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.

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