User kenso - MathOverflow

Discreteness of a group of hyperbolic isometries

2012-06-21T07:15:08Z

Referring to http://mathoverflow.net/questions/100130/a-question-about-hyperbolic-double-torus, there is non-discrete $\Gamma= \left< a,b,c,d~~|~~[a,b][c,d] \right> \subset PSL_{2}(\mathbb{R})$ where $a,b,c,d$ are hyperbolic elements. I wondered what the sufficient and necessary conditions of the representations of $a,b,c,d$ for the discreteness of $\Gamma$. This may be related to some concepts of Teichmuller space...

A question about hyperbolic double torus

2012-06-20T13:03:07Z

I have a question about hyperbolic 2-torus, from now on donoted by $\Sigma_{2}$

Actually I've tried to prove that for a group $\Gamma \subset \textrm{Isom}^{+}(\mathbb{H}^{2})$ represented by $\Gamma = \left< a,b,c,d~~|~~[a,b][c,d]=1 \right>$, $a$,$b$,$c$, and $d$ are hyperbolic element in $\textrm{Isom}^{+}(\mathbb{H}^{2})$(which fix 2 points only on $\partial \mathbb{H}^{2}$.) if and only if $\mathbb{H}^{2}/\Gamma =\Sigma_{2}$. "If"-part is easy so I mainly consider "only if"-part.

It can be verified that one could make $\Sigma_{2}$ by pairing sides of a hyperbolic regular octagon which has angle $\frac{\pi}{4}$. Let the pairing isometries be $\alpha$, $\beta$, $\gamma$ and $\delta$$\in \textrm{Isom}^{+}(\mathbb{H}^{2})$. Then they generate a holonomy group, or deck transformation group of $\mathbb{H}^{2}$ with the relation $[\alpha,\beta][\gamma,\delta]=1$, where $[A,B]=ABA^{-1}B^{-1}$. Note that $\alpha$, $\beta$, $\gamma$ and $\delta$ should be hyperbolic elements.

Now the question is following :

For $\Gamma$, is there ALWAYS an 'corresponding' octagon(not necessarily regular) which is a fundamental domain for $\Sigma_{2}$?? I mean the side-pairing of the octagon induces $\Sigma_{2}$ and the paring maps are exactly $a$,$b$,$c$, and $d$.

Or, is it possible that the octagon is weird(?), for example, some of its edges intersect each other not at a vertex but at an middle of an edge? (in this case, the boundary of the octagon has self intersection.)

(I might have heard that sometimes the 'weird case' occurs but I'm not sure.)

Thanks in advance.

Embedding Again

2012-06-01T10:26:39Z

Let $S=[(x,y)\in\mathbb{H}^{2}:0< x< 2\pi]$ where $\mathbb{H}^{2}$ is a hyperplane with standard metric. I.e., a strip whose boundary components are geodesics, both approaching a common infinite point.

My simple question is about whether it can be isometrically embedded in $\mathbb{R}^{3}$.

I don't remember exactly but I read some articles about an isometric embedding from a portion of $\mathbb{H}^{2}$ into $\mathbb{R}^{3}$. For example, it might have been proved that infinite polygons of some type can be isometrically embedded but I could not catch the meaning of the type the author said about. Or, I read in another article that every(?) equidistant strip in $\mathbb{H}^{2}$ can be embedded in $\mathbb{R}^{3}$. If this is true, the answer to my question would be positive.

Is there anyone who know about this content precisely?

How about $T=[(x,y)\in\mathbb{H}^{2}:0<\sqrt{{x}^{2}+{y}^{2}}<{e}^{2\pi}]$? This strip also has geodesic boundary components but both components are approaching different infinite point.

A question about embedding hyperbolic space onto pseudosphere

2012-05-28T10:00:39Z

I have a difficulty with hyperbolic geometry.

Let $\mathbb{H}^{2}$ be a 2-dimensional hyperbolic plane.

(i.e., upper half plane in $\mathbb{R}^{2}$ with a metric $\frac{ds}{y}$)

(or, upper half plane in $\mathbb{C}$ with a metric $\frac{|dz|}{\textrm{Im}(z)}$ )

You may have heard about pseudosphere in $\mathbb{R}^{3}$. Let's denote the half-pseudosphere by $P$

This can be obtained by glueing both side($x=0$ and $x=2\pi$ parts) of

$\left[(x,y)\in\mathbb{H}^{2} : 0\leq x\leq 2\pi, y>1 \right]$

Denoting this quotient space by $A$(Note that $A$ is homeomorphic to a cylinder), we can now get a "globally isometrically embedding" map $\rho:A \rightarrow \mathbb{R}^{3}$ with $\rho(A)=P$

To be specific, $\rho(x,y) = (t-\tanh(t), \textrm{sech}(t)\ \cos(x), \textrm{sech}(t)\sin(x))$ where $t=\textrm{arccosh}(y)$

Now the question I have is following

: Is there another possible (global and isometric)embedding $\rho$ from $A$ into $\mathbb{R}^{3}$ ?

Actually I'm interested in $\rho(A)$ and by calculating, one can easily find that if $\rho(A)$ is a "surface of revolution", then it should be $P$(up to isometry of $\mathbb{R}^3$"

Thus I'm looking for $\rho(A)$ which is different from a surface of revolution.

Any idea?

Comment by KENSO

2012-06-21T05:57:23Z

@Misha: Can you explain a bit more about the relation of torsion-freeness and discreteness of $\Gamma$?

Comment by KENSO

2012-06-21T05:39:49Z

@Misha: I see. Can you give me URL address for your paper mentioned above?

Comment by KENSO

2012-06-20T19:29:13Z

Sorry for additional questions. What is the definition of nonelementary representation??

Comment by KENSO

2012-06-20T19:01:42Z

@Misha and Lee: I've got your points.......... ;_;

Comment by KENSO

2012-06-20T18:42:57Z

To sum, my question is whether $\Sigma_{2}$ can be obtained from 'a geodesic octagon' in $\mathbb{H}^{2}$ by USING the generators $a$~$d$ of $\Gamma$ to side-pair, if $\Sigma_{2}=\mathbb{H}^{2}/ \Gamma$.

Comment by KENSO

2012-06-20T18:34:22Z

@Lee: I can't catch your point. So what do you mean by "to arrange"? Anyway, do you mean that nonabelian $\Gamma$ whose generators are all hyperbolic can be indiscrete?

Comment by KENSO

2012-06-20T18:32:45Z

@Misha: That was the fact that I used to prove the "easy" part.

Comment by KENSO

2012-06-20T18:14:09Z

@Misha: If I'm not mistaken, I was confusing you because I did not mention that my $\Sigma_{2}$ is smooth. For easy part, I think $\Gamma$ does not have any elliptic elements(candidates of finite order) since it is the deck transformation group for $\Sigma$. For the "fundamental domain polygon", I meant that polygon whose edges are all geodesics in $\mathbb{H}^{2}$, which makes $a$~$d$ be the side-pairings to get $\Sigma_{2}$. Since $\Sigma_{2}$ should be smooth, the angle sum should be $2\pi$.

Comment by KENSO

2012-06-20T16:36:01Z

@Lee: I see. So the space given by side-pairing of a fundamental domain(polygon) always equals to the space given by a quotient space over the group generated by side-pairing maps, right? And what about my comment for your last paragraph? something false?

Comment by KENSO

2012-06-20T15:54:09Z

@Lee: Hmm... for the last paragraph of your comment, I'm gonna tell you that there's a theorem(maybe due to Nielsen) which says that if $\Gamma\subset\textrm{Isom}^{+}(\mathbb{H}^{2})$ which consists only of the identity and hyperbolic elements is NONabelian, then $\Gamma$ is discrete. So I'm not sure about your comment, especially about the 'continuous perturbation'. For the first paragraph, you approach this problem more roughly. (I would think more about "When you straighten the four circles α,β,γ,δ in this new hyperbolic structure,"......)

Comment by KENSO

2012-06-01T15:35:44Z

@O'Rourke: Yeah, pablo seems to give a nice answer. But I think the pictures there are not that helpful to me actually... Thanks.

Comment by KENSO

2012-06-01T15:31:50Z

@Bonnot: I see what you mean, especially the type ${M}_{1}$ of polygon. And sorry that there is a typo in my question: $T=[(x,y)\in\mathbb{H}^{2}:1<\sqrt{{x}^{2}+{y}^{2}}<{e}^{2\pi}]$. If I'm not mistaken, $T$ is also able to be embedded in ${E}^{3}$, right? Then one more question. What is the definition of polygons in $\mathbb{H}^{2}$ in general? Do they allow the ideal parts to be an edge??

Comment by KENSO

2012-05-31T09:46:52Z

@Bryant: However, one can find that the strip which is a part of a sphere cannot be isometrically twisted, observing that a surface with constant positive curvature should be on a sphere. I think the situation of a pseudosphere-strip is not much different from that of this sphere-strip. So if $A$ is embedded onto a subset which is not a pseudosphere, then we should be able to twist the pseudosphere to make the subset but I don't believe this is not possible.

Comment by KENSO

2012-05-31T09:40:07Z

@Bryant: I can't clarify the anticipated "evidence" so I've come here. But the intuition underlying my thought is following. Consider 3 tiny strips(surfaces of revolution) in $\mathbb{R}^{3}$, homeomorphic to a cylinder, each of which is a part of a cylinder, a sphere, and a pseudosphere, respectively. Now let me isometrically twist them. It is easy to imagine that the strip which is a part of a cylinder is well twisted. For example, if the projecting image of the original strip toward its meridian direction is a circle, it can be a ellipse after isometrically twisting it.

Comment by KENSO

2012-05-30T05:37:49Z

@Agol: I think Dini's surface is even not homeomorphic to a cylinder.