MathOverflow is a question and answer site for professional mathematicians. 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

A simple fact: Given a vector field on a compact manifold with boundary, if the vector field points inward along the boundary, then it must vanish somewhere in the interior. (EDIT: As pointed out in the accepted answer and in a comment, the Euler characteristic must be nonzero for this to be true.)

My question: Is there an analog of this fact in infinite dimensions? Perhaps for Banach manifolds?

share|cite|improve this question
I would bet against because compactness isn't available. Closed balls aren't compact. – Ben McKay Jul 6 '13 at 20:52
This depends on the topology though. In general, closed balls can be compact if the model space is not Banach. – Matthias Ludewig Jul 6 '13 at 20:58
I know that the answer can't be super simple since there is no compactness, but maybe there are hypotheses that make it work. (For example, something like the Palais-Smale condition.) Even one specific situation in the literature where such an argument was used would be interesting to me. – Dan Lee Jul 7 '13 at 0:35
Dan, what you claim as a "simple fact" is not true. Take for example $S^1 \times [0,1]$, this has an inward-pointing everywhere non-zero vector field. The Poincare-Hopf index theorem is what tells you when you have to have a zero, and that's given in terms of the Euler characteristic of the manifold. – Ryan Budney Jul 7 '13 at 10:40
Given how embarrassingly wrong my original premise was, it's a bit awkward to ask this, but I still wonder if something interesting can be said about the unit ball in Banach space. – Dan Lee Jul 8 '13 at 4:43
up vote 5 down vote accepted

The simple fact in question is false in any dimension greater than one.

Consider the strip $ \mathbb{R} \times [-\pi/2,\pi/2] \subset \mathbb{R}^2$. At a point $(x, y)$ take the vector $(-sin(y), cos(y))$. This does not depend on $x$ so descends to a vector field on the annulus $\mathbb{R}/\mathbb{Z} \times [-\pi/2, \pi/2]$. It is obviously nowhere vanishing and points inwards at the boundaries.

share|cite|improve this answer
Thanks. The "reasoning" I had in mind was just totally bogus. – Dan Lee Jul 8 '13 at 4:20

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.