Lower estimate on the covariant derivative of a section of a curved vector bundle

Question

Let $(M,g)$ be a Riemannian manifold, and let $E\to M$ be a vector bundle with metric $h$ and connection $\nabla$. Assume that its curvature $F\in\Omega^2(M;\mathrm{End}E)$ is nonzero at $e\in E_p$, $p\in M$, in the sense that one finds $v$, $w\in T_pM$ such that $F(v,w)e\ne 0$. Then it is impossible to find a parallel section $s$ of $E$ defined near $p$ with $s(p)=e$. The best one can hope for in general would be $\nabla s|_p=0$.

My question is: can this be quantified? That is, are there suitable geometrically motivated norms (say, constructed from $g$, $h$ and $\nabla$, and equivalent to some standard norm, e.g. Sobolev or Hölder) such that one has a general estimate of the kind $$\|\nabla s\|\ge c\,\|F(-,-)s\|$$ for some $c>0$?

Answer 1

The exterior covariant derivative $d^\nabla:\Omega^{1}(M;E)\rightarrow \Omega^{2}(M;E)$ is a linear map that satisfies $d^{\nabla}\nabla=F$, and extends to a continous linear map on all Sobolev/Holder versions of vector valued forms (regardless of the fiber metric on $E$). Hence, by continuity, for example as a mapping between $W^{s+1,p}$ and $W^{s,p}$ sections, $$ \|F(\cdot,\cdot)s\|_{s,p}=\|d^{\nabla}\nabla s\|_{s,p}\leq C\|\nabla s\|_{s+1,p}. $$

If you happen to need a reference for this, see for example the very last exercises in chapter 9 in Peter's Peterson book Riemannian Geometry, 3rd edition.

---

EDIT at the request of @Sebastian Goette.

I think the following might work as a way to obtain a $W^{s,p}\rightarrow W^{s,p}$ estimate, at least in the case $M$ is compact (possibly with boundary). Maybe you can loosen the compactness assumption, but I don't know how to work with noncompact manifolds.

The covariant derivative as a linear map, $\nabla :W^{s+1,p}(E)\rightarrow W^{s,p}(E\otimes T^{*}M)$, has a closed range, and $\ker(\nabla)$ is finite dimensional. This can be proved using elliptic theory of the operator $\nabla^{*}\nabla$ with boundary conditions on $\nabla_{n}v=0$ on the boundary, where one requires compactness. Let us denote this range by $\mathcal{R}^{s,p}(\nabla)$ when it is equipped with the $W^{s,p}$ norm. It is also possible to show that if $s\leq r$ then $\mathcal{R}^{r,p}(\nabla)$ is dense in $\mathcal{R}^{s,p}(\nabla)$ with this norm.

Now, $d^{\nabla}:\mathcal{R}^{s+1,p}(\nabla)\rightarrow W^{s,p}(E\otimes \Lambda^{2}T^{*}M)$ satisfies the estimate, $$ \|d^{\nabla}\nabla v\|_{s,p}=\|F(\cdot,\cdot)v\|_{s,p}\le C\|v\|_{s,p}. $$ Where we used the fact that $F$ maps $W^{s,p}\rightarrow W^{s,p}$ continously (Here too a compactness assumption on $M$ enters, since $F$ is a bundle map).

If one replaces $v$ with its projection onto the complement of $\ker\nabla$, let us denote it by $\tilde{v}$, then $\nabla v=\nabla \tilde{v}$ and since $\nabla:W^{s+1,p}\rightarrow W^{s,p}$ has a closed range it is bounded from below when restricted to this complement, which is to say, $$ \|d^{\nabla}\nabla v\|_{s,p}=\|d^{\nabla}\nabla \tilde{v}\|_{s,p}\leq C\|\tilde{v}\|_{s,p}\leq C\|\tilde{v}\|_{s+1,p}\leq C\|\nabla{\tilde{v}}\|_{s,p}=C\|\nabla{v}\|_{s,p} $$ where I did not rename the constants. Thus, by the density of $\mathcal{R}^{s+1,p}(\nabla)$ in $\mathcal{R}^{s,p}(\nabla)$ in the $W^{s,p}$-norm, $d^{\nabla}$ extends to a continous linear map $d^{\nabla}:\mathcal{R}^{s,p}(\nabla)\rightarrow W^{s,p}(\nabla)$, with an estimate, $$ \|F(\cdot,\cdot)v\|_{s,p}=\|d^{\nabla}\nabla v\|_{s,p}\le C\|\nabla v\|_{s,p} $$ as required.