My question is related to this question. There are three or four other questions on Killing Vector Fields here, however none of them that I have seen address my question.


I've been studying some Differential Geometry, and have been thinking about the Killing Vector Fields.


In Stan Liou's answer he mentions cyclic coordinates. In the geodesic equation

$$ \ddot{y} + \Gamma^y_{xx}\dot{x}\dot{x} + \Gamma^y_{xy}\dot{x}\dot{y} + \Gamma^y_{yx}\dot{y}\dot{x} +\Gamma^y_{yy}\dot{y}\dot{y} = 0 $$

we see that x is a cyclic cooordinate. Moreover, he mentions Killing Vector Fields.


I am fimilar with the concept of cyclic coordinates giving an integral of the motion, as discussed in Landau Vol. 1, for example.

Here, for $q_i$ a cyclic coordinate the Euler-Lagrange equation for $q_i$

$$ \frac{d}{dt} \left( \frac{ \partial L}{ \partial \dot{q}_i} \right) - \frac{ \partial L}{ \partial q_i} = 0 $$

reduces to

$$ \frac{d}{dt} \left( \frac{ \partial L}{ \partial \dot{q}_i} \right) = 0 $$


$$ \frac{ \partial L}{ \partial \dot{q}_i} = E_i $$

say, is an integral of the motion.


Moreover, we know that a Killing Vector Field $K$ is an isometry of the metric tensor $g$ such that

$$ \mathcal{L}_{\small K} g = 0 $$

That is, $K$ is a symmetry of the metric tensor $g$. So for diffeomorphisms $\phi : M \rightarrow M$ which 'move us along the integral curves of $K$' (I don't know the best way to phrase this!) the metric tensor $g$ will remain the same.


However, when we write out the geodesic equations, are these two things going to be the same in some way, such that we we find

$$ K \sim q $$

for $K$ a Killing Vector and $q$ a cyclic coordinate?

It seems like in the cyclic coordinate case we have a hypersurface $\Sigma \subset \mathbb{R}^4$ with Cartesian coordinates $x^{i} = (x,y,z)$ with $t$ our 'extra' coordinate that we 'move along' and the integral of the motion stays the same, where $(M,g)$ here is our Lorentzian manifold.


Furthermore, I am quite fimilar with the Killing Form in the Theory of Lie Groups, as a symmetric bilinear form give by

$$ K(X,Y) = \mbox{tr}(\mbox{ad}_X \mbox{ad}_Y) $$

where $X, Y$ $\in \mathfrak{g}$ for some Lie algebra $g$, with $\mbox{ad}_X$ the adjoint representation of $X \in \mathfrak{g}$. Then this Killing form is bi-invariant under the action of the Lie Group G, and has many other nice properties like non-singularity and being negative-definite for semi-simple compact groups.


We can also use the Killing Form to define a metric on the underlying manifold of our Lie Group $G$, so part of me feels that these ideas are connected, but thus far I cannot merge them together in my head.


So in brief, my question is, are Killing Vector Fields 'simply' cyclic coordinates? (I use simply here loosely) If not, what exactly is the difference?



  • $\begingroup$ I realise that this question got a bit on the lengthy side, but I just wanted to show that I have tried to think about it before asking. Thanks! $\endgroup$
    – Flint72
    May 17, 2014 at 21:27

1 Answer 1


By the straightening lemma it is always possible to find a chart $(x, y, \ldots)$ such that the Killing vector is given by $$K = \partial/\partial x.$$ This is called straightening the vector field since in this coordinate frame, the field appears straight.

Now using the formula for the Lie derivative of a tensor you see that for $K = \partial/\partial x$ we have $$(\mathcal L_K g)_{\mu\nu} = \frac{\partial g_{\mu\nu}}{\partial x}$$ so that $K$ is a Killing vector if and only if $K = \partial /\partial x$ in some coordinates, such that $x$ is cyclic.

The obstruction to straightening several vector fields is the Lie bracket. It is possible to straighten $K_1$ and $K_2$ simultaneously if and only iff $$[K_1, K_2] = 0.$$ You can realize that this is an obstruction since the Lie bracket is coordinate independent, and so if $K_1$ and $K_2$ have expressions of the form $$K_i = \partial/\partial x_i$$ their bracket must vanish in any coordinates. It is less trivial that this is the only obstruction.

To illustrate with an example in the Schwarzschild metric, $t$ and $\varphi$ are both cyclic coordinates. The Killing vector $\partial/\partial \varphi$ corresponds to rotations around some axis. There are two other (independent) rotations and you can find expressions for the Killing vector fields associated with them, however they will not commute with $\partial/\partial \varphi$, and so one cannot write the Schwarzschild metric with three (or four) cyclic coordinates. This shows that the set of Killing vectors contains more information than the set of cyclic coordinates when the isometry group is not Abelian.

As for your phrasing question, I would phrase

diffeomorphisms $\varphi : M \to M$ which 'move us along the integral curves of $K$'.


the one-parameter subgroup of the isometry group, corresponding to the flow of the vector field $K$.

You could also say

the isometries generated by $K$


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