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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?

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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
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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
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Yes, that's correct. See the picture here: en.wikipedia.org/wiki/Tubular_neighborhood . 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

2 Answers 2

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.

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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?

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