A regular weight-one axis has zero logarithmic angle defect

Fix an interior axis value $z_0$. Restricting the metric to the fixed-$\tau ,z,\varphi$ meridian and the axial orbit gives exactly $L$ and $C$. Continuity gives $L(\rho ,z_0)\mathrm{/}\rho \to e^{\alpha (0,z_0)-U(0,z_0)}$: the difference between this average integral and its limiting integrand is bounded by the supremum of the integrand's difference on $[0,\rho ]$. Also $C(\rho ,z_0)\mathrm{/}(2\pi \rho )\to e^{-U(0,z_0)}>0$. Thus the ratio in the statement tends to $e^{\alpha (0,z_0)}$. It remains to show this is one.

Write $Z$ for the smooth axial generator and $A$ for the fixed axis. Near the axis point under consideration choose coordinates $(x^0,x^1,y^1,y^2)$ with $A=\{y=0\}$ and with ${\mathrm{\partial }}_{y^1},{\mathrm{\partial }}_{y^2}$ an orthonormal normal pair along $A$. Here is a local construction of the latter refinement of a slice chart. The tangential metric block along $A$ is invertible, because the normal plane is nondegenerate and $g$ is nondegenerate. Subtract the tangential projections of the two transverse coordinate vectors to obtain a normal basis $V_1,V_2$. Put $e_1=V_1\mathrm{/}\sqrt{g(V_1,V_1)}$, $W_2=V_2-g(V_2,e_1)e_1$, and $e_2=W_2\mathrm{/}\sqrt{g(W_2,W_2)}$. All denominators are smooth and positive. In the original chart map $(x,y)$ to $(x,0)+y^1{e}_1(x)+y^2{e}_2(x)$, where the vectors are expressed in that chart. Its differential at $y=0$ maps to a tangential basis and this normal pair, hence is invertible; [F2] makes this a local chart. In these refined coordinates $g_{ab}(x,0)={\delta }_{ab}$ and $g_{ia}(x,0)=0$ for normal indices $a,b$ and tangential indices $i$.

At $y=0$ the generator vanishes. Differentiating the action in its parameter and in the coordinates, with the derivative interchange in [F2], shows that its linearization is zero tangentially and is the generator $J$ of the weight-one rotation on the normal plane; $J^{\mathsf{T}}J=I$. The sign of $J$ can be fixed locally and does not affect the calculation. With $r=\mathrm{\mid }y\mathrm{\mid }$, Taylor expansion gives tangential components $Z^i=O(r^2)$ and normal components $Z^a=(Jy{)}^a+O(r^2)$. Consequently, for $f=g(Z,Z)$, Taylor expansion of $f$ and its first derivatives gives $f=r^2+O(r^3)$, ${\mathrm{\partial }}_{y^a}f=2y^a+O(r^2)$ and ${\mathrm{\partial }}_{x^i}f=O(r^2)$. These estimates are uniform on a smaller chart: bound the finitely many required smooth derivatives on a closed coordinate box using [F4]. For the positive orbit radius $R=\sqrt{f}$ off the axis, it follows that $R=r+O(r^2)$, ${\mathrm{\partial }}_{y^a}R=y^a\mathrm{/}r+O(r)$ and ${\mathrm{\partial }}_{x^i}R=O(r)$. The smooth inverse metric has normal block $I+O(r)$, mixed block $O(r)$ and bounded tangential block. Thus the invariant scalar $Q:=g^{-1}(dR,dR)=1+O(r)$ tends to one, uniformly in a local axis neighborhood. This uses smooth axis geometry, without asserting that the Weyl meridian has a smooth limiting tangent.

The axial generator used above is the one measured by the reconstructed angular coordinate: differentiating the equivariance identity in $s$ at zero, off the axis, gives $d\iota ({\mathrm{\partial }}_{\varphi })=Z\circ \iota$. Therefore $g(Z,Z)\circ \iota =\mathbf{g}({\mathrm{\partial }}_{\varphi },{\mathrm{\partial }}_{\varphi })={\rho }^2{e}^{-2U}$. The isometric identification and the chain rule also identify the inverse-metric norm of the differential of this orbit radius on the two open manifolds: in local charts, put $P=D\iota$, let $G$ be the matrix of $g$, and let $\lambda$ be the column of components of $dR$. The pulled-back metric matrix and covector column are $P^{\mathsf{T}}GP$ and $P^{\mathsf{T}}\lambda$; since $P$ is invertible, $(P^{\mathsf{T}}\lambda {)}^{\mathsf{T}}(P^{\mathsf{T}}GP{)}^{-1}(P^{\mathsf{T}}\lambda )={\lambda }^{\mathsf{T}}{G}^{-1}\lambda$. To check the approach to the axis, fix ${\tau }_0\in J$, ${\varphi }_0\in S^1$, $z_0\in I$, and put $p_0=p({\tau }_0,z_0)$. For $K=\{\xi \in {\mathbb{R}}^2:\mathrm{\mid }\xi -(1,0)\mathrm{\mid }\le 1\mathrm{/}4\}$, define $q_{ϵ}(\xi )=\iota ({\tau }_0,ϵ{\xi }_1,z_0+ϵ{\xi }_2,{\varphi }_0)$. The parameter points defining $q_{ϵ}$ are at Euclidean $(\rho ,z)$ distance at most $5ϵ\mathrm{/}4$ from $(0,z_0)$, since $\mathrm{\mid }\xi \mathrm{\mid }\le 5\mathrm{/}4$ on $K$. By continuity of $\bar{\iota }$ at $({\tau }_0,0,z_0,{\varphi }_0)$, every neighborhood of $p_0$ therefore contains all $q_{ϵ}(K)$ for sufficiently small $ϵ$. In the smaller smooth chart of step 2.1, the normal coordinates of these points consequently tend to zero uniformly, and the uniform estimate $Q=1+O(r)$ there yields $Q(q_{ϵ}(\xi ))\to 1$ uniformly on $K$. This establishes the needed coordinate-to-extension limit without differentiating the boundary extension.

By the generator identification in step 3.1, in the reconstructed coordinates $R=\rho e^{-U}=C\mathrm{/}(2\pi )$, independently of $\tau ,\varphi$. Inverting the meridian block of the displayed reconstructed metric therefore gives $Q=e^{2U-2\alpha }(R_{\rho }^2+R_z^2)$. In particular $\mathrm{\mid }DR\mathrm{\mid }=e^{\alpha -U}\sqrt{Q}$, where $D$ and its norm are Euclidean in $(\rho ,z)$. Set $B=e^{-U(0,z_0)}>0$ and $H=e^{\alpha (0,z_0)-U(0,z_0)}>0$. On the closed Euclidean disk $K=\{\xi \in {\mathbb{R}}^2:\mathrm{\mid }\xi -(1,0)\mathrm{\mid }\le 1\mathrm{/}4\}$ define $F_{ϵ}(\xi )=R(ϵ{\xi }_1,z_0+ϵ{\xi }_2)\mathrm{/}ϵ$ for positive sufficiently small $ϵ$. This disk stays off the axis. Continuity of $U,\alpha$ at $(0,z_0)$ gives $F_{ϵ}\to B{\xi }_1$ uniformly on $K$. Step 3.1 gives $Q\to 1$ uniformly on these shrinking coordinate disks. The chain rule and the inverse-block identity then give $\mathrm{\mid }D{F}_{ϵ}\mathrm{\mid }\to H$ uniformly on $K$. No convergence of derivatives of $U$ or $\alpha$ has been assumed.

To compare these two limits, fix $\eta >0$ and set $h_{ϵ}(\xi )=F_{ϵ}(\xi )-B{\xi }_1$ and ${\xi }_{\ast }=(1,0)$. The continuous function $h_{ϵ}(\xi )-\eta \mathrm{\mid }\xi -{\xi }_{\ast }{\mathrm{\mid }}^2$ attains a maximum on $K$ by [F4]. For small $ϵ$ it is smaller on the boundary than at ${\xi }_{\ast }$: the boundary penalty is $\eta \mathrm{/}16$, while $2{\sup }_K\mathrm{\mid }{h}_{ϵ}\mathrm{\mid }\to 0$. Hence a maximizer ${\xi }_{ϵ}$ is interior, and Fermat's theorem gives $D{F}_{ϵ}({\xi }_{ϵ})=B(1,0)+2\eta ({\xi }_{ϵ}-{\xi }_{\ast })$. The reverse triangle inequality implies $\mid \mathrm{\mid }D{F}_{ϵ}({\xi }_{ϵ})\mathrm{\mid }-B\mid \le \eta \mathrm{/}2$. Uniform gradient-norm convergence in step 4.1 now implies $\mathrm{\mid }H-B\mathrm{\mid }\le \eta \mathrm{/}2$. Since $\eta >0$ was arbitrary, $H=B$.

The equality $H=B$ is $e^{\alpha (0,z_0)-U(0,z_0)}=e^{-U(0,z_0)}$, so $e^{\alpha (0,z_0)}=1$ and $\alpha (0,z_0)=0$. Step 1.1 then gives $\mathbf{b}(z_0)=\alpha (0,z_0)=0$. As $z_0$ was arbitrary in the open interval, this proves the full statement.