Proof. The scheme $\mathcal {Y}$ is a closed subscheme of $\mathcal {X}\times \mathcal {F}$ and $\mathcal {F}$ is projective, thus $\mathcal {Y} \to \mathcal {X}$ is projective. Let U denote the open subset of $\mathcal {X}$ on which N is regular or l is invertible. We write down an inverse to f on U by writing down the required flag for each R-point $(\Phi , N)$ of U:

• • When N is regular (that is, its value at each point of $\operatorname {Spec} R$ is regular), take $F_i = \ker (N^{n-i})$ ;

• • When l is invertible, take $F_i=\bigoplus _{j=i}^{n-1} \ker (\Phi -q^i)$ .

The relation $\Phi N \Phi ^{-1} = qN$ implies that these agree on the locus where N is regular and l is invertible, and one can check that this defines a two-sided inverse $U \to f^{-1}(U) \subset \mathcal {Y}$ of $f|_{f^{-1}(U)}$ .

Proof.

1. 1. We may cover $\mathcal {F}$ by open affine subschemes U, with $U=\operatorname {Spec}(A)$ , such that the projection $\mathrm {GL}_n \rightarrow \mathcal {F}$ has a section $\gamma :U\rightarrow \mathrm {GL}_n$ . Notice that $\gamma \in GL_n(A)$ , so the universal pair $(\Phi ,N)$ on $\pi _{\mathcal {Z}}^{-1}(U)$ takes the form

   $$\begin{align*}(\gamma(\Phi_0+M)\gamma^{-1},\gamma N \gamma^{-1})\end{align*}$$
   with $\Phi _0=\textrm {diag}(q^{n-1},...,q,1)$ and $M,N\in \mathfrak {n}$ . It is now easy to see that $\pi _{\mathcal {Z}}^{-1}(U)\cong U\times \mathfrak {n}^2$ ; examining the behaviour of M under conjugation by B gives that $\mathcal {Z}$ is an affine bundle.

2. 2. When $n=2$ , $\mathcal {Y}=\mathcal {Z}$ , so we are done.

   For $n=3$ , the argument of part 1 gives similarly that $\mathcal {Y} \times _{\mathcal {F}}U\cong U\times \mathcal {C}(\mathfrak {n})$ , where

   $$\begin{align*}\mathcal{C}(\mathfrak{n})=\{(M,N)\in \mathfrak{n}\times \mathfrak{n}| (\Phi_0+M)N-qN(\Phi_0+M)=0\}\end{align*}$$
   Let
   $$\begin{align*}M=\begin{pmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{pmatrix}\end{align*}$$
   and
   $$\begin{align*}N=\begin{pmatrix} 0 & d & e \\ 0 & 0 & f \\ 0 & 0 & 0 \end{pmatrix}.\end{align*}$$
   Then the equation defining $\mathcal {C}(\mathfrak {n})$ is $(q^2-1)e+af-dc=0$ . This equation is not divisible by a uniformiser of $\mathcal {O}$ , so $\mathcal {C}(\mathfrak {n})$ and hence $\mathcal {Y}$ is $\mathcal {O}$ -flat. Now, $\mathcal {C}(\mathfrak {n})_{\mathbb {F}}$ is the affine hypersurface in $\operatorname {Spec} \mathbb {F}[a,b,c,d,e,f]$ defined by $af - cd = 0$ . Since $af-cd$ is irreducible, the rest of part (2) follows.

Proof. Part (2) of Theorem 2.4 follows from part (1) by Lemma 2.1.4 of [Reference Snowden21] and the fact that Y is Cohen–Macaulay. We will prove part (1) in the next section.

Proof. Using the short exact sequence $0 \to \mathcal {O}_{\mathcal {Y}} \xrightarrow {\times \varpi } \mathcal {O}_{\mathcal {Y}} \to \mathcal {O}_{Y} \to 0$ it suffices to show that $H^1(\mathcal {Y}, \mathcal {O}_{\mathcal {Y}}) = 0$ . Using that $\mathcal {X}$ is affine, this is equivalent to showing that $R^1f_{*}(\mathcal {O}_{\mathcal {Y}}) = 0$ . By Theorem 2.4, $R^if_{*}(\mathcal {O}_{Y}) = 0$ for $i \ge 1$ and, in particular, $(R^1f_{\ast } \mathcal {O}_{\mathcal {Y}})\otimes \mathbb {F} = 0$ (using the short exact sequence again). But we also have $R^1f_{\ast }\mathcal {O}_{\mathcal {Y}} \otimes E = 0$ by flat base change and the fact that f is an isomorphism after inverting l. Now $R^1f_{\ast }\mathcal {O}_{\mathcal {Y}}$ is a finitely generated A-module M (as f is proper) such that $M \otimes _{\mathcal {O}} \mathbb {F} = M \otimes _{\mathcal {O}} E = 0$ , from which it follows that $M = 0$ .

Proof. We know the proposition after inverting l. We claim that $A \to B$ is surjective. After $\otimes \mathbb {F}$ , this follows as, by the previous lemma and Theorem 1.1 (2), $B \otimes _{\mathcal {O}} \mathbb {F} = \Gamma (Y, \mathcal {O}_Y) = (A\otimes _{\mathcal {O}} \mathbb {F})^{\operatorname {red}}$ . But then the cokernel, a finite A-module, vanishes after $\otimes E$ and $\otimes \mathbb {F}$ and so must be zero, as at the end of the previous proof.

Now, $A \to B$ is a surjective map of flat $\mathcal {O}$ -algebras that is an isomorphism after inverting l, and is therefore an isomorphism, as required.

Proof of Theorem 1.1

Since A is $\varpi $ -torsion free, $\varpi $ is a regular element of A. By Proposition 2.6, $A \otimes _{\mathcal {O}} \mathbb {F} = B \otimes _{\mathcal {O}} \mathbb {F} = \Gamma (Y, \mathcal {O}_Y)$ . This algebra is reduced since Y is reduced, and so $X = \operatorname {Spec}(A \otimes _{\mathcal {O}} \mathbb {F})$ . By Theorem 2.4, X is Cohen–Macaulay. Finally, we will show below in Lemma 6.3 that the singular locus of X has codimension 2, and so X is normal and reduced by Serre’s criterion and the fact that it is Cohen–Macaulay.

Proof. It follows from the above that Y is irreducible. Since $\mathcal {Y} \to \mathcal {X}$ is proper, surjective over E, and $\mathcal {X}$ is flat over $\mathcal {O}$ , we see that f is surjective. As $n \le 3$ , Y is irreducible, and so the same is true for X. Moreover, $\dim Y = n^2 - 1$ (again, for $n \le 3$ ), and this is also the dimension of the set of points $(M,N,F)\in Y$ such that N is regular. But f is an isomorphism on this locus, and is therefore birational.

Proof.

1. 1. We have that Z is the total space of the G-equivariant vector bundle $\mathfrak {n}^2$ , and $\mathfrak {n}^{\ast } \cong \mathfrak {g}/\mathfrak {b}$ . Therefore

   $$\begin{align*}Z=\underline{\operatorname{Spec}}_{F}(\operatorname{Sym}[(\mathfrak{g}/\mathfrak{b})^2]),\end{align*}$$
   and the result follows.

2. 2. Recall that $\omega _Z\cong \omega _{Z/F}\otimes _{\mathcal {O}_Z} \pi _Z^{\ast }\omega _F$ because $\pi _Z$ and F are smooth. We also recall from [Reference Jantzen14] II. §4.2 that $\omega _F\cong \mathcal {O}(-2\rho )$ . It remains to calculate $\omega _{Z/F}$ . Notice that the relative tangent bundle has $T_{Z/F}=\pi _Z^{\ast }((\mathfrak {n})^2)$ , so $\omega _{Z/F}\cong \pi _Z^{\ast }\det ((\mathfrak {g}/\mathfrak {b})^2)=\pi _Z^{\ast }(\mathcal {O}(4\rho ))$ . Therefore

   $$\begin{align*}\omega_Z\cong \pi_Z^{\ast}(\mathcal{O}(-2\rho)\otimes\mathcal{O}(4\rho))=\pi_Z^{\ast}(\mathcal{O}(2\rho)) = \mathcal{O}_Z(2\rho).\end{align*}$$

3. 3. Write a general point of $\mathfrak {n}^2$ as

   $$\begin{align*}(M, N) = \left(\begin{pmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & d & e \\ 0 & 0 & f \\ 0 & 0 & 0 \end{pmatrix}\right)\end{align*}$$
   as in Lemma 2.3. Then $a, b, \ldots , f$ are a basis for $(\mathfrak {n}^2)^{\ast }$ . The ideal sheaf of $C(\mathfrak {n})$ inside $\operatorname {Sym}[(\mathfrak {n}^2)^{\ast }]$ is principal, generated by $af - dc$ , which is an element of $\operatorname {Sym}^2[(\mathfrak {n}^2)^{\ast }]$ of weight $\rho $ .Footnote ¹ In other words,
   $$\begin{align*}\mathcal{I}_{C(\mathfrak{n})} \cong \mathcal{O}_{\mathfrak{n} \times \mathfrak{n}}(\rho)\end{align*}$$
   as B-equivariant sheaves on $\mathfrak {n}^2$ . Since $Y = G\times ^B C(\mathfrak {n})\subset G \times ^B\mathfrak {n}^2 = Z$ we get that
   $$\begin{align*}\mathcal{I}_Y \cong \mathcal{O}_Z(\rho)\end{align*}$$
   as required.

4. 4. This follows from the adjunction formula (see [22, Section 0AU3 (7)])

   $$\begin{align*}\omega_Y = i^{\ast}(\omega_Z\otimes_{\mathcal{O}_Z} \mathcal{I}_Y^\vee)\end{align*}$$
   and parts (2) and (3).

Proof of Theorem 2.4 when $n=2$

In this case $F=\mathbb {P}^1$ and $Y=Z$ is the total space of the vector bundle $\mathfrak {n}^2$ . As line bundles on $\mathbb {P}^1$ , we see that

$$\begin{align*}\mathfrak{g}/\mathfrak{b} \cong \mathfrak{n}^{\ast}\cong \mathcal{O}(2)\end{align*}$$

and $\mathcal {O}(\rho )\cong \mathcal {O}(1)$ , so that

$$\begin{align*}H^1(\mathbb{P}^1,\operatorname{Sym}^r\left[(\mathfrak{g}/\mathfrak{b})^2\right])=H^1(\mathbb{P}^1,\operatorname{Sym}^r\left[(\mathfrak{g}/\mathfrak{b})^2\right](2\rho))=0\end{align*}$$

for every $r\geq 0$ . By Proposition 3.2,

$$\begin{align*}H^1(Y,\mathcal{O}_Y)=H^1(F,\pi_{\ast}\mathcal{O}_Y)=0\end{align*}$$

and

$$\begin{align*}H^1(Y,\omega_Y)=H^1(F,\pi_{\ast}\omega_Y)=0.\end{align*}$$

Since X is affine, $H^1(Y,\mathcal {O}_Y) = H^0(X, R^1f_{\ast }\mathcal {O}_Y)$ vanishes and so

$$\begin{align*}R^1f_{\ast}\mathcal{O}_Y = 0.\end{align*}$$

Similarly, $R^1f_{\ast }\omega _Y= 0$ .

It remains to show that $f^{\ast }:H^0(X,\mathcal {O}_X)\hookrightarrow H^0(Y,\mathcal {O}_Y)$ is an isomorphism. For convenience, we set $R=H^0(X,\mathcal {O}_X)$ and $\tilde {R}=H^0(Y,\mathcal {O}_Y)$ . We note that the natural map

$$\begin{align*}H^0(F,\operatorname{Sym}\left[\mathfrak{g}^2\right]) = \mathbb{F}[\mathfrak{g}^2]\rightarrow R\end{align*}$$

is surjective because X is defined as a closed subscheme of $\mathfrak {g}^2$ (and recall that we are identifying $\mathfrak {g} \cong \mathfrak {g}^{\ast }$ via the trace pairing). Let I be the kernel of this surjection. The composite

$$\begin{align*}H^0(F,\operatorname{Sym}\left[\mathfrak{g}^2\right])\rightarrow R\xrightarrow{f^{\ast}}\tilde{R}=H^0(F,\operatorname{Sym}\left[(\mathfrak{g}/\mathfrak{b})^2\right])\end{align*}$$

is the morphism induced by the natural surjective map of coherent sheaves $\mathfrak {g}^2\rightarrow (\mathfrak {g}/\mathfrak {b})^2$ .

Letting $S=\mathbb {F}[\mathfrak {g}^2]$ and $S_+$ be the irrelevant ideal of S, Proposition 2.1.5 of [Reference Snowden21] gives us that

$$\begin{align*}\operatorname{Tor}^S_0(R,\mathbb{F})=\tilde{R}/S_+\tilde{R}\cong H^0(F, \mathcal{O}_F)\oplus H^1(F, \mathfrak{b}^2).\end{align*}$$

Because of the (nonequivariant) isomorphism of vector bundles $\mathfrak {b}\cong \mathcal {O}(-1)^2$ , this shows that $\tilde {R}/S_+\tilde {R}$ is $1$ -dimensional. Hence the composite $\mathbb {F}[\mathfrak {g}^2] \rightarrow R \rightarrow \tilde {R}$ is surjective. It follows that the map $H^0(X,\mathcal {O}_X)\rightarrow H^0(Y,\mathcal {O}_Y)$ is an isomorphism as required.

Proof of Theorem 2.4 when $n=3$

By Proposition 4.15 below,

$$\begin{align*}H^i(Y,\mathcal{O}_Y)=H^i(Y,\omega_Y)=0\end{align*}$$

for all $i>0$ . To prove the theorem, the only thing that remains to check is that the natural morphism $f^{\ast }H^0(X,\mathcal {O}_X)\hookrightarrow H^0(Y,\mathcal {O}_Y)$ , injective since $Y \to X$ is birational, is an isomorphism.

As in the case $n = 2$ , define $S=\mathbb {F}[\mathfrak {g}^2]=\operatorname {Sym}[\mathfrak {g}^2]$ , with irrelevant ideal $S_+$ . Let $\tilde {R}=H^0(Z,\mathcal {O}_Z)$ for Z the total space of the vector bundle $\mathfrak {n}^2$ on $G/B$ . By Lemma 4.12, $H^i(Z, \mathcal {O}_Z) = 0$ for $i> 0$ . We may therefore apply Proposition 2.1.5 of [Reference Snowden21] to deduce that

$$\begin{align*}\tilde{R}/S_+\tilde{R}=\operatorname{Tor}^S_0(\tilde{R},\mathbb{F})=\bigoplus H^i(F,\Lambda^i[\mathfrak{b}^2])[i].\end{align*}$$

By Calculation 4.4 below, we know that $H^i(F,\Lambda ^i[\mathfrak {b}^2])=0$ unless $i=0$ , when

$$\begin{align*}H^0(F,\Lambda^0[\mathfrak{b}^2])=H^0(F,\mathbb{F})=\mathbb{F}.\end{align*}$$

Thus the map $S\rightarrow \tilde {R}$ is surjective and, as $H^1(Z, \mathcal {I}_Y) = 0$ by Proposition 3.2 and Lemma 4.12, the map $\tilde {R} = H^0(Z, \mathcal {O}_Z)\rightarrow H^0(Y,\mathcal {O}_Y)$ is surjective. The composite map

$$\begin{align*}S \to H^0(Z, \mathcal{O}_Z) \to H^0(Y, \mathcal{O}_Y)\end{align*}$$

is equal to the composite

$$\begin{align*}S \to H^0(X, \mathcal{O}_X) \xrightarrow{f^{\ast}}H^0(Y,\mathcal{O}_Y),\end{align*}$$

and so $f^{\ast }$ is surjective as required.

Calculation 4.3. Suppose that $l \ge 5$ . Then the G-representations $H^i(\Lambda ^j\mathfrak {b})$ are as shown in Table 1. Moreover, $H^i(\mathfrak {b} \otimes \mathfrak {b}) = H^i(\Lambda ^2\mathfrak {b})$ for all i.

Table 1

The cohomology groups $H^i(\Lambda ^j\mathfrak {b})$ .

Table 1 Long description

The table is organized with columns labeled by the index i (0, 1, 2, 3) and rows labeled by the index j (0, 1, 2, 3, 4, 5). The top-left header cell contains j backslash i.

* Row j equals 0: Column i equals 0 contains blackboard bold F. Columns i equals 1, 2, and 3 contain a centered dot.

* Row j equals 1: All columns (i equals 0, 1, 2, 3) contain a centered dot.

* Row j equals 2: Column i equals 1 contains blackboard bold F. Columns i equals 0, 2, and 3 contain a centered dot.

* Row j equals 3: Column i equals 2 contains blackboard bold F. Columns i equals 0, 1, and 3 contain a centered dot.

* Row j equals 4: All columns (i equals 0, 1, 2, 3) contain a centered dot.

* Row j equals 5: Column i equals 3 contains blackboard bold F. Columns i equals 0, 1, and 2 contain a centered dot.

Proof. If $l \ge 5$ then each of the representations $\mathfrak {b}$ , $\mathfrak {g}/\mathfrak {b}$ , $\mathfrak {b} \otimes \mathfrak {b}$ , and $\mathfrak {b} \otimes \mathfrak {g}/\mathfrak {b}$ is BWB-good. We may therefore discuss the potential support of these representations and apply Lemma 4.2 to them.

1. 0. When $j=0$ , $\Lambda ^j\mathfrak {b}=\mathbb {F}$ . By Theorem 4.1, $H^i(\mathbb {F})=0$ unless $i=0$ , in which case $H^0(\mathbb {F})=\mathbb {F}$ .

2. 1. When $j=1$ , we note that $\operatorname {psupp}^2(\mathfrak {b})=\operatorname {psupp}^3(\mathfrak {b})=\emptyset $ , so

   $$\begin{align*}H^2(\mathfrak{b})=H^3(\mathfrak{b})=0.\end{align*}$$
   We have
   $$\begin{align*}\operatorname{psupp}^0(\mathfrak{b}) = \operatorname{psupp}^1(\mathfrak{b}) = \{0, 0\}\end{align*}$$
   and so all composition factors of $H^0(\mathfrak {b})$ and $H^1(\mathfrak {b})$ are trivial as G-representations.

   Further, there is a short exact sequence

   $$\begin{align*}0\rightarrow\mathfrak{b}\to \mathfrak{g}\to \mathfrak{g}/\mathfrak{b}\to 0\end{align*}$$
   which gives a long exact sequence in cohomology
   
   As $\mathfrak {g}$ is a G-representation, we see from part (0) that $H^0(\mathfrak {g})=\mathfrak {g}$ and $H^1(\mathfrak {g})=0$ . For $\mathfrak {g}/\mathfrak {b}$ we have $\operatorname {psupp}^0(\mathfrak {g}/\mathfrak {b})=\{\rho \}$ and $\operatorname {psupp}^i(\mathfrak {g}/\mathfrak {b}) = 0$ for $i> 0$ . Thus $H^0(\mathfrak {g}/\mathfrak {b})\subset V(\rho ) = \mathfrak {g}$ . As all subquotients of $H^0(\mathfrak {b})$ and $H^1(\mathfrak {b})$ are the trivial representation, we must have $H^0(\mathfrak {b}) = H^1(\mathfrak {b}) = 0$ and $H^0(\mathfrak {g}/\mathfrak {b}) = \mathfrak {g}$ .

3. 2. When $j=2$ , consider the exact sequence

   $$\begin{align*}0\rightarrow\mathfrak{b}\otimes\mathfrak{b}\to \mathfrak{g}\otimes\mathfrak{b}\to \mathfrak{g}/\mathfrak{b}\otimes\mathfrak{b}\to 0,\end{align*}$$
   from which we get a long exact sequence with parts
   $$\begin{align*}H^i(\mathfrak{g}\otimes\mathfrak{b})\to H^i(\mathfrak{g}/\mathfrak{b}\otimes\mathfrak{b})\to H^{i+1}(\mathfrak{b}\otimes\mathfrak{b})\to H^{i+1}(\mathfrak{g}\otimes\mathfrak{b}).\end{align*}$$
   As $H^i(\mathfrak {b})=0$ for all i, we obtain isomorphisms $H^i(\mathfrak {b}\otimes \mathfrak {g}/\mathfrak {b})\cong H^{i+1}(\mathfrak {b}\otimes \mathfrak {b})$ for all i. Thus, $H^0(\mathfrak {b}\otimes \mathfrak {b})=0$ and $H^3(\mathfrak {b}\otimes \mathfrak {b})=H^2(\mathfrak {g}/\mathfrak {b}\otimes \mathfrak {b})=0$ , since $\operatorname {psupp}^2(\mathfrak {g}/\mathfrak {b}\otimes \mathfrak {b})=\emptyset $ .

   To show that $H^2(\mathfrak {b}\otimes \mathfrak {b}) = 0$ , consider first the exact sequence

   $$\begin{align*}0\rightarrow\mathfrak{b}\otimes\mathfrak{g}/\mathfrak{b}\to \mathfrak{g}\otimes\mathfrak{g}/\mathfrak{b}\to \mathfrak{g}/\mathfrak{b}\otimes\mathfrak{g}/\mathfrak{b}\to 0\end{align*}$$
   giving rise to the long exact sequence
   $$\begin{align*}0\to H^0(\mathfrak{b}\otimes\mathfrak{g}/\mathfrak{b})\to \mathfrak{g} \otimes \mathfrak{g}\to H^{0}(\mathfrak{g}/\mathfrak{b}\otimes\mathfrak{g}/\mathfrak{b})\to H^{1}(\mathfrak{b}\otimes\mathfrak{g}/\mathfrak{b})\to 0.\end{align*}$$
   Here we have used that $H^1(\mathfrak {g}\otimes \mathfrak {g}/\mathfrak {b})\cong \mathfrak {g} \otimes H^1(\mathfrak {g}/\mathfrak {b})=0$ , as $\operatorname {psupp}^1(\mathfrak {g}/\mathfrak {b}) = \emptyset $ , and $H^0(\mathfrak {g}\otimes \mathfrak {g}/\mathfrak {b})\cong \mathfrak {g} \otimes H^0(\mathfrak {g}/\mathfrak {b})\cong \mathfrak {g}\otimes \mathfrak {g}$ . Thus
   $$\begin{align*}H^2(\mathfrak{b} \otimes \mathfrak{b}) \cong H^1(\mathfrak{b}\otimes \mathfrak{g}/\mathfrak{b})\cong\operatorname{coker}(\mathfrak{g}\otimes\mathfrak{g}\to H^0(\mathfrak{g}/\mathfrak{b}\otimes \mathfrak{g}/\mathfrak{b})).\end{align*}$$
   Since $\operatorname {psupp}^2(\mathfrak {b} \otimes \mathfrak {b}) = \{0,0,0,0\}$ , every irreducible constituent of $H^2(\mathfrak {b}\otimes \mathfrak {b})$ is trivial. So it is enough to show that $H^0(\mathfrak {g}/\mathfrak {b}\otimes \mathfrak {g}/\mathfrak {b})$ does not have the trivial representation as a subquotient. If $l \ge 7$ , this follows as $\mathfrak {g}/\mathfrak {b} \otimes \mathfrak {g}/\mathfrak {b}$ is BWB-good and $0 \not \in \operatorname {psupp}^0(\mathfrak {b} \otimes \mathfrak {g}/\mathfrak {b})$ . If $l = 5$ then every weight of $\mathfrak {g}/\mathfrak {b} \otimes \mathfrak {g}/\mathfrak {b}$ except for $2\rho $ is in the BWB locus and nonzero, so cannot give rise to a trivial subquotient of $H^0(\mathfrak {g}/\mathfrak {b} \otimes \mathfrak {g}/\mathfrak {b})$ . Every other composition factor of $H^0(\mathfrak {g}/\mathfrak {b}\otimes \mathfrak {g}/\mathfrak {b})$ is isomorphic to a composition factor of $H^0(\mathbb {F}(2\rho ))$ . By the strong linkage principle of [Reference Jantzen14, Proposition II.6.13], these are of the form $V(\rho )$ or $L(2\rho )$ (the irreducible representation with highest weight $2\rho $ ), and are therefore nontrivial as required.

   Finally, we compute

   $$\begin{align*}\chi(\mathfrak{b}\otimes\mathfrak{b})=4[\mathbb{F}]-4[\mathbb{F}]-4[\mathbb{F}]+2[\mathbb{F}]+2[\mathbb{F}]-[\mathbb{F}]=-[\mathbb{F}].\end{align*}$$
   Since $H^i(\mathfrak {b}\otimes \mathfrak {b}) = 0$ for $i \ne 1$ , we deduce that $H^1(\mathfrak {b}\otimes \mathfrak {b})=\mathbb {F}$ . As $\Lambda ^2\mathfrak {b}$ is a direct summand of $\mathfrak {b}\otimes \mathfrak {b}$ (as $l> 2$ ), it follows that $H^i(\Lambda ^2\mathfrak {b})=0$ when $i\neq 1$ and, as
   $$\begin{align*}\chi(\Lambda^2\mathfrak{b})=[\mathbb{F}]-4[\mathbb{F}]+2[\mathbb{F}]=-[\mathbb{F}],\end{align*}$$
   $H^1(\Lambda ^2\mathfrak {b})= \mathbb {F} = H^1(\mathfrak {b} \otimes \mathfrak {b})$ .

   For later use, also note that we have shown $H^i(\mathfrak {b} \otimes \mathfrak {g}/\mathfrak {b}) = 0$ for $i> 0$ ; an Euler characteristic calculation then shows that $H^0(\mathfrak {b} \otimes \mathfrak {g}/\mathfrak {b}) = \mathbb {F}$ .

4. 3. For $j = 3, 4, 5$ we have a B-equivariant pairing

   $$\begin{align*}\Lambda^j\mathfrak{b} \times \Lambda^{5-j}\mathfrak{b} \to \Lambda^5 \mathfrak{b} \cong \mathbb{F}(-2\rho)\end{align*}$$
   and so
   $$\begin{align*}(\Lambda^j\mathfrak{b})^{\ast} \otimes \mathbb{F}(-2\rho) \cong \Lambda^{5-j}\mathfrak{b}.\end{align*}$$
   Since $\omega _{G/B} \cong \mathbb {F}(-2\rho )$ , Serre duality gives
   $$\begin{align*}H^i(\Lambda^j\mathfrak{b}) \cong H^{3-i}(\Lambda^{5-j}\mathfrak{b})^{\ast}\end{align*}$$
   and the result follows from parts 0–2.

Calculation 4.4. Suppose that $l \geq 5$ . Then

$$\begin{align*}H^i(\mathfrak{b} \otimes \Lambda^2\mathfrak{b}) = \begin{cases} \mathfrak{g}^2 \oplus \mathbb{F} & \text{for } i=2 \\ 0 & \text{otherwise.} \end{cases}\end{align*}$$

Proof. Firstly, we observe that $\mathfrak {b}\otimes \mathfrak {b} \otimes \mathfrak {g}$ , and therefore any subquotient of it, is BWB-good for $l \ge 5$ . This justifies the use of potential supports in the following calculation.

Considering the long exact sequence associated to

$$\begin{align*}0\to\Lambda^2\mathfrak{b}\otimes \mathfrak{b}\to \Lambda^2\mathfrak{b}\otimes\mathfrak{g}\to\Lambda^2\mathfrak{b}\otimes \mathfrak{g}/\mathfrak{b}\to 0\end{align*}$$

along with the isomorphism $H^i(\mathfrak {g}\otimes \Lambda ^2\mathfrak {b})=\mathfrak {g}\otimes H^i(\Lambda ^2\mathfrak {b})$ and the results of Calculation 4.3, we obtain an exact sequence

$$\begin{align*}0\to H^0(\Lambda^2\mathfrak{b}\otimes\mathfrak{g}/\mathfrak{b})\to H^1(\Lambda^2\mathfrak{b}\otimes\mathfrak{b})\to \mathfrak{g} \to H^1(\Lambda^2\mathfrak{b}\otimes\mathfrak{g}/\mathfrak{b})\to H^2(\Lambda^2\mathfrak{b}\otimes\mathfrak{b})\to 0,\end{align*}$$

an isomorphism

$$\begin{align*}H^2(\Lambda^2\mathfrak{b}\otimes\mathfrak{g}/\mathfrak{b})\cong H^3(\Lambda^2\mathfrak{b}\otimes\mathfrak{b}),\end{align*}$$

and $H^0(\Lambda ^2\mathfrak {b}\otimes \mathfrak {b})=0$ . As $\operatorname {psupp}^2(\Lambda ^2\mathfrak {b}\otimes \mathfrak {g}/\mathfrak {b})=\emptyset $ , we see that $H^3(\Lambda ^2\mathfrak {b}\otimes \mathfrak {b})=0$ .

We calculate (using superscripts to denote multiplicity)

$$\begin{align*}\operatorname{psupp}^i(\Lambda^2\mathfrak{b}\otimes\mathfrak{b}) = \begin{cases} \{0^2\} & i = 0 \\ \{0^{10}\} & i=1 \\ \{0^{14}, \rho^2\} & i=2 \\ \{0^7\} & i=3 \end{cases}\end{align*}$$

and so

$$\begin{align*}\chi(\Lambda^2\mathfrak{b} \otimes \mathfrak{b}) = 2[\mathfrak{g}] - [\mathbb{F}].\end{align*}$$

Since $\rho $ appears only in $\operatorname {psupp}^2$ , we obtain that

$$\begin{align*}H^1(\Lambda^2\mathfrak{b}\otimes\mathfrak{b}) \cong \mathbb{F}^n\end{align*}$$

and

$$\begin{align*}H^2(\Lambda^2\mathfrak{b}\otimes\mathfrak{b})\cong\mathfrak{g}^2\oplus \mathbb{F}^{n+1}\end{align*}$$

for some integer $n \ge 0$ . It follows that

$$\begin{align*}H^0(\Lambda^2\mathfrak{b} \otimes \mathfrak{g}/\mathfrak{b})\cong \mathbb{F}^n\end{align*}$$

since the map $H^1(\Lambda ^2\mathfrak {b}\otimes \mathfrak {b})\to \mathfrak {g}$ must be zero.

The B-representation $\Lambda ^2\mathfrak {b}\otimes \mathfrak {g}/\mathfrak {b}$ is a direct summand of $\mathfrak {b}\otimes \mathfrak {b}\otimes \mathfrak {g}/\mathfrak {b}$ , which fits into the short exact sequence

$$\begin{align*}0\to\mathfrak{b}\otimes \mathfrak{b}\otimes \mathfrak{g}/\mathfrak{b}\to \mathfrak{g}\otimes \mathfrak{b}\otimes\mathfrak{g}/\mathfrak{b}\to\mathfrak{g}/\mathfrak{b}\otimes \mathfrak{b}\otimes \mathfrak{g}/\mathfrak{b}\to 0\end{align*}$$

and so we have an exact sequence

$$\begin{align*}0 \to H^0(\mathfrak{b}\otimes \mathfrak{b}\otimes \mathfrak{g}/\mathfrak{b})\hookrightarrow \mathfrak{g}\otimes H^0(\mathfrak{b}\otimes\mathfrak{g}/\mathfrak{b})\cong \mathfrak{g}\end{align*}$$

(the last isomorphism appearing in part 2 of the proof of Calculation 4.3). As $H^0(\Lambda ^2\mathfrak {b}\otimes \mathfrak {g}/\mathfrak {b})\cong \mathbb {F}^n$ and $\mathbb {F}$ is not a subrepresentation of $\mathfrak {g}$ , we deduce that $H^0(\Lambda ^2\mathfrak {b}\otimes \mathfrak {g}/\mathfrak {b})=H^1(\Lambda ^2\mathfrak {b}\otimes \mathfrak {b})=0$ .

Calculation 4.5.

1. 1. Suppose that $l \ge 5$ . Then

   $$\begin{align*}H^3(\Lambda^3\mathfrak{b} \otimes \mathfrak{b}) = 0.\end{align*}$$

2. 2. Suppose that $l \ge 5$ . Then

   $$\begin{align*}H^3(\Lambda^2 \mathfrak{b} \otimes \Lambda^2 \mathfrak{b}) = 0.\end{align*}$$

Proof.

1. 1. We first show that $H^3(\Lambda ^3\mathfrak {b} \otimes \mathfrak {b}) = 0$ . By Serre duality, the pairing $\Lambda ^3 \mathfrak {b} \times \Lambda ^2\mathfrak {b} \to \Lambda ^5\mathfrak {b} \cong \mathbb {F}(-2\rho )$ , and the isomorphism $\omega _F \cong \mathcal {O}_F(-2\rho )$ , we have

   $$\begin{align*}H^3(\Lambda^3\mathfrak{b} \otimes \mathfrak{b}) \cong H^0(\Lambda^2\mathfrak{b} \otimes \mathfrak{b}^{\ast})^{\ast}.\end{align*}$$
   Since $\mathfrak {b}^{\ast } \cong \mathfrak {g}/\mathfrak {n}$ , we wish to show that $H^0(\Lambda ^2\mathfrak {b} \otimes \mathfrak {g}/\mathfrak {n}) = 0$ . The last line of the proof of Calculation 4.4 shows that $H^0(\Lambda ^2\mathfrak {b} \otimes \mathfrak {g}/\mathfrak {b}) = 0$ , and by Calculation 4.3 $H^0(\Lambda ^2\mathfrak {b}) = 0$ . The result follows by tensoring the short exact sequence of B-representations
   $$\begin{align*}0 \to \mathfrak{b}/\mathfrak{n} \cong \mathbb{F}^2 \to \mathfrak{g}/\mathfrak{n} \to \mathfrak{g}/\mathfrak{b} \to 0\end{align*}$$
   with $\Lambda ^2\mathfrak {b}$ and applying $H^0(\cdot )$ .

2. 2. If $l\geq 5$ then $\Lambda ^2\mathfrak {b} \otimes \Lambda ^2\mathfrak {b}$ is BWB-good, which justifies the following argument in characteristic l. Note that the only weights in $\operatorname {psupp}^3(\Lambda ^2\mathfrak {b} \otimes \Lambda ^2\mathfrak {b})$ are $0$ and $\rho $ , so that $H^3(\Lambda ^2\mathfrak {b} \otimes \Lambda ^2\mathfrak {b})$ is a direct sum of copies of $\mathbb {F}$ and $\mathfrak {g}$ .

   Let $e_{\pm \alpha }$ , $e_{\pm \rho }$ and $e_{\pm \beta }$ be root vectors in $\mathfrak {g}$ (chosen to be elementary matrices) and let P be the parabolic subgroup of G whose Lie algebra is spanned by $\mathfrak {b}$ and $e_{\alpha }$ , with unipotent radical $U_P$ and Levi quotient $M = P/U_P \cong GL_2$ . We will use the following lemma of Demazure.

The next lemma allows us to recognize certain representations that extend to P.

We first show that $H^3(\Lambda ^2\mathfrak {b} \otimes \Lambda ^2\mathfrak {b})$ does not contain the trivial representation. Let $V_1 \subset \Lambda ^2\mathfrak {b} \otimes \Lambda ^2\mathfrak {b}$ be the $\mathbb {F}$ -span of all weight vectors of weights in

$$\begin{align*}\{-\rho - 2\alpha, -2\rho-2\alpha,-2\rho-\alpha,-3\rho,-2\rho-\beta,-2\rho-2\beta, -\rho-2\beta\}\end{align*}$$

and let $V_2$ be the $\mathbb {F}$ -span of $V_1$ and all weight vectors of weights in $\{-2\rho , -\rho - \alpha , -\rho -\beta \}$ . Then $V_1$ and $V_2$ are B-subrepresentations of $V_3 = \Lambda ^2\mathfrak {b} \otimes \Lambda ^2\mathfrak {b}$ . Let $V = V_2/V_1$ . Because $0$ is not contained in $\operatorname {psupp}^3(V_1)$ or $\operatorname {psupp}^3(V3/V_2)$ , to show that $H^3(V_3)$ contains no copy of the trivial representation it is enough to show that $H^3(V) = 0$ .

Consider the weight space $V_{-\rho -\beta }\subseteq V$ . The B-representation generated by $V_{\rho -\beta }$ is $V^{\beta }:=V_{-\rho -\beta }\oplus e_{-\alpha }(V_{-\rho -\beta })$ , and this representation splits as a direct sum into copies of $\mathbb {F}(-\rho -\beta )$ and $2$ -dimensional representations satisfying the hypotheses of Lemma 4.7. Thus, $H^3(V^{\beta })=0$ by Lemma 4.6. Analogously, consider the B-subrepresentation $V^{\alpha }\subseteq V/V^{\beta }$ generated by $V_{-\rho -\alpha }$ . In the exact same way (switching the roles of $\alpha $ and $\beta $ in Lemmas 4.6 and 4.7), we see that $H^3(V^{\alpha })=0$ . To show that $H^3(V)=0$ , it remains simply to show that $V^{\alpha }=V/V^{\beta }$ , or equivalently, that $V_{-2\rho }=e_{-\alpha }(V_{-\rho -\beta })+e_{-\beta }(V_{-\rho -\alpha })$ .

This can be shown via a direct calculation by showing that each of the $17$ simple tensors which span $V_{-2\rho }$ are in this joint image. We outline this calculation. Set $f_{\gamma }=e_{-\gamma }\in \mathfrak {b}$ (reserving $e_{-\gamma }=[f_{\gamma },\_]$ for the linear map) and set $t_{\alpha },t_{\beta }\in \mathfrak {t}$ a basis of weight $0$ vectors such that $e_{-\nu }(t_{\mu })=\delta _{\nu ,\mu }f_{\nu }$ in $\mathfrak {b}$ for $\nu ,\mu \in \{\alpha ,\beta \}$ . (Such a basis exists when the characteristic $l\neq 3$ ). Then

$$ \begin{align*} (f_{\alpha}\wedge f_{\rho})\otimes(t_{\alpha}\wedge f_{\beta})&=e_{-\beta}\Big[(f_{\alpha}\wedge f_{\rho})\otimes(t_{\alpha}\wedge t_{\beta})\Big] \\ (f_{\alpha}\wedge f_{\rho})\otimes(t_{\beta}\wedge f_{\beta})&=\frac{1}{2}e_{-\alpha}\Big[(t_{\alpha}\wedge f_{\rho})\otimes(t_{\beta}\wedge f_{\beta}) \\ &\quad \color{red}{+\,(f_{\beta}\wedge f_{\alpha})\otimes(t_{\beta}\wedge f_{\beta})} +(f_{\beta}\wedge t_{\alpha})\otimes(t_{\beta}\wedge f_{\rho})\Big] \\ (t_{\alpha}\wedge f_{\rho})\otimes(f_{\alpha}\wedge f_{\beta})&=\frac{1}{2}e_{-\beta}\Big[\color{red}{(t_{\alpha}\wedge f_{\alpha})\otimes(t_{\beta}\wedge f_{\rho})} \\ &\quad +(t_{\alpha}\wedge f_{\alpha})\otimes(f_{\alpha}\wedge f_{\beta}) \color{red}{+(t_{\alpha}\wedge f_{\rho})\otimes(f_{\alpha}\wedge t_{\beta})}\Big]. \end{align*} $$

These three equations give us $12$ linearly independent basis elements in the image when we consider the two independent symmetries of swapping the left and right sides of the tensor, and the symmetry interchanging $e_{-\alpha }$ and $-e_{-\beta }$ , the effect of which is to flip the signs of the terms in red. We have further:

$$ \begin{align*} (t_{\alpha}\wedge f_{\rho})\otimes(t_{\alpha}\wedge f_{\rho})&=e_{-\beta} \Big[\color{red}{(t_{\alpha}\wedge f_{\alpha})\otimes(t_{\alpha}\wedge f_{\rho})}\Big] \\ (t_{\alpha}\wedge f_{\rho})\otimes(t_{\beta}\wedge f_{\rho})&=\frac{1}{2}e_{-\beta}\Big[\color{red}{(t_{\alpha}\wedge f_{\alpha})\otimes(t_{\beta}\wedge f_{\rho})} \\ &\quad +(t_{\alpha}\wedge f_{\alpha})\otimes(f_{\alpha}\wedge f_{\beta}) \color{red}{-(t_{\alpha}\wedge f_{\rho})\otimes(f_{\alpha}\wedge t_{\beta})}\Big] \end{align*} $$

again with the two conjugates (swapping $\alpha \leftrightarrow \beta $ and flipping the signs of the terms in red) giving us in total $16$ of the simple tensor spanning $V_{-2\rho }$ .

The last basis element can be expressed as

$$ \begin{align*} (f_{\alpha}\wedge f_{\beta})\otimes (f_{\alpha}\wedge f_{\beta})&=e_{-\beta}\Big[(f_{\alpha}\wedge f_{\beta})\otimes(f_{\alpha}\wedge t_{\beta})\Big] \\ &\quad-(f_{\alpha}\wedge f_{\beta})\otimes (f_{\rho}\wedge t_{\beta})-(f_{\rho}\wedge f_{\beta})\otimes(f_{\alpha}\wedge t_{\beta}) \end{align*} $$

with the last two terms inside $e_{-\alpha }V_{-\rho -\beta }+e_{-\beta }(V_{\rho -\alpha })$ as previously considered. Since $V_{-2\rho }$ is a $17$ dimensional vector space, this shows that $V_{-2\rho }=e_{-\alpha }(V_{-\rho -\beta })+e_{-\beta }(V_{-\rho -\alpha })$ and, consequently, ${H^3(V)=0}$ .

We have now shown that $H^3(\Lambda ^2\mathfrak {b} \otimes \Lambda ^2\mathfrak {b})\cong \mathfrak {g}^s$ for some $s \ge 0$ . The only weight of $\Lambda ^2\mathfrak {b} \otimes \Lambda ^2\mathfrak {b}$ remaining that can contribute to $H^3$ is $-3\rho $ , which occurs with multiplicity two; thus $s \le 2$ and we wish to show that $s = 0$ .

Let $\mu = -2\beta - \rho $ . In $\Lambda ^2 \mathfrak {b} \otimes \Lambda ^2\mathfrak {b}$ let

$$\begin{align*}v = (e_{-\rho}\wedge e_{-\beta}) \otimes (e_{-\beta}\wedge t_{\alpha})\end{align*}$$

and

$$\begin{align*}v' = (e_{-\beta}\wedge t_{\alpha})\otimes (e_{-\rho}\wedge e_{-\beta}).\end{align*}$$

Let $\tilde {W}$ be the B-representation generated by v, and let $W = \tilde {W}_{U_P}$ be its $U_P$ -coinvariants. The weights of W are a subset of

$$\begin{align*}\{\mu = -2\beta - \rho, \mu - \alpha, \mu - 2\alpha = -3\rho\}. \end{align*}$$

Since

$$\begin{align*}e_{-\alpha}^2(v) = 2(e_{-\rho} \wedge e_{-\beta}) \otimes (e_{-\rho} \wedge e_{-\alpha})\ne 0\end{align*}$$

and $\left \langle \mu + L_1, \alpha \right \rangle = 2$ , Lemma 4.7 applies to show that $W(L_1)$ extends to a representation of P and hence that $H^i(W) = 0$ for all i.

Now, $\operatorname {psupp}^3(\ker (\tilde {W}\rightarrow W)) = \emptyset $ and so, as $H^i(W) = 0$ for all i,

$$\begin{align*}H^3(\tilde{W}) = 0.\end{align*}$$

Constructing $\tilde {W}'$ similarly from $v'$ we see that $H^3(\tilde {W}') = 0$ . Let

$$\begin{align*}Q = (\Lambda^2\mathfrak{b} \otimes \Lambda^2\mathfrak{b})/(\tilde{W} + \tilde{W}').\end{align*}$$

Then

$$\begin{align*}H^3(Q) \cong H^3(\Lambda^2\mathfrak{b} \otimes \Lambda^2\mathfrak{b}) \cong \mathfrak{g}^s.\end{align*}$$

However, $\tilde {W} + \tilde {W}'$ contains the entire two-dimensional weight space of $\Lambda ^2\mathfrak {b} \otimes \Lambda ^2\mathfrak {b}$ of weight $-3\rho $ , and so $\rho \not \in \operatorname {psupp}^3(Q)$ . Thus $s = 0$ , as required.

Calculation 4.8. Suppose that $l \ge 5$ , $0 \le i \le 3$ , and $0 \le j \le 4$ . Then the cohomology group $H^i(\Lambda ^j(\mathfrak {b} \oplus \mathfrak {b}))$ is as given in Table 2.

Table 2

The group $H^i(\Lambda ^j[\mathfrak {b} \oplus \mathfrak {b}])$ . Cohomology groups which are zero are denoted by a dot, while those which are unknown are denoted by a question mark.

Table 2 Long description

The table is organized with index i as columns (0 to 3) and index j as rows (0 to 4).

* Row j equals 0: Column i equals 0 contains blackboard bold F. Columns 1, 2, and 3 contain dots.

* Row j equals 1: Columns 0, 1, 2, and 3 all contain dots.

* Row j equals 2: Column i equals 1 contains blackboard bold F super 3. Columns 0, 2, and 3 contain dots.

* Row j equals 3: Column i equals 2 contains fraktur g super 4 direct sum blackboard bold F super 4. Columns 0, 1, and 3 contain dots.

* Row j equals 4: Columns 0, 1, and 2 contain question marks. Column 3 contains a dot.

Proof. This is immediate from Calculations 4.3, 4.4 and 4.5 and the decompositions

$$ \begin{align*} \Lambda^2(\mathfrak{b} \oplus \mathfrak{b}) &= \left(\Lambda^2\mathfrak{b}\right)^2 \oplus \mathfrak{b} \otimes \mathfrak{b},\\ \Lambda^3(\mathfrak{b} \oplus \mathfrak{b}) &= \left(\Lambda^3 \mathfrak{b}\right)^2 \oplus \left(\Lambda^2\mathfrak{b} \otimes \mathfrak{b}\right)^2\end{align*} $$

and

$$ \begin{align*} \Lambda^4(\mathfrak{b} \oplus \mathfrak{b}) &= \left(\Lambda^4 \mathfrak{b}\right)^2 \oplus \left(\Lambda^3\mathfrak{b} \otimes \mathfrak{b}\right)^2 \oplus \left(\Lambda^2\mathfrak{b} \otimes \Lambda^2\mathfrak{b}\right) \end{align*} $$

which hold for $l \ge 5$ .

Calculation 4.10. Let $\lambda \in X(T)_+$ be a nonzero dominant weight. Suppose that $l \ge 5$ . Then:

1. 0. $H^i(\mathbb {F}(\lambda )) = 0$ for all $i> 0$ ;

2. 1. $H^i(\mathfrak {b}(\lambda )) = 0$ if $i> 1$ , and $H^1(\mathfrak {b}(\lambda )) = 0$ if $\lambda - \rho $ is dominant or if $l \ge 2 + \max (\left \langle \lambda , \alpha ^\vee \right \rangle , \left \langle \lambda , \beta ^\vee \right \rangle )$ ;

3. 2. $H^i(\Lambda ^2[\mathfrak {b} \oplus \mathfrak {b}](\lambda )) = 0$ if $i \ge 2$ ;

4. 3. $H^3(\Lambda ^3[\mathfrak {b} \oplus \mathfrak {b}](\lambda )) = 0$ .

Proof. Recall that $H^i(\mathbb {F}(\lambda )) = 0$ for all $i> 0$ by Kempf’s vanishing theorem [Reference Jantzen14, Proposition II.4.5]. This deals immediately with part (0). Recall also from [Reference Jantzen14, Proposition II.5.4 (a)] that, if $\mu \in X(T)$ with $\left \langle \mu , \kappa ^\vee \right \rangle = -1$ for some simple root $\kappa $ , then $H^i(\mathbb {F}(\mu )) = 0$ for all $i \ge 0$ . In particular, this implies that $H^i(\mathbb {F}(\lambda )) = 0$ for all $\lambda \in -\rho + X(T)_+$ and all $i>0$ .

For part (1), if $\lambda - \rho $ is dominant, then every weight of $\mathfrak {b}(\lambda )$ lies in $-\rho + X(T)_+$ and so $H^i(\mathfrak {b}(\lambda )) = 0$ for all $i> 0$ . Otherwise, we have (without loss of generality) that $\lambda = aL_1$ ; assume this now.

For every weight $\mu $ of $\mathfrak {b}(\lambda )$ , we have $\mu \in -\rho + X(T)^+$ , or $\mu = -\beta + aL_1$ and $\left \langle \mu , \beta ^\vee \right \rangle = -2$ . We wish to show that $H^i(\mathbb {F}(\mu ))$ vanishes in the latter case for $i> 1$ . By [Reference Jantzen14, Proposition II.5.4 (d)], we have

$$\begin{align*}H^i(\mathbb{F}(\mu)) \cong H^{i-1}(\mathbb{F}(s_\beta\cdot \mu))= 0\end{align*}$$

for all $i> 1$ Footnote ². It follows that $H^i(\mathfrak {b}(\lambda )) = 0$ for all $i\ge 2$ .

Suppose now that $l \ge 2 + a$ . Then $\mathfrak {b}(\lambda )$ is BWB-good and $\operatorname {psupp}^1(\mathfrak {b}(\lambda )) = \{\lambda \}$ . On the other hand, from the short exact sequence

$$\begin{align*}0 \to \mathfrak{b}(\lambda) \to \mathfrak{g}(\lambda) \to (\mathfrak{g}/\mathfrak{b})(\lambda) \to 0\end{align*}$$

we obtain a surjection $H^0((\mathfrak {g}/\mathfrak {b})(\lambda )) \twoheadrightarrow H^1(\mathfrak {b}(\lambda ))$ . But, using that $\lambda = aL_1$ , $\operatorname {psupp}^0((\mathfrak {g}/\mathfrak {b})(\lambda )) = \{\rho + \lambda , \beta + \lambda \}$ which is disjoint from $\{\lambda \}$ , and so

$$\begin{align*}H^1(\mathfrak{b}(\lambda)) = 0\end{align*}$$

as required.

For part (2), as before, it is enough to consider weights $\mu $ of $\Lambda ^2[\mathfrak {b} \oplus \mathfrak {b}](\lambda )$ with $\left \langle \mu + \rho , \alpha ^\vee \right \rangle < 0$ . Any such weight $\mu $ either lies in $s_\alpha \cdot (-\rho + X(T)_+)$ , or is equal to $-2\rho - L_3$ and so lies on the line $\left \langle \mu , \beta ^\vee \right \rangle = -1$ . In the latter case, $H^i(\mathbb {F}(\mu )) = 0$ for all i; in the former, as $\left \langle \mu + \rho , \alpha ^\vee \right \rangle \ge -3$ and $l \ge 3$ , [Reference Jantzen14, Proposition II.5.4 (d)] implies that

$$\begin{align*}H^i(\mathbb{F}(\mu)) \cong H^{i-1}(\mathbb{F}(s_\alpha \cdot \mu)) = 0\end{align*}$$

for all $i \ge 2$ . Thus $H^i(\Lambda ^2[\mathfrak {b} \oplus \mathfrak {b}](\lambda )) = 0$ for all $i \ge 2$ .

Finally, for part (3), suppose $\mu $ is a weight of $\Lambda ^3[\mathfrak {b}\oplus \mathfrak {b}](\lambda )$ . By inspection of the weights of $\Lambda ^3[\mathfrak {b} \oplus \mathfrak {b}]$ and the assumption that $l \ge 5$ we see that either:

1. 1. $\mu $ lies on one of the lines $\left \langle \mu + \rho , \kappa \right \rangle = 0$ for some root $\kappa $ ;

2. 2. $\mu \in C^2_{\mathbb {Z}}$ ;

3. 3. $\mu \in C^1_{\mathbb {Z}}$ and $0 \le \left \langle \mu +\rho , \kappa \right \rangle \le 4$ for some simple root $\kappa $ ; or

4. 4. $\mu \in C^0_{\mathbb {Z}}$ .

In each case we have $H^3(\mathbb {F}(\mu )) = 0$ by [Reference Jantzen14]: Proposition II.5.4 (a) in case (1), Corollary II.5.5 (b) in case (2), Proposition II.5.4 (d) in case (3) and Theorem II.4.5 in case (4).

Calculation 4.11. Let $\lambda \in X(T)_+$ . If $l \ge 5$ then

$$\begin{align*}H^i(\Lambda^j[\mathfrak{b}\oplus \mathfrak{b}](\lambda + \alpha)) =H^i(\Lambda^j[\mathfrak{b}\oplus \mathfrak{b}](\lambda + \beta))= 0\end{align*}$$

whenever $i> j$ .

If $l \ge 7$ and $i \ne 1$ , then

$$\begin{align*}H^i(\Lambda^i[\mathfrak{b} \oplus\mathfrak{b}](\alpha)) = H^i(\Lambda^i[\mathfrak{b} \oplus\mathfrak{b}](\beta))= 0,\end{align*}$$

while

$$\begin{align*}H^1(\mathfrak{b}(\alpha)) \cong H^1(\mathfrak{b}(\beta)) \cong \mathfrak{g}.\end{align*}$$

Proof. The statement in the case $i> j$ can be proved by applying [Reference Jantzen14, Proposition II.5.4] exactly as in Calculation 4.10, and we omit the details. We therefore focus on the case $i = j$ ; by symmetry, it is enough to deal with $\alpha $ . The reader may check that, in the calculations below, every representation to which we apply $\operatorname {psupp}$ is BWB-good for $l\ge 7$ .

If $i = 0$ then $H^0(\mathbb {F}(\alpha )) = 0$ as $\alpha $ is on the boundary of $\overline {C}_{\mathbb {Z}}$ .

We have already shown that $H^i(\mathfrak {b}(\alpha )) = 0$ for $i> 1$ . From the inclusion $\mathfrak {b} \hookrightarrow \mathfrak {g}$ we obtain an inclusion

$$\begin{align*}H^0(\mathfrak{b}(\alpha)) \hookrightarrow \mathfrak{g}\otimes H^0(\mathbb{F}(\alpha)) = 0\end{align*}$$

and so $H^0(\mathfrak {b}(\alpha )) = 0$ . Computing the Euler characteristic shows that $H^1(\mathfrak {b}(\alpha )) \cong \mathfrak {g}$ .

We next show that $H^1((\mathfrak {g}/\mathfrak {b})(\alpha )) = 0$ . Indeed, this sits in a short exact sequence

$$\begin{align*}\mathfrak{g}\otimes H^1(\mathbb{F}(\alpha)) \to H^1((\mathfrak{g}/\mathfrak{b})(\alpha)) \to H^2(\mathfrak{b}(\alpha))\end{align*}$$

and the outer terms are both zero.

Now, from the exact sequence

$$\begin{align*}0 \to \mathfrak{b} \otimes \mathfrak{b}(\alpha) \to \mathfrak{b} \otimes \mathfrak{g}(\alpha)\to (\mathfrak{b} \otimes \mathfrak{g}/\mathfrak{b})(\alpha) \to 0\end{align*}$$

and the fact that $H^2(\mathfrak {b}(\alpha )) = 0$ , we obtain a surjection

$$\begin{align*}H^1((\mathfrak{b} \otimes \mathfrak{g}/\mathfrak{b})(\alpha)) \twoheadrightarrow H^2(\mathfrak{b} \otimes \mathfrak{b}(\alpha))\end{align*}$$

As $\operatorname {psupp}^2(\mathfrak {b}\otimes \mathfrak {b}(\alpha )) = \{0\}$ we have that $H^2(\mathfrak {b}\otimes \mathfrak {b}(\alpha )) \subset \mathbb {F}$ .

From the short exact sequence $0 \to (\mathfrak {b} \otimes \mathfrak {g}/\mathfrak {b})(\alpha ) \to (\mathfrak {g} \otimes \mathfrak {g}/\mathfrak {b})(\alpha )\to (\mathfrak {g}/\mathfrak {b} \otimes \mathfrak {g}/\mathfrak {b})(\alpha ) \to 0$ and the vanishing of $H^1((\mathfrak {g}/\mathfrak {b})(\alpha ))$ we get a surjection

$$\begin{align*}H^0((\mathfrak{g}/\mathfrak{b} \otimes \mathfrak{g}/\mathfrak{b})(\alpha)) \twoheadrightarrow H^1(\mathfrak{g}/\mathfrak{b} \otimes \mathfrak{b}(\alpha))\end{align*}$$

and hence a surjection

$$\begin{align*}H^0((\mathfrak{g}/\mathfrak{b} \otimes \mathfrak{g}/\mathfrak{b})(\alpha)) \twoheadrightarrow H^2(\mathfrak{b} \otimes \mathfrak{b}(\alpha)).\end{align*}$$

But $\operatorname {psupp}^0(\mathfrak {g}/\mathfrak {b} \otimes \mathfrak {g}/\mathfrak {b}(\alpha ))$ does not contain $\{0\}$ , so we must have

$$\begin{align*}H^2(\mathfrak{b} \otimes \mathfrak{b}(\alpha))= 0\end{align*}$$

from which we can obtain as in Calculation 4.8

$$\begin{align*}H^2(\Lambda^2[\mathfrak{b}\oplus \mathfrak{b}](\alpha))=0.\end{align*}$$

Finally we do the case $i = j = 3$ . Here $\operatorname {psupp}^3(\Lambda ^3\mathfrak {b}(\alpha )) = \emptyset $ so it is enough to show that

$$\begin{align*}H^3(\Lambda^2\mathfrak{b} \otimes \mathfrak{b}(\alpha)) = 0.\end{align*}$$

From the usual exact sequence and the vanishing of $H^3(\Lambda ^2\mathfrak {b}(\alpha ))$ we obtain a surjection

$$\begin{align*}H^2((\Lambda^2\mathfrak{b} \otimes \mathfrak{g}/\mathfrak{b})(\alpha)) \twoheadrightarrow H^3(\Lambda^2\mathfrak{b} \otimes \mathfrak{b}(\alpha)).\end{align*}$$

But $\operatorname {psupp}^2(\Lambda ^2\mathfrak {b} \otimes \mathfrak {g}/\mathfrak {b}(\alpha )) = \emptyset $ and so

$$\begin{align*}H^2((\Lambda^2\mathfrak{b} \otimes \mathfrak{g}/\mathfrak{b})(\alpha)) = H^3(\Lambda^2\mathfrak{b} \otimes \mathfrak{b}(\alpha)) = 0.\\[-34pt] \end{align*}$$

Proof. Because the map $\pi _Z:Z\rightarrow F$ is affine, from Proposition 3.2 we have the following isomorphisms of cohomology groups:

$$\begin{align*}H^i(Z,\mathcal{O}_Z(\lambda))=H^i(F,\pi_{Z,*}[\mathcal{O}_Z(\lambda)])=H^i(F, \operatorname{Sym}([\mathfrak{g}/\mathfrak{b}]^2)(\lambda))\end{align*}$$

So the question reduces to the calculation of the groups $H^i(F,\operatorname {Sym}^n[\mathfrak {g}/\mathfrak {b}^2])$ .

As in Section 4 of [Reference Vilonen and Xue23], one can consider the Koszul complex of $0\to \mathfrak {b}^2 \to \mathfrak {g}^2\to \mathfrak {g}/\mathfrak {b}^2\to 0$ and twist by $\lambda $ , giving us a resolution

$$\begin{align*}...\to \Lambda^i[\mathfrak{b}^2]\otimes \operatorname{Sym}^{n-i}[\mathfrak{g}^2](\lambda) \to ...\to\operatorname{Sym}^n[\mathfrak{g}^2](\lambda)\to \operatorname{Sym}^n[\mathfrak{g}/\mathfrak{b}^2](\lambda)\to 0\end{align*}$$

We therefore have a spectral sequence

$$\begin{align*}E_1^{-i,i+k} = H^{i+k}(\Lambda^i[\mathfrak{b}^2](\lambda))\otimes \operatorname{Sym}^{n-i}[\mathfrak{g}^2]\Longrightarrow H^k(\operatorname{Sym}^n[\mathfrak{g}/\mathfrak{b}^2](\lambda)).\end{align*}$$

By Calculation 4.8 in the case of $\lambda =0$ , and Calculations 4.10 and 4.11 in the other cases, we see that $H^{k+i}(\Lambda ^i[\mathfrak {b}^2](\lambda ))=0$ for any $k>0$ and any $i \ge 0$ , and therefore $H^k(\operatorname {Sym}[(\mathfrak {g}/\mathfrak {b})^2](\lambda ))=0$ for all $k>0$ .

Proof. From Proposition 3.2 that $\mathcal {I}_Y\cong \mathcal {O}_Z(\rho )$ and $\omega _Z\cong \mathcal {O}_Z(2\rho )$ . The lemma now follows from Lemma 4.12.

Proof. The first claim follows from Lemmas 4.12 and 4.14 and the long exact sequence in cohomology arising from

$$\begin{align*}0\to \mathcal{I}_Y(\lambda)\to\mathcal{O}_Z(\lambda)\to \mathcal{O}_Y(\lambda)\to 0.\end{align*}$$

The second claim follows from the isomorphism $\omega _Y \cong \mathcal {O}_Y(\rho )$ proved in Proposition 3.2.

Proof. The statement for $\mathcal {X}_{\operatorname {St}}$ follows from that for $\mathcal {X}$ and the discussion following Assumption 2.1.

To show that $\mathcal {X}$ is cut out by the given equations in $GL_2 \times GL_2$ , we need only check this after $\otimes _{\mathcal {O}} \mathbb {F}$ and $\otimes _{\mathcal {O}} E$ . After $\otimes \mathbb {F}$ , and writing $\Phi = I + M$ , $\Sigma = I + N$ , we obtain exactly the equations generating the ideal $\tilde {I}$ above; thus we are done by Theorem 5.3.

By Proposition 5.1, the first three equations alone already define $\mathcal {X}\otimes _{\mathcal {O}} E$ , and this is a smooth variety. Since every point of $\mathcal {X}(\overline {E})$ has $\Phi \sim \operatorname {diag}(q,1)$ , and the fourth equation is easily seen to hold at such points, we have that this equation is also contained in the defining ideal of $\mathcal {X} \otimes _{\mathcal {O}} E$ .

Proof. By Proposition 5.2, along with Calculation 4.8 and 4.5, we see that:

$$ \begin{align*} I/S^+I &\cong \bigoplus_i H^{i-1}(F,\Lambda^i[\mathfrak{b}\oplus \mathfrak{b}])[i] \\ & =\mathbb{F}^3[2]\oplus (\mathfrak{g}^4\oplus \mathbb{F}^{4})[3]. \end{align*} $$

Hence we see that,

$$\begin{align*}\dim(I/S_+I)_k = \begin{cases} 3 & \mathrm{if}\ k = 2 \\ 36 & \mathrm{if}\ k = 3 \\ 0 & \mathrm{otherwise}. \\ \end{cases}\end{align*}$$

Let $\tilde {I}$ be the ideal of S generated by the elements listed in the statement of the theorem. Since all the elements of $\tilde {I}$ vanish for $M, N \in \mathfrak {n}$ , $\tilde {I}\subset I$ . We show that they are equal.

As $\operatorname {tr}(M^2), \operatorname {tr}(MN)$ , and $\operatorname {tr}(N^2)$ are all linearly independent, we see that they span the degree 2 subspace of $\tilde {I}/S_+\tilde {I}$ . By a computer calculationFootnote ³ we see that the entries of $M^2N$ and $NM^2$ span a 17-dimensional space modulo the subspace spanned by entries of $\operatorname {tr}(M)^2N$ , $M\operatorname {tr}(MN)$ , and $N \operatorname {tr}(M^2)$ so long as $l> 2$ (this is just a matter of computing the invariant factors of an explicit finitely generated $\mathbb {Z}$ -module). This same calculation also shows that the entries of $MNM$ are in $\tilde {I}$ . Therefore, using that there is a natural bigrading on S with respect to which $\tilde {I}$ is bihomogeneous, we have

$$\begin{align*}\dim(\tilde{I}/S^+\tilde{I})_3 = 1 + 1 + 17 + 17 = 36 = \dim(I/S^+I)_3.\end{align*}$$

Thus $\dim (\tilde {I}/S_+\tilde {I})_k = \dim (I/S_+I)_k$ for all $k \ge 0$ , and so $I = \tilde {I}$ .

Proof. Let $i : Y \to Z$ be the inclusion. Then we have that $H^0(X, \mathcal {O}_X) = H^0(Z, i_{\ast }\mathcal {O}_Y)$ and a short exact sequence

$$\begin{align*}0 \to H^0(Z,\mathcal{I}_Y) \to H^0(Z, \mathcal{O}_Z) \to H^0(Z, i_{\ast}\mathcal{O}_Y) \to 0\end{align*}$$

since, by Lemmas 2.3 and 4.12, $H^1(Z, \mathcal {I}_Y)$ vanishes.

We claim that $I_X = H^0(Z, \mathcal {I}_Y)$ is generated, as a (homogeneous) ideal of $R = H^0(Z, \mathcal {O}_Z)$ , by the entries of $MN - NM$ . These entries give eight linearly independent elements of $I_X$ and it suffices to show that

$$\begin{align*}\dim(I_{X}\otimes_{S} \mathbb{F})_i = \begin{cases} 8 & \mathrm{if}\ i = 2 \\ 0 & \mathrm{otherwise.} \\ \end{cases} \end{align*}$$

By Lemma 2.3, $\mathcal {I}_Y \cong \mathcal {O}_Z(\rho )$ , locally (on F) generated in degree 2; hence $I_X \cong H^0(Z, \mathcal {O}_Z(\rho ))(-2)$ as a graded S-module, where $(-2)$ is the shift operator. By Proposition 5.2,

$$\begin{align*}(I_X \otimes_S \mathbb{F})_i = \dim H^{i}(F, \mathcal{O}(\rho) \otimes \Lambda^i[\mathfrak{b} \oplus \mathfrak{b}])[i+2]\end{align*}$$

for $i \ge 0$ . The claim then follows from Calculation 4.10.

As J is the preimage of $I_X$ under $S \to R$ , Theorem 5.7 then follows from Theorem 5.5.

Proof. The statement for $\mathcal {X}_{\operatorname {St}}$ follows from that for $\mathcal {X}$ and the discussion following Assumption 2.1.

To show that $\mathcal {X}$ is cut out by the given equations in $GL_3 \times GL_3$ , we need only check this after $\otimes _{\mathcal {O}} \mathbb {F}$ and $\otimes _{\mathcal {O}} E$ . After $\otimes \mathbb {F}$ , and writing $\Phi = I + M$ , $\Sigma = I + N$ , we obtain exactly the equations in Theorem 5.7, as required.

By Proposition 5.1, the first three equations alone already generate $\mathcal {X}\otimes _{\mathcal {O}} E$ , and this is a smooth variety. Since every point of $\mathcal {X}(\overline {E})$ has $\Phi \sim \operatorname {diag}(q^2,q,1)$ , and the fourth, fifth, and sixth equations are easily seen to hold at such points, we have that these equations are also contained in the defining ideal of $\mathcal {X} \otimes _{\mathcal {O}} E$ .

Proof. This is [Reference Fossum and Iversen8] Proposition 1.1, noting that a normal integral scheme of finite type over a field is Krull.

The maps $\mathrm {Cl}(F) \to \mathrm {Cl}(E)$ and $\mathrm {Cl}(E) \to \mathrm {Cl}(C)$ are the functorial maps (see [Reference Fossum and Iversen8]). The map $\delta $ in the exact sequence is constructed as follows. Let $u \in \Gamma (C, \mathcal {O}_C^\times )$ . Then u determines an element of the function field of E and hence a divisor $\operatorname {\mathrm {Div}}(u)$ on E, which one can show is the pullback of a divisor D on F. We define $\delta (u) = \mathcal {O}(D) \in \mathrm {Cl}(X)$ .

Proof. If $D\subset F$ is a closed integral subscheme of codimension 1 with generic point d, and if we choose a trivialization $\mathcal {L}_d \cong \operatorname {Spec} \mathcal {O}_{F,d}[s]$ such that $s = 0$ is the zero section then we have

$$\begin{align*}t = \kappa_D s\end{align*}$$

for some $\kappa _D \in K$ . We have that $v_d(\kappa _D) = 0$ for all but finitely many D and, by definition,

$$\begin{align*}\delta(t) = \mathcal{O}\left(\sum_{D}v_d(\kappa_D)D\right).\end{align*}$$

On the other hand, regard $\mathcal {L}$ as a sheaf on X that is locally free of rank 1, and let $t^\vee \in \Gamma (\operatorname {Spec}(K), \mathcal {L}_K)$ correspond to the section $t^\vee : \operatorname {Spec}(K) \to \operatorname {Spec} K[t]$ such that $(t^\vee )^{\ast }(t) = 1$ . Similarly, we have $s^\vee \in \Gamma (X_{d}, \mathcal {L}_d)$ , and the relation $t^\vee = \kappa _D^{-1} s^\vee $ for each D as above. Then we have

$$\begin{align*}\mathcal{L} = \mathcal{O}\left(\sum_D v_d(\kappa_D^{-1})D\right),\end{align*}$$

from which the result follows.

Proof. If $n = 2$ , then $C(\mathfrak {n}) = \mathfrak {n} \times \mathfrak {n} \cong \mathbb {A}^2$ and $C(\mathfrak {n})^{\mathrm {reg}} = \mathbb {A}^2 \setminus \{(0,0)\}$ . Therefore

$$\begin{align*}\mathrm{codim}(Y \setminus V \subset Y) = \mathrm{codim}(\{0,0\} \subset \mathbb{A}^2) = 2.\end{align*}$$

Now

$$\begin{align*}U = \pi(V) = \{(0,0)\} \in X \subset \mathfrak{g} \times \mathfrak{g}\end{align*}$$

and X has dimension $3$ , so $\mathrm {codim}(X\setminus U \subset X) = 3$ . This proves part (1).

Now suppose that $n = 3$ . We write

$$\begin{align*}C(\mathfrak{n})=\left\{\left(\begin{pmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & d & e \\ 0 & 0 & f \\ 0 & 0 & 0 \end{pmatrix} \right):af=cd\right\},\end{align*}$$

which has dimension 5, and see that

$$ \begin{align*}C(\mathfrak{n}) \setminus C(\mathfrak{n})^{\mathrm{reg}} &= \mathbb{V}(af-cd, ac,df) \\ &= \mathbb{V}(a,d)\cup \mathbb{V}(c,f).\end{align*} $$

Therefore

$$\begin{align*}\mathrm{codim}(Y \setminus V \subset Y) = \mathrm{codim}(C(\mathfrak{n})\setminus C(\mathfrak{n})^{\mathrm{reg}} \subset C(\mathfrak{n})) = 1.\end{align*}$$

We have an isomorphism

$$ \begin{align*} \mathbb{A}^2\setminus \{(0,0)\}\times \mathbb{G}_m\times \mathbb{A}^2&\rightarrow C(\mathfrak{n})^{\mathrm{reg}} \\ (a,d,\lambda,b, e)&\mapsto (a,b,c = \lambda a, d,e,f = \lambda d) \end{align*} $$

with inverse defined by $\frac {c}{a}\mapsto \lambda $ when $a\neq 0$ , and $\frac {f}{d}\mapsto \lambda $ when $d\neq 0$ .

Finally, we compute the dimension of $X \setminus U$ . Since the locus $M = N = 0$ has codimension 8 in X, it is enough to compute the dimension of $\{(M, N) \in X \setminus U : M \ne 0\}$ . The projection map from this subset to $\{M \in \mathfrak {g} : M^2 = 0, M \ne 0\}$ is a $GL_3$ -equivariant surjection. Its image is the orbit in $\mathfrak {g}$ of the matrix $M_0 = \begin {pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end {pmatrix}$ whose centralizer consists of all matrices of the form

$$\begin{align*}\begin{pmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & a \end{pmatrix}\end{align*}$$

and whose orbit in $\mathfrak {g}$ has dimension $4$ (by the orbit-stabilizer theorem). The fibre in $X\setminus U$ of $M_0$ is then

$$\begin{align*}\left\lbrace\begin{pmatrix} 0 & b & c \\ 0 & 0 & e \\ 0 & 0 & 0 \end{pmatrix}: be=0\right\rbrace\end{align*}$$

which has dimension $2$ , and so

$$\begin{align*}\dim(X \setminus U) = 4 + 2 = 6 = \dim(X) - 2\end{align*}$$

as required.

Proof. The case $n = 2$ follows quickly from Lemma 6.3 part (1): we have

$$\begin{align*}\mathrm{Cl}(X) = \mathrm{Cl}(U) = \mathrm{Cl}(V) = \mathrm{Cl}(Y) = \mathrm{Cl}(F)\end{align*}$$

since $X \setminus U$ has codimension 2 in X, $Y \setminus V$ has codimension 3 in Y, and Y is a vector bundle over F and so $\mathrm {Cl}(Y) \cong \mathrm {Cl}(F)$ (e.g., by Lemma 6.1). As is well-known,

$$\begin{align*}\mathrm{Cl}(F) = \operatorname{\mathrm{Pic}}(F) = X^{\ast}(T) \cong \mathbb{Z}.\end{align*}$$

Suppose now that $n = 3$ . By Lemma 6.3 part (2), as the codimension of $X\setminus U$ in X is 2, we have

$$\begin{align*}\mathrm{Cl}(X) = \mathrm{Cl}(U) = \mathrm{Cl}(V).\end{align*}$$

Let K be the function field of F. By Lemma 6.1 and Lemma 6.3 part (2), we have an exact sequence

$$\begin{align*}\Gamma(C(\mathfrak{n})^{\mathrm{reg}}_K, \mathcal{O}^\times_{C(\mathfrak{n})^{\mathrm{reg}}_K}) \xrightarrow{\delta} \mathrm{Cl}(F) \to \mathrm{Cl}(V) \to \mathrm{Cl}(C(\mathfrak{n})^{\mathrm{reg}}_K) \to 0.\end{align*}$$

Since $C(\mathfrak {n})^{\mathrm {reg}}_K$ is an open subscheme of $\mathbb {A}^5_K$ , $\mathrm {Cl}(C(\mathfrak {n})^{\mathrm {reg}}_K) = 0$ . We have

$$\begin{align*}\mathrm{Cl}(F) = \operatorname{\mathrm{Pic}}(F) = X^{\ast}(T) \cong \mathbb{Z}^2.\end{align*}$$

It remains to compute the image of $\delta $ .

By Lemma 6.3, we have a morphism $h : C(\mathfrak {n})_K \to \operatorname {Spec} K[\lambda , \lambda ^{-1}]$ that induces an isomorphism

$$\begin{align*}\Gamma(C(\mathfrak{n})^{\mathrm{reg}}_K, \mathcal{O}^\times_{C(\mathfrak{n})^{\mathrm{reg}}_K}) \xrightarrow{\sim} \{\lambda^n : n \in \mathbb{Z}\}.\end{align*}$$

This morphism is B-equivariant if $\lambda = \frac {c}{a} = \frac {f}{d}$ is given weight $L_1 - L_2 - (L_2 - L_3) = 3(L_1 + L_3)$ , and so we obtain a morphism

$$\begin{align*}V \to \mathcal{O}(-3(L_1 + L_3)) \setminus \{\text{zero section}\}\end{align*}$$

whose restriction to the generic fibres is h. We may therefore apply Lemma 6.2 and deduce that

$$\begin{align*}\delta(\lambda) = \mathcal{O}(3(L_1 + L_3)).\end{align*}$$

The result follows.

Proof. In either case, by Theorem 2.4, we have that $\omega _X=f_{\ast }\omega _Y$ .

If $n = 2$ then we have $\omega _Y = \pi ^{\ast }\mathcal {O}(2\rho )$ by Proposition 3.2. We therefore have

$$\begin{align*}\omega_X = f_{\ast}\pi^{\ast}\mathcal{O}(2\rho),\end{align*}$$

but also

$$\begin{align*}j^{\ast}\nu(2\rho) = f_{\ast}(\pi^{\ast}\mathcal{O}(2\rho)|_V) = j^{\ast}(f_{\ast}\pi^{\ast}\mathcal{O}(2\rho))\end{align*}$$

as $f : V \to U$ is an isomorphism. Thus $\omega _X= \nu (2\rho )$ .

If $n = 3$ we again apply Proposition 3.2 to obtain

$$\begin{align*}\omega_X = f_{\ast} \pi^{\ast}\mathcal{O}(\rho)\end{align*}$$

but also

$$\begin{align*}j^{\ast}\nu(\rho) = f_{\ast}(\pi^{\ast}\mathcal{O}(\rho)|_V) = j^{\ast}(f_{\ast}\pi^{\ast}\mathcal{O}(\rho))\end{align*}$$

as in the case $n = 2$ . Thus $\omega _X = \nu (\rho )$ .

Proof. The map $\nu $ is easily seen to be $\iota $ -equivariant. If $\mathcal {M} = \nu (aL_1 - bL_3)$ we have

$$\begin{align*}{\mathcal{H}\! \mathit{om}}_X(\mathcal{M}, \omega_X) = \nu(L_1 - L_3 - (aL_1 - bL_3))\end{align*}$$

and

$$\begin{align*}\iota^{\ast}\mathcal{M} = \nu(bL_1 - aL_3).\end{align*}$$

These are equal if and only if

$$\begin{align*}L_1 - L_3 = (a + b)(L_1 - L_3) + 3k(L_1 + L_3)\end{align*}$$

for some $k \in \mathbb {Z}$ . This holds if and only if $a + b = 1$ and $k = 0$ . The different possibilities for $b \bmod 3$ give the result.

Proof. That $f_{\ast }\mathcal {O}(w)$ is Cohen–Macaulay is [Reference Hacon and Witaszek10] Proposition 2.2. It is therefore reflexive. The final claim follows as, if $j : U \hookrightarrow X$ is the natural inclusion, then $X \setminus U$ has codimension 2 and $j^{\ast }f_{\ast }\mathcal {O}(w) = j^{\ast }\nu (w)$ by construction.

Proof. By Proposition 4.15, $H^i(Y, \mathcal {O}_Y(\lambda ))$ and $H^i(Y, \mathcal {O}_Y(\rho - \lambda ))$ vanish for each of these values of $\lambda $ and for all $i> 0$ . Since X is affine, we obtain exactly the vanishing of $R^if_{\ast }$ required to apply Lemma 6.7 to these $\lambda $ , and the Proposition follows.

Proof. Let $\lambda $ be one of the weights occurring in the proposition and let $M = H^0(Z, \mathcal {O}_Z(\lambda ))$ and $N = H^0(Y, \mathcal {O}_Y(\lambda ))$ . As in Section 5, by Lemma 4.12 we are in the situation of Proposition 5.2 with F the flag variety of $\mathrm {GL}_3$ , the short exact sequence

$$\begin{align*}0 \to \xi \to \epsilon \to \eta \to 0\end{align*}$$

being

$$\begin{align*}0\to\mathfrak{b}^2\to \mathfrak{g}^2\to (\mathfrak{g}/\mathfrak{b})^2\to 0,\end{align*}$$

and $\mathcal {L} = \mathcal {O}_F(\lambda )$ . We define $S=\operatorname {Sym}[(\mathfrak {g}^{*})^2]$ as in Proposition 5.2, and define

$$\begin{align*}R=H^0(Z,\mathcal{O}_Z)=H^0(F,\operatorname{Sym}[(\mathfrak{g}/\mathfrak{b})^2])\end{align*}$$

Thus (and recalling that, as in the proof of Theorem 5.7, $H^1(Z, \mathcal {I}_Y)$ vanishes) we have morphisms of graded algebras

$$\begin{align*}S\twoheadrightarrow R\twoheadrightarrow H^0(Y,\mathcal{O}_Y)\xrightarrow{\sim} H^0(X,\mathcal{O}_X)\end{align*}$$