On a manifold, is the $L^p$ space of vector fields complete?

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.

To be specific, with the equivalence relation between $p$-integrable vector-fields via $Y\equiv X$ if $\|X-Y\|_p=0$, where $$\|X\|_p:=\sqrt[p]{\int_{M} g(X,X)^{p/2}\ dVol}$$ $L^p$ is the space of equivalence classes, and it is a complete normed vectorspace.

Look specifically at $M=\mathbb R^n$, where you have a global chart and can identify all tangent spaces. Here the norm is: $$\|X\|_p=\sqrt[p]{\int_{\mathbb R^n} (X(x)^T\cdot g(x)\cdot X(x))^{p/2}\ |g|\, d^nx}\tag{1}$$ Let $a$ be a hermitian root of the positive matrix $g$. Asking whether vector fields are complete wrt this norm is the same as asking whether or not functions that have a representant of the form $Y=\sqrt[p]{|g|^2}a\,X$ for some $X$ are complete sub-set of the regular $L^p$ space.

But $a$ is invertible because $g$ is, and $\sqrt[p]{|g|^2}$ is never zero so for any representant $Y$ you have $$Y=\sqrt[p]{|g|^2}a\left( \frac{a^{-1}}{\sqrt[p]{|g|^2}}Y\right)$$ and you have that this space is actually the regular $L^p$ space of vector fields on $\mathbb R^n$.

If you have a manifold $M$ that is covered by a finite amount of charts $C_i$, you have $$\|\cdot\|_{C_i}≤\|\cdot\|_M≤\sum_i \|\cdot\|_{C_i}$$ where $\|\cdot\|_{C_i}$ is the norm in ($1$). So anything Cauchy in $M$ is Cauchy on all $C_i$ and thus convergent in these and the last inequality makes it convergent on $M$ too.

In general you have a countable cover and not a finite one, I'm not sure the details work out but I would bet on it. I would guess a partition of unity will induce a map $\mathcal L^p(M)\to \overline{\bigoplus_{\alpha} L_n^p(\mathbb R_\alpha^n)}$ where $X$ is sent too $\sum_\alpha \underbrace{\varphi_\alpha X}_{\in L_n^p(\mathbb R^n_\alpha)}$. This should be a continuous linear map with continuous inverse.

• Perhaps one can use a partition of unity in the case when $M$ is noncompact. Your norm in (1) is missing a power in the integrand. I'll think about this... Dec 19 '16 at 19:46

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.