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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?

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    $\begingroup$ I would bet against because compactness isn't available. Closed balls aren't compact. $\endgroup$
    – Ben McKay
    Commented Jul 6, 2013 at 20:52
  • $\begingroup$ This depends on the topology though. In general, closed balls can be compact if the model space is not Banach. $\endgroup$
    – Matthias Ludewig
    Commented Jul 6, 2013 at 20:58
  • $\begingroup$ 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. $\endgroup$
    – Dan Lee
    Commented Jul 7, 2013 at 0:35
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    $\begingroup$ 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. $\endgroup$
    – Ryan Budney
    Commented Jul 7, 2013 at 10:40
  • $\begingroup$ 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. $\endgroup$
    – Dan Lee
    Commented Jul 8, 2013 at 4:43

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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.

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  • $\begingroup$ Thanks. The "reasoning" I had in mind was just totally bogus. $\endgroup$
    – Dan Lee
    Commented Jul 8, 2013 at 4:20

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