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Put $\mu =r_{+}(r_{+}^2+a^2)$. At large $r$, the one-form dual to $K$ hasOnly the $dr\wedge dt$ term contributes to the constant-$t,r$ Komar integral, andWith the stated orientation, the asymptotic Hodge star operator givesThe angular integral is the area of the unit four-sphere,ConsequentlyThis is the Komar mass of the Singly rotating six-dimensional Myers-Perry black hole in the units of the question.

Write $g=\mathrm{Re}h$ for the underlying Riemannian metric of the Hermitian manifold. Its fundamental form of a Hermitian manifold isIt is real and skew-symmetric. In a unitary coframe it is $\omega ={\sum }_{j=1}^n{e}^j\wedge J{e}^j$, which also shows that it has type $(1,1)$ and thatThe Hodge star operator is characterized, after complex-linear extension, byExpanding in the same unitary coframe gives

The Hodge Laplacian and Dolbeault Laplacian areThe Dolbeault Hodge decomposition on a compact Hermitian manifold says that every Dolbeault class has a unique ${\Delta }_{\bar{\partial }}$-harmonic representative andIf ${\Delta }_{\bar{\partial }}\eta =0$, thenso $\eta$ is $\bar{\partial }$-closed and ${\bar{\partial }}^{\ast }$-closed. If also $\eta =\bar{\partial }\xi$, then $\Vert \eta {\Vert }_2^2=⟨{\bar{\partial }}^{\ast }\eta ,\xi ⟩=0$, hence $\eta =0$.

Now suppose $X$ is compact and Kähler. With $d^c=i(\bar{\partial }-\partial )$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$, the Kähler identities make the mixed anticommutators vanish and give ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Consequently

Let $\alpha =d^c\gamma$ and $d\alpha =0$. The Kähler Laplacian identity implies that the $d$-Laplacian commutes with $d^c$. Since a harmonic form is $d^{c\ast }$-closed, $\alpha$ is orthogonal to every harmonic form. If $G$ is the Green operator of the Hodge Laplacian, thenwhere the $d^{\ast }dG\alpha$ term vanishes because $dG\alpha =Gd\alpha =0$. The Green operator commutes with $d^c$, and the anticommutation identity just proved givesTherefore, for the $(k-2)$-form $\beta =-d^{\ast }G\gamma$,This is the d d c lemma in the form required here.

Varying the inverse metric in $H_{abc}{H}^{abc}$ produces three identical terms. With $T_{ab}=-(2/\sqrt{-g})\delta S_H/\delta g^{ab}$,In four dimensions use the Hodge star operator to define the dual covector $v^a=(1/3!){ϵ}^{abcd}{H}_{bcd}$. Then $H^2=-6v^2$ and $H_{acd}{H}_b{}^{cd}=2(v_a{v}_b-g_{ab}{v}^2)$, so equivalently

On an oriented pseudo-Riemannian vector space $({\mathbb{R}}^n,\eta ,\mathrm{vol})$, the Hodge star operator is the unique linear mapsatisfyingfor all $p$-forms $\alpha ,\beta$. If $s$ is the number of negative metric directions, thenon $p$-forms under this convention. For the two-forms in the question,whereasThe induced inner product is symmetric, so these expressions are negatives of one another. Hence a self-dual differential form and an anti-self-dual differential form are orthogonal and

Write $w=x^1+i{x}^2$ and $z=x^3+i{x}^4$. The metric is conformal to the standard Euclidean metric, and the Hodge star on middle-degree differential forms is conformally invariant. Taking $e^{ij}=d{x}^i\wedge d{x}^j$, the three real forms arebecause ${\omega }_1+i{\omega }_2=dw\wedge dz$. For the orientation specified by
$dw\wedge dz\wedge d\bar{w}\wedge d\bar{z}$, one hasThe corresponding relations for the complementary basis forms immediately giveThus these forms give a real basis of the self-dual two-forms.

Let $D_{\alpha }={\partial }_{\alpha }+A_{\alpha }$ be the gauge covariant derivative. In these complex coordinates the Anti-self-dual Yang-Mills equations areIntroduce the spectral parameter $\lambda$ and the linear operatorsTheir commutator isTherefore the Lax pair for the anti-self-dual Yang-Mills equationsis compatible for every $\lambda$ exactly when the ASDYM equations hold.

In particular, $F_{wz}=0$ says that the connection restricted to each $(w,z)$ surface is a flat connection. On a simply connected coordinate patch, the compatible equationshave an invertible solution $g$. Applying the associated gauge transformation setsThis conclusion is local; global topology can obstruct a single such gauge over the whole space.

On an oriented Euclidean four-space with its metric volume form, the Hodge star operator is defined byfor forms $\alpha$ and $\beta$ of the same degree. In $n$ Euclidean dimensions it satisfiesHence on two-forms in four dimensions,The normalization of $\mathrm{vol}$ as the metric volume form is essential; reversing its orientation reverses a single Hodge star but leaves its square unchanged.

The metric induces an inner product on decomposable $r$-forms byextended bilinearly. On an oriented Riemannian manifold, the Hodge star operator is characterized byOn $r$-forms in dimension $n$,

The curvature of $D=d+A$ isor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }+[A_{\mu },A_{\nu }]$. The covariant exterior derivative and the Jacobi identity give the Bianchi identity

Since the Hodge star operator satisfies ${\ast }^2=1$ on Euclidean two-forms,Using the opposite choice of sign as well givesEquality holds precisely for a self-dual or anti-self-dual Yang-Mills instanton, with the sign selected by that of $k$.

Now assume ${\partial }_0{A}_{\mu }=0$ and write $\Phi =A_0$. ThenThe spatial Bianchi identity and the self-duality equations becomewhere the upper four-dimensional sign gives the first displayed reduced sign under the orientation used here. Consequently

Gauge invariance of the inner product givesFor $w=1-|\Phi {|}^2$,Since $w\to 0$ at infinity, the weak maximum principle for elliptic operators excludes a negative interior minimum. Hence

With $d^{4}x=d{x}^0{d}^{3}x$, direct decomposition givesBecause $D_i{B}_i=0$,so the Pontryagin density reduces to the surface charge $4\pi N$ per unit $x^0$. The conventional three-dimensional energyobeys the Bogomolny bound $E_3\ge 4\pi |N|$, saturated by the Bogomolny-Prasad-Sommerfield monopole equation. The four-dimensional action density per unit $x^0$ is $2E_3$; signs relating $k$ and $N$ depend on the self-duality and orientation convention.

For the hedgehog ansatz for a monopole in the question, direct differentiation with $[T_a,T_b]=-{ϵ}_{abc}{T}_c$ givesOn a sphere of radius $r$,ThereforeThe boundary conditions $f\to 1$ and $r^2\alpha \to 1$ make the integrand equal to $r^{-2}+o(r^{-2})$. Thus

Write $F_{12}={\partial }_1{A}_2-{\partial }_2{A}_1$. Up to a total derivative, the covariant-gradient identity and completion of the magnetic and potential terms giveFor positive flux, equality in the Bogomolny bound therefore gives the Bogomolny vortex equationsThe opposite simultaneous signs describe negative winding.

For the orientation $d{x}^1\wedge d{x}^2>0$, the Hodge star operator isThus

A radial vortex of winding $N>0$ haswith $f(0)=a(0)=0$, $f(1)=1$, $a(1)=N$, and equations

For $e^{2\rho }=8/(1-|z{|}^2{)}^2$, put $u=\log |\Phi |$. Away from zeros, the first vortex equation givesSincethe field $\psi =u-\log (1-z\bar{z})+\log 2$ satisfies the Liouville equationFor holomorphic $g:D\to D$, direct use of the Cauchy-Riemann equations verifiesChoosing $g(z)=z^{N+1}$ gives the radial Witten hyperbolic vortexFor $N=1$ these reduce toSince $Bd{\mu }_g=dA=a_1^{\prime }(r)dr\wedge d\theta$,This directly verifies the Abelian Higgs vortex flux relation for unit winding.

The metric on covectors is induced by the inverse matrix $g^{-1}$, and on $p$-forms by the determinant pairingThe Riemannian volume form is the unique positive top form taking value one on every oriented orthonormal frame. The Hodge star operator is uniquely determined byNondegeneracy of the wedge pairing proves existence and uniqueness pointwise, and the smooth metric dependence makes $\ast$ a well-defined smooth bundle map.

On compactly supported forms, Stokes theorem and the graded Leibniz rule givewhere $m=\dim M$; this is the formal $L^2$ adjoint of $d$. The Hodge Laplace-Beltrami operator is

For $g_f=e^{2f}g$, the covector metric scales by $e^{-2f}$, the $p$-form metric by $e^{-2pf}$, and the volume form by $e^{mf}$. ThereforeIf $f$ is constant, the two star factors in the codifferential contribute $e^{-2f}$, so ${\delta }_f=e^{-2f}\delta$. Since $d$ is metric-independent,

The first vacuum Maxwell equations equation is preserved for every vector field because part a givesOn two-forms in four dimensions, the mixed volume tensor $T_{ab}{}^{cd}$ represents twice the Hodge star operator. Part b therefore says that a conformal Killing field commutes with the Hodge star:It follows thatHence $\tilde{F}={\mathcal{L}}_{X}F$ satisfies both vacuum Maxwell equations whenever $F$ does.

The Hodge star operator is defined byOn two-forms in oriented Euclidean four-space, ${\star }^2=1$ and the wedge product is symmetric. ThereforeFor an anti-self-dual curvature $F=-\star F$, the Yang-Mills instanton action becomes purely topological. Withthe standard positive-action convention gives

An orientation selects the positive ordered bases in each tangent space. On an oriented $n$-dimensional Riemannian manifold, the Riemannian volume form ${\omega }_g$ is the unique smooth $n$-form satisfyingfor every positively oriented orthonormal frame. In positively oriented local coordinates,

The metric induces an inner product on the bundle ${\Lambda }^p{T}^{\ast }M$ of $p$-forms. The Hodge star operator is the unique linear mapsuch thatfor all $p$-forms $\alpha ,\beta$. With the codifferential $\delta$, the Laplace-Beltrami operator on differential forms is

The Hodge decomposition theorem says that on a compact oriented Riemannian manifold,an $L^2$-orthogonal direct sum, where ${\mathcal{H}}^p(M)=\ker \Delta$ is the finite-dimensional space of harmonic $p$-forms. Every de Rham cohomology class has exactly one harmonic representative.

On $p$-forms in dimension $n$, the defining identity for the Hodge star operator givesFor $n=4$ and $p=2$, therefore, ${\ast }^2=1$. For every $\alpha \in {\Omega }^2(N)$ defineThen $\ast {\alpha }_{+}={\alpha }_{+}$, $\ast {\alpha }_{-}=-{\alpha }_{-}$, and $\alpha ={\alpha }_{+}+{\alpha }_{-}$. The two eigenspaces of the involution $\ast$ have zero intersection, which proves uniqueness. They are respectively the spaces of self-dual and anti-self-dual two-forms.

Now suppose $N$ is compact and let $\beta$ be an exact three-form, say $\beta =d\theta$. Apply the Hodge decomposition theorem to the two-form $\theta$:Set $a=\delta \psi$. Then $da=d\theta =\beta$ and $\delta a={\delta }^2\psi =0$. For a two-form in dimension four, $\delta =-\ast d\ast$, so $d\ast a=0$. The self-dual formsatisfiesThus every exact three-form is the exterior derivative of a self-dual two-form.

The Hodge star operator is defined byfor forms of the same degree. In four dimensions, under $\hat{g}=e^{2\psi }g$, the inner product on $p$-forms scales by $e^{-2p\psi }$ and the volume form scales by $e^{4\psi }$. HenceFor $p=2$ the exponent vanishes, so the Hodge star on two-forms is conformally invariant.

For a function $f$ and one-form $\alpha$, the codifferential obeys $\delta (f\alpha )=f\delta \alpha -⟨df,\alpha {⟩}_g$. This follows from the graded Leibniz rule and the definition of the Hodge star operator. Applying it to $d(fh)=fdh+hdf$ gives the product formula. In particular ${\Delta }_{+}(f^2)=2f{\Delta }_{+}f-2|df{|}_g^2$. The cross-term sign reverses for the opposite Laplace-Beltrami operator convention.

On a compact oriented smooth manifold $N$ of dimension four without boundary, set $Q([\alpha ],[\beta ])={\int }_N\alpha \wedge \beta$ for closed real two-forms. The Stokes theorem makes the expression independent of representatives, and the even degrees make it symmetric. It is the real de Rham cohomology version of the intersection form. The Hodge decomposition theorem and Hodge star operator express it as $Q(h,k)=(h,\ast k{)}_{L^2}$ on harmonic representatives.

On an oriented Riemannian four-manifold, a two-form $\alpha$ is self-dual when $\ast \alpha =\alpha$.

On an oriented Riemannian four-manifold, a two-form is self-dual when $\ast \alpha =\alpha$. Here $\ast$ is the Hodge star operator.

The self-dual Yang-Mills equations require a gauge curvature two-form to have positive Hodge star operator eigenvalue on an oriented Riemannian four-manifold. By the Bianchi identity, they imply the full second-order Yang-Mills equations. Smooth finite-action solutions saturate the Yang-Mills instanton Bogomolny bound. Reversing orientation exchanges this equation with the Anti-self-dual Yang-Mills equations.

On a closed oriented Riemannian manifold of dimension four, the Hodge star operator preserves harmonic two-forms and splits them into orthogonal spaces ${\mathcal{H}}^{+}$ and ${\mathcal{H}}^{-}$ with eigenvalues $+1$ and $-1$. The real de Rham intersection form in dimension four equals the positive $L^2$ inner product on ${\mathcal{H}}^{+}$ and its negative on ${\mathcal{H}}^{-}$. It is nondegenerate, since $Q(h,\ast h)=\Vert h{\Vert }_2^2$. Its signature pair is $(\dim {\mathcal{H}}^{+},\dim {\mathcal{H}}^{-})$.

For a compact connected oriented boundaryless $n$-dimensional smooth manifold, choose a Riemannian metric. Harmonic functions are constant because $⟨f,\Delta f⟩=\Vert df{\Vert }^2$, and Hodge star commutes with the Hodge Laplacian. Thus the Hodge star operator identifies the constant functions with the harmonic differential forms of top degree, exactly $\mathbb{R}{\omega }_g$. The Hodge decomposition theorem identifies these with the top de Rham cohomology. The metric volume form has nonzero class by Stokes theorem. The harmonic-representative theorem is stated in Denis Auroux's Hodge theory lecture.