Global section of differential operators on moduli stack of G bundles

Question

I know the definition of the ring of differential operators, by using the smooth coverings. But I do not know how to calculate the global section.

Can anyone help explain that why the global section of twisted differential operators on $Bun_{G}$ are locally constant functions except when it is twisted by $\omega^{1/2}$?

Here $Bun_{G}$ is the moduli stack of $G$-bundles on a smooth algebraic curve $X$ over $\mathbb{C}$, $G$ is semisimple. $\omega$ is the canonical line bundle. I do not know whether this is true for other algebraically closed field

Answer 1

A naive way to see why we need twist by square root of $\omega_{S}$ is the following. Given $L:\omega_{S}^{1/2}\rightarrow \omega_{S}^{1/2}$, we can form its adjoint $L^{t}:\omega_{S}^{1/2}\rightarrow \omega_{S}^{1/2}$, thus we have a map $L\rightarrow L^{t}, \forall L\in \mathcal{D(\omega_{S}^{1/2})}$. In order to get global differential operator, we consider the following: $0\rightarrow \mathcal{D}_{n-1}/\mathcal{D}_{n-2}\rightarrow \mathcal{D}_{n}/\mathcal{D}_{n-2}\rightarrow \mathcal{D}_{n}/\mathcal{D}_{n-1}\rightarrow 0$, consider the map $L\rightarrow L^{t}$, then $\mathcal{D}_{n-1}/\mathcal{D}_{n-2}, \mathcal{D}_{n}/\mathcal{D}_{n-1}$ are eigenspace of eigenvalue $(-1)^{n-1}$ and $(-1)^{n}$, thus the exact sequence split, intuitively, we can construct consecutively a differential operator from symbols through this canonical splitting.