# Finding killing vector

I am introducing myself to the topic of killing vectors and therefore, after doing some reading, I try to solve some easy problems. For simplicity, I do my first steps in 2D.

First, I chose the 2D-Minkowski metric, so $$ds^2 = -dt^2+dx^2$$ holds. The Killing-equation $$\nabla_a \xi_b+ \nabla_b \xi_a = 0$$ simplifies to $$\partial_a\xi_b+\partial_b\xi_a =0$$. By differentiating this equation, permuting the indices and adding/subtracting the permuted equations I get a quite easy differential-equation. It's solution reads $$\xi_a = A_{ab}X^b+B v_a$$ where $$A \in \mathbb{R}^{2\times2}$$ and $$B \in \mathbb{R}^{2\times1}$$. Since the Killing-equation has to hold for this solution $$C$$ has to be anti-symmetric and thus we get the three well known Killing-vector fields $$\xi_0 = \partial_t$$, $$\xi_1 = \partial_x, \xi_2 = -\partial_t+\partial_x$$ for the Minkowski-Metric.

Although it wasn't too difficult for me to find the Killing-vectors for Minkowski-metric I struggle when it comes to more sophisticated metrics because I don't see a similar way to find a solution.

Therefore, I chose for my second toy-example the metric associated with $$ds^2 = -\frac{1}{x^2}dt^2+\frac{1}{x^2}dx^2$$. Since it is time-independent one Killing-vector, $$\xi_o = \partial_t$$ , can be read out easily. The metric is still diagonal but it depends on $$x$$ so the Killing-equation doesn't simplify as in the first example. So, things turn out to be much more difficult than in the previous toy-example.

Although I've already spent some time, I haven't figured out a way how to derive some useful equation that helps me finding Killing-vectors.

In the first example (Minkowski-metric) I started from the Killing-equation and by using some information about the given metric I derived all possible Killing-vectors. So the maximum number of linearly independent Killing-vectors was "naturally included" in the solution I derived. In the second problem, I even don't know anything about the number of Killing-vectors that I should expect (I've read in some textbook that $$\frac{n(n-1)}{2}$$ gives the number of Killing-vectors, but this was stated regarding to a specific class of metrics, so I'm not sure whether this is applicable here).

I tried to find the Killing-vectors as similar as possible to my first approach (maybe that's the problem?). Is there any general way how one can find Killing-vectors? Does anyone have an idea/a hint how I can find the Killing-vectors at least for this specific metric?

Thank's a lot for your help!

PS: My background is solid physics and maths knowledge but only basic knowledge in general-relativity (that's what I want to improve by dealing with this topic).

PPS: I already took a look at this post but since this question is even more general than mine here it didn't help me to proceed.

• I know this is an old question but I was also looking for something similar and someone had recommended sec 7.1 of this book
– M111
Jan 27, 2023 at 21:32

Quoting Sean Carroll's conclusion ("An Introduction to General Relativity - Spacetime and Geometry", Chapter 3, "Curvature", Section 3.8, "Symmetries and Killing vectors"),

The problem of finding all of the Killing vectors of a metric is therefore somewhat tricky, as it is not always clear when to stop looking.

That being said, I think some aspects deserve a little more clarification:

A. In principle, you can always find all Killing vectors for a given metric tensor $$g_{\mu\nu}$$ by solving the set of partial differential equations

$$\nabla_{\mu}\xi_{\nu} + \nabla_{\nu}\xi_{\mu} = 0.$$

Of course, this route is safe but lengthy if the number of spatial dimensions exceeds two.

B. Another route to obtain some Killing vectors is to look for the symmetries of your given metric tensor. As you correctly pointed out, if $$g_{\mu\nu}$$ is independent of some coordinate $$x^{\mu}$$ (as time, in both examples you wrote), then the associated basis vector is a Killing one. However, you must be careful with this path since some symmetries can be only spotted in certain coordinates.

You must also be aware, however, that the number of Killing vectors vary from metric to metric. Some are even said to have no Killing vectors at all!.

For the second metric you write, $$ds^{2} = (-dt^{2} + dx^{2})/x^{2}$$, let me go for the first route to obtain the Killing vectors:

1. Expand the Killing vector equation,

$$\partial_{\mu}\xi_{\nu} + \partial_{\nu}\xi_{\mu} - 2\Gamma^{\sigma}_{\ \mu\nu}\xi_{\sigma} = 0.$$

Here, I employed the definition of covariant derivative,

$$\nabla_{\mu}\xi_{\nu} = \partial_{\mu}\xi_{\nu} - \Gamma^{\sigma}_{\ \mu\nu}\xi_{\sigma},$$

with $$\Gamma^{\sigma}_{\ \mu\nu}$$ the Christoffel symbols, and used the fact $$\Gamma^{\sigma}_{\ \mu\nu} = \Gamma^{\sigma}_{\ \nu\mu}$$. Since your metric is 2D, you only have to solve three equations:

$$\partial_{t}\xi_{t} + \partial_{t}\xi_{t} - 2\Gamma^{\sigma}_{\ tt}\xi_{\sigma} = 0$$

$$\partial_{t}\xi_{x} + \partial_{x}\xi_{t} - 2\Gamma^{\sigma}_{\ tx}\xi_{\sigma} = 0$$

$$\partial_{x}\xi_{x} + \partial_{x}\xi_{x} - 2\Gamma^{\sigma}_{\ xx}\xi_{\sigma} = 0.$$

1. Compute the Christoffel symbols. In two dimensions, and due to their symmetry property in the lower indices, you have 6 independent symbols. However, from them the only nonzero are: $$\Gamma^{t}_{\ tx}$$ = $$\Gamma^{x}_{\ tt}$$ = $$\Gamma^{x}_{\ xx}$$ = $$-1/x$$.

2. Plug the Christoffel symbols into the equations to get

$$\partial_{t}\xi_{t} + \frac{1}{x}\xi_{x} = 0\ \ \ (1)$$

$$\partial_{t}\xi_{x} + \partial_{x}\xi_{t} + \frac{2}{x}\xi_{t} = 0\ \ \ (2)$$

$$\partial_{x}\xi_{x} + \frac{1}{x}\xi_{x} = 0\ \ \ (3)$$

1. Eq.(3) can also be expressed as $$\partial_{x}(x\xi_{x}) = 0$$, indicating the general solution is $$\xi_{x} = f_{1}(t)/x$$. Plugging this into (1) and (2) leads to

$$\partial_{t}\xi_{t} + \frac{f_{1}(t)}{x^{2}} = 0\ \ \ (1')$$

$$x\frac{df_{1}(t)}{dt} + \partial_{x}(x^{2}\xi_{t}) = 0.\ \ \ (2')$$

Eq. (1') suggests us to look for $$\xi_{t} = f_{2}(t)/x^{2}$$, which leads to a final system

$$\frac{df_{2}(t)}{dt} + f_{1}(t) = 0\ \ \ (1'')$$

$$\frac{df_{1}(t)}{dt} = 0\ \ \ (2'')$$

whose solution is $$f_{1}(t) = a_{1}$$, $$f_{2}(t) = -a_{1}t + a_{2}$$. Therefore, the general solution to the Killing equation for the given metric is

$$\left(\begin{array}{c}\xi_{t}\\ \xi_{x}\end{array}\right) = a_{1}\left(\begin{array}{c} -t/x^{2} \\ 1/x \end{array}\right) + a_{2}\left(\begin{array}{c} 1/x^{2} \\ 0 \end{array}\right),$$

i.e. the solution space is spanned by two Killing vectors. Notice the second one emerges from symmetry considerations since, denoting it as $$\xi^{(2)}_{\mu}$$, we have

$$\xi^{\mu}_{(2)} = g^{\mu\nu}\xi^{(2)}_{\nu} = (g^{tt}\xi^{(2)}_{t},0) = (-x^{2}\cdot x^{-2},0) = (-1,0)$$

i.e. the basis vector associated to the $$t$$ coordinate.

Carroll's textbook is excellent, so for a more profound discussion I suggest you to continue there. Hope this helps!