# A Proof of Trigonometric Formulas in the Plane of Constant Curvature


1. Induced Connection Along a Mapping Suppose $M$ and $N$ be two smooth manifolds, and $\phi\mathpunct{:}N\to M$ is a smooth mapping. A vector field $X$ along $\phi$ is an assignment which corresponding each $x\in N$ to a vector $X(x)\in T_{\phi(x)}M$. In particular, for any vector field $V$ on $N$, $\phi_\ast V$ may not be a vector field on $M$, but it is a vector field along $\phi$. Clearly, the collection of vector fields along $\phi$ is a vector space, with the natural defined addtion and scalar multiplication.

We can extend a connection of vector fields on $M$ to a connection of vector fields along $\phi$. Let $\nabla$ be a connection of $M$, and $\set{E_i}_{i=1}^{\mathrm{dim} M}$ be a local frame in a neighborhood $\mathcal U$ of $\phi(x)$. Then the vector field $X$ along $\phi$ can be expressed as
$X(x)=X^i(x)E_i(\phi(x)),$
and we call $X$ is smooth if all the component functions $X^i$ are smooth. Now, taken a $v\in T_xN$, we define a mapping, which is called a covariant derivative, $\tilde\nabla$ from the vector fields along $\phi$ to $T_{\phi(x)}M$ as:
$\tilde\nabla_{v}X=(vX^i)E_i(\phi(x))+X^i(x)\nabla_{\phi_\ast v}E_i,$
a directly verification will show that it is independent on the choice of the frame. Let $V$ be a smooth vector field of $N$, then we obtain a smooth vector field along $\phi$ as
$(\tilde\nabla_VX)(x)=\tilde\nabla_{V(x)}X.$
The mapping $\tilde\nabla$ which assign a vector field $V$ of $N$ and a vector field $X$ along $\phi$ to the vector field along $\phi$, i.e., $\tilde\nabla_V X$, is called induced connection.

Particularly, if $\gamma$ is a curve on $M$, then the induced connection is commonly referred to as the connection along a curve [1].

An important property of the induce connection is that, even though we will not use it, if $\nabla$ is the Riemannian connection then $\tilde\nabla$ will also be Riemannian” in the following sense\footnote{the second one is a litter trick, for a proof, see [1]}:
\begin{align*}
v\langle X,Y\rangle&=\langle\tilde\nabla_vX,Y\rangle+\langle X,\tilde\nabla_vY\rangle,\\
\phi_\ast([V,W])&=\tilde\nabla_V\phi_\ast W+\tilde\nabla_W\phi_\ast V,
\end{align*}
where $v\in T_xN$, $X,Y$ are vector fields along $\phi$ and $V,W$ are vector fields on $N$.

2. The Geodesics in Geodesic Polar Coordinates At once one have a connection along a curve, then we can define the concept of parallel transformation along the curve, and obtain the geodesics of a manifold as the self-parallel curve. More precisely, if $\gamma\mathpunct{:}[0,1]\to M$ is a curve, and the tangent vector field of $\gamma(t)$ will be denoted as $\dot\gamma(t)=\gamma_\ast\left(\dt\right)$, which is clearly a vector field along $\gamma$, then
$\nabla_{\dot\gamma}\dot\gamma=0.$
The above equation is a second-order non-linear ordinary differential equation (in a local frame), by the theory of ordinary differential equations, it will have an unique solution for the following initial values
$\begin{cases} \gamma(0)=p\\ \dot\gamma(0)=v, \end{cases}$
where $p\in M, v\in T_pM$. What is more, the fundamental theory of ordinary differential equation tells us that, for any $\eps>0$, by rescalling the time parameter $t$ then we can suppose that $\eps>1$, there exists a neighborhood $\mathcal U$ of $p$ and a $\delta>0$, such that for any $q\in\mathcal U$ and any $v\in T_qM$ with $|v|<\delta$, the geodesic equation will have a unique solution on $(-\eps,\eps)$ which satisfy the following initial condition $\begin{cases} \gamma(0)=q\\ \dot\gamma(0)=v. \end{cases}$ This derive a map $\exp_p$ form a neighborhood of $T_pM$ to $M$, and in fact it is a locally diffeomorphism, called exponential map. Hence we can obtain a coordinate system in a neighborhood of $M$, which is induced from the coordinates of $T_pM$ by the exponential map. Especially, when we taken the polar coordinate systems on $T_pM$ then the corresponding coordinate will be called geodesic polar coordinates. From now on, let $M$ be a two dimensional manifold of constant curvature, $\nabla$ be the Riemannian connection (Levi-Civita connection). The polar coordinates of $M$ at $p$ will be donoted as $(r,\theta)$, as usual, $\set{\frac{\partial}{\partial r},\frac{\partial}{\partial \theta}}$ denote the local natural basis. Then a curve $\gamma(t)$ on $M$ can be written as $(r(t),\theta(t))$, where $t$ is the arc-lenth parameter, and the tangent vector of $\gamma$ is $\dot\gamma(t)=\dot r(t)\pr+\dot\theta(t)\pt$. Now, if $\gamma(t)$ is a geodesic, then \begin{align} 0&=\nabla_{\dot\gamma(t)}\dot\gamma(t) =\nabla_{\dot\gamma(t)}\left(\dot r(t)\pr\right)+ \nabla_{\dot\gamma(t)}\left(\dot\theta(t)\pt\right)\notag\\ &=\dt\dot r(t)\cdot\pr+\dot r(t)\cdot\nabla_{\dot\gamma(t)}\prr +\dt\dot\theta(t)\cdot\pt+\dot\theta(t)\cdot\nabla_{\dot\gamma(t)}\ptt\notag\\ &=\ddot r(t)\pr+\ddot\theta(t)\pt+\dot r(t)\left(\dot r(t)\nabla_{\pr}\prr+\dot\theta(t)\nabla_{\pt}\prr\right)\notag\\ &\qquad+\dot\theta(t)\left(\dot r(t)\nabla_{\pr}\ptt+\dot\theta(t) \nabla_{\pt}\ptt\right)\notag\\ &=\ddot r(t)\pr+\ddot\theta(t)\pt+2\dot r(t)\dot\theta(t)\nabla_{\pt}\prr+\dot\theta(t)\dot\theta(t) \nabla_{\pt}\ptt\label{eq:1}. \end{align} The last equation obtained, since the $r$-curves are geodesics, thus $\nabla_{\prr}\prr=0$ and note also that $[\prr,\ptt]=0$, hence $\nabla_{\ptt}\prr=\nabla_{\prr}\ptt$. the Gauss lemms assert that $\langle\prr,\ptt\rangle=0$, then the metric of $M$ can be written as $$\label{eq:2} (\mathrm ds)^2=(\mathrm d r)^2+f^2(r,\theta)(\mathrm d\theta)^2,$$ where $f(r,\theta)$ is a positive function. Using the compatibility and torsion free again, we have \begin{alignat*}{2} \langle\nabla_{\ptt}\prr,\prr\rangle&=0,\quad&\langle\nabla_{\ptt}\prr,\ptt\rangle&=\langle\nabla_{\prr}\ptt,\ptt\rangle=ff_r,\\ \langle\nabla_{\ptt}\ptt,\prr\rangle&=-ff_r,\quad& \langle\nabla_{\ptt}\ptt,\ptt\rangle&=ff_\theta. \end{alignat*} Inserting these relations into \eqref{eq:1}, $0=\left\{\ddot r(t)-\dot\theta(t)\dot\theta(t)ff_r\right\}\pr+ \left\{\ddot\theta(t)+2\dot r(t)\dot\theta(t)ff_r+\dot\theta(t)\dot\theta(t)ff_\theta\right\}\pt.$ Thus, the geodesic equation is $$\label{eq:3}\begin{cases} 0&=\ddot r(t)-\dot\theta(t)\dot\theta(t)ff_r\\ 0&=\ddot\theta(t)+2\dot r(t)\dot\theta(t)ff_r+\dot\theta(t)\dot\theta(t)ff_\theta, \end{cases}$$ where the $f,f_r$ should be evaulated at $\gamma(t)=(r(t),\theta(t))$. To get the relation of $f$ should be satisfied as $M$ is of constant curvature, we can calculate the Christoffel symbols, but a more efficient way is the method moving frame. Set $\omega^1=\rd r$, $\omega^2=f\rd\theta$, then \eqref{eq:2} can be re-written as $(\rd s)^2=\omega^1\omega^1+\omega^2\omega^2.$ A directly calculation will show that $\begin{cases} \rd\omega^1&=\rd(\rd r)=0\\ \rd\omega^2&=\rd(f\rd\theta)=f_r\rd r\wedge\rd\theta, \end{cases}$ thus, if we set $\omega_1^2=f_r\rd\theta=-\omega_2^1$, then $\begin{cases} \rd\omega^1&=\omega^2\wedge\omega_2^1\\ \rd\omega^2&=\omega^1\wedge\omega_1^2， \end{cases}$ thus the Cartan's Lemma asserts that $\omega^2_1$ is the connection 1-form and the Gauss equation says that $\rd\omega_1^2=-K\omega^1\wedge\omega^2,$ where $K$ is the Gauss curvature of $M$, which is a constant by assumption. We conclude that for a 2-dimensional manifold $M$, it has constant sectional curvature if and only if the function $f$ must satisfy the following differential equation $$\label{eq:4} f_{rr}+K f=0.$$ Hence, since $t$ is arc-length parameter, the geodesic equations are $$\begin{cases} 0&=\ddot r(t)-\dot\theta(t)\dot\theta(t)ff_r\\ 0&=\ddot\theta(t)+2\dot r(t)\dot\theta(t)ff_r+\dot\theta(t)\dot\theta(t)ff_\theta\\ 1&=\dot r(t)^2+f^2\dot\theta(t)^2\\ 0&=f_{rr}+K f. \end{cases}$$