Proof. The map $p_{\lambda }$ can be presented as the map of quotient stacks $(V^{\lambda \geqslant 0}\times ^{P_{\lambda }} G)/G\rightarrow V/G$ coming from the G-equivariant map

which can be factored as a closed immersion followed by a projective map

$$\begin{align*}\begin{matrix} V^{\lambda\geqslant 0}\times^{P_{\lambda}}G&\rightarrow&V\times G/P_{\lambda}&\rightarrow&V\\ (v,g)&\mapsto&(g\cdot v,gP_{\lambda})&&\\ &&(v,gP_{\lambda})&\mapsto&v \end{matrix} \end{align*}$$

and is therefore projective. Therefore, $p_{\lambda }$ is representable and projective.

The fact that $q_{\lambda }$ is smooth, surjective of relative dimension $\#\mathcal {W}^{\lambda>0}(V)-\mathcal {W}^{\lambda >0}(\mathfrak {g})$ is easily seen. That it is in addition a vector bundle stack is a rather classical fact (which is not used in this paper). It comes from the facts that $V^{\lambda \geqslant 0}\rightarrow V^{\lambda }$ is a vector bundle (by the Białinicky-Birula decomposition [Reference Białynicki-BirulaBia73], but this can be seen directly in the case of linear actions) and the kernel of the map $P_{\lambda }\rightarrow G^{\lambda }$ is the unipotent radical of $P_{\lambda }$ .

Proof. The arguments are parallel to that of the proof [Reference Meinhardt and ReinekeMR19, Lemma 2.1]. Namely, we consider the commutative diagram

where (see §1.4), so that we have an equivalence of stacks $V^{\lambda \geqslant 0}/P_{\lambda }\simeq \widetilde {V}^{\lambda \geqslant 0}/G$ . We now explain the definitions of the maps in this diagram.

The map $q_{\lambda }$ is given by the equivariant map $(V^{\lambda \geqslant 0},P_{\lambda })\rightarrow (V^{\lambda },G^{\lambda })$ , $(v,g)\mapsto(\lim \limits _{t\rightarrow 0} \lambda (t)\cdot v,\lim \limits _{t\rightarrow 0}\lambda (t)g\lambda (t)^{-1})$ . Therefore, the pullback of functions by $q_{\lambda }$ induces a morphism between rings of invariant functions

$$\begin{align*}\overline{q}_{\lambda}^*\colon\mathbf{C}[V^{\lambda}]^{G^{\lambda}}\rightarrow \mathbf{C}[\widetilde{V}^{\lambda\geqslant 0}]^{G} \end{align*}$$

whose dual is the morphism between affine schemes

$$\begin{align*}\overline{q}_{\lambda}\colon \mathrm{Spec}(\mathbf{C}[\widetilde{V}^{\lambda\geqslant 0}]^{G})\rightarrow\mathrm{Spec}(\mathbf{C}[V^{\lambda}]^{G^{\lambda}})=V^{\lambda}/\!\!/ G^{\lambda}\,. \end{align*}$$

The map $\jmath _{\lambda }$ is given by the equivariant morphism $(V^{\lambda },G^{\lambda })\rightarrow (V^{\lambda \geqslant 0},P_{\lambda })$ , $(v,g)\mapsto (v,g)$ which induces the morphism of algebras

$$\begin{align*}\overline{\jmath}_{\lambda}^*\colon \mathbf{C}[\widetilde{V}^{\lambda\geqslant 0}]^{G}\rightarrow\mathbf{C}[V^{\lambda}]^{G^{\lambda}} \end{align*}$$

and dually the morphism of schemes

$$\begin{align*}\overline{\jmath}_{\lambda}\colon\mathrm{Spec}(\mathbf{C}[V^{\lambda}]^{G^{\lambda}})\rightarrow\mathrm{Spec}(\mathbf{C}[\widetilde{V}^{\lambda\geqslant 0}]^{G})\,. \end{align*}$$

Since $q_{\lambda }\circ \jmath _{\lambda }=\mathrm {id}$ , we have $\overline {q}_{\lambda }\circ \overline {\jmath }_{\lambda }=\mathrm {id}$ . On the ring of functions, we have $\overline {\jmath }_{\lambda }^*\circ \overline {q}_{\lambda }^*=\mathrm {id}$ and so $\overline {\jmath }_{\lambda }^*$ is surjective. This tells us that $\overline {\jmath }_{\lambda }$ is a closed immersion. Moreover, $\imath _{\lambda }=\overline {p}_{\lambda }\circ \overline {\jmath }_{\lambda }$ and so it suffices to prove that $\overline {p}_{\lambda }$ is a finite map. Since it is a morphism of finite type complex schemes, it suffices to prove that it is an integral morphism [Sta24, Lemma 01WJ].

Since $\widetilde {p}_{\lambda }\colon \widetilde {V}^{\lambda \geqslant 0}\rightarrow V$ is a projective morphism (Proof of Lemma 2.1), $(\widetilde {p}_{\lambda })_*\mathcal {O}_{\widetilde {V}^{\lambda \geqslant 0}}$ is a coherent $\mathcal {O}_{V}$ -module. Therefore, the map $\mathbf {C}[V]\rightarrow \mathbf {C}[\widetilde {V}^{\lambda \geqslant 0}]$ is finite and hence integral [Sta24, Lemma 01WJ]. We deduce that the map

$$\begin{align*}\overline{p}_{\lambda}^*\colon\mathbf{C}[V]^G\rightarrow \mathbf{C}[\widetilde{V}^{\lambda\geqslant 0}]^G \end{align*}$$

is integral, and hence finite, as follows. Let $a\in \mathbf{C}[\widetilde{V}^{\lambda\geqslant 0}]^G $ . It is a solution of a monic polynomial with coefficients in $\mathbf {C}[V]$ . By applying the Reynolds operator to the equation, we obtain a monic equation with coefficients in $\mathbf {C}[V]^G$ of which a is a zero. This concludes the proof.

Proof. We have

$$\begin{align*}\mathcal{W}(V)=\mathcal{W}^{\lambda>0}(V)\sqcup\mathcal{W}^{\lambda=0}(V)\sqcup\mathcal{W}^{\lambda<0}(V) \end{align*}$$

and by weak symmetry (Definition 2.3), $\#\mathcal {W}^{\lambda>0}(V)=\#\mathcal {W}^{\lambda <0}(V)$ . Therefore, $\#\mathcal {W}(V)=2\#\mathcal {W}^{\lambda>0}(V)+\#\mathcal {W}^{\lambda =0}(V)$ . These equalities are also valid for $V=\mathfrak {g}$ under the adjoint action of G (which is a symmetric representation, Example 2.4). We obtain the lemma by combining them together.

Proof. The induction at the level of cohomology vector spaces is given by applying the derived global section functor to the morphism of complexes (2.4) (the sheafified induction). Moreover, by Lemma 2.6, $d/2=r_{\lambda }+d_{\lambda }/2$ . Therefore, in the weakly symmetric case, the right-hand side of (2.4) is precisely $\pi _*\mathbb {D}\mathbf {Q}_{V/G}^{\mathrm {vir}}$ . Moreover, $\mathbb {D}\mathbf {Q}_{V/G}^{\mathrm {vir}}\cong \mathbf {Q}_{V/G}^{\mathrm {vir}}$ and $\mathbb {D}\mathbf {Q}_{V^{\lambda }/G^{\lambda }}^{\mathrm {vir}}\cong \mathbf {Q}_{V^{\lambda }/G^{\lambda }}^{\mathrm {vir}}$ since $V/G$ and $V^{\lambda }/G^{\lambda }$ are smooth and the chosen shifts/Tate twists (by the dimension) make the considered constant sheaves Verdier self-dual.

Proof. The cocharacters $\lambda ,\mu $ define a morphism

$$\begin{align*}\lambda\times\mu\colon \mathbf{G}_{\mathrm{m}}^{\times2}\rightarrow T\,. \end{align*}$$

We obtain $\nu $ by composing a general one-parameter subgroup $\zeta \colon \mathbf {G}_{\mathrm {m}}\rightarrow \mathbf {G}_{\mathrm {m}}^{\times 2}$ , $t\mapsto (t^a,t^b)$ with $\lambda \times \mu $ . This guarantees the first two equalities of the Lemma. To guarantee the last two, it suffices to take a general one-parameter subgroup with the constraint $b>0$ by replacing $\zeta $ by $\zeta ^{-1}$ if necessary.

Proof. The equality classically follows by base-change in the commutative diagram

(2.5)

Proof. It suffices to prove that $W^{\lambda }$ sends $\mathcal {W}^{\lambda <0}(V)\subseteq \mathcal {W}(V)$ to itself, and similarly for $\mathcal {W}^{\lambda <0}(\mathfrak {g})$ . This comes from the invariance of the pairing between characters and cocharacters: for any $w\in W$ , $\langle w\cdot \lambda ,w\cdot \alpha \rangle =\langle \lambda ,\alpha \rangle $ and if $w\in W^{\lambda }$ , $w\cdot \lambda =\lambda $ . Therefore, for any $\alpha \in X^*(T)$ , $\langle \lambda ,\alpha \rangle <0$ if and only if $\langle \lambda ,w\cdot \alpha \rangle <0$ .

Proof. The second equality follows from the $W^{\lambda }$ -invariance of both f (since $f\in \mathrm {H}^*(V^{\lambda }/G^{\lambda })=\mathrm {H}^*(V^{\lambda }/T)^{W^{\lambda }}$ ) and $k_{\lambda }$ (Lemma 2.10). The first equality is a computation of Euler classes exactly as in the proof of [Reference Kontsevich and SoibelmanKS11, Theorem 2] in the case of quiver representations.

Proof. This follows immediately from Proposition 2.11 applied to $V^{\mu }$ as a representation of $G^{\mu }$ and the cocharacter $\lambda $ of T, seen as a maximal torus of $G^{\mu }$ , and the identification of the Weyl group of $(G^{\mu })^{\lambda }=G^{\lambda }\cap G^{\mu }$ with $W^{\lambda }\cap W^{\mu }$ .

Proof. It is immediate to check that $\mathcal {W}'(V^{\lambda })=\{\alpha \in \mathcal {W}'(V)\mid \langle \lambda ,\alpha \rangle =0\}$ and for any $\alpha \in \mathcal {W}'(V^{\lambda })$ , we can identify the weight spaces $V^{\lambda }_{\alpha }=V_{\alpha }$ . Therefore, the (weak) symmetry of V implies that of $V^{\lambda }$ .

Proof. We prove the statement for V. The statement for $\mathfrak {g}$ is obtained by specialising $V=\mathfrak {g}$ , since $\mathfrak {g}$ under the adjoint action of G is a symmetric representation of G (by the existence of invariant pairings, nondegenerate up to the center of $\mathfrak {g}$ , Example 2.4).

Since V is weakly symmetric, we have $\#\mathcal {W}^{\lambda>0}(V)=\#\mathcal {W}^{\lambda <0}(V)$ and moreover, $\#\mathcal {W}(V)=\#\mathcal {W}^{\lambda>0}(V)+\#\mathcal {W}^{\lambda <0}(V)+\mathcal {W}(V^{\lambda })$ . Since we have the same equalities with $\lambda $ replaced by $\mu $ and $\mathcal {W}^{\lambda =0}(V)=\mathcal {W}^{\mu =0}(V)$ , we obtain $\#\mathcal {W}^{\lambda>0}(V)=\#\mathcal {W}^{\mu >0}(V)$ . The other equalities follow.

Proof. If $\lambda \sim \mu $ , we have $\mathcal {W}^{\lambda>0}(V)\sqcup \mathcal {W}^{\lambda <0}(V)=\mathcal {W}^{\mu>0}(V)\sqcup \mathcal {W}^{\mu <0}(V)$ and $\mathcal {W}^{\lambda>0}(\mathfrak {g})\sqcup \mathcal {W}^{\lambda <0}(\mathfrak {g})=\mathcal {W}^{\mu>0}(\mathfrak {g})\sqcup \mathcal {W}^{\mu <0}(\mathfrak {g})$ and since V and $\mathfrak {g}$ are weakly symmetric representations of G, $\overline {\mathcal {W}^{\lambda>0}(V)}=-\overline {\mathcal {W}^{\lambda <0}(V)}$ , $\overline {\mathcal {W}^{\mu>0}(V)}=-\overline {\mathcal {W}^{\mu <0}(V)}$ and similarly for $\mathfrak {g}$ . Therefore, for any $\alpha \in \mathcal {W}^{\lambda <0}(V)$ (resp. $\mathcal {W}^{\lambda <0}(\mathfrak {g})$ ), there exists $\xi _{\alpha }\in \mathbf {Q}^*$ (resp. $\zeta _{\alpha }\in \mathbf {Q}^*$ ) such that

$$\begin{align*}\begin{matrix} \mathcal{W}^{\lambda<0}(V)&\rightarrow&\mathcal{W}^{\mu<0}(V)\\ \alpha&\mapsto &\xi_{\alpha}\alpha \end{matrix} \end{align*}$$

and

$$\begin{align*}\begin{matrix} \mathcal{W}^{\lambda<0}(\mathfrak{g})&\rightarrow&\mathcal{W}^{\mu<0}(\mathfrak{g})\\ \alpha&\mapsto &\zeta_{\alpha}\alpha \end{matrix} \end{align*}$$

are bijections. Then,

$$\begin{align*}\frac{\prod_{\alpha\in\mathcal{W}^{\lambda<0}(V)}\xi_{\alpha}}{\prod_{\alpha\in\mathcal{W}^{\lambda<0}(\mathfrak{g})}\zeta_{\alpha}} k_{\lambda}=k_{\mu}\,. \end{align*}$$

We may therefore set .

Proof. We use the explicit formula for the induction (Proposition 2.11). Let $f\in \mathrm {H}^*(V^{\lambda }/G^{\lambda })=\mathrm {H}^*(V^{\mu }/G^{\mu })$ . We have

$$\begin{align*}\mathrm{Ind}_{\lambda}(f)=\sum_{w\in W/W^{\lambda}}w\cdot(fk_{\lambda}),\quad \mathrm{Ind}_{\mu}(f)=\sum_{w\in W/W^{\mu}}w\cdot(fk_{\mu})\,. \end{align*}$$

Moreover, $G^{\lambda }=G^{\mu }$ , $W^{\lambda }=W^{\mu }$ and by Lemma 2.15, $k_{\lambda }=\xi k_{\mu }$ . This concludes the proof.

Proof. By Lemma 2.16, we may assume that for some $w\in W$ , $\mu =w\cdot \lambda $ . Then, $\mathcal {W}^{\mu <0}(V)=w\cdot \mathcal {W}^{\lambda <0}(V)$ and $\mathcal {W}^{\mu <0}(\mathfrak {g})=w\cdot \mathcal {W}^{\lambda <0}(\mathfrak {g})$ . Therefore, $w(k_{\lambda })=k_{\mu }$ . Using the explicit formula for the induction product (Proposition 2.11), we obtain

$$\begin{align*}\mathrm{Ind}_{\mu}(f)=\mathrm{Ind}_{\lambda}(w^{-1}(f))\,, \end{align*}$$

and therefore, since $w^{-1}$ induces an isomorphism $\mathrm {H}^*(V^{\mu }/G^{\mu })=\mathrm {H}^*(V^{w\cdot \lambda }/G^{w\cdot \lambda })\cong \mathrm {H}^*(V^{\lambda }/G^{\lambda })$ , the images of $\mathrm {Ind}_{\lambda }$ and $\mathrm {Ind}_{\mu }$ coincide.

Proof. The group $G_{\lambda }$ is a subgroup of the center of $G^{\lambda }$ , and so it is a diagonalisable algebraic group. Therefore, its neutral component is a torus. We have a splitting $G\cong G_{\lambda }^{\circ }\times \pi _0(G_{\lambda })$ [Reference HumphreysHum12, Theorem p.104] and so, by the Künneth formula,

$$\begin{align*}\mathrm{H}^*(\mathrm{pt}/G_{\lambda})\cong\mathrm{H}^*(\mathrm{pt}/G_{\lambda}^{\circ})\otimes\mathrm{H}^*(\mathrm{pt}/\pi_0(G_{\lambda}))\,. \end{align*}$$

Since $\pi _0(G_{\lambda })$ is a finite group, $\mathrm {H}^*(\mathrm {pt}/\pi _0(G_{\lambda }))\cong \mathrm {H}^*(\mathrm {pt})^{\pi _0(G_{\lambda })}\cong \mathbf {Q}$ , which concludes.

Proof. We let $\dot {w}\in N_G(T)$ be a lift of $w\in W$ . Then, $V^{w\cdot \lambda }=\dot {w}V^{\lambda }$ . The first statement of the lemma follows then immediately from the fact that if $w,w'\in W$ , and $\dot {w}, \dot {w'}\in N_G(T)$ are respective lifts, then, $\dot {w}\dot {w'}$ is a lift of $ww'$ and $\dot {w}^{-1}$ a lift of $w^{-1}$ .

We prove that $W^{\lambda }\subseteq W_{\lambda }$ . Let $w\in W^{\lambda }$ and $\dot {w}\in N_{G^{\lambda }}(T)$ be a lift of w. Then, $\dot {w}G^{\lambda }\dot {w}^{-1}=G^{w\cdot \lambda }=G^{\lambda }$ and $\dot {w}V^{\lambda }=V^{w\cdot \lambda }=V^{\lambda }$ and so $w\in W_{\lambda }$ . This proves the inclusion.

We now show that $W^{\lambda }$ is a normal subgroup of $W_{\lambda }$ . It suffices to check that if $w\in W_{\lambda }$ and $w'\in W^{\lambda }$ , and $\dot {w}\in N_G(T)$ , $\dot {w'}\in N_{G^{\lambda }}(T)$ are any lifts, then $\dot {w}\dot {w'}\dot {w}^{-1}\in G^{\lambda }$ . We have $\dot {w}\dot {w'}\dot {w}^{-1}\in \dot {w}G^{\lambda }\dot {w}^{-1}=G^{w\cdot \lambda }=G^{\lambda }$ . This concludes the proof.

Proof. The Weyl group W of G acts on $\mathrm {H}^*(\mathrm {pt}/T)$ . We just have to check that the induced action of the subgroup $W_{\lambda }$ preserves $W^{\lambda }$ -invariants. This is a consequence of Lemma 4.2.

Proof. By definition, $G_{\lambda }$ is preserved by $W_{\lambda }$ , as $G_{\lambda }$ only depends on $\overline {\lambda }\in \mathscr {P}_V$ and $W_{\lambda }$ is the stabiliser of $\overline {\lambda }$ . The second statement follows from the fact that $G_{\lambda }$ is contained in the center of $G^{\lambda }$ .

Proof. We have $\mathrm {H}^*(V^{\lambda }/\overline {G^{\lambda }})\cong \mathrm {H}^*(\mathrm {pt}/(T/G_{\lambda }^{\circ }))^{W^{\lambda }}$ . The restriction to $W_{\lambda }$ of the W action on T preserves $G_{\lambda }$ since $G_{\lambda }$ is in the center of $G^{\lambda }$ . Therefore, we obtain a $W_{\lambda }$ -action on $T/G_{\lambda }^{\circ }$ . It induces a $W_{\lambda }$ -action on $\mathrm {H}^*(\mathrm {pt}/(T/G_{\lambda }^{\circ }))$ . The fact that $W_{\lambda }$ normalises $W^{\lambda }$ (Lemma 4.2) implies that $\mathrm {H}^*(\mathrm {pt}/(T/G_{\lambda }^{\circ }))^{W^{\lambda }}$ is preserved by the $W_{\lambda }$ -action.

The tensor product decomposition of the action follows immediately from this and Lemma 4.4.

Proof. It suffices to prove the existence and unicity of the rational number $\varepsilon _{V,\lambda }(w)\in \mathbf {Q}^*$ satisfying the equality of the proposition, for any $w\in W_{\lambda }$ . The multiplicativity then follows from the unicity of this sign.

For $w\in W_{\lambda }$ , the cocharacters $\lambda ,w^{-1}\cdot \lambda $ of T are such that $\overline {\lambda }=\overline {w^{-1}\cdot \lambda }$ . Therefore, by Lemma 2.15, there exists $\xi \in \mathbf {Q}^*$ such that $w(k_{\lambda })=k_{w^{-1}\cdot \lambda }=\xi k_{\lambda }$ . We set . This proves the existence. The unicity of $\xi $ is straightforward.

Proof. By Proposition 2.11, we have

$$\begin{align*}\begin{aligned} \mathrm{Ind}_{\lambda}(w\cdot f)&=\frac{1}{\# W^{\lambda}}\sum_{w'\in W}w'(wf)w'(k_{\lambda})\\ &=\frac{1}{\# W^{\lambda}}\sum_{w"\in W}(w"f)w"(w^{-1}k_{\lambda})\\ &=\frac{1}{\# W^{\lambda}}\sum_{w"\in W}(w"f)w"(\varepsilon_{V,\lambda}(w)k_{\lambda}) \quad\text{ by Proposition}~4.6 \end{aligned} \end{align*}$$

which concludes.

Proof. This is [Reference SerreSer71, §2.6, Théorème 8].

Proof. If $\mu \prec \lambda $ , then we have $\dim V^{\mu }\leqslant \dim V^{\lambda }$ and $\dim \mathfrak {g}^{\mu }\leqslant \dim \mathfrak {g}^{\lambda }$ with at least one strict inequality (where the order relation $\preceq $ is defined in §2.6). If by contradiction we have $\lambda \preceq \mu $ , then we have

$$\begin{align*}\dim V^{\lambda}\leqslant \dim V^{\mu}\,, \quad \dim\mathfrak{g}^{\lambda}\leqslant\dim \mathfrak{g}^{\mu}\,. \end{align*}$$

Therefore, if $\mu \prec \lambda $ , then $\lambda \not \preceq \mu $ and by definition, $k_{\mu ,\lambda }\in J_{\lambda }$ .

Conversely, if $\lambda \not \preceq \mu $ , we let $\nu \in X_*(T)$ be such that $V^{\nu }=V^{\lambda }\cap V^{\mu }$ , $\mathfrak {g}^{\nu }=\mathfrak {g}^{\lambda }\cap \mathfrak {g}^{\mu }$ , $(V^{\lambda })^{\nu \geqslant 0}=(V^{\lambda })^{\mu \geqslant 0}$ and $(\mathfrak {g}^{\lambda })^{\nu \geqslant 0}=(\mathfrak {g}^{\lambda })^{\mu \geqslant 0}$ (Lemma 2.8). We have $V^{\nu }\subseteq V^{\lambda }$ and $\mathfrak {g}^{\nu }\subseteq \mathfrak {g}^{\lambda }$ with at least one strict inclusion. Therefore, $\nu \prec \lambda $ and moreover, $k_{\nu ,\lambda }=k_{\mu ,\lambda }$ . This proves the lemma.

Proof. This comes from the fact that the induction product §2.2 is well-defined on the cohomology, without requiring localisation, even though the explicit formula (Proposition 2.11) involves rational fractions. More precisely, $J_{\lambda }^{W^{\lambda }}$ is linearly generated by the elements

$$\begin{align*}\sum_{w\in W^{\lambda}}w\cdot(fk_{\mu,\lambda}) \end{align*}$$

for $f\in \mathcal {A}_{\lambda }^{\mathrm {prim}}$ and $\mu \prec \lambda $ , by Lemma 4.9. These are averages over $W^{\lambda }$ of elements of $J_{\lambda }$ . We can rewrite this sum as

$$\begin{align*}\sum_{\overline{w'}\in W^{\lambda}/W^{\mu}}w'\cdot\left[\left(\sum_{w\in W^{\mu}}(w\cdot f)\right)k_{\mu,\lambda}\right] \end{align*}$$

since $k_{\mu ,\lambda }$ is $W^{\mu }$ -invariant (Lemma 2.10). This is (Proposition 2.11) the formula for

Since $\sum _{w\in W^{\mu }}(w\cdot f)$ is polynomial, its induction I is also polynomial. Therefore, it belongs to $\mathcal {H}_{\lambda }$ . Moreover, $I\in J_{\lambda }\subseteq \mathcal {A}_{\lambda }^{\mathrm {prim}}[\prod _{\alpha \in \mathcal {W}(\mathfrak {g}^{\lambda })\smallsetminus \{0\}}\alpha ^{-1}]$ by $W^{\lambda }$ -invariance of $J_{\lambda }$ . Therefore, $I\in \mathcal {H}_{\lambda }\cap \mathcal {A}_{\lambda }^{\mathrm {prim}}[\prod _{\alpha \in \mathcal {W}(\mathfrak {g}^{\lambda })\smallsetminus \{0\}}\alpha ^{-1}]=\mathcal {H}_{\lambda }^{\mathrm {prim}}$ .

For the second claim of the lemma, we have obviously $J_{\lambda }^{W^{\lambda }}\subseteq J_{\lambda }$ . Combined with the first part of the lemma, this gives the reverse inclusion $J_{\lambda }^{W^{\lambda }}\subseteq J_{\lambda }\cap \mathcal {H}_{\lambda }^{\mathrm {prim}}$ . For the direct inclusion, if $f\in J_{\lambda }$ is also in $\mathcal {H}_{\lambda }^{\mathrm {prim}}=(\mathcal {A}_{\lambda }^{\mathrm {prim}})^{W^{\lambda }}$ , it clearly is $W^{\lambda }$ -invariant: $f\in J_{\lambda }^{W^{\lambda }}$ .

Proof. Let $w\in W_{\lambda }$ . We let . Then, $J_{\lambda }^w$ is $W^{\lambda }$ -stable by Lemma 4.2. It is an $\mathcal {A}_{\lambda }^{\mathrm {prim}}$ -submodule of $\mathcal {A}_{\lambda }^{\mathrm {prim}}[\prod _{\alpha \in \mathcal {W}(\mathfrak {g}^{\lambda })\smallsetminus \{0\}}(w\cdot \alpha )^{-1}]=\mathcal {A}_{\lambda }^{\mathrm {prim}}[\prod _{\alpha \in \mathcal {W}(\mathfrak {g}^{\lambda })\smallsetminus \{0\}}\alpha ^{-1}]$ . Moreover, it contains

$$\begin{align*}k_{w\cdot\mu,w\cdot\lambda} \end{align*}$$

for $\lambda \not \preceq \mu $ . Now, we use that by Lemma 2.15, for $w\in W_{\lambda }$ , $\{\pm k_{\mu ,\lambda }\colon \lambda \not \preceq \mu \}=\{\pm k_{w\cdot \mu ,w\cdot \lambda }\mid \lambda \not \preceq \mu \}$ since for $w\in W_{\lambda }$ , $\overline {w\cdot \lambda }=\overline {\lambda }$ . By minimality of $J_{\lambda }$ , we deduce that $J_{\lambda }\subseteq J_{\lambda }^w$ . By symmetry of the argument, we conclude $J_{\lambda }^w=J_{\lambda }$ .

Proof. This follows directly from Lemmas 4.2 and 4.11.

Proof. We only prove the second part of the lemma, as the second part can be proved in the exact same way by using Lemma 4.9.

This is a calculation very similar to that in the proof of Lemma 4.11. Namely, if $f\in \mathcal {H}_{\mu ,\lambda }$ and $g\in \mathrm {H}^*(\mathrm {pt}/G_{\lambda })$ , $\mathrm {Ind}_{\mu ,\lambda }(fg)=\mathrm {Ind}_{\mu ,\lambda }(f)g$ by $W^{\lambda }$ -invariance of g (Lemma 4.4). Moreover, $fg\in \mathcal {H}_{\mu ,\lambda }$ . Therefore, a spanning subset of $J_{\lambda }^{W^{\lambda }}\otimes \mathrm {H}^*(\mathrm {pt}/G_{\lambda })$ can be obtained in this way by considering all $\mu \prec \lambda $ . Conversely, if $f\in \mathcal {H}_{\mu ,\lambda }\subseteq \mathcal {A}_{\mu ,\lambda }=\mathcal {A}_{\lambda }=\mathrm {H}^*(\mathrm {pt}/T)$ for $\mu \prec \lambda $ , we write $f=\sum _{i=1}^Nf_i\otimes g_i$ with $f_i\in \mathcal {A}_{\lambda }^{\mathrm {prim}}$ and $g_i\in \mathrm {H}^*(\mathrm {pt}/G_{\lambda })$ so that

$$\begin{align*}\mathrm{Ind}_{\mu,\lambda}(f)=\sum_{i=1}^N\mathrm{Ind}_{\mu,\lambda}(f_i)\otimes g_i \end{align*}$$

indeed belongs to $J_{\lambda }^{W^{\lambda }}\otimes \mathrm {H}^*(\mathrm {pt}/G_{\lambda })$ .

Proof. We proceed by induction on $\lambda \in X_*(T)$ . If $\lambda $ is minimal, then $J_{\lambda }=\{0\}$ (by Lemma 4.9) and so $\mathcal {P}_{\lambda }=\mathrm {H}^*(V^{\lambda }/(G^{\lambda }/G_{\lambda }))$ . By Proposition 3.7, the map

$$\begin{align*}\mathcal{P}_{\lambda}\otimes\mathrm{H}^*(\mathrm{pt}/G_{\lambda})\rightarrow\mathrm{H}^*(V^{\lambda}/G^{\lambda}) \end{align*}$$

of the lemma is indeed surjective (as it even is an isomorphism).

Now, let $\lambda \in X_*(T)$ . We assume that the map of the lemma is surjective for any $\lambda '\prec \lambda $ . By associativity of the induction (Proposition 2.9), the image of the map of the lemma coincides with the image of the composition

$$\begin{align*}\left(\bigoplus_{\mu\prec\lambda}\bigoplus_{\mu'\preceq\mu}\mathcal{P}_{\mu'}\otimes\mathrm{H}^*(\mathrm{pt}/G_{\mu'})\right)\oplus \left(\mathcal{P}_{\lambda}\otimes\mathrm{H}^*(\mathrm{pt}/G_{\lambda})\right)\xrightarrow{I\oplus\mathrm{id}} \left(\bigoplus_{\mu\prec\lambda}\mathcal{H}_{\mu}\right)\oplus\left(\mathcal{P}_{\lambda}\otimes\mathrm{H}^*(\mathrm{pt}/G_{\lambda})\right)\rightarrow\mathcal{H}_{\lambda} \end{align*}$$

where the morphism I is $\bigoplus _{\mu \prec \lambda }\bigoplus _{\mu '\preceq \mu }\mathrm {Ind}_{\mu ',\mu }$ . By induction hypothesis,

$$\begin{align*}\left(\bigoplus_{\mu\prec\lambda}\bigoplus_{\mu'\preceq\mu}\mathcal{P}_{\mu'}\otimes\mathrm{H}^*(\mathrm{pt}/G_{\mu'})\right)\rightarrow\left(\bigoplus_{\mu\prec\lambda}\mathcal{H}_{\mu}\right) \end{align*}$$

is surjective and by Lemma 4.13, the image of

$$\begin{align*}\left(\bigoplus_{\mu\prec\lambda}\mathcal{H}_{\mu}\right)\rightarrow\mathcal{H}_{\lambda} \end{align*}$$

is $J_{\lambda }^{W^{\lambda }}\otimes \mathrm {H}^*(\mathrm {pt}/G_{\lambda })$ , of which $\mathcal {P}_{\lambda }\otimes \mathrm {H}^*(\mathrm {pt}/G_{\lambda })$ is a direct sum complement by definition of $\mathcal {P}_{\lambda }$ . This proves the surjectivity.

Proof. We may assume that $y\neq 0$ . Via the natural embeddings $X_*(T)\subseteq \mathfrak {t}_{\mathbf {Q}}$ and $X^*(T)\subseteq \mathfrak {t}_{\mathbf {Q}}^*$ , the natural pairing $X_*(T)\times X^*(T)\rightarrow \mathbf {Z}$ corresponds to the natural pairing $\langle -,-\rangle \colon \mathfrak {t}\times \mathfrak {t}^*\rightarrow \mathbf {C}$ (see also §1.4). We let

and

The rational vector space $\cap _{\alpha \in \mathcal {W}_1}\alpha ^{\perp }\subseteq \mathfrak {t}_{\mathbf {Q}}$ is non-zero, as it is defined by the rational equations $\langle -,\alpha \rangle =0$ and its complexification contains y. The complement of the finite union of rational hyperplanes is therefore also non-empty. We let $y'\in U_{y}$ be a rational point ( $y\in \mathfrak {t}_{\mathbf {Q}}$ ). For some $N\geqslant 1$ , $Ny'\in X_*(T)\subseteq \mathfrak {t}$ . We let . It satisfies the property of the lemma.

Proof. By possibly replacing V by $V^{\lambda }$ and G by $G^{\lambda }$ , we may assume that $\lambda =0$ . It suffices to prove that $J_{\lambda }^{W^{\lambda }}\subseteq \mathcal {H}_{\lambda }^{\mathrm {prim}}$ has finite codimension. If we replace V by $V\times \mathfrak {g}$ , then the corresponding module $J_{\lambda }\subseteq \mathcal {A}^{\mathrm {prim}}_{\lambda }[\prod _{\alpha \in \mathcal {W}(\mathfrak {g})\smallsetminus \{0\}}\alpha ^{-1}]$ is smaller than the one for V (by the very definition of $J_{\lambda }$ at the beginning of §4.2, since when replacing V by $V\times \mathfrak {g}$ , the new $k_{\lambda }$ is equal to the numerator of the old $k_{\lambda }$ ), and therefore so is the corresponding module of invariants $J_{\lambda }^{W^{\lambda }}$ , and therefore its codimension is bigger. Moreover, in this case, $J_{\lambda }\subseteq \mathcal {A}_{\lambda }^{\mathrm {prim}}$ , i.e. there is no need to localise.

We have an inclusion

$$\begin{align*}\mathcal{H}_{\lambda}^{\mathrm{prim}}/J_{\lambda}^{W^{\lambda}}\rightarrow\mathcal{A}_{\lambda}^{\mathrm{prim}}/J_{\lambda}\,, \end{align*}$$

because $\mathcal {H}_{\lambda }^{\mathrm {prim}}\cap J_{\lambda }=J_{\lambda }^{W^{\lambda }}$ (second part of Lemma 4.10). Therefore, it suffices to prove that the ideal $J_{\lambda }\subseteq \mathcal {A}_{\lambda }^{\mathrm {prim}}$ has finite codimension. We see $\mathcal {A}_{\lambda }^{\mathrm {prim}}$ as the ring of regular functions on the affine algebraic variety $\mathfrak {t}/\mathfrak {g}_{\lambda }$ . We prove that the closed subscheme defined by $J_{\lambda }$ is supported on $\{0\}$ . If the ideal $J_{\lambda }$ vanishes on $y\in (\mathfrak {t}/\mathfrak {g}_{\lambda })_{\overline {\mathbf {Q}}}\smallsetminus \{0\}$ (where $\overline {\mathbf {Q}}$ is the field of algebraic numbers) we let $y'\in \mathfrak {t}$ be a lift of y and $\nu $ be given by Lemma 4.15. It satisfies $\mathcal {W}^{\nu =0}(V)=\{\alpha \in \mathcal {W}(V)\mid \langle y,\alpha \rangle =0\}\subseteq \mathcal {W}(V)$ and $\mathcal {W}^{\nu =0}(\mathfrak {g})=\{\alpha \in \mathcal {W}(\mathfrak {g})\mid \langle y,\alpha \rangle =0\} \subseteq \mathcal {W}(\mathfrak {g})$ . Moreover, at least one of these inequalities is strict. Indeed, we have $y'\not \in \mathfrak {g}_{\lambda }$ , which means that either $y'$ is not in the center of $\mathrm {Lie}(G^{\lambda })$ and so $\nu $ defines a strict Levi $G^{\nu }=G^y$ of $G^{\lambda }$ or $y'$ is not in the Lie algebra of the kernel of the action $G^{\lambda }\rightarrow \mathrm {GL}(V^{\lambda })$ in which case $V^{\nu }=V^{y}\subsetneq V^{\lambda }$ . Therefore, $\nu \prec \lambda =0$ , i.e $\overline {\nu }$ is nontrivial. By definition, we have $\prod _{\alpha \in \mathcal {W}^{\nu <0}(V^{\lambda })}\alpha \in J_{\lambda }$ .

We also have $\prod _{\alpha \in \mathcal {W}^{\nu <0}(V^{\lambda })}\langle y,\alpha \rangle \neq 0$ : indeed, by choice of $\nu $ , for any $\alpha \in \mathcal {W}^{\nu <0}(V^{\lambda })$ , $\langle \nu ,\alpha \rangle =0$ if and only if $\langle y,\alpha \rangle =0$ . This is a contradiction since we found an element $\prod _{\alpha \in \mathcal {W}^{\nu <0}(V^{\lambda })}\alpha \neq 0$ of $J_{\lambda }$ not vanishing on y. Therefore, $J_{\lambda }$ defines a subscheme of $\mathfrak {t}/\mathfrak {g}_{\lambda }$ supported on $0$ .

Proof. Let $\alpha \in \mathcal {W}^{\lambda <0}(V)\cup \mathcal {W}^{\lambda <0}(\mathfrak {g})$ . We see $\alpha \in \mathfrak {t}^*$ via the natural inclusion $X^*(T)\subseteq \mathfrak {t}^*$ and $\langle \lambda ,\alpha \rangle \neq 0$ implies that $\alpha $ does not vanish on $\mathfrak {g}_{\lambda }$ since it takes a nonzero value on $\left (\frac {\mathrm {d}}{\mathrm {d}t}\lambda (t)\right )_{t=1}\in \mathfrak {g}_{\lambda }\subseteq \mathfrak {t}$ . Therefore, the projection $\alpha '\in \mathfrak {g}_{\lambda }^*$ of $\alpha $ with respect to the chosen (or any) decomposition $\mathfrak {t}^*\cong (\mathfrak {t}/\mathfrak {g}_{\lambda })^*\oplus \mathfrak {g}_{\lambda }^*$ is nonzero.

We choose a basis of $\mathfrak {g}_{\lambda }^*$ such that $x_1=\alpha '$ and we order the monomials by the lexicographic order so that . Let $g\in \mathcal {A}_{\lambda }^{\mathrm {loc}}\cong \mathcal {A}_{\lambda }^{\mathrm {prim},\mathrm {loc}}\otimes \mathrm {H}^*(\mathrm {pt}/G_{\lambda })$ . It may be written in a unique way (see Lemma 4.1)

(4.1)

with $g_{\nu }\in \mathcal {A}_{\lambda }^{\mathrm {prim},\mathrm {loc}}$ . We have an equivalence between the following two statements:

1. (1) $g\not \in \widetilde {J}_{\lambda }$ ,

2. (2) $g_{\nu }\not \in J_{\lambda }$ for some $\nu \in \mathbf {N}^{\dim \mathfrak {g}_{\lambda }}$ .

The implication $(1)\Rightarrow (2)$ is immediate as if all $g_{\nu }$ belong to $J_{\lambda }$ , then g would be in $\widetilde {J}_{\lambda }$ . The reverse implication $(2)\Rightarrow (1)$ is also true by the unicity of the decomposition (4.1) of g.

Assume that $g\not \in \widetilde {J}_{\lambda }$ . We need to show that for $\alpha \in \mathcal {W}^{\lambda <0}(V)\cup \mathcal {W}^{\lambda <0}(\mathfrak {g})$ , $\alpha g\not \in \widetilde {J}_{\lambda }$ . We let $\nu =\max \{\nu '\in \mathbf {N}^{\dim \mathfrak {g}_{\lambda }}\colon g_{\nu '}\neq 0\}$ (the maximum is taken with respect to the lexicographic order). We may assume that $g_{\nu }\not \in J_{\lambda }$ . We have

$$\begin{align*}\begin{aligned} \alpha g&=(\alpha-\alpha')g+\alpha'g\\ &=(\alpha-\alpha')g+x_1g \end{aligned} \end{align*}$$

and $\alpha -\alpha '\in (\mathfrak {t}/\mathfrak {g}_{\lambda })^*\subseteq \mathrm {Sym}((\mathfrak {t}/\mathfrak {g}_{\lambda })^*)$ . Therefore, the term of highest degree in the variables in $\alpha g$ is $g_{\nu }x_1x^{\nu }$ and $g_{\nu }\not \in J_{\lambda }$ . By the equivalence above, $\alpha g\not \in \widetilde {J}_{\lambda }$ . This concludes the proof.

Proof. This follows from Lemma 4.17 as we localise by a set of nonzero divisors of the quotient $\mathcal {A}_{\lambda }$ -module $\mathcal {A}_{\lambda }^{\mathrm {loc}}/\widetilde {J}_{\lambda }$ in $\mathcal {A}_{\lambda }$ . More precisely, the first localisation morphism $L\colon \mathcal {A}_{\lambda }^{\mathrm {loc}}/\widetilde {J}_{\lambda }\rightarrow \mathcal {A}_{\lambda }^{\prime \mathrm {loc}}/\widetilde {J}_{\lambda }'$ is the localisation with respect to $\prod _{\alpha \in \mathcal {W}^{\lambda <0}(V)\cup \mathcal {W}^{\lambda <0}(\mathfrak {g})}\alpha $ for the $\mathcal {A}_{\lambda }$ -module $\mathcal {A}_{\lambda }^{\mathrm {loc}}/\widetilde {J}_{\lambda }$ , and the localisation element is not a zero-divisor of $\mathcal {A}_{\lambda }^{\mathrm {loc}}/\widetilde {J}_{\lambda }$ . Therefore, the first localisation morphism is injective.

Now, the two localisation morphisms of the lemma fit into the commutative diagram

(4.2)

The vertical map a is induced by the inclusion $\mathcal {H}_{\lambda }=\mathcal {H}_{\lambda }^{W^{\lambda }}\subseteq \mathcal {A}_{\lambda }\subseteq \mathcal {A}_{\lambda }^{\mathrm {loc}}$ . The map a is injective since $\mathcal {H}_{\lambda }\cap \widetilde {J}_{\lambda }=(\widetilde {J}_{\lambda })^{W^{\lambda }}$ , which follows from the arguments of the proof of Lemma 4.10. Therefore, the second localisation map (the top line of (4.2)) must be injective.

Proof. By the properties satisfied by the cocharacter $\nu $ given by Lemma 2.8, we have $\mathcal {W}^{\mu <0}(V)=\mathcal {W}^{\nu <0}(V)\cap \mathcal {W}^{\lambda =0}(V)\sqcup \mathcal {W}^{\mu <0}(V)\cap \mathcal {W}^{\lambda \neq 0}(V)$ and similarly for V and $\mathfrak {g}$ exchanged. Therefore,

$$\begin{align*}\begin{aligned} k_{\mu}&=\frac{\prod_{\alpha\in\mathcal{W}^{\mu<0}(V)}\alpha}{\prod_{\alpha\in\mathcal{W}^{\mu<0}(\mathfrak{g})}\alpha}\\ &=\frac{\prod_{\alpha\in\mathcal{W}^{\nu<0}(V)\cap\mathcal{W}^{\lambda=0}(V)}\alpha}{\prod_{\alpha\in\mathcal{W}^{\nu<0}(\mathfrak{g})\cap\mathcal{W}^{\lambda=0}(\mathfrak{g})}\alpha}\frac{\prod_{\alpha\in\mathcal{W}^{\mu<0}(V)\cap\mathcal{W}^{\lambda\neq0}(V)}\alpha}{\prod_{\alpha\in\mathcal{W}^{\mu<0}(\mathfrak{g})\cap\mathcal{W}^{\lambda\neq0}(\mathfrak{g})}\alpha}\\ &=k_{\nu,\lambda}\frac{\prod_{\alpha\in\mathcal{W}^{\mu<0}(V)\cap\mathcal{W}^{\lambda\neq0}(V)}\alpha}{\prod_{\alpha\in\mathcal{W}^{\mu<0}(\mathfrak{g})\cap\mathcal{W}^{\lambda\neq0}(\mathfrak{g})}\alpha}\,. \end{aligned} \end{align*}$$

Proof. Of course, the presence of W is superfluous since $w k_{\mu }=k_{w^{-1}\cdot \mu }$ for any $w\in W$ and it suffices to prove that for any $\lambda ,\mu \in X_*(T)$ such that $\lambda \not \preceq \mu $ , $k_{\mu }\in \widetilde {J}^{\prime }_{\lambda }$ .

We let $\nu \in X_*(T)$ be such that $V^{\nu }=V^{\lambda }\cap V^{\mu }$ , $\mathfrak {g}^{\nu }=\mathfrak {g}^{\lambda }\cap \mathfrak {g}^{\mu }$ , $(V^{\lambda })^{\nu \geqslant 0}=(V^{\lambda })^{\mu \geqslant 0}$ and $(\mathfrak {g}^{\lambda })^{\nu \geqslant 0}=(\mathfrak {g}^{\lambda })^{\mu \geqslant 0}$ (Lemma 2.8). By assumption, $\nu \prec \lambda $ .

We can write, by Lemma 4.19,

$$\begin{align*}k_{\mu}=k_{\nu,\lambda}\underbrace{\frac{\prod_{\alpha\in\mathcal{W}^{\mu<0}(V)\cap\mathcal{W}^{\lambda\neq0}(V)}\alpha}{\prod_{\alpha\in\mathcal{W}^{\mu<0}(\mathfrak{g})\cap\mathcal{W}^{\lambda\neq0}(\mathfrak{g})}\alpha}}_{=k^{\prime}_{\mu,\lambda}\in\mathcal{A}^{\prime}_{\lambda}}\,. \end{align*}$$

This element is in $\widetilde {J}^{\prime }_{\lambda }$ since $k_{\nu ,\lambda }$ is in ${J}_{\lambda }$ , by Lemma 4.9.

Proof. The lemma follows from the inclusions $V^{\lambda }\subseteq V^{w\cdot \lambda }$ and $\mathfrak {g}^{\lambda }\subseteq \mathfrak {g}^{w\cdot \lambda }$ and the equalities of dimensions $\dim V^{\lambda }=\dim V^{w\cdot \lambda }$ and $\dim \mathfrak {g}^{\lambda }=\dim \mathfrak {g}^{w\cdot \lambda }$ .

Proof. We calculate:

$$\begin{align*}\begin{aligned} &L\mathrm{Res}_{\lambda}\left(\sum_{\widetilde{\mu}\in\mathscr{P}_V/W}\mathrm{Ind}_{\mu}(f_{\mu})\right)\\ &\quad =L\mathrm{Res}_{\lambda}\left( \sum_{\widetilde{\mu}\in\mathscr{P}_V/W}\frac{1}{\# W^{\mu}}\sum_{w\in W}(w\cdot f_{\mu})(w\cdot k_{\mu})\right)\quad\text{by Proposition}~2.11\\ &\quad \equiv\frac{1}{\# W^{\lambda}}\sum_{\substack{w\in W\\ w\overline{\lambda}=\overline{\lambda}}}(w\cdot f_{\lambda})(w\cdot k_{\lambda})\quad \pmod{\widetilde{J}_{\lambda}'}\quad\text{by Lemmas}~4.20 \text{ and } 2.21\\ &\quad = \frac{1}{\# W^{\lambda}}k_{\lambda}\cdot\sum_{w\in W_{\lambda}}\varepsilon_{V,\lambda}(w)^{-1}(w\cdot f_{\lambda})\quad\text{by definition of the character} \varepsilon_{V,\lambda}, \text{ Proposition}~4.6\,. \end{aligned}\\[-18pt] \end{align*}$$

Proof. Let $f_{\mu }\in (\mathcal {P}_{\mu }\otimes \mathrm {H}^*(\mathrm {pt}/G_{\mu }))^{\varepsilon _{V,\mu }}$ , $\widetilde {\mu }\in \mathscr {P}_V/W$ . We let . We assume that not all of the $f_{\mu }$ ’s are zero, and we have to prove that $U\neq 0$ . We let $\widetilde {\lambda }\in \mathscr {P}_V/W$ be such that $f_{\lambda }\neq 0$ and for any $\widetilde {\mu }\in \mathscr {P}_V/W$ such that $f_{\mu }\neq 0$ , for any $w\in W$ , $\lambda \not \preceq w\cdot \mu $ (otherwise, we may replace $\widetilde {\lambda }$ by $\widetilde {\mu }$ , etc.).

By Corollary 4.22, we have

$$\begin{align*}\begin{aligned} L\mathrm{Res}_{\lambda}\left(\sum_{\widetilde{\mu}\in\mathscr{P}_V/W}\mathrm{Ind}_{\mu}(f_{\mu})\right)&\equiv k_{\lambda}\cdot\sum_{w\in W_{\lambda}/W^{\lambda}}\varepsilon_{V,\lambda}(w)^{-1}(w\cdot f_{\lambda})\quad \pmod{\widetilde{J}^{\prime}_{\lambda}}\\ &\equiv k_{\lambda}\# (W_{\lambda}/W^{\lambda}) f_{\lambda}\quad \pmod{\widetilde{J}^{\prime}_{\lambda}}\quad\text{by Proposition } 4.8\,. \end{aligned} \end{align*}$$

and moreover, $f_{\lambda }\not \in \widetilde {J}^{\prime }_{\lambda }$ since $f_{\lambda }\in \mathcal {P}_{\lambda }\otimes \mathrm {H}^*(\mathrm {pt}/G_{\lambda })$ , and $k_{\lambda }$ is invertible in $\mathcal {A}_{\lambda }'$ . We obtain that $\mathrm {Res}_{\lambda }(U)\neq 0$ . Therefore, $U\neq 0$ , proving injectivity.

Proof. The injectivity is Proposition 4.23. The surjectivity comes from Lemmas 4.14 and 2.17. Indeed, by Lemma 4.14, any element $f\in \mathrm {H}^*(V/G)$ has the form $f=\sum _{\mu \preceq 0}\mathrm {Ind}_{\mu }(g_{\mu })$ for some $g_{\mu }\in \mathcal {P}_{\mu }\otimes \mathrm {H}^*(\mathrm {pt}/G_{\mu })$ . By Lemma 2.17, we may rewrite $f=\sum _{\widetilde {\mu }\in \mathscr {P}_{V}/W}\mathrm {Ind}_{\mu }(\widetilde {g}_{\mu })$ . We have the formula

$$\begin{align*}\mathrm{Ind}_{\mu}(\widetilde{g}_{\mu})=\frac{1}{\# W^{\mu}}\sum_{\overline{w'}\in W/W_{\mu}}w'\cdot\left(\sum_{w\in W_{\mu}}\varepsilon_{V,\mu}(w)^{-1}w\cdot \widetilde{g}_{\mu} \right)(w'\cdot k_{\mu}) \end{align*}$$

for $\widetilde {g}_{\mu }\in \mathcal {H}_{\mu }$ , deduced from Propositions 2.11 and 4.6 which implies, using Proposition 4.8, that $\mathrm {Ind}_{\mu }(\widetilde {g}_{\mu })$ vanishes if f is not in the $\varepsilon _{V,\mu }$ -isotypic component of $\mathcal {P}_{\mu }\otimes \mathrm {H}^*(\mathrm {pt}/G_{\mu })$ for the $W_{\mu }$ -action. Therefore, we may assume that $\widetilde {g}_{\mu }$ is $\varepsilon _{V,\lambda }$ -invariant. This concludes the proof of the corollary.

Proof. By possibly replacing V by $V^{\lambda }$ and G by $G^{\lambda }$ , we may assume that $\lambda $ is the trivial cocharacter of T. If $\ker (G\rightarrow \mathrm {GL}(V))^{\circ }\not \subseteq Z(G)$ , then there exists a cocharacter $\mu \colon \mathbf {G}_{\mathrm {m}}\rightarrow \ker (G\rightarrow \mathrm {GL}(V))\subseteq T$ which does not factor through $Z(G)$ . We have $\overline {\mu }\in \mathscr {P}_V\smallsetminus \{\overline {0}\}$ . Then, the induction diagram reads

and the induction map $\mathrm {Ind}_{\mu }\colon \mathrm {H}^*(V/G^{\mu })=\mathrm {H}^*(\mathrm {pt}/T)^{W^{\mu }}\rightarrow \mathrm {H}^*(V/G)=\mathrm {H}^*(\mathrm {pt}/T)$ is the map

$$\begin{align*}f\mapsto \sum_{w\in W/W^{\mu}}w\cdot\left(\frac{f}{\prod_{\alpha\in\mathcal{W}^{\mu<0}(\mathfrak{g})}\alpha}\right)\,. \end{align*}$$

It is surjective. Hence, $\mathcal {P}_0=\{0\}$ .