Let $\mathscr{C}\subseteq \mathbb{R}^3$ be a cone with half angle $\alpha$ (we exclude the vertex of the cone). This cone can (almost) be covered by an isometric chart given simply by the development of the cone on the plane. This isometry is constructed as follows:
$$F:\{(x,y):\varphi(x,y)\in (0,2\pi \sin(\alpha))\}\to \mathscr{C}\subseteq \mathbb{R}^3 $$
$$(x,y)\equiv (\varphi,r)\mapsto (r\sin(\alpha)\cos(\varphi/\sin(\alpha)),r\sin(\alpha)\sin(\varphi/\sin(\alpha)),r\cos(\alpha))\equiv (\varphi/\sin(\alpha),r)$$
Where I explicitated the map both in cartesian coordinates and in polar ones.
Let $P_0\in \mathscr{C}$ and let $\gamma$ be the "parallel" of the cone passing through $P_0=(x_0,y_0,z_0)\equiv (\theta_0,r_0)$ (i.e. the circumference with constant $z$ coordinate passing through $P_0$ and contained in the cone). Basically in polar coordinates:
$$\gamma(\theta)=(\theta_0+\theta,r_0).$$
I'd like to calculate the parallel transport of the tangent versor to this curve in $P_0$ along the curve $\gamma$.
My attempt
I tried using the fact that $F$ is a local isometry to transform this into a trivial problem in an open subset of $\mathbb{R}^3$. So I managed to construct the parallel transport in the domain of $F$:
$$\tilde{U}=(\cos(\varphi_0),\sin(\varphi_0))$$
where $\varphi_0=\theta_0\sin(\alpha)$. Now I should simply pushforward this vector field through $F$:
$$U_\theta=d_{(r_0\cos(\varphi),r_0\sin(\varphi))} F(\cos(\varphi_0),\sin(\varphi_0))=$$
$$=\frac{d}{dt}|_{t=0}F(r_0\cos(\varphi)+t\cos(\varphi_0), r_0\sin(\varphi)+t\sin(\varphi_0))$$
where $\varphi=\theta\sin(\alpha)$. And now things get pretty messy because explicitating $F$ is excruciating. Is there a better way?
Best Answer
Maybe you can use the local isometry $F$ but the expression is pretty ugly and it is not globally defined. $F$ is definitely the best way to understand what happens since parallel transport on parts of a plane is trivial and I suggest that you draw it to see how your tangent vectors are twisted after one trun around the cone.
But to get a closed expression, I think it is easier (at least for me) to consider the global diffeomorphism, $$ \psi : \left\{\begin{array}{rcl} \mathbb{R}^2\backslash\{0\} & \rightarrow & \mathcal{C} \subset \mathbb{R}^3 \\ z & \mapsto & (z,\|z\|\mathbb{cotan}(\alpha)) \end{array}\right., $$ where $\mathbb{cotan} = \frac{1}{\tan}$.
$\phi$ is of course not an isometry when we take the canonical metric on $\mathbb{R}^2\backslash\{0\}$ (except in the trivial case $\alpha = \frac{\pi}{2}$). To solve this, let us change the metric on the plane with $h = \phi^*g$ where $g$ is the metric on $\mathcal{C}$ inherited from the canonical metric on $\mathbb{R}^3$.
The use advantage on working on the plane is that it is parallelizable with a canonical isomorphism $T\mathbb{C} \cong \mathbb{C}^2$. Let $z \in \mathbb{R}^2\backslash\{0\}$ and $(v,w)$ two elements in $T_z\mathbb{R}^2\backslash\{0\} \cong \mathbb{R}^2\backslash\{0\} \times \mathbb{R}^2$.
We have $d\psi(z)v = \left(v,\frac{\left<z,v\right>}{\|z\|}\mathrm{cotan}(\alpha)\right)$ thus, \begin{align*} h_z(v,w) & = g_{\psi(z)}(d\phi(z)v,d\phi(z)w)\\ & = \left<\left(v,\frac{\left<z,v\right>}{\|z\|}\mathrm{cotan}(\alpha)\right),\left(w,\frac{\left<z,w\right>}{\|z\|}\mathrm{cotan}(\alpha)\right)\right>\\ & = \left<v,w\right> + \frac{\left<z,v\right>\left<z,w\right>}{\|z\|^2}\mathrm{cotan}^2(\alpha). \end{align*} Be careful, the scalar product $\left<,\right>$ is sometimes taken in $\mathbb{R}^2$, sometimes in $\mathbb{R}^3$, I use the same notation for both.
Now, take your smooth curve $\gamma : \theta \mapsto (r_0e^{i(\theta_0 + \theta)},r_0\mathrm{cotan}(\alpha))$ where I identify $\mathbb{R}^3$ with $\mathbb{C} \times \mathbb{R}$ for simplicity of notation. Similarly, I identify $\mathbb{R}^2\backslash\{0\}$ with $\mathbb{C}^*$ and we have $\psi^{-1} \circ \gamma : \theta \mapsto r_0e^{i(\theta_0 + \theta)}$. We identify by abuse of notation $\gamma$ and $\psi^{-1} \circ \gamma$.
With the parallelization $T\mathbb{C}^* \cong \mathbb{C}^* \times \mathbb{C}$, the parallel transport from time $-\theta_0$ to time $\theta$, $L = \Gamma(\gamma)_{-\theta_0}^\theta$ can be seen as an $\mathbb{R}$-linear automorphism of $\mathbb{C}$ and it must satistfy for all $\theta,v,w$, $$ L\dot\gamma(-\theta_0) = \dot\gamma(\theta), \qquad h_{\gamma(\theta)}(L v,L w) = h_{\gamma(-\theta_0)}(v,w). $$ For all $\theta$, $\dot\gamma(\theta) = ir_0e^{i(\theta + \theta_0)}$ so the first condition is $L(ir_0) = ir_0e^{i(\theta_0 + \theta)}$ or equivalently, $L(i) = ie^{i(\theta_0 + \theta)} = -\sin(\theta_0 + \theta) + i\cos(\theta_0 + \theta)$.
We want now to use the second relation at points $(v,w) = (1,1)$ and $(v,w) = (1,i)$ to determine $L(1)$. Let $L(1) = a + ib$. We have, \begin{align*} h_{\gamma(\theta)}(L(1),L(1)) & = \left<L(1),L(1)\right> + \frac{\left<L(1),\gamma(\theta)\right>^2}{\|\gamma(\theta)\|^2}\mathrm{cotan}^2(\alpha)\\ & = a^2 + b^2 + (a\cos(\theta_0 + \theta) + b\sin(\theta_0 + \theta))\mathrm{cotan}^2(\alpha). \end{align*} And, \begin{align*} h_{\gamma(-\theta_0)}(1,1) & = \left<1,1\right> + \frac{\left<1,\gamma(-\theta_0)\right>^2}{\|\gamma(-\theta_0)\|^2}\mathrm{cotan}^2(\alpha)\\ & = 1 + \mathrm{cotan}^2(\alpha). \end{align*} For $(v,w) = (1,i)$, we obtain, \begin{align*} h_{\gamma(\theta)}(L(1),L(i)) & = \left<L(1),L(i)\right> + \frac{\left<L(1),\gamma(\theta)\right>\left<L(i),\gamma(\theta)\right>}{\|\gamma(\theta)\|^2}\mathrm{cotan}^2(\alpha)\\ & = \left<L(1),L(i)\right> \textrm{ because $L(i)$ and $\gamma(\theta)$ are orthogonal,}\\ & = -a\sin(\theta_0 + \theta) + b\cos(\theta_0 + \theta). \end{align*} And, \begin{align*} h_{\gamma(-\theta_0)}(1,i) & = \left<1,i\right> + \frac{\left<1,\gamma(-\theta_0)\right>\left<i,\gamma(-\theta_0)\right>}{\|\gamma(-\theta_0)\|^2}\mathrm{cotan}^2(\alpha)\\ & = 0 \end{align*} We deduce by the second relation on parallel transport that, \begin{align*} a^2 + b^2 + (a\cos(\theta_0 + \theta) + b\sin(\theta_0 + \theta))\mathrm{cotan}^2(\alpha) & = 1 + \mathrm{cotan}^2(\alpha) & (1)\\ -a\sin(\theta_0 + \theta) + b\cos(\theta_0 + \theta) & = 0. & (2) \end{align*} $(1) + i(2)$ gives, $$ \|L(1)\|^2 + L(1)e^{-i(\theta_0 + \theta)}\mathrm{cotan}^2(\alpha) = 1 + \mathrm{cotan}^2(\alpha). $$ This is a real number hence $L(1) = \rho e^{i(\theta_0 + \theta)}$ for some (not necessarily non-negative) real number $\rho$ and $\rho^2 + \rho\mathrm{cotan}^2(\alpha) = 1 + \mathrm{cotan}^2(\alpha)$ thus $\rho = 1$ or $-1 - \mathrm{cotan}^2(\alpha)$ but at $\theta = -\theta_0$, $L = \mathrm{Id}$ so $\rho = 1$. By continuity, we have $\rho = 1$ for all $\theta$ hence $L(1) = e^{i(\theta_0 + \theta)}$.
We deduce form the expressions of $L(1)$ and $L(i)$ that $L$ is simply the rotation of angle $\theta_0 + \theta$ and by composition, $\Gamma(\gamma)_{\theta_1}^{\theta_2} = \Gamma(\gamma)_{-\theta_0}^{\theta_2} \circ \Gamma(\gamma)_{\theta_1}^{-\theta_0}$ is the rotation of angle $\theta_2 - \theta_1$.
When we come back in the cone via $\psi$, we have for all $\theta$, $T_{\gamma(\theta)}\mathcal{C} = \mathrm{Im}(d\psi(\gamma(\theta))) = \left\{\left(v,w,\frac{v\cos(\theta_0 + \theta) + w\sin(\theta_0 + \theta)}{r_0}\mathrm{cotan}^2(\alpha)\right)\middle|(v,w) \in \mathbb{R}^2\right\}$ and $\Gamma(\gamma)_{\theta_1}^{\theta_2}$ is given by, $$ \left\{\begin{array}{rcl} T_{\gamma(\theta_1)}\mathcal{C} & \rightarrow & T_{\gamma(\theta_2)}\mathcal{C} \\ \left(v,w,\frac{v\cos(\theta_0 + \theta_1) + w\sin(\theta_0 + \theta_1)}{r_0}\mathrm{cotan}^2(\alpha)\right) & \mapsto & \left(v\cos(\theta_2 - \theta_1) - w\sin(\theta_2 - \theta_1),v\sin(\theta_2 - \theta_1) + w\cos(\theta_2 - \theta_1),\frac{v\cos(\theta_0 + \theta_2) + w\sin(\theta_0 + \theta_2)}{r_0}\mathrm{cotan}^2(\alpha)\right) \end{array}\right. $$ In particular, notice that $\Gamma(\gamma)_0^{2\pi} = \mathrm{Id}$ and $\pi_1(\mathcal{C},\gamma(0))$ is generated by $[\gamma_{|[0,2\pi]}]$. Since $\mathcal{C}$ is flat (via the local isometry you gave), its holonomy group is trivial.