# Character which defines canonical bundle on flag variety

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.

• One trick is to work out the details completely for the projective line, where you have the Euler sequence. Sep 9, 2021 at 12:55
• 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?
– KKD
Sep 9, 2021 at 13:14

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...)

• Isn't it $\mathfrak{b}^{\bot} \cong (\mathfrak{g}/\mathfrak{b})^*= (\bigoplus_{\alpha \in \Delta_-} \mathfrak{g}_\alpha)^*\cong \bigoplus_{\alpha \in \Delta_+} \mathfrak{g}_\alpha$?
– KKD
Sep 9, 2021 at 17:54
• Oh yes sorry! I got confused, I'm editing. Sep 9, 2021 at 17:55
• 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.
– KKD
Sep 9, 2021 at 17:58
• I think yes ultimately Sep 9, 2021 at 18:08

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.

• 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?
– KKD
Sep 9, 2021 at 17:14