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Working on a problem in Differential Geometry, which is quite far away from my area of expertise, I was recently led to consider the class of those smooth, $n$-dimensional embedded submanifolds $M \subset \mathbb{R}^N$ such that the following condition is satisfied:

the manifold $M$ possesses an oriented atlas $\{(U_{\alpha}, \, \phi_{\alpha})\}$, where $\phi_{\alpha} \colon U_{\alpha} \stackrel{\cong}{\longrightarrow} V_{\alpha} \subset \mathbb{R}^{n}$, such that, denoting by $i_{\alpha} \colon U_{\alpha} \hookrightarrow \mathbb{R}^N$ the inclusion map, the composition $$i_{\alpha} \circ \phi_{\alpha}^{-1} \colon \, V_{\alpha} \longrightarrow \mathbb{R}^N$$
has bounded partial derivatives up to order $k$, uniformly on $\alpha$.

My question is now the following:

Does the condition above have a name? In this case, is there any characterization of the manifolds $M$ satisfying it in terms of the usual Riemannian invariants (for instance, curvature)?

Any answer or reference to the relevant literature will be greatly appreciated.

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    $\begingroup$ Are you sure your question has sense? $\endgroup$ Nov 2, 2017 at 23:25
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    $\begingroup$ Not sure if it's cheating, but if I replace $\phi_\alpha$ by $C_\alpha\phi_\alpha$, then I can choose $C_\alpha$ large enough to kill any non-uniformity. $\endgroup$
    – Fan Zheng
    Nov 3, 2017 at 5:11
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    $\begingroup$ jaapeldering.nl/files/NHIM-noncompact-book.pdf I think you should have a look at definition 2.21 . $\endgroup$
    – Thomas Rot
    Nov 3, 2017 at 8:43
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    $\begingroup$ I see you are an algebraic geometer: do you happen to be specifically interested in the situation when $M$ is a complex submanifold of $\mathbb{R}^N=\mathbb{C}^{N/2}$ and the parametrizations $i_{\alpha} \circ \phi_{\alpha}^{-1}$ are holomorphic? If so, note that there should be stronger results available in this more specific setting. I am not sure where to look for the holomorphic version of the extrinsic results mentioned by Rbega, but ... $\endgroup$
    – macbeth
    Nov 3, 2017 at 17:16
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    $\begingroup$ ... you can consult [Tian-Yau][1] Proposition 1.2, [Hein][2] Lemma 2.7 for the holomorphic version of the intrinsic results I mentioned. [1]: mathscinet.ams.org/mathscinet-getitem?mr=1040196 [2]: mathscinet.ams.org/mathscinet-getitem?mr=2869021 $\endgroup$
    – macbeth
    Nov 3, 2017 at 17:16

4 Answers 4

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After writing this out I noticed Thomas Rot's link above which does the same thing but with more detail...maybe this will still be helpful.

I'm not entirely sure if the following condition is equivalent, but it certainly implies what you want:

Say a submanifold $M$ is graphical to order $k$, $k\geq 2$, on scale $r$ if for each $p\in M$ one can express the component of $C_{r}(T_p M, p)\cap M$ containing $p$ as the graph of a function $$U_{p,r}: D_r(T_p M,p)\to \mathbb{R}^{N-n}$$ that satisfies $$ \sup_{0\leq i \leq k, q\in D_r(T_p M,p) } r^{i-1} |D^i U_{p,r}(q)|\leq 1. $$ The derivatives $D$ are the standard euclidean ones on $T_p M$.

Here $$ D_r(P, p) \mbox{ is the disk of (euclidean) radius $r$ in the plane $P$ centered at $p\in P$}, $$ $$ C_r(P, p) \mbox{ is the cylinder of height $r$ over $D_r(P, p)$} $$ and the graph of $U_{p,r}$ is (when $T_pM=\mathbb{R}^n\times\{0\}\subset \mathbb{R}^{N}$), $$\Gamma_{U_{p,r}}=\{(q, U_{p,r}(q)): q\in D_r(T_p M,p) \} $$ The more general case is obtained by rotation.

It's not hard to see (for instance this is done in Colding and Minicozzi's book on minimal surfaces in the codimension one case) that there is an $\epsilon_2>0$ so if $$ \sup_{M} |A_M|\leq \epsilon_2 $$ then $M$ is graphical to order $2$ on scale $1$. Here $A_M$ is the second fundamental form of $M$. There is a more scaling invariant formulation of this I will leave to you.

In fact, once you are in the order $2$ situation you have uniform two sided bounds on the metric and on the Christoffel symbols and so covariant derivatives are ``the same" as euclidean ones. In particular, for $k\geq 2$, there is an $\epsilon_k>0$ so that if $$ \sup_{M} \sum_{i=2}^k |\nabla^{i-2}_M A_M | \leq \epsilon_k $$ then $M$ is graphical to order $k$ on scale $1$. Here $\nabla_M$ is the induced connection on $M$.

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  • $\begingroup$ In fact, the second half of this nice answer does not appear in the Rot link.... $\endgroup$
    – macbeth
    Nov 3, 2017 at 15:55
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Posting this a an answer since it's too long for a comment.

I have a feeling that the question as formulated might be ill posed, or at least not quite what you'd expect in the sense that all embedded manifolds $M \subset \mathbb{R}^N$ satisfy it, as already noted by Deane Yang and Fan Zheng: you didn't fix an atlas on $M$, so you can always stretch charts.

Possibly a better condition would be to consider if the embedding $\iota: M \to \mathbb{R}^N$ is a $C^k$ bounded map in the sense of bounded geometry, e.g. with Definition 2.9 from my book (that Thomas Rot already linked to) and with $M$ having its metric induced by the Euclidean metric of $\mathbb{R}^N$. Modulo some issues with loss of smoothness due to curvature conditions in (that Rbega seems able to address) this should be equivalent to $C^k$ boundedness of $M$ in terms of local graphs.

The fact that your original condition would be satisfied by any $C^k$ embedding $M$ seems to match with the fact that any smooth manifold possesses a metric of bounded geometry (see Greene, "Complete metrics of bounded curvature on noncompact manifolds", Arch. Math. (31) 1978), as mentioned by Deane Yang.

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  • $\begingroup$ Thank you for your answer. When you write "all embedded manifolds $M \subset \mathbb{R}^N$ satisfy it", you mean (for instance) that if we have $$||i_{\alpha} \circ \phi_{\alpha}^{-1}||_{\infty} \leq M_{\alpha} \quad \mathrm{on} \; \; V_{\alpha}$$ then we can replace every $\phi_{\alpha}$ with $M_{\alpha} \phi_{\alpha}$ in order to obtain the uniform bound $$||i_{\alpha} \circ \phi_{\alpha}^{-1}||_{\infty} \leq 1 ?$$ $\endgroup$ Nov 3, 2017 at 22:08
  • $\begingroup$ @FrancescoPolizzi: indeed. So if you're ok with any atlas on $M$, then your condition can always be satisfied. Using bounded geometry essentially means that you're restricting to atlases generated by exponential maps. I also looked a bit into an alternative definition of uniformity that's only expressed in terms of the atlas, see my notes jaapeldering.nl/files/uniform-mflds.pdf. This definition is essentially equivalent to bounded geometry wrt. a class of metrics on $M$. $\endgroup$ Nov 4, 2017 at 12:23
  • $\begingroup$ Do you mean any atlas compatible with the original one, right? Of course, I want that $M$ remains the same embedded submanifold of $\mathbb{R}^N$ as before. $\endgroup$ Nov 4, 2017 at 12:55
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    $\begingroup$ Yes, any compatible atlas. Also, as a side remark: an atlas satisfying your condition can never be maximal, see also my remark under definition 4 in my notes. $\endgroup$ Nov 4, 2017 at 13:25
  • $\begingroup$ ok, thank you again. This is very useful. $\endgroup$ Nov 4, 2017 at 14:41
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Update: Everything I wrote before is fine, but here's a specific result, from Gilbarg-Trudinger Corollary 16.7 (2001 edition), as an idea of what's possible. Let $\Omega\subseteq \mathbb{R}^n$, and let $u:\Omega\to\mathbb{R}$. Let $H$ denote the mean curvature of the graph of $u$ (this graph is a hypersurface in $\mathbb{R}^{n+1}$). Then for all $y\in\Omega$, $$|D^{k+1}u|_y|\leq C(n,k,\sup|H|,\ldots \sup |D^kH|,\operatorname{dist}(y,\partial\Omega),\sup|u|).$$

Old: I think you may wish to speak with a geometric measure theorist. A few remarks:

  1. You're looking for a kind of bound on "parametrization badness", and this is perfectly reasonable, even if tricky to formulate.

  2. An $n$-submanifold of $\mathbb{R}^N$ is locally a graph of a function $f:\mathbb{R}^n\to\mathbb{R}^{N-n}$ (by selecting a suitable $n$-subspace of the ambient space). It is probably equivalent to bound the "badness" of these functions $f$.

  3. I would guess that bounds on mean curvature (and its higher derivatives) imply bounds on "graph badness"/"parametrization badness". For example, Propositions 3.2 and 5.1 of these notes on the Allard regularity theorem seem sort of similar (but again, talk to a geometric measure theorist!)

  4. I am an intrinsic geometer, not an extrinsic geometer, so my intuition here is coming from the corresponding theorems in the intrinsic case. A good reference is Section 1.2 of Hebey's book Sobolev Spaces on Riemannian Manifolds. Briefly, one has an abstract (not embedded) Riemannian manifold $(M,g)$, and one formulates a notion of "chart badness" as follows: for a chart $\varphi:V\to B$ (with $V\subseteq \mathbb{R}^n$) whose image is some geodesic ball $B=B_x(r)\subseteq M$, the metric tensor is $C^{1,\alpha}$-controlled by $\epsilon$ in that chart if, letting $g_{ij}$ be the components of the metric tensor with respect to the chart,

    • $(1-\epsilon)g_{ij}\leq \delta_{ij}\leq (1+\epsilon)g_{ij}$ as bilinear forms
    • $r \sup|Dg_{ij}|\leq \epsilon$
    • $r^{1+\alpha}\sup_{x\neq y}\frac{|Dg_{ij}|_x-Dg_{ij}|_y|}{d(x,y)^\alpha}<\epsilon$

    (Ugh! The point is, getting the right definition is hard.) Then it is a theorem of Anderson, Anderson-Cheeger that upper and lower bounds on Ricci curvature imply the existence of $C^{1,\alpha}$ charts with uniform size $r$ and badness $\epsilon$. The proof is by taking harmonic co-ordinates. If you also have bounds on the derivatives of the Ricci tensor, then you get higher-order bounds on the metric tensor in the chart.

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    $\begingroup$ Mean curvature isn't strong enough (e.g. one can have catenoidal necks) $\endgroup$
    – Rbega
    Nov 3, 2017 at 11:56
  • $\begingroup$ Oh, interesting. So (as I am sure @Rbega knows) I actually omitted an important hypothesis for the intrinsic case -- one needs bounded injectivity radius, or bounded-below volume of balls of a fixed radius, in order to get the theorem "bounded Ricci implies good charts". Is there a similar geometric hypothesis in the extrinsic case to get "bounded mean curvature implies graph of good function"? $\endgroup$
    – macbeth
    Nov 3, 2017 at 15:47
  • $\begingroup$ If the manifold is graphical to order 1 on scale 1 (in the notation from my answer), then the mean curvature of the graph satisfies a uniformly elliptic quasi-linear PDE on a fixed size domain so one can use elliptic estimates to bootstrap up the regularity to that of H. In practice, to show that the manifold is graphical to order 1, one generally uses the second fundamental form condition described in my answer. $\endgroup$
    – Rbega
    Nov 3, 2017 at 17:30
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Without "uniformly on $\alpha$" and assuming $M$ is a smooth manifold (i.e., the maps $\phi_\alpha\circ\phi^{-1}_\beta$, where defined, are smooth), this appears to be the definition of a $C^k$ embedding of $M$ into $\mathbb{R}^N$.

I don't know of any standard terminology if "uniformly on $\alpha$" is assumed, but one could call such an embedding a "uniformly $C^k$ embedding".

Any compact manifold $M$ certainly admits such an embedding, and, if you put no constraint on the dimension $N$, I'm pretty sure any manifold $M$ has a uniformly $C^k$ embedding.

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    $\begingroup$ Thank you for your interesting remarks. I'm primarily interested in the non-compact case. Do you think that my condition can be somehow related to the concept of manifold with $C^k$-bounded geometry? $\endgroup$ Nov 2, 2017 at 21:40
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    $\begingroup$ I've never studied carefully Riemannian manifolds with $C^k$-bounded geometry. Is it not true that any open smooth manifold has a Riemannian metric with $C^k$-bounded geometry? $\endgroup$
    – Deane Yang
    Nov 2, 2017 at 22:14
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    $\begingroup$ If you assume $M$ is a Riemannian manifold with $C^k$-bounded geometry, then you can turn your question into a global question: Does there exist a smooth embedding $u: $M$ \rightarrow \mathbb{R}^N$ and constant $C > 0$ such that $\|\nabla^ju\|_\infty < C$ for $1 \le j \le k$. $\endgroup$
    – Deane Yang
    Nov 2, 2017 at 22:17

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