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A surface is a generalization of a plane which needs not be flat, that is, the curvature is not necessarily zero. This is analogous to a curve generalizing a straight line
7
votes
Accepted
Construction of a linear Weingarten surface from a space curve
You'll find a discussion of the analysis of linear Weingarten surfaces via exterior differential systems in these lecture notes of mine. Particularly look at Section 5.1, where it is discussed at len …
5
votes
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When does a spherical curve equal its tangent indicatrix?
Your question is not very clearly phrased, which may explain why you didn't get any answers on MSE.
When you say, "spherical curve $\gamma$ whose tangent indicatrix is the same as the original curve, …
6
votes
Minimal surface as varities
I assume that you are asking for a proof of the Weierstrass-Enneper representation theorem that, roughly speaking, tells you how to express solutions of the minimal surface equation in terms of holomo …
5
votes
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The class of uniformly accelerated curves and surfaces
I don't know what you might mean by 'uniformly accelerated surface', but I think that, by 'uniformly accelerated curve', you mean a curve in the plane parametrized in such a way that its velocity at t …
4
votes
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Umbilic points on Euclidean hypersurfaces
Because people have asked for it, I thought I would supply an example of what I mentioned in my comment above, an immersion of the $3$-sphere into $\mathbb{R}^4$ that has three distinct principal curv …
14
votes
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A variant of the Monge-Cayley-Salmon theorem?
Setting aside the assumption that $\phi$ be a polynomial mapping for the moment (however, see below for a construction of a large family of polynomial solutions), if one makes the 'nondegeneracy' assu …
8
votes
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homogeneous surface in $\mathbb{R}^4$
I'm rearranging my answer a little bit because I realized that I overlooked an apparent possibility (that turns out not to occur), and I didn't want my answer to be misleading:
If the surface in Eucl …
12
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Geometric meaning of derivatives of the curvature
This is not really an answer, but it's too long for a comment, so I'm posting it this way. One way in which the expression you are considering has appeared in recent years is via the heat equation sh …
18
votes
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How useful/pervasive are differential forms in surface theory?
Well, there's really not a whole lot more to say beyond what Deane already wrote. He certainly hit the main points, but maybe I can expand a bit on what he wrote and comment on my own experience over …
3
votes
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Classes of curves closed under Minkowsky sum
There are lots of finite dimensional curve families that are closed under Minkowski sum with circular disks. Here's a way to construct examples:
First, choose a finite dimensional space $\mathcal{C}$ …
3
votes
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Requirement of parametrization of surfaces
You can always do this, but it's not as simple as using some kind of ODE (such a flow of vector fields) to construct such charts.
First, assume that your surface is connected and simply-connected. Th …
3
votes
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Planar curves in $M^{m}$ vs curves in $M^{2}$
Here are some comments about the OP's question that don't give a definitive answer to the final question (although the answer may well be 'no', see below), but do provide more information, at least in …
2
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Parametrization of $k$-ruled submanifold: can we choose the base to be orthogonal to the rul...
The answer is already 'no' in the first nontrivial case: A 3-manifold in $\mathbb{R}^d$ (where $d>3$) that is ruled by lines (i.e., $k=1$).
One can see this as follows: As the OP notes, one can writ …
18
votes
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When does the shape operator commute with a derivative?
The question is essentially equivalent to the following classical question: Given a smooth surface $S$ and a bundle map $L:TS\to TS$, when does there exist an immersion $x:S\to \mathbb{R}^3$ such tha …
6
votes
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Can every smooth space curve be realized as an origami curved crease?
Note: I'm revising my answer to make the argument/construction more transparent. In the previous version, I stated an existence result about flat surfaces, but didn't indicate a proof (because, at th …