Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

do non-zero global section always exist in a manifold $M$? If $M$ is compact I think they do because taking a partition of unity $\rho_{\alpha}$ subordinated to a finite covering, and defining local sections $s_{\alpha}$ in this finite covering I can take $$s:=\sum s_{\alpha} \rho_{\alpha}$$ Is this argument right? I guess it is not true for general $M$ thanks

share|cite|improve this question
Non-zero global section of what vector bundle exactly? It's certainly not true for every vector bundle - consider the Mobius bundle over the circle. (You can draw it in $\mathbb{R}^3$). The problem is that your partition of unity argument doesn't guarantee things won't cancel out at some points. – Jason DeVito Aug 28 '12 at 17:37
up vote 5 down vote accepted

You do not need compactness nor even paracompactness (i.e. no partition of unity is necessary).
Take any nowhere zero continuous section of the vector bundle on an open trivializing subset $U$ for the vector bundle , multiply it by a continuous plateau function with compact support in $U$ and extend by zero to the whole manifold: this yields a non-identically zero continuous section of the vector bundle.

NB I have interpreted your question as asking for non-identically zero continuous sections:they always exist.
In general it is however impossible to find a nowhere zero section of an arbitrary vector bundle on an arbitrary manifold, as shown in other answers.
But sometimes it is possible: on a contractible manifold like $\mathbb R^n$ all vector bundles are trivial and thus they certainly admit of nowhere zero continuous sections.

share|cite|improve this answer

No, every even dimensional sphere is a counter-example (cf. Hairy ball theorem).

Moreover, a closed orientable manifold admits a nowhere zero section of its tangent bundle iff its Euler class (and therefore also its Euler characteristic) vanishes.

share|cite|improve this answer

Absolutely not! Take the tangent bundle over a manifold. A globally defined non-zero section is a non-singular vector field. The Poincaré-Hopf Theorem relates the topology of your surface with the existence of non-singular vector field. Consider, e.g. a sphere, there are no continuous, non-zero vector fields on the sphere. This is called the Hairy Ball Theorem. This is true of any compact, orientable surface with non-zero Euler Characteristic.

For further reading, take a look at Chern Classes (in the complex case) and Stiefel–Whitney classes (in the real case).

share|cite|improve this answer

In general it's not true.

For instance take the tangent bundle of $\mathbb S^2$. Then you can't have a non-vanishing vector field defined globally.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.