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I'm trying to understand the proof of the Beilinson-Bernstein localisation theorem at the moment, but there's just one point where I'm having a mental block, and was wondering if anybody could clarify things for me.

Specifically, it's this (not quite general) theorem which I'm trying to prove:

Let $G$ be a semisimple algebraic group over $\mathbb{C}$, $B$ a Borel subgroup and $X=G/B$ the flag variety. Let $\mathfrak{g}$ be the Lie algebra of $G$, $\mathfrak{b}$ the Lie algebra of $B$, $\mathfrak{h}$ the Cartan subalgebra contained in $\mathfrak{b}$, $\mathfrak{b}=\mathfrak{h}\oplus\mathfrak{n}$ ($\mathfrak{n}$ nilpotent), $\mathfrak{g}=\mathfrak{n}^-\oplus\mathfrak{h}\oplus\mathfrak{n}$ (you get the picture). Writing $U(\mathfrak{g})$ for the universal enveloping algebra of $\mathfrak{g}$ we have $U(\mathfrak{g})=U(\mathfrak{h})\oplus(\mathfrak{n}^-U(\mathfrak{g})+U(\mathfrak{g})\mathfrak{n})$, by PBW.

Let $\lambda:B\to \mathbb{C}^\times$ be a character, and let $\mathcal{L}^\lambda$ denote the $G$-equivariant invertible sheaf on $X$ with fiber $\mathbb{C}^{-\lambda}$ at $eG$. That is, $B$ acts on the trivial $\mathbb{C}^{-\lambda}$-bundle $G\times\mathbb{C}^{-\lambda}$ by $b(g,m)=(gb^{-1},(-\lambda)(b)m)=(gb^{-1},\lambda(b^{-1})m)$, and $B\backslash G\times(\mathbb{C}^{-\lambda})$ is a $G$-equivariant $\mathbb{C}^{-\lambda}$-bundle on $X$; $\mathcal{L}^\lambda$ is then its sheaf of sections.

Since $\mathcal{L}^\lambda$ is $G$-equivariant we obtain a homomorphism $\alpha^\lambda:U(\mathfrak{g})\to\Gamma(X,\mathcal{D}_X^\lambda)$ where $\mathcal{D}_X^\lambda=\mathcal{L}^\lambda\otimes\mathcal{D}_X\otimes\mathcal{L}^{-\lambda}$ is the sheaf of differential operators on $\mathcal{L}^\lambda$. (tensor products taken over ${\mathcal{O}_X}$).

Then, what I would like to prove is that the restriction of $\alpha_\lambda$ to the centre $Z(\mathfrak{g})$ of $U(\mathfrak{g})$ factors through the character $\chi_\lambda$ (i.e. the map $Z(\mathfrak{g})\to U(\mathfrak{h})$ coming from the direct sum decomposition above, composed with the map $\lambda:U(\mathfrak{h})=\operatorname{Sym}(\mathfrak{h})\to\mathbb{C}$).

Of course, this isn't the whole theorem, but it's the only part I'm having trouble with.

I believe I am correct in thinking that $\mathcal{L}^\lambda$ is nothing more than the pushforward from $G$ of a certain subsheaf of $\mathcal{O}_G$, namely the one whose sections $f$ are those satisfying $f(gb)=\lambda(b)f(g)$ for $g\in G,b\in B$. So it should be enough to show that $Z(\mathfrak{g})$ acts on that in the right way. (For some reason, I find the action of $\mathfrak{g}$ on $\mathcal{O}_G$ much easier to think about than its action on $\mathcal{L}^\lambda$.)

But I am really stuck. I've looked in the book "D-modules, Perverse Sheaves and Representation Theory" by Hotta et al., where they seem to prove this on pages 278-279, but only found it confusing (and they gave the wrong definition of the Harish-Chandra homomorphism, which I found off-putting). I've also looked in Gaitsgory's notes (http://www.math.harvard.edu/~gaitsgde/267y/catO.pdf) where he seems to prove this (at least, the case $\lambda=0$) on pages 42-43, but that's also confusing. What's worse, is that apparently Gaitsgory's proof makes no use of the algebraic geometry of $\mathfrak{g}$, whereas Hotta et al. appear the Springer resolution of the nilpotent cone in an important way.

If anyone could enlighten me at all about this, I would be extremely thankful!

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One suggestion is to be more careful about the formulation of the Harish-Chandra homomorphism, due to the varying notational conventions of your sources. The shift by $-\rho$ is sometimes hidden, but is essential. –  Jim Humphreys May 1 '13 at 14:11
    
Good point - I am keeping track of my rhos though, I believe! I was mistaken when I said the Hotta et al. book had an error, I just misread is all. –  user30576 May 2 '13 at 5:21
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up vote 3 down vote accepted

This may not be the most efficient way to get to the result, but here's how I would think about it:

The universal enveloping algebra can be identified with right $G$-invariant differential operators on the group $G$, via the map sending an element of the Lie algebra to the corresponding left translation vector field (and vice versa with left and right switched). In particular, the center of $U(\mathfrak{g})$ is given by bi-invariant differential operators on the group, in two different ways.

If I want to understand how elements of the center act on functions that satisfy $f(gb)=\lambda(b)f(g)$, then I should write them as $z=h(z) + n_1m_1+ \cdots+n_km_k$ where $m_i\in U(\mathfrak{n})\mathfrak{n}$ which is possible by the PBW theorem and the fact that central elements have weight 0 (here $h\colon Z(\mathfrak{g})\to U(\mathfrak{h})$ is the Harish-Chandra homomorphism). Thus, $z\cdot f= d\lambda (h(z)) f$; here I'm using $d\lambda$ to distinguish between characters of the group and Lie algebra.

Ok, I'm basically done, but I cheated a little here, since here I was using the map of the center to bi-invariant operators for the right action, and you really want the one that comes from the left action, which a priori might be different. Let me be lazy, and note that we've now shown that the map of the center factors through some character, and that this is induced by some ring map $Z(\mathfrak{g}) \to U(\mathfrak{h})$ (maybe not the HC homomorphism). Thus, it suffices to check it at a Zariski dense set of points; this follows from Borel-Weil, since we know how the center acts on the sections of $\mathcal{L}^\lambda$.

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Thanks very much Ben, I understand it now! Wow, I really did have a mental block about it and now it seems clear. –  user30576 May 2 '13 at 5:19
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