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7

Yes you can do this by small cancellation. The presentation satisfies $C'(1/6)$ (in fact $C'(1/7)$), so by Greendlinger's lemma any word equal to the identity in the group must have a subword consisting of more than half of a cyclic conjugate of a relator or its inverse. Since $[a,b]^k$ does not contain such a subword for any $0 \ne k \in {\mathbb Z}$, it ...


6

You can also do an ad-hoc approach (without small cancelation). We take two (more or less) arbitrary noncommuting matrices $A:=\begin{pmatrix} 1&1\\0&1 \end{pmatrix}$ and $B:=\begin{pmatrix} 1&0\\1&1 \end{pmatrix}$ (living over, say, the rationals). The map $a\mapsto A,b\mapsto B, c\mapsto B^{-1}, d \mapsto A^{-1}$ extends to a ...


3

If you don't mind "a hammer"(although not really much more difficult than small cancellation machinery), you can use some one relator theory, and apply the Freiheitssatz, to get that $\langle a,b\rangle$ generates a free subgroup on two generators in the whole group. This result can be found in both of "the" Combinatorial Group Theory books (one by Lyndon ...


2

Here is a topological/geometric proof. Label the sides of an octagon by this relator. You will see that by gluing opposite sides there is a single vertex in the quotient. The quotient space $S$ is a surface, and your group is isomorphic to $\pi_1 S$. The element $[a,b]$ is represented by a simple closed curve $c$ on $S$ which separates $S$ into two one-...


2

This is one of those things where as you learn more about non-euclidean geometry the "evidence" just accumulates. The distance formula is $$ds^2=\cosh^2 y dx^2+dy^2$$ and we see that $\cosh y\to 1$ as $y\to 0$ another formula is the pythagorean formula $\cosh c=\cosh a\cosh b$ limits to $c^2=a^2+b^2$ for small $a$ and $b$. Good refs are Carslaw, and ...


2

The fibers are precisely the orbits of the natural action of $SL_2(\mathcal{O}_K)$ on $\mathbb{P}^1(K)$. See this note by Keith Conrad for details.


2

Yes, it is certainly true. The ray-casting algorithm should work for any uniquely geodesic metric space which is diffeomorphic to $\mathbb R^2$. Not sure exactly how rigorous you want it, but I find this proof adequate: Outline of proof: First suppose that the point $P$ does not lie on the extension of any of the geodesic edges. Then when a ray from $P$ ...


2

The computation looks fine. And yes, there is a tesselation of type $\{5,n\}$ for every $n \ge 4$. To see why, you need the fact that if $\alpha_0$ is the interior angle of a regular Euclidean pentagon then for any $\alpha < \alpha_0$ one can construct a regular hyperbolic pentagon $P$ having interior angles equal to $\alpha$. One then computes $\...


1

All correct (I think I wrote that on wikipedia) $$d(O, Q) = \frac{1}{2} \ln \frac{1+r}{1-r} = \operatorname{artanh} r$$ see also the formulas for tanh and artanh at en.wikipedia.org/wiki/Hyperbolic_function I have added it to the wikipedia page ps related: For the Poincare disk model he hyperbolic distance between O the center of unit disk and a point ...


1

HINT: You should have a formula for geodesic curvature just in terms of the first fundamental form (and the curve, of course) if you're working in an orthogonal parametrization. (You don't say what material you know and what tools you have at your disposal. It's also a very straightforward computation using differential forms and moving frames.)



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