It is a classical result of Wakimoto that the sheaf of chiral differential operators $D\_{ch}$ on $\mathbf{P}^1$ has global sections
$$D\_{ch}(\mathbf{P}^1)\ \simeq\ L\_{-2}(\mathfrak{sl}\_2)$$
the simple quotient affine vertex algebra at critical level $-2$. Apparently the map
$$L\_{-2}(\mathfrak{sl}\_2)\ \to\ D\_{ch}(\mathbf{A}^1)\ =\ \mathbf{C}[x\_0,x\_{-1},...,\partial\_{-1},\partial\_{-2},...\ ]$$
sends the fields $e(z),h(z),f(z)$ to $\partial(z)$, to $-2:x(z)\partial(z):-2$ and to $:x(z)^2\partial(z):+x'(z)$, where we have the fields
$$x(z)\ =\ \sum\_{n\in\mathbf{Z}} x\_nz^{-n},\ \ \ \partial(z)\ =\ \sum\_{n\in\mathbf{Z}} \partial\_nz^{-n-1}.$$
**Question:** I am having trouble showing that $-2:x(z)\partial(z):-2$, or equivalently just $\alpha=:x(z)\partial(z):$, glues to a global section of $D\_{ch}$.


Indeed, the transition function $\tau: D\_{ch}(\mathbf{C}^\times)\to D\_{ch}(\mathbf{C}^\times)$, where $D\_{ch}(\mathbf{C}^\times)=D\_{ch}(\mathbf{A}^1)[x\_0^{-1}]$, sends
$$x(z)\ \mapsto\ x(z)^{-1}, \ \ \ \partial(z)\ \mapsto \ :x(z)^2\partial(z):+2x'(z)$$
and so it sends
$$\alpha\ =\ :x(z)\partial(z):\ \mapsto\ :x(z)\partial(z):+2x'(z)x(z)^{-1}.$$
Thus $\alpha$ glues to a global section iff $\tau \alpha$ has no pole at $x\_0=0$.


But my **problem** is that $x'(z)x(z)^{-1}$ has a pole! For instance,
$$x'(z)x(z)^{-1}|0\rangle\ =\ -x\_{-1}x\_0^{-1}|0\rangle \text{ mod }z.$$



Confusion about Wakimoto's chiral differential operators on $\mathbf{P}^1$
There are a couple of errors in the statement of the question. First in the transformation formula that I suppose you took from MSV "Chiral de Rham complex". Let us denote $\tilde{x} = x^{-1}$ and $\tilde{\partial} = x^2 \partial + 2 x'$. We assume the standard OPE $\partial(z) \cdot x(w) \sim \frac{1}{z-w}$ as the only non-zero OPE among generators.


In general for a function $f = f(x)$, denoting by $f(z)$ the corresponding field, we have $\partial(z) \cdot f(w) \sim \frac{f'(w)}{z-w}$


Notice that $x(z) \cdot \partial(w) \sim - \frac{1}{z-w}$ and in general $f(z) \cdot \partial(w) \sim - \frac{f'(w)}{z-w}$.


Now we compute $x^{-1}(z) \cdot \partial(w) \sim \frac{x^{-2}(w)}{z-w}$, from where we obtain
$$
\tilde{x}(z) \cdot \tilde{\partial}(w) = x^{-1}(z) \cdot x^2(w)\partial(w) \sim \frac{1}{z-w} \neq - \frac{1}{z-w} \sim x(z) \cdot \partial(w)
$$


So these don't glue to give a sheaf. The second error is that even assuming your transformation formula, the transformation for $\alpha$ is not correctly computed. I assume here that you missed quasi-associativity. Namely


$$
:x^{-1}(z) :x^2(z) \partial(z): :\ \neq\ ::x^{-1}(z) x^2(z): \partial(z):\ =\ :x(z) \partial(z):
$$


When you fix both issues everything will be fine.


