Let $H = \{z \in \mathbf{C}: \operatorname{Im} z > 0\}$ be the upper half-plane, and let $D = H^2 \setminus \Delta$, where $\Delta = \{(u,u) \in H^2\}$ is the diagonal. Define $\varphi: D \to \mathbf{R}^4$ by $$(u,v) \mapsto (2\operatorname{Arg}(u), 2\operatorname{Arg}(u-1), \operatorname{Arg}\left(\frac{u-v}{u-\overline{v}}\right), 2\operatorname{Arg}(v-1)).$$

What is the image of $\varphi$?

Write $(\varphi_{u0},\varphi_{u1}, \varphi_{vu}, \varphi_{v1})$ for coordinates in the image. Clearly $0 < \varphi_{u0} < \varphi_{u1} < 2\pi$ because $0 < \operatorname{Arg}(u) < \pi$, but what about the rest?

Some additional information:

This is to compute the weights in the Kontsevich quantization formula. For $u \neq v$, $\operatorname{Arg}\left(\frac{u-v}{u-\overline{v}}\right)$ is in fact $\phi(v,u)$ in the notation of Wikipedia, i.e. the angle at $v$ formed by two geodesics (w.r.t. the hyperbolic metric) through $v$, namely a vertical line and a circular arc that passes also through $u$ (the angle is measured counterclockwise from the vertical). The components of $\varphi$ above are $\phi(u,0), \phi(u,1), \phi(v,u)$ and $\phi(v,1)$. Maybe this helps.

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    $\begingroup$ What kind of description of the image of $\varphi$ you are seeking? $\endgroup$ – Alex Ravsky Sep 2 '15 at 7:44

Looks like bonus math homework or some sort of puzzle.

Write the components of $\phi = (p,q,r,s)$. It's pretty easy to see that to each $u$ there corresponds exactly one $(p,q)$ and vice-versa. This is because $p$ is essentially the argument as seen from the origin and $q$ is the argument as seen from $1$ so locating $u$ from those two angles is a ``triangulation'' problem.

I think that, for any fixed $u$, as $v$ ranges in the upper half-plane, $s$ ranges in $(0,2\pi)$ and $r$ ranges either in $(-\pi,0)$ or $(0,\pi)$ depending on certain relations between $u$ and $s$. I now give a proof sketch. To that end, fix any $s \in (0,2\pi)$.

You can check that $r = Arg[(u-v)(\bar{u}-v)] = Arg \, f(v)$. The $f(v)$ inside is a quadratic polynomial in $v$ and its only roots are at $v=u,\bar{u}$ which are not in our domain so the Arg is well-defined. Furthermore, $f(v) = |u|^2 - 2 (\Re u)v + v^2 \geq 0$ whenever $v$ is real (in retrospect this is perhaps irrelevant).

For the next bit it is easier to think in terms of $h = 1-u$ and $w = v-1$. Writing $w = te^{i\theta} = tz$ (with obviously $s = 2\theta$ forcing the value of $\theta$ but leaving $t$ free) and letting $t$ range in $(0,\infty)$ we arrive at $Arg \, f(v) = Arg \, [(\Im h)^2 + (\Re h + zt)^2] = Arg \, J(t)$. Then, if $h$ and $z$ are in the same quadrant, $Arg \, J(t)$ ranges in $(0,\pi)$ while otherwise it ranges in $(-\pi,0)$. I think.

Edit: actually there seems to be maybe four cases, depending on the two possible quadrants for $h$ and $z$.

  • $\begingroup$ The behavior seems more complicated than that: for different fixed $h,z$ in the right quadrant, the range of $\operatorname{Arg} J(t)$ can be $(0,\pi)$ but also something else, e.g. for $z = (2i + 1)/\sqrt{5}$ and $h = 10 + 6i$ the plot of $\operatorname{Arg} J(t)$ is as follows: i.imgur.com/ugkB3N2.png $\endgroup$ – Ricardo Buring Jul 6 '15 at 14:46

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