1) Assume that $E\to M$ is a smooth real vector bundle and $\nabla$ is a connection. (We do not assume any metric compatibility since we do not fix a metric on $E$). Assume that for every vector field $X$ on $M$ with a periodic orbit $\gamma$ and for every $s\in \Gamma^{\infty} (E)$, the section $\nabla_X s$ vanishes on at least one point of $\gamma$. Does this implies that $E$ is a (trivial) line bundle?

The motivation: The Rolle theorem is not valid in dimension greater than one.

2) Let $\ell$ be the canonical line bundle on $\mathbb{C}P^{1}\simeq S^2$, thought of as a smooth complex line bundle. Is there a connection $\nabla$ such that for every section $s$ and every periodic orbit $\gamma$ of $X$, the section $\nabla_{X}^{s}$ vanishes on at least one point of $\gamma$?

I ask the above question because I searched for some unusual differential operator associated with a vector field such that these operators can count the number of attractors of a vector field $X$. As a related post, please see: Elliptic operators corresponds to non vanishing vector fields

  • $\begingroup$ I don't quite understand the role of your motivation. Do you perhaps mean Rolle's theorem? $\endgroup$ – Joonas Ilmavirta Oct 6 '14 at 9:49
  • $\begingroup$ @JoonasIlmavirta yes thanks The Rolle theorem. I rvise it $\endgroup$ – Ali Taghavi Oct 6 '14 at 10:10
  • 4
    $\begingroup$ Counterexample to question 1: $E\to S^1$ is the Moebius band. Then each continuous section has a zero. The only possible closed orbit is $S^1$. $\endgroup$ – Peter Michor Oct 6 '14 at 11:40
  • $\begingroup$ @PeterMichor Very nice example. However it is a (nontrivial) line bundle. What about hiher dim. bundle and higher dimensional manifold $M$. $\endgroup$ – Ali Taghavi Oct 6 '14 at 13:26

As a first observation, if $(E,\nabla)$ has this Rolle property then if $i: S^1 \to M$ is any embbeding, $(i^*E,i^*\nabla)$ has this property as well (this is just a tubular neighborhood + cutoff argument). In particular we need to understand what possible such vector bundles with connection appear over $S^1$. It is fairly clear that the only possibilities are line bundles. If $i^*E$ is the Mobius bundle, then any connection has this property. If it is the trivial bundle, then the connection must have trivial holonomy (in other words there must be a trivialization for which the connection $i^*\nabla$ is the canonical one associated to that trivialization). In particular the connection must be flat.

For $\dim M > 2$ this classifies all such bundles rather satisfactorily: they are precisely the line bundles with a metric, equipped with the unique connection that preserves this metric. This is because any embedded circle in $\dim \geq 3$ admits a family of embeddings $i_n:S^1 \to M$ converging to the immersion of multiplicity two $z \mapsto i(z^2)$. The holonomy along these converges to the holonomy along the multiplicity two embedding, so the holonomy along $i$ must be $-1$ (a priori it could have been any negative scaling).

In dimension two, the above argument breaks down, and indeed there are line bundles on $S^1 \times \mathbb{R}$ that have holonomy a negative number $\neq 1$. However, there is clearly some reasonable casework that could be done: on $\mathbb R P^2$ all line bundles satisfying the condition are metric compatible: all homotopically nontrivial embedded circles have nontrivial normal bundle, so we may find embeddings $i_n$ converging to $i$ with multiplicity two. In the torus, we simply see that there are no holonomy maps $\mathbb{Z}^2 \to \mathbb{R}^{\times}$ which take on values besides $\{ \pm 1\}$ but also take on negative values or $1$ at every class representable by an embedding (namely if some class takes on a value beside $\pm 1$, some class representing an embedding takes on a positive value not equal to $1$), so again all line bundles satisfying our condition are metric compatible. (this section was edited)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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