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I am reading a paper Desingularisation des varietes de Schubert generalisees by Demazure. I am interested in Lemma 3 on page 58. In particular, I would like to know whether the lemma is true and how to prove it in detail. I have talked to several people about it, and they either think the lemma is false, or find Demazure's proof very unconvincing.

Lemma: the pullback sheaf $\sigma^*\mathcal{L}_{X'}(D)$ is isomorphic to $\mathcal{L}_X(-\alpha)$.

I first quote Demazure's proof and pinpoint the problems with it. Notation and assumptions are explained at the bottom.

Demazure's proof: We identify $P/B'$ with $\mathbb{P}^1$ so that $\infty$ is the fixed point of $B$. Then $B$ acts on $\mathbb{P}^1$ by $b.z = \alpha(b)z + u(b)$, where $u$ is a function on $B$. Since $\mathcal{L}_{\mathbb{P}^1}(\infty)$ is generated by the rational function $z$, one can deduce the lemma without difficulty.

Question 1: OK, $f$ is a locally trivial fibration with fibres isomorphic to $\mathbb{P}^1$ (see below for the definition of $f$). But how is Demazure's discussion of $\mathbb{P}^1$ exactly related to the lemma?

Question 2: $\mathcal{L}_{\mathbb{P}^1}(\infty)$ is a sheaf corresponding to a divisor defined by a homogeneous polynomial of degree $1$ ($z_0 = 0$). So it is isomorphic to the sheaf $\mathcal{O}(1)$. But we have $\alpha =2$ or $-2$ and I believe that we have $\mathcal{L}_{\mathbb{P}^1}(-\alpha) = \mathcal{O}(\langle - \alpha^\vee, - \alpha \rangle) = 2$. Isn't this a contradiction?

Question 3: What exactly happens when we pull-back?

Setup and notation: Let $G$ be a reductive linear algebraic group, with some fixed maximal torus $T$, and a (minimal nontrivial) parabolic subgroup containing two Borel subgroups $B,B'$. Let $L$ denote the Levi factor of $P$. Then the root system of the pair $(L,T)$ contains only two roots $\alpha, -\alpha$, and the Weyl group is $\{e,s_\alpha\}$.

Now let $X$ be a quasi-projective $k$-scheme with a right $B$-action such that the quotient $X/B$ exists and the canonical projection $X \to X/B$ is a locally trivial principal fibration. Let $B \times B'$ act on $X \times P$ by $(x,p).(b,b') = (xb,b^{-1}pb')$. Let $X':=(X \times P)/B$ and $X'/B' = (X \times P)/(B \times B')$.

We have an obvious projection $f: X'/B' \to X/B$, which is a locally trivial fibration with fibre $P/B' \cong \mathbb{P}^1$. Choose a representative $n_\alpha$ of $s_\alpha$ in $P$. Then $\sigma : X/B \to X'/B'$ sending $\overline{x}$ to $\overline{(x,n_\alpha)}$ is a section of $f$. The image of $\sigma$ in $X'/B'$ is an effective divisor. Let us denote it by $D$ and the associated sheaf by $\mathcal{L}_{X'}(D)$.

Now let $\lambda$ be a character of the torus $T$. Then $X \times^B k \to X/B$, where $B$ acts on $X \times^B k$ by $(x,v).b = (xb,\lambda(b)^{-1} v)$ is a vector bundle. Let $\mathcal{L}_X(\lambda)$ denote its sheaf of sections.

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  • $\begingroup$ What is the first part that confuses you? $\endgroup$
    – S. Carnahan
    Apr 9, 2014 at 13:38
  • $\begingroup$ @S.Carnahan Well I don't really see the connection between what Demazure says in the proof and the lemma. I think his strategy is to somehow reduce the problem to invertible sheaves on the projective line (which is the fibre of $f$), but I don't understand this reduction. $\endgroup$ Apr 9, 2014 at 14:57

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