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original poster
hbghlyj
posted 2025-5-22 00:49
- 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 are given
\[
H^2=\{(x_1, x_2)\in \mathbb{R}^2: x_2>0\},\quad g=\frac{dx_1^2+dx_2^2}{x_2^2},
\]
with the (only nonzero) Christoffel symbols (with respect to the coordinate vector fields $\partial_1,\partial_2$)
\[
\Gamma_{11}^2=-\Gamma_{22}^2=\frac{1}{x_2},\quad \Gamma_{12}^1=-\frac{1}{x_2},
\]
and the curvature
\[
R(\partial_1,\partial_2,\partial_2,\partial_1)=-\frac{1}{x_2^4}.
\]
The geodesic under consideration is
\[
\gamma(t)=(0,e^t),\quad t\in [0,\infty).
\]
Since $e^t>0$, $\gamma$ is well-defined.
- 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 $k=1,2$.
Step 1. Choose a convenient frame along $\gamma$.
Along $\gamma(t)=(0,e^t)$:
- The tangent vector is
\[
\gamma'(t)=(0,e^t) = e^t\partial_2.
\] - A unit vector field perpendicular to $\gamma'$ is obtained by parallel transporting the vector $\partial_1$ (after normalization). Notice that
\[
g(\partial_1,\partial_1) = \frac{1}{x_2^2}=\frac{1}{e^{2t}},
\]
so a unit vector perpendicular to $\gamma'(t)$ is
\[
E_1(t) = e^t\partial_1.
\]
(One may check that $g(E_1,E_1)=1$ and $g(E_1,\gamma'(t))=0$.)
Step 2. The two cases:
Case 1: $J$ perpendicular to $\gamma'$ (with initial direction $\partial_1$).
Write
\[J_1(t) = f(t)E_1(t),\]
where $f(t)$ is a scalar function to be determined. The Jacobi equation along $\gamma$ (for a direction normal to $\gamma'$) reads
\[f''(t)+K(t) f(t)=0,\]
where $K(t)$ is the sectional curvature in the plane spanned by $\gamma'(t)$ and $E_1(t)$. For the hyperbolic plane, the curvature is constant:
\[
K \equiv -1.
\]
Thus the equation becomes
\[
f''(t) - f(t)=0.
\]
The general solution is
\[
f(t) = A e^t + B e^{-t}.
\]
The initial conditions are:
- $J_1(0)=0$ implies $f(0)=A+B=0$ so that $B=-A$.
- $J'_1(0)=\partial_1$.
To compute the covariant derivative $J'_1(t) = \nabla_{\gamma'(t)}J_1(t)$, use that $E_1(t)$ is parallel along $\gamma$ (which may be verified using the Christoffel symbols and the symmetry of $H^2$); therefore
\[
\nabla_{\gamma'}J_1(t) = f'(t)E_1(t).
\]
At $t=0$, $E_1(0) = e^0\partial_1=\partial_1$ so that
\[
J'_1(0) = f'(0)\partial_1 = \partial_1.
\]
Hence $f'(0)=1$. Now, since
\[
f(t)=A(e^t - e^{-t}),
\]
we have
\[
f'(t) = A(e^t + e^{-t}),
\]
and thus
\[
f'(0) = 2A=1 \quad \Longrightarrow \quad A=\frac{1}{2}.
\]
Therefore, the Jacobi field is
\[
J_1(t) = \frac{1}{2}(e^t-e^{-t}) E_1(t)
= \frac{1}{2}(e^t-e^{-t})e^t\partial_1 = e^t \sinh(t)\partial_1.
\]
Case 2: $J$ tangent to $\gamma'$ (with initial direction $\partial_2$).
In any Riemannian manifold, the tangent (or "trivial") Jacobi field along a geodesic is given by the variation by re-parameterization. That is,
\[
J_2(t) = t\gamma'(t).
\]
Check that:
- $J_2(0) = 0$.
- $J'_2(t) = \gamma'(t) + t\nabla_{\gamma'}\gamma'(t) = \gamma'(t)$ (because $\gamma$ is a geodesic so $\nabla_{\gamma'}\gamma'(t)=0$). In particular, $J'_2(0)=\gamma'(0)=\partial_2$ (since at $t=0$ we have $\gamma(0)=(0,1)$ and $\gamma'(0)=(0,1)=\partial_2$ in these coordinates).
Thus, the two Jacobi fields are:
- For $k=1$: $J_1(t) = e^t \sinh(t)\partial_1$
- For $k=2$: $J_2(t) = te^t\partial_2$
- Deduction that the conjugate locus in $(H^2,g)$ is empty:
Because hyperbolic space $(H^2,g)$ is a model of a complete, simply connected surface of constant curvature $K=-1$, the exponential map at any point $p\in H^2$ is a diffeomorphism. Equivalently, along any geodesic $\gamma$ emanating from $p$ there is no nontrivial Jacobi field $J$ (other than the trivial re-parameterization field) that vanishes at $p$ and then again at some time $t>0$.
In our explicit computation, we can verify this by examining the zeros of the Jacobi fields:
- $J_1(t)\ne0\forall t\ne0$
- $J_2(t)\ne0\forall t\ne0$
By the homogeneity of $H^2$ (using the given isometry assumption), this result generalizes to any geodesic starting from any point in $H^2$. Thus, there are no conjugate points along any geodesic in $H^2$, and the conjugate locus of any point is empty.
This result aligns with the geometric intuition: in spaces of negative curvature like $H^2$, geodesics tend to diverge rather than converge, preventing the formation of conjugate points.
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