# 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... – Ryan Unger Dec 19 '16 at 19:46