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!

`$-\rho$`

is sometimes hidden, but is essential. $\endgroup$ – Jim Humphreys May 1 '13 at 14:11