Local weight-(1,0) one-form coefficient on the sphere

Give $\widehat{\mathbb{C}}=\mathbb{C}\cup \{\mathrm{\infty }\}$ the one-point compactification topology: finite points have their usual neighborhoods, and a neighborhood of $\mathrm{\infty }$ has compact complement in $\mathbb{C}$. Any open cover has a member containing $\mathrm{\infty }$; its compact complement is covered by finitely many other members, so $\widehat{\mathbb{C}}$ is compact. Two finite points have disjoint Euclidean disk neighborhoods, while a finite point and $\mathrm{\infty }$ are separated by a disk with compact closure and the complement of that closure. Hence it is Hausdorff. A countable base consists of disks with rational centers and radii together with $N_m=\{\mathrm{\infty }\}\cup \{z\in \mathbb{C}:\mathrm{\mid }z\mathrm{\mid }>m\}$ for positive integers $m$: every compact subset of $\mathbb{C}$ is bounded, so some $N_m$ lies in any prescribed neighborhood of $\mathrm{\infty }$. The plane is path-connected and dense in its one-point compactification; if the compactification had a separation, density would make both separating open sets meet the plane, contradicting its connectedness. Thus $\widehat{\mathbb{C}}$ is connected. Since it is Hausdorff, the chart domains $U_z=\widehat{\mathbb{C}}\setminus \{\mathrm{\infty }\}=\mathbb{C}$ and $U_{\zeta }=\widehat{\mathbb{C}}\setminus \{0\}$ are open. They map to $\mathbb{C}$ by $z\mapsto z$ and by $\zeta (z)=1\mathrm{/}z$ for finite $z\mathrm{\ne }0$, $\zeta (\mathrm{\infty })=0$. The first is the identity homeomorphism. The second is bijective with inverse $\zeta \mapsto 1\mathrm{/}\zeta$ for $\zeta \mathrm{\ne }0$ and $0\mapsto \mathrm{\infty }$. For $r>0$, ${\zeta }^{-1}(D(0,r))=\{\mathrm{\infty }\}\cup \{z:\mathrm{\mid }z\mathrm{\mid }>1\mathrm{/}r\}$ is an open neighborhood of $\mathrm{\infty }$. Conversely, if $O$ is a neighborhood of $\mathrm{\infty }$ with compact complement $K\subset \mathbb{C}$, choose $M$ with $K\subset \{\mathrm{\mid }z\mathrm{\mid }\le M\}$; the inverse chart maps $D(0,1\mathrm{/}(M+1))$ into $O$. Thus the chart map and inverse are continuous at $\mathrm{\infty }$ and $0$, respectively, and are continuous elsewhere by the ordinary reciprocal map. On the overlap $\mathbb{C}\setminus \{0\}$ the transition maps $\zeta =1\mathrm{/}z$ and $z=1\mathrm{/}\zeta$ are holomorphic, so these charts give a holomorphic atlas. Therefore this is a compact connected Hausdorff second-countable Riemann surface, and the centered disk $U$ at $p=0$ is the chart required by [F3].

For $\varphi \mathrm{\ne }0$ supported in a compact $K⋐U_z$, choose a bounded rectangle $Q\subset {\mathbb{R}}^2$ containing $K$. Since $K$ is compactly contained in $U_z$, the zero extension of $\varphi$ vanishes near the boundary of $U_z$ and is continuous on ${\mathbb{R}}^2$, hence Borel and Lebesgue measurable by [F12] and [F13]. The assignment $\varphi \mapsto {\mathcal{O}}_z(\varphi )$ is complex-linear and each weak coefficient is complex-linear in $a,b$. Since $\mathrm{\mid }\varphi \mathrm{\mid }\le p_{K,0}(\varphi ){\mathbf{1}}_Q$, [F10] and [F11] make $\mathrm{\mid }\varphi \mathrm{\mid }$ integrable, so [F8] defines the complex integral. Writing $\varphi =u+iv$, each of the four positive and negative parts of $u,v$ is bounded by $p_{K,0}(\varphi ){\mathbf{1}}_Q$; the triangle inequality and the real positive/negative-part definition then give $\mathrm{\mid }({\mathcal{O}}_z(\varphi )a)(b)\mathrm{\mid }\le 4\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }b\mathrm{\mid }{\lambda }_2(Q)p_{K,0}(\varphi )$. This is a continuous bound on every fixed-support test-function stage, hence every weak coefficient is continuous in the inductive-limit topology specified in [F2].

By [F5], $Df(z):\mathbb{C}\to \mathbb{C}$ is complex-linear, so $Df(z)(h)=hDf(z)(1)$. Its defining linear approximation gives $\frac{f(z+h)-f(z)}{h}=Df(z)(1)+\frac{r(h)}{h}$; because $\mathrm{\mid }r(h)\mathrm{\mid }\mathrm{/}\mathrm{\mid }h\mathrm{\mid }\to 0$, [F15] identifies $Df(z)(1)=f^{\mathrm{\prime }}(z)$. Thus the real differential is multiplication by $f^{\mathrm{\prime }}(z)=\alpha +i\beta$, with matrix $\left(\begin{array}{cc}\alpha & -\beta \\ \beta & \alpha \end{array}\right)$ and determinant ${\alpha }^2+{\beta }^2=\mathrm{\mid }{f}^{\mathrm{\prime }}(z){\mathrm{\mid }}^2>0$. Both $f$ and its inverse are smooth by [F4], so $f$ is a $C^1$ diffeomorphism. By [F6] applied to $f^{-1}\circ f=\mathrm{id}$, $(f^{-1}{)}^{\mathrm{\prime }}(f(z))f^{\mathrm{\prime }}(z)=1$, hence $(f^{-1}{)}^{\mathrm{\prime }}(f(z))=1\mathrm{/}{f}^{\mathrm{\prime }}(z)$.

For $\psi \mathrm{\ne }0$ supported in a compact $L⋐U_w$, the function $(f^{-1}{)}^{\mathrm{\prime }}$ is smooth by [F4] and bounded on $L$. The product $\psi (f^{-1}{)}^{\mathrm{\prime }}$ has compact support in $U_w$, so its zero extension vanishes near the boundary and is continuous, hence Lebesgue measurable by [F12] and [F13]. Choose a bounded rectangle $Q_L$ containing $L$. Applying the same four-positive-part estimate as in step 1.2 gives $\mathrm{\mid }({\mathcal{O}}_w(\psi )a)(b)\mathrm{\mid }\le 4\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }b\mathrm{\mid }{\lambda }_2(Q_L){\sup }_L\mathrm{\mid }(f^{-1}{)}^{\mathrm{\prime }}\mathrm{\mid }{p}_{L,0}(\psi )$, so the transformed weak coefficient is also a distribution on $U_w$.

Under $\xi =g(w)$ the inverse-derivative multiplier is $(g^{-1}{)}^{\mathrm{\prime }}(\xi )(f^{-1}{)}^{\mathrm{\prime }}(g^{-1}(\xi ))$. By [F6] this is $(f^{-1}\circ g^{-1}{)}^{\mathrm{\prime }}(\xi )=((g\circ f{)}^{-1}{)}^{\mathrm{\prime }}(\xi )$, so the pointwise coefficient composes with the exact weight-$(1,0)$ law [F1].

For $\psi \in \mathcal{D}(U_w)$, the function $H(w)=\psi (w)(f^{-1}{)}^{\mathrm{\prime }}(w)$ is smooth and compactly supported in $U_w$, so its zero extension and the real and imaginary parts and their positive and negative parts are continuous and Lebesgue measurable by [F12] and [F13]. Apply [F7] to the positive and negative parts of $\mathrm{Re}H$ and $\mathrm{Im}H$; [F8] then gives ${\int }_{U_w}\psi (w)(f^{-1}{)}^{\mathrm{\prime }}(w)d^{2}w={\int }_{U_z}\psi (f(z))(f^{-1}{)}^{\mathrm{\prime }}(f(z))\mathrm{\mid }{f}^{\mathrm{\prime }}(z){\mathrm{\mid }}^2{d}^{2}z={\int }_{U_z}\frac{\mathrm{\mid }{f}^{\mathrm{\prime }}(z){\mathrm{\mid }}^2}{f^{\mathrm{\prime }}(z)}\psi (f(z))d^{2}z$. The pulled-back test function is smooth because $\mathrm{\mid }{f}^{\mathrm{\prime }}{\mathrm{\mid }}^2\mathrm{/}{f}^{\mathrm{\prime }}=\overline{f^{\mathrm{\prime }}}$ and compactly supported because $f^{-1}$ is continuous. Multiplying by $ab$ proves the exact weak primary test-function identity from [F2]; pointwise, $(f^{-1}{)}^{\mathrm{\prime }}(f(z)){\iota }_{\mathrm{b}\mathrm{i}\mathrm{l}}=(f^{\mathrm{\prime }}(z){)}^{-1}{\mathcal{O}}_z(z)$, the weight-$(1,0)$ primary law [F1].

Applying the weak identity of step 3.1 first to $g$ and then to $f$ gives the test multiplier $\frac{\mathrm{\mid }{f}^{\mathrm{\prime }}(z){\mathrm{\mid }}^2}{f^{\mathrm{\prime }}(z)}\frac{\mathrm{\mid }{g}^{\mathrm{\prime }}(f(z)){\mathrm{\mid }}^2}{g^{\mathrm{\prime }}(f(z))}\chi (g(f(z)))=\frac{\mathrm{\mid }(g\circ f{)}^{\mathrm{\prime }}(z){\mathrm{\mid }}^2}{(g\circ f{)}^{\mathrm{\prime }}(z)}\chi ((g\circ f)(z))$. The equality follows from [F6] and $\mathrm{\mid }(g\circ f{)}^{\mathrm{\prime }}{\mathrm{\mid }}^2=\mathrm{\mid }{f}^{\mathrm{\prime }}{\mathrm{\mid }}^2\mathrm{\mid }{g}^{\mathrm{\prime }}\circ f{\mathrm{\mid }}^2$. Each successively transformed function remains a test function by the compact-support and smoothness argument in step 3.1. Hence the weak smearing law for the composite coordinate is exactly the law assigned directly to $((g\circ f{)}^{-1}{)}^{\mathrm{\prime }}(\xi ){\iota }_{\mathrm{b}\mathrm{i}\mathrm{l}}$.

Since $dw=f^{\mathrm{\prime }}(z)dz$, the local one-form coefficient satisfies ${\mathcal{O}}_w(w)dw=(f^{-1}{)}^{\mathrm{\prime }}(w){\iota }_{\mathrm{b}\mathrm{i}\mathrm{l}}{f}^{\mathrm{\prime }}(z)dz={\iota }_{\mathrm{b}\mathrm{i}\mathrm{l}}dz={\mathcal{O}}_z(z)dz$. The weight is the integer $1$, so no branch is required [F14]. These coefficients are defined only on the chosen patch; no global extension across $\mathrm{\infty }$ or full CFT structure is asserted.