The zero-parameter spatial cylinder and its limiting area

Let $F(y)=\mathrm{\mid }y{\mathrm{\mid }}^2$ on ${\mathbb{R}}^3$. On $F^{-1}(a^2)$, $a>0$, its differential is $d{F}_y(v)=2y\cdot v$, which is nonzero since $y\mathrm{\ne }0$; the fibre is nonempty. By [F3], the unit sphere ${\mathbb{S}}^2=F^{-1}(1)$ and, for $R>R_0$, the coordinate sphere $S_R=F^{-1}(R^2)$ in the spatial end are smooth embedded surfaces. Their tangent spaces are $y^{\perp }$: differentiating the level equation gives containment, and both spaces have dimension two. The inclusion $j:{\mathbb{S}}^2\hookrightarrow {\mathbb{R}}^3$ has injective differential, and ${\iota }_R:{\mathbb{S}}^2\to E$, ${\iota }_R(\omega )=R\omega$, has differential $v\mapsto Rv$, also injective, with smooth inverse $x\mapsto x\mathrm{/}R$ onto $S_R$. Thus [F4] constructs the round metric ${\gamma }_{{\mathbb{S}}^2}=j^{\ast }\delta$ and the induced sphere metric $k_R={\iota }_R^{\ast }{h}_0$, both smooth and positive definite. Give ${\mathbb{S}}^2$ the orientation in [F5] for which $(v,w)$ is positive precisely when $\det (\omega ,v,w)>0$. This determinant is a smooth nowhere-zero alternating form on each tangent plane, so the orientation is well defined; transport it to $S_R$ by ${\iota }_R$. The positive metrics and this orientation meet the hypotheses of [F2].

Use the six hemisphere charts $H_{i,s}=\{\omega :s{\omega }_i>0\}$, $i\in \{1,2,3\}$ and $s\in \{-1,1\}$, with coordinates given by the other two components, ordered positively for the orientation in step 1.1. Fix $\chi (u)=0$ for $u\le 1\mathrm{/}4$ and $\chi (u)=\exp (-1\mathrm{/}(u-1\mathrm{/}4))$ for $u>1\mathrm{/}4$, and put ${\psi }_{i,s}(\omega )=\chi (s{\omega }_i)\mathrm{/}{\sum }_{j,t}\chi (t{\omega }_j)$. The denominator is positive because ${\max }_i\mathrm{\mid }{\omega }_i\mathrm{\mid }\ge 1\mathrm{/}\sqrt{3}$. Thus these six smooth nonnegative weights sum to one, with each support compactly contained in its hemisphere. Write $a=(i,s)$ and $I_a$ for the chart integral in [F6]. For any smooth two-form $\eta$ on the sphere, define $J(\eta )={\sum }_a{I}_a({\psi }_a\eta )$; each summand meets [F6]'s compact-support hypothesis. Define the induced coordinate-sphere area by ${\mathrm{Area}}_{h_0}(S_R)=J({\mathrm{vol}}_{k_R})$, using the oriented parametrization ${\iota }_R$ from step 1.1. The independence argument below proves that this definition is independent of chart and partition choices.

Differentiate $x=e^{\rho }\omega$ using [F9] and [F10]: $dx=e^{\rho }(\omega d\rho +d\omega )$, with $\omega \cdot d\omega =0$. Thus $r^{-2}{\delta }_{ij}d{x}^{i}d{x}^j=d{\rho }^2+{\gamma }_{{\mathbb{S}}^2}$. In the error term the two differential factors cancel $r^{-2}$, leaving coefficients obtained by contracting $E(e^{\rho }\omega )$ with the smooth vectors $\omega$ and angular derivatives of $\omega$. On a compact angular chart domain with coordinates $u$, [F9] gives ${\mathrm{\partial }}_{\rho }(E\circ f)=x^i({\mathrm{\partial }}_{i}E)\circ f$ and ${\mathrm{\partial }}_{u^a}(E\circ f)=r({\mathrm{\partial }}_{u^a}{\omega }^i)({\mathrm{\partial }}_{i}E)\circ f$. Repeated use of [F9] and the coordinatewise product rule [F10] shows by induction that each mixed derivative of a contracted error coefficient of total order $k\le 5$ is a finite sum of terms $r^{j}c(u)({\mathrm{\partial }}^{\beta }{E}_{il})(r\omega (u))$, where $\mathrm{\mid }\beta \mathrm{\mid }=j\le k$ and $c$ is smooth and bounded on that compact domain. At order zero this follows from the displayed contraction. A further derivative either differentiates $r^{j}c(u)$, retaining that form with the same $j$, or differentiates the composed spatial derivative, adding one factor $r$, one bounded angular factor and one spatial derivative, giving $j+1$. All needed derivatives of the angular factors are bounded on the compact domain. By [F1], every resulting term is bounded by $C{r}^j{r}^{-1-j}=C{r}^{-1}=C{e}^{-\rho }$. The six hemisphere charts in step 2.1, restricted to $s{\omega }_i>1\mathrm{/}3$, still cover the sphere because ${\max }_i\mathrm{\mid }{\omega }_i\mathrm{\mid }\ge 1\mathrm{/}\sqrt{3}$; their closures are compact inside their original charts. Taking the largest of finitely many constants proves the asserted bounds through order five on a finite cover, and hence the precise cylinder statement.

First let a smooth two-form $\zeta$ have compact support in the overlap of two positive charts. In the first chart, a sufficiently fine fixed rectangular grid gives finitely many closed squares meeting that support, all contained in the overlap: the compact support has positive distance from the complement of this open coordinate overlap. Their union $K$ is compact Jordan and contains the support in its interior. The transition map on the open overlap is injective, smooth and has invertible differential with positive determinant. The top-form transformation law and [F7], applied on $K$ to the bounded smooth coefficient in the second chart, equate the two integrals; the integrands vanish outside $K$ and its image. The zero extensions in [F6] and finite additivity in [F8] therefore give equality of the two chart integrals of $\zeta$. Now let $(V_b,{\theta }_b)$ be any finite positive-chart decomposition with smooth nonnegative weights summing to one and each support compactly contained in $V_b$. Each ${\psi }_a{\theta }_b\eta$ has compact support in $H_a\cap V_b$. Using the equality just proved and finite linearity in [F8] gives $J(\eta )={\sum }_{a,b}{I}_a({\psi }_a{\theta }_b\eta )={\sum }_{a,b}{I}_b({\psi }_a{\theta }_b\eta )={\sum }_b{I}_b({\theta }_b\eta )$. This also applies to a locally finite subordinate partition: compactness of the sphere reduces its nonzero weights to finitely many, by taking a finite subcover of neighborhoods each meeting only finitely many supports. If $\eta$ itself has compact support in one positive chart $V$, the same overlap equality gives $J(\eta )={\sum }_a{I}_V({\psi }_a\eta )=I_V(\eta )$. Thus $J$ is the global oriented integral, agreeing with chart integrals and independent of partition. Negative charts give the same values because their orientation sign in [F6] cancels the negative transition determinant. A change of oriented parametrization pulls back the form and its chart decomposition to positive charts with the same local coefficients, so it also preserves $J$. Hence step 2.1 defines the intrinsic induced area.

In the notation of step 1.1, restriction of step 3.1 to $r=R$ gives $k_R={\gamma }_{{\mathbb{S}}^2}+{ϵ}_R$ with $\mathrm{\parallel }{\gamma }_{{\mathbb{S}}^2}^{-1}{ϵ}_R\mathrm{\parallel }\le C\mathrm{/}R$ uniformly. By [F2], the ratio of its positive area form to the round area form is $a_R=\sqrt{\det (I+{\gamma }_{{\mathbb{S}}^2}^{-1}{ϵ}_R)}=1+O(R^{-1})$ uniformly: determinant and square root are smooth near the positive identity, and the uniform smallness keeps the matrices in that neighborhood. In the global integral established in steps 2.1 and 3.2, ${\mathrm{Area}}_{h_0}(S_R)=J(a_R{\mathrm{vol}}_{{\gamma }_{{\mathbb{S}}^2}})$. Since the six weights are nonnegative, [F8] applied to their chart integrals and $\mathrm{\mid }{a}_R-1\mathrm{\mid }\le C\mathrm{/}R$ gives $\mathrm{\mid }{\mathrm{Area}}_{h_0}(S_R)-J({\mathrm{vol}}_{{\gamma }_{{\mathbb{S}}^2}})\mathrm{\mid }\le (C\mathrm{/}R)J({\mathrm{vol}}_{{\gamma }_{{\mathbb{S}}^2}})$. Each round chart integrand is smooth with compact support, so this finite round area is well defined.

In spherical coordinates $\omega =(\sin \vartheta \cos \phi ,\sin \vartheta \sin \phi ,\cos \vartheta )$, $0<\vartheta <\pi$, $0<\phi <2\pi$, the pullback of the round metric is $d{\vartheta }^2+{\sin }^2\text{\hspace{0.17em}⁣}\vartheta d{\phi }^2$ and this coordinate order is positive for the orientation of step 1.1. The omitted meridian and poles have zero area: they form finitely many smooth arcs and points, whose portions in the six compact chart supports have two-dimensional content zero: a smooth arc is locally Lipschitz, and subdividing its bounded parameter interval into intervals of length $\eta$ covers its image by $O({\eta }^{-1})$ squares of area $O({\eta }^2)$. The smooth area densities on the compact supports are bounded. Apply [F7] on compact Jordan pieces away from the meridian and poles, where spherical coordinates and hemisphere-chart coordinates are related by smooth diffeomorphisms with invertible differential. Letting the excluded coordinate strips and caps shrink adds zero in the limit by the preceding bounded-density estimate. Thus changing coordinates in the finite integrals of step 4.1 computes the round area as ${\int }_0^{2\pi }{\int }_0^{\pi }\sin \vartheta d\vartheta d\phi =4\pi$. Step 4.1 now gives ${\mathrm{Area}}_{h_0}(S_R)=4\pi +O(R^{-1})\to 4\pi$. The coefficient is one in the round metric because the mass-one spatial leading term in [F1] is exactly $r^{-2}\delta$, as used in step 3.1.