More generally, looking at "A proof of Jantzen conjectures" (section 1.8), BB mention that for equivariant $\mathcal{L}$ one obtains a lifting of the map $\alpha:\mathfrak{g}\to\mathcal{T}_X$ to $\alpha_\mathcal{L}:\mathfrak{g}\to\text{End}_\mathbb{C}\mathcal{L}$ such that $\alpha_\mathcal{L}(a)(fs)=f\alpha_\mathcal{L}(a)(s)+\alpha(a)(f)\cdot s$, for $f\in\mathcal{O}_X, s\in\mathcal{L}$. Is there a construction that BB are referring to that is not HTT's formula? Or how can we see that they agree? (Note that here $\mathcal{L}$ is only assumed to be quasicoherent, not necessarily locally free.)

Sorry, I don't quite follow -- in the algebraic case, we don't have an exponential map, right? So "calculus" doesn't apply? Instead, one looks at the expression $\varphi^{-1}(\sigma^*(f\Phi(a)s))-\varphi^{-1}(\sigma^*(\Phi(a)(fs)))$. How does this simplify such that $a$ differentiates $f$? I'm also confused because $a$ is viewed as a vector field on $G$, not $X$.

Thanks, I think I see how $\varphi$ captures these notions now. Actually, I was having trouble seeing that $\Phi(\mathfrak{g})\subset F^1\mathscr{D}(\mathcal{L})$... intuitively this is clear, due to how the action is defined. But how does one show that given $f\in\Gamma(X,\mathcal{O}_X)$, we have $\Phi(a)(fs)-f\Phi(a)s=gs$ for some $g\in\Gamma(X,\mathcal{O}_X)$, without an explicit expression for, say, $\varphi^{-1}$ from $\mathbb{C}[G]\otimes_\mathbb{C}\Gamma(X,\mathcal{L})$ to itself?

Sorry, is HTT just working with the vector bundle associated to $\mathcal{L}$? In that case, I understand what the pullbacks mean, but then it seems to me that then there should be a much more straightforward way of formulating the action, without mentioning $\varphi$. Does that make sense? If we stick with the sheaf language, I'm unsure about how to compute anything with this definition.

Dumb question: how is $\sigma^*s$ defined, exactly?

Thanks a bunch for the references - glad to see there's a nice way to do this in general.