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.

Solution 1:

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.

Solution 2:

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.