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Let $(M,g)$ be an orientable Riemannian manifold $U \subset M$ with chart

$$ \varphi: U \rightarrow V \subset \mathbb{R}^n$$

I wonder if one can say that

(1) "If $u \in H^k(U)$ then $v:=(u \circ \varphi^{-1}) \in H^k(V)$ and vice versa."

Here, by $H^k$ I denote the $2,k$ Sobolev space.

I tried calculating this when $k=1$:

$$\Vert u \Vert^2_{H^1} = \int\limits_U \vert u \vert^2 + \int\limits_U g^{ij} \partial_i u \partial_j u $$

and

$$ \Vert v \Vert^2_{H^1}=\int\limits_V \vert u \vert^2 + \int\limits_V (\partial_i u )^2 $$

As far as I see, (1) should hold, if $g$ is bounded. Is that correct?

User1
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  • How do you know that $g^{ij}$ is bounded in your chart? – Kajelad May 08 '21 at 23:58
  • If you fix an atlas of coordinate charts, this works. If you have a smooth Riemannian metric, you can just define the Sobolev norm to be $$|u|{H^k}^2 = \sum{j=0}^{k} \int_M |\nabla^ju|^2,dV_g$$ – Deane May 09 '21 at 01:27
  • @Kajelad I meant that if we assume that $g$ is bounded – User1 May 09 '21 at 06:53
  • @User1 What does mean it mean to say that $g$ is bounded? There are no norms available except those induced by $g$ itself. – Kajelad May 09 '21 at 06:57
  • I though about $\sup\limits_{M} g_{ij}$ – User1 May 09 '21 at 06:58
  • @User1 What does that mean? The components $g_{ij}$ aren't defined without first choosing a coordinate chart, and even then they won't be defined on all of $M$. – Kajelad May 09 '21 at 12:48
  • I am also not quite sure what this means, but in "Wave maps in general relativity" it says "Suppose $g$ is smooth with uniformly bounded derivatives of arbitrary order." – User1 May 09 '21 at 12:51

1 Answers1

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From the point of view of a pedantic mathematician, the answer to your question is no. From the point of view of a physicist interested in Wave maps in general relativity however, you can certainly make use of the intuition that a positive answer would provide, and this intuition will be sufficient 'for all practical purposes' and not too misleading.

There are a lot of technical issues about Sobolev spaces on manifolds, even on charts. In order to keep the discussion simple, let me assume that $U$ has smooth boundary (depending on whom you ask to, this might already be part of the definition of chart on a smooth manifold). By this I mean that there exists an atlas of $M$ which, when restricted to $\partial U$, turns $\partial U$ into a smooth manifold.

What the OP probably has in mind is that $U$ is a bounded chart (it is not part of the standard definition of chart that the domain of the chart be a bounded set). If so, then the metric $g$ is bounded (by compactness of $\overline U$ and since $g$ extends continuously beyond $U$). Clearly the same holds for all the derivatives of $g$. This means that for every multi-index $\alpha$ there exists a constant $C_{|\alpha|}>0$ so that $$\sup_{x\in U} \|D^\alpha g(x)\| \leq C_{|\alpha|}$$ where the norm denotes the standard norm in a suitable (depending on $\alpha$) space of tensors (equivalent to the standard norm on some Euclidean space), and $|\alpha|$ is the length of the multi-index $\alpha$. In the case $\alpha=0$ we let $D^\alpha\equiv \mathbb 1$, and the norm is simply any norm on the space of $n\times n$ matrices (all of these norms are equivalent). Incidentally (to clarify the comments to the OP), the metric $g$ is (globally) bounded on $M$ if the constants $C_{|\alpha|}$ can be chosen independently of the chart(s) $U$. In this case – and only in this case –, the intuition behind (1) in the OP is correct, as discussed in some detail here.

It is instructive to see what goes wrong if $U$ is not a bounded chart. The first problem is that now $U$ can be the whole of $M$. E.g., $\mathbb R^n$ (with any metric $g$) admits a unique chart. In the rest of the discussion we can therefore assume that $U=M$, since this does not in fact make things more complicated than they already are for unbounded $U$. This is a problem because now the very definition of Sobolev space is in question, and there are (at least) three main competing ones:

  1. closure of smooth compactly supported functions $\mathring H^k:= \overline{\{\mathcal C^\infty_c(M)\}}^{\|\cdot\|_{H^k}}$;

  2. closure of smooth (possibly unbounded) functions sufficiently integrable and with sufficiently integrable derivatives $H^k := \overline{\{f\in \mathcal C^\infty(M): \|f\|_{H^k}<\infty\}}^{\|\cdot\|_{H^k}}$

  3. distributions the derivatives of which are identifiable with $ L^2$-functions $W^k:= \{ f \text{ distribution on } M : D^\alpha f \in L^2, |\alpha|\leq k\}$

  4. Bessel potential spaces

The $H^k$-norm is defined as in the OP.

Fact: these spaces do not coincide!

Without going into the details here, counterexamples can be found in [A] (below). This suggests that, in defining Sobolev spaces on manifolds, one should take a global perspective rather than a local one. Indeed, the previous fact has dire consequences:

Fact: most of the standard results on Sobolev spaces (e.g. Sobolev/Rellich–Kondrashov embedding, etc.) do not hold in this generality, and they actually might fail very badly, e.g. at all integrability scales $p$ (here $p=2$), and at every order $k$.

This can happen even on $3$-dimensional manifolds with Ricci curvature bounded below by 0.

A solution to these problems is to assume

Definition (bounded geometry) $(M,g)$ has bounded geometry if (a) $(M,g)$ has injectivity radius bounded away from $0$ (i.e. $\mathrm{inj}_g M>0$) and (b) $g$ (and all its derivatives) are globally bounded.

In this case, the above definitions of Sobolev spaces coincide. Note that (a) is necessary to have well-behaved Sobolev spaces. Indeed (a) and (b) rule out two undesirable behaviours at infinity. (a) tells that the manifold cannot accumulate too much topology at infinity at small scales, while (b) tells that the manifold cannot explode (metrically), i.e. that it is uniformly controlled at large scales. Both of these assumptions are important, and (b) is not sufficient to rule out strange phenomena in full generality.

[A] E. Hebey, Sobolev Spaces on Riemannian Manifolds, 1996

AlephBeth
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