Recently I started working on a problem in Differential Geometry (where I'm not a specialist, so I apologize if this question turns out to have a trivial answer) and I had to consider the following situation.

Let $M \subset \mathbb{R}^n$ be a smooth submanifold of dimension $n-k$. By a normal tube of radius $\varepsilon$ centered at $M$, I mean a tubular neighborhood of $M$ given by a disjoint union $$\mathscr{B}(M, \, \varepsilon):=\bigsqcup_{p \in M} B(p, \, \varepsilon),$$ where $B(p, \, \varepsilon)$ is a $k$-dimensional ball of radius $\varepsilon$ centred at $p \in M$ and contained in the (embedded) normal subspace $N_pM \subset \mathbb{R}^n$.

Q. Under which conditions on $M$ does a normal tube $\mathscr{B}(M, \, \varepsilon)$ exist (for $\varepsilon$ sufficiently small)?

It seems to me that this is always the case when $M$ is compact, because then we can identify $\mathscr{B}(M, \, \varepsilon)$ with the open neighborhood $B_M(\varepsilon)$ of $M$ in $\mathbb{R}^n$ given by $$B_M(\varepsilon) := \{x \in \mathbb{R}^n \, | \, d(x, \, M) < \varepsilon\}.$$ On the other hand, there are also examples where $\mathscr{B}(M, \, \varepsilon)$ exists even if $M$ is not compact, for instance in the case where $M$ is a linear subspace (in this case, in fact, the normal spaces to $M$ are pairwise disjoint).

My feeling is that $\mathscr{B}(M, \, \varepsilon)$ should always exist when the curvature of $M$ is bounded in some suitable sense, but I'm not able to specify it better.

Any answer, example/counterexample and reference to the relevant literature will be greatly appreciated.

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    $\begingroup$ A necessary and sufficient condition (with epsilon constant) is that the injectivity radius from M is bounded from below. $\endgroup$
    – Raziel
    Oct 14, 2017 at 11:28
  • $\begingroup$ @Raziel: this is interesting. Could you please be so kind and expand your comment in a more detailed answer? $\endgroup$ Oct 14, 2017 at 11:33
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    $\begingroup$ Think of the curve $y=1/|x|$, $0<|x|<1$ in the plane. Curvature is bounded, yet $\varepsilon$-tubes do not exist. $\endgroup$ Oct 14, 2017 at 12:40
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    $\begingroup$ Let $M$ be the set of all lines in the plane parallel to a given line and of rational distance from that given line. Then $M$ is a flat submanifold, but has no normal tubes. $\endgroup$
    – Ben McKay
    Oct 14, 2017 at 12:49
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    $\begingroup$ @Deane Yang comment. Think at $X$ defined by the union of the graph of $e^x$ and its reflection w.r.t. the x axis in $R^2$. The ambient curvature is zero and the second fundamental form of the submanifold is bounded from above and below. Indeed you have a uniform lower bound on the conjugate radius and the exponential map is an immersion on a uniform neighborhood,, but the injectivity radius from X is non bounded below, hence you don't have a uniform tubular neighborhood. $\endgroup$
    – Raziel
    Oct 16, 2017 at 6:34

4 Answers 4


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.

  • $\begingroup$ Indeed, Lee is a very good reference. In the second edition this is Theorem 6.24. As for Gray's book, he does not give the proof of the existence of a tubular neighborhood but refers to the book by O'Neill I have cited in my answer. $\endgroup$ Oct 14, 2017 at 16:38
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    $\begingroup$ I read both O'Neill and Lee. It seems to me that the injectivity radius you are referring to is the injectivity radius $\rho$ of the normal exponential map $$\mathrm{exp}^{\perp} \colon NM \to \mathbb{R}^n, \quad (x, \, v) \mapsto x+v,$$ namely the function $\rho \colon M \to \mathbb{R}$ such that $\rho(x)$ is the supremum of all $\delta$ such that $$V_{\delta}(x)=\{(x', \, \delta') \in NM \; : \;|x-x'| < \delta, \, |v'| < \delta \}$$ is sent by $\mathrm{exp}^{\perp}$ diffeomorphically onto its image in $\mathbb{R}^n$. Am I right? $\endgroup$ Oct 19, 2017 at 14:00
  • $\begingroup$ Yes, that's exact?! $\endgroup$
    – Raziel
    Oct 19, 2017 at 19:26

You are right that a tubular $\varepsilon$-neighborhood exists, if $M$ is compact. Local existence follows from the properties of the exponential map, global existence by compactness argument, see for example O'Neill, Semi-Riemannian geometry, Proposition 7.26.

Boundedness of principal curvatures is indeed necessary for the existence of a tubular $\varepsilon$-neighborhood, but not sufficient. Consider, for example, a spiral in $\mathbb{R}^3$ with the distance between coils tending to $0$: $$ x(t) = \cos t, \quad y(t) = \sin t, \quad z(t) = \operatorname{arsinh} t. $$

If you don't need $\varepsilon$ to be constant, but allow it to tend to $0$, then a tubular neighborhood exists, this is again O'Neill, Proposition 7.26.

EDIT: The book of John M. Lee cited by Raziel in another answer to this question gives a very clear account on tubes around submanifolds of $\mathbb{R}^n$. O'Neill deals with submanifolds of Riemannian manifolds.

  • $\begingroup$ Thank you for the interesting answer and for the example. $\endgroup$ Oct 14, 2017 at 11:32

I think you will find your answer in the book "Normally hyperbolic invariant manifolds — the noncompact case," of Jaap Eldering, freely available here. In particular have a look at Theorem 2.33 and the surrounding discussion.


If you take the ambient manifold to be $\mathbb{C}^n$ with the Euclidean metric and take X as a graph $z_n=f(z_1,..,z_{n-1})$ where f is a polynomial, then it has a uniform tubular neighbourhood. This fact is stated without proof in http://www.math.stonybrook.edu/~dror/paper-OSV.pdf (Remark on page 7).

While I don't have a full proof yet, the point is that we need to prove that there exists an epsilon so that for every point p on X there is no other point q on X such that $q+N(q)t=p+N(p)t$ for all $|t|<\epsilon$ where $N(x)$ is the unit normal at x.

An application of the fundamental theorem of calculus shows that it is enough to bound the derivative of N(x). This further reduces the problem to bounding $\frac{D^2f}{\sqrt{1+|Df|^2}}$ from above uniformly. I can see this when f is a polynomial of 1 variable but it is not clear in the more general case.

EDIT : I think the above "fact" is false in general if f is a polynomial of more than one variable. For example, $z_3=z_1 z_2^2$ does not admit a uniform tubular neighbourhood.

  • $\begingroup$ Could you please add more details? $\endgroup$ Oct 14, 2017 at 18:00
  • $\begingroup$ I qualified my assertion and proved it in one case. I suspect this should go through more generally. Sorry for being terse. $\endgroup$
    – Vamsi
    Oct 14, 2017 at 20:04
  • $\begingroup$ You are trying to prove that the injectivity radius from $X$ has a uniform lower bound given by $\varepsilon$. Your claim seems true to me in general dimensions, passing in polar coordinates and usin.g the fact that $D^2f$ has degree $d-2$, while $Df$ has degree $d-1$ $\endgroup$
    – Raziel
    Oct 15, 2017 at 8:38
  • $\begingroup$ Actually I don't think my statement is completely correct (if f involves more than one variable). $\endgroup$
    – Vamsi
    Oct 15, 2017 at 14:14

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