4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ What kind of description of the image of $\varphi$ you are seeking? $\endgroup$ – Alex Ravsky Sep 2 '15 at 7:44
1
$\begingroup$

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 $z$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.