Holonomy group of Poincaré metric

Question

I am following the book "Geometry, Topology, and Physics" by Nakahara. In section 7.5 exercise 7.14, the author asks us to calculate the holonomy group of the Poincaré metric. The Poincaré metric is given by $$ g=\frac{dx \otimes dx + dy \otimes dy}{y^2} $$ The Christoffel symbols I calculated and verified are $$\Gamma ^ x _{xy} = \Gamma ^ x _{yx} = -\frac{1}{y}$$ $$\Gamma^y_{xx}=-\Gamma^y_{yy}=\frac{1}{y}$$ This yields the following equations for parallel transporting a vector along a curve.

$$\frac{dX^x}{dt}-\frac{1}{y} \left(\frac{dx}{dt} X^y + \frac{dy}{dt} X^x\right) = 0\\ \frac{dX^y}{dt}+\frac{1}{y} \left(\frac{dx}{dt} X^x - \frac{dy}{dt} X^y\right) = 0$$ Then I parallel transported the vector along the rectangle defined by the following four curves. \begin{align} c(t) &= (\pi t, y_0)\\ c(t) &= (\pi, y_0+\epsilon t)\\ c(t) &= (\pi (1-t), y_0+\epsilon )\\ c(t) &= (0, y_0+\epsilon (1-t)) \end{align} For the first (and third) curve, the parallel transport solution I got was

\begin{align*} X^x_1 = -X^x_0 \\ X^y_1 = -X^y_0 \end{align*}

Then for the second curve, the parallel transport solution I got was \begin{align*} X^x(t) = \frac{y_0+\epsilon t}{y_0} X^x_0 \\ X^y(t) = \frac{y_0+\epsilon t}{y_0} X^y_0 \end{align*}

Then for the fourth curve, the parallel transport solution I got was \begin{align*} X^x(t) = \frac{y_0-\epsilon t}{y_0} X^x_0 \\ X^y(t) = \frac{y_0-\epsilon t}{y_0} X^y_0 \end{align*}

Overall, if I parallel transport the vector (1,0) in the "rectangle" defined by the four curves, I get back the same (1,0) vector.

The holonomy group of the Poincaré metric is supposed to be $SO(2)$. Is my choice of curve for parallel transporting wrong? Or is the calculation wrong? Thank you :)

Answer 1

There are several possible proofs, depending on how much do you know:

1. Suppose you already know the Gauss-Bonnet formula in the hyperbolic plane, in the form that the angle deficit of each hyperbolic triangle $T$ equals its hyperbolic area. (The angle deficit of a triangle with angles $\alpha, \beta, \gamma$ is $\pi- (\alpha+\beta+\gamma)$.) Then, by looking at how a tangent vector $v$ at a vertex $p$ of $T$ parallel-translates along $T$, you see that the holonomy along $T$ equals the angle deficit, i.e. the area of $T$. Then, by a direct construction, one proves the existence of hyperbolic triangles with arbitrary angle deficit in the interval $(0,\pi)$. From this, it follows that the holonomy group is isomorphic to $SO(2)$. (Since the hyperbolic plane is orientable, the holonomy group is contained in $SO(2)$.)

2. Suppose you already know that if $\gamma_s$ is a 1-parameter family of loops in a Riemannian manifold $(M,g)$ based at a point $p\in M$, then the holonomy $Hol_{\gamma_s}$ along $s$ depends continuously on $s$. (This is just a fact from ODEs: Solution of a linear ODE depends smoothly on the coefficients of the ODE and the initial conditions.) Suppose you also know that if a Riemannian manifold $(M,g)$ has trivial holonomy group, then $(M,g)$ is (locally) flat. Suppose you also know that the hyperbolic plane ${\mathbb H}^2$ is simply-connected and has nonzero curvature. Given this, let us prove that the holonomy group of the hyperbolic plane is $SO(2)$. First, since the hyperbolic plane is not flat, there is a loop $\gamma$ based at $p$ such that $Hol_\gamma$ is a nontrivial rotation by some angle $\alpha$. Since ${\mathbb H}^2$ is simply-connected, there is a 1-parameter family of smooth loops $\gamma_s$ based at $p$ such that $\gamma_1=\gamma$ and $\gamma_0$ is the constant map to $p$. Since the holonomy of $\gamma_0$ is the rotation by the zero angle, by the intermediate value theorem, every angle $\beta\in [0,\alpha]$ is the holonomy of some loop $\gamma_s, s\in [0,1]$. Since $\alpha\ne 0$, the subgroup of $SO(2)$ generated by the rotations by all angles $\beta\in [0,\alpha]$ is the entire $SO(2)$. Thus, again, the holonomy group of ${\mathbb H}^2$ is $SO(2)$.

3. Suppose, you know none of these, but you know how to compute the parallel transport (of the hyperbolic metric) along horizontal and vertical line (Euclidean) segments in the hyperbolic plane (the upper half-plane model). From this, you see that the holonomy along such rectangles gives you rotations by arbitrary angles.