Geodesics and the Energy Functional

Let $(M,g)$ be a smooth Riemannian manifold. Write $\nabla$ for its Levi–Civita connection and $T_pM$ for the tangent space at $p$. The coordinate conventions are those in tangent spaces and derivations.

We now apply concepts developed in Section Tangent Spaces and Riemannian Metrics to study properties of smooth curves on differentiable manifolds $M$. Let $[a,b] \subset \mathbb R$, and $\gamma : [a,b] \to M$ be a smooth curve. Define the length of $\gamma$ as

\[ L(\gamma) = \int_a^b \lVert \gamma' (t) \rVert \dif t\tag{1}\]

and the energy of $\gamma$ as

\[ E(\gamma) = \frac12 \int_a^b \lVert \gamma' (t) \rVert^2 \dif t ~.\tag{2}\]

We can calculate both length and energy in local coordinates. To do so, let $(x^1,\cdots,x^d) = (x^1(\gamma(t)),\cdots,x^d(\gamma(t)))$. Then, $\dot{x}^i (t) = \dod{}{t} x^i (\gamma(t))$, and

\[\begin{aligned} L(\gamma) &= \int_a^b \sqrt{g_{ij}(x(\gamma(t))) \dot{x}^i (t) \dot{x}^j (t)} \dif t ~, \\ E(\gamma) &= \frac12 \int_a^b {g_{ij}(x(\gamma(t))) \dot{x}^i (t) \dot{x}^j (t)} \dif t ~. \end{aligned}\tag{3}\]

By way of notation, when $\gamma(t)$ is a single-variable function, $\gamma'(t)$ denotes its derivative. When a single-variable function is indexed with superscript (or sometimes subscripts), we use the time derivative notation to denote its derivative.

Assume $M$ is connected for the distance discussion. For any two points $p,q \in M$, define the distance as

\[ d(p,q) = \inf \left\{ L(\gamma) : \gamma:[a,b] \to M \text{ where } \gamma(a) = p, \gamma(b) = q, \text{ and $\gamma$ is piecewise smooth} \right\} ~.\tag{4}\]

Connectedness and local coordinate balls imply that any two points can be joined by a piecewise smooth curve: the set reachable from a fixed point is open, and its complement is open by the same local argument. Thus $d(p,q)$ is finite. Reversing and concatenating curves prove symmetry and the triangle inequality.

For positivity and the topology, take coordinates $x(p)=0$ and a closed coordinate ball $\overline B_{2r}$ contained in the chart. Compactness and positive definiteness give constants $0<c\leq C$ such that $c|v|^2\leq g_x(v,v)\leq C|v|^2$ there. Straight coordinate segments give $d(p,q)\leq\sqrt C\,|x(q)|$ for $q\in B_r$. Any curve from $p$ that first reaches the coordinate sphere of radius $\rho\leq r$ has length at least $\sqrt c\,\rho$, by integrating its Euclidean speed up to first exit. A curve to $q\in B_r$ either stays in the ball, with length at least $\sqrt c|x(q)|$, or exits it, with length at least $\sqrt c r$. These estimates give $d(p,q)>0$ for $p\ne q$ and show that small coordinate balls and metric balls contain one another. Therefore $d$ is a metric inducing the manifold topology.

Lemma 1 (Reparametrization). Let $\psi:[\alpha,\beta]\to[a,b]$ be a smooth diffeomorphism of intervals, increasing or decreasing. Then $L(\gamma\circ\psi)=L(\gamma)$.

Now, by way of notation, we write $g^{ij} = (g_{ij})^{-1}$. By direct calculations, we have $g^{i\ell} g_{\ell j} = \delta^i_j$. Furthermore, we write $g_{j\ell,k} = \dpd{}{{x^{k}}} g_{j\ell}$.

Definition 2. The Christoffel symbol is defined as

\[\Gamma_{jk}^{i} = \frac12 g^{i\ell} (g_{j\ell,k} + g_{k\ell,j} - g_{jk,\ell})\]

in local coordinates.

Length is unchanged by reparametrization, whereas energy also depends on how fast the curve is traversed. We obtain the geodesic equation by varying the energy with the endpoints fixed. The Euler–Lagrange calculation below turns this variational condition into a system of ordinary differential equations.

Proposition 3. In a coordinate domain $\Omega$, let $L(x,v)=\frac12g_{ij}(x)v^iv^j$ and

\[I[x]=\int_a^b L(x(t),\dot x(t))\,dt.\tag{5}\]

A $C^2$ curve with fixed endpoints is stationary for all smooth endpoint-zero variations if and only if

\[\ddot x^i+\Gamma^i_{jk}(x)\dot x^j\dot x^k=0,\qquad 1\leq i\leq d.\tag{6}\]

In particular every local energy minimizer is stationary and satisfies this equation. Stationarity alone does not assert global minimization.

Definition 4. A smooth curve $\gamma : [a,b] \to M$ solving (6) is called geodesic.

Proof of Proposition 3. Let $v:[a,b]\to\mathbb R^d$ be smooth and zero at both endpoints. For sufficiently small $\varepsilon$, $x+\varepsilon v$ stays in the coordinate domain, by compactness of the curve image and openness of the domain. Differentiation under the integral is valid for the smooth integrand on this compact parameter set. Thus

\[\left.\frac d{d\varepsilon}I[x+\varepsilon v]\right|_0 =\int_a^b\left(\frac{\partial L}{\partial x^k}v^k+ \frac{\partial L}{\partial\dot x^k}\dot v^k\right)dt =\int_a^b\left(\frac{\partial L}{\partial x^k}-\frac d{dt}\frac{\partial L}{\partial\dot x^k}\right)v^kdt.\]

The boundary term vanishes because $v(a)=v(b)=0$. If the integral is zero for every such $v$, each continuous coefficient must vanish: a coefficient of fixed nonzero sign on a small interval can be tested against a nonnegative bump function supported there. This is the fundamental lemma of the calculus of variations, and gives

\[\frac{\partial L}{\partial x^k}-\frac d{dt}\frac{\partial L}{\partial\dot x^k}=0.\tag{7}\]

Here $\partial L/\partial\dot x^k=g_{kj}\dot x^j$ and $\partial L/\partial x^k=\frac12(\partial_k g_{ij})\dot x^i\dot x^j$. Substituting gives

\[g_{kj}\ddot x^j+(\partial_i g_{kj})\dot x^i\dot x^j-\frac12(\partial_k g_{ij})\dot x^i\dot x^j=0.\]

Symmetrize the middle coefficient in $i,j$, then multiply by $g^{\ell k}$. The coefficient of $\dot x^i\dot x^j$ becomes $\Gamma^\ell_{ij}$, proving (6). Reversing the integration-by-parts calculation proves the converse. For a curve crossing several charts, use variations supported in individual coordinate intervals; the resulting equation is the coordinate form of $\nabla_{\dot\gamma}\dot\gamma=0$, so it agrees on overlaps. ◻

Example 5 (Geodesics of a round sphere). For a sphere of radius $R>0$, use

\[X(\theta,\varphi)=(R\sin\theta\cos\varphi,R\sin\theta\sin\varphi,R\cos\theta),\quad0<\theta<\pi.\]

The coordinate vectors have inner products $\langle X_\theta,X_\theta\rangle=R^2$, $\langle X_\theta,X_\varphi\rangle=0$, and $\langle X_\varphi,X_\varphi\rangle=R^2\sin^2\theta$. Therefore

\[ds^2=R^2(d\theta^2+\sin^2\theta\,d\varphi^2),\qquad (g_{ij})=\begin{pmatrix}R^2&0\\0&R^2\sin^2\theta\end{pmatrix},\quad (g^{ij})=\begin{pmatrix}R^{-2}&0\\0&R^{-2}\sin^{-2}\theta\end{pmatrix}.\]

The only nonzero first-kind symbols are $\Gamma_{\theta\varphi,\varphi}=\Gamma_{\varphi\theta,\varphi}=R^2\sin\theta\cos\theta$ and $\Gamma_{\varphi\varphi,\theta}=-R^2\sin\theta\cos\theta$. Raising the last index gives $\Gamma^\theta_{\varphi\varphi}=-\sin\theta\cos\theta$ and $\Gamma^\varphi_{\theta\varphi}=\Gamma^\varphi_{\varphi\theta}=\cot\theta$. Thus

\[\ddot\theta-\sin\theta\cos\theta\,\dot\varphi^2=0,\qquad \ddot\varphi+2\cot\theta\,\dot\theta\dot\varphi=0.\tag{8}\]

The second equation, without division by $\dot\varphi$, is

\[\frac d{dt}(\sin^2\theta\,\dot\varphi)=0,\qquad\sin^2\theta\,\dot\varphi=C.\]

The squared speed $c^2=R^2(\dot\theta^2+\sin^2\theta\,\dot\varphi^2)$ is constant along a geodesic. When $c>0$, rescale the affine parameter to arrange $c=1$. On an interval where $\dot\theta\ne0$, this yields

\[\dot\theta=\pm R^{-1}\sqrt{1-R^2C^2/\sin^2\theta},\qquad \frac{d\varphi}{d\theta}=\pm\frac{RC}{\sin\theta\sqrt{\sin^2\theta-R^2C^2}}.\]

For $0<|RC|<1$, substitute $u=RC\cot\theta/\sqrt{1-R^2C^2}$. Integration gives $\varphi=\varphi_0\pm\arccos u$, so

\[\sin\theta\cos(\varphi-\varphi_0)=\frac{RC}{\sqrt{1-R^2C^2}}\cos\theta.\]

In Cartesian coordinates this is the equation of a plane through the origin. Its intersection with the sphere is a great circle. The omitted cases are also great circles: $C=0$ gives meridians, while $|RC|=1$ forces $\theta=\pi/2$ and gives the equator. A zero-speed geodesic is constant.

A coordinate-independent check completes the argument through poles and turning points. For an induced metric, the first variation gives that an energy-stationary curve has ambient acceleration perpendicular to the sphere: test against all endpoint-zero tangent fields and integrate the Euclidean energy by parts. Therefore $\gamma''=\lambda\gamma$. Differentiating $|\gamma|^2=R^2$ twice gives $\lambda=-c^2/R^2$, while $\langle\gamma'',\gamma'\rangle=0$ gives constant speed. Solving this constant-coefficient equation gives, for $c>0$,

\[\gamma(t)=\cos(ct/R)\,\gamma(0)+\frac R c\sin(ct/R)\,\gamma'(0).\]

It stays in the two-plane spanned by its initial position and velocity. Conversely this formula has normal acceleration and hence satisfies the geodesic equation. Thus every nonconstant geodesic is an affinely parametrized great circle, not necessarily a unit-speed or globally minimizing arc.

Example 6 (Poincaré Half-Plane). The Poincaré half-plane, otherwise known as the upper-half plane, is

\[\mathbb{H} = \left\{ (x,y) \in \mathbb R^2 : y > 0 \right\} ~.\]

The Poincaré half-plane admits the metric

\[\dif s^2 = \frac{\dif x^2 + \dif y^2}{y^2} ~.\]

Reading off the metric coefficients, we have

\[g_{ij} = \begin{pmatrix} \frac{1}{y^2} & 0 \\ 0 & \frac{1}{y^2} \end{pmatrix} ~,~ g^{ij} = \begin{pmatrix} y^2 & 0 \\ 0 & y^2 \end{pmatrix} ~.\]

Using the standard notation $i,j \in \{1,2\}$, the Christoffel symbols of the first kind are

\[\begin{aligned} \Gamma_{11,1} &= \frac12 (0 + 0 - 0) = 0 \\ \Gamma_{11,2} &= \frac12 \left(0 + 0 - \left( -\frac{2}{y^3} \right) \right) = \frac{1}{y^3} \\ \Gamma_{12,1} &= -\frac{1}{y^3} \\ \Gamma_{12,2} &= 0 \\ \Gamma_{22,1} &= 0 \\ \Gamma_{22,2} &= -\frac{1}{y^3} \end{aligned}\]

and some symbols are omitted here by a symmetry argument. Raising the third index with the inverse metric, we obtain the Christoffel symbols of the second kind $\Gamma_{ij}^k = g^{kl} \Gamma_{ij,l}$:

\[\begin{aligned} \Gamma_{12}^1 = \Gamma_{21}^1 &= g^{11} \Gamma_{12,1} = y^2 \left( -\frac{1}{y^3} \right) = -\frac{1}{y} \\ \Gamma_{11}^2 &= g^{22} \Gamma_{11,2} = y^2 \left( \frac{1}{y^3} \right) = \frac{1}{y} \\ \Gamma_{22}^2 &= g^{22} \Gamma_{22,2} = y^2 \left( -\frac{1}{y^3} \right) = -\frac{1}{y} \end{aligned}\]

with all other symbols being zero. For $k=1,2$, we obtain the following geodesic equations from $\Ddot{x}^k + \Gamma_{ij}^k \dot{x}^i \dot{x}^j = 0$:

\[\begin{aligned} \Ddot{x} + \Gamma_{12}^1 \dot{x} \dot{y} + \Gamma_{21}^1 \dot{x} \dot{y} &= 0 \\ \Ddot{y} + \Gamma_{11}^2 \dot{x}^2 + \Gamma_{22}^2 \dot{y}^2 &= 0 \end{aligned}\]

i.e.

\[\begin{aligned} \Ddot{x} - \frac{2 \dot x \dot y}{y} &= 0 \\ \Ddot{y} + \frac{\dot{x}^2 - \dot{y}^2}{y} &= 0 \end{aligned}\tag{9}\]

There are two cases. If $\dot x = 0$, then (9) is trivially satisfied, and (9) reads $\Ddot{y} - \frac{\dot{y}^2}{y} = 0$. This means $x = \text{constant}$, i.e. one type of geodesic is straight vertical lines in the upper-half plane.

Alternatively, if $\dot x \neq 0$, then (9) can be rewritten as $\dod{}{t} \left( \frac{\dot x}{y^2} \right) = 0$. This means $\dot x = a y^2$ for some constant $a \neq 0$. Note that $a = 0$ recovers the previous case. After rescaling a nonconstant geodesic to be parametrized by arclength, we obtain

\[\frac{\dot x^2 + \dot y^2}{y^2} = 1\]

(because this chosen parametrization has unit speed). Therefore, by the chain rule,

\[\left( \dod{y}{x} \right)^2 = \frac{\dot y^2}{\dot x^2} = \frac{y^2 - \dot{x}^2}{\dot{x}^2}\]

which, by substituting $\dot x^2 = a^2 y^4$, is equivalent to

\[\left( \dod{y}{x} \right)^2 = \frac{y^2 - a^2 y^4}{a^2 y^4} = \frac{1 - a^2 y^2}{a^2 y^2} \implies \dod{y}{x} = \pm \frac{\sqrt{1 - a^2 y^2}}{a y} ~.\]

Separation of variables and integrating directly gives

\[\begin{aligned} \frac{\pm a y}{\sqrt{1 - a^2 y^2}} \dif y &= \dif x \\ \implies \mp \frac1a \sqrt{1-a^2 y^2} &= x + b \\ \implies \frac{1}{a^2} (1 - a^2 y^2) &= (x+b)^2 \\ \implies (x+b)^2 + y^2 &= \frac{1}{a^2} \end{aligned}\]

which is the equation of a circle in $\mathbb R^2$. In the context of $\mathbb{H}$, this represents a semicircle. Therefore, another type of geodesic in $\mathbb{H}$ is semicircles centred anywhere on the $x$-axis.

The vertical case also determines the parameter: $(\log y)''=0$, so $y(t)=y_0e^{ct}$ and $x(t)=x_0$; unit speed means $|c|=1$. In the circular case, the integrations are local between turning points of $y$. A global unit-speed parametrization of the upper semicircle of radius $r$ centred at $b$ is

\[x(t)=b+r\tanh(t-t_0),\qquad y(t)=r\operatorname{sech}(t-t_0).\]

Direct differentiation gives $\dot x/y^2=1/r$ and $(\dot x^2+\dot y^2)/y^2=1$, and substitution verifies both geodesic equations, including the top point where $\dot y=0$. Reversing the parameter covers the other direction. The geodesic initial-value uniqueness theorem then shows that these curves, the vertical curves and the constant curves exhaust all possibilities.

The geodesic equation is a coupled second-order ODE. Written as a first-order system for $(x,\dot x)$, its right-hand side is smooth, so each initial pair $(p,v)$ determines a unique maximal geodesic $c_v$, smoothly depending on its initial data wherever defined. Affine rescaling and uniqueness give

\[c_{\lambda v}(t)=c_v(\lambda t)\]

whenever both sides are defined. Local existence and compactness of the unit sphere in $T_pM$ supply a common small existence time for unit initial velocities. This time cannot be chosen arbitrarily close to one on an incomplete manifold. Instead, rescaling shows that sufficiently small initial velocities exist through time one; this is what makes the exponential map defined near zero.

Definition 7. For $p\in M$, let $V_p\subset T_pM$ be the set of initial velocities whose maximal geodesics exist through time one. The exponential map is

\[\exp_p:V_p\longrightarrow M,\qquad\exp_p(v)=c_v(1).\]

The maximal-flow domain is open, so $V_p$ is open; it contains zero and a neighbourhood of zero. It need not be all of $T_pM$ without a completeness assumption.

Remark 8. The zero initial velocity produces the constant curve $c_0(t)=p$, defined for all times, so $0\in V_p$ and $\exp_p(0)=p$. Smooth dependence for the geodesic ODE makes $\exp_p$ smooth on $V_p$. For $v\in V_p$, the scaling identity gives $c_v(t)=\exp_p(tv)$ at every time for which the radial expression is defined.

Lemma 9. The exponential map at $p \in M$ maps a neighbourhood of $0 \in T_p M$ diffeomorphically onto a neighbourhood of $p \in M$.

Proof. Identify $T_0(T_pM)$ canonically with the vector space $T_pM$. Since $\exp_p(tv)=c_v(t)$ near zero,

\[(d\exp_p)_0(v)=\left.\frac d{dt}\right|_{t=0}\exp_p(tv)=c_v'(0)=v.\]

Thus the differential is the identity from $T_0(T_pM)$ to $T_pM$, hence invertible. Smoothness of $\exp_p$, already supplied by ODE dependence, allows application of the inverse function theorem. It yields neighbourhoods of zero and $p$ on which $\exp_p$ is a diffeomorphism. ◻

Lemma 9 gives us the ability to define normal coordinates. Because $\exp_p$ is a diffeomorphism from a neighbourhood $U$ of $0 \in T_p M$ to neighbourhood $V$ of $p \in M$, and that $T_p M$ can be identified with the Euclidean space, $\{\{U,V\}, \exp^{-1}\}$ is a local chart of $M$ near $p \in M$. Now, fix an orthonormal basis of $T_p M$, i.e. let $\{X^i\}$ be coordinates for $U$.

Definition 10. The local chart $\{U, \{x^1,\cdots,x^n\} \}$ is the normal coordinates system near $p \in M$.

Normal coordinates make geodesics through the centre into radial straight lines in the coordinate domain. More precisely, let $X_p = x^i \pd{}{{x^i}}$ be a tangent vector in $T_p M$, and consider the geodesic $\gamma(t) = \exp_p (t X_p)$. Then, in normal coordinates, we can express $\gamma(t) = (tx^1,\cdots,tx^n)$, i.e. this is called the radial geodesic.

Lemma 11 (Properties of Normal Coordinates). Let $\{U, \{x^1,\cdots,x^n\} \}$ be a normal coordinates system near $p \in M$. Then,

$g_{ij}(p) = \delta_{ij}$.
$\Gamma_{ij}^k(p) = 0$ for all $i,j,k$.
$\pd{}{{x^k}} g_{ij}(p) = 0$.

Proof. Choose the normal coordinates using an orthonormal basis $e_1,\ldots,e_d$ of $T_pM$. The coordinate vector at the centre is $(d\exp_p)_0e_i=e_i$, so $g_{ij}(p)=\delta_{ij}$.

For every initial vector $v$, the radial geodesic has coordinates $tv$. At $t=0$, its geodesic equation gives $\Gamma^k_{ij}(p)v^iv^j=0$. The Christoffel coefficients are symmetric in $i,j$. Taking $v=e_i$ gives the diagonal coefficients zero; taking $v=e_i+e_j$ then gives twice each off-diagonal coefficient zero. Hence all $\Gamma^k_{ij}(p)$ vanish.

Metric compatibility gives the coordinate identity

\[\partial_k g_{ij}=g_{\ell j}\Gamma^\ell_{ki}+g_{i\ell}\Gamma^\ell_{kj}.\]

It can also be checked directly by inserting the Christoffel formula and cancelling the three metric derivatives. At $p$, both terms vanish, yielding $\partial_k g_{ij}(p)=0$. ◻

Remark 12 (Riemannian polar coordinates). Let $M$ be a differentiable manifold. Take any $p \in M$, and $r > 0$ small enough such that the restriction of $\exp_p$ to $B_r(0) \subset T_p (M)$ is a diffeomorphism onto an open subset $U \subset M$. Let $(U, \{x^1,\cdots,x^n\})$ be the corresponding normal coordinates, and $(V, \{\varphi^1,\cdots,\varphi^{n-1}\})$ be a chart on $\mathbb S^{n-1}$, and define the “polar coordinates chart” $(\mathbb R^+ V, \{y^1,\cdots,y^n\})$ on $\mathbb R^n$ by

\[\begin{aligned} \mathbb R^+ V \triangleq \{ rv : r \in \mathbb R^+, v \in V \} \text{ and } (y^1,\cdots,y^n) = \left( \lVert x \rVert, \varphi \left( \frac{x}{\lVert x \rVert} \right) \right) ~. \end{aligned}\]

By putting $U' = x^{-1} (\mathbb R^+ V \cap B_r (0))$, we have that $(U', y \circ x)$ is a chart on $M$. We can explicit construct the Riemannian metric in this coordinates system $(U', y \circ x)$.

To do so, first, appealing to the definition of the exponential map, for any vector $v \in T_p M$, the curve

\[\begin{aligned} x(t) = tv \text{ for } |t| < \frac{r}{\lVert v \rVert} \end{aligned}\]

is a geodesic in $M$ with respect to the chart $(U,x)$. For all $v \in V$ such that $\lVert v \rVert = 1$, we have that the curve $x(t)$ reduces to $y(t) = (t, \varphi(x))$ in the chart $(U', y \circ x)$ (note such a curve $y(t)$ is well-defined for $t \in (0,r)$).

With respect to the chart $(U', y \circ x)$, let $\Gamma^k_{ij} (y)$ denote the Christoffel symbols, and $(h_{ij}(y))$ denote the matrix representing the Riemannian metric. Then, the geodesic equation reads

\[\begin{aligned} \ddot y^k (t) + \Gamma^k_{ij} (y(t)) \dot y^i (t) \dot y^j (t) = 0 \quad \text{for } t \in (0,r) \text{ and } i \in \{1, \cdots, n\} ~. \end{aligned}\]

Furthermore,

\[\begin{aligned} \Gamma^k_{ij} (y) = \frac12 h^{k\ell} (y) (h_{i\ell, j} (y) + h_{j\ell, i} (y) - h_{ij,\ell} (y)) \quad \text{for } y \in (0,r) ~. \end{aligned}\]

It is clear from definition and direct calculations that $\Gamma^k_{11} (y(t)) = 0$ for all $t \in (0,r)$ and $i \in \{1,\cdots,n\}$. This means that, by the definition of the Christoffel symbol, we have

\[\begin{aligned} h^{k\ell} (y) (2 h_{1\ell, 1}(y) - h_{11,\ell}(y)) = 0 \quad \text{for } y \in \Omega ~, \end{aligned}\]

where $\Omega = (0,r) \times \varphi(V) \subset \mathbb R^n$. Note that the conclusion above holds because, by unfixing $v \in V$, $\Gamma^k_{11} (y) = 0$ for all $y \in \Omega$. Multiplying the Christoffel symbol equation by the metric $h_{jk}$, summing over $k$, and using the fact that the product of a metric and inverse metric is the Kronecker delta, we have

\[\begin{aligned} 2 h_{1j, 1}(y) - h_{11,j}(y) = 0 \quad \text{for } y \in \Omega \text{ and } j \in \{1, \cdots, n\} ~. \end{aligned}\]

For $j=1$, this means $h_{11,1} (y) \equiv 0$. However, note that, by the change of coordinates formula, for all $y \in \Omega$,

\[\begin{aligned} h_{11}(y) = \dpd{{x^i}}{{y^1}} \dpd{{x^\ell}}{{y^1}} g_{i\ell} (x) ~. \end{aligned}\]

Separate the radial variable from the angular direction by writing

\[\begin{aligned} h_{11}(y) = z^i z^\ell g_{i\ell} (x) ~, \end{aligned}\]

where $z = \varphi^{-1} (y^2, \cdots, y^n) \in \mathbb S^{n-1} \subset \mathbb R^n$. Note the following features of $h_{11}(y)$: it is a constant as $y^1$ varies on $(0,r)$, but $(y^2, \cdots, y^n) \in \varphi(V)$ is fixed. However, for any fixed $(y^2, \cdots, y^n)$, letting $y^1 \to 0^+$ implies $x \to 0 \in \mathbb R^n$, so $g_{i\ell} (x) \to \delta_{i\ell}$. Because $z$ is also fixed, we obtain

\[\begin{aligned} \lim\limits_{y^1 \to 0^+} h_{11}(y) = z^i z^\ell \lim\limits_{x \to 0} g_{i\ell}(x) = \sum_{1 \leq i \leq n} (z^i)^2 = 1 ~. \end{aligned}\]

Therefore, $h_{11}(y) = 1$ for all $y \in \Omega$. We can now extend this argument for all $i = k \geq 2$, and we get

\[\begin{aligned} \dpd{}{{y^1}} h_{1j} (y) = 0 \quad \text{for } y \in \Omega \text{ and } j \in \{2, \cdots, n\} ~. \end{aligned}\]

Using the same arguments as before, we have, for $w = \pd{}{{y^j}} \varphi^{-1} (y^2, \cdots, y^n) \in \mathbb R^n$, that

\[\begin{aligned} h_{1j} (y) = z^i y^1 w^\ell g_{i\ell} (x) ~. \end{aligned}\]

Fixing $(y^2, \cdots, y^n) \in \varphi(V)$ and letting $y^1 \to 0^+$, we have that $z$ and $w$ are fixed as $g_{i\ell} (x) \to \delta_{i\ell}$. As such,

\[\begin{aligned} h_{1j}(y) = 0 \quad \text{for } y \in \Omega \text{ and } j \in \{2, \cdots, n\} ~. \end{aligned}\]

Collecting everything above, and appealing to the symmetry $h_{i\ell} \equiv h_{\ell i}$, we have

\[(h_{ij}(y)) = \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & h_{22}(y) & \cdots & h_{2n}(y) \\ \vdots & \vdots & \ddots & \vdots \\ 0 & h_{n2}(y) & \cdots & h_{nn}(y) \end{pmatrix} ~,\]

The radial coefficient is one and the mixed radial-angular coefficients vanish. Thus the radial direction is orthogonal to the coordinate spheres.

The limit at radial coordinate zero is taken in the original normal chart, where the metric is smooth; the polar chart itself does not contain the centre. In particular, the angular coordinate vector contains the factor $y^1$, so the mixed term tends to zero even though an angular coordinate is undefined at the centre. Since the mixed term is constant along each fixed ray, that limit proves it vanishes everywhere on the punctured chart. This is the local Gauss lemma: radial speed is one and radial and angular directions are orthogonal.

1. The Hamiltonian Formulation of Geodesics

The first variation of the energy functional gives us Proposition 3. This system admits another formulation on the cotangent bundle. Let $M$ be a differentiable manifold such that $\dim M = d$. By first locally trivializing $T^* M$ by a chart, i.e. such that

\[\begin{aligned} T^* M\big|_{U} : U \times \mathbb R^d \to M \end{aligned}\]

with coordinates $(x^1,\cdots,x^d,p_1,\cdots,p_d)$, we can then define the Hamiltonian

\[ H(x,p) = \frac12 g^{ij} (x) p_i p_j ~.\tag{10}\]

Theorem 13. The geodesic equations are equivalent to the following system of $2d$ equations:

\[\begin{aligned} \dot x^i &= \dpd{H}{{p_i}} ~~, \quad i \in \llbracket d \rrbracket ~, \\ \dot p_i &= -\dpd{H}{{x^i}} ~~, \quad i \in \llbracket d \rrbracket ~, \end{aligned}\tag{11}\]

Here $\partial H/\partial p_i=g^{ij}p_j$ and $\partial H/\partial x^i=\frac12(\partial_i g^{jk})p_jp_k$. The two terms from differentiating the symmetric quadratic form cancel the factor $1/2$ in the momentum derivative.

Proof. The first Hamilton equation gives $\dot x^i=g^{ij}p_j$, equivalently $p_i=g_{ij}\dot x^j$. Differentiating the identity $g^{jk}g_{k\ell}=\delta^j_\ell$ gives

\[\partial_i g^{jk}=-g^{ja}(\partial_i g_{ab})g^{bk}.\]

Substitute this into the second Hamilton equation, and also differentiate $p_i=g_{ij}\dot x^j$ along the curve:

\[\dot p_i=\frac12(\partial_i g_{ab})\dot x^a\dot x^b =g_{ij}\ddot x^j+(\partial_k g_{ij})\dot x^k\dot x^j.\]

Multiplying by $g^{\ell i}$ and symmetrizing the coefficient of $\dot x^a\dot x^b$ yields

\[\ddot x^\ell+\frac12g^{\ell i} (\partial_a g_{ib}+\partial_b g_{ia}-\partial_i g_{ab})\dot x^a\dot x^b=0.\]

The coefficient is $\Gamma^\ell_{ab}$, so this is the geodesic equation. Conversely, given a geodesic $x(t)$, define $p_i=g_{ij}\dot x^j$. The first Hamilton equation holds by construction, and reversing the displayed calculation proves the second. Thus the equivalence holds in both directions. ◻

Definition 14. The system determined by (11) and (11) is called the cogeodesic flow on $T^* M$.

Hamilton’s equations also show that the energy is constant along each integral curve:

\[\dod{H}{t} = H_{x^i} \dot x^i + H_{p_i} \dot p_i = - \dot p_i \dot x^i + \dot x^i \dot p_i = 0 ~.\]

Furthermore, note that the cogeodesic flow maps $\mathcal E_\lambda$ to $\mathcal E_\lambda$ for every $\lambda \geq 0$, where

\[ \mathcal E_\lambda \triangleq \left\{ (x,p) \in T^* M : H(x,p) = \lambda \right\} ~.\tag{12}\]

(12) is often referred to as the level set of the Hamiltonian energy.

Remark 15. If $M$ is compact, then $\mathcal E_\lambda$ is compact. To see this, this is equivalent to showing the sphere bundle $\pi : SM \to M$ is compact for compact manifolds $M$. Let $\{x_j,v_j\}\subset SM$ denote any sequence. By compactness of $M$, there is a subsequence and a limit point, i.e. some subsequence such that $x_{j_k} \to x$. Let $U$ be a trivializing neighbourhood of $x$, i.e. $\pi^{-1}(U) \cong U \times \mathbb S^{n-1}$. It follows that $(x_j,v_j)\in U \times \mathbb S^{n-1}$, by the fact that $x_j \to x \in U$ for infinitely many $j$. Repeat the same argument for the compact manifold $\mathbb S^{n-1}$, we have a limit point and another subsequence such that $v_{j_k} \to v$. Therefore, any arbitrary sequence $\{x_j,v_j\}\subset SM$ has an accumulation point in $SM$, hence $SM$ is compact.

For $\lambda>0$, the metric identifies the covector level $H=\lambda$ with a sphere bundle of radius $\sqrt{2\lambda}$, so the subsequence argument applies after this identification. For $\lambda=0$, positive definiteness makes the level the zero section, which is homeomorphic to compact $M$. In the sphere-bundle argument, use a smooth local orthonormal frame (obtained by Gram–Schmidt from a coordinate frame); this identifies the varying metric unit spheres with one fixed $S^{d-1}$. The coordinate subsequence converges in that trivialization to a vector in the fibre over the limiting base point. Thus no uniform choice of global coordinates is being assumed.

Reading: Carmo (1992), Riemannian geometry.