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Let $G$ be a reductive group over a field $k$ of characteristic zero with maximal split torus $T$ and Borel $B \supset T$ defining a set of simple roots $\Delta$. Additionally let $\rho$ be the half sum of all positive roots with respect to $B$. For a character $\lambda \in X^*(T)$ we define a line bundle $L_\lambda$ on $G/B$ by

$L_\lambda(U)=\{f: \pi^{-1}(U) \rightarrow k \mid f(gb)=\lambda(b)f(g) \text{ for all }g \in \pi^{-1}(U)(\bar{k}), b \in B(\bar{k})\}$

for $U \subset G/B$ open and $\pi:G\rightarrow G/B$ the natural projection.

Now I'm interested which character $\lambda$ defines the canonical bundle $\omega_{G/B}$ on $G/B$. I figured by several sources out that its either $-2\rho$ or $2\rho$. But as already mentioned in https://math.stackexchange.com/questions/654793/confused-about-borel-weil-theorem, the literature seems somehow inconsistent and I got stuck.

I will greatly appreciate help with this.

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  • $\begingroup$ One trick is to work out the details completely for the projective line, where you have the Euler sequence. $\endgroup$
    – Ben McKay
    Sep 9, 2021 at 12:55
  • $\begingroup$ I thought in this direction but then it seems for me that I have to translate line bundles defined by characters into $\mathcal{O}(n)$ and there I also stuck at the moment but like to learn more about this. Or is it possible without? $\endgroup$
    – KKD
    Sep 9, 2021 at 13:14

2 Answers 2

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I think the most direct way comes from the following fact. There is a natural $G$-equivariant isomorphism $$ T^*(G/B) \cong G \times_{B} \mathfrak{b}^{\bot}$$ where $\mathfrak{b}^{\bot}$ is the sub-Lie algebra of $\mathfrak{g}^*$ given by the annihilator of $\mathfrak{b}$. This can be found in " Representation theory and complex geometry" by Ginzburg Chriss,chapter 1, page 45-46.

This is true in general for Lie groups. Now, in the reductive group context you have an explicit way of describing $\mathfrak{b}^{\bot}$. You start from the decomposition $$\mathfrak{g}=\mathfrak{t} \bigoplus_{\alpha \in \Delta_+}\mathfrak{g}_{\alpha} \bigoplus_{\alpha \in \Delta_{-}} \mathfrak{g}_{\alpha} .$$

Here $\Delta_{+}$ is the set of positive roots chosen so that $$\mathfrak{b}= \mathfrak{t} \bigoplus_{\alpha \in \Delta_+} \mathfrak{g}_{\alpha}.$$ In this way you can see that $$\mathfrak{b}^{\bot} \cong \bigoplus_{\alpha \in \Delta_{+}} \mathfrak{g}_{\alpha}$$ in such a way that $$T^*(G/B) \cong G \times_{B} \mathfrak{b}^{\bot} \cong \bigoplus_{\alpha \in \Delta_+} \mathcal{L}(\alpha) .$$

Taking determinants now you should find that $$\omega_{G/B}=-2 \rho $$ I think (hoping not havinh some signs messing up...)

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  • $\begingroup$ Isn't it $\mathfrak{b}^{\bot} \cong (\mathfrak{g}/\mathfrak{b})^*= (\bigoplus_{\alpha \in \Delta_-} \mathfrak{g}_\alpha)^*\cong \bigoplus_{\alpha \in \Delta_+} \mathfrak{g}_\alpha$? $\endgroup$
    – KKD
    Sep 9, 2021 at 17:54
  • $\begingroup$ Oh yes sorry! I got confused, I'm editing. $\endgroup$ Sep 9, 2021 at 17:55
  • $\begingroup$ No problem. So this leads also to $-2\rho$ (in my notation) as the sections of $G\times_B \mathbb{C}_\lambda$ comes with a negative sign. See my comment to the other answer. $\endgroup$
    – KKD
    Sep 9, 2021 at 17:58
  • $\begingroup$ I think yes ultimately $\endgroup$ Sep 9, 2021 at 18:08
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The confusing part here is that there is a subtlety when one passes from characters to line bundles. Let $\lambda$ be a dominant character of T. Consider the line bundle L given by $G \times_B \mathbb{C}_{\lambda}$, where U acts trivially on this 1 dimensional vector space and T acts via the character $\lambda$. It's not unreasonable to call this $L_{\lambda}$. Now we have a Plucker map $p: G/B \rightarrow \mathbb{P}(V_{\lambda})$, and it turns out that $L=p^* O(-1)$, and so has no global sections, whereas it's dual $L^*$ has global sections giving the Plucker map (up to maybe a second dualization that I am forgetting). So now the name $L_{\lambda}$ appears unfortunate, because now only antidominant weights will give rise to line bundles with global sections. I think this may be the source of the differing notations you see: how one names the line bundle $L$ is somewhat variable.

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  • $\begingroup$ Thats not directly an answer to my question. But let me see if I got it correctly. By Jantzen "Representation of algebraic groups" I. 5.8 and 5.15 we have that for $U \subset G/B$ $\Gamma(U, G \times_B \mathbb{C}_\lambda)=\{f: \pi^{-1}(U) \rightarrow \mathbb{C} \mid f(gb)=\lambda(b)^{-1}f(g) \ldots \}$. Then by your arguments this implies that $\omega_{G/B}$ is defined (in my notation) by $-2\rho$ as the canonical bundle has no global sections? $\endgroup$
    – KKD
    Sep 9, 2021 at 17:14

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