Proof. Follows from the fact that $\mathrm{Sym}^n(V)^{\vee} = \Gamma^n(V^{\vee})$ and $\Lambda^n(V)^{\vee} =\Lambda^n(V^{\vee})$ .

Proof. Write $\lambda = (k_1 \geq \cdots \geq k_r > 0)$ . By Proposition 3.14, it is enough to prove that $S_{\lambda}V$ is a direct summand of $V^{\otimes |\lambda |}$ in the category of $\mathrm{GL}(V)$ -modules for one k-vector space of dimension greater than $|\lambda |$ . Consider a vector space V of dimension n. Note that the surjection

\[\nabla^{\otimes r} : V^{\otimes |\lambda |} \rightarrow \mathrm{Sym}^{\lambda} V := \bigotimes_{1 \leq i\leq r} \mathrm{Sym}^{k_i} V\]

admits a section when $p > \max_i k_i = k_1$ . Indeed, we define it as $s = s_1 \otimes \cdots\otimes s_r$ where

\[s_i(v_1v_2\cdots v_r) = \frac{1}{k_i!}\sum_{\sigma \in \mathfrak{S}_{k_i}}v_{\sigma(1)}\otimes\cdots \otimes v_{\sigma(k_i)}.\]

By definition, $S_{\lambda}V$ is a sub- $\mathrm{GL}(V)$ -module of $\mathrm{Sym}^{\lambda} V$ and we would like to find a condition on p that guarantees it is also a direct summand. The following exact sequence of $\mathrm{GL}(V)$ -modules

is split if we can show that $\mathrm{Ext}^1_{\mathrm{GL}(V)}(\mathrm{Sym}^{\lambda} V/S_{\lambda}V,S_{\lambda}V)$ vanishes. Let $\tilde{\lambda}$ denote the character $(|\lambda |, 0, \ldots, 0)$ of the standard maximal torus of $\mathrm{GL}(V)$ . Since $\mathrm{Sym}^{\lambda}V$ is of highest weight $\tilde{\lambda}$ , the $\mathrm{GL}(V)$ -modules $S_{\lambda}V$ and $\mathrm{Sym}^{\lambda} V/S_{\lambda}V$ are filtered by simple modules $L(\nu)$ with $\nu \leq \tilde{\lambda}$ . Under the assumption that $\tilde{\lambda}$ is p-small, Proposition 2.6 implies that we have isomorphisms

\[L(\nu) = \nabla(\nu) = \Delta(\nu)\]

for all characters $\nu$ satisfying $\nu \leq \tilde{\lambda}$ . Since $\mathrm{Ext}^1_{\mathrm{GL}(V)}(\mathrm{Sym}^{\lambda} V/S_{\lambda}V,S_{\lambda}V)$ is the limit of a spectral sequence involving Ext groups

\[\mathrm{Ext}^1_{\mathrm{GL}(V)}(L(\nu),L(\nu^\prime)) = \mathrm{Ext}^1_{\mathrm{GL}(V)}(\Delta(\nu),\nabla(\nu^\prime)),\]

that vanishes by Proposition 2.16, $\mathrm{Ext}^1_{\mathrm{GL}(V)}(\mathrm{Sym}^{\lambda}V/S_{\lambda}V,S_{\lambda}V)$ must vanish. In conclusion, we have the desired splitting, provided that $\tilde{\lambda}$ is p-small, i.e.

\begin{align*}p \geq \max_{\alpha \in \Phi^+} \langle \tilde{\lambda} + \rho, \alpha^{\vee} \rangle &= \max_{ 1\leq i < j \leq n} \langle \tilde{\lambda} + \rho, \varepsilon_i -\varepsilon_j \rangle\\&= \max_{ 1 \leq i < j \leq n} \biggr\langle \biggr(|\lambda |+\frac{n-1}{2}, \frac{n-3}{2}, \ldots,-\frac{n-1}{2} \biggr), \varepsilon_i -\varepsilon_j \biggr\rangle\\&= |\lambda |+n-1.\end{align*}

Since our argument is valid only when $n \geq |\lambda |$ , we get the bound $p \geq 2| \lambda|-1$ .

Proof of Proposition 3.16. By Proposition 3.14, it is enough to show that $S_{\lambda}\circ S_{\mu} (V)$ admits a $\nabla$ -filtration as a $\mathrm{GL}(V)$ -module where V is one vector space of dimension greater that $|\lambda |\,|\mu |$ . Consider a vector space V of dimension $n\geq |\lambda |\,|\mu |$ . By Lemma 3.17, $S_{\lambda}(S_{\mu}V)$ is a direct summand of $(S_{\mu}V)^{\otimes |\lambda |}$ as $\mathrm{GL}(S_{\mu}V)$ -modules. After restriction to the category of $\mathrm{GL}(V)$ -modules through the map $\mathrm{GL}(V) \rightarrow \mathrm{GL}(S_{\mu}V)$ induced by $S_{\mu}$ , $S_{\lambda}(S_{\mu}V)$ is again a direct summand of $(S_{\mu}V)^{\otimes |\lambda |}$ . By Corollary 2.15, the $\mathrm{GL}(V)$ -module $(S_{\mu}V)^{\otimes |\lambda |}$ admits a $\nabla$ -filtration. By Corollary 2.17, the $\mathrm{GL}(V)$ -module $S_{\lambda}\circ S_{\mu} (V)$ admits a $\nabla$ -filtration.

Proof. Left to the reader.

Proof. For the equivalence of the definitions, see [Reference HartshorneHar66, Prop 1.1/1.2/1.4].

Proof. For the equivalence of the definitions, see [Reference LazarsfeldLaz04b, Theorem 1.7.6] or [Sta21, Lemma 02O1].

Proof. See [Reference HartshorneHar66, Proposition 3.2/3.3] for a complete proof of

\[(1) \Leftrightarrow (2) \Leftrightarrow (3).\]

For $(1) \Leftrightarrow (4)$ , this numerical criterion is due to Barton [Reference BartonBar71].

Proof. See [Reference HartshorneHar66, Proposition 2.2/Corollary 2.5] for assertions (1), (3) and (4), [Reference HartshorneHar66, Corollary 3.4] for assertion (4) and [Reference BartonBar71, Theorem 3.3] for assertion (5).

Proof. Assume that $\mathcal{E}^{\otimes n}$ is ample. As a quotient of $\mathcal{E}^{\otimes n}$ , $\mathrm{Sym}^n\mathcal{E}$ is ample and we conclude with [Reference HartshorneHar66, Proposition 2.4].

Proof. See [Reference LazarsfeldLaz04a, Proposition 1.2.9] and [Reference LazarsfeldLaz04a, Corollary 1.2.24].

Proof. Since the Frobenius map is finite surjective, it follows from the previous proposition.

Proof. Choose $m \geq 0$ large enough such that $\mathcal{F}(m)$ is globally generated over X. We get a surjective morphism

\[\mathcal{O}_X^{\oplus s} \rightarrow \mathcal{F}(m)\]

for some $s \geq 1$ and then we tensor by $\mathcal{O}_X(-m)$ .

Proof. We use Lemma 4.12 to write $\mathcal{F}$ as a quotient of $\mathcal{O}_X(-m)^{\oplus s}$ for a suitable $s \geq 1$ . Take n large enough such that $\mathcal{O}_X(-m) \otimes \mathcal{E}^{(p^r)}$ is globally generated. Since the quotient of a globally generated sheaf is globally generated, we get that $\mathcal{F} \otimes\mathcal{E}^{(p^r)}$ is globally generated.

Proof. Choose n large enough such that $\mathcal{E}^{(p^n)}(-1)$ is globally generated. We deduce that $\mathcal{E}^{(p^n)}$ is quotient of $\mathcal{O}_X(1)^{\oplus s}$ for a suitable $s\geq 1$ . By assertion (3) of Proposition 4.7, $\mathcal{E}^{(p^n)}$ is ample and by Corollary 4.10, $\mathcal{E}$ is ample.

Proof. See [Reference HartshorneHar66, Proposition 6.4/Corollary 6.7] for assertions (1), (2) and (4). Hartshorne does not state assertion (3), so we give a proof. Let $\mathcal{F}$ be a coherent sheaf and $r_0 \geq 1$ be an integer such that $\mathcal{F} \otimes \mathcal{E}^{(p^r)}$ is globally generated for all $r \geq r_0$ . For all $r \geq r_0$ , the surjective morphism $\mathcal{E} \rightarrow \mathcal{E}^\prime$ induces a surjective morphism of $\mathcal{O}_X$ -modules

\[\mathcal{F} \otimes \mathcal{E}^{(p^r)}\rightarrow \mathcal{F} \otimes {(\mathcal{E}^\prime)}^{(p^r)}\]

and from assertion (2) of Lemma 4.2, the module $\mathcal{F} \otimes {(\mathcal{E}^\prime)}^{(p^r)}$ is globally generated over X.

Proof. See [Reference KleimanKle69, Proposition 9].

Proof. The proof follows directly from the isomorphism

\[ H^i(X,\mathcal{F}\otimes (\mathcal{E} \oplus \mathcal{E}^\prime)^{(p^r)}) = H^{i}(X,\mathcal{F}\otimes \mathcal{E}^{(p^r)}) \oplus H^{i}(X,\mathcal{F}\otimes {\mathcal{E}^{\prime}}^{(p^r)}) \]

Proof. Let $\mathcal{F}$ be a coherent sheaf on Y. Since f is finite, the Leray spectral sequence degenerates at page 2 and we have isomorphisms

\[H^i(X,f_*(\mathcal{F} \otimes f^*\mathcal{E}^{(p^r)})) = H^i(Y,\mathcal{F}\otimes f^*\mathcal{E}^{(p^r)})\]

for all $i \geq 0$ and $r \geq 0$ . Since f is finite, the pushforward $f_*\mathcal{F}$ is a coherent $\mathcal{O}_X$ -module and the projection formula implies that

\[f_*(\mathcal{F} \otimes f^*\mathcal{E}^{(p^r)}) = f_*\mathcal{F} \otimes \mathcal{E}^{(p^r)}.\]

Since $\mathcal{E}$ is cohomologically p-ample on X, there is an integer $r_0 \geq 1$ such that

\[H^i(X, f_*\mathcal{F} \otimes \mathcal{E}^{(p^r)}) = 0 = H^i(Y,\mathcal{F}\otimes (f^*\mathcal{E})^{(p^r)})\]

for all $i>0$ and $r\geq r_0$ . In particular, $f^*\mathcal{E}$ is cohomologically p-ample on Y.

Proof. Note that $\mathcal{L}^{(p^r)} = \mathcal{L}^{\otimes p^r}$ for all $r \geq 0$ . Assume that $r_0 \geq 1$ is an integer such that $\mathcal{L}^{\otimes p^r}(-D)$ is ample for all $r \geq r_0$ . If $\mathcal{L}$ was not nef, we could find a subcurve $C \subset X$ such that the intersection product

\[c_1(\mathcal{L})\cdot [C]\]

is negative. It would imply that the intersection product

\[c_1(\mathcal{L}^{\otimes p^r}(-D)) \cdot [C] = p^r \underbrace{\bigl( c_1(\mathcal{L})\cdot[C]\bigr)}_{<0} - {D}\cdot [C]\]

is negative for some $r \geq r_0$ large enough, which contradicts the ampleness of $\mathcal{L}^{\otimes p^r}(-D)$ . Inversely, we assume that $\mathcal{L}$ is nef and there exists an integer $n_0 \geq 1$ such that $\mathcal{L}^{\otimes n_0}(-D)$ is ample. Let r be an integer such that

\[r \geq \log_pn_0\]

and consider

\[\mathcal{L}^{(p^r)}(-D) = \mathcal{L}^{\otimes p^r}(-D) = \mathcal{L}^{\otimes n_0}(-D) \otimes \mathcal{L}^{\otimes p^r-n_0},\]

which is ample as the tensor product of an ample line bundle with a nef line bundle.

Proof. Assume that there exists an effective Cartier divisor H on X such that $\mathcal{L}$ is H-ample. We have seen in Proposition 4.22 that $\mathcal{L}$ is nef and there is an integer $n_0\geq 1$ such that $\mathcal{L}^{\otimes n_0}(-H)$ is ample. Moreover, since we can write $\mathcal{L}^{\otimes n_0}$ as a tensor product

\[\mathcal{L}^{\otimes n_0} = \mathcal{L}^{\otimes n_0}(-H) \otimes \mathcal{O}_X(H)\]

of an ample line bundle with an effective line bundle, $\mathcal{L}$ is big. We are left to show the implication $\rm (1) \Rightarrow (2)$ . Since $\mathcal{L}$ is big, there exists an integer $n_0\geq 1$ and an ample line bundle $\mathcal{A}$ such that $\mathcal{L}^{\otimes n_0}\otimes\mathcal{A}^{-1} = \mathcal{O}_X(H)$ with H an effective divisor. In particular, the line bundle $\mathcal{L}^{\otimes n_0}(-H)$ is ample. We conclude with Proposition 4.22.

Proof of Lemma 4.25. For condition (1), take a line bundle $\mathcal{E} \twoheadrightarrow \mathcal{L}^{\prime}$ such that $\delta(\mathcal{E}) = \deg\mathcal{L}^\prime$ . If we tensor it by $\mathcal{L}$ , we get $\delta(\mathcal{E} \otimes \mathcal{L}) \leq \deg \mathcal{L}^\prime + \deg\mathcal{L} = \delta(\mathcal{E}) + \deg \mathcal{L}$ . The same argument applied to $\mathcal{E} \otimes \mathcal{L}^{-1}$ shows the reverse inequality. For condition (2), take a line bundle $\mathcal{E} \twoheadrightarrow \mathcal{L}^{\prime}$ such that $\delta(\mathcal{E}) = \deg \mathcal{L}^\prime$ . The pullback $f^*$ induces a quotient map $f^*\mathcal{E} \twoheadrightarrow f^*\mathcal{L}^{\prime} = {\mathcal{L}^{\prime}}^{\otimes d}$ which shows that $\delta(f^*\mathcal{E}) \leq d\deg\mathcal{L}^{\prime} = d\delta(\mathcal{E})$ .

Proof. Assume that $\mathcal{E}$ is $(\varphi,D)$ -ample and consider $r_0 \geq 1$ such that

\[\mathcal{E}^{(p^r)}(-D)\]

is ample for all $r \geq r_0$ . By Barton’s numerical criterion of ampleness recalled in assertion (4) of Proposition 4.6, for all $r \geq r_0$ , we have a real number $\varepsilon_r > 0$ such that for all finite morphism $g : C \rightarrow X$ where C is a smooth projective curve over k, we have

\[\delta(g^*\mathcal{E}^{(p^r)}(-D)) \geq \varepsilon_r\lVert g_*C \rVert,\]

which is equivalent to

\[\delta(g^*\mathcal{E}^{(p^r)}) -D\cdot C \geq \varepsilon_r\lVert g_*C \rVert,\]

where $D\cdot C$ is the degree of the line bundle $g^*\mathcal{O}_X(D) = \mathcal{O}_X(D)_{|C}$ (it is also equal to the intersection number of D with C). If $D\cdot C\leq 0$ , then

\begin{align*}\delta(g^*\mathcal{E}^{(p^r)}(-nD)) &= \delta(g^*\mathcal{E}^{(p^r)})-D\cdot C - (n-1)D\cdot C \\&\geq \delta(g^*\mathcal{E}^{(p^r)})-D\cdot C \\&\geq \varepsilon_r\lVert g_*C \rVert,\end{align*}

for all $r\geq r_0$ . If $D\cdot C > 0$ , we take $r_1 \geq r_0$ such that $r_1 \geq r_0 + \log_p(n)$ and let $r\geq r_1$ be an integer. Since $\mathcal{E}^{(p^r)}(-D)$ is ample, the bundle

\[(\mathcal{E}^{(p^{r_0})}(-D))^{(p^{r-r_0})} = \mathcal{E}^{(p^r)}(-p^{r-r_0}D)\]

is ample and we have

\[\delta(g^*\mathcal{E}^{(p^r)}(-p^{r-r_0}D)) \geq \varepsilon_r^\prime \lVert g_*C \rVert\]

for some real number $\varepsilon_r^\prime > 0$ . Thus,

\begin{align*}\delta(g^*\mathcal{E}^{(p^r)}(-nD)) &= \delta(g^*\mathcal{E}^{(p^r)}(-p^{r-r_0}D)) +(p^{r-r_0}-n)D\cdot C \\&\geq \delta(g^*\mathcal{E}^{(p^r)}(-p^{r-r_0}D)) \\&\geq \varepsilon_r^\prime \lVert g_*C \rVert.\end{align*}

In conclusion, we have

\[\delta(g^*\mathcal{E}^{(p^r)}(-nD)) \geq \min(\varepsilon_r,\varepsilon_r^\prime) \lVert g_*C \rVert\]

for all $r \geq r_1$ and all $g : C \rightarrow X$ , which means that $\mathcal{E}$ is $(\varphi,nD)$ -ample. Inversely, consider an integer $r_0 \geq 1$ such that for all $r \geq r_0$ , we have a real number $\varepsilon_r > 0$ such that for all finite morphisms $g : C \rightarrow X$ where C is a smooth projective curve over k, we have

\[\delta(g^*\mathcal{E}^{(p^r)}) -nD\cdot C \geq \varepsilon_r\lVert g_*C \rVert.\]

If $D\cdot C\geq 0$ , we have

\begin{align*}\delta(g^*\mathcal{E}^{(p^r)}(-D)) &= \delta(g^*\mathcal{E}^{(p^r)}(-nD)) +(n-1)D\cdot C \\&\geq \delta(g^*\mathcal{E}^{(p^r)}(-nD)) \\&\geq \varepsilon_r\lVert g_*C \rVert\end{align*}

for all $r\geq r_0$ . Consider an integer $r_1 \geq r_0 + \log_pn$ . If $D\cdot C< 0$ , we have

\begin{align*}p^{r-r_0}\delta(g^*\mathcal{E}^{(p^{r_0})}(-D)) &\geq \delta(g^*\mathcal{E}^{(p^r)}(-p^{r-r_0}D)) \\&\geq \delta(g^*\mathcal{E}^{(p^r)}(-nD)) + (n-p^{r-r_0})D\cdot C \\&\geq \delta(g^*\mathcal{E}^{(p^r)}(-nD)) \\&\geq \varepsilon_r\lVert g_*C \rVert\end{align*}

for all $r \geq r_1$ . In conclusion, we have

\[\delta(g^*\mathcal{E}^{(p^r)}(-D)) \geq \frac{\varepsilon_r}{p^{r-r_0}} \lVert g_*C \rVert\]

for all $r \geq r_1$ and all $g : C \rightarrow X$ , which means that $\mathcal{E}$ is $(\varphi,D)$ -ample.

Proof. Let $r_0 \geq 1$ be an integer such that for all $r \geq r_0$ , the bundles $\mathcal{E}^{(p^r)}(-D)$ and ${(\mathcal{E}^\prime)}^{(p^r)}(-D)$ are ample. For all $r \geq r_0$ , we have

\[(\mathcal{E} \oplus \mathcal{E}^{\prime})^{(p^r)}(-D) = \mathcal{E}^{(p^r)}(-D) \oplus {(\mathcal{E}^\prime)}^{(p^r)}(-D),\]

which is ample by assertion (1) of Proposition 4.7.

Proof. On a regular scheme, the Frobenius morphism is flat by [Reference KunzKun69] or [Sta21, Lemma 0EC0]. As a consequence, we have an integer $r_0\geq 1$ and an exact sequence

of vector bundles on X where $ (\mathcal{E}_1)^{(p^r)}(-D) $ and $(\mathcal{E}_2)^{(p^r)}(-D)$ are ample for all $r\geq r_0$ . We conclude with assertion (2) of Proposition 4.7.

Proof. Let $r_0 \geq 1$ be an integer such that for all $r \geq r_0$ , the bundle $\mathcal{E}^{(p^r)}(-D)$ is ample. For all $r\geq r_0$ , the surjective morphism $\mathcal{E} \rightarrow \mathcal{E}^\prime$ induces a surjection

\[\mathcal{E}^{(p^r)}(-D) \rightarrow {(\mathcal{E}^\prime)}^{(p^r)}(-D)\]

and we conclude with assertion (3) of Proposition 4.7.

Proof. Let $r_0 \geq 1$ be an integer such that for all $r \geq r_0$ , the bundles $\mathcal{E}^{(p^r)}(-D)$ and ${(\mathcal{E}^\prime)}^{(p^r)}(-D)$ are ample. For all $r\geq r_0$ , we have

\[(\mathcal{E} \otimes \mathcal{E}^{\prime})^{(p^r)}(-2D) = \mathcal{E}^{(p^r)}(-D) \otimes {(\mathcal{E}^\prime)}^{(p^r)}(-D)\]

which is ample by assertion (5) of Proposition 4.7. It shows that $\mathcal{E}\otimes\mathcal{E}^{\prime}$ is $(\varphi,2D)$ -ample and we conclude with Proposition 4.26.

Proof. Assume that $\mathcal{E}^{\otimes n}$ is $(\varphi,D)$ -ample. By Proposition 4.26, $\mathcal{E}^{\otimes n}$ is also $(\varphi,nD)$ -ample. Let $r_0 \geq 1$ be an integer such that for all $r \geq r_0$ , the bundle ${(\mathcal{E}^{\otimes n})}^{(p^r)}(-nD)$ is ample. For all $r \geq r_0$ , the bundle

\[{(\mathcal{E}^{(p^r)}(-D))}^{\otimes n} = {(\mathcal{E}^{\otimes n})}^{(p^r)}(-nD)\]

is ample. Thus, the bundle $\mathcal{E}^{(p^r)}(-D)$ is ample for all $r \geq r_0$ by Proposition 4.8.

Proof. Let $r_0 \geq 1$ be an integer such that for all $r \geq r_0$ , the bundle $\mathcal{E}^{(p^r)}(-D)$ is ample. For all $r\geq r_0$ , the bundle

\[f^*(\mathcal{E}^{(p^r)}(-D)) = {(f^*\mathcal{E})}^{(p^r)}(-f^{-1}D)\]

is ample by Proposition 4.9. If f is assumed surjective, the converse holds by Proposition 45 again.

Proof. The first implication follows directly from [Reference BartonBar71, Proposition 3.1]. Now, assume that $\mathcal{E}$ is $(\varphi,D)$ -ample and consider the universal line bundle $\mathcal{O}(1)$ on the projective bundle $\mathbb{P}(\mathcal{E})$ . We have a surjective map $\pi^*\mathcal{E} \rightarrow \mathcal{O}(1)$ and since $(\varphi,D)$ -ampleness is stable under quotient by Proposition 4.29, $\mathcal{O}(1)$ is, in particular, nef and big by Proposition 4.24. It shows that $\mathcal{E}$ is nef and L-big. Take $r \geq 1$ such that $\mathcal{E}^{(p^r)}(-D)$ is ample. We deduce that there is an integer $n \geq 1$ such that

\[\mathrm{Sym}^n(\mathcal{E}^{(p^r)}(-D)) \otimes \mathcal{O}_X(-1) = \mathrm{Sym}^n(\mathcal{E}^{(p^r)})(-nD) \otimes \mathcal{O}_X(-1)\]

is globally generated. Since $\mathrm{Sym}^n(\mathcal{E}^{(p^r)}) \otimes \mathcal{O}_X(-1)$ can be expressed as a tensor product of a globally generated vector bundle with $\mathcal{O}_X(nD)$ , it is globally generated on the complementary open subset of the support of D. It implies that the augmented base locus of $\mathcal{E}^{(p^r)}$ is not equal to X, i.e. that $\mathcal{E}^{(p^r)}$ is V-big.

Proof. See [Sta21, Lemma 01F2].

Proof. See [Sta21, Lemma 01E8].

Proof. See [Reference AlexandreAle24, Lemma 3.19].

Proof. This is a general result on associated sheaves [Reference JantzenJan03, Part 1, Chapter 5].

Proof. The commutativity of the lower square follows directly from the definitions. We focus on the commutativity of the upper square. Consider a representation V of P. We have a cartesian diagram

where the map $\zeta$ is induced by E, the map $\zeta_P$ is induced by the universal P-reduction of E on $\mathcal{F}_P(E)$ and the vertical arrow $\tilde{\pi}$ between the classifying stacks is induced by the inclusion $P \subset G$ . Denote by $\tilde{\mathcal{L}}(V)$ the vector bundle on the classifying stack of P associated to V and $\tilde{\mathcal{W}}(V)$ the vector bundle on the classifying stack of G associated to the G-module $\mathrm{Ind}_P^G (V)$ . It follows directly from the definitions that

\[\left\{\begin{aligned}&\tilde{\pi}_* \tilde{\mathcal{L}}(V) = \tilde{\mathcal{W}}(V), \\&\zeta_{P}^*\tilde{\mathcal{L}}(V) = \mathcal{L}(V), \\&\zeta^*\tilde{\mathcal{W}}(V) = \mathcal{W}(V),\end{aligned}\right.\]

as sheaves on the stack $ \lfloor G \backslash * \rfloor$ . Since $\zeta$ is a flat morphism of algebraic stacks, the base change theorem in the derived category of quasi-coherent sheaves over X tells us that the map

(2) \begin{equation}\zeta^* \circ {R}\tilde{\pi}_* \tilde{\mathcal{L}}(V) \xrightarrow{\scriptstyle\sim} {R}\pi_*\circ\zeta_{P}^*\tilde{\mathcal{L}}(V)\end{equation}

is an isomorphism. Taking global sections in (2) yields the following isomorphism over X.

Now assume that $\lambda$ is P-dominant. By Kempf’s vanishing theorem from Proposition 2.3 combined with Lemma 5.3, we deduce

\[\left\{\begin{aligned}&{R}\pi_* \mathcal{L}_{\lambda} = \pi_* \mathcal{L}_{\lambda},\\&{R}\tilde{\pi}_* \mathcal{L}_{\lambda} = \tilde{\pi}_* \mathcal{L}_{\lambda},\end{aligned}\right.\]

and we get an isomorphism

\[R\pi_* \mathcal{L}_{\lambda} \simeq \mathcal{W}_{\lambda} [0].\]

Proof. This is a direct consequence of Propositions 2.10 and 5.10.

Proof. From the cartesian diagram

we deduce an isomorphism $\tilde{\zeta}^*\Omega^1_{\tilde{\pi}} = \Omega^1_{\pi}$ . We know that

\[\Omega^{1}_{\tilde{\pi}} \simeq \mathcal{L}(\mathrm{Lie}(G)/\mathrm{Lie}(P_I)^{\vee}),\]

hence

\[\Omega^{top}_{\tilde{\pi}} \simeq \mathcal{L}(\Lambda^{top}\mathrm{Lie}(G)/\mathrm{Lie}(P_I)^{\vee}).\]

The weights of the T-action on $\mathrm{Lie}(G)/\mathrm{Lie}(P_I)$ are the roots $-\Phi^+_I$ , so $\Lambda^{top}\mathrm{Lie}(G)/\mathrm{Lie}(P_I)$ is a one-dimensional module of weight $-2\rho_I$ and by taking the linear dual, we get an isomorphism $\Omega^{top}_{\pi} =\tilde{\zeta}^*\Omega^{top}_{\tilde{\pi}} =\mathcal{L}_{-2\rho_I}$ .