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Let $U$ be a bounded domain in $R^2$ and let $n : U \to S^2$. Which (necessary/sufficient) conditions must $n$ satisfy in order that there exist an immersion $f : U \to R^3$ such that $n(x)$ is the normal to $f$ at the point $f(x)$, for all $x\in U$ ? The ideal answer would be of the form: "such $f$ exists if and only if $n$ satisfies the following PDE: ..."

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2 Answers 2

up vote 5 down vote accepted

As Alexandre Eremenko points out, there's no PDE that $n$ would have to satisfy, at least for local solvability, which is believable when you think of it in heuristic terms: There are $3$ unknowns involved in specifying an immersion $f:U\to\mathbb{R}^3$ and only $2$ arbitrary functions needed to specify $n:U\to S^2$, so it's an underdetermined problems and there 'ought' to be many local solutions, in particular, there shouldn't be any equations that $n$ would have to satisfy in order to be the Gauss mapping of some immersion.

More explicitly, you can see how to write down all the possible $f$s for a given $n$ in the 'generic' case in which $n:U\to S^2$ is itself an immersion. [As Eremenko points out, in this case, a particular solution is just to take $f = n$.]

Here's how to get the 'general' solution: Let $x$ and $y$ be the standard coordinates on $\mathbb{R}^2$ and note that the assumption that $n$ be an immersion is equivalent to the condition that $n_x$ and $n_y$ be linearly independent on $U$. Now, consider a possible solution $f$ and set $c = f\cdot n$, so that $f = c\ n + g$, where $g:U\to\mathbb{R}^3$ satisfies $n\cdot g = 0$. Then one has $$ 0 = n\cdot df = n\cdot (c\ dn + n\ dc + dg) = dc + n\cdot dg = dc - g\cdot dn + d(g\cdot n) = dc - g\cdot dn $$ (note that I have used that $n\cdot dn = \tfrac12d(n\cdot n) = 0$). This latter equation uncouples as the pair of equations $$ g\cdot n_x = c_x\qquad\text{and}\qquad g\cdot n_y = c_y\ ,\tag{1} $$ which, together with $g\cdot n = 0$, determines $g$ completely in terms of $c$.

Conversely, if we start with any function $c:U\to \mathbb{R}$ and define $g:U\to\mathbb{R}^3$ to be the unique vector field that satisfies $g\cdot n = 0$ and the equations (1), then setting $f = c\ n + g$ will yield a mapping $f:U\to\mathbb{R}^3$ with the property that $n\cdot df = 0$. Thus, wherever $f$ is an immersion, its Gauss map will be $n$. It is easy to see that for any $p\in U$, the generic choice of $c$ on a neighborhood of $p$ will give an $f$ that is an immersion, so there are many local solutions to this problem, and they essentially depend on one arbitrary function (i.e., $c$) of $2$ variables.

How to determine the conditions that $c$ defined on $U$ should satisfy such that the resulting $f$ will be an immersion everywhere on $U$ is a global question that doesn't have anything to do with solving PDE.

In the case that $n$ is not an immersion, there are singularity issues where $dn$ changes rank, but the cases when the rank is constant are easy to deal with.

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Got it. Thank you very much for the great answer. –  Michael Dec 11 '12 at 11:08

There is no PDE. Let $n: U\to S^2$ be a given smooth map, and suppose that the Jacobian is not $0$ at some point. Let $f$ be the composition of $n$ with the standard embedding of $S^2\to R^3$ as the unit sphere. Then the Gauss map of this $f$ is exactly $n$. So there cannot be any local condition, that is no PDE.

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Thanks very much! –  Michael Dec 11 '12 at 11:13

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