect}_k(X)$ denotes isomorphism classes of real rank $k$ vector bundles on $X$ and the square brackets denote homotopy classes of maps. In particular,

$$\operatorname{Vect}_1(X) \cong [X, \operatorname{Gr}_1(\mathbb{R}^{\infty})] = [X, \mathbb{RP}^{\infty}] = [X, K(\mathbb{Z}_2, 1)] \cong H^1(X; \mathbb{Z}_2).$$

Under this identification, the isomorphism class of a line bundle $L$ on $X$ is identified with its first Stiefel-Whitney class $w_1(L)$. So $L$ is trivial if and only if $w_1(L) = 0$ (i.e. $L$ is orientable).

Every open subset of $\mathbb{R}^n$ is paracompact (metric spaces are paracompact), so we can use the above characterisation in order to answer your question.

The open set $U = \mathbb{R}^2\setminus\{(0,0)\}$ in $\mathbb{R}^2$ deformation retracts onto $S^1$, so

$$\operatorname{Vect}_1(U) \cong H^1(U; \mathbb{Z}_2) \cong H^1(S^1; \mathbb{Z}_2) \cong \mathbb{Z}_2.$$

Therefore, there is a non-trivial line bundle on $U$.

However, if we take $U$ to be contractible, then the claim holds.

Note, there are non-contractible open sets of $\mathbb{R}^n$ such that every vector bundle on them is trivial. For example, $U = \mathbb{R}^4\setminus\{(0,0,0,0)\}$ which deformation retracts onto $S^3$. Either by the above characterisation or by using the clutching construction, we see that $\operatorname{Vect}_k(U) \cong \pi_2(O(k))$. As the second homotopy group of any Lie group is zero, we see that every vector bundle on $U$ is trivial.

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  • $\begingroup$ I can't really evaluate this answer for I lack some background on Homology. It is nevertheless appreciated. $\endgroup$
    – Lolman
    Jul 31 '16 at 10:11

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