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For a surface of revolution whose induced metric is $g=E(\rho )d{\rho }^2+{\rho }^{2}d{\varphi }^2$, with $\rho >0$ and $E>0$ smooth on a coordinate interval, define $\sigma (\rho )={\int }_{{\rho }_{\ast }}^{\rho }\sqrt{E(r)}dr/r$. Then ${\sigma }^{\prime }>0$ and the conformal factor ${\rho }^{-1}$ gives ${\rho }^{-2}g=d{\sigma }^2+d{\varphi }^2$. This proves the rescaled metric is locally the Euclidean metric without solving a curvature equation. Poles and places where $\rho$ is not a coordinate must be treated in other charts; the angular coordinate is also understood locally.

A flat torus is ${\mathbb{R}}^d/\Lambda$ with its quotient Euclidean metric, where $\Lambda$ is a full-rank Euclidean lattice. Its volume is the lattice covolume. The dual lattice indexes its complete Fourier series of Laplacian eigenfunctions.

On an oriented Euclidean metric four-space, the Hodge star operator on two-forms is an orthogonal involution. Its two eigenspaces each have dimension three. A self-dual two-form has eigenvalue $+1$, while an anti-self-dual two-form has eigenvalue $-1$. The splitting uses the metric-normalized Riemannian volume form; an arbitrarily rescaled volume in the defining wedge identity would rescale the operator itself.

Write $w=x^1+i{x}^2$, $z=x^3+i{x}^4$. The Euclidean metric is ${\sum }_a(d{x}^a{)}^2$, and the printed four-form is $4d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d{x}^4$, so it specifies the usual positive orientation. The normalized metric volume form is one quarter of that expression. The self-dual frame in complex Euclidean coordinates isThese are real and have the required complex combinations. For the Hodge star operator with this orientation,and applying $\ast$ again gives the reverse relations. Thus $\ast {\omega }_i={\omega }_i$. They are linearly independent, while the $+1$ eigenspace of $\ast$ on 2-forms has dimension three, proving they span ${\Lambda }_{+}^2$. The TeX aid incorrectly reads the subscript as $1$.

The ASDYM equations require the self-dual projection of the gauge field strength to vanish. Orthogonality to ${\omega }_1,{\omega }_2$ sets its $(2,0)$ and $(0,2)$ parts to zero; orthogonality to ${\omega }_3$ removes the trace of its $(1,1)$ part. Explicitly, if $F={\sum }_{a<b}{f}_{ab}d{x}^a\wedge d{x}^b$, these conditions are $f_{13}-f_{24}=0$, $f_{14}+f_{23}=0$ and $f_{12}+f_{34}=0$. Sinceand the conjugate equation supplies the other complex component for a real curvature form of a connection, the equivalent system is

To obtain the complex potential reduction of anti-self-dual Yang-Mills, set $D_w={\partial }_w+A_w$, $D_z={\partial }_z+A_z$. The first equation is the integrability condition $[D_w,D_z]=0$. Locally it allows an invertible complex matrix $s$ satisfying ${\partial }_{w}s=-A_{w}s$ and ${\partial }_{z}s=-A_{z}s$. The Yang-Mills gauge transformation $A\mapsto s^{-1}As+s^{-1}ds$ consequently gives $A_w=A_z=0$. This is a complex gauge; a real compact gauge group alone generally cannot implement it.

In this gauge the second equation becomesThus the one-form $A_{\bar{w}}dz-A_{\bar{z}}dw$ is closed with respect to the exterior derivative in the $(w,z)$ directions. The local complex version of the Poincare lemma gives a potential $K$ such thatBecause the gauge transformation is complex, $K$ is generally valued in the complexification of a Lie algebra ${\mathfrak{g}}_{\mathbb{C}}$; the printed $\mathfrak{g}$ must be understood in that sense. The elementary reduction is local, and the transformed fields retain a reality condition inherited from the original real connection. For the usual compact matrix gauge groups and smooth fields on all of ${\mathbb{R}}^4={\mathbb{C}}^2$, the gauge and potential can also be chosen globally if no condition at infinity is imposed. The flat partial connection defines a holomorphic principal bundle on the conjugate complex space. That base is a contractible Stein manifold, so the Oka-Grauert principle gives a global trivialization and hence a global complex gauge. After that trivialization, the conjugate of Stein vanishing for the Dolbeault cohomology of functions gives a global primitive $K$, component by component in ${\mathfrak{g}}_{\mathbb{C}}$. Prescribed framing or decay at infinity requires a separate compatibility check and is not automatically preserved by this gauge.

Substitute the potential into the remaining ASDYM equations component:Therefore all three ASDYM equations reduce to the single ASDYM potential equationThe sign follows directly from $A_{\bar{z}}=-K_w$; changing a potential convention would change the displayed commutator sign. Conversely, this equation and the displayed gauge reconstruction make all three curvature conditions vanish, subject to the inherited reality condition when a real gauge field is required.

Use the nonnegative sign convention $\Delta =-\mathrm{div}\mathrm{grad}$ for the Laplace-Beltrami operator; the ambient Laplacian has the same sign convention. Changing both signs changes the signs of the eigenvalues below. Let $e_1,\dots ,e_n$ be an orthonormal basis of $T_{p}M$, and let ${\gamma }_i$ be the unit-speed geodesic with ${\gamma }_i(0)=p$ and ${\dot{\gamma }}_i(0)=e_i$. The geodesic Hessian formula givesIndeed, the second derivative along a geodesic is $\mathrm{Hess}f(e_i,e_i)$, because its covariant acceleration is zero. Taking the trace of the Hessian matrix in this orthonormal basis gives the displayed Laplace-Beltrami operator.

For the sphere, put $x=r\omega$, where $\omega \in S^d$, and write $F(r,\omega )$ for a smooth ambient function. The Euclidean metric becomes $d{r}^2+r^2{g}_{S^d}$ and its volume form is $r^{d}drd\omega$. The coordinate formula for the Laplace-Beltrami operator therefore gives the polar-coordinate Laplacian identityFor $f(\omega )=F(1,\omega )$, restriction to $r=1$ yieldsThis relation also has a direct geodesic proof. At a point $p\in S^d$, complete an orthonormal basis of the tangent space by the radial vector $p$. Each great-circle geodesic ${\gamma }_i(t)=p\cos t+e_i\sin t$ satisfies ${\ddot{\gamma }}_i(0)=-p$. The chain rule gives $(F\circ {\gamma }_i{)}^{\prime \prime }(0)=D^{2}F(e_i,e_i)-{\partial }_{r}F$. Summing and comparing with the ambient trace gives precisely the radial correction above.

Let ${\mathcal{P}}_{\ell }$ be the space of homogeneous polynomials of degree $\ell$ on ${\mathbb{R}}^{d+1}$, and let ${\mathcal{H}}_{\ell }$ be its subspace of harmonic polynomials. For $H\in {\mathcal{H}}_{\ell }$, homogeneity gives $H(r\omega )=r^{\ell }H(\omega )$. Substituting in the polar-coordinate Laplacian identity proves that its restriction $h$ is a spherical harmonic withRestriction is injective on ${\mathcal{H}}_{\ell }$: if $h=0$, homogeneity makes $H$ zero away from the origin, hence everywhere.

To count and exhaust these eigenfunctions, use the harmonic decomposition of homogeneous polynomialsHere is an algebraic proof rather than an assumption about the spectrum. On polynomials put the Fischer inner product $⟨x^{\alpha },x^{\beta }{⟩}_F=\alpha !{\delta }_{\alpha \beta }$. Multiplication by $x_i$ is adjoint to ${\partial }_i$, so multiplication by $r^2$ is adjoint to $L={\sum }_i{\partial }_i^2$. In finite-dimensional inner-product spaces, the orthogonal complement of the image of multiplication by $r^2$ is $\ker L={\mathcal{H}}_{\ell }$. This proves the decomposition. Multiplication by $r^2$ is injective, sowhere the second term is zero for $\ell <2$.

Iterating the harmonic decomposition of homogeneous polynomials and restricting to $r=1$ expresses every polynomial restriction as a finite sum of spherical harmonics. Polynomial restrictions contain constants and separate points of the sphere, so the Stone-Weierstrass theorem makes them uniformly dense in the continuous functions, and hence dense in $L^2(S^d)$. The Laplace-Beltrami operator is self-adjoint, and eigenfunctions with different eigenvalues are orthogonal. If a smooth eigenfunction had an eigenvalue not in the list, it would be orthogonal to a dense subspace and would vanish. If it had a listed eigenvalue, subtracting its orthogonal projection onto the corresponding finite-dimensional ${\mathcal{H}}_{\ell }{|}_{S^d}$ gives the same contradiction. Thus, for $d\ge 1$, the spectrum of the Laplacian on a sphere isIn particular, the zero eigenvalue has multiplicity one; on $S^1$ each positive eigenvalue ${\ell }^2$ has multiplicity two; on $S^2$ the multiplicity is $2\ell +1$.

A flat torus is the quotient ${\mathbb{R}}^d/\Lambda$, where $\Lambda$ is a full-rank Euclidean lattice, with the Riemannian metric induced from the Euclidean metric. Translation by a lattice vector is a Riemannian isometry, so this metric is well defined and has zero curvature. Its volume is the covolume of $\Lambda$.

The dual lattice is ${\Lambda }^{\vee }=\{w:⟨w,v⟩\in \mathbb{Z}\text{ for all }v\in \Lambda \}$. For $w\in {\Lambda }^{\vee }$, the function $e_w(x)=\exp (2\pi i⟨w,x⟩)$ descends to the flat torus, and direct differentiation givesConversely, write $\Lambda =B{\mathbb{Z}}^d$ and $x=By$. Ordinary Fourier series on ${\mathbb{R}}^d/{\mathbb{Z}}^d$ have frequencies $n\in {\mathbb{Z}}^d$, corresponding to $w=B^{-\mathsf{T}}n\in {\Lambda }^{\vee }$. After normalization they form a complete orthonormal basis in $L^2$, and the Fourier coefficients of a smooth function decay rapidly. Applying the Laplace-Beltrami operator term by term shows that a smooth eigenfunction with eigenvalue $\lambda$ has nonzero coefficients only where $4{\pi }^2|w{|}^2=\lambda$. Thus the spectrum of a flat torus isThe formula counts multiplicity over either the real or complex numbers: the two frequencies $w,-w$ give the real sine and cosine functions.

For the rigidity assertion in dimension two, it suffices to reconstruct a rank-two Euclidean lattice $L={\Lambda }^{\vee }$ from its vector-length multiset. The multiset first determines its covolume $A$. Indeed, a bounded fundamental parallelogram, and comparison of the cells meeting a disk with slightly larger and smaller disks, give the lattice-point asymptotics by fundamental cellsThe spectrum determines $N_L(R)$, including multiplicities, so it determines $A$.

Let $a$ be the smallest nonzero vector length, and choose $v\in L$ with $|v|=a$. This vector is primitive: $v=ku$ with $|k|>1$ would contradict minimality. From the length multiset subtract the two vectors $kv,-kv$ at every positive length $ka$, for $k=1,2,\dots$. The shortest remaining length $b$ is exactly the length of a shortest vector $w\notin \mathbb{Z}v$. This subtraction is legitimate even if several directions have length $a$: it subtracts just two copies, and then $b=a$. It also proves that $b$ is independent of our choice of shortest direction. Replace $w$ by $w-kv$ to arrangeThe replacement stays outside $\mathbb{Z}v$ and cannot be shorter than $b$, so a reduced choice still has length $b$.

The shortest-independent-vector basis lemma says that $v,w$ are a basis of $L$. To prove it, choose coordinates with $v=(a,0)$. Because $v$ is primitive, $L/\mathbb{Z}v\cong \mathbb{Z}$; let $h>0$ be the smallest positive vertical coordinate in $L$. Write $w=(x,mh)$, with $m\ge 1$ after changing its sign. If $m\ge 2$, a vector $u$ of vertical coordinate $h$, reduced horizontally modulo $v$, hasThis contradicts the definition of $b$, since $u\notin \mathbb{Z}v$. Hence $m=1$ and $v,w$ generate $L$.

The Gram matrix of this basis is $\left(\begin{array}{cc}{a}^2& c\\ c& b^2\end{array}\right)$, with $c=v\cdot w$, and its determinant is $A^2$. ThereforeChanging the sign of one basis vector allows $c\ge 0$. The numbers $a,b,A$, all audible from the spectrum, consequently determine the Gram matrix, and hence $L$, up to an orthogonal transformation. Taking dual lattices gives the same conclusion for $\Lambda$. Isospectral flat two-dimensional tori are isometric. The argument uses the special basis property of rank-two Euclidean lattices, so it does not claim higher-dimensional rigidity.

Use the orientation selected by the volume form, and let ${\mathrm{vol}}_g$ be its normalized Riemannian volume form. The Euclidean metric induces an inner product on the exterior algebra of covectors. The Hodge star operator is the unique map from $p$-forms to $(4-p)$-forms satisfyingFor an oriented orthonormal coframe $e^1,\dots ,e^4$, it sends a basis wedge to the complementary wedge with the sign of the permutation that restores $e^1\wedge \cdots \wedge e^4$. Applying the Hodge star operator twice exchanges blocks of $p$ and $4-p$ covectors, soHere the supplied volume is interpreted in the usual metric-normalized sense. If instead one defines the operator with an arbitrary unnormalized $\mathrm{vol}=f{\mathrm{vol}}_g$, that operator is $f{\ast }_g$ and its square on two-forms is $f^2$. The usual self-duality statements use the metric-normalized Hodge star operator, with the supplied volume specifying orientation.

The projections $P_{\pm }=(1\pm \ast )/2$ give the Hodge splitting of Euclidean two-forms:Both spaces have dimension three. With $e^4=dt$, bases areThe Hodge star operator is an orthogonal involution on two-forms and therefore is self-adjoint. For a self-dual two-form $H$ and an anti-self-dual two-form $G$,ConsequentlyThis is the wedge orthogonality of opposite-duality two-forms.

For the Yang-Mills action, take an anti-Hermitian special unitary group connection and the fundamental matrix trace, so the positive invariant pairing on its Lie algebra is $⟨X,Y⟩=-\mathrm{tr}(XY)$. Write its gauge curvature as $F=F_{+}+F_{-}$ using the Hodge splitting of Euclidean two-forms, and defineCross terms vanish by the wedge orthogonality of opposite-duality two-forms. ThusWith this anti-Hermitian convention the Second Chern number isFor example, this normalization follows by expanding $\det (1+iF/(2\pi ))$ and using $\mathrm{tr}F=0$. Interpreting the integral as an integer Second Chern number on ${\mathbb{R}}^4$ assumes the usual decay and gauge behavior that allow extension over the point at infinity. The following norm inequality itself does not require integrality:The Yang-Mills instanton Bogomolny bound is saturated precisely when one component vanishes: $F=\ast F$ or $F=-\ast F$. In the stated convention the self-dual case has $k\le 0$ and the anti-self-dual case has $k\ge 0$. Reversing orientation, or defining topological charge with the opposite sign, reverses this assignment while leaving the absolute-value bound unchanged.

For the self-dual Yang-Mills equations in temporal gauge, choose ${\mathrm{vol}}_g=d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge dt$ and ${ϵ}_{123}=1$. The relevant Hodge star operator identities areWriting $F=F_{ti}dt\wedge d{x}^i+\frac{1}{2}{F}_{jk}d{x}^j\wedge d{x}^k$, the self-dual Yang-Mills equations $F=\ast F$ giveIn temporal gauge, $A_t=0$, so the gauge curvature component is $F_{ti}={\partial }_t{A}_i-{\partial }_i{A}_t+[A_t,A_i]={\partial }_t{A}_i$. HenceThe minus sign follows from placing $dt$ last in the orientation and first in the mixed curvature component.

The induced metric is obtained from the ambient Euclidean metric by the pullback of a Riemannian metric operation. Differentiating $x^2+a^2{\rho }^2=1$ on a branch with $x\ne 0$ gives $dx=-a^2\rho d\rho /x$. Also $d{y}^2+d{z}^2=d{\rho }^2+{\rho }^{2}d{\varphi }^2$. ConsequentlyThe assumption that $(\rho ,\varphi )$ are coordinates already excludes the equator $x=0$ as well as the poles $\rho =0$. On such a chart, $0<\rho <1/|a|$ and $E>0$.

The conformal flattening of a surface of revolution supplies the particularly simple conformal factorSince ${\sigma }^{\prime }>0$, these are isothermal coordinates with a flat rescaled metric. An angular coordinate is understood on a local branch; the flat metric may equally be viewed locally as a cylinder metric. The apparent divergence of $E$ at the equator is a coordinate failure, not a singularity of the induced metric. For example, $(x,\varphi )$ give $g=[1+x^2/(a^2(1-x^2))]d{x}^2+(1-x^2)d{\varphi }^2/a^2$, regular at $x=0$.

For $w=x^1+i{x}^2$, $z=x^3+i{x}^4$ and the standard oriented Euclidean metric, the real and imaginary parts of $dw\wedge dz$, together with $i(dw\wedge d\bar{w}+dz\wedge d\bar{z})$, span the self-dual two-forms. The printed complex four-form $dw\wedge dz\wedge d\bar{w}\wedge d\bar{z}$ is four times the metric volume form and gives the same orientation.