Kerr Geometry, Separability and Twistors

Question

One of the remarkable properties of the Kerr black hole geometry is that scalar field equations separate and are exactly solvable (reducible to quadrature), even though naively it does not have enough symmetries to justify this statement (there is no maximal set of Killing vectors, but apparently there is a higher rank covariantly constant tensor which does the job). I once overheard some vague statement to the effect that this simplicity is somehow related to, or more manifest in the language of twistors in this geometry, or relatedly to properties of the massless Dirac equation in this background. I am wondering if someone can help me pin down the precise statement, or provide me with entry point to the literature.

Answer 1

The relation with twistors follows by taking a further square root of Urs's answer.

If $(M^n,g)$ is an $n$-dimensional spin manifold with spinor bundle $S$, we have a natural conformally-invariant operator $P: \Omega^1(S) \to C^\infty(S)$, where $C^\infty(S)$ are the smooth sections of $S$ (i.e., smooth spinor fields) and $\Omega^1(S)$ are the smooth 1-forms on $M$ with values in $S$. The operator $P$ is the "gamma-traceless" part of the spin connection. In other words, if $X$ is any vector field and $\psi$ any spinor field, one has $$ P_X \psi = \nabla_X \psi + \frac1n X \cdot D \psi~, $$ where $D$ is the Dirac operator and $X \cdot$ means Clifford product. Relative to a basis and using the conventions $\Gamma_a \Gamma_b + \Gamma_b \Gamma_a = - 2 \eta_{ab} \mathbf{1}$, one has $$ P_a = \nabla_a + \frac1n \Gamma_a D~. $$ The gamma-traceless condition is precisely $\Gamma^aP_a = 0$.

Now a spinor field $\psi$ satisfying $P_a \psi = 0$ is called a twistor or a conformal Killing spinor field. $P$ is a conformally invariant operator, whence twistors in two conformally related spin manifolds correspond.

If $\psi_1$ and $\psi_2$ are twistor spinors (not necessarily distinct) their tensor product $\psi_1 \otimes \psi_2$ is a linear combination of differential $p$-forms $$ \psi_1 \otimes \psi_2 = \sum_p \omega_p $$ where depending on the signature/dimension of the spacetime only some $p$ may appear.

The point is that the $\omega_p$ are (special) conformal Killing forms, which may be squared to make the Killing-Yano tensors in Urs's answer.

Perhaps a good reference is §6.7 in volume II of Spinors and Spacetime by Penrose and Rindler. There is a section precisely about the Kerr black hole.

Answer 2

That higher rank tensor which you have in mind is called a (conformal) Killing-Yano tensor .

These are skew-symmetric tensors (differential forms) that are covariantly constant in a suitable sense and that serve as "square roots" of Killing tensors in direct analogy to how a vielbein serves as a sqare root for a metric (which is the canonical rank-2 Killing tensor).

For every Killing tensor on a spacetime the relativistic particle propagating on that manifold has an extra conserved quantity. (For the metric itself this is its Hamiltonian). For every refinement of a Killing tensor to the square of a Killing-Yano tensor, also the spinning particle or superparticle has an extra odd conserved quantity (in the case of the metric this is an extra worldline supersymmetry, an extra Dirac operator).

Analogs of all this hold for the case of "conformal" Killing-Yano tensors. The Kerr-spacetime famously admits such.

See for instance

Jacek Jezierski, Maciej Łukasik, Conformal Yano-Killing tensor for the Kerr metric and conserved quantities (arXiv:gr-qc/0510058),

where the relation to twistor geometry is briefly mentioned on p. 4, together with a bunch of references.