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With an understanding of how the laws of physics adapt to curved spacetime, it is undeniably tempting to start in on applications. However, a few extra mathematical techniques will simplify our task a great deal, so we will pause briefly to explore the geometry of manifolds some more.

When we discussed manifolds in section 2, we introduced maps between two different manifolds and how maps could be composed. We now turn to the use of such maps in carrying along tensor fields from one manifold to another. We therefore consider two manifolds M and N, possibly of different dimension, with coordinate systems x$\scriptstyle \mu$ and y$\scriptstyle \alpha$, respectively. We imagine that we have a map $ \phi$ : M $ \rightarrow$ N and a function f : N $ \rightarrow$ $ \bf R$.

Figure 5.1

It is obvious that we can compose $ \phi$ with f to construct a map (fo$ \phi$) : M $ \rightarrow$ $ \bf R$, which is simply a function on M. Such a construction is sufficiently useful that it gets its own name; we define the pullback of f by $ \phi$, denoted $ \phi_{*}^{}$f, by

Equation 5.1 (5.1)

The name makes sense, since we think of $ \phi_{*}^{}$ as "pulling back" the function f from N to M.

We can pull functions back, but we cannot push them forward. If we have a function g : M $ \rightarrow$ $ \bf R$, there is no way we can compose g with $ \phi$ to create a function on N; the arrows don't fit together correctly. But recall that a vector can be thought of as a derivative operator that maps smooth functions to real numbers. This allows us to define the pushforward of a vector; if V(p) is a vector at a point p on M, we define the pushforward vector $ \phi^{*}_{}$V at the point $ \phi$(p) on N by giving its action on functions on N:

Equation 5.2 (5.2)

So to push forward a vector field we say "the action of $ \phi^{*}_{}$V on any function is simply the action of V on the pullback of that function."

This is a little abstract, and it would be nice to have a more concrete description. We know that a basis for vectors on M is given by the set of partial derivatives $ \partial_{\mu}$ = $ {{\partial}\over{\partial x^\mu}}$, and a basis on N is given by the set of partial derivatives $ \partial_{\alpha}$ = $ {{\partial}\over{\partial y^\alpha}}$. Therefore we would like to relate the components of V = V$\scriptstyle \mu$$ \partial_{\mu}$ to those of ($ \phi^{*}_{}$V) = ($ \phi^{*}_{}$V)$\scriptstyle \alpha$$ \partial_{\alpha}$. We can find the sought-after relation by applying the pushed-forward vector to a test function and using the chain rule (2.3):

Equation 5.3 (5.3)

This simple formula makes it irresistible to think of the pushforward operation $ \phi^{*}_{}$ as a matrix operator, ($ \phi^{*}_{}$V)$\scriptstyle \alpha$ = ($ \phi^{*}_{}$)$\scriptstyle \alpha$$\scriptstyle \mu$V$\scriptstyle \mu$, with the matrix being given by

Equation 5.4 (5.4)

The behavior of a vector under a pushforward thus bears an unmistakable resemblance to the vector transformation law under change of coordinates. In fact it is a generalization, since when M and N are the same manifold the constructions are (as we shall discuss) identical; but don't be fooled, since in general $ \mu$ and $ \alpha$ have different allowed values, and there is no reason for the matrix $ \partial$y$\scriptstyle \alpha$/$ \partial$x$\scriptstyle \mu$ to be invertible.

It is a rewarding exercise to convince yourself that, although you can push vectors forward from M to N (given a map $ \phi$ : M $ \rightarrow$ N), you cannot in general pull them back - just keep trying to invent an appropriate construction until the futility of the attempt becomes clear. Since one-forms are dual to vectors, you should not be surprised to hear that one-forms can be pulled back (but not in general pushed forward). To do this, remember that one-forms are linear maps from vectors to the real numbers. The pullback $ \phi_{*}^{}$$ \omega$ of a one-form $ \omega$ on N can therefore be defined by its action on a vector V on M, by equating it with the action of $ \omega$ itself on the pushforward of V:

Equation 5.5 (5.5)

Once again, there is a simple matrix description of the pullback operator on forms, ($ \phi_{*}^{}$$ \omega$)$\scriptstyle \mu$ = ($ \phi_{*}^{}$)$\scriptstyle \mu$$\scriptstyle \alpha$$ \omega_{\alpha}^{}$, which we can derive using the chain rule. It is given by

Equation 5.6 (5.6)

That is, it is the same matrix as the pushforward (5.4), but of course a different index is contracted when the matrix acts to pull back one-forms.

There is a way of thinking about why pullbacks and pushforwards work on some objects but not others, which may or may not be helpful. If we denote the set of smooth functions on M by $ \cal {F}$(M), then a vector V(p) at a point p on M (i.e., an element of the tangent space TpM) can be thought of as an operator from $ \cal {F}$(M) to $ \bf R$. But we already know that the pullback operator on functions maps $ \cal {F}$(N) to $ \cal {F}$(M) (just as $ \phi$ itself maps M to N, but in the opposite direction). Therefore we can define the pushforward $ \phi_{*}^{}$ acting on vectors simply by composing maps, as we first defined the pullback of functions:

Figure 5.2

Similarly, if TqN is the tangent space at a point q on N, then a one-form $ \omega$ at q (i.e., an element of the cotangent space Tq*N) can be thought of as an operator from TqN to $ \bf R$. Since the pushforward $ \phi^{*}_{}$ maps TpM to T$\scriptstyle \phi$(p)N, the pullback $ \phi_{*}^{}$ of a one-form can also be thought of as mere composition of maps:

Figure 5.3

If this is not helpful, don't worry about it. But do keep straight what exists and what doesn't; the actual concepts are simple, it's just remembering which map goes what way that leads to confusion.

You will recall further that a (0, l ) tensor - one with l lower indices and no upper ones - is a linear map from the direct product of l vectors to $ \bf R$. We can therefore pull back not only one-forms, but tensors with an arbitrary number of lower indices. The definition is simply the action of the original tensor on the pushed-forward vectors:

Equation 5.7 (5.7)

where T$\scriptstyle \alpha_{1}$ ... $\scriptstyle \alpha_{l}$ is a (0, l ) tensor on N. We can similarly push forward any (k, 0) tensor S$\scriptstyle \mu_{1}$ ... $\scriptstyle \mu_{k}$ by acting it on pulled-back one-forms:

Equation 5.8 (5.8)

Fortunately, the matrix representations of the pushforward (5.4) and pullback (5.6) extend to the higher-rank tensors simply by assigning one matrix to each index; thus, for the pullback of a (0, l ) tensor, we have

Equation 5.9 (5.9)

while for the pushforward of a (k, 0) tensor we have

Equation 5.10 (5.10)

Our complete picture is therefore:

Figure 5.4

Note that tensors with both upper and lower indices can generally be neither pushed forward nor pulled back.

This machinery becomes somewhat less imposing once we see it at work in a simple example. One common occurrence of a map between two manifolds is when M is actually a submanifold of N; then there is an obvious map from M to N which just takes an element of M to the "same" element of N. Consider our usual example, the two-sphere embedded in $ \bf R^{3}_{}$, as the locus of points a unit distance from the origin. If we put coordinates x$\scriptstyle \mu$ = ($ \theta$,$ \phi$) on M = S2 and y$\scriptstyle \alpha$ = (x, y, z) on N = $ \bf R^{3}_{}$, the map $ \phi$ : M $ \rightarrow$ N is given by

Equation 5.11 (5.11)

In the past we have considered the metric ds2 = dx2 + dy2 + dz2 on $ \bf R^{3}_{}$, and said that it induces a metric d$ \theta^{2}_{}$ + sin2$ \theta$ d$ \phi^{2}_{}$ on S2, just by substituting (5.11) into this flat metric on $ \bf R^{3}_{}$. We didn't really justify such a statement at the time, but now we can do so. (Of course it would be easier if we worked in spherical coordinates on $ \bf R^{3}_{}$, but doing it the hard way is more illustrative.) The matrix of partial derivatives is given by

Equation 5.12 (5.12)

The metric on S2 is obtained by simply pulling back the metric from $ \bf R^{3}_{}$,

Equation 5.13 (5.13)

as you can easily check. Once again, the answer is the same as you would get by naive substitution, but now we know why.

We have been careful to emphasize that a map $ \phi$ : M $ \rightarrow$ N can be used to push certain things forward and pull other things back. The reason why it generally doesn't work both ways can be traced to the fact that $ \phi$ might not be invertible. If $ \phi$ is invertible (and both $ \phi$ and $ \phi^{-1}_{}$ are smooth, which we always implicitly assume), then it defines a diffeomorphism between M and N. In this case M and N are the same abstract manifold. The beauty of diffeomorphisms is that we can use both $ \phi$ and $ \phi^{-1}_{}$ to move tensors from M to N; this will allow us to define the pushforward and pullback of arbitrary tensors. Specifically, for a (k, l ) tensor field T$\scriptstyle \mu_{1}$ ... $\scriptstyle \mu_{k}$$\scriptstyle \nu_{1}$ ... $\scriptstyle \mu_{l}$ on M, we define the pushforward by

Equation 5.14 (5.14)

where the $ \omega^{(i)}_{}$'s are one-forms on N and the V(i)'s are vectors on N. In components this becomes

Equation 5.15 (5.15)

The appearance of the inverse matrix $ \partial$x$\scriptstyle \nu$/$ \partial$y$\scriptstyle \beta$ is legitimate because $ \phi$ is invertible. Note that we could also define the pullback in the obvious way, but there is no need to write separate equations because the pullback $ \phi_{*}^{}$ is the same as the pushforward via the inverse map, [$ \phi^{-1}_{}$]*.

We are now in a position to explain the relationship between diffeomorphisms and coordinate transformations. The relationship is that they are two different ways of doing precisely the same thing. If you like, diffeomorphisms are "active coordinate transformations", while traditional coordinate transformations are "passive." Consider an n-dimensional manifold M with coordinate functions x$\scriptstyle \mu$ : M $ \rightarrow$ $ \bf R^{n}_{}$. To change coordinates we can either simply introduce new functions y$\scriptstyle \mu$ : M $ \rightarrow$ $ \bf R^{n}_{}$ ("keep the manifold fixed, change the coordinate maps"), or we could just as well introduce a diffeomorphism $ \phi$ : M $ \rightarrow$ M, after which the coordinates would just be the pullbacks ($ \phi_{*}^{}$x)$\scriptstyle \mu$ : M $ \rightarrow$ $ \bf R^{n}_{}$ ("move the points on the manifold, and then evaluate the coordinates of the new points"). In this sense, (5.15) really is the tensor transformation law, just thought of from a different point of view.

Figure 5.5

Since a diffeomorphism allows us to pull back and push forward arbitrary tensors, it provides another way of comparing tensors at different points on a manifold. Given a diffeomorphism $ \phi$ : M $ \rightarrow$ M and a tensor field T$\scriptstyle \mu_{1}$ ... $\scriptstyle \mu_{k}$$\scriptstyle \nu_{1}$ ... $\scriptstyle \mu_{l}$(x), we can form the difference between the value of the tensor at some point p and $ \phi_{*}^{}$[T$\scriptstyle \mu_{1}$ ... $\scriptstyle \mu_{k}$$\scriptstyle \nu_{1}$ ... $\scriptstyle \mu_{l}$($ \phi$(p))], its value at $ \phi$(p) pulled back to p. This suggests that we could define another kind of derivative operator on tensor fields, one which categorizes the rate of change of the tensor as it changes under the diffeomorphism. For that, however, a single discrete diffeomorphism is insufficient; we require a one-parameter family of diffeomorphisms, $ \phi_{t}^{}$. This family can be thought of as a smooth map $ \bf R$ x M $ \rightarrow$ M, such that for each t $ \in$ $ \bf R$ $ \phi_{t}^{}$ is a diffeomorphism and $ \phi_{s}^{}$o$ \phi_{t}^{}$ = $ \phi_{s+t}^{}$. Note that this last condition implies that $ \phi_{0}^{}$ is the identity map.

One-parameter families of diffeomorphisms can be thought of as arising from vector fields (and vice-versa). If we consider what happens to the point p under the entire family $ \phi_{t}^{}$, it is clear that it describes a curve in M; since the same thing will be true of every point on M, these curves fill the manifold (although there can be degeneracies where the diffeomorphisms have fixed points). We can define a vector field V$\scriptstyle \mu$(x) to be the set of tangent vectors to each of these curves at every point, evaluated at t = 0. An example on S2 is provided by the diffeomorphism $ \phi_{t}^{}$($ \theta$,$ \phi$) = ($ \theta$,$ \phi$ + t).

Figure 5.6

We can reverse the construction to define a one-parameter family of diffeomorphisms from any vector field. Given a vector field V$\scriptstyle \mu$(x), we define the integral curves of the vector field to be those curves x$\scriptstyle \mu$(t) which solve

Equation 5.16 (5.16)

Note that this familiar-looking equation is now to be interpreted in the opposite sense from our usual way - we are given the vectors, from which we define the curves. Solutions to (5.16) are guaranteed to exist as long as we don't do anything silly like run into the edge of our manifold; any standard differential geometry text will have the proof, which amounts to finding a clever coordinate system in which the problem reduces to the fundamental theorem of ordinary differential equations. Our diffeomorphisms $ \phi_{t}^{}$ represent "flow down the integral curves," and the associated vector field is referred to as the generator of the diffeomorphism. (Integral curves are used all the time in elementary physics, just not given the name. The "lines of magnetic flux" traced out by iron filings in the presence of a magnet are simply the integral curves of the magnetic field vector B.)

Given a vector field V$\scriptstyle \mu$(x), then, we have a family of diffeomorphisms parameterized by t, and we can ask how fast a tensor changes as we travel down the integral curves. For each t we can define this change as

Equation 5.17 (5.17)

Note that both terms on the right hand side are tensors at p.

Figure 5.7

We then define the Lie derivative of the tensor along the vector field as

Equation 5.18 (5.18)

The Lie derivative is a map from (k, l ) tensor fields to (k, l ) tensor fields, which is manifestly independent of coordinates. Since the definition essentially amounts to the conventional definition of an ordinary derivative applied to the component functions of the tensor, it should be clear that it is linear,

Equation 5.19 (5.19)

and obeys the Leibniz rule,

Equation 5.20 (5.20)

where S and T are tensors and a and b are constants. The Lie derivative is in fact a more primitive notion than the covariant derivative, since it does not require specification of a connection (although it does require a vector field, of course). A moment's reflection shows that it reduces to the ordinary derivative on functions,

Equation 5.21 (5.21)

To discuss the action of the Lie derivative on tensors in terms of other operations we know, it is convenient to choose a coordinate system adapted to our problem. Specifically, we will work in coordinates x$\scriptstyle \mu$ for which x1 is the parameter along the integral curves (and the other coordinates are chosen any way we like). Then the vector field takes the form V = $ \partial$/$ \partial$x1; that is, it has components V$\scriptstyle \mu$ = (1, 0, 0,..., 0). The magic of this coordinate system is that a diffeomorphism by t amounts to a coordinate transformation from x$\scriptstyle \mu$ to y$\scriptstyle \mu$ = (x1 + t, x2,..., xn). Thus, from (5.6) the pullback matrix is simply

Equation 5.22 (5.22)

and the components of the tensor pulled back from $ \phi_{t}^{}$(p) to p are simply

Equation 5.23 (5.23)

In this coordinate system, then, the Lie derivative becomes

Equation 5.24 (5.24)

and specifically the derivative of a vector field U$\scriptstyle \mu$(x) is

Equation 5.25 (5.25)

Although this expression is clearly not covariant, we know that the commutator [V, U] is a well-defined tensor, and in this coordinate system

Equation 5.26 (5.26)

Therefore the Lie derivative of U with respect to V has the same components in this coordinate system as the commutator of V and U; but since both are vectors, they must be equal in any coordinate system:

Equation 5.27 (5.27)

As an immediate consequence, we have £VS = - £WV. It is because of (5.27) that the commutator is sometimes called the "Lie bracket."

To derive the action of £V on a one-form $ \omega_{\mu}^{}$, begin by considering the action on the scalar $ \omega_{\mu}^{}$U$\scriptstyle \mu$ for an arbitrary vector field U$\scriptstyle \mu$. First use the fact that the Lie derivative with respect to a vector field reduces to the action of the vector itself when applied to a scalar:

Equation 5.28 (5.28)

Then use the Leibniz rule on the original scalar:

Equation 5.29 (5.29)

Setting these expressions equal to each other and requiring that equality hold for arbitrary U$\scriptstyle \mu$, we see that

Equation 5.30 (5.30)

which (like the definition of the commutator) is completely covariant, although not manifestly so.

By a similar procedure we can define the Lie derivative of an arbitrary tensor field. The answer can be written

Equation 5.31 (5.31)

Once again, this expression is covariant, despite appearances. It would undoubtedly be comforting, however, to have an equivalent expression that looked manifestly tensorial. In fact it turns out that we can write

Equation 5.32 (5.32)

where $ \nabla_{\mu}^{}$ represents any symmetric (torsion-free) covariant derivative (including, of course, one derived from a metric). You can check that all of the terms which would involve connection coefficients if we were to expand (5.32) would cancel, leaving only (5.31). Both versions of the formula for a Lie derivative are useful at different times. A particularly useful formula is for the Lie derivative of the metric:

Equation 5.33 (5.33)

where $ \nabla_{\mu}^{}$ is the covariant derivative derived from g$\scriptstyle \mu$$\scriptstyle \nu$.

Let's put some of these ideas into the context of general relativity. You will often hear it proclaimed that GR is a "diffeomorphism invariant" theory. What this means is that, if the universe is represented by a manifold M with metric g$\scriptstyle \mu$$\scriptstyle \nu$ and matter fields $ \psi$, and $ \phi$ : M $ \rightarrow$ M is a diffeomorphism, then the sets (M, g$\scriptstyle \mu$$\scriptstyle \nu$,$ \psi$) and (M,$ \phi_{*}^{}$g$\scriptstyle \mu$$\scriptstyle \nu$,$ \phi_{*}^{}$$ \psi$) represent the same physical situation. Since diffeomorphisms are just active coordinate transformations, this is a highbrow way of saying that the theory is coordinate invariant. Although such a statement is true, it is a source of great misunderstanding, for the simple fact that it conveys very little information. Any semi-respectable theory of physics is coordinate invariant, including those based on special relativity or Newtonian mechanics; GR is not unique in this regard. When people say that GR is diffeomorphism invariant, more likely than not they have one of two (closely related) concepts in mind: the theory is free of "prior geometry", and there is no preferred coordinate system for spacetime. The first of these stems from the fact that the metric is a dynamical variable, and along with it the connection and volume element and so forth. Nothing is given to us ahead of time, unlike in classical mechanics or SR. As a consequence, there is no way to simplify life by sticking to a specific coordinate system adapted to some absolute elements of the geometry. This state of affairs forces us to be very careful; it is possible that two purportedly distinct configurations (of matter and metric) in GR are actually "the same", related by a diffeomorphism. In a path integral approach to quantum gravity, where we would like to sum over all possible configurations, special care must be taken not to overcount by allowing physically indistinguishable configurations to contribute more than once. In SR or Newtonian mechanics, meanwhile, the existence of a preferred set of coordinates saves us from such ambiguities. The fact that GR has no preferred coordinate system is often garbled into the statement that it is coordinate invariant (or "generally covariant"); both things are true, but one has more content than the other.

On the other hand, the fact of diffeomorphism invariance can be put to good use. Recall that the complete action for gravity coupled to a set of matter fields $ \psi^{i}_{}$ is given by a sum of the Hilbert action for GR plus the matter action,

Equation 5.34 (5.34)

The Hilbert action SH is diffeomorphism invariant when considered in isolation, so the matter action SM must also be if the action as a whole is to be invariant. We can write the variation in SM under a diffeomorphism as

Equation 5.35 (5.35)

We are not considering arbitrary variations of the fields, only those which result from a diffeomorphism. Nevertheless, the matter equations of motion tell us that the variation of SM with respect to $ \psi^{i}_{}$ will vanish for any variation (since the gravitational part of the action doesn't involve the matter fields). Hence, for a diffeomorphism invariant theory the first term on the right hand side of (5.35) must vanish. If the diffeomorphism in generated by a vector field V$\scriptstyle \mu$(x), the infinitesimal change in the metric is simply given by its Lie derivative along V$\scriptstyle \mu$; by (5.33) we have

Equation 5.36 (5.36)

Setting $ \delta$SM = 0 then implies

Equation 5.37 (5.37)

where we are able to drop the symmetrization of $ \nabla_{(\mu}^{}$V$\scriptstyle \nu$) since $ \delta$SM/$ \delta$g$\scriptstyle \mu$$\scriptstyle \nu$ is already symmetric. Demanding that (5.37) hold for diffeomorphisms generated by arbitrary vector fields V$\scriptstyle \mu$, and using the definition (4.70) of the energy-momentum tensor, we obtain precisely the law of energy-momentum conservation,

Equation 5.38 (5.38)

This is why we claimed earlier that the conservation of T$\scriptstyle \mu$$\scriptstyle \nu$ was more than simply a consequence of the Principle of Equivalence; it is much more secure than that, resting only on the diffeomorphism invariance of the theory.

There is one more use to which we will put the machinery we have set up in this section: symmetries of tensors. We say that a diffeomorphism $ \phi$ is a symmetry of some tensor T if the tensor is invariant after being pulled back under $ \phi$:

Equation 5.39 (5.39)

Although symmetries may be discrete, it is more common to have a one-parameter family of symmetries $ \phi_{t}^{}$. If the family is generated by a vector field V$\scriptstyle \mu$(x), then (5.39) amounts to

Equation 5.40 (5.40)

By (5.25), one implication of a symmetry is that, if T is symmetric under some one-parameter family of diffeomorphisms, we can always find a coordinate system in which the components of T are all independent of one of the coordinates (the integral curve coordinate of the vector field). The converse is also true; if all of the components are independent of one of the coordinates, then the partial derivative vector field associated with that coordinate generates a symmetry of the tensor.

The most important symmetries are those of the metric, for which $ \phi_{*}^{}$g$\scriptstyle \mu$$\scriptstyle \nu$ = g$\scriptstyle \mu$$\scriptstyle \nu$. A diffeomorphism of this type is called an isometry. If a one-parameter family of isometries is generated by a vector field V$\scriptstyle \mu$(x), then V$\scriptstyle \mu$ is known as a Killing vector field. The condition that V$\scriptstyle \mu$ be a Killing vector is thus

Equation 5.41 (5.41)

or from (5.33),

Equation 5.42 (5.42)

This last version is Killing's equation. If a spacetime has a Killing vector, then we know we can find a coordinate system in which the metric is independent of one of the coordinates.

By far the most useful fact about Killing vectors is that Killing vectors imply conserved quantities associated with the motion of free particles. If x$\scriptstyle \mu$($ \lambda$) is a geodesic with tangent vector U$\scriptstyle \mu$ = dx$\scriptstyle \mu$/d$ \lambda$, and V$\scriptstyle \mu$ is a Killing vector, then

Equation 5.43 (5.43)

where the first term vanishes from Killing's equation and the second from the fact that x$\scriptstyle \mu$($ \lambda$) is a geodesic. Thus, the quantity V$\scriptstyle \mu$U$\scriptstyle \mu$ is conserved along the particle's worldline. This can be understood physically: by definition the metric is unchanging along the direction of the Killing vector. Loosely speaking, therefore, a free particle will not feel any "forces" in this direction, and the component of its momentum in that direction will consequently be conserved.

Long ago we referred to the concept of a space with maximal symmetry, without offering a rigorous definition. The rigorous definition is that a maximally symmetric space is one which possesses the largest possible number of Killing vectors, which on an n-dimensional manifold is n(n + 1)/2. We will not prove this statement, but it is easy to understand at an informal level. Consider the Euclidean space $ \bf R^{n}_{}$, where the isometries are well known to us: translations and rotations. In general there will be n translations, one for each direction we can move. There will also be n(n - 1)/2 rotations; for each of n dimensions there are n - 1 directions in which we can rotate it, but we must divide by two to prevent overcounting (rotating x into y and rotating y into x are two versions of the same thing). We therefore have

Equation 5.44 (5.44)

independent Killing vectors. The same kind of counting argument applies to maximally symmetric spaces with curvature (such as spheres) or a non-Euclidean signature (such as Minkowski space), although the details are marginally different.

Although it may or may not be simple to actually solve Killing's equation in any given spacetime, it is frequently possible to write down some Killing vectors by inspection. (Of course a "generic" metric has no Killing vectors at all, but to keep things simple we often deal with metrics with high degrees of symmetry.) For example in $ \bf R^{2}_{}$ with metric ds2 = dx2 + dy2, independence of the metric components with respect to x and y immediately yields two Killing vectors:

Equation 5.45 (5.45)

These clearly represent the two translations. The one rotation would correspond to the vector R = $ \partial$/$ \partial$$ \theta$ if we were in polar coordinates; in Cartesian coordinates this becomes

Equation 5.46 (5.46)

You can check for yourself that this actually does solve Killing's equation.

Note that in n $ \geq$ 2 dimensions, there can be more Killing vectors than dimensions. This is because a set of Killing vector fields can be linearly independent, even though at any one point on the manifold the vectors at that point are linearly dependent. It is trivial to show (so you should do it yourself) that a linear combination of Killing vectors with constant coefficients is still a Killing vector (in which case the linear combination does not count as an independent Killing vector), but this is not necessarily true with coefficients which vary over the manifold. You will also show that the commutator of two Killing vector fields is a Killing vector field; this is very useful to know, but it may be the case that the commutator gives you a vector field which is not linearly independent (or it may simply vanish). The problem of finding all of the Killing vectors of a metric is therefore somewhat tricky, as it is sometimes not clear when to stop looking.

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