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Killing Vectors

A Killing vector field (or simply, Killing vector) is a vector field that leaves the metric invariant under a change of coordinates. If the Lie derivative of the metric tensor with respect to a vector field vanishes, the vector field is a Killing vector field:

Lugμν=∂gμν∂xσuσ+∂uα∂xμgαν+∂uβ∂xνgμβ=0.\mathcal{L}_{\boldsymbol{u}} g_{\mu \nu} = \frac{\partial g_{\mu \nu}}{\partial x^{\sigma}} u^{\sigma} + \frac{\partial u^{\alpha}}{\partial x^{\mu}} g_{\alpha \nu} + \frac{\partial u^{\beta}}{\partial x^{\nu}} g_{\mu \beta} = 0.

In terms of the Levi-Civita connection, the following may be obtained:

Lugμν=∂gμν∂xσuσ+∂uα∂xμgαν+∂uβ∂xνgμβ=∂gμν∂xσuσ+∂uα∂xμgαν+uα∂gαν∂xμ−uα∂gαν∂xμ+∂uβ∂xνgμβ+uβ∂gμβ∂xν−uβ∂gμβ∂xν=∂gμν∂xσuσ+∂∂xμ(uαgαν)−uα∂gαν∂xμ+∂∂xν(uβgμβ)−uβ∂gμβ∂xν=∂gμν∂xσuσ+∂uν∂xμ−uα∂gαν∂xμ+∂uμ∂xν−uβ∂gμβ∂xν=∂uν∂xμ+∂uμ∂xν+uσ∂gμν∂xσ−uσ∂gσν∂xμ−uσ∂gμσ∂xν=∂uν∂xμ+∂uμ∂xν−uσ(∂gσν∂xμ+∂gμσ∂xν−∂gμν∂xσ)=∂uν∂xμ+∂uμ∂xν−2uλ12gσλ(∂gσμ∂xν+∂gσν∂xμ−∂gμν∂xσ)=∂uν∂xμ+∂uμ∂xν−2uλΓλμν=∂uν∂xμ−uλΓλμν+∂uμ∂xν−uλΓλνμ=∇μuν+∇νuμ=0, \begin{align*} \mathcal{L}_{\boldsymbol{u}} g_{\mu \nu} &= \frac{\partial g_{\mu \nu}}{\partial x^{\sigma}} u^{\sigma} + \frac{\partial u^{\alpha}}{\partial x^{\mu}} g_{\alpha \nu} + \frac{\partial u^{\beta}}{\partial x^{\nu}} g_{\mu \beta} \\ &= \frac{\partial g_{\mu \nu}}{\partial x^{\sigma}} u^{\sigma} + \frac{\partial u^{\alpha}}{\partial x^{\mu}} g_{\alpha \nu} + u^{\alpha} \frac{\partial g_{\alpha \nu}}{\partial x^{\mu}} - u^{\alpha} \frac{\partial g_{\alpha \nu}}{\partial x^{\mu}} + \frac{\partial u^{\beta}}{\partial x^{\nu}} g_{\mu \beta} + u^{\beta} \frac{\partial g_{\mu \beta}}{\partial x^{\nu}} - u^{\beta} \frac{\partial g_{\mu \beta}}{\partial x^{\nu}} \\ &= \frac{\partial g_{\mu \nu}}{\partial x^{\sigma}} u^{\sigma} + \frac{\partial}{\partial x^{\mu}} (u^{\alpha} g_{\alpha \nu}) - u^{\alpha} \frac{\partial g_{\alpha \nu}}{\partial x^{\mu}} + \frac{\partial}{\partial x^{\nu}} (u^{\beta} g_{\mu \beta}) - u^{\beta} \frac{\partial g_{\mu \beta}}{\partial x^{\nu}} \\ &= \frac{\partial g_{\mu \nu}}{\partial x^{\sigma}} u^{\sigma} + \frac{\partial u_{\nu}}{\partial x^{\mu}} - u^{\alpha} \frac{\partial g_{\alpha \nu}}{\partial x^{\mu}} + \frac{\partial u_{\mu}}{\partial x^{\nu}} - u^{\beta} \frac{\partial g_{\mu \beta}}{\partial x^{\nu}} \\ &= \frac{\partial u_{\nu}}{\partial x^{\mu}} + \frac{\partial u_{\mu}}{\partial x^{\nu}} + u^{\sigma} \frac{\partial g_{\mu \nu}}{\partial x^{\sigma}} - u^{\sigma} \frac{\partial g_{\sigma \nu}}{\partial x^{\mu}} - u^{\sigma} \frac{\partial g_{\mu \sigma}}{\partial x^{\nu}} \\ &= \frac{\partial u_{\nu}}{\partial x^{\mu}} + \frac{\partial u_{\mu}}{\partial x^{\nu}} - u^{\sigma} \left(\frac{\partial g_{\sigma \nu}}{\partial x^{\mu}} + \frac{\partial g_{\mu \sigma}}{\partial x^{\nu}} - \frac{\partial g_{\mu \nu}}{\partial x^{\sigma}}\right) \\ &= \frac{\partial u_{\nu}}{\partial x^{\mu}} + \frac{\partial u_{\mu}}{\partial x^{\nu}} - 2 u_{\lambda} \frac{1}{2} g^{\sigma \lambda} \left(\frac{\partial g_{\sigma \mu}}{\partial x^{\nu}} + \frac{\partial g_{\sigma \nu}}{\partial x^{\mu}} - \frac{\partial g_{\mu \nu}}{\partial x^{\sigma}}\right) \\ &= \frac{\partial u_{\nu}}{\partial x^{\mu}} + \frac{\partial u_{\mu}}{\partial x^{\nu}} - 2 u_{\lambda} \Gamma^{\lambda}{}_{\mu \nu} \\ &= \frac{\partial u_{\nu}}{\partial x^{\mu}} - u_{\lambda} \Gamma^{\lambda}{}_{\mu \nu} + \frac{\partial u_{\mu}}{\partial x^{\nu}} - u_{\lambda} \Gamma^{\lambda}{}_{\nu \mu} \\ &= \nabla_{\mu} u_{\nu} + \nabla_{\nu} u_{\mu} \\ &= 0, \end{align*}

where

∇μuν+∇νuμ=0\nabla_{\mu} u_{\nu} + \nabla_{\nu} u_{\mu} = 0

is called the Killing equation. u\boldsymbol{u} is a killing vector if it satisfies the Killing equation.

When a manifold has more than one Killing vectors, any linear combination of them is also a Killing vector.

Consider the 2D Cartesian coordinates in flat space. The metric and its inverse are as follows:

gμν=[1001],gμν=[1001], \begin{align*} g_{\mu \nu} &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \\ g^{\mu \nu} &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \end{align*}

and the Levi-Civita coefficients are all zero. This simplifies the Killing equation:

∂uν∂xμ+∂uμ∂xν=0,\frac{\partial u_{\nu}}{\partial x^{\mu}} + \frac{\partial u_{\mu}}{\partial x^{\nu}} = 0,

or written out:

∂ux∂x+∂ux∂x=2∂ux∂x=0,∂ux∂y+∂uy∂x=0,∂uy∂x+∂ux∂y=0,∂uy∂y+∂uy∂y=2∂uy∂y=0. \begin{align*} \frac{\partial u_x}{\partial x} + \frac{\partial u_x}{\partial x} &= 2 \frac{\partial u_x}{\partial x} = 0, \\ \frac{\partial u_x}{\partial y} + \frac{\partial u_y}{\partial x} &= 0, \\ \frac{\partial u_y}{\partial x} + \frac{\partial u_x}{\partial y} &= 0, \\ \frac{\partial u_y}{\partial y} + \frac{\partial u_y}{\partial y} &= 2 \frac{\partial u_y}{\partial y} = 0. \end{align*}

The second and the third equations are the same, so we have the following:

∂ux∂x=∂uy∂y=0,∂ux∂y=−∂uy∂x. \begin{align*} \frac{\partial u_x}{\partial x} = \frac{\partial u_y}{\partial y} = 0, \\ \frac{\partial u_x}{\partial y} = -\frac{\partial u_y}{\partial x}. \end{align*}

The two trivial solutions are:

u1=ex,u2=ey. \begin{align*} \boldsymbol{u_1} &= \boldsymbol{e_x}, \\ \boldsymbol{u_2} &= \boldsymbol{e_y}. \end{align*}

There is another solution, from the first equation, we can obtain the following:

∂2ux∂x∂y=0,∂2uy∂x∂y=0, \begin{align*} \frac{\partial^2 u_x}{\partial x \partial y} &= 0, \\ \frac{\partial^2 u_y}{\partial x \partial y} &= 0, \end{align*}

and from the second equation:

∂∂y∂ux∂y=−∂∂y∂uy∂x,∂2ux∂y2=−∂2uy∂x∂y,∂2ux∂y2=0,∂∂x∂ux∂y=−∂∂x∂uy∂x,∂2ux∂x∂y=−∂2uy∂x2,∂2uy∂x2=0, \begin{align*} \frac{\partial} {\partial y} \frac{\partial u_x}{\partial y} &= - \frac{\partial}{\partial y} \frac{\partial u_y}{\partial x}, \\ \frac{\partial^2 u_x}{\partial y^2} &= - \frac{\partial^2 u_y}{\partial x \partial y}, \\ \frac{\partial^2 u_x}{\partial y^2} &= 0, \\[5ex] \frac{\partial} {\partial x} \frac{\partial u_x}{\partial y} &= - \frac{\partial}{\partial x} \frac{\partial u_y}{\partial x}, \\ \frac{\partial^2 u_x}{\partial x \partial y} &= - \frac{\partial^2 u_y}{\partial x^2}, \\ \frac{\partial^2 u_y}{\partial x^2} &= 0, \end{align*}

implying the following partial derivatives:

∂ux∂y=A+f(x),∂uy∂x=B+h(y), \begin{align*} \frac{\partial u_x}{\partial y} &= A + f(x), \\ \frac{\partial u_y}{\partial x} &= B + h(y), \end{align*}

and the following second order partial derivatives:

∂2ux∂x∂y=dfdx=0,∂2uy∂x∂y=dhdy=0, \begin{align*} \frac{\partial^2 u_x}{\partial x \partial y} = \frac{d f}{d x} &= 0, \\ \frac{\partial^2 u_y}{\partial x \partial y} = \frac{d h}{d y} &= 0, \end{align*}

implying that ff and hh are linear functions:

∂ux∂y=A+αx,∂uy∂x=B+βy. \begin{align*} \frac{\partial u_x}{\partial y} &= A + \alpha x, \\ \frac{\partial u_y}{\partial x} &= B + \beta y. \end{align*}

The equation

∂ux∂y=−∂uy∂x\frac{\partial u_x}{\partial y} = -\frac{\partial u_y}{\partial x}

implies the following:

A+αx=−B−βy.A + \alpha x = -B - \beta y.

Taking the following derivatives, yields:

∂∂x(A+αx)=∂∂x(−B−βy),α=0,∂∂y(A+αx)=∂∂y(−B−βy),β=0, \begin{align*} \frac{\partial}{\partial x} (A + \alpha x) &= \frac{\partial}{\partial x} (- B - \beta y), \\ \alpha &= 0, \\[3ex] \frac{\partial}{\partial y} (A + \alpha x) &= \frac{\partial}{\partial y} (- B - \beta y), \\ \beta &= 0, \end{align*}

also implying A=−BA = -B.

The current form of the equations is as follows:

∂ux∂y=A,∂uy∂x=−A, \begin{align*} \frac{\partial u_x}{\partial y} &= A, \\ \frac{\partial u_y}{\partial x} &= -A, \end{align*}

implying:

ux=Ay+f(x),uy=−Ax+h(y), \begin{align*} u_x &= Ay + f(x), \\ u_y &= -Ax + h(y), \end{align*}

note: the functions ff and hh are different from before.

Since we know the following:

∂ux∂x=∂uy∂y=0,\frac{\partial u_x}{\partial x} = \frac{\partial u_y}{\partial y} = 0,

we find that the function ff and hh are constants:

dfdx=dhdy=0,f(x)=α,h(y)=β, \begin{align*} \frac{d f}{d x} = \frac{d h}{d y} &= 0, \\ f(x) &= \alpha, \\ h(y) &= \beta, \\ \end{align*}

note: similarly, the constants α\alpha and β\beta are different from before. This gives us the following form of the vector's coordinates:

ux=Ay+α,uy=−Ax+β, \begin{align*} u_x &= Ay + \alpha, \\ u_y &= -Ax + \beta, \end{align*}

or when the indices are raised:

ux=Ay+α,uy=−Ax+β. \begin{align*} u^x &= Ay + \alpha, \\ u^y &= -Ax + \beta. \end{align*}

The vector may be written as a linear combination of its components:

u=uxex+uyey=(Ay+α)ex+(−Ax+β)ey=Ayex+αex−Axey+βey=A(yex−xey)+αu1+βu2, \begin{align*} \boldsymbol{u} &= u^x \boldsymbol{e_x} + u^y \boldsymbol{e_y} \\ &= (Ay + \alpha) \boldsymbol{e_x} + (-Ax + \beta) \boldsymbol{e_y} \\ &= Ay \boldsymbol{e_x} + \alpha \boldsymbol{e_x} -Ax \boldsymbol{e_y} + \beta \boldsymbol{e_y} \\ &= A (y \boldsymbol{e_x} - x \boldsymbol{e_y}) + \alpha \boldsymbol{u_1} + \beta \boldsymbol{u_2}, \end{align*}

where u1\boldsymbol{u_1} and u2\boldsymbol{u_2} are the trivial solutions mentioned above. Recall that another Killing vector is formed by a linear combination of Killing vectors, implying that u\boldsymbol{u} is either independent Killing vector or a linear combination of Killing vectors.

αu1+βu2\alpha \boldsymbol{u_1} + \beta \boldsymbol{u_2} is a linear combination of Killing vectors, meaning this expression is also a Killing vector. Meaning the rest of u\boldsymbol{u}

A(yex−xey)=Au3A (y \boldsymbol{e_x} - x \boldsymbol{e_y}) = A \boldsymbol{u_3}

is also a Killing vector, where

u3=yex−xey\boldsymbol{u_3} = y \boldsymbol{e_x} - x \boldsymbol{e_y}

is the third independent Killing vector. Together, the three Killing vectors are:

u1=ex,u2=ey,u3=yex−xey. \begin{align*} \boldsymbol{u_1} &= \boldsymbol{e_x}, \\ \boldsymbol{u_2} &= \boldsymbol{e_y}, \\ \boldsymbol{u_3} &= y \boldsymbol{e_x} - x \boldsymbol{e_y}. \end{align*}

This is consistent with a theorem from classical geometry: a tranformation in the Euclidean plane that preserves distances can be represented as either translation (u1,u2\boldsymbol{u_1}, \boldsymbol{u_2}) or rotation about some point (u3\boldsymbol{u_3}). Rotation about some other point may be represented as translation, then rotation and then translation back.

The coordinate system may also be reflected. This change, however, is not continuous but descrete.

Consider the Levi-Civita connection, a vector UνU^{\nu} tangent to a geodesic and a Killing field KνK_{\nu}. The quantity UνKνU^{\nu} K_{\nu} is conserved along the geodesic:

Uμ∇μ(UνKν)=0.U^{\mu} \nabla_{\mu} (U^{\nu} K_{\nu}) = 0.

Recall that the parallel transport of a vector along itself is a geodesic:

Uμ∇μUν=0,U^{\mu} \nabla_{\mu} U^{\nu} = 0,

simplifying the above assumption:

UμUν∇μKν=0,U^{\mu} U^{\nu} \nabla_{\mu} K_{\nu} = 0,

and since the indices are just dummy indices, we can freely swap them:

UμUν∇μKν=UνUμ∇νKμ.U^{\mu} U^{\nu} \nabla_{\mu} K_{\nu} = U^{\nu} U^{\mu} \nabla_{\nu} K_{\mu}.

From the Killing equation, we obtain:

∇μKν=−∇νKμ.\nabla_{\mu} K_{\nu} = -\nabla_{\nu} K_{\mu}.

Finally, substituting, we prove the conservation:

UμUν∇μKν=UνUμ∇νKμ,UμUν∇μKν=−UμUν∇μKν,UμUν∇μKν=0, \begin{align*} U^{\mu} U^{\nu} \nabla_{\mu} K_{\nu} &= U^{\nu} U^{\mu} \nabla_{\nu} K_{\mu}, \\ U^{\mu} U^{\nu} \nabla_{\mu} K_{\nu} &= -U^{\mu} U^{\nu} \nabla_{\mu} K_{\nu}, \\ U^{\mu} U^{\nu} \nabla_{\mu} K_{\nu} &= 0, \\ \end{align*}

So the following is true:

UνKν=constant.U^{\nu} K_{\nu} = \textrm{constant}.

For a curve parametrized by λ\lambda, the above may also be expressed as follows:

ddλ(UνKν)=0.\frac{d}{d\lambda} (U^{\nu} K_{\nu}) = 0.

Recall the symmetrization of a tensor:

T(μ1μ2⋯μn)=1n!(Tμ1μ2⋯μn+sum over permutations of μ indices).T_{(\mu_1 \mu_2 \cdots \mu_n)} = \frac{1}{n!} (T_{\mu_1 \mu_2 \cdots \mu_n} + \textrm{sum over permutations of \(\mu\) indices}).

The original Killing equation could then be written as:

∇(μKν)=0.\nabla_{(\mu} K_{\nu)} = 0.

The equation above may be generalized to a higher order symmetric Killing tensor Kν1ν2⋯νn=K(ν1ν2⋯νn)K_{\nu_1 \nu_2 \cdots \nu_n} = K_{(\nu_1 \nu_2 \cdots \nu_n)}:

∇(μKν1ν2⋯νn)=0.\nabla_{(\mu} K_{\nu_1 \nu_2 \cdots \nu_n)} = 0.

Every Killing vector corresponds to a symmetry in the manifold. However, Killing tensors lack a similar geometric interpretation.

Similarly to Killing vectors, a Killing tensor corresponds to a conserved quantity along geodesics. For a vector UμU^{\mu} tangent to a geodesic, the following holds true:

Uμ1Uμ2⋯UμnKμ1μ2⋯μn=constant,U^{\mu_1} U^{\mu_2} \cdots U^{\mu_n} K_{\mu_1 \mu_2 \cdots \mu_n} = \textrm{constant},

or:

ddλ(Uμ1Uμ2⋯UμnKμ1μ2⋯μn)=0.\frac{d}{d\lambda} \left(U^{\mu_1} U^{\mu_2} \cdots U^{\mu_n} K_{\mu_1 \mu_2 \cdots \mu_n}\right) = 0.