25

Let me elaborate on Sam Hopkins' comment.
The main reason that makes this and other problems on continuous curves so hard is that a "simple closed curve" or "Jordan curve", i.e. a non-self-intersecting continuous loop in the plane, can be a horrible object, for instance a nowhere differentiable curve such as the Koch snowflake and other fractal curves. ...

answered Jul 31 '15 at 21:27

Francesco Polizzi

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Recall the quartic version of the Cayley-Bacharach theorem:
Theorem: Consider two quartics in general position, which intersect in $16$ points (by Bezout's theorem). Then if a third quartic passes through $13$ of those points, it also passes through the other $3$.
This can be used to establish a simpler result:
Lemma: Let $\Gamma$ be a cubic, and let $T_1,...

answered Feb 28 '16 at 18:02

Adam P. Goucher

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I've added a few sentences to my answer to clarify something that some readers may be wondering about, which is why there isn't as simple an answer for surfaces in $3$-space as there is for curves in $3$-space.
First, a word of caution about the so-called 'Fundamental Theorem of Space Curves': You need to assume that the curvature $\kappa(s)$ is nowhere ...

answered Jan 26 '14 at 12:39

Robert Bryant

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Edit: See the end for a summary of this answer
I disagree with the statement "One can construct an appropriate surface patch locally in a neighborhood of each point". In fact, there are local obstructions to the existence of the desired surface. Let $\gamma:(-\epsilon,\epsilon) \rightarrow \mathbb{R}^3$ be an embedded arc parameterized proportional to arc ...

answered May 23 '15 at 5:29

Andy Putman

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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 the years both learning and teaching differential geometry.
Maybe I should say a little bit about my own education: Originally, I learned differential ...

answered Mar 9 '14 at 13:11

Robert Bryant

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There is a straightforward way to deduce necessary conditions for a space curve to lie on an ellipsoid, and it's really a matter of calculation to make these conditions explicit in terms of the curvature and torsion. I'll describe how to do this and the result of the calculation below, but first let me insert a note of caution about the 'necessary and ...

answered Mar 27 '18 at 9:21

Robert Bryant

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I guess the most conceptual proof is the one using Morse theory:
Take a Morse function on the (closed, orientable) surface S. If it has no saddle points, then (using the gradient flow) $S\cong S^2$. Assume by induction that a surface with k-1 saddle points is $S^2$ with finitely many handles added. For the inductive step, consider a Morse function with k ...

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16

If you just calculate using the moving frame, you'll get the answer for the variation of the principal curvatures in a few lines:
$$
\delta\kappa_i = \mathrm{Hess}(u)(e_i,e_i) + \kappa_i^2\,u .
$$
Here, $\delta\kappa_i$ is the first $t$-derivative of $\kappa_i$ at $t=0$ (i.e., the 'first variation of $\kappa_i$'), $\mathrm{Hess}(u)$ is the quadratic form ...

answered Oct 7 '14 at 15:14

Robert Bryant

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To answer Joseph's questions:
First, it's not impossible to integrate the geodesic flow of the hyperbolic plane in these coordinates, but the formulae I got aren't very nice, so I'm not going to type them in unless I can find a better way to express them. It's probably easier than I got on a first pass through, but I don't have time to work on simplifying ...

answered Nov 25 '13 at 14:09

Robert Bryant

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As others have pointed out, it's not hard to show that any function $F(\kappa_1,\kappa_2)$ that is intrinsic to the surface metric must be a function of $K = \kappa_1\kappa_2$, so that settles what one might call the 'lowest-order' case. However, there are certainly higher-order versions. For example, the expression $|\nabla K|^2$ is an intrinsic invariant,...

answered Jun 6 '16 at 11:35

Robert Bryant

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### Besides the tracioid are there other surfaces of revolution that have a constant negative curvature?

There are many examples of surfaces in $\mathbb{R}^3$ with constant negative curvature. They can be described by using the so-called parametrization by Chebyshev nets. Have a look at the paper by Robert McLachlan
A gallery of constant-negative-curvature surfaces,
The Mathematical Intelligencer 16 (1994), 31-37.
However (and this answers your question) ...

answered Sep 15 '14 at 12:48

Francesco Polizzi

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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' assumptions
$\mathrm{dim}\ \mathrm{span}\bigl( \phi_s(s,t), \phi_t(s,t)\bigr) =2 $,
$\mathrm{dim}\ \mathrm{span}\bigl( \phi_s(s,t), \phi_t(s,t), \phi_{ss}(s,t), \...

answered Jan 27 '17 at 2:18

Robert Bryant

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Already Riemann in his famous "On the Hypotheses Which Lie at the Bases of Geometry" concludes that the spaces of constant curvature are precisely those in which figures can move without distortion. However, as you correctly remarked, a free mobility of rigid bodies is a rather subtle notion. For historical perspective on this problem see https://arxiv.org/...

12

We all should wait for Robert Bryant to answer this question since he's by far the most qualified person to do so.
But let me just say that differential forms are useful not only in global differential geometry but also local differential geometry. One reason for this is that the Maurer-Cartan equations for a Lie group have a beautiful clean formulation ...

12

EDIT: Brendle-Margalit have released their paper. See here.
One should observe that these are not all examples of Ivanov's metaconjecture (for instance, the automorphism group of the disk complex is the handle body group, not the whole mapping class group). In any case, Dan Margalit and Tara Brendle are currently writing a paper that proves a version of ...

answered Mar 19 '15 at 5:54

Andy Putman

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12

When a definition applies to a much larger class of spaces $\ X\subseteq \mathbb R^3,\ $ then such definition rather is not directly related to any parametrization for the smaller class of the parametrized curves (which can be rectified). Several of variants of Hausdorff dimension for dimension 1 would be a possible answer. Another could be
$$\lim_{r\...

answered May 18 '15 at 8:03

Włodzimierz Holsztyński

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12

As per the original poster's suggestion, here is a slightly expanded version of what I posted as a comment:
If we consider the envelope of the family of lines $L_\theta$ which connect the points with arguments $\theta$ and $2\theta$ on the unit circle, we get a cardioid, as explained in the question. (This fact is, for example, mentioned on this page where ...

12

This is more an extended comment than an answer to the question. The first thing to note is that there are different strenghts of the classification theorem for surfaces. Of course, there are the differentiable, triangulated and topological setting. But even if we choose such a setting, there are two statements one has to prove (at least in one approach):
...

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11

The answer is rewritten.
First you need to find the analogs of natural parametrization, curvature and torsion for surfaces; the same can be done for all submanifolds, say of dimesion $m$.
Let us denote by $S^k(\mathbb R^m)$ the space of homogenious symmetric polynomials of degree $k$ on $\mathbb R^m$.
Assume that the submanifold is given locally by a ...

answered Jan 26 '14 at 3:51

Anton Petrunin

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11

I think the relevant location is item 23, page 352, but what Hadamard aims to is stated as follows:
A smooth, co-orientable surface of $\mathbb{R}^3$ with Gauss curvature bounded below by some $\kappa >0$ is simply connected. (implicitly, the surface is compact without boundary)
("Or une surface à deux côtés et sans points singuliers, à courbure ...

answered Nov 22 '18 at 21:28

Benoît Kloeckner

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11

Chern explains moving frames in a masterful 4 pages: (1990, pp. 210–213).
Further nice book-length treatments using them:
Valiron (1950, Chap XIII–XIV) — last of the classic Cours, translated, very complete.
Guggenheimer (1963, Chap. 10–11) — actually says “Darboux frames” throughout.
O’Neill (1966, Chap. VI–VII) — standard choice.
Cartan (1967, Chap. III)...

answered Oct 7 '19 at 9:37

Francois Ziegler

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The proof of Zeeman described in this note is by a substantial margin the easiest and most conceptual proof I know. To simplify the exposition I restrict to orientable surfaces in the note, but it is trivial to also do the non-orientable case (and see the edit below for one description of how to arrange this to avoid using the fact that three cross caps is ...

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answered Feb 20 '20 at 17:44

Andy Putman

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In question (1), if you allow $g$ to vary, then this is answered positively by Hardt and Simon (see also).
The answer to question (2) is no. Almgren and Thurston construct unknots which do not bound an embedded disk in their convex hull. If $\mathscr{F}$ (fixing $g=0$) contained a minimal area member, then it would have to be a minimal surface. However, a ...

10

As indicated in my comments, the Teichmüller question is a duplicate of this question.
For the measured lamination case, the fact that $6g-5$ curves suffice was shown by Hamenstädt.
Hamenstädt, Ursula, Parametrizations of Teichmüller space and its Thurston boundary., Hildebrandt, Stefan (ed.) et al., Geometric analysis and nonlinear partial differential ...

10

I don't know if it's exactly what you're looking for, but line integration is the unique way to assign a real number $I(\omega,c)\in\mathbb{R}$ to every pair of a smooth $1$-form $\omega$ on a smooth manifold $M$ with boundary and smooth path $c\colon[0,1]\to M$ such that:
(adjunction) if $f\colon M\to N$ is a smooth map of smooth manifolds with boundary, $\...

9

Yes, it is true. This is easily seen by looking for example at the Picard group, generated by $C$ and $D$, the two fibres of the projections. Since $C^2=D^2=0$ and$C\cdot D=1$, the only curves of self intersection $0$ are multiple of $C$ or $D$, and the irreducible ones are equivalent to $C$ or $D$. Composing an automorphism by the exchange of coordinates ...

9

The flow consists of a sequence of surgeries using one fixed oriented arc $\alpha$ to cut (and isotope) all other arcs $\beta$ to remove one point of $\alpha\cap\beta$ at a time. Each surgery cuts one arc into two arcs. Say $\beta$ is cut into $\beta'$ and $\beta''$. It can happen that one of these two arcs, say $\beta'$, is trivial, cutting off a disk $D$ ...

9

There is a very general discussion of moving frames in Clelland's book, From Frenet to Cartan: The Method of Moving Frames, and this in particular includes discussion of Darboux's use of frames on surfaces. Darboux's own 4 volume Lecons sur la Theorie Generale des Surfaces at les Applications Geometriques du Calcul Infinitesimal is quite readable, although ...

9

Some examples and references are mentioned here Examples of plane algebraic curves. You can find many Jordan curves in the family $e^{it}+re^{int}, 0\leq t\leq 2\pi,$ by choosing parameters properly.
To generalize this, take any polynomial $P$ which is univalent (=injective) in the closed unit disk. The image of the unit circle is a Jordan curve. By Riemann'...

answered Feb 27 '20 at 12:45

Alexandre Eremenko

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No, for any knotted curve. Such a curve would bound an embedded disk on either side, and, therefore, would be unknotted.

answered May 23 '15 at 1:04

Igor Rivin

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