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?

  • $\begingroup$ 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 $a$". $\endgroup$ Commented Mar 27, 2015 at 3:34
  • $\begingroup$ Dumb question: how is $\sigma^*s$ defined, exactly? $\endgroup$ Commented Mar 27, 2015 at 3:44
  • $\begingroup$ $\sigma^* s$ 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 $\sigma^* \mathcal{L}$. $\endgroup$ Commented Mar 27, 2015 at 3:55
  • $\begingroup$ 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. $\endgroup$ Commented Mar 27, 2015 at 4:08
  • $\begingroup$ 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. $\endgroup$ Commented Mar 27, 2015 at 4:25

1 Answer 1


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).

The filtration being preserved is just the fact that an element of the Lie algebra (something "first order") is sent to a derivation (also "first order").

  • $\begingroup$ 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? $\endgroup$ Commented Mar 27, 2015 at 15:47
  • $\begingroup$ @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. $\endgroup$
    – Ben Webster
    Commented Mar 27, 2015 at 21:08
  • $\begingroup$ 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 $X$. $\endgroup$ Commented Mar 29, 2015 at 14:48
  • $\begingroup$ 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.) $\endgroup$ Commented Mar 29, 2015 at 14:53
  • $\begingroup$ See section 6 of Gaitsgory's notes math.harvard.edu/~gaitsgde/267y/catO.pdf for a reasonably rigorous treatment of this question. Also, Proposition II.4.4.4 of Demazure-Gabriel's book "Introduction to Algebraic Geometry and Algebraic Groups" might be helpful. $\endgroup$
    – user91132
    Commented Nov 27, 2015 at 20:51

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