Killing Tensors and Hidden Symmetry

A geodesic can possess a conserved quantity that is polynomial in its tangent even when no ordinary isometry generates it. The coefficient is a symmetric Killing tensor. Products of known Killing fields give reducible examples; irreducible Killing tensors encode hidden symmetry and can organize integrability and variable separation without appearing as a one-parameter motion of spacetime points.

Required background. Tensor Fields and Natural Operations supplies symmetric tensors and contractions; Affine Connections and Covariant Derivatives supplies the tensor derivative in the defining equation.

Helpful background. Calculus and Linear Algebra for Relativity supplies multilinear and bilinear algebra; the polynomial argument is developed here. Affine and Variational Geodesics supplies the affine geodesic equation; the polynomial first integral is derived on this page.

Symmetric Killing tensors

A rank-$r$ Killing tensor is a smooth symmetric covariant tensor $K_{a_1\ldots a_r}$ satisfying

$$\nabla_{(b}K_{a_1\ldots a_r)}=0.$$

For an affinely parametrized geodesic with tangent $u^a$, form

$$J_K=K_{a_1\ldots a_r}u^{a_1}\cdots u^{a_r}.$$

Then

$$\frac{dJ_K}{d\lambda} =u^b\nabla_bK_{a_1\ldots a_r}u^{a_1}\cdots u^{a_r}=0,$$

because the product of $r+1$ tangent vectors selects the fully symmetric derivative. Rank one recovers a Killing field. Rank two is the most common hidden-symmetry case:

$$\nabla_{(a}K_{bc)}=0, \qquad J_K=K_{ab}u^au^b.$$

Frolov, Krtouš, and Kubizňák give the tensor hierarchy and its relation to geodesic integrability in Frolov, Krtouš, and Kubizňák 2017, §§2.1–2.2, pp. 14–19.

In Hamiltonian language, $J_K$ is homogeneous of degree $r$ in the momenta. Its Poisson bracket with the geodesic Hamiltonian is proportional to the fully symmetrized covariant derivative of $K$. Thus the tensor equation is equivalent to $\{J_K,H\}=0$. This offers an independent check and makes clear why affine geodesic flow, rather than a force-driven trajectory, is the domain of the theorem.

Reducible and hidden integrals

If $X$ and $Y$ are Killing fields, then

$$K_{ab}=X_{(a}Y_{b)}$$

is a Killing tensor, and $J_K=(X\cdot u)(Y\cdot u)$ is simply a product of two linear integrals. The metric itself is another Killing tensor because $\nabla g=0$; its integral is the constant tangent norm. Linear combinations of symmetrized products of Killing fields and metrics are called reducible.

An irreducible Killing tensor cannot be expressed that way using the available Killing fields and metric. Its conserved polynomial is therefore not generated by an ordinary spacetime isometry. “Hidden” describes this algebraic distinction, not invisibility or gauge dependence. Determining irreducibility requires knowing the actual isometry algebra; a component expression that looks complicated is not evidence by itself.

Reducibility is also local versus global. A tensor may admit a decomposition on a highly symmetric patch even when the candidate Killing fields do not extend globally. Statements about a global hidden integral must specify where the tensor and geodesic flow are defined. Likewise, adding a multiple of the metric changes $J_K$ by a multiple of the already conserved norm; it may simplify a representative without creating a new independent integral.

For geodesic flow on a $2n$-dimensional cotangent bundle, Liouville integrability requires $n$ functionally independent integrals, including the Hamiltonian, in mutual Poisson commutation on the regular region considered. A single Killing tensor need not supply enough independent integrals or guarantee complete separation of the Hamilton–Jacobi equation. The additional compatibility conditions for a separability structure are reviewed in Frolov, Krtouš, and Kubizňák 2017, §2.3, pp. 20–22, PDF.

Killing–Yano forms

A Killing–Yano $p$-form $f_{a_1\ldots a_p}=f_{[a_1\ldots a_p]}$ obeys

$$\nabla_a f_{b_1\ldots b_p} =\nabla_{[a}f_{b_1\ldots b_p]},$$

equivalently $\nabla_{(a}f_{b)c_2\ldots c_p}=0$. Its derivative is totally antisymmetric. A quadratic contraction of such a form produces a symmetric tensor,

$$K_{ab} =\frac{1}{(p-1)!} f_{a c_2\ldots c_p}f_b{}^{c_2\ldots c_p},$$

and the Killing–Yano equation implies that $K_{ab}$ is Killing. A short transport proof explains why. Along an affine geodesic, set $q_{c_2\ldots c_p}=u^a f_{a c_2\ldots c_p}$. Then

$$\frac{Dq_{c_2\ldots c_p}}{d\lambda} =u^bu^a\nabla_bf_{a c_2\ldots c_p}=0,$$

because the derivative is antisymmetric in $a,b$. Thus $q$ is parallel, and metric compatibility makes its squared norm constant. That norm divided by $(p-1)!$ is exactly $K_{ab}u^au^b$. Since this works for every initial tangent, polarization gives the Killing-tensor equation. In Lorentzian signature the squared norm need not be positive; its conservation is the relevant fact. The converse does not generally hold: not every Killing tensor possesses a Killing–Yano square root. See Frolov, Krtouš, and Kubizňák 2017, §2.4, pp. 22–26, PDF.

Scope and discriminating checks

The invariant test for a proposed rank-two tensor is direct: symmetrize $\nabla_aK_{bc}$ over all three indices. The geodesic test is independent: differentiate $K_{ab}u^au^b$ and use only affine motion. Passing both tests establishes the local first integral; it does not establish global completeness of the geodesic flow, independence from other integrals, or separability in a chosen coordinate system.

This page stops at classical geometric structure. Symmetry operators for wave or Dirac equations require ordering and curvature corrections, and concrete spacetime classifications require additional analysis.

Two failure cases keep the interpretation honest. A symmetric tensor with $\nabla_{(a}K_{bc)}\ne0$ can still give a constant on a specially chosen geodesic, because that tangent samples only one cubic contraction. And a genuine Killing tensor can yield an integral functionally dependent on the Hamiltonian and linear Killing integrals. The tensor equation proves conservation for all affine geodesics, not novelty or complete integrability.

Exercises

Let $X$ and $Y$ be Killing fields. Prove directly that $K_{ab}=X_{(a}Y_{b)}$ is a Killing tensor, and explain why the associated geodesic integral is reducible.

Solution

Using the product rule,

$$\nabla_{(a}K_{bc)} =\nabla_{(a}\bigl(X_bY_{c)}\bigr) =Y_{(c}\nabla_aX_{b)}+X_{(b}\nabla_aY_{c)}=0,$$

because each fully symmetrized term contains the Killing equation for $X$ or $Y$. Contracting with $u^au^bu^c$ gives the same conclusion. More transparently,

$$K_{ab}u^au^b=(X_au^a)(Y_bu^b),$$

a product of two linear Killing integrals. That explicit factorization is exactly the reducible case.

References

• Valeri P. Frolov, Pavel Krtouš, and David Kubizňák, “Black Holes, Hidden Symmetries, and Complete Integrability,” Living Reviews in Relativity 20, 6 (2017), arXiv.