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.
Tubular Neighborhood Theorem for $C^1$ Submanifold – Differential Geometry
dg.differential-geometrymanifoldssmooth-manifolds
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I suggest to look at the very clear proof of the tubular neighborhood theorem in
Lee, John M., Introduction to smooth manifolds, Graduate Texts in Mathematics. 218. New York, NY: Springer. xvii, 628 p. (2002). ZBL1030.53001., pag 253, I think it's exactly what you need.
He proves the existence of a tubular neighborhood for any embedded submanifold of $\mathbb{R}^n$ (but indeed the proof is the same for any embedded submanifold of a smooth manifold, it just makes use of an auxiliary Riemannian structure to define "normal lines"). Notice that his construction works also in the non-compact case, but in this case the tubular neighborhood has non-constant radius.
If you want $\varepsilon$ to be constant, then looking into the proof you just need a lower bound to the injectivity radius from the submanifold, which you always have if the submanifold is compact (and vice-versa, if you have a tubular neighborhood with constant $\epsilon$, then the injectivity radius is $\geq \varepsilon$.
In any way, curvature has no relation whatsoever with existence of tubular neighborhoods (more precisely, you cannot give sufficient conditions for existence of a uniform tubular neighborhood in terms of curvature bounds, I'm general). Your problem is deeply related with the regularity of the distance from the submanifold. In fact, you can easily build the tubular neighborhoods using the gradient flow of the distance function from the submanifold, which is as smooth as the submanifold in a neighborhood of the latter. To this regard, I can also suggest you to give a look at this paper : Foote, Robert L., Regularity of the distance function, Proc. Am. Math. Soc. 92, 153-155 (1984). ZBL0528.53005.
Ah, and indeed, the magnificent
Gray, Alfred, Tubes, Redwood City, CA etc.: Addison-Wesley Publishing Company. xii, 283 p. (1990). ZBL0692.53001.
The condition $\mathcal P(\epsilon)$ is needlessly complicated. It's always true without passing to components.
Let $U$ be a tubular neighborhood of $S$ diffeo to $[-1,1]\times S$. Let $\pi: U\to \mathbb R$ be the projection onto the first factor. Let $V=f^{-1}(U)$. Then $V$ is compact and the transversality assumption is equivalent to $h=\pi\circ f$ not having critical points in $V$. Therefore $V=h^{-1}([-1,1])$ is a product by standard Morse theory. The same works for $h^{-1}([0,\epsilon])$ for any $\epsilon$.
Best Answer
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 $M$.