Given a Riemannian manifolds $(M,g)$, a Killing vector field $X$ on $M$, and a geodesic $\gamma: K \rightarrow M$ defined on an interval $K \subseteq \mathbb{R}$, how does one show that $X \circ \gamma$ is a Jacobi field along $\gamma$?
2026-05-16 05:24:50.1778909090
Killing vector fields restricted to geodesics
1.8k Views Asked by user20609 https://math.techqa.club/user/user20609/detail At
1
There are 1 best solutions below
Related Questions in RIEMANNIAN-GEOMETRY
- What is the correct formula for the Ricci curvature of a warped manifold?
- How to show that extension of linear connection commutes with contraction.
- geodesic of infinite length without self-intersections
- Levi-Civita-connection of an embedded submanifold is induced by the orthogonal projection of the Levi-Civita-connection of the original manifold
- Geodesically convex neighborhoods
- The induced Riemannian metric is not smooth on the diagonal
- Intrinsic vs. Extrinsic notions of Harmonic maps.
- Equivalence of different "balls" in Riemannian manifold.
- Why is the index of a harmonic map finite?
- A closed manifold of negative Ricci curvature has no conformal vector fields
Trending Questions
- Induction on the number of equations
- How to convince a math teacher of this simple and obvious fact?
- Find $E[XY|Y+Z=1 ]$
- Refuting the Anti-Cantor Cranks
- What are imaginary numbers?
- Determine the adjoint of $\tilde Q(x)$ for $\tilde Q(x)u:=(Qu)(x)$ where $Q:U→L^2(Ω,ℝ^d$ is a Hilbert-Schmidt operator and $U$ is a Hilbert space
- Why does this innovative method of subtraction from a third grader always work?
- How do we know that the number $1$ is not equal to the number $-1$?
- What are the Implications of having VΩ as a model for a theory?
- Defining a Galois Field based on primitive element versus polynomial?
- Can't find the relationship between two columns of numbers. Please Help
- Is computer science a branch of mathematics?
- Is there a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?
- Identification of a quadrilateral as a trapezoid, rectangle, or square
- Generator of inertia group in function field extension
Popular # Hahtags
geometry
circles
algebraic-number-theory
functions
real-analysis
elementary-set-theory
proof-verification
proof-writing
number-theory
elementary-number-theory
puzzle
game-theory
calculus
multivariable-calculus
partial-derivative
complex-analysis
logic
set-theory
second-order-logic
homotopy-theory
winding-number
ordinary-differential-equations
numerical-methods
derivatives
integration
definite-integrals
probability
limits
sequences-and-series
algebra-precalculus
Popular Questions
- What is the integral of 1/x?
- How many squares actually ARE in this picture? Is this a trick question with no right answer?
- Is a matrix multiplied with its transpose something special?
- What is the difference between independent and mutually exclusive events?
- Visually stunning math concepts which are easy to explain
- taylor series of $\ln(1+x)$?
- How to tell if a set of vectors spans a space?
- Calculus question taking derivative to find horizontal tangent line
- How to determine if a function is one-to-one?
- Determine if vectors are linearly independent
- What does it mean to have a determinant equal to zero?
- Is this Batman equation for real?
- How to find perpendicular vector to another vector?
- How to find mean and median from histogram
- How many sides does a circle have?
Wlog $\gamma = \gamma(t)$ is parameterized by arclength. Then this is shown by differentiating $X\circ \gamma$ twice wRt $t$, using the defining equations for a geodesic and Killing field and the rules for interchanging covariant derivates -- these, you need to know, of course. Differentiating once results in: $$\nabla_{\frac{\partial}{\partial t}} X\circ\gamma = \nabla_{\gamma^{\prime}} X = \nabla_X \gamma^{\prime}$$ the last equality being true cause $X$ is killing (implying that the Lie derivative $[X,\gamma^{\prime}]$ vanishes). Hence $$\nabla_{\frac{\partial}{\partial t}} \nabla_{\frac{\partial}{\partial t}} X\circ\gamma = \nabla_{\gamma^{\prime}} \nabla_X \gamma^{\prime} = \nabla_X \nabla_{\gamma^{\prime}} \gamma^{\prime} + R(\gamma^{\prime},X)\gamma^\prime $$ (depending on the sign conventions you are using for the curvature tensor the last term may appear with a different sign.) In the last expression the first term vanishes, cause $\gamma$ is a geodesic, and the term involving the Lie derivate does not appear, again because X is Killing. So you are done.