The relation with twistors follows by taking a further square root of Urs's answer.
If (Mn,g) is an n-dimensional spin manifold with spinor bundle S, we have a natural conformally-invariant operator P:Ω1(S)→C∞(S), where C∞(S) are the smooth sections of S (i.e., smooth spinor fields) and Ω1(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 ψ any spinor field, one has
PXψ=∇Xψ+1nX⋅Dψ ,
where
D is the Dirac operator and
X⋅ means Clifford product. Relative to a basis and using the conventions
ΓaΓb+ΓbΓa=−2ηab1, one has
Pa=∇a+1nΓaD .
The gamma-traceless condition is precisely
ΓaPa=0.
Now a spinor field ψ satisfying Paψ=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 ψ1 and ψ2 are twistor spinors (not necessarily distinct) their tensor product ψ1⊗ψ2 is a linear combination of differential p-forms
ψ1⊗ψ2=∑pωp
where depending on the signature/dimension of the spacetime only some
p may appear.
The point is that the ωp 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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