Can anyone reference/disprove the theorem in the case where the embedded submanifold is merely $C^1$ instead of smooth? I have a compact $C^1$ embedded submanifold of $\mathbb{R}^n$ without boundary that I want to show there exists a tubular neighborhood of (with the radius of the tube not necessarily constant). Actually, I want to be able to create one piecewise using a finite covering. Much obliged.
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2 Answers
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The answer to your question depends on whether you are looking for a tubular neighborhood in the general differential topological sense or the more restrictive geometric sense. The answer in the topological sense is Yes, but in the geometric sense the answer in general is No. These two conceptions may coincide for $C^2$ submanifolds, but for $C^1$ submanifolds the construction is a bit more subtle, or non-canonical, as I will elaborate below.
A $C^1$ submanifold $M\subset R^n$ may not have any neighborhoods fibrated by normal vectors. So $C^1$ submanifolds do not in general have canonical tubular neighborhoods generated by the exponential map, and "radius" of the neighborhood does not really make sense; however, they still have a tubular neighborhood in the topological sense, i.e., a $C^1$ embedding of the normal bundle which extends that of $M$, as the zero section of the bundle. These may be constructed as follows.
The fastest way to construct a topological tubular neighborhood for $M$ is by noting that there exists a $C^1$ diffeomorphism $\phi\colon R^n\to R^n$ such that $\phi(M)$ is $C^\infty$ (e.g., see Thm. 3.6 in Chap. 2 of Hirsch's book). Then $\phi(M)$ is going to have a tubular neighborhood $U$ in the standard sense, obtained by applying the inverse function theorem to the exponential map (or taking the union of all sufficiently short normal vectors), and $\phi^{-1}(U)$ yields the desired tubular neighborhood of $M$ (which will be $C^1$-diffeomorphic to the normal bundle).
An example of a $C^1$ submanifold without a neighborhood fibrated by normals may be constructed as follows. Take an ellipse in $R^2$, which is not a circle, and let $M$ be the inner parallel curve of the ellipse which passes through the foci. Then $M$ will be $C^1$, but it will not have a tubular neighborhood in the geometric sense. Indeed, the nearest point projection map into $M$ will not be one-to-one in any neighborhood of $M$.
answered Nov 20, 2017 at 16:49
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$\begingroup$ I'd not use the expression "in geometric sense" for a tubular nbd with leaves orthogonal to the submanifold. Maybe "Riemannian sense". (But that doesn't seem to be the issue of the OP, as it is quite obvious that for $C^1$, in general there is no tub nbd in this sense). $\endgroup$ Nov 21, 2017 at 9:35
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2$\begingroup$ The term "Riemannian" here did not seem to me to be preferable to "geometric" because the question is stated in the context of the Euclidean space $R^n$, and I did not want to unduly expand the context at the risk of creating confusion. When I say "geometric" I mean involving the notion of metric. Of course things are obvious to experts, but in my experience many people are confused about the notion of the tubular neighborhood of $C^1$ submanifolds. I was also confused about this not long ago. I do not think that standard texts such as Hirsch make this issue sufficiently clear. $\endgroup$ Nov 21, 2017 at 16:07
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$\begingroup$ This is also true, then maybe for submanifolds of $\mathbb{R}^{n}$ "canonical vs. non canonical" would do. (Also, the term "topological" suggests a non differentiable object) $\endgroup$ Nov 21, 2017 at 18:06
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$\begingroup$ OK, I incorporated the term canonical in the answer. Thanks. $\endgroup$ Nov 21, 2017 at 18:19
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$\begingroup$ (that was just my opinion --I didn't mean to ask you to change) $\endgroup$ Nov 23, 2017 at 22:46
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Yes, you can; you don't need compactness of $N$ nor finite dimension: the key word here is paracompactness. I wish to do a general remark and recall some basic facts, since this seems to be a recurrent topic. Let $N$ be this embedded sumbanifold of $M$. You want a homeomorphism (maybe with more regularity and with some special feature) between the normal bundle $E$ of $N$ in $M$ and some open nbd of your submanifold $N\subset M$, that extends the natural homeomorphism $j$ between the zero-section $E_0$ of $E$, and $N$. The typical situation is that you already have some map $\Phi:E\to M$ that extends $j$ and which is a local homeo at any point of $E_0$ (a local condition that it is usually easy to check e.g. by the local inverse theorem), and you want to show it subordinates a global homeo between an open nbd $U$ of $E_0$ in $E$, and an open nbd $V$ of $N$ in $M$. Clearly, $\Phi$ is an open map on some nbd of $S_0$ so all you need is to have $\Phi_U$ injective on some nbd, which is not completely obvious; then you may further restrict it to an open nbd $U'$ diffeomorphic to $E$. The key lemma to get injectivity is the following topological "extension of homeomorphisms" lemma:
Lemma. Let $E$ and $F$ be paracompact spaces; let $X$ be a closed subset of $E$ and $Y$ a closed subset of $F$. Let $f:E\to F$ be a continuous map such that:
(i) $f$ is a local homeo at any $x\in X$ (meaning that for any $x\in X$ there is an open nbd $B_x$ of $x$ in $E$ such that $f(B_x)$ is open in $F$ and $f_{|B_x}:B_x\to f(B_x)$ is homeo);
(ii) $f_{|X}:X\to Y$ is bijective (hence a homeo).
Then there are open nbds $U$ of $X$ in $E$ and $V$ of $Y$ in $F$ such that $f_{|U}:U\to V$ is a homeo.
answered Nov 20, 2017 at 15:16
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$\begingroup$ I do not follow how the "local inverse theorem" mentioned above can be applied when the manifold is only C^1, because then the exponential map will be only C^0 in general. $\endgroup$ Nov 21, 2017 at 3:09
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2$\begingroup$ Dear Mohammed, of course, but I am not talking about the exponential map. Local Inverse Theorem: if the differential of a $C^1$ map is invertible in a point, the map is a local diffeo at that point- (You can find this material in any textbook) . Nor I am talking about a tubular nbd which is a foliation in orthogonal disks to $N$ (it is trivial that for a $C^1$ submanifold in general this will not give a foliation: think to the graph of $|x|^{3/2}|$ in $\mathbb{R}^2$). But orthogonality of the leaves is not required for a tubular nbd. $\endgroup$ Nov 21, 2017 at 7:14
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$\begingroup$ If not the exponential map, then which map are you using here? $\endgroup$ Nov 21, 2017 at 12:20
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2$\begingroup$ @MohammadGhomi At least for me, a tubular neighbourhood is any embedding of the normal bundle that extends the embedding of the submanifold. There is a contractible space of those, so it makes sense to speak of "the" tubular neighbourhood. $\endgroup$ Nov 21, 2017 at 15:12
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1$\begingroup$ @ Mohammad Ghomi this is also as I understand the question. In my experience, such a map $\Phi$ may come from various sources or constructions (e.g. by integrating a vector field, or even projecting some $\mathbb{R}^{n+h}$-valued map to $\mathbb{R}^{n}$: this happens in the method of characteristic lines for PDE). While in almost all these situations the map $\Phi$ is naturally a local diffeo at points of $N$ by construction, the common problem is to restrict $\Phi$ to some smaller open nbd where it is a global diffeo. $\endgroup$ Nov 21, 2017 at 18:03