Considering two arbitrary solutions of the Ernst equations, Robinson was able to construct an
identity [152], the integration of which proved the uniqueness of the Kerr metric. The complicated nature of
the Robinson identity dashed the hope of finding the corresponding electrovac identity by trial and error
methods.^{68}
In fact, the problem was only solved when Mazur [131, 133] and Bunting [25] independently
succeeded in deriving the desired divergence identities by using the distinguished structure of
the EM equations in the presence of a Killing symmetry. Bunting’s approach, applying to a
general class of harmonic mappings between Riemannian manifolds, yields an identity which
enables one to establish the uniqueness of a harmonic map if the target manifold has negative
curvature.^{69}

The Mazur identity (34) applies to the relative difference of
two arbitrary hermitian matrices. If the latter are solutions of a -model with
symmetric target space of the form , then the identity
implies^{70}

The reduction of the EM equations with respect to the axial Killing field yields the coset (see Section 4.5), which, reduces to the vacuum coset (see Section 4.3). Hence, the above formula applies to both the axisymmetric vacuum and electrovac field equations, where the Laplacian and the inner product refer to the pseudo-Riemannian three-metric defined by Eq. (55). Now using the existence of the stationary Killing symmetry and the circularity property, one has , which reduces Eq. (72) to an equation on . Integrating over the semi-strip and using Stokes’ theorem immediately yields

where and are the volume form and the Hodge dual with respect to . The uniqueness of the Kerr–Newman metric follows from the facts that- the integrand on the RHS is non-negative.
- The LHS vanishes for two solutions with the same mass, electric charge and angular momentum.

The RHS is non-negative because of the following observations: First, the inner product is definite, and is a positive volume-form, since is a Riemannian metric. Second, the factor is non-negative in , since is the image of the upper half-plane, . Last, the one-forms and are space-like, since the matrices depend only on the coordinates of .

In order to establish that on the boundary of the semi-strip, one needs the
asymptotic behavior and the boundary and regularity conditions of all potentials. A careful
investigation^{71}
then shows that vanishes on the horizon, the axis and at infinity, provided that the solutions
have the same mass, charge and angular momentum.

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