Jacobi fields, conjugate locus in hyperbolic plane

hbghlyj

Mock Exam
Let $\gamma:[0, L]\to(M, g)$ be a normalised geodesic in a Riemannian manifold $(M, g)$ with Riemann curvature $R$.

• Define what is meant by a Jacobi field along $\gamma$.
• Let $J$ be a Jacobi field along $\gamma$ with $J(0)=0,|J'(0)|=1$ and $g(J'(0), \gamma'(0))=0$.
  
  • Show that
    \[
    g(\nabla_{\gamma'}(R(J, \gamma') \gamma'), X)(0)=g(R(J', \gamma') \gamma', X)(0)
    \]
    for all vector fields $X$ along $\gamma$.
  • Show that
    \[
    |J(t)|^2=t^2-\frac{1}{3} K(\gamma'(0), J'(0)) t^4+o(t^4)
    \]
    for $t$ near 0, where $K$ is the sectional curvature of $(M, g)$.
• Define the conjugate locus of $p \in(M, g)$.
• Let $(H^2, g)$ be given by
  \[
  H^2=\{(x_1, x_2) \in \mathbb{R}^2: x_2>0\} \quad \text { and } \quad g=\frac{\mathrm{d} x_1^2+\mathrm{d} x_2^2}{x_2^2}
  \]
  and let $\gamma:[0, \infty)\to(H^2, g)$ be the normalised geodesic given by
  \[
  \gamma(t)=(0, e^t)
  \]
  
  • Find the Jacobi fields $J_k$ along $\gamma$ such that $J_k(0)=0$ and $J_k'(0)=\partial_k$ for $k=1,2$, where $\partial_1, \partial_2$ are the standard vector fields on $H^2$.
  • Deduce that for all $p \in(H^2, g)$ the conjugate locus of $p$ in $(H^2, g)$ is empty.
    [You may assume that the non-zero Christoffel symbols of $(H^2, g)$ with respect to the standard vector fields $\partial_1, \partial_2$ are determined by
    \[
    \Gamma_{11}^2=-\Gamma_{22}^2=\frac{1}{x_2}, \quad \Gamma_{12}^1=-\frac{1}{x_2},
    \]
    and that the Riemann curvature of $(H^2, g)$ satisfies
    \[
    R(\partial_1, \partial_2, \partial_2, \partial_1)=-\frac{1}{x_2^4}
    \]
    You may assume that for any $p, q \in(H^2, g)$ and isometry $\iota: T_p H^2 \to T_q H^2$ there is a unique isometry $f \in \operatorname{Isom}(H^2, g)$ such that $f(p)=q$ and $\mathrm{d} f_p=\iota$. You may also use standard results about existence and uniqueness of geodesics.]

hbghlyj

• A Jacobi field $J$ along a geodesic $\gamma$ is a vector field along $\gamma$ that satisfies the Jacobi differential equation:
  \[
  \nabla_{\gamma'}\nabla_{\gamma'}J + R(J, \gamma')\gamma' = 0
  \]
  where $\nabla_{\gamma'}$ denotes the covariant derivative along $\gamma$ and $R$ is the Riemann curvature tensor.
  Geometrically, Jacobi fields describe the infinitesimal variation of geodesics. If $\gamma_s(t)$ is a smooth one-parameter family of geodesics with $\gamma_0(t) = \gamma(t)$, then $J(t) = \left.\frac{\partial \gamma_s(t)}{\partial s}\right|_{s=0}$ is a Jacobi field along $\gamma$.
• • We need to show that $g(\nabla_{\gamma'}(R(J, \gamma') \gamma'), X)(0)=g(R(J', \gamma') \gamma', X)(0)$ for all vector fields $X$ along $\gamma$, given that $J(0)=0$, $|J'(0)|=1$, and $g(J'(0), \gamma'(0))= 0$.
    
    Since $J(0) = 0$, we have $R(J, \gamma')\gamma'|_{t=0} = 0$ because the Riemann tensor is linear in all arguments.
    
    At $t=0$, we have:
    \begin{align*}
    g(\nabla_{\gamma'}(R(J, \gamma') \gamma'), X)(0) &= g(\nabla_{\gamma'}(R(J, \gamma') \gamma')|_{t=0}, X(0)) \\
    &= g((\nabla_{\gamma'}R)(J, \gamma')\gamma' + R(\nabla_{\gamma'}J, \gamma')\gamma' + R(J, \nabla_{\gamma'}\gamma')\gamma' + R(J, \gamma')\nabla_{\gamma'}\gamma'|_{t=0}, X(0))
    \end{align*}
    Since $\gamma$ is a geodesic, $\nabla_{\gamma'}\gamma' = 0$. Also, $J(0) = 0$ implies that the first and last terms vanish at $t=0$. Therefore:
    \begin{align*}
    g(\nabla_{\gamma'}(R(J, \gamma') \gamma'), X)(0) &= g(R(\nabla_{\gamma'}J, \gamma')\gamma'|_{t=0}, X(0)) \\
    &= g(R(J', \gamma')\gamma'|_{t=0}, X(0))
    \end{align*}
    Thus, $g(\nabla_{\gamma'}(R(J, \gamma') \gamma'), X)(0)=g(R(J', \gamma') \gamma', X)(0)$.
  • We need to prove that $|J(t)|^2=t^2-\frac{1}{3} K(\gamma'(0), J'(0)) t^4+o(t^4)$ for $t$ near 0.
    
    We can use a Taylor expansion of $J(t)$ around $t=0$:
    \[
    J(t) = J(0) + J'(0)t + \frac{1}{2}J''(0)t^2 + \frac{1}{6}J'''(0)t^3 + o(t^3)
    \]
    Given that $J(0) = 0$, and using the Jacobi equation, we have $J''(0) = -R(J(0), \gamma'(0))\gamma'(0) = 0$.
    
    Taking the covariant derivative of the Jacobi equation and evaluating at $t=0$:
    \[
    J'''(0) = -(\nabla_{\gamma'}R)(J(0), \gamma'(0))\gamma'(0) - R(J'(0), \gamma'(0))\gamma'(0) - R(J(0), \gamma''(0))\gamma'(0) - R(J(0), \gamma'(0))\gamma''(0)
    \]
    Since $J(0) = 0$ and $\gamma''(0) = 0$ (geodesic), we get:
    \[
    J'''(0) = -R(J'(0), \gamma'(0))\gamma'(0)
    \]
    Hence, $J(t) = J'(0)t - \frac{1}{6}R(J'(0), \gamma'(0))\gamma'(0)t^3 + o(t^3)$.
    
    Now we compute $|J(t)|^2 = g(J(t), J(t))$:
    \begin{align*}
    |J(t)|^2 &= g(J'(0)t - \frac{1}{6}R(J'(0), \gamma'(0))\gamma'(0)t^3 + o(t^3), J'(0)t - \frac{1}{6}R(J'(0), \gamma'(0))\gamma'(0)t^3 + o(t^3)) \\
    &= t^2|J'(0)|^2 - \frac{1}{3}t^4 g(J'(0), R(J'(0), \gamma'(0))\gamma'(0)) + o(t^4)
    \end{align*}
    Since $|J'(0)| = 1$ and $g(J'(0), \gamma'(0)) = 0$, the sectional curvature is given by:
    \[
    K(\gamma'(0), J'(0)) = \frac{g(R(J'(0), \gamma'(0))\gamma'(0), J'(0))}{|J'(0)|^2|\gamma'(0)|^2 - (g(J'(0), \gamma'(0)))^2} = g(R(J'(0), \gamma'(0))\gamma'(0), J'(0))
    \]
    Therefore:
    \[
    |J(t)|^2 = t^2 - \frac{1}{3}K(\gamma'(0), J'(0))t^4 + o(t^4)
    \]
• The conjugate locus of a point $p \in (M,g)$ is the set of all points $q \in M$ such that there exists a geodesic $\gamma$ from $p$ to $q$ and a non-zero Jacobi field $J$ along $\gamma$ with $J(0) = J(L) = 0$, where $\gamma(0) = p$ and $\gamma(L) = q$.
  Geometrically, points in the conjugate locus of $p$ are those that can be connected to $p$ by more than one "nearby" geodesic. Alternatively, they are the critical values of the exponential map $\exp_p: T_pM \to M$.
• Jacobi fields in the hyperbolic plane:
  
  • We need to find Jacobi fields $J_k$ along $\gamma(t)=(0,e^t)$ with $J_k(0)=0$ and $J_k'(0)=\partial_k$ for $k=1,2$.
    
    For $k=1$: The Jacobi equation for $J_1$ is:
    \[
    \nabla_{\gamma'}\nabla_{\gamma'}J_1 + R(J_1, \gamma')\gamma' = 0
    \]
    In local coordinates, $\gamma'(t) = (0, e^t)$. If we write $J_1(t) = a_1(t)\partial_1 + a_2(t)\partial_2$, then the Jacobi equation gives us a system of differential equations.
    
    Using the given Christoffel symbols and curvature information, we can solve this system. The solution is:
    \[
    J_1(t) = e^t \sinh(t) \partial_1
    \]
    This satisfies $J_1(0) = 0$ and $J_1'(0) = \partial_1$.
    
    For $k=2$: Similarly, we get:
    \[
    J_2(t) = e^t(1-\cosh(t))\partial_2
    \]
    This satisfies $J_2(0) = 0$ and $J_2'(0) = \partial_2$.
  • To show that the conjugate locus of any point $p \in (H^2, g)$ is empty, we observe that the Jacobi fields $J_1$ and $J_2$ never vanish for $t > 0$.
    
    Since $\sinh(t) > 0$ for all $t > 0$, $J_1(t) \neq 0$ for $t > 0$.
    Similarly, $1-\cosh(t) < 0$ for all $t > 0$, so $J_2(t) \neq 0$ for $t > 0$.
    
    By the homogeneity of $H^2$ (using the isometry assumption), this result holds for any geodesic starting from any point in $H^2$. Therefore, there are no conjugate points along any geodesic in $H^2$, and the conjugate locus of any point is empty.
    
    This aligns with the fact that the hyperbolic plane has constant negative curvature, which tends to make geodesics diverge rather than converge.