Formal coordinate-change action of a conformal vertex algebra

Every germ has a unique formal factorization $\rho =s_a\circ f$, with $a={\rho }^{\mathrm{\prime }}(0)\mathrm{\ne }0$ and $f^{\mathrm{\prime }}(0)=1$. At each jet order the triangular recursion of [F2] gives the unique coefficients $c_1,\dots ,c_d$ in $f=\exp (\sum c_n{t}^{n+1}{\mathrm{\partial }}_t)t$. The recursions are compatible as $d$ increases; this establishes the infinite formal expression. It does not assert that its logarithmic vector field converges as a holomorphic vector field. On $E_d=V_{\le d}$ each mode with $n>d$ vanishes and every product of more than $d$ positive modes vanishes. Hence the displayed operator is well-defined there and equals $A(a)R_d(f)$, where $A(a){\mathrm{\mid }}_{V_m}=a^{m}I$. The finite restrictions agree and therefore define an operator on the algebraic direct sum $V$.

For $b\mathrm{\ne }0$ put $f^{[b]}=s_b^{-1}\circ f\circ s_b$. Conjugation changes $t^{n+1}{\mathrm{\partial }}_t$ to $b^n{t}^{n+1}{\mathrm{\partial }}_t$ in the exponential expression for $f^{[b]}$. Weight lowering gives $A(b{)}^{-1}L(n)A(b)=b^{n}L(n)$, so $R_d(f)A(b)=A(b)R_d(f^{[b]})$. If ${\rho }_1=s_a\circ f$ and ${\rho }_2=s_b\circ g$, their composite is $s_{ab}\circ f^{[b]}\circ g$. Applying [F2] and this conjugation identity gives $U({\rho }_1)U({\rho }_2)=A(ab)R_d(f^{[b]})R_d(g)=U({\rho }_1\circ {\rho }_2)$ on $E_d$. Compatibility proves the identity on $V$. Since $U(\mathrm{i}\mathrm{d})=I$ and a germ with nonzero derivative has an inverse germ, $U({\rho }^{-1})$ is the two-sided inverse. Alternatively invertibility on each jet follows from the finite polynomial inverse and $A(a{)}^{-1}=A(a^{-1})$.

For a holomorphic family of germs, every Taylor coefficient and the nonvanishing derivative $a$ are holomorphic in the parameter. Division by $a$ gives holomorphic coefficients of $f$, and the finite triangular recursion makes every $c_n$ holomorphic. On $E_d$ the finite exponential is polynomial in $c_1,\dots ,c_d$ and $A(a)$ has entries $a^m$, $0\le m\le d$. Thus every matrix entry is holomorphic. No branch of $\log a$ is needed because the weights are integers. This proves all the assertions.