Proof (1) and (2) These two statements can be found in [Reference Vinberg and Popov30, Theorems 1 and 2].

(3) We have $O_{\lambda }^U\subset V^U={\mathbb {K}^{*}} v\cup \{0\}$ ; hence; $O_{\lambda }^U\subset {\mathbb {K}^{*}} v$ . They are equal because $O_{\lambda }$ is a cone. Since $G_v = G_w$ for all $w \in {\mathbb {K}^{*}} v$ , we see that $G_v = \operatorname {\mathrm {Cent}}_{G}(O_{\lambda }^{U})$ and $V^{G_v} \supseteq \mathbb {K} v$ . Now, the second claim follows because $U \subseteq G_v^\circ \subseteq G_v$ , and so $V^{G_v} \subseteq V^{G_v^\circ } \subseteq V^U = \mathbb {K} v$ .

(4) G acts on the projective space ${\mathbb P}(V)$ , and the projection $p\colon V\setminus \{0\} \to {\mathbb P}(V)$ is G-equivariant and sends closed cones to closed subsets. In particular, $p(O_{\lambda })=G\,p(v)$ is closed, and so $P_{\lambda } := G_{p(v)}=\operatorname {\mathrm {Norm}}_G(\mathbb {K} v)= \operatorname {\mathrm {Norm}}_G(O_\lambda ^U)\subset G$ is a parabolic subgroup normalizing $G_v$ . If $g \in G$ normalizes $G_v^\circ $ , then $G_{g v}^\circ =G_v^\circ $ , and so $g v \in {\mathbb {K}^{*}} v=O_\lambda ^U$ by (3). Hence, $\operatorname {\mathrm {Norm}}_G(G_v) \subseteq \operatorname {\mathrm {Norm}}_G(G_v^\circ ) \subseteq \operatorname {\mathrm {Norm}}(O_\lambda ^U) = P_\lambda \subseteq \operatorname {\mathrm {Norm}}_G(G_v)$ .

(5) By (1) and (2), we have $\operatorname {\mathrm {Aut}}_{G}(\overline {O_{\lambda }})=\operatorname {\mathrm {Aut}}_{G}(O_{\lambda })$ . Since $\overline {O_{\lambda }}$ is a cone, we have an inclusion ${\mathbb {K}^{*}}\hookrightarrow \operatorname {\mathrm {Aut}}_{G}(\overline {O_{\lambda }})$ . Any $\sigma \in \operatorname {\mathrm {Aut}}_{G}(\overline {O_{\lambda }})$ is U-equivariant and hence preserves ${\overline {O_{\lambda }}}^U = V^{U}$ as well as $\{0\}\in V^{U}$ , and the claim follows.

(6) Let $Y:=\overline {G v} \subset \mathcal N_{V}$ , which implies that $0\in Y$ . Since Y is irreducible, the fixed point set $Y^{U}$ does not contain isolated points (see, e.g., [Reference Kraft20, Section III.5, Theorem 5.8.8]), and so $Y^U \neq \{0\}$ . Hence, Y contains a highest weight vector, and so $Y \supset O_{\lambda }$ .

(7) The tangent space $T_0\overline {O_{\lambda }}$ is a nontrivial submodule of $V_\lambda $ , hence equal to $V_\lambda $ . If $\overline {O_{\lambda }}$ is smooth, then $\dim \overline {O_{\lambda }} = \dim T_0\overline {O_{\lambda }} = \dim V_\lambda $ and so $\overline {O_{\lambda }} = V_\lambda $ . The other implication is clear.

Proof Since $\varphi _{k}^{-1}(0) = \{0\}$ , the homogeneous morphism $\varphi _{k}$ is finite, the image $\varphi _{k}(V_{\lambda })$ is closed, and the fibers of $\varphi _{k}$ are the $\mu _{k}$ -orbits. This yields the first statement. The last statement follows from the fact that $\overline {O_{k\lambda }}$ is normal, by Lemma 2.2(1).

Proof If $v:=p(w)\neq 0$ , then $\dim G w \geq \dim G v> 0$ . Hence, we can assume that $W = V$ is a simple G-module and p the identity map.

Given a closed subset $Y \subset V$ of a vector space, one defines the associated cone $\mathcal C Y \subset V$ to be the zero set of the functions $\operatorname {\mathrm {gr}} f$ , $f \in I(Y)\subset \mathcal O(V)$ , where $\operatorname {\mathrm {gr}} f$ denotes the homogeneous term of f of maximal degree. If Y is irreducible, G-stable and belongs to a fiber $\pi ^{-1}(z)$ of the quotient morphism $\pi \colon V \to V /\!\!/ G$ , then $\mathcal C Y \subseteq \mathcal N_{V}$ , and $\mathcal C Y$ is G-stable and equidimensional of dimension $\dim Y$ (see [Reference Borho and Kraft3, Section 3]). Lemma 2.2(6) now implies that the highest weight orbit $O \subset V$ belongs to $\mathcal C Y$ , and the claim follows.

Proof (1) It is a standard fact that the intersection of a lattice with a line is a sublattice of rank 1 generated by any of the two indivisible elements.

(2a) Consider the covariant $\varphi _k\colon V_{\lambda _0}\rightarrow V_{\lambda }$ . We have

$$\begin{align*}\varphi_k^{-1}(\mathbb{K} v)=\varphi_k^{-1}(V_{\lambda}^U)=V_{\lambda_0}^U=\mathbb{K} v_0, \end{align*}$$

so by Lemma 2.2(4), we obtain that

$$\begin{align*}P_{\lambda}=\operatorname{\mathrm{Norm}}_{G}(\mathbb{K} v)=\operatorname{\mathrm{Norm}}_G(\mathbb{K} v_0)=P_{\lambda_0}. \end{align*}$$

(2b) and (2c) Since $P_{\lambda }/G^{\circ }_\lambda \to P_{\lambda }/G_\lambda \simeq {\mathbb {K}^{*}}$ is a finite connected cover of ${\mathbb {K}^{*}}$ , we have $G_{\lambda }^{\circ }=\ker (\lambda _1)$ for some character $\lambda _1\colon P_{\lambda }\to {\mathbb {K}^{*}}$ , and $\lambda =l\lambda _1$ , where $l=|G_\lambda /G^{\circ }_\lambda |$ . Furthermore, $\lambda _1$ is a dominant weight because $\lambda $ is a dominant weight. Since $G^{\circ }_\lambda $ has no finite index subgroup, it follows that $\lambda _1$ is indivisible, and so $\lambda _1=\lambda _0$ . This yields (2b) and also implies (2c).

(2d) follows from Lemma 2.2(7) and Remark 2.4(3).

(3) For $w \in O$ and $v=\varphi (w) \in O_{\lambda }$ , we get a finite covering $G/G_w=O \to O_\lambda = G/G_v$ , and hence $G_v^\circ \subseteq G_w \subseteq G_v$ . By (2b), we have $G/G_v^\circ = O_{\lambda _0}$ , where $\lambda = k\lambda _0$ for an integer $k\geq 1$ , and the composition $G/G_{\lambda _0} = O_{\lambda _0} \to G/G_w = O \to G/G_v = O_{\lambda }$ is a cyclic covering of degree k. Therefore, $O_{\lambda _0} \to O$ and $O \to O_{\lambda }$ are both cyclic, of degree m and $\ell $ , respectively, and $k = \ell m$ . Hence, $O \simeq O_{m\lambda _0}$ and $\ell (m\lambda _0) = \lambda $ .

Proof Choose an isomorphism $\nu \colon O_\lambda \simeq O$ . Since $\mathcal O(O_\lambda ) = \mathcal O(\overline {O_\lambda })$ by Lemma 2.2(2), the morphism $\nu $ extends to a G-equivariant morphism $\bar \nu \colon \overline {O_\lambda } \to \overline {O}^{\,n}$ . We claim that $\bar \nu $ is an isomorphism. Then the lemma follows from Lemma 2.2(1) and (2).

It remains to see that $\mathcal O(\overline {O}^{\,n}) = \mathcal O(O)$ . By Lemma 2.2(2), we have $\mathcal O(O) \simeq \bigoplus _{k\geq 0} V_{k\lambda ^{\!\vee }}$ , and so the G-stable subalgebra $\mathcal O(\overline {O}^{\,n}) \subseteq \mathcal O(O)$ is a direct sum of some of the $V_{k\lambda ^{\!\vee }}$ . This implies that a power of every element from $V_{\lambda ^{\!\vee }}$ belongs to $\mathcal O(\overline {O}^{\,n})$ . Hence, $V_{\lambda ^{\!\vee }} \subset \mathcal O(\overline {O}^{\,n})$ and the claim follows.

Proof (1) Let $\overline {O}= O\cup \{x\}$ for some fixed point $x\in Z$ . If $\eta \colon Y \to \overline {O}$ is the normalization, then $\eta ^{-1}(O) \to O$ is an isomorphism because O is normal. Since $\eta ^{-1}(x)$ is finite and G-stable, it must be a single fixed point $y \in Y$ . Moreover, $Y\setminus \eta ^{-1}(O)=\{y\}$ has codimension $\geq 2$ in Y because a semisimple group does not have one-dimensional quasi-affine orbits. (In fact, the only simple groups having one-dimensional orbits are $\operatorname {\mathrm {SL}}_2$ and $\operatorname {\mathrm {PSL}}_2$ [Reference Dynkin7], and their orbits are projective.) It follows that $\mathcal O(Y)=\mathcal O(O)$ .

(2) Since $\mathcal O(O) \xrightarrow {\sim } \mathcal O(Y)$ by (1) and X is affine, the G-equivariant morphism $\varphi \colon O \to X$ induces a G-equivariant morphism $\tilde \varphi \colon Y \to X$ . There is a closed G-equivariant embedding of X into a G-module W, $X \hookrightarrow W$ , and a linear projection $\operatorname {\mathrm {pr}}_{V_\lambda }\colon W \to V_{\lambda }$ onto a simple G-module $V_{\lambda }$ such that $\varphi (O)$ is not in the kernel of $\operatorname {\mathrm {pr}}_{V_\lambda }$ .

Set $\psi :=\operatorname {\mathrm {pr}}_{V_\lambda }\circ \tilde \varphi \colon Y \to V_{\lambda }$ . Since a unipotent group U does not have isolated fixed points on an irreducible affine U-variety (see, e.g., [Reference Kraft20, Theorem 5.8.8]), we get $O^{U}\neq \emptyset $ , and so $\psi (O)^{U}\neq \emptyset $ . This implies that $\psi (O) = O_{\lambda }$ and $\psi (Y) = \overline {O_{\lambda }}$ . We have $\psi ^{-1}(0) = \{y\}$ , and so $\psi $ is finite and surjective. In particular, O is a minimal orbit of the same type as $O_{\lambda }$ , by Lemma 2.7(3). From the factorization

we see that both maps are finite, and so $\varphi (O)$ is a minimal orbit as well, of the same type as $O_{\lambda }$ , again by Lemma 2.7(3).

Proof (1) This follows from Proposition 2.11(2) and the fact that $\mathcal O(\overline {O}^{\,n}) = \mathcal O(O)$ .

(2) If $\nu \colon O \to O'$ is another G-equivariant morphism, then, for a given $v \in O^U$ , we have $\nu (v)=t_0\cdot \eta (v)$ for a suitable $t_0\in {\mathbb {K}^{*}}$ . Since the G-action commutes with the scalar multiplication, we get $\nu (gv) = g\nu (v) = g(t_0\cdot \eta (v)) = t_0\cdot g\eta (v) = t_0\cdot \eta (gv)$ , and the claim follows.

(3) Choose $v \in O^U$ and set $v':=\eta (v) \in {O'}^U$ . With respect to the scalar multiplication, we have $O^U = {\mathbb {K}^{*}}\cdot v$ and ${O'}^U = {\mathbb {K}^{*}}\cdot v'$ . Since $\eta (O^U) = {O'}^U$ , this implies that $\eta (t\cdot v) = t^k \cdot v' = t^k\cdot \eta (v)$ for a suitable $k \in {\mathbb Z}$ . By (1), $\eta $ extends to $\overline {O}$ ; hence, $k\geq 1$ by the definition of the scalar multiplication. Since the G-action commutes with the scalar multiplication, the formula $\eta (t\cdot v) = t^k \cdot \eta (v)$ holds for any $v \in O$ , and $k=\deg \eta $ .

(4) For $s \in T$ , $v \in O_\lambda ^U$ , and $v'\in O_{\lambda '}^U$ , we have $sv=\lambda (s)\cdot v$ and $sv' = \lambda '(s)\cdot v'$ . By (2) and the G-equivariance of $\eta $ , we get

$$ \begin{align*}\eta(sv) = \eta(\lambda(s)\cdot v) = \lambda(s)^k\cdot \eta(v) = s\eta(v), \end{align*} $$

where $k=\deg \eta $ , and so $\lambda ' = k \lambda $ . The last statement follows from Proposition 2.11(3).

(5) This is clear from (3) and (4): the scalar multiplication on $V_\lambda $ induces the multiplication by $t^{-n}$ on the homogeneous component of $\mathcal O(V_\lambda )$ of degree n.

Proof (1) The first claim follows from Lemma 2.13(5) and obviously implies the second.

(2) This is an immediate consequence of Lemma 2.13, statements (4) and (3).

(3) Since $\mathcal O(O) = \mathcal O(\overline {O}^{\,n})$ , the claim holds if the closure $\overline {O}$ is normal. By (1), the canonical ${\mathbb {K}^{*}}$ -action on $\overline {O}^{\,n}$ corresponds to the grading of the coordinate ring $\mathcal O(\overline {O}^{\,n}) \simeq \bigoplus _{k\geq 0} V_{k\lambda ^{\!\vee }}$ . In the general case, $\mathcal O(\overline {O})$ is a G-stable subalgebra of $\mathcal O(\overline {O}^{\,n})$ . Since the homogeneous components $V_{k\lambda ^{\!\vee }}$ are simple and pairwise nonisomorphic G-modules, we see that $\mathcal O(\overline {O})$ is a graded subalgebra, hence stable under the canonical ${\mathbb {K}^{*}}$ -action.

(4) This obviously holds for the scalar multiplication on $O_\lambda \subset V_\lambda $ , hence in the case where $\overline {O}$ is normal. By (3), it is true in general.

(5) We have $\operatorname {\mathrm {Norm}}_{G}(O_\lambda ^{\,U}) = P_{\lambda }$ and $p x = \lambda (p)\cdot x$ for $p\in P_{\lambda }$ , $x \in O_{\lambda }^{\,U}$ (cf. Lemma 2.2(4)). This shows that the action of $P_{\lambda }$ on $O^{U}$ is given by the canonical ${\mathbb {K}^{*}}$ -action. Since $\overline {O^U} \setminus O^U$ is the unique fixed point of $\overline {O}$ under G, we have $\operatorname {\mathrm {Norm}}_{G}(\overline {O}^{U}) = \operatorname {\mathrm {Norm}}_{G}(O^{U})$ .

Proof For the isotypic component $R_{\lambda }$ of R, denote by $R_{\lambda }' \subset R_{\lambda }$ the direct sum of all weight spaces of weight different from the lowest weight. Then $R_{\lambda }=(R_{\lambda })^{U^{-}} \oplus R_{\lambda }'$ . Since $R_{\lambda }\cdot R_{\mu } \subseteq R_{\lambda +\mu }$ , we get $R_{\lambda }\cdot R_{\mu }' \subset R_{\lambda +\mu }'$ because the lowest weight of $R_{\lambda +\mu }$ is equal to the sum of the lowest weights of $R_{\lambda }$ and $R_{\mu }$ . It follow that $\bigoplus _{\mu }R_{\mu }' = \ker \pi _U \subset R$ is an ideal and that the induced linear isomorphism $R^{U^{-}} \xrightarrow {\sim } R_{U}$ is an isomorphism of $\mathbb {K}$ -algebras.

Proof Let $V \subset R$ be a simple submodule. Then either $\varphi (V)=(0)$ or $\varphi |_V : V \to \varphi (V)$ is an isomorphism. If $\varphi ^U$ or $\varphi _U$ is injective, then we are in the second case and so $\varphi $ is injective. If $W \subset S$ is a simple submodule which is not contained in the image of $\varphi $ , then $W \cap \varphi (R) = (0)$ and so $W^U$ and $W_U$ are not in the image of $R^U$ (resp. $R_U$ ). This proves the first part of the lemma and (1). As for (2), we simply remark that $\varphi $ is zero in case $\varphi ^U$ (or $\varphi _U$ ) is zero.

Proof We first consider the evaluation map $\tilde \varepsilon \colon \mathcal O(G) \to \mathbb {K}$ , $f\mapsto f(e)$ , which is the comorphism of the inclusion $\tilde \iota \colon \{e\}\hookrightarrow G$ . We claim that on the isotypic components $V_\lambda \otimes V_\lambda ^\vee $ of $\mathcal O(G)$ , the map $\tilde \varepsilon $ is given by the formula $\tilde \varepsilon (v\otimes \sigma ) = \sigma (v)$ . Indeed, let $\rho _\lambda \colon G \to \operatorname {\mathrm {GL}}(V_\lambda ) \subset \operatorname {\mathrm {End}}(V_\lambda )$ denote the representation on $V_\lambda $ . Then the composition $\rho _\lambda \circ \tilde \iota $ sends e to $\operatorname {\mathrm {id}}_{V_\lambda }$ ; hence, the comorphism $\operatorname {\mathrm {End}}(V_\lambda )^\vee \to \mathbb {K}$ is given by $\ell \mapsto \ell (\operatorname {\mathrm {id}}_{V_\lambda })$ . We have mentioned above that the isomorphism $V \otimes V^\vee \xrightarrow {\sim } \operatorname {\mathrm {End}}(V)^\vee $ is defined by $(v\otimes \sigma )(\varphi ) := \sigma (\varphi (v))$ . This implies that $\tilde \varepsilon \colon V_\lambda \otimes V_\lambda ^\vee \xrightarrow {\sim } \operatorname {\mathrm {End}}(V)^\vee \to \mathbb {K}$ is given by $v\otimes \sigma \mapsto \sigma (v)$ as claimed.

For the restriction $\varepsilon $ of $\tilde \varepsilon $ to $\mathcal O(G/U)=\mathcal O(G)^U$ , we thus find for $v \in V_\lambda \simeq \mathcal O(G/U)_\lambda $ that $\varepsilon (v) = \sigma _0(v)$ , where $\sigma _0$ is a highest weight vector in $V_\lambda ^\vee $ . As a consequence, $\varepsilon (v) \neq 0$ if v has weight $-\lambda ^\vee $ , i.e., if $v \in \mathcal O(G/H)^{U^-}$ . Now, the claims follow from Example 3.2.

Proof The action of T on $\mathcal O(G/U)_\mu \otimes \operatorname {\mathrm {Hom}}_G(\mathcal O(G/U)_\lambda ,R)$ is by scalar multiplica tion with the character $\mu ^\vee - \lambda ^\vee $ ; hence, $(\mathcal O(G/U)_\mu \otimes \operatorname {\mathrm {Hom}}_G(\mathcal O(G/U)_\lambda ,R))^T = 0$ unless $\mu = \lambda $ . For $\mu =\lambda $ , the torus T acts trivially and so $(\mathcal O(G/U)_\lambda \otimes \operatorname {\mathrm {Hom}}_G(\mathcal O(G/U)_\lambda ,R))^T\xrightarrow {\sim } R_\lambda $ as we have seen above. Thus, the left-hand side is $\bigoplus _{\lambda \in \Lambda _G} \mathcal O(G/U)_\lambda \otimes \operatorname {\mathrm {Hom}}_G(\mathcal O(G/U)_\lambda ,R)^T$ , which is canonically isomorphic to $\bigoplus _{\lambda \in \Lambda _G} R_\lambda = R$ .

Proof (1) We first show that for every dominant weight $\lambda $ , there is a canonical isomorphism $\varphi _\lambda \colon \operatorname {\mathrm {Hom}}_G(\mathcal O(G/U)_\lambda ,R) \xrightarrow {\sim } (R_\lambda )_U$ . For $\rho \in \operatorname {\mathrm {Hom}}_G(\mathcal O(G/U)_\lambda ,R)$ , consider the composition $\pi _{U,\lambda }\circ \rho \colon \mathcal O(G/U)_\lambda \to R_\lambda \to (R_\lambda )_U$ , where $\pi _{U,\lambda } \colon R_\lambda \to (R_\lambda )_U$ is the universal U-projection (see Remark 3.2). From the universal property of $\varepsilon _\lambda \colon \mathcal O(G/U)_\lambda \to \mathbb {K}$ (Lemma 3.7), we obtain a unique factorization

It is easy to see that the map $\varphi _\lambda \colon \operatorname {\mathrm {Hom}}_G(\mathcal O(G/U)_\lambda ,R) \to (R_\lambda )_U$ defined by $\rho \mapsto \bar \rho (1)$ has the required properties.

(2) Next, we show that for every dominant weight $\lambda $ , there is a canonical isomorphism $\psi _\lambda \colon \operatorname {\mathrm {Hom}}_G(\mathcal O(G/U)_\lambda ,R) \xrightarrow {\sim } (R_\lambda )^{U^-}$ . Here, we use the elements $f_\lambda :=\bar \varepsilon _\lambda ^{\,-1}(1)$ defined after Lemma 3.7, and set $\psi _\lambda (\rho ):=\rho (f_\lambda )$ . Now, the claim follows from (1) because $\varepsilon _\lambda (f_\lambda ) = 1$ and so $\pi _{U,\lambda }(\rho (f_\lambda )) = \tilde \rho (1)$ , i.e., $\bar \pi _{U,\lambda }\circ \psi _\lambda = \varphi _\lambda $ , where $\bar \pi _{U,\lambda }\colon R_\lambda ^{U^-} \xrightarrow {\sim } (R_\lambda )_U$ is the T-equivariant isomorphism induced by $\pi _{U,\lambda }$ (see Lemma 3.7).

Proof The definition of the multiplication on $\operatorname {\mathrm {gr}} R$ implies that the subalgebra $(\operatorname {\mathrm {gr}} R)^{U^-}\subset \operatorname {\mathrm {gr}} R$ is equal to the subalgebra $R^{U^-} \subset R$ since one has $R_\mu ^{U^-} \cdot R_\lambda ^{U^-} \subseteq R_{\mu +\lambda }^{U^-}$ . Applying Theorem 3.8 to the isotypically graded G-algebra $\operatorname {\mathrm {gr}} R$ , we get

$$ \begin{align*}(\mathcal O(G/U)\otimes R^{U^{-}})^{T} = (\mathcal O(G/U)\otimes (\operatorname{\mathrm{gr}} R)^{U^{-}})^{T} \xrightarrow{\sim} \operatorname{\mathrm{gr}} R, \end{align*} $$

hence the claim.

Proof On $\Lambda _{G}$ , we have a partial ordering

$$ \begin{align*}\mu \preceq \lambda :\!\!\iff \lambda - \mu \text{ is a sum of positive roots}, \end{align*} $$

which has the following property: if $V_\lambda , V_\mu $ are simple G-modules of highest weight $\lambda $ and $\mu $ , then $V_\lambda \otimes V_\mu = V_{\lambda + \mu } \oplus W$ where the simple summands of W have highest weights $\prec \lambda + \mu $ .

The cone ${\mathbb Q}_{\geq 0}\cdot \Phi _G^+ \subset {\mathbb Q}\otimes \Lambda _G$ generated by the positive roots $\Phi _G^+$ is pointed and contains ${\mathbb Q}_{\geq 0}\cdot \Lambda _{G}$ . Therefore, we can find a ${\mathbb Q}$ -linear function $p\colon {\mathbb Q}\otimes \Lambda _G \to {\mathbb Q}$ such that the following holds:

1. (1) $p(\lambda )\in {\mathbb N}$ for all dominant weights $\lambda \in \Lambda _G$ .

2. (2) $p(\alpha ) \in {\mathbb N}_{>0}$ for all positive roots $\alpha \in \Phi _G^+$ .

In particular, if $\mu \prec \lambda $ , then $p(\mu )<p(\lambda )$ . Setting $R_n:=\bigoplus _{p(\lambda )\leq n} R_\lambda $ for $n\geq 0$ , we get $R_n \cdot R_m\subseteq R_{n+m}$ . It follows that the G-stable subspace

$$ \begin{align*}\tilde R:=\textstyle{\bigoplus_{n\geq 0}} \mathbb{K} s^n \otimes R_n \subseteq \mathbb{K}[s]\otimes R \end{align*} $$

is a subalgebra containing $\mathbb {K}[s]$ . Since the isotypic component $\tilde R_\lambda $ is given by

$$ \begin{align*}\tilde R_\lambda = \textstyle{\bigoplus_{n\geq p(\lambda)}} \mathbb{K} s^n\otimes R_\lambda, \end{align*} $$

we see that $\tilde R$ is a free $\mathbb {K}[s]$ -module, proving (1). Moreover,

$$ \begin{align*}s(\mathbb{K} s^n \otimes R_n) = \mathbb{K} s^{n+1}\otimes R_n \subseteq \mathbb{K} s^{n+1} \otimes R_{n+1}, \end{align*} $$

and hence $\tilde R / s \tilde R = \bigoplus _{n\geq 0} R_n/R_{n-1}$ where we set $R_{-1}=(0)$ . From the canonical decomposition $R_n = (\bigoplus _{p(\lambda )=n} R_\lambda ) \oplus R_{n-1}$ , we see that $\tilde R / s\tilde R = R = \operatorname {\mathrm {gr}} R$ as a G-module, and the multiplication of $R_\lambda \subset R_n/R_{n-1}$ with $R_\mu \subset R_m/R_{m-1}$ is given by the product in R followed by the projection onto $R_{n+m}/R_{n+m-1}$ . We have $R_\lambda \cdot R_\mu = V \oplus W$ , where $V \subseteq R_{\lambda +\mu }$ and all summands of W have highest weights $\rho \prec \lambda + \mu $ . This implies that $p(\rho ) < p(\lambda +\mu ) = n+m$ and so $W \subseteq R_{n+m-1}$ . Hence, the product of $R_\lambda $ and $R_\mu $ in $\tilde R / s\tilde R$ coincides with the product in $\operatorname {\mathrm {gr}} R$ , proving (2).

Finally, the subalgebra $\tilde R_s \subseteq \mathbb {K}[s,s^{-1}]\otimes R$ contains $\mathbb {K} s^\ell \otimes R_n$ for all $\ell \in {\mathbb Z}$ and all $n\in {\mathbb N}$ , and hence is equal to $\mathbb {K}[s,s^{-1}] \otimes R$ , proving the last claim (3).

Proof If $R^U$ has property $\mathcal P$ , then so does $R^{U^-}$ . Hence, assumptions (ii)–(vi) imply that $(\mathcal O(G/U)\otimes R^{U^-})^T$ has property $\mathcal P$ . In particular, $\operatorname {\mathrm {gr}} R$ has property $\mathcal P$ by Proposition 3.12. Now, (i) implies that R has property $\mathcal P$ as well.

Proof By definition, the comorphism $\Phi ^{*}\colon \mathcal O(X) \longrightarrow (\mathcal O(G/U)\otimes \mathcal O(X^{U}))^{T}$ is given as follows: if $\Phi ^{*}(f) = \sum _{j} f_{j}\otimes h_{j}$ , then $\Phi ^{*}(f)(g U,x)=f(g x) = \sum _{j}f_{j}(g U) h_{j}(x)$ . Consider the evaluation map $\varepsilon \colon \mathcal O(G/U)\to \mathbb {K}$ , $f\mapsto f(e U)$ . Then $f(x)=\sum _{j}\varepsilon (f_{j})h_{j}(x)$ for all $x\in X^U$ , which shows that the diagram

commutes, where $\rho $ is the restriction map, i.e., $\rho (f) = \sum _{j}\varepsilon (f_{j})h_{j} = (\varepsilon \otimes \operatorname {\mathrm {id}})(\Phi ^{*}(f))$ . Since $\mathcal O(X)$ is an isotypically graded G-algebra, it follows from Remark 3.4 that the restriction map $\rho $ is equal to the universal U-projection $\pi _{U}\colon \mathcal O(X) \to \mathcal O(X)_{U}$ . If we show that $\varepsilon \otimes \operatorname {\mathrm {id}}$ is also equal to the U-projection ${\pi }_U\colon (\mathcal O(G/U)\otimes \mathcal O(X^{U}))^{T}\to \left ((\mathcal O(G/U)\otimes \mathcal O(X^{U}))^{T}\right )_U$ , then $\Phi ^*$ is an isomorphism by Lemma 3.5. We have

$$ \begin{align*}(\mathcal O(G/U)\otimes \mathcal O(X^{U}))^{T} = \textstyle \bigoplus_{\lambda\in\Lambda_G}\mathcal O(G/U)_\lambda \otimes \mathcal O(X^U)_{[-\lambda^\vee]}, \end{align*} $$

where $ \mathcal O(X^U)_{[\mu ]}$ is the T-weight space of $\mathcal O(X^U)$ of weight $\mu $ . Since the evaluation map $\varepsilon _\lambda \colon \mathcal O(G/U)_\lambda \to \mathbb {K}$ , $f\mapsto f(e U)$ , is the universal U-projection (Lemma 3.7), we see that the linear map $\mathcal O(G/U)_\lambda \otimes \mathcal O(X^U)_{[-\lambda ^\vee ]} \to \mathcal O(X^U)_{[-\lambda ^\vee ]}$ , $\sum _j f_j\otimes h_j \mapsto \sum _j\varepsilon (f_j)h_j$ , is the U-projection as well, and the claim follows.

It remains to see that $\Phi ^*$ is equal to the inverse of $\Psi $ from Theorem 3.8. Using again Lemma 3.5, it suffices to show that the diagram

commutes. This is stated in Remark 3.11.

Proof (1) By hypothesis, any nontrivial orbit $O \subset X$ is minimal, so $\overline {O}=O\cup \{x_{0}\}$ , where $x_{0}\in X^{G}$ by Proposition 2.11(1). In particular, the G-action is fix-pointed.

We can assume that X is a closed G-stable subvariety of a G-module W (see, for example, [Reference Kraft20, Corollary 2.3.5]). Let $O \subset X$ be a nontrivial orbit. There is a linear projection $p\colon W \to V$ onto a simple G-module V of highest weight $\lambda $ such that $O \nsubseteq \ker p$ . Proposition 2.11(2) implies that $p(O) = O_{\lambda }$ and that O is of the same type as $O_{\lambda }$ . The same is true for all orbits $O'$ from the open subset $X':= X \setminus \ker p$ of X. Since X is irreducible, all minimal orbits are of type $\lambda $ . It follows from Example 3.1 that $\mathcal O(O)$ is of type $\lambda ^\vee $ .

(2) Since X is small, we have $X = G\cdot X^U$ , showing that the morphism $G/U \times X^U \to X$ is surjective. Thus, we obtain a G-equivariant inclusion $\mathcal O(X) \hookrightarrow \mathcal O(G/U) \otimes \mathcal O(X^U)$ where the G-algebra on the right is isotypically graded (by (*) and Example 3.1). Hence, it follows from (1) that $\mathcal O(X)$ is isotypically graded of type $\lambda ^\vee $ .

(3) The first part follows from Remark 3.4. It also shows that if X is normal, then $X^U$ is normal. The other implication follows from the isomorphism $\Phi $ in Proposition 4.2 because $G/\!\!/ U$ is normal (cf. Example 3.15). The second equivalence is clear.

Proof (1) By Proposition 4.3(2), $\mathcal O(X)$ is an isotypically graded G-algebra of type $\lambda ^{\vee }$ . If we define the ${\mathbb {K}^{*}}$ -action on $\mathcal O(X)$ such that the isotypic component of type $n \lambda ^{\vee }$ has weight $-n$ , then this action is fix-pointed and restricts to the canonical ${\mathbb {K}^{*}}$ -action on the closure of each minimal orbit (Proposition 2.14(3)). Since X is the union of the closures of the minimal orbits, this ${\mathbb {K}^{*}}$ -action is unique and commutes with the G-action. We have $X^G= (X^U)^{{\mathbb {K}^{*}}}$ and the ${\mathbb {K}^{*}}$ -action on $X^U$ is fix-pointed, since this holds for the closure of a minimal orbit. This implies that $X^G = (X^U)^{{\mathbb {K}^{*}}} \xrightarrow {\sim } X^U/\!\!/ {\mathbb {K}^{*}}$ , and $X^G \xrightarrow {\sim } X/\!\!/ G$ since the G-action is fix-pointed, which yields the remaining claims.

(2) Choose $x_{0} \in O_{\lambda }^{U}$ and consider the G-equivariant morphism $\eta \colon G/\!\!/ U \to \overline {O_{\lambda }}$ induced by $g U\mapsto g x_0$ . For $t \in T$ , we get $\eta (gt^{-1}U) = gt^{-1}x_0 = \lambda (t)^{-1}\cdot gx_0$ . This shows that the T-action on $G/\!\!/ U$ by right multiplication induces a T-action on $\overline {O_{\lambda }}$ which factors through $\lambda \colon T \to {\mathbb {K}^{*}}$ , and the induced ${\mathbb {K}^{*}}$ -action is the inverse of the scalar multiplication, i.e., the inverse of the canonical ${\mathbb {K}^{*}}$ -action.

Define $D:=\ker \lambda \subset T$ . We claim that $\eta $ is the algebraic quotient under the action of D. In fact, the action of $t\in T$ on $\mathcal O(G/U)_\mu $ is by scalar multiplication with $\mu ^\vee (t)$ (Remark 3.6(1)). Hence, the action of D is trivial if and only if $\mu $ is a multiple of $\lambda ^{\!\vee }$ . This implies that

$$ \begin{align*}\mathcal O(G/U)^{D} = \textstyle\bigoplus_{\mu\in\Lambda_G} \mathcal O(G/U)_\mu^D = \bigoplus_{k\geq 0} V_{k\lambda^\vee} \simeq \mathcal O(\overline{O_{\lambda}}) \end{align*} $$

(see Lemma 2.2(2)). Since X is of type $\lambda $ , it follows that $D = \ker \lambda $ acts trivially on $X^U$ . Therefore, $(G/\!\!/ U \times X^U)/\!\!/ D = \overline {O_{\lambda }}\times X^U$ , and so

$$ \begin{align*}G/\!\!/ U \star^T X^U = (G/\!\!/ U \star^D X^U)/\!\!/ T = (\overline{O_{\lambda}}\times X^U)/\!\!/ T. \end{align*} $$

By construction, the T-action on $\overline {O_{\lambda }}\times X^U$ is given by $t(v,x)= (\lambda (t)^{-1}\cdot v, t x)$ , i.e., by the inverse of the canonical ${\mathbb {K}^{*}}$ -action on $\overline {O_{\lambda }}$ and the given action on $X^U$ . Hence, $ (\overline {O_{\lambda }}\times X^U)/\!\!/ T = \overline {O_{\lambda }} \star ^{{\mathbb {K}^{*}}} X^U$ , and the claim follows from Proposition 4.2.

(3) Consider the action of $P_{\lambda }$ on $G \times X^{U}$ given by $p(g,x) = (g p^{-1},p x)$ . Then the action map $G \times X^{U}\to X$ , $(g,x)\mapsto g x$ , factors through the geometric quotient

$$ \begin{align*} G\times^{P_{\lambda}}X^{U}:=(G\times X^{U})/ P_{\lambda}. \end{align*} $$

For $\Theta $ , we have the following factorization:

where the first map is a closed immersion and the second an isomorphism. Since $G/P_{\lambda }$ is complete, it follows that $\Theta $ is proper. Moreover, $\Theta $ is surjective because every G-orbit meets $X^{U}$ . We claim that $\Theta $ induces a bijection $G\times ^{P_{\lambda }}(X^{U}\setminus X^{G}) \to X\setminus X^{G}$ , which implies that $\Theta $ is birational. Indeed, if $x\in X^{U}\setminus X^{G}$ , then $x \in O^{U}$ for a minimal orbit $O \subset X$ . If $g x = g' x'$ for some $x'\in X^{U}$ , $g'\in G$ , then $x'\in O^{U}$ , and hence $x' = q x$ for some $q \in P_{\lambda }$ because the action of $P_{\lambda }$ on $O^{U}$ is transitive. It follows that $g^{-1}g' q\in G_{x}\subset P_{\lambda }$ , and hence $p:=g^{-1} g'\in P_{\lambda }$ . Thus, $[g',x']=[g p,x']=[g,p x'] = [g,x]$ .

It remains to see that the comorphism of $\Theta $ is an isomorphism on the global functions. Let $K_{\lambda }$ be the kernel of the character $\lambda \colon P_{\lambda }\to {\mathbb {K}^{*}}$ . Then $G/K_\lambda \simeq O_\lambda $ , and the action of $P_{\lambda }$ on G by right multiplication induces an action of ${\mathbb {K}^{*}} = P_{\lambda }/K_{\lambda }$ on $G/K_{\lambda }$ by right multiplication corresponding to the canonical action on $O_\lambda $ . This gives the G-equivariant isomorphisms

$$\begin{align*}G\times^{P_{\lambda}}X^{U} \xrightarrow{\sim} G/K_{\lambda}\star^{{\mathbb{K}^{*}}} X^{U} \xrightarrow{\sim} O_\lambda \star^{{\mathbb{K}^{*}}} X^U, \end{align*}$$

and the claim follows from (2).

Proof By definition, X is an affine G-variety. For $x = [v,y]\in \overline {O_{\lambda }}\star ^{{\mathbb {K}^{*}}}Y$ , $v\neq 0$ , the G-orbit $G x \subset X$ is the image of $O_{\lambda } \times \{y\}$ in X, hence a minimal orbit of type $\lambda $ or a point (Proposition 2.11(2)). As a consequence, X is a small G-variety of type $\lambda $ . Furthermore, since the canonical ${\mathbb {K}^{*}}$ -action on $\overline {O_{\lambda }}$ commutes with the G-action, we have

$$ \begin{align*}X^{U}=(\overline{O_{\lambda}}\star^{{\mathbb{K}^{*}}}Y)^{U} = \overline{O_{\lambda}}^{U}\star^{{\mathbb{K}^{*}}}Y \simeq \mathbb{K}\star^{{\mathbb{K}^{*}}} Y \xrightarrow{\sim} Y, \end{align*} $$

where the last morphism is given by $[t,y]\mapsto t y$ which is an isomorphism, as explained in Remark 4.6.

Proof (of Corollary 1.4)

This corollary follows from Proposition 4.4(2) and Lemma 4.7.