# Prove Killing vector is an affine collineation

• thecommexokid
In summary, a Killing vector is a vector field that preserves the metric tensor on a manifold, representing a symmetry of the manifold. An affine collineation is a transformation that preserves collinearity and ratio of distances, representing a symmetry of affine space. To prove that a Killing vector is an affine collineation, one can use the properties of the metric tensor and the Lie derivative. This is important because it allows for the use of powerful tools in studying the symmetries of a manifold, with applications in fields such as physics, mathematics, engineering, and computer graphics.
thecommexokid

## Homework Statement

Given a manifold M with metric gab and associated derivative operator a, let ξ a be a Killing vector on M. Prove that ξ a is an affine collineation for a.

## Homework Equations

(a) For a vector ξ a to be an affine collineation for a derivative operator a, it must satisfy
ab ξ c = Rmabc ξ m.​

(b) We are told that ξ a is a Killing vector, so
(a ξ b) = 0.​

(c) The action of the Riemann tensor on a vector is
Rabcd ξ c = –2 ∇[ab] ξ d.​

## The Attempt at a Solution

I've gone down one blind alley after another. Below I reproduce what seemed like the most promising path I've explored so far. Can somebody help me with the next step, or suggest an alternative path that could be more fruitful?

Rmabc ξ m = gcp Rmabp ξ m.​
It's a symmetry property of the Riemann tensor that I can switch the first pair of indices with the second:
gcp Rmabp ξ m = gcp Rbpma ξ m = gcp gaq Rbpmq ξ m.​
We've now finagled this into the form of (c):
gcp gaq Rbpmq ξ m = –2 gcp gaq [bp] ξ q = –2 gcp[bp] ξ a.​
Performing the antisymmetrization,
–2 gcp[bp] ξ a = – gcp (∇bp ξ a – ∇pb ξ a) .​
Now it's easy to use (b) on the first of these two terms:
gcp (∇bp ξ a – ∇pb ξ a) = + gcp (∇ba ξ p + ∇pb ξ a).​
But I still have that second term sitting there! What do I do about it?

Let me start you off on an easier path. We have by definition ##\nabla_a \nabla_b \xi_c - \nabla_b \nabla_a \xi_c = R_{abc}{}{}^{d}\xi_d## which we can rewrite using ##\nabla_{(a}\xi_{b)} = 0## as ##\nabla_a \nabla_b \xi_c + \nabla_b \nabla_c \xi_a = R_{abc}{}{}^{d}\xi_d##. Now perform cyclic permutations of the indices ##(abc)## and combine the resulting equations in the appropriate manner in order to get the result ##\nabla_a\nabla_b \xi_c = -R_{bca}{}{}^{d}\xi_d##. Your desired result is then immediate.

1 person
So as you say, I have the 3 equations:$$\left\{ \begin{array}{l} R_{abc}{}^{d}\xi_d = \nabla_a\nabla_b\xi_c+\nabla_b\nabla_c\xi_a \\ R_{cab}{}^{d}\xi_d = \nabla_c\nabla_a\xi_b+\nabla_a\nabla_b\xi_c \\ R_{bca}{}^{d}\xi_d = \nabla_b\nabla_c\xi_a+\nabla_c\nabla_a\xi_b. \end{array}\right.$$If I add together the first two equations and subtract the third, I get
$$(R_{abc}{}^{d}+R_{cab}{}^{d}-R_{bca}{}^{d})\xi_d = 2\nabla_a\nabla_b\xi_c.$$The Riemann tensor antisymmetrizes to 0 in its first 3 indices:$$R_{[abc]}{}^{d} = \tfrac13(R_{abc}{}^{d}+R_{bca}{}^{d}+R_{cab}{}^{d})=0.$$So the $R_{abc}{}^{d}+R_{cab}{}^{d}$ in my equation can be replaced by a $-R_{bca}{}^{d}$:$$-2R_{bca}{}^{d}\xi_d = 2\nabla_a\nabla_b\xi_c.$$I can exchange the first pair of indices in the Riemann tensor with the second pair for free:$$-R_{ambc}\xi^m = \nabla_a\nabla_b\xi_c.$$But swapping the order of the first two costs a minus sign:$$+R_{mabc}\xi^m = \nabla_a\nabla_b\xi_c.$$And finally raising c on both sides completes the proof.

Cool. Cool cool cool. Thanks for the help!

No problem! The utilization of ##\epsilon_{abcd}##, ##R_{[abc]d}## (usually in conjunction with the former) and index permutation constitute very standard techniques that you will make use of over and over when performing these kinds of calculations so it's good to internalize them.

1 person

## What is a Killing vector?

A Killing vector is a vector field on a manifold that preserves the metric tensor, meaning it preserves the distance and angle measurements between points. In other words, it represents a symmetry of the manifold.

## What is an affine collineation?

An affine collineation is a transformation that preserves collinearity (the property of points lying on a straight line) and the ratio of distances along a line. In other words, it represents a symmetry of the affine space.

## How do you prove that a Killing vector is an affine collineation?

To prove that a Killing vector is an affine collineation, we need to show that it preserves the ratio of distances along a line. This can be done by using the properties of the metric tensor and the Lie derivative, which measures the change in a vector field along another vector field.

## Why is it important to prove that a Killing vector is an affine collineation?

Proving that a Killing vector is an affine collineation is important because it allows us to use the powerful tools of differential geometry to study the symmetries of a manifold. This can help us understand the underlying structure and properties of the manifold, which can have important implications in various fields of science.

## What are some applications of the concept of "Prove Killing vector is an affine collineation"?

The concept of proving that a Killing vector is an affine collineation has applications in various fields such as physics, mathematics, and engineering. For example, it is used in general relativity to study the symmetries of spacetime, in fluid dynamics to analyze the symmetries of flows, and in computer graphics to generate realistic animations.

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