Beilinson-Bernstein localization: $\mathfrak{g}$ action on $G$-equivariant sheaf

I have a few elementary questions related to Beilinson-Bernstein localization.

Let $G$ be a semisimple algebraic group over $\mathbb{C}$ with Lie algebra $\mathfrak{g}$. Consider the setup of Beilinson-Bernstein localization: we have a sheaf of (twisted) differential operators $$\mathscr{D}(\mathcal{L})=\mathcal{L}\otimes\mathscr{D}_X\otimes\mathcal{L}^\vee$$ acting on a locally free $G$-equivariant sheaf $\mathcal{L}$ on a smooth variety $X$. We would like to build a homomorphism $\Phi:\mathfrak{g}\to\Gamma(X,\mathscr{D}(\mathcal{L}))$ using the equivariance condition $\varphi:p_X^*\mathcal{L}\xrightarrow{\sim}\sigma^*\mathcal{L}$, where $\sigma:G\times X\to X$ and $p_X:G\times X\to X$ are the $G$-action on $X$ and the projection, respectively. HTT, in "D-Modules, Perverse Sheaves, and Representation Theory" give the following homomorphism for the algebraic category. For $s\in\Gamma(X,\mathcal{L}),a\in\mathfrak{g}$, they write $$\varphi^{-1}(\sigma^*(\Phi(a)s))=(a\otimes 1)\cdot \varphi^{-1}(\sigma^*s).$$ Here $\mathfrak{g}$ is acting as a right-invariant vector field on $G$, since $\Gamma(G\times X,p_2^*\mathcal{L})\cong\mathbb{C}[G]\otimes_\mathbb{C}\Gamma(X,\mathcal{L}).$

I'm not sure what is meant here by $\sigma^*s$ (unless we are working instead with the underlying bundles), and more importantly, what the intuitive picture behind this formulation is. Moreover, [HTT] uses this formula to claim a number of facts, e.g. that $\Phi$ is filtration-preserving (on $\mathcal{U}\mathfrak{g}$ and $\mathscr{D}(\mathcal{L})$) or that $\Phi$ is equivariant, which I don't know how to obtain from such a formal definition. A worked-out/elementary reference to this construction or an explanation would be much appreciated.

More generally, is there some appropriate notion encapsulating this idea of "differentiating" the equivariant structure of a sheaf?

Edit: Let me try to make my first question more precise. I don't really understand/see how to compute with the formula above. Is there a more straightforward definition? In particular, does the following work? Replace $\mathcal{L}$ with its underlying vector bundle $\pi:L\to X$. The equivariant structure on $\mathcal{L}$ is then just a fancy way of saying that we have a $G$-action on the total space $L$ such that $\pi$ is $G$-linear. For any fixed $l\in L$ we have $\sigma_l:G\to L$ taking $g\mapsto g\cdot l$. Passing to tangent spaces and varying $l$ yields a map $\mathfrak{g}\to\Gamma(L,TL)$. Does this give a map $\mathfrak{g}\to\Gamma(X,\mathscr{D}(\mathcal{L}))$? If so, is this construction any easier to work with?

• I'm not an expert but I feel that a more comfortable definition for me is to let $i: X\rightarrow G\times X$ by $x\mapsto (e,x)$ and define the action by $\Phi(a)s=i^*[(a\otimes 1)\cdot \phi^{-1}(\sigma^*s)]$. This illustrate the idea of "infinitesimal action of $a$". – Zhaoting Wei Mar 27 '15 at 3:34
• Dumb question: how is $\sigma^*s$ defined, exactly? – Nilay Kumar Mar 27 '15 at 3:44
• $\sigma^* s$ is defined simply by "composition with $\sigma$". We can see that if $s$ is a section of $\mathcal{L}$ then the composition gives a section of $\sigma^* \mathcal{L}$. – Zhaoting Wei Mar 27 '15 at 3:55
• 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. – Nilay Kumar Mar 27 '15 at 4:08
• Actually $\varphi$ gives the "$G$-action" on $\mathcal{L}$ and this is what does it mean by $G$-equivariant vector bundles or more generally, $G$-equivariant sheaves: We such a map $\varphi$ which satisfies some properties. See for example Kashiwara's paper kurims.kyoto-u.ac.jp/~kenkyubu/kashiwara/sd.pdf page 22-23. – Zhaoting Wei Mar 27 '15 at 4:25

If you want to "take derivative" with respect to an element $Y$ of a Lie algebra, you want to take a Newton quotient as usual. That means that you pick a path $\gamma(t)\colon [0,\epsilon) \to G$ in your Lie group such that $\dot{\gamma}(0)=Y$. If you want to differentiate a function with respect to this, for example, you consider $$\frac{d}{dY}f(x)=\lim_{h\to 0}\frac{f(\gamma(h)\cdot x)-f(x)}{h}.$$ Differential geometers would say that this directional derivative defines a vector field, (since it gives a derivation on the sheaf of differentiable functions). However, if you're looking not at a function, but a section of some line bundle, you're in trouble: $f(\gamma(h)\cdot x)-f(x)$ makes no sense, since these are elements of different fibers of the line bundle $\mathcal{L}_{\gamma(h)\cdot x}$ and $\mathcal{L}_x$. The usual solution to this is to pick a connection (which again, people usually define as the rule for taking derivative of sections of your bundle), but because our vector field comes from a group, we have another solution. If $\mathcal{L}$ is equivariant, we can identify fibers using the action of the group; that is, we can look at $$\frac{d}{dY}\sigma(x)=\lim_{h\to 0}\frac{\gamma(h)^*\sigma(\gamma(h)\cdot x)-f(x)}{h}.$$ Here, $\gamma(h)^*$ means applying the isomorphism $\gamma(h)^*\colon \mathcal{L}_{\gamma(h)\cdot x}\to\mathcal{L}_x$ which comes from the equivariant structure. For functions (and the usual equivariant structure), this is just identifying $\mathbb{C}\cong \mathbb{C}$ by the identity, so this includes the formula above. This is what HTT are writing, but in a more abstract way (so for example, you can think about algebraic differential operators).
• 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? – Nilay Kumar Mar 27 '15 at 15:47
• @NilayKumar The function $g$ is the derivative of f with respect to $a$. This is proven exactly like the multiplication rule in first year calculus. – Ben Webster Mar 27 '15 at 21:08
• 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 $X$. – Nilay Kumar Mar 29 '15 at 14:48
• 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.) – Nilay Kumar Mar 29 '15 at 14:53