Mapping the Poincar Disk to Hyperbolic Surfaces in $mathbbR^3$.

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ewcommandRealsmathbfR$If $S$ is a surface in $Reals^3$ of constant curvature $-1$, then sufficiently small pieces of $S$ are indeed locally isometric to small pieces of the hyperbolic plane, and the universal cover of $S$ is globally isometric to a proper open subset of the hyperbolic plane. (There's no local isometry from the entire hyperbolic plane to $S$ by Hilbert's theorem. Equivalently, $S$ is not geodesically complete.)Methods for finding a "hyperbolic parametrization" of $S$ depend on $S$. For the Dini-type surface in the question, there are formulas depending on definite integrals of explicit elementary functions, similar to the standard descriptions of a surface of rotation of constant Gaussian curvature.If you will excuse the self-promotion, in "action-angle" coordinates, the Gaussian curvature formally becomes $-frac12$ times the second derivative of a "profile" function defining the metric. These coordinates are straightforwardly adapted to surfaces invariant under a helical group of Euclidean isometries. For a metric of curvature $-1$, the profile is a quadratic polynomial $varphi(u) = a u^2$. If $a = 0$, the profile is a tractrix; if $a > 0$ (diagram below), one gets hyperbolic surfaces with helical symmetry, and with "U"-shaped profile curves resembling the profiles of constant-curvature surfaces of rotation.Regarding uniqueness, if $U_1$ and $U_2$ are regions in the hyperbolic plane, and if $Phi_1:U_1 to S$ and $Phi_2:U_2 to S$ are hyperbolic parametrizations, then $Phi_2^-1 circ Phi_1$ is a hyperbolic isometry everywhere it's defined.

1. Quotient of space and a group of maps, Riemann surfaces

Suppose a group $G$ acts (on the left) on a set $X$, so we have a map beginalign* G times X &to X (g,x) &mapsto gx , . endalign* Given $x in X$, consider its orbit $Gx = gx : g in G$. These orbits partition $X$, so we can define the quotient $G backslash X = Gx : x in X$ as the set of orbits. (I write $G backslash X$ rather than $X/G$ because $G$ is acting on the left.)If $X$ is a topological space and $G$ is a topological group and acting continuously on $X$, i.e., the map beginalign* G times X &to X (g,x) &mapsto gx endalign* is continuous, then we can upgrade $G backslash X$ to a topological space by giving it the quotient topology as follows. We have a quotient map beginalign* pi: X &to G backslash X x &mapsto Gx endalign* that sends $x$ to its orbit, and we define the topology on $G backslash X$ to be the strongest topology such that $pi$ is continuous. That is, $U subseteq G backslash X$ is open iff $pi^-1(U) subseteq X$ is open.Let's look at a geometric-flavored example. The group $textPSL_2(mathbbR)$ acts on the upper half-plane by Mbius transformations: beginalign* textPSL_2(mathbbR) times mathcalH &to mathcalH beginpmatrix a & b c & d endpmatrix cdot z &= fracaz bcz d , . endalign* Then $$ beginpmatrix 1 & 1 0 & 1 endpmatrix cdot z = z 1 $$ so it acts as $T(z) = z1$. I claim that $langle T

angle backslash mathcalH$ is homeomorphic to an infinite cylinder. To see this, we find a fundamental domain for the action, i.e., a set of representative points in $mathcalH$ so that each orbit has exactly one representative. In this case $z in mathcalH : 0 leq Re(z)

2. parametrization of the intersection of two surfaces

With $x=rcosphi$ and $y=rsinphi$, the equations become $r^2cosphisinphirz(cosphisinphi)=0$ and $r^2-z^2=0$. Thus we have $r=pm z$ and therefore $cosphisinphipm(cosphisinphi)=0$. With $phi=thetafracpi4$ and thus $cosphi=(costheta-sintheta)/sqrt2$ and $sinphi=(costhetasintheta)/sqrt2$, this becomes $(cos^2theta-sin^2theta)/2pmcostheta/sqrt2=0$. Using $sin^2theta=1-cos^2theta$, solving the resulting quadratic equation for $costheta$ and keeping only the solutions with $|costheta|le1$ leads to $costheta=pm(1-1/sqrt2)$ and thus $phi=fracpi4pmarccos(pm(1-1/sqrt2))$, with the inner sign determined by the sign of $z$. From this $cosphi$ and $sinphi$ can be obtained as algebraic constants, and the intersection takes the form of two lines through the origin, parametrized by $z$

3. canonical "simplest” cobordism for fixed boundary

Uh...your comment about $n

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