42

Well, all I did was a search on "homeomorphism history", but... I tried to extract some points that are made in conjunction to your question (Riemann, Möbius, Jordan), though feel free to edit it down if it is too long (and apologies to those who think this should be remapped to a History of Math Q/A).
The evolution of the concept of homeomorphism, by ...

18

Here are explicit examples when $M$ is compact, connected, and $\chi(M)\le0$.
Orientable case: Let $M$ be the 1-point compactification of the hyperelliptic Riemann surface defined in the affine plane $\mathbb{C}^2$ by
$$
y^2 = x^{2g+1}-1.
$$
This is a smooth Riemann surface of genus $g\ge1$ and hence $\chi(M) = 2-2g$. The holomorphic $1$-form
$$
\omega = \...

answered Dec 14 '19 at 21:23

Robert Bryant

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16

The answer is already given in the comments (by Ryan Budney and Mizar). But I think it makes sense to clear this confusing point. The classical Gauss-Bonnet formula is [e.g. https://en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem ]
$$\int_M K dA+\int_{\partial M} k_g ds=2\pi \chi(M).$$
In this formula nothing requires orientation of $M$! $dA$ is the ...

15

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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15

Yes. 2. Yes. (I suppose that the surfaces are "the same" if they are homeomorphic).
For 1, it is sufficient to check the definition of surface: that every point has a neigborhood homeomorphic to the disc. For interior points of the polygon, and for points on
the sides, this is evident, and for the corners this is easy.
For 2, just recall classification of ...

answered Jun 27 '14 at 12:46

Alexandre Eremenko

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15

I'll assume we're talking about complex functions; if real, tensor with $\mathbb{C}$. Now pass to the group of units. With the topology given by spectral radius (this is an algebraic description of the C-* topology), the group of connected components of the group of units is $H^1(X, \mathbb{Z})$ which of course knows the genus.
If you really like ...

answered Jan 28 '16 at 22:48

Qiaochu Yuan

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15

First, for simple closed curves, this was known long before Freedman-Hass-Scott. For closed surfaces, it was first proved by Baer in
Baer, R., Kurventypen auf Flächen. J. reine angew. Math., 156 (1927), 231–246.
and
Baer, R., Isotopie von Kurven auf orientierbaren, geschlossenen Flächen und ihr Zusammenhang mit der topologischen Deformation der Flächen. ...

14

### 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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14

In general, surfaces in $\mathbb{E}^3$ for which the principal curvatures satisfy a given functional relation $F(\kappa_1,\kappa_2)=0$ are said to be Weingarten surfaces (of type $F$), and the condition for a graph $z = f(x,y)$ to be a Weingarten surface of type $F$ is a single second order PDE for the function $f(x,y)$. The general theory tells you that, ...

answered Jul 27 '17 at 11:44

Robert Bryant

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13

Also, not an answer but some comments. When one learns about the geometry of smooth surfaces in $\mathbb{R}^3$, the question of rigidity and flexibility arises quite naturally. And, at first sight, it is plausible that there should be some characterization of these properties in terms of geometric invariants, especially the second fundamental form.
However, ...

12

Any smooth compact surface smoothly embedded in $\mathbb{R}^3$ that is not the $2$-sphere must have an infinite fundamental group and hence must have infinitely many distinct (in your sense) geodesics joining any two distinct points.
This result follows from Morse theory: If $S$ is the surface and $a$ and $b$ are points on it, then each fixed-endpoint ...

answered Jan 10 '15 at 3:54

Robert Bryant

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12

This is a particular case of Corollary 1.1 of Edwards, Robert D.; Kirby, Robion C. Deformations of spaces of imbeddings. Ann. of Math. (2) 93 (1971), 63--88. MR0283802, which says that the group of homeomorphisms of any compact manifold is locally contractible.

11

The special feature of $X$, a sphere with three or more punctures, that is being used here is that the space $E(X)$ of all homotopy equivalences $X\to X$ has $\pi_1 E(X)=0$. (Here we take the identity map of $X$ as the basepoint of $E(X)$ for computing $\pi_1 E(X)$.) The corresponding statement when $X$ is an annulus is not true, since $\pi_1 E(X)={\mathbb Z}...

11

The number of the orbits is infinite.
Consider the upper central series, that is a sequence of derived subgroups: $G^1=[G,G]$ and $G^{i+1}=[G^{i},G^{i}]$. All subgroups $G^i$ are normal in the group $G$. Since $G^1$ is free of infinite rank, the sequence $\{G^i\}_{i=1,\ldots,\infty}$ is a sequence of free groups of countable rank that does not stabilize, i....

11

Problem 2 in the list of open problems that Douglas Zare linked to answers the question (namely that there is a standard candidate, and it is even called the standard triple bubble).
I quote it here with a few interspersed comments of my own.
Problem 2 (Sullivan) We construct the standard clusters of k bubbles in $\mathbb{R}^n$ ($k\leq n+1$) as follows....

11

No, there doesn't exist such a foliation. The existence of any foliation would mean the Euler characteristic is zero, so the surface must be either a torus or a Klein bottle. Foliations for these surfaces are understood well enough to rule out having both dense and non-dense leaves.
Any foliation will contain a "Reeb component" (for which no leaf is dense) ...

11

For any topological group $G$, there is a classifying space $BG$ and a principal $G$-bundle $EG \to BG$ called the universal principal $G$-bundle which is determined up to isomorphism by the fact that $EG$ is weakly contractible. On a paracompact topological space $X$, any principal $G$-bundle $P \to X$ admits a map $f : X \to BG$, called a classifying map, ...

answered Jan 3 '17 at 11:10

Michael Albanese

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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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10

No. Consider the case of an ellipsoid with three distinct axes, and remove the four umbilic points. Then you cannot find such vector fields on a (punctured) neighborhood of the deleted umbilics. Have a look at this reference on umbilics and try drawing the vector field on such a punctured neighborhood, and you'll see why.

answered Sep 27 '15 at 10:30

Robert Bryant

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10

The torus has two functions $f$ and $g$ which are (1) relatively prime, (2) each have two square roots, and (3) whose product has $4$ square roots. For instance take two functions which vanish on disjoint loops which are not null-homologous. There sphere does not have two such functions because (1) implies that $V(f)$ and $V(g)$ are disjoint and (2) implies ...

10

Counterexamples are easily constructed using the Thurston norm. In fact, any example of a fibered, oriented, closed 3-manifold $M$, with a fiber of genus $\ge 2$ and with pseudo-Anosov monodromy, and with 2nd homology of rank $\ge 2$, gives counterexamples.
The Thurston norm on $H_2(M;\mathbb{R})$ has a polyhedral unit ball, and there is a symmetric set of ...

10

It seems that such pill exists.
Take a ball and drill a hole through it, so you get a solid torus;
we assume it has smooth boundary $\Sigma$.
By Gauss--Bonnet formula, we gave
$$\int\limits_\Sigma G=0,$$
where $G$ denotes Gauss curvature.
Denote by $H$ the mean curvature of $\Sigma$;
it is mostly very negative in the surface of the hole.
It is easy to ...

answered Mar 4 '17 at 20:21

Anton Petrunin

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10

The second statement ought to be in the literature somewhere but I don't know a reference so I'll give an argument.
The result can be rephrased in terms of graphs. Let $S$ be a compact connected surface with non-empty boundary and let $P$ be a non-empty finite set of points in the interior of $S$. Consider finite connected graphs $X$ in $S$ with $P$ as ...

10

I would recommend looking at the work of Moira Chas to start.
Here are two interesting papers of hers to read:
The Goldman bracket and the intersection of curves on surfaces.
Combinatorial Lie bialgebras of curves on surfaces
She even has an app on her website that computes the bracket for you: Goldman Bracket.
How to use the app: in the first box you ...

9

This is asked on MSE, and answered (see Jim Belk's answer, which is NOT the accepted answer).

answered Nov 19 '15 at 14:11

Igor Rivin

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9

The group $F$ is isomorphic to the symmetric group $S_5$.
In fact, since $N_5$ is non-orientable of genus $5$, both $F$ and the extended group $F^*$ (of order twice the order of $F$) act on its orientable double cover, that has genus $4$. In Conder's database, this is expressed by saying that the action of $F$ in genus $4$ is reflexible and that there is a ...

answered Jul 1 at 19:48

Francesco Polizzi

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8

I think Myers only considered analytic metrics, see his papers "Connections between differential geometry and topology I and II", Duke Math. J. 1 (1935), 376-391, and 2 (1936), 95-102.
For arbitrary metrics on $S^2$ the cut locus is indeed a tree. This can be deduced from e.g. in [Shiohama and Tanaka, Cut loci and distance spheres on Alexandrov surfaces] ...

answered Dec 29 '13 at 18:36

Igor Belegradek

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8

McMullen and Taubes 4-manifolds with inequivalent symplectic forms and 3-manifolds with inequivalent fibrations constructs 3-manifolds $N$ with different fibrations, whose Euler classes do not lie in the same $Diff(N)$-orbit.
The idea of the proof is that two fibrations can not be in the same $Diff(N)$-orbit if the Poincaré duals of their fibers belong to ...

8

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 Euclidean $\mathbb{R}^4$ has positive Gauss curvature and is homogeneous, it will be complete and hence compact. Hence the group of ambient symmetries will have to ...

answered Jul 25 '17 at 14:42

Robert Bryant

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8

But the existence of an umbilic point on the sphere follows from topological considerations: The sum of the Hopf indices of the umbilics is 1 (by a theorem of Hopf) so there has to be at least one umbilic.
Put another way: If there were no umbilics, then union of the principal directions at each point would define a 4-fold covering space of the sphere, ...

answered Oct 16 '17 at 1:19

Robert Bryant

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