= Solution
Use $G=c=1$, the <Einstein field equations> with zero <cosmological constant>, and <metric signature> $(-,+,+,+)$. Write $m=m_a\,dx^a$ for the <differential one-form> dual to the axial <Killing vector field>. To fix the sign of the volume formula, use the component <Hodge star operator> convention $(\star j)_{abc}=\epsilon_{abcd}j^d$ and $(\star F)_{ab}=\tfrac12\epsilon_{abcd}F^{cd}$. The <Killing equation> gives $(dm)_{ab}=2\nabla_am_b$, and its contracted curvature identity gives
$$
\nabla^b(dm)_{ab}=2R_{ab}m^b,\qquad
d\star dm=-2\star(R_{ab}m^b\,dx^a).
$$
To see the curvature step, tracing the <Killing equation> gives $\nabla_bm^b=0$. The <second covariant derivative of a Killing vector> then gives $\nabla^b\nabla_bm_a=-R_{ab}m^b$, while commuting the <covariant derivatives> gives $\nabla^b\nabla_am_b=R_{ab}m^b$. Subtracting these two terms is precisely $\nabla^b(dm)_{ab}$ above. The component <Hodge star operator> converts this divergence to $d\star dm$ with the displayed minus sign.
In a <vacuum spacetime> region $R_{ab}=0$, so $d\star dm=0$. If $S_1$ and $S_2$ are enclosing <spacelike submanifolds> in the same <homology class> bounding a vacuum three-dimensional region $W$, <Stokes theorem> yields
$$
\int_{S_2}\star dm-\int_{S_1}\star dm=\int_Wd\star dm=0.
$$
Thus \b[the <Komar angular momentum> is independent of the enclosing vacuum spacelike two-manifold], provided the surfaces have the same <orientation> and enclose the same sources and inner boundaries. The vacuum region need not be stationary: an axial <Killing vector field> suffices.
For a regular filling <hypersurface> $\Sigma$ with $\partial\Sigma=S$, the <Einstein field equations> give
$$
d\star dm=-16\pi\star\left[J'-\frac12Tm\right],\qquad
J'_a=T_{ab}m^b.
$$
The pullback of $\star m$ to $\Sigma$ vanishes because $m^a$ is tangent to $\Sigma$: the dual three-form measures the normal component, which is zero. Therefore
$$
\boxed{J=-\int_\Sigma\star J'.}
$$
This is the <stress-energy current from a Killing vector> integrated over the slice, with the signs fixed by the stated <Hodge star operator> convention.
There is a necessary boundary qualification omitted from the printed formulation. If the slice has inner boundaries $S_i$, orient them so $\partial\Sigma=S-\bigcup_iS_i$. The actual <Komar angular momentum with inner boundaries> identity is
$$
J-\sum_iJ_i=-\int_\Sigma\star J',\qquad
J_i=\frac1{16\pi}\int_{S_i}\star dm.
$$
For a vacuum <Kerr black hole>, $T_{ab}=0$ on the exterior slice but $J\ne0$; its horizon supplies precisely the inner boundary contribution. Thus the matter-only formula is false for arbitrary exterior slices. It is valid when a nonsingular filling with no inner boundaries exists, or when all omitted inner charges vanish. The <Komar angular momentum> independence likewise concerns homologous surfaces in the same vacuum region, not an unrestricted comparison of differently enclosed objects.
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