# A Question about Killing vector fields

1. Jul 26, 2016

### lichen1983312

I am trying to follow Nakahara's book. From the context, it seems that the author is trying to say if moving a point along a flow always give a isometry, the corresponding vector field $$X$$ is a Killing vector field. am I right?

then the book gives a proof. It only considers a linear approximation
$$f:{x^\mu } \mapsto {x^\mu } + \varepsilon {X^\mu }$$

in each step ignoring terms containing higher orders of $$\varepsilon$$

$$\begin{array}{l} {g_{\mu \nu }}(x) = \frac{{\partial ({x^\kappa } + \varepsilon {X^\kappa })}}{{\partial {x^\mu }}}\frac{{\partial ({x^\lambda } + \varepsilon {X^\lambda })}}{{\partial {x^\nu }}}{g_{\kappa \lambda }}(x + \varepsilon X)\\ = (\delta _\mu ^\kappa + \varepsilon {\partial _\mu }{X^\kappa })(\delta _\nu ^\lambda + \varepsilon {\partial _\nu }{X^\lambda })[{g_{\kappa \lambda }}(x) + \varepsilon {X^\xi }{\partial _\xi }{g_{\kappa \lambda }}(x)]\\ \approx (\delta _\mu ^\kappa \delta _\nu ^\lambda + \delta _\mu ^\kappa \varepsilon {\partial _\nu }{X^\lambda } + \varepsilon {\partial _\mu }{X^\kappa }\delta _\nu ^\lambda )[{g_{\kappa \lambda }}(x) + \varepsilon {X^\xi }{\partial _\xi }{g_{\kappa \lambda }}(x)]\\ \approx \delta _\mu ^\kappa \delta _\nu ^\lambda {g_{\kappa \lambda }}(x) + \delta _\mu ^\kappa \delta _\nu ^\lambda \varepsilon {X^\xi }{\partial _\xi }{g_{\kappa \lambda }}(x) + \delta _\mu ^\kappa \varepsilon {\partial _\nu }{X^\lambda }{g_{\kappa \lambda }}(x) + \varepsilon {\partial _\mu }{X^\kappa }\delta _\nu ^\lambda {g_{\kappa \lambda }}(x)\\ \approx {g_{\mu \nu }}(x) + \varepsilon {X^\xi }{\partial _\xi }{g_{\mu \nu }}(x) + \varepsilon {\partial _\nu }{X^\lambda }{g_{\mu \lambda }}(x) + \varepsilon {\partial _\mu }{X^\kappa }{g_{\kappa \nu }}(x) \end{array}$$

then we obtain the Killing equation
$${X^\xi }{\partial _\xi }{g_{\mu \nu }}(x) + {\partial _\nu }{X^\lambda }{g_{\mu \lambda }}(x) + {\partial _\mu }{X^\kappa }{g_{\kappa \nu }}(x) = 0$$

I feel uncomfortable here because the Killing equation only looks a necessary condition for the equation

$${g_{\mu \nu }}(x) = \frac{{\partial ({x^\kappa } + \varepsilon {X^\kappa })}}{{\partial {x^\mu }}}\frac{{\partial ({x^\lambda } + \varepsilon {X^\lambda })}}{{\partial {x^\nu }}}{g_{\kappa \lambda }}(x + \varepsilon X)$$
to be true, how about the terms contianing higher order of $$\varepsilon$$?

2. Jul 26, 2016

### MarcusAgrippa

The Killing equation is certainly a necessary condition, as you note. The converse of the theorem is more difficult to prove, and a theorem may not admit an exact converse but only a partial converse. In fact, most theorems that involve differentiation usually only admit partial converses, since differentiation probes only properties in the infinitesimal neighbourhood of a point, while the original statement is valid in a finite domain. The finite domain may be topologically complicated. All information about global topology is lost in the differentiation procedure. So a converse usually needs additional information about global topology to be supplied.

The higher powers of epsilon? Try writing out a few more terms in your expansions. The remaining terms, which you have not written out above, all contain epsilon squared or higher powers. In accordance with the methods of the differential calculus, after taking out the term common to both sides (the undifferentiated g's with no epsilon coefficient) you are left with only with terms that has factors that are powers of epsilon. You may then divide by epsilon (in a limiting procedure, epsilon is never permitted to be zero and so there is no danger of a zero division). After dividing by epsilon, take the limit of the remaining equation as epsilon goes to zero. This knocks out all the terms except the ones that you have written out explicitly above.

3. Jul 26, 2016

### lichen1983312

Thanks very much for the detailed explanation. In fact I am not at the point worrying about the strictness of the proof, like what you described in the first paragraph.

I think I did not state my question well. If I understand right, the Killing equation should guarantee that (at least in a good neighborhood), moving a point along a flow is an isometry
$$f:x \mapsto \exp (\varepsilon X)x$$
if denote
$$x' = \exp (\varepsilon X)x$$
then it seems that I should show directly
$${g_{\mu \nu }}(x) = {g_{\alpha \beta }}(x')\frac{{\partial x{'^\alpha }}}{{\partial {x^u}}}\frac{{\partial x{'^\beta }}}{{\partial {x^\nu }}}$$
the proof in the books seems first write
$$x' = {x^\mu } + \varepsilon {X^\mu } + O({\varepsilon ^2})$$
and start from
$${g_{\mu \nu }}(x) = \frac{{\partial ({x^\kappa } + \varepsilon {X^\kappa })}}{{\partial {x^\mu }}}\frac{{\partial ({x^\lambda } + \varepsilon {X^\lambda })}}{{\partial {x^\nu }}}{g_{\kappa \lambda }}(x + \varepsilon X)$$
if I put the higher order terms back, this should reach
$$0 = \varepsilon {X^\xi }{\partial _\xi }{g_{\mu \nu }}(x) + \varepsilon {\partial _\nu }{X^\lambda }{g_{\mu \lambda }}(x) + \varepsilon {\partial _\mu }{X^\kappa }{g_{\kappa \nu }}(x) + O({\varepsilon ^2})$$
I think what you are saying in the second paragraph is, since the Killing equation bring the first order to zero, we have

$$\mathop {\lim }\limits_{\varepsilon \to \infty } \frac{{\varepsilon {X^\xi }{\partial _\xi }{g_{\mu \nu }}(x) + \varepsilon {\partial _\nu }{X^\lambda }{g_{\mu \lambda }}(x) + \varepsilon {\partial _\mu }{X^\kappa }{g_{\kappa \nu }}(x) + O({\varepsilon ^2})}}{\varepsilon } = 0$$

here is my confusion, it seems there is a missing step to explain how above relationship guarantee ?
$${g_{\mu \nu }}(x) = {g_{\alpha \beta }}(x')\frac{{\partial x{'^\alpha }}}{{\partial {x^u}}}\frac{{\partial x{'^\beta }}}{{\partial {x^\nu }}}$$
since the higher order terms only vanish in the limit eposlon approaches zero.
is my understanding right?

4. Jul 27, 2016

### MarcusAgrippa

1. Yes, that is the limiting process that I described. I assume you know that you need to cancel the epsilon from the first three terms in the numerator of your limit.
2. Yes, you are correct in saying that Killing's equation is a necessary consequence of the isometry condition. Your final question is equivalent to asking, is it also a sufficient condition to guarantee that
is an isometry. In other words, you are asking for a proof of the converse of the proposition that you have proved.

As I tried to point out, the proof of sufficiency takes very much more work than the proof of necessity - this is a common feature of proofs. You will find a proof in Abraham and Marsden, Foundation of Mechanics, Second Edition, proposition 2.7.10, page 151-152, "A vector field is a Killing field if and only if its flow F_t consists of isometries." However, you will require to do a considerable amount of work before you will understand their proof. I do not know any simpler or more transparent proof, though they might exist in other books. Console yourself with the fact that in practice one hardly ever needs to use finite isometries - the infinitesimal ones are usually sufficient.

Hope this helps.

5. Jul 27, 2016

### lichen1983312

Thanks very much for your explanation. Since there is no simple way to understand the strict proof, I will let it pass. After all, I am learning this part to study physics.