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.



Character which defines canonical bundle on flag variety
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...)


