Rational highest-weight modules from adjoint Plücker vectors

Identify $G$ with a closed subgroup scheme of $GL(V)$ by [F6]. Then $\mathfrak{g}\subseteq \mathfrak{g}\mathfrak{l}(V)=\mathrm{End}(V)$, and for $g\in G$ the conjugation $C_g$ is the restriction to $G$ of the ambient conjugation $c_g:GL(V)\to GL(V)$, $c_g(u)=gu{g}^{-1}$, which is given by polynomial formulas in the matrix entries of $g,g^{-1},u$. For $X\in \mathfrak{g}\mathfrak{l}(V)$ the matrix exponential satisfies $g\exp (tX)g^{-1}=\exp (tgX{g}^{-1})$, which is [F5] applied to the automorphism $c_g$ of $GL(V)$; differentiating at $t=0$ gives $d(c_g{)}_{I}X=gX{g}^{-1}$, so by [F4] ${\mathrm{Ad}}_g(X)=d(C_g{)}_{e}X=gX{g}^{-1}(X\in \mathfrak{g})\mathrm{.}$ Hence $(g,X)\mapsto {\mathrm{Ad}}_g(X)$ is the restriction of the morphism $GL(V)\times \mathfrak{g}\mathfrak{l}(V)\to \mathfrak{g}\mathfrak{l}(V)$ to the closed subvariety $G\times \mathfrak{g}$, so $\mathrm{Ad}:G\to GL(\mathfrak{g})$ is a morphism of varieties, and by [F7] so is $g\mapsto {\bigwedge }^k{\mathrm{Ad}}_g$ for every $k$, a homomorphism of abstract groups by [F4]; thus ${\bigwedge }^k\mathfrak{g}$ is a rational representation of $G$ in the sense of [F6].

For every $u\in U$ the operator ${\mathrm{Ad}}_u$ on $\mathfrak{g}$ is unipotent. Indeed, $z\mapsto u_{\beta }(z)={\exp }_G(z{e}_{\beta })$ has polynomial matrix entries in the faithful matrix realization of [F2], so ${\sum }_{j\ge 0}{z}^j{e}_{\beta }^j\mathrm{/}j!$ is a polynomial in $z$ and $e_{\beta }\in \mathfrak{g}\mathfrak{l}(V)$ is nilpotent; the operators $X\mapsto e_{\beta }X$ and $X\mapsto X{e}_{\beta }$ on $\mathfrak{g}\mathfrak{l}(V)$ commute and are nilpotent, so their difference induces the nilpotent operator ${\mathrm{ad}}_{e_{\beta }}$ on the invariant subspace $\mathfrak{g}$, and by [F4], ${\mathrm{Ad}}_{u_{\beta }(z)}=\exp (z{\mathrm{ad}}_{e_{\beta }})$ is unipotent. To justify the product, order the finite adjoint weights by a linear functional positive on every positive root. Each ${\mathrm{ad}}_{e_{\beta }}$, $\beta >0$, strictly raises this common weight filtration, so every ${\mathrm{Ad}}_{u_{\beta }(z)}$ is upper triangular with diagonal entries $1$ in one weight-compatible basis. Their product is upper triangular with the same diagonal and therefore unipotent. The same common filtration restricts to the invariant subspaces $\mathfrak{b}$ and ${\mathfrak{p}}_{\alpha }$.

Let $n=\dim \mathfrak{h}$ and $m=\mathrm{\mid }{\mathrm{\Phi }}^{+}\mathrm{\mid }$, so $\dim \mathfrak{b}=n+m$ by [F1] and $\dim {\mathfrak{p}}_{\alpha }=n+m+1$ by [F3]. By [F9] the adjoint action of $\mathfrak{h}$ has weights $0$ on $\mathfrak{h}$ and $\gamma$ on the one-dimensional space ${\mathfrak{g}}_{\gamma }$; hence the $n+m$ vectors $h_1,\dots ,h_n,e_{{\gamma }_1},\dots ,e_{{\gamma }_m}$ (the positive roots in any order) are a basis of $\mathfrak{b}$ of $\mathfrak{h}$-eigenvectors, and $\mathbb{C}{v}_B={\bigwedge }^{n+m}\mathfrak{b},H\cdot v_B=\left({\sum }_{\gamma \in {\mathrm{\Phi }}^{+}}\gamma \right)(H)v_B=2\rho (H)v_B(H\in \mathfrak{h})$ by [F8] and [F10]. Likewise ${\mathfrak{p}}_{\alpha }=\mathfrak{b}\oplus {\mathfrak{g}}_{-\alpha }$ is a direct sum of $\mathfrak{h}$-stable subspaces by [F3] and [F9], and with $0\mathrm{\ne }{f}_{\alpha }\in {\mathfrak{g}}_{-\alpha }$ one has ${\bigwedge }^{n+m+1}{\mathfrak{p}}_{\alpha }=\mathbb{C}(v_B\wedge f_{\alpha })$, so $H\cdot v_{\alpha }=(2\rho -\alpha )(H)v_{\alpha }$ for all $H\in \mathfrak{h}$.

The $2\rho$-weight space of ${\bigwedge }^{n+m}\mathfrak{g}$ is $\mathbb{C}{v}_B$. Extend $h_1,\dots ,h_n,e_{\gamma }$ $(\gamma \in {\mathrm{\Phi }}^{+})$ by $f_{{\gamma }^{\mathrm{\prime }}}=e_{-{\gamma }^{\mathrm{\prime }}}$ $({\gamma }^{\mathrm{\prime }}\in {\mathrm{\Phi }}^{+})$ to a basis of $\mathfrak{g}$ of $\mathfrak{h}$-eigenvectors of weights $0,\gamma ,-{\gamma }^{\mathrm{\prime }}$; by [F7] the wedges over $(n+m)$-element index sets $I$ form a basis of ${\bigwedge }^{n+m}\mathfrak{g}$ of $\mathfrak{h}$-eigenvectors of weight ${\lambda }_I={\sum }_{i\in I}{\mu }_i$. Let $P\subseteq {\mathrm{\Phi }}^{+}$ and $N\subseteq {\mathrm{\Phi }}^{+}$ be the sets of positive and negated negative roots selected by $I$. If ${\lambda }_I=2\rho ={\sum }_{\gamma \in {\mathrm{\Phi }}^{+}}\gamma$, then ${\sum }_{\gamma \in {\mathrm{\Phi }}^{+}\setminus P}\gamma +{\sum }_{\gamma \in N}\gamma =0.$ Both sums are sums of positive roots, hence nonnegative integral combinations of simple roots by [F9], so both are $0$; since a positive root has a nonzero coefficient at some simple root, this forces ${\mathrm{\Phi }}^{+}\setminus P=N=\mathrm{\varnothing }$. Thus $I$ is exactly the set of the $n$ Cartan indices and the $m$ positive roots, so $e_I=\pm v_B$ and the weight space is one-dimensional spanned by $v_B$.

Similarly the $(2\rho -\alpha )$-weight space of ${\bigwedge }^{n+m+1}\mathfrak{g}$ is $\mathbb{C}{v}_{\alpha }$. For an index set $I$ of size $n+m+1$ the weight condition reads ${\sum }_{\gamma \in {\mathrm{\Phi }}^{+}\setminus P}\gamma +{\sum }_{\gamma \in N}\gamma =\alpha \mathrm{.}$ By [F9] both sums are nonnegative integral combinations of simple roots whose total is the simple root $\alpha$, so they equal $c\alpha$ and $d\alpha$ with $c,d\in {\mathbb{Z}}_{\ge 0}$ and $c+d=1$. If $c=1$, then the first sum equals $\alpha$, which forces ${\mathrm{\Phi }}^{+}\setminus P=\{\alpha \}$ (a sum of distinct positive roots is a simple root only when it is that root), so $\mathrm{\mid }I\mathrm{\mid }=(m-1)+0+n<n+m+1$, a contradiction. Hence $c=0$ and $d=1$: $P={\mathrm{\Phi }}^{+}$, $N=\{\alpha \}$, and the cardinality of $I$ forces all $n$ Cartan indices to be selected. Thus $e_I=\pm v_{\alpha }$ and the weight space is one-dimensional spanned by $v_{\alpha }$.

The weights $2\rho$ and $2\rho -\alpha$ are dominant integral: $⟨2\rho ,{\beta }^{\vee }⟩=2⟨\rho ,{\beta }^{\vee }⟩=2\in {\mathbb{Z}}_{\ge 0}$ for every simple root $\beta$ by [F10]; and for $\beta \mathrm{\ne }\alpha$ simple one has $⟨\alpha ,{\beta }^{\vee }⟩\le 0$, since otherwise $s_{\beta }(\alpha )=\alpha -⟨\alpha ,{\beta }^{\vee }⟩\beta$ would be a root by [F9] whose $\alpha$-coordinate is $1>0$ and whose $\beta$-coordinate is negative, contradicting the one-sign property of [F9]. Hence $⟨2\rho -\alpha ,{\alpha }^{\vee }⟩=2-2=0$ and $⟨2\rho -\alpha ,{\beta }^{\vee }⟩=2-⟨\alpha ,{\beta }^{\vee }⟩\ge 2>0$ for $\beta \mathrm{\ne }\alpha$, so $2\rho -\alpha$ is dominant integral by [F10].

The differentiated action on ${\bigwedge }^k\mathfrak{g}$ is $X\mapsto {\bigwedge }^k{\mathrm{ad}}_X$: for $B\in \mathfrak{g}\mathfrak{l}(V)$ one has $(I+sB{)}^{\wedge k}=I+s{\bigwedge }^{k}B+O(s^2)$ in the matrix algebra, since on a decomposable wedge the coefficient of $s$ is ${\sum }_i{x}_1\wedge \cdots \wedge B{x}_i\wedge \cdots \wedge x_k=({\bigwedge }^{k}B)(x_1\wedge \cdots \wedge x_k)$, so the chain rule with $B={\mathrm{ad}}_X$ gives $\frac{d}{dt}{\mid }_{t=0}{\bigwedge }^k{\mathrm{Ad}}_{{\exp }_G(tX)}=\frac{d}{dt}{\mid }_{t=0}{\bigwedge }^k\text{\hspace{0.17em}⁣}\left(e^{t{\mathrm{ad}}_X}\right)={\bigwedge }^k{\mathrm{ad}}_X,$ using ${\mathrm{Ad}}_{{\exp }_G(tX)}=e^{t{\mathrm{ad}}_X}$ from [F4]. This is exactly the diagonal $\mathfrak{g}$-module structure of [F7], so the differentiated module of the rational representation of step 1.1 is ${\bigwedge }^k\mathfrak{g}$ with $X\cdot w={\bigwedge }^k({\mathrm{ad}}_X)w$.

The lines $\mathbb{C}{v}_B$ and $\mathbb{C}{v}_{\alpha }$ are $B$-stable. For $u\in U$ one has $C_u(B)=B$, because $B$ is a subgroup containing $u$, so ${\mathrm{Ad}}_u\mathfrak{b}=\mathrm{Lie}{C}_u(B)=\mathfrak{b}$ by [F4]; hence $u$ preserves ${\bigwedge }^{\dim \mathfrak{b}}\mathfrak{b}=\mathbb{C}{v}_B$ and acts there by the top exterior power of the unipotent operator ${\mathrm{Ad}}_u{\mathrm{\mid }}_{\mathfrak{b}}$ of step 1.2, which is unipotent and therefore the identity on a one-dimensional space. So $U$ fixes $v_B$, and $\mathbb{C}{v}_B$ is $B$-stable with $T$-weight $2\rho$ by step 1.3. Replacing $\mathfrak{b}$ by ${\mathfrak{p}}_{\alpha }$ and $B$ by $P_{\alpha }$, which contains $U$ by [F3], the same computation gives $U{v}_{\alpha }=v_{\alpha }$, so $\mathbb{C}{v}_{\alpha }$ is $B$-stable with $T$-weight $2\rho -\alpha$.

The analogous statement for the minimal parabolic holds by the same computation with ${\mathfrak{p}}_{\alpha }$, $v_{\alpha }$ and $2\rho -\alpha$ in place of $\mathfrak{b}$, $v_B$ and $2\rho$: by step 2.2 the group $U$ fixes $v_{\alpha }$, so ${\mathfrak{n}}^{+}\cdot v_{\alpha }=0$ and $v_{\alpha }$ is a highest weight vector of weight $2\rho -\alpha$ in $M_{\alpha }=U(\mathfrak{g})v_{\alpha }$ by [F8] and steps 1.3 and 1.5; in any decomposition $M_{\alpha }=N_1\oplus \cdots \oplus N_k$ into irreducible submodules given by [F11] the components of $v_{\alpha }$ are again annihilated by ${\mathfrak{n}}^{+}$ and have $\mathfrak{h}$-weight $2\rho -\alpha$, and they all lie in the one-dimensional space $(M_{\alpha }{)}_{2\rho -\alpha }=\mathbb{C}{v}_{\alpha }$ by step 1.5, so at most one of them is nonzero and $M_{\alpha }=N_j$ for that $j$; thus $M_{\alpha }$ is an irreducible highest weight module of highest weight $2\rho -\alpha$, isomorphic to $L(2\rho -\alpha )$ by [F11] and step 1.6.

Let $M:=U(\mathfrak{g})v_B\subseteq {\bigwedge }^{n+m}\mathfrak{g}$, the smallest $\mathfrak{g}$-submodule containing $v_B$, and let $H=\{g\in G:gM=M\}$ be its stabilizer in $G$, a subgroup of $G$. For $X\in \mathfrak{g}$ one has $X\cdot M\subseteq M$, so every power of the endomorphism ${\bigwedge }^{n+m}({\mathrm{ad}}_X)$ preserves $M$ and hence so does its exponential; by steps 1.1 and 2.1, ${\exp }_G(tX)\cdot w=e^{t{\bigwedge }^{n+m}({\mathrm{ad}}_X)}w$ lies in $M$ for every $w\in M$ and every $t$, so ${\exp }_G(tX)\in H$. Thus ${\exp }_G(\mathfrak{g})\subseteq H$; by [F12] the image of a suitable open neighborhood of $0$ is an open neighborhood of $e$ in $G$, so $H$ contains an open neighborhood of $e$ and, being a subgroup, is open in $G$; an open subgroup of the connected group $G$ is all of $G$, so $H=G$ and $M$ is $G$-stable. Hence $M$ is the smallest $G$-stable subspace containing $v_B$, and with the restricted action of the rational representation of step 1.1 it is a finite-dimensional rational subrepresentation of ${\bigwedge }^{n+m}\mathfrak{g}$ whose differentiated module is $M$. The same argument with ${\mathfrak{p}}_{\alpha }$ and $v_{\alpha }$ in place of $\mathfrak{b}$ and $v_B$ shows that $M_{\alpha }:=U(\mathfrak{g})v_{\alpha }$ is $G$-stable and is the smallest $G$-stable subspace containing $v_{\alpha }$.

$M$ is irreducible with highest weight $2\rho$. By step 2.2 the Lie algebra ${\mathfrak{n}}^{+}=\mathrm{Lie}U$ of [F1] annihilates $v_B$: differentiating the trivial action of $U$ on the line $\mathbb{C}{v}_B$ at the identity gives $X\cdot v_B=0$ for $X\in {\mathfrak{n}}^{+}$. So $v_B$ is a highest weight vector of the $\mathfrak{g}$-module $M$ of weight $2\rho$ by [F8], and $M=U(\mathfrak{g})v_B$ is generated by it. By [F11] $M=N_1\oplus \cdots \oplus N_k$ is a direct sum of irreducible submodules; the components $v_j\in N_j$ of $v_B={\sum }_j{v}_j$ are again annihilated by ${\mathfrak{n}}^{+}$ (the projections commute with $\mathfrak{g}$) and have $\mathfrak{h}$-weight $2\rho$, so each nonzero $v_j$ is a highest weight vector of weight $2\rho$ in $N_j$. Since $M\subseteq {\bigwedge }^{n+m}\mathfrak{g}$, step 1.4 gives $M_{2\rho }=\mathbb{C}{v}_B$, so all components $v_j$ are multiples of $v_B$; if $v_j\mathrm{\ne }0$ then $N_j=U(\mathfrak{g})v_j=U(\mathfrak{g})v_B=M$. Hence at most one component is nonzero and $M=N_j$ for that $j$: $M$ is irreducible, and it is a finite-dimensional simple highest weight module of highest weight $2\rho$. By [F11] and step 1.6 it is isomorphic to $L(2\rho )$.

The only $B$-stable line in $M$ is $\mathbb{C}{v}_B$. Let $\mathbb{C}w\subseteq M$ be a $B$-stable line. The torus $T$ acts on it by a character, whose differential is a functional $\mu \in {\mathfrak{h}}^{\ast }$ with $H\cdot w=\mu (H)w$ for $H\in \mathfrak{h}$; the unipotent group $U$ acts on the one-dimensional space $\mathbb{C}w$ by a character, whose image is a unipotent subgroup of the torus ${\mathbb{C}}^{\times }=G{L}_1(\mathbb{C})$ and hence trivial, so $U$ acts trivially; differentiating at the identity gives ${\mathfrak{n}}^{+}\cdot w=0$, so $w$ is a highest weight vector of weight $\mu$ in $M$. By [F11] applied to the highest weight module $M=U(\mathfrak{g})v_B$ of weight $2\rho$ (step 3.3), every weight of $M$ is $\le 2\rho$ and $M_{2\rho }=\mathbb{C}{v}_B$; since $\mu$ is a weight of $M$, $\mu \le 2\rho$. Conversely, $w$ generates $M$ (it is a highest weight vector of the irreducible module $M$, [F11]), so every weight of $M$ is $\le \mu$, in particular $2\rho \le \mu$. Antisymmetry of the root order [F11] gives $\mu =2\rho$, and then $M_{2\rho }=\mathbb{C}{v}_B$ is one-dimensional with $w\in M_{2\rho }$ nonzero, so $\mathbb{C}w=\mathbb{C}{v}_B$.

The same argument as step 4.1 with $M_{\alpha }$, $v_{\alpha }$ and $2\rho -\alpha$ in place of $M$, $v_B$ and $2\rho$, using step 3.1 for the irreducibility of $M_{\alpha }$, shows that the only $B$-stable line in $M_{\alpha }$ is $\mathbb{C}{v}_{\alpha }$.

Collecting steps 1.3, 1.4, 1.5, 1.6, 2.2, 3.1, 3.2, 3.3, 4.1 and 5.1 proves (i) and (ii): $v_B$ and $v_{\alpha }$ span the respective weight spaces, the lines they span are $B$-stable of $T$-weights $2\rho$ and $2\rho -\alpha$, the $G$-spanning modules $W_B=M$ and $W_{\alpha }=M_{\alpha }$ are finite-dimensional rational subrepresentations with differentiated modules $L(2\rho )$ and $L(2\rho -\alpha )$, and the $B$-stable lines are unique. The Axiom of Choice is used exactly through [F13] and the suppliers [F11] of the highest weight classification, the countable-choice interfaces of [F4] and [F12], and the finite-dimensional linear algebra of [F7]; the argument itself chooses only the fixed simple root $\alpha$, the finitely many basis vectors $h_1,\dots ,h_n,e_{\gamma },f_{\alpha }$ and the faithful representation of [F6].