# The “Rolle theorem” for sections of a vector bundle

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

• I don't quite understand the role of your motivation. Do you perhaps mean Rolle's theorem? – Joonas Ilmavirta Oct 6 '14 at 9:49
• @JoonasIlmavirta yes thanks The Rolle theorem. I rvise it – Ali Taghavi Oct 6 '14 at 10:10
• 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 $S^1$. – Peter Michor Oct 6 '14 at 11:40
• @PeterMichor Very nice example. However it is a (nontrivial) line bundle. What about hiher dim. bundle and higher dimensional manifold $M$. – 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)