7
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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... $\endgroup$
    – Ryan Unger
    Dec 19 '16 at 19:46
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.