Brian Bi

## What is the significance of Killing vector/tensor fields?

I'm not proficient in GR, but nobody else wrote an answer to this question.

I assume proficiency in special relativity.

So the motivation goes something like this. In special relativity, space-time has the Poincaré symmetry group, which is a precise way of stating that it's homogeneous and isotropic. These symmetries are pretty simple, although of course their consequences are profound. For example, you will not hear people being confused much about what exactly translational symmetry is. It's very simple. You just pick some four-vector $$a^\mu$$ and then you have an infinitesimal symmetry operation given by $x^\mu \to x^\mu + \epsilon a^\mu$

The fact that these symmetries take such a simple "form" is attributable to our choice of an inertial frame. Consider what happens if, instead, I'm sitting at the origin rotating at a constant rate, so that my coordinate axes rotate relative to those of an inertial frame. If a free particle travels at a constant speed along the z-axis (which we'll take as my axis of rotation) in the inertial frame, I will also observe that particle to travel at a constant speed along my z-axis. In the inertial frame, performing a translation operation would yield a free particle travelling at a constant speed parallel to the z-axis, which is another perfectly valid trajectory for a free particle. But if I did the same in my frame, I would not get another valid free particle trajectory in general. For a straight line parallel to the z-axis in my frame would, in the inertial frame, be a circular helix with its axis coincident with the z-axis, which is obviously not the trajectory of a free particle. Conversely, the correct symmetry operation—which is a simple translation in the inertial frame—appears to produce a helical path in my frame. So symmetries have become harder to describe in my frame simply by the virtue of my choice of non-inertial coordinates.

Non-inertial frames are not very important in special relativity—they only arise because we choose our coordinates poorly. But in general relativity, they are very important, because once you have gravity, it's no longer possible to have an inertial frame. So if we want to study symmetries in general relativity, we must find some coordinate-independent (or "generally covariant") formalism for space-time symmetries. [1]

The way to do this is to represent the generator of each symmetry as a vector field.

Vector fields have an existence independent of the coordinates used to represent them. The symmetry operations themselves are obtained by taking events and moving them along the integral curves of the vector field.

The ten generators of the Poincaré group can be expressed as vector fields with the following coordinate representations in an inertial frame: \begin{align*} P_0 &= (1, 0, 0, 0) \\ P_1 &= (0, 1, 0, 0) \\ P_2 &= (0, 0, 1, 0) \\ P_3 &= (0, 0, 0, 1) \\ J^{01} &= (x^1, x^0, 0, 0) \\ J^{02} &= (x^2, 0, x^0, 0) \\ J^{03} &= (x^3, 0, 0, x^0) \\ J^{12} &= (0, -x^2, x^1, 0) \\ J^{23} &= (0, 0, -x^3, x^2) \\ J^{31} &= (0, x^3, 0, -x^1) \end{align*} (These are vector fields, so they can vary by position—in an inertial frame, the P's don't, but you can see that the J's do.)

In the specific case of translation in the x direction generated by $$P_1$$, we see for example that the integral curves are given by $x^\mu(\sigma) = x^\mu(0) + (0, \sigma, 0, 0)$ This is a finite [2] translation. To construct the same finite translation in the rotating frame, all we have to do is transform the vector field $$P_1$$ in the usual fashion. If at time t = 0 my axes coincide with those of the inertial frame and I rotate around the z-axis with angular velocity $$\omega$$, taking my coordinate axes with me, then \begin{align*} t' &= t \\ x' &= x \cos \omega t + y \sin \omega t \\ y' &= -x \sin \omega t + y \cos \omega t \\ z' &= z \end{align*} where $$(t', x', y', z')$$ are the coordinates of an event in my frame and $$(t, x, y, z)$$ are the coordinates in the inertial frame. Then we can compute the Jacobian, which I'll write in primed coordinates: \begin{bmatrix} 1 & 0 & 0 & 0 \\ \omega y' & \cos \omega t & \sin \omega t & 0 \\ -\omega x' & -\sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} This is the matrix we use to transform the coordinates of $$P_1$$, so we can see that the primed coordinates are simply the second column: $P_1 = (0, \cos \omega t, -\sin \omega t, 0)$ Now consider a particle located on the z axis at some time t, so that its coordinates are $x^{\mu'}(0) = (t, 0, 0, z)$ To apply the symmetry transformation generated by $$P_1$$ with finite parameter $$\sigma$$ we have to solve the initial value problem $\frac{\mathrm{d}x^{\mu'}(\sigma)}{\mathrm{d}\sigma} = (0, \cos \omega t, -\sin\omega t, 0)$ with $$x^{\mu'}(0) = (t, 0, 0, z)$$. It's easy to see that the solution is \begin{align*} t'(\sigma) &= t \\ x'(\sigma) &= \sigma \cos \omega t \\ y'(\sigma) &= -\sigma \sin \omega t \\ z'(\sigma) &= z \end{align*} so the trajectory along the z-axis given by \begin{align*} x'(t, \sigma = 0) &= 0 \\ y'(t, \sigma = 0) &= 0 \\ z'(t, \sigma = 0) &= z_0 + vt \end{align*} is simply taken to \begin{align*} x'(t, \sigma) &= \sigma \cos \omega t \\ y'(t, \sigma) &= \sigma \sin \omega t \\ z'(t, \sigma) &= z_0 + vt \end{align*} which shows that the new trajectory is a helix. We already knew that, of course, it was obvious—but the point of this example was to demonstrate how once the vector field $$P_1$$ is written in the rotating system, it tells us how to perform the symmetry transformation in that coordinate system! Doing so gave the correct answer: it mapped a valid trajectory along the z-axis into another valid trajectory, which is helical in our coordinate system (but straight in the inertial system).

These vector fields that generate symmetries are the Killing vector fields of special relativity. As I said before, in special relativity you can—and should—always choose to work in an inertial system, so Killing vector fields aren't important, and they don't show up explicitly. But in general relativity that luxury doesn't exist and you have to work with the Killing vector fields directly.

But space-time does not necessarily have the symmetry of the Poincaré group in general relativity; it isn't homogeneous and isotropic anymore, because we consider the metric to be part of the description of space-time itself, and the metric may vary with position, so space-time itself is different at different points. So if we want symmetries and conservation laws in the presence of gravity, we have to find vector fields such that when we flow all events along them, each neighbourhood is mapped onto another neighbourhood in such a way that the metric is the same between corresponding points. [3] Such fields are the Killing vector fields [5] It's clear that if the metric is arbitrary, then in general you can't expect there are any Killing vector fields at all. Thus, energy, momentum, and angular momentum are not conserved in general!

Mathematically, the condition that a Killing field must satisfy is that the Lie derivative of the metric along the field vanishes everywhere. This guarantees that when you follow the integral curves—or "Lie drag" points as one might say—a point is always mapped onto another point where the metric is the same, as zero derivative means no change [4]. That is $\mathcal{L}_X g = 0$ A version of Noether's theorem adapted to curved space-time then gives the following as a conserved current, $J^\alpha = T^{\alpha\beta} X_\beta$ Note that in an inertial frame, where each constant vector field is a Killing field, this just reduces to ordinary conservation of energy and momentum.

Again, these conservation laws can only be found when Killing vectors really do exist—we have ten in flat space but in curved space we're lucky to even have one. Because they can simplify the study of the system under consideration greatly, it's useful to know what they are.

For example, the Schwarzschild solution has four Killing vector fields: one timelike, since the solution is stationary, and three spacelike, since the solution is spherically symmetric. To the timelike Killing field we can associate a conserved energy; to the three spacelike Killing fields we also associate conserved quantities, and since they are associated with rotational symmetry we call them angular momenta. These conservation laws greatly simplify the analysis of the geodesics of the Schwarzschild spacetime; for example, the conservation of angular momentum implies here, as it does in classical mechanics, that the motion of a particle will lie in some plane; and because of the rotational symmetry, we can simply take that plane to be the xy plane without loss of generality.

By the way, note that since Killing vectors are the generators of continuous symmetry, they live in a Lie algebra. This implies two facts, one of which was probably already obvious, and one of which may not be. First, any linear combination of Killing vectors is a Killing vector, so the Poincaré Killing vectors previously given are part of a ten-dimensional vector space of Killing vectors for Minkowski space. Second, given two Killing vectors $$X$$ and $$Y$$, we can construct another Killing vector, $$[X, Y]$$ (the Lie bracket of $$X$$ and $$Y$$). This essentially describes flowing along $$X$$, then $$Y$$, then backward along $$X$$, then backward along $$Y$$. In some cases this takes you back to where you started, like if you took $$[P_0, P_1]$$, so the Lie bracket is 0, and you don't get a new Killing vector. In other cases the two vector fields fail to commute in this way, and get a "new" Killing vector, which is not a linear combination of the two original Killing vectors. For example, one can show that $$[J^{12}, J^{23}] = -J^{31}$$.

Bottom line: Killing vector fields generate symmetries in curved space-time. They are needed because symmetries are not easy to write down in curved space-time like they are in flat space-time, and their existence cannot be taken for granted either; where they do exist, they provide a way to access the insight that having symmetry brings to a problem.

[1] Gauge symmetries are a different story. In general relativity the basic formalism is unchanged from special relativity because the internal degrees of freedom on which gauge symmetries operate are independent of space-time.
[2] "Finite" is used here to mean "not infinitesimal", as is standard practice in physics.
[3] Note that we're not considering the metric itself as a dynamical field here: we're going to pick up and move all the particles and electromagnetic fields, but leave the metric where it is. If you're allowed to also pick up and move the metric like everything else, then you can choose any smooth vector field whatsoever as a generator of symmetry. This is the so-called diffeomorphism invariance of general relativity.
[4] This doesn't mean the components of the metric don't change. If you take the metric at one point and you drag it to another point, its new coordinates will be expressed in terms of the local coordinate basis at the new point. So when calculating the Lie derivative explicitly in coordinates there are additional terms to "correct" for this.
[5] Sometimes instead of saying "Killing vector field" one might simply say "Killing vector", but you should read this as "Killing vector field" when this is required by the context.