Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $U \subseteq \mathbb{R}^2$ be an open set and let $\sigma : U \rightarrow \mathbb{R}^3$ be a parametrization of an oriented surface $S$ embedded in $\mathbb{R}^3$ whose unit normal in $\sigma (u,v)$ is $N(u,v)$. $\forall \delta \in \mathbb{R}$ we define the map $\tau_\delta : U \rightarrow \mathbb{R}^3$ as

$\tau_\delta (u,v) := \sigma (u,v)+ \delta N(u,v)$

Prove that, if $V \subseteq U$ is an open set with compact closure in $U$, there exist $\epsilon > 0$ such that $\tau_\delta|_V$ is the parametrization of a regular embedded surface $\forall \delta \in (-\epsilon , \epsilon)$.

How can I approach this kind of problem?

share|cite|improve this question
Do you understand why $\epsilon$ can't be arbitrary in general (i.e. must be small for certain surfaces)? Understanding this point will take you half-way to the answer. – Marek Jan 28 '13 at 11:09
Actually, I can't picture in my mind how should the surface parametrized by $\tau_\delta$ be, but I think the problem is that for too big value of $\delta$ the surface can have self intersections. Am I on the right way? – Frankenstein Jan 28 '13 at 11:28
Yes, that's correct. See the picture here: . Now, try to think of conditions that guarantee that self-intersections don't occur. The (relative) compactness, (i.e. boundedness) is essential here. – Marek Jan 28 '13 at 11:35
Well, I have to verify that there exists $\epsilon > 0$ s.t. $\forall \delta \in (-\epsilon , \epsilon)$ we have: $1_$ $\tau_\delta |_V$ is a $C^\infty$ map (and this is obvious $\forall \delta$); $2_$ $\tau_\delta |_V$ is a homeomorphism onto its image; $3_$ $(D \tau_\delta |_V)_p : \mathbb{R}^2 \rightarrow \mathbb{R}^3$ is a one-to-one linear map $\forall p = \sigma (u,v)$; I'm quite convinced that I have to work with the condition $3_$, but I only know that $(D \sigma |_V)_p$ is one to one and I don't know how to continue... Any help? – Frankenstein Jan 28 '13 at 13:25
sorry, I don't have time right now, that's why I'm only posting comments. Suppose for contradiction that no such $\epsilon$ exists. Produce a sequence of points $(u,v)$ that map under $\tau_{\delta}$ to a singular point. Now, you can pick a convergent subsequence in $\bar V$ and this will under $\sigma|_{\bar V}$ map to a singular point of $\sigma(\bar V)$, a contradiction. Now the general case follows by restricting to an open subset $V$ of $\bar V$. – Marek Jan 28 '13 at 16:16

Consider the map $F(u,v,t)=\sigma (u,v)+t N(u,v)$. If $\sigma $ is $C^2$ smooth, then $N$, and consequently $F$, is $C^1$ smooth. I claim that the Jacobian matrix $DF$ is invertible at every point with $t=0$. Indeed, at such a point $F_u$ and $F_v$ are linearly independent tangent vectors, while $F_t=N(u,v)$, a normal vector. These three are linearly independent, proving the claim.

Use the inverse function theorem to cover the surface with 3D balls $B_i$ in which $F$ is a diffeomorphism. Choose a finite subcover of $\overline{V}$. By the Lebesgue number lemma there exists $\delta>0$ such that every subset of $\overline{V}$ with diameter $\le 2\delta$ is contained in some $B_i$. The surface $\sigma_\delta$ has no self-intersections: if $\sigma (u,v)+\delta N(u,v)=\sigma (u',v')+\delta N(u',v')$, then $\sigma (u,v)$ and $\sigma (u',v')$ are at distance at most $2\delta$ from each other, contradicting the choice of $\delta$. This and $F$ being a diffeomorphism imply that $\sigma_\delta$ is regular.

share|cite|improve this answer

Let's see if it works...

Let $N_S(\eta)$ be a tubular neighborhood of $S$ (where $\eta:S\rightarrow(0,+\infty)$ is a continuous function) and let $N_{\sigma(\overline V)}(\eta)$ be the restriction of this tubular neighborhood to the set $\sigma(\overline V)$ (observe that $\overline {N_{\sigma(\overline V)}(\eta)}$ is a compact set).

Suppose for contradiction that $\forall\epsilon>0$ exists $\delta\in(-\epsilon,\epsilon)$ such that $\tau_\delta(V)$ has a singular point. We choose $n\in\mathbb{Z}_+$ and set $\epsilon=\frac{1}{n}$, then there is $\delta_n$ with $|\delta_n|<\frac{1}{n}$ such that $\tau_{\delta_n}(V)$ has a singular point, let's call it $p_n$.

Now the sequence $\{p_n\}_{n\in\mathbb{Z}_+}$ is definitively contained in the compact set $\overline {N_{\sigma(\overline V)}(\eta)}$, so there is a subsequence $\{p_{n_k}\}$ which converges in $\overline {N_{\sigma(\overline V)}(\eta)}$ to a certain point $\overline p$; also $\overline p\in\sigma(\overline V)\subseteq\sigma(U)$.

Let $A\subseteq\sigma(U)$ be any open neighborhood of $\overline p$, let $a$ be a point of $A$ and let $I_S(a,\eta(a))$ be the segment $a+(-\eta(a),\eta(a))N(a)$ of lenght $2\eta(a)$ centered in $a$ and normal to $T_PS$. Then in $A$ there are at least two points $x,y$ such that $I_S(x,\eta(x))\cap I_S(y,\eta(y))\neq\emptyset$, contradicting the fact that $N_S(\eta)$ is a tubular neighborhood of $S$.

Do you think it is correct?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.