TITLE: On a manifold, is the $L^p$ space of vector fields complete?
QUESTION [7 upvotes]: If $(M,g)$ is a Riemannian manifold, let $\mathcal L^p(M)$ denote the set of vector fields $X$ whose norm $|X|$ is an $L^p(M)$ function. Is this complete? The usual proof fails miserably because of off-diagonal terms in the metric.

REPLY [2 votes]: Very late answer, but maybe still interesting to some:
Let $\pi\colon E\to M$ be a Hermitian vector bundle, i.e. a (continuous) vector bundle with an inner product on the fibers such that the local trivialization are fiberwise linear isometries (let's say the inner product on $\mathbb{R}^n$ is the standard one, but that does not really matter). We endow $E$ with its Borel $\sigma$-algebra.
The space $L^p(M;E)$ consist of all a.e.-equivalence classes of measurable section $X\colon M\to E$ such that $\int_M \langle X_x,X_x\rangle_x^{p/2}\,d\mathrm{vol}(x)<\infty$. I claim that $L^p(M;E)$ is isometrically isomorphic to $L^p(M;\mathbb{R}^n)$, which is of course complete.
Indeed, since $M$ is Lindelöf (and this is the only topological assumption we need, i.e. $M$ doesn't have to be a manifold), there exists a countable cover $(U_k)$ of $M$ by open sets and local trivializations $\phi_k\colon\pi^{-1}(U_k)\to U_k\times\mathbb{R}^n$. From these we can construct a global trivialization, albeit only in the measurable category:
Let $A_{k}=U_{k}\setminus \bigcup_{j=1}^{k-1} U_j$. These form a measurable partition of $M$. Let $\phi\colon E\to M\times \mathbb{R}^n$ be the map that coincides with $\phi_k$ on $\pi^{-1}(A_k)$. It is obviously measurable with measurable inverse and $\mathrm{pr}_2\circ\phi$ restricts to an isometry from $\pi^{-1}(x)$ to $\mathbb{R}^n$ for every $x\in M$. Thus the map
$$
TX=\mathrm{pr}_2\circ\phi\circ X
$$
maps measurable sections of $E$ to measurable maps from $M$ to $\mathbb{R}^n$. Moreover,
$|TX(x)|=|X_x|_x$. Hence $T$ maps $L^p(M;E)$ isometrically into $L^p(M;\mathbb{R}^n)$. Finally, to see that $T$ is surjective, it suffices to notice that
$$
S\colon L^p(M;\mathbb{R}^n)\to L^p(M;E),\,(Sf)_x=\phi^{-1}(x,f(x))
$$
is an inverse to $T$.
In the particular case of this question, one can take $E=TM$ to get the desired conclusion.