Finite-jet integration of positive Virasoro operators

Put $A_d=\mathbb{C}[t]\mathrm{/}(t^{d+2})$ and $D_n=t^{n+1}{\mathrm{\partial }}_t$ for $1\le n\le d$. These are well-defined derivations of $A_d$, and $[D_m,D_n]=(n-m)D_{m+n}$, with $D_k=0$ for $k>d$. Any product of $d+1$ such derivations is zero: each factor raises the degree of a nonconstant monomial by at least one and kills constants. Thus $\exp D$ is a finite sum for $D=\sum c_n{D}_n$. The repeated Leibniz identity $D^k(uv)={\sum }_{j=0}^k\left(\genfrac{}{}{0px}{}{k}{j}\right)D^{j}u{D}^{k-j}v$ implies that $\exp D$ is an algebra automorphism, with inverse $\exp (-D)$. Its image of $t$ has coefficient $c_n$ in degree $n+1$, plus a polynomial in $c_1,\dots ,c_{n-1}$, since a product of at least two derivations contributing to that degree has positive indices summing to $n$. Recursive coefficient comparison therefore gives unique $c_n$ for each $f$, polynomial in $b_1,\dots ,b_n$. The substitution operator $S_f:h\mapsto h\circ f$ consequently equals $\exp D$.

On $E_d$, $L(n)$ lowers weight by $n$, so $L(n)=0$ for $n>d$ and every product of $d+1$ positive modes vanishes. Define the linear map $J(D_n)=L(n){\mathrm{\mid }}_{E_d}$. It reverses brackets: $J([X,Y])=-[J(X),J(Y)]$. This follows on basis elements from the commutators, including the zero terms when $m+n>d$, and extends by bilinearity.

We prove integration of this map directly. The coefficients $b_1,\dots ,b_d$ identify $G_d$ with ${\mathbb{C}}^d$. Along any piecewise smooth path $f(s)$ starting at the identity, set $X=S_f^{-1}{\mathrm{\partial }}_s{S}_f$. Differentiating the multiplicativity of $S_f$ shows that $X$ is a derivation. It raises monomial degree and hence equals $\sum x_n{D}_n$: a derivation is determined by its value on $t$, which lies in $t^2{A}_d$. Define $U$ by ${\mathrm{\partial }}_{s}U=J(X)U$, $U(0)=I$. This equation has the explicit solution $I+{\sum }_{k=1}^d{\int }_{0<s_k<\cdots <s_1<s}J(X(s_1))\cdots J(X(s_k))d{s}_k\cdots d{s}_1$. Differentiation verifies the equation; the same finite iteration applied to the difference of two solutions proves uniqueness.

This solution depends only on the endpoint. Indeed, take a smooth fixed-endpoint homotopy $f(s,u)$ and set $Y=S_f^{-1}{\mathrm{\partial }}_u{S}_f$, $A=J(X)$, $B=J(Y)$. Differentiating inverses gives ${\mathrm{\partial }}_{u}X-{\mathrm{\partial }}_{s}Y=[X,Y]$, and therefore ${\mathrm{\partial }}_{u}A-{\mathrm{\partial }}_{s}B=-[A,B]$. The defect $W={\mathrm{\partial }}_{u}U-BU$ satisfies ${\mathrm{\partial }}_{s}W=AW$ by substitution and this last identity. At $s=0$, $Y=0$ and $U=I$, so $W=0$ there; uniqueness gives $W=0$ throughout. At $s=1$, $Y=0$, so ${\mathrm{\partial }}_{u}U(1,u)=0$. Any two smooth paths with the same endpoints are joined by their linear interpolation in the coefficient coordinates ${\mathbb{C}}^d$; piecewise smooth paths can be reparametrized with vanishing derivatives at joins and the argument applied to the pieces. The iterated-integral solution justifies every parameter differentiation here, without any analytic ODE existence theorem.

Write the resulting endpoint operator as $R(f)$. For $f,g\in G_d$, first follow a path from the identity to $g$, then the path $g$ to $f\circ g$ obtained from a path $f(s)$ by $f(s)\circ g$. Substitution reverses composition, so $S_{f(s)\circ g}=S_g{S}_{f(s)}$ and its left logarithmic derivative is exactly $S_{f(s)}^{-1}{\mathrm{\partial }}_s{S}_{f(s)}$. The second segment consequently transports $R(g)$ to $R(f)R(g)$. Endpoint independence proves $R(f\circ g)=R(f)R(g)$. Along the path $f(s)=\exp (sD)t$, its substitution is $\exp (sD)$, so $X=D$ and the explicit solution is $\exp (sJ(D))$. Thus $R=R_d$ as stated.

The exponential formula and the triangular polynomial recursion prove polynomial dependence on the jet coefficients. Restricting from $E_{d+1}$ kills $L(d+1)$ and leaves all other positive modes unchanged; coefficients $c_n$ with $n\le d$ depend only on the corresponding truncated jet. Thus restrictions agree. All arguments use finite sums, finite-dimensional matrices, explicit polynomial recursions and finite integrals, so require neither convergence of the infinite vector field nor any additional choice axiom.