Proof. Under the identification of $\mathfrak{gl}(N,\mathbb{C})$ with the space of vector fields spanned by $\{U_{i}\unicode[STIX]{x2202}_{U_{j}}\}$ , assertion (1) follows from [Reference Goodman and WallachGW98, §2.3.1, p. 70], while (2) is in [Reference HoweHow85, §6].◻

Notation 3.9. For the rest of the section assume (for simplicity) that $\mathbb{K}=\mathbb{C}$ and fix $r\in \mathbb{N}^{\ast }=\mathbb{N}\smallsetminus \{0\}$ . Recall from §2 that the (abstract) Lie algebra $\mathfrak{so}(n+2,\mathbb{C})$ is denoted by $\mathfrak{g}$ . Let $\widetilde{\unicode[STIX]{x1D713}}=\widetilde{\unicode[STIX]{x1D713}}_{d}:\mathfrak{g}\rightarrow \widetilde{\mathfrak{g}}_{d}\subset {\mathcal{D}}(A)$ be an isomorphism defined by Proposition 3.8(2); an explicit map can be found in Corollary 4.4.

Proof. We first show that for $k\geqslant 0$ ,

(3.13) $$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{1}(u\mathtt{F}^{k})=\unicode[STIX]{x1D6E5}_{1}(u)\mathtt{F}^{k}+2k\mathtt{E}_{1}(u)\mathtt{F}^{k-1}+k(n+2(k-1))u\mathtt{F}^{k-1}.\end{eqnarray}$$

To see this, note that $\unicode[STIX]{x2202}_{X_{j}}(u\mathtt{F}^{k})=\unicode[STIX]{x2202}_{X_{j}}(u)\mathtt{F}^{k}+2kX_{j}u\mathtt{F}^{k-1}$ and hence

$$\begin{eqnarray}\unicode[STIX]{x2202}_{X_{j}}^{2}(u\mathtt{F}^{k})=\unicode[STIX]{x2202}_{X_{j}}^{2}(u)\mathtt{F}^{k}+2kX_{j}\unicode[STIX]{x2202}_{X_{j}}(u)\mathtt{F}^{k-1}+2k(u\mathtt{F}^{k-1}+X_{j}\unicode[STIX]{x2202}_{X_{j}}(u)\mathtt{F}^{k-1}+2(k-1)X_{j}^{2}u\mathtt{F}^{k-2}).\end{eqnarray}$$

Equation (3.13) follows by summing over $j$ , and (3.11) then follows by an elementary computation. Taking $d=n/2-r$ and $k=r$ in (3.11) immediately gives (3.12).◻

Proof. Set ${\mathcal{B}}=\mathbb{I}_{{\mathcal{D}}(A)}(\mathtt{F}^{r}{\mathcal{D}}(A))$ in the notation of (3.2). Combining Lemmas 3.6 and 3.10 shows that $\widetilde{\mathfrak{g}}_{d}\subset {\mathcal{B}}$ ; equivalently, by (3.3), the elements of $\widetilde{\mathfrak{g}}_{d}$ induce differential operators on $R_{r}$ . Now apply Proposition 3.8(2).◻

Proof. Since $R\subset \operatorname{Im}(\unicode[STIX]{x1D713}_{r})\subset {\mathcal{D}}(R)$ , one has $R=U(\mathfrak{g}).1$ . By construction, $D_{s,t}.1=P_{j}.1=0$ and so the choice of Chevalley system in Lemma 4.3 implies that $\mathfrak{n}^{+}.1=0$ . Moreover, $h_{j}.1=0$ for $2\leqslant j\leqslant \ell$ whereas $h_{1}.1=-I(d).1=-d.1=-d\unicode[STIX]{x1D700}_{1}(h_{1}).1$ . The choice of $\mathsf{B}\subset \unicode[STIX]{x1D6F7}^{+}$ yields $\unicode[STIX]{x1D71B}_{1}=\unicode[STIX]{x1D700}_{1}$ ; see [Reference BourbakiBou81, Planche II, pp. 252–253, and Planche IV, pp. 256–257]. Hence $R$ is a highest-weight module with highest weight $\unicode[STIX]{x1D706}=-d\unicode[STIX]{x1D71B}_{1}$ .◻

Proof. Since $\bigoplus _{j=v}^{t}H(m-2j)\mathtt{F}^{j}$ is a multiplicity-free, semi-simple $U(\mathfrak{k})$ -module, the claim follows from Jacobson’s density theorem; see [Reference BourbakiBou73, §4, no. 2, Corollaire 2].◻

Proof. Let $0\neq \overline{M}\subseteq N(\unicode[STIX]{x1D706})$ be a $\mathfrak{g}$ -submodule. Since $R_{r}\subset \operatorname{Im}(\unicode[STIX]{x1D713}_{r})$ , the module $\overline{M}$ is also an ideal of $R_{r}$ . Therefore $\overline{M}=M/(\mathtt{F}^{r})$ for some $\widetilde{\mathfrak{g}}_{d}$ -stable ideal $M$ of $A$ . In particular, by Proposition 3.8, $(\mathtt{E}_{1}+d)(M)\subseteq M$ , whence $\mathtt{E}_{1}(M)\subseteq M$ , and so $M=\bigoplus _{m}M\cap A(m)$ is homogeneous.

Let $p\in M\cap A(m)$ with $p\notin (\mathtt{F}^{r})$ , and write $p=\sum _{k=v}^{t}p_{k}\mathtt{F}^{k}$ for $p_{k}\in H(m-2k)$ with $p_{v}\neq 0\neq p_{t}$ . Note that $t<r$ . We want to simplify our choice of $p$ . Recall that $\mathfrak{g}\supset \mathfrak{k}\cong \mathfrak{so}(n,\mathbb{C})$ . Thus, by Corollary 4.11, $p_{k}\mathtt{F}^{k}\in M\cap H(m-2k)\mathtt{F}^{k}$ for all $k$ and so, as $H(m-2k)\mathtt{F}^{k}$ is an irreducible $\mathfrak{k}$ -module, $H(m-2k)\mathtt{F}^{k}\subset M$ for all $k$ such that $p_{k}\neq 0$ . In particular, $H(m-2v)\mathtt{F}^{v}\subset M$ and we can replace $p$ by $p_{v}\mathtt{F}^{v}$ . There exists $u\in U(\mathfrak{k})$ such that $u.p_{v}=\unicode[STIX]{x1D709}^{m-2v}=(X_{1}-iX_{\ell })^{m-2v}$ . From $D_{s,t}.\mathtt{F}^{k}=0$ we then deduce that $u.p=(u.p_{v})\mathtt{F}^{v}\in M$ and, upon replacing $p$ by $u.p$ , we may therefore assume that $p_{v}=\unicode[STIX]{x1D709}^{m-2v}$ . Finally, we assume that the harmonic valuation $v=v(p)$ is as small as possible among such choices of $p$ .

Now apply the operator $P_{\ell +1}=P_{\ell +1}(d)$ to $p$ ; from Lemma 4.12 and $\unicode[STIX]{x2202}_{X_{\ell +1}}(p_{v})=0$ we obtain

(4.15) $$\begin{eqnarray}P_{\ell +1}.p=(P_{\ell +1}(2v+d).p_{v})\mathtt{F}^{v}+v(n-2v-2d)X_{\ell +1}p_{v}\mathtt{F}^{v-1}\end{eqnarray}$$

where $P_{\ell +1}(2v+d).p_{v}\in H(m-2v-1)$ and $X_{\ell +1}p_{v}\in H(m-2v+1)$ . Since $P_{\ell +1}.p\in M\smallsetminus (\mathtt{F}^{r})$ and $X_{\ell +1}p_{v}\neq 0$ , the minimality of $v$ , combined with (4.15), forces $v(n-2v-2d)=0$ . In other words, either $v=0$ or $n-2v-2d=0$ . The latter is equivalent to $v=r$ , which is excluded. Thus we have proved that $M$ contains a nonzero element $p=p_{0}\in H(m)$ . In particular, $H(m)A\subseteq M$ .

Now pick $0\neq p=\unicode[STIX]{x1D709}^{m}\in M\cap H(m)$ with $m$ minimal. As $p$ is homogeneous, if $m=0$ then $1\in M$ and $\overline{M}=N(\unicode[STIX]{x1D706})$ . In the $m\geqslant 1$ case, applying $P_{1}=X_{1}\unicode[STIX]{x1D6E5}_{1}-(\mathtt{E}_{1}+d)\unicode[STIX]{x2202}_{X_{1}}$ gives

$$\begin{eqnarray}P_{1}.\unicode[STIX]{x1D709}^{m}=-m(\mathtt{E}_{1}+d)(\unicode[STIX]{x1D709}^{m-1})=-m(m-1+d)\unicode[STIX]{x1D709}^{m-1}\in ~M\cap H(m-1).\end{eqnarray}$$

The minimality of $m$ then implies that $m(m-1+d)=0$ , whence $m-1+d=0$ . Thus, we have shown that $\overline{M}=N(\unicode[STIX]{x1D706})$ unless $d$ satisfies $d=-m+1$ for some $m\geqslant 1$ . In the latter case $0\geqslant d=n/2-r$ . Since $r,m\in \mathbb{N}$ , this is equivalent to $r$ being even with $r\geqslant n/2$ . This proves (1).

It remains to prove (2), where $r=n/2+m-1$ for $m=1-d\geqslant 1$ . Note that, in this case, we have shown that every nonzero $\mathfrak{g}$ -submodule $\overline{M}$ of $R_{r}=N(\unicode[STIX]{x1D706})$ contains the image $S$ of the ideal $\mathfrak{J}=H(m)A$ .

Since $\unicode[STIX]{x1D706}=-d\unicode[STIX]{x1D71B}_{1}=(m-1)\unicode[STIX]{x1D71B}_{1}\in \mathsf{P}^{++}$ is a dominant integral weight, $L(\unicode[STIX]{x1D706})$ is finite-dimensional. Moreover, it is isomorphic to the $\mathfrak{g}$ -module of harmonic polynomials of degree $m-1$ in $n+2$ variables (apply Theorem 4.10 to $\mathfrak{so}(n+2,\mathbb{C})$ ).

Since $D_{s,t}(H(m))\subseteq H(m)$ and $\mathfrak{J}$ is homogeneous, certainly $\unicode[STIX]{x1D6E9}.\,\mathfrak{J}\subseteq \mathfrak{J}$ for $\unicode[STIX]{x1D6E9}=D_{s,t},X_{j}$ and $(\mathtt{E}_{1}+d)$ . Therefore, in order to show that $\mathfrak{J}$ is a $\mathfrak{g}$ -module, it remains to prove that $P_{j}.\,\mathfrak{J}\subseteq \mathfrak{J}$ or, equivalently, $P_{j}.X^{\unicode[STIX]{x1D6FC}}f\in \mathfrak{J}$ for any $f\in H(m)$ and monomial $X^{\unicode[STIX]{x1D6FC}}=X_{1}^{\unicode[STIX]{x1D6FC}_{1}}\cdots X_{n}^{\unicode[STIX]{x1D6FC}_{n}}$ . We argue by induction on $|\unicode[STIX]{x1D6FC}|=\sum _{i}\unicode[STIX]{x1D6FC}_{i}$ . If $|\unicode[STIX]{x1D6FC}|=0$ , we have

(4.16) $$\begin{eqnarray}P_{j}.f=-(m-1+d)\unicode[STIX]{x2202}_{X_{j}}(f)=0\end{eqnarray}$$

by the choice of $m$ . If $|\unicode[STIX]{x1D6FC}|\geqslant 1$ , write $X^{\unicode[STIX]{x1D6FC}}=X_{k}X^{\unicode[STIX]{x1D6FD}}$ for $|\unicode[STIX]{x1D6FD}|=|\unicode[STIX]{x1D6FC}|-1$ . Then

$$\begin{eqnarray}P_{j}.X^{\unicode[STIX]{x1D6FC}}f=[P_{j},X_{k}].X^{\unicode[STIX]{x1D6FD}}f+X_{k}P_{j}.X^{\unicode[STIX]{x1D6FD}}f.\end{eqnarray}$$

By induction, $P_{j}.X^{\unicode[STIX]{x1D6FD}}f\in \mathfrak{J}$ , while $[P_{j},X_{k}]=D_{j,k}-\unicode[STIX]{x1D6FF}_{j,k}(\mathtt{E}_{1}+d)$ from Proposition 3.8(b), which implies that $[P_{j},X_{k}].X^{\unicode[STIX]{x1D6FD}}f\in \mathfrak{J}$ . Hence $P_{j}.X^{\unicode[STIX]{x1D6FC}}f\in \mathfrak{J}$ and $\mathfrak{J}$ is indeed a $\mathfrak{g}$ -module. Since $H(m)\not =A$ and $H(m)\not \subseteq (\mathtt{F}^{r})$ , the image $S$ of $\mathfrak{J}$ in $R_{r}$ is a nontrivial $\mathfrak{g}$ -submodule of $R_{r}=N(\unicode[STIX]{x1D706})$ . However, we have already noted that $S$ is contained in every nonzero $\mathfrak{g}$ -submodule $\overline{M}\subseteq N(\unicode[STIX]{x1D706})$ ; in other words, $S=\operatorname{Soc}(N(\unicode[STIX]{x1D706}))=Z_{r}$ .

Recall that $H(m)$ contains all the functions $\unicode[STIX]{x1D701}^{m}$ associated to the isotropic vectors $\unicode[STIX]{x1D701}\in \mathbb{C}^{n}$ . Since $\mathbb{C}^{n}$ has a basis $(\unicode[STIX]{x1D701}_{j})_{j}$ of such vectors, the ideal $\sum _{j=1}^{n}\unicode[STIX]{x1D701}^{m}A$ has finite codimension in  $A$ . Therefore, $\dim R_{r}/Z_{r}\leqslant \dim A/H(m)A<\infty$ . Thus, on the one hand, $N(\unicode[STIX]{x1D706})/Z_{r}$ is finite-dimensional and hence completely reducible; but on the other hand, as it is a factor of the Verma module $M(\unicode[STIX]{x1D706})$ , it has a unique simple quotient. This forces $N(\unicode[STIX]{x1D706})/Z_{r}\cong L(\unicode[STIX]{x1D706})$ .

Finally, we need to compute the highest weight of $Z_{r}$ . Recall that the $\mathfrak{k}$ -module $H(m)$ is generated by the function $\unicode[STIX]{x1D709}^{m}$ associated to the isotropic vector $\unicode[STIX]{x1D709}=v_{1}-iv_{\ell }\in \mathbb{C}^{n}$ . We have shown in (4.16) that $P_{j}.\unicode[STIX]{x1D709}^{m}=0$ for all $j$ and hence $\mathfrak{r}^{+}.\unicode[STIX]{x1D709}^{m}=0$ . Since $\unicode[STIX]{x1D709}^{m}=U_{2}^{m}$ , Lemma 4.3 shows that $x_{\unicode[STIX]{x1D700}_{a+1}\pm \unicode[STIX]{x1D700}_{b+1}}.\unicode[STIX]{x1D709}^{m}=0$ for $1\leqslant a<b\leqslant \ell ^{\prime }$ . Hence $\mathfrak{n}^{\prime }.\unicode[STIX]{x1D709}^{m}=0$ (recall that $n$ is even) and so $\mathfrak{n}^{+}.\unicode[STIX]{x1D709}^{m}=0$ . Moreover, from our choice of $h_{j}=iD_{j,j+\ell ^{\prime }}$ one obtains

$$\begin{eqnarray}h_{1}.\unicode[STIX]{x1D709}^{m}=-\unicode[STIX]{x1D709}^{m},\quad h_{2}.\unicode[STIX]{x1D709}^{m}=m\unicode[STIX]{x1D709}^{m}\quad \text{and}\quad h_{j}.\unicode[STIX]{x1D709}^{m}=0\text{ for }j\geqslant 3.\end{eqnarray}$$

Thus $\unicode[STIX]{x1D709}^{m}\in Z_{r}$ is a highest-weight vector, with weight $\unicode[STIX]{x1D707}=-\unicode[STIX]{x1D700}_{1}+m\unicode[STIX]{x1D700}_{2}$ . Hence $Z_{r}\cong L(\unicode[STIX]{x1D707})$ . That $\unicode[STIX]{x1D707}$ equals $\unicode[STIX]{x1D706}-m\unicode[STIX]{x1D6FC}_{1}$ and satisfies (4.14) are easy exercises using [Reference BourbakiBou81, Planche IV].◻

Proof. (1) By Theorem 4.13, $Z_{r}=\operatorname{Soc}N(\unicode[STIX]{x1D706})\cong L(\unicode[STIX]{x1D714})$ while $N(\unicode[STIX]{x1D706})/Z_{r}$ is finite-dimensional (possibly zero). Hence, by [Reference JantzenJan83, Lemma 8.14], $\operatorname{ann}_{U(\mathfrak{g})}Z_{r}=\operatorname{ann}_{U(\mathfrak{g})}N(\unicode[STIX]{x1D706})$ ; equivalently, $J_{r}=I(\unicode[STIX]{x1D706})=I(\unicode[STIX]{x1D714})$ .

(2) Since $\operatorname{GKdim}L(\unicode[STIX]{x1D714})=\operatorname{GKdim}R_{r}=n-1$ , [Reference JantzenJan83, Satz 10.9] implies that $\operatorname{GKdim}U(\mathfrak{g})/J(\unicode[STIX]{x1D714})=2(n-1)$ . Thus ${\mathcal{V}}(J_{r})=\overline{\mathbf{O}}$ for a nilpotent orbit $\mathbf{O}$ with $\dim \mathbf{O}=2(n-1)$ . This forces $\mathbf{O}=\mathbf{O}_{\text{min}}$ .◻

Proof. (1) This is obvious.

(2) Observe that $1-d<0\;\Longleftrightarrow \;r<n/2-1\;\Longleftrightarrow \;r\leqslant \ell -2=(n-3)/2$ . Let $w^{-1}$ be the cycle $(\ell -r,\ell -r-1,\ldots ,2,1)\in W$ . Then $\unicode[STIX]{x1D708}=w^{-1}(\unicode[STIX]{x1D706}+\unicode[STIX]{x1D70C})$ has the desired form.

(3) Here, $1-d<0\;\Longleftrightarrow \;r<\ell -2$ . If $w^{-1}$ is the cycle $(\ell -r-1,\ell -r-2,\ldots ,2,1)\in W$ , the weight $\unicode[STIX]{x1D708}=w^{-1}(\unicode[STIX]{x1D706}+\unicode[STIX]{x1D70C})$ has the desired form.◻

Proof. Since ${\mathcal{V}}(J_{r})=\overline{\mathbf{O}}_{\text{min}}$ , any proper factor of $U(\mathfrak{g})/J_{r}$ must be finite-dimensional, and that factor then has to be unique. On the other hand, if $U(\mathfrak{g})/J_{r}$ is a simple ring, then it cannot have a nonzero finite-dimensional module. Thus part (i) already follows from Theorem 4.13.

Conversely, suppose that $J_{r}=I(\unicode[STIX]{x1D706})$ is not maximal; then it has a finite-dimensional quotient. This implies that there exists $\unicode[STIX]{x1D708}\in \mathsf{P}_{++}\subset \overline{\mathsf{C}}$ such that $\unicode[STIX]{x1D708}\in W(\unicode[STIX]{x1D706}+\unicode[STIX]{x1D70C})$ (see [Reference BourbakiBou90, ch. VIII.7.2, Corollaire 1, and VIII.8.5, Corollaire 1]). Proposition 4.18 gives the unique weight $\unicode[STIX]{x1D708}$ in $W(\unicode[STIX]{x1D706}+\unicode[STIX]{x1D70C})\cap \overline{\mathsf{C}}$ . By inspection, $\unicode[STIX]{x1D708}\in \mathsf{P}_{++}$ if and only if $1-d\in \mathbb{N}^{\ast }$ . As usual, this forces $n$ to be even and $r-n/2\geqslant 0$ .◻

Hypotheses 5.1. Let $\mathfrak{g}$ be a finite-dimensional complex reductive Lie algebra with triangular decomposition $\mathfrak{g}=\mathfrak{n}^{-}\boldsymbol{\oplus }\mathfrak{h}\boldsymbol{\oplus }\mathfrak{n}^{+}$ , and assume that $R$ is a commutative, finitely generated $\mathbb{C}$ -algebra that is a $\mathfrak{g}$ -module via a morphism $\unicode[STIX]{x1D712}:U(\mathfrak{g})\rightarrow {\mathcal{D}}(R)$ . Set $I=\operatorname{Ker}(\unicode[STIX]{x1D712})$ and identify $\overline{U}=U(\mathfrak{g})/I$ with its image $\unicode[STIX]{x1D712}(U(\mathfrak{g}))$ in ${\mathcal{D}}(R)$ . Assume moreover that:

1. (1) under the above action, $R$ is a quasi-simple highest-weight $\mathfrak{g}$ -module;

2. (2) $R$ is generated by $\unicode[STIX]{x1D712}(\mathfrak{r})$ for some Lie subalgebra $\mathfrak{r}\subseteq \mathfrak{n}^{-}$ ;

3. (3) if $N=\operatorname{Soc}_{U(\mathfrak{g})}(R)$ , then $\operatorname{End}_{R}(N)=R$ .

Proof. The hypothesis on $C$ implies that $R$ is Cohen–Macaulay and that the grade of the $R$ -module $R/M$ is at least $2$ ; see [Reference MatsumuraMat80, (16.A) and (16.B)]. In other words, $\operatorname{Hom}_{R}(R/M,R)=\operatorname{Ext}_{R}^{1}(R/M,R)=0$ . Applying $\operatorname{Hom}_{R}(-,R)$ to the short exact sequence $0\longrightarrow M\longrightarrow R\longrightarrow R/M\longrightarrow 0$ then gives $R=\operatorname{End}_{R}(R)=\operatorname{Hom}_{R}(M,R)$ . It follows that $\operatorname{Hom}_{R}(M,R)=\operatorname{End}_{R}(M)$ and so $\operatorname{End}_{R}(M)=R$ .◻

Proof. (i) By induction, assume that ${\mathcal{A}}_{k-1}(M,N)$ is an $R$ -bimodule. Let $\unicode[STIX]{x1D703}\in {\mathcal{A}}_{k}(M,N)$ , $y\in R$ and $x\in \mathfrak{r}$ . Then, by induction and the fact that $R$ is commutative,

$$\begin{eqnarray}[x,\unicode[STIX]{x1D703}y]=x\unicode[STIX]{x1D703}y-\unicode[STIX]{x1D703}xy=(x\unicode[STIX]{x1D703}-\unicode[STIX]{x1D703}x)y\in {\mathcal{A}}_{k-1}(M,N)R={\mathcal{A}}_{k-1}(M,N).\end{eqnarray}$$

Hence $\unicode[STIX]{x1D703}y\in {\mathcal{A}}_{k}(M,N)$ and, by symmetry, the result follows.

(ii) As $\mathfrak{n}^{-}$ acts ad-nilpotently on any finite-dimensional $\mathfrak{g}$ -module and hence on ${\mathcal{L}}(M,N)$ , Hypothesis 5.1(2) implies that ${\mathcal{A}}(M,N)\supseteq {\mathcal{L}}(M,N)$ . Since the inclusion ${\mathcal{A}}_{k}(M,N)\supseteq {\mathcal{D}}_{k}(M,N)$ is obvious, we are reduced to proving that ${\mathcal{A}}_{k}(M,N)\subseteq {\mathcal{D}}_{k}(M,N)$ for each $k$ . By induction and for some $m$ , assume that the result holds for all $k<m$ and fix $\unicode[STIX]{x1D703}\in {\mathcal{A}}_{m}(M,N)$ .

Filter $R=\bigcup \unicode[STIX]{x1D6EC}_{p}R$ by taking $\unicode[STIX]{x1D6EC}_{0}=\mathbb{C}$ and $\unicode[STIX]{x1D6EC}_{p}R=\unicode[STIX]{x1D712}(\mathfrak{r})^{p}+\unicode[STIX]{x1D6EC}_{p-1}R$ for $p>0$ . For some $q$ , assume that $[\unicode[STIX]{x1D703},z]\in {\mathcal{D}}_{m-1}(M,N)$ for all $z\in \unicode[STIX]{x1D6EC}_{s}R$ and $s<q$ . (The case $s=1$ holds automatically by the definition of ${\mathcal{A}}(M,N)$ , so the induction does start.) Let $y\in \unicode[STIX]{x1D6EC}_{q}R$ and suppose that $y=x_{1}x_{2}$ for $x_{1}\in \unicode[STIX]{x1D712}(\mathfrak{r})$ and $x_{2}\in \unicode[STIX]{x1D6EC}_{q-1}R$ . Then, by part (i) and induction on $q$ ,

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D703}y-y\unicode[STIX]{x1D703}=\unicode[STIX]{x1D703}x_{1}x_{2}-x_{1}x_{2}\unicode[STIX]{x1D703} & = & \displaystyle \unicode[STIX]{x1D703}x_{1}x_{2}-x_{1}\unicode[STIX]{x1D703}x_{2}+x_{1}\unicode[STIX]{x1D703}x_{2}-x_{1}x_{2}\unicode[STIX]{x1D703}\nonumber\\ \displaystyle & = & \displaystyle (\unicode[STIX]{x1D703}x_{1}-x_{1}\unicode[STIX]{x1D703})x_{2}+x_{1}(\unicode[STIX]{x1D703}x_{2}-x_{2}\unicode[STIX]{x1D703})\nonumber\\ \displaystyle & \in & \displaystyle {\mathcal{A}}_{m-1}(M,N)x_{2}+x_{1}{\mathcal{A}}_{m-1}(M,N)\subseteq {\mathcal{A}}_{m-1}(M,N).\nonumber\end{eqnarray}$$

Hence $[\unicode[STIX]{x1D703},y]\in {\mathcal{A}}_{m-1}(M,N)$ . By linearity and our inductive hypotheses, $[\unicode[STIX]{x1D703},z]\in {\mathcal{A}}_{m-1}(M,N)={\mathcal{D}}_{m-1}(M,N)$ for all $z\in R$ . Hence ${\mathcal{A}}_{m}(M,N)\subseteq {\mathcal{D}}_{m}(M,N)$ .

(iii) Since $R$ is a quasi-simple $U(\mathfrak{g})$ -module, [Reference JantzenJan83, Lemma 8.14] implies that $\operatorname{ann}_{U(\mathfrak{g})}(M)=\operatorname{ann}_{U(\mathfrak{g})}(R)$ and so $\overline{U}{\hookrightarrow}{\mathcal{L}}(M,M)$ by [Reference JantzenJan83, §6.8(5)]. By definition, ${\mathcal{M}}={\mathcal{L}}(M,M)$ is a locally finite $\mathfrak{g}$ -module under the adjoint action and hence is semi-simple. Thus, we may write ${\mathcal{M}}=\overline{U}\oplus P$ for some complementary $(\operatorname{ad}\mathfrak{g})$ -module $P$ . Assume that $P\not =0$ .

Pick $p\in P\smallsetminus \{0\}$ with $p\in {\mathcal{A}}_{k}(M,M)$ for $k$ as small as possible. Then, for $x\in \mathfrak{r}$ , we have both $[x,p]\in {\mathcal{A}}_{k-1}(M,M)$ by the definition of the order filtration (5.4) and $[x,p]\in P$ as $P$ is an  $(\operatorname{ad}\mathfrak{r})$ -module. Hence, by the choice of $k$ we have $[x,p]=0$ for all $x\in \mathfrak{r}$ . Therefore, by hypothesis and part (ii),

$$\begin{eqnarray}p\in {\mathcal{A}}_{0}(M,M)={\mathcal{D}}_{0}(M)=\operatorname{End}_{R}(M)=R.\end{eqnarray}$$

However, by Hypothesis 5.1(2), $R\subseteq \overline{U}$ as subrings of ${\mathcal{D}}(R)$ . Hence $p\in \overline{U}$ , giving the required contradiction.◻

Proof. Both $M=R$ and $M=N$ satisfy the hypotheses of Proposition 5.5 and so the displayed equation follows. The final assertion then follows from [Reference Joseph and StaffordJS84, Theorem 2.9].◻

Example.

Take $A=\mathbb{C}[x,y,xz,yz,z^{2},z^{3}]\subset \mathbb{C}[x,y,z]=C$ ; then $C=A+\mathbb{C}z$ . Set $I=A\cap z^{2}C$ and $R=A/I$ . Clearly the nilradical $N(R)$ equals $(zC\cap A)/I=((xz,yz)+I)/I$ , and so $R/N(R)\cong \mathbb{C}[x,y]$ . However, $N(R)/I=(xz,yz)\cong (x,y)$ as modules over $k[x,y]\cong R/N(R)$ .

Now let $M=(x,y,zx,zy,z^{2},z^{3})/I$ be the augmentation ideal of $R$ . The maximal ring of fractions $\text{Quot}(R)$ clearly contains $z=(zx)x^{-1}$ . However, $z\not \in R$ , yet by the computations of the last paragraph one has $zM\subseteq R$ . Indeed, $zM\subseteq M$ and so $\operatorname{End}_{R}(M)\supsetneq R$ .

Proof. Suppose first that $\unicode[STIX]{x1D719}\in {\mathcal{D}}(A)$ satisfies $\unicode[STIX]{x1D719}(\mathtt{F}^{r}A)\subseteq \mathtt{F}^{r-1}A$ and so induces a function $\widetilde{\unicode[STIX]{x1D719}}:R_{r}\rightarrow R_{r-1}$ . We claim that $\widetilde{\unicode[STIX]{x1D719}}\in {\mathcal{D}}(R_{r},\,R_{r-1})$ . To see this, assume that $\unicode[STIX]{x1D719}\in {\mathcal{D}}_{t}(A)$ and that the analogous statement is true for all $\unicode[STIX]{x1D719}^{\prime }\in {\mathcal{D}}_{t-1}(A)$ . Then, for $a\in A$ , the function $\unicode[STIX]{x1D719}^{\prime }=[\unicode[STIX]{x1D719},a]$ also satisfies $\unicode[STIX]{x1D719}^{\prime }(\mathtt{F}^{r}A)\subseteq \mathtt{F}^{r-1}A$ , but now $\unicode[STIX]{x1D719}^{\prime }\in {\mathcal{D}}_{t-1}(A)$ . Hence $\unicode[STIX]{x1D719}^{\prime }\in {\mathcal{D}}_{t-1}(R_{r},\,R_{r-1})$ . Thus $\unicode[STIX]{x1D719}\in {\mathcal{D}}_{t}(R_{r},\,R_{r-1})$ , proving the claim.

In order to prove the lemma, it therefore suffices to show that each of our elements $\unicode[STIX]{x1D719}$ satisfies $\unicode[STIX]{x1D719}(\mathtt{F}^{r}A)\subseteq \mathtt{F}^{r-1}A$ . This is obvious for $1$ and $\unicode[STIX]{x2202}_{i}$ , while for $\unicode[STIX]{x1D6E5}_{1}$ it follows from (3.13).◻

Notation 6.2. For a fixed value of $r$ , the images of $\unicode[STIX]{x1D6E5}_{1}$ , $\unicode[STIX]{x2202}_{j}$ and $1$ in ${\mathcal{D}}(R_{r},\,R_{r-1})$ will be written, respectively, as $\overline{\unicode[STIX]{x1D6E5}}_{1}$ , $\overline{\unicode[STIX]{x2202}}_{j}$ and $\overline{1}$ . For concreteness, we note that the adjoint action $\widetilde{\unicode[STIX]{x1D713}}$ of $\mathfrak{g}=\mathfrak{so}(n+2,\mathbb{C})$ on ${\mathcal{D}}={\mathcal{D}}(R_{r},\,R_{r-1})$ is defined by $\widetilde{\unicode[STIX]{x1D713}}(x)(\unicode[STIX]{x1D703})=\unicode[STIX]{x1D713}_{r-1}(x)\circ \unicode[STIX]{x1D703}-\unicode[STIX]{x1D703}\circ \unicode[STIX]{x1D713}_{r}(x)$ for $x\in \mathfrak{g}$ and $\unicode[STIX]{x1D703}\in {\mathcal{D}}$ .

Proof. By the definition of ${\mathcal{L}}(R_{r},\,R_{r-1})$ it suffices to prove that $V$ is a $\mathfrak{g}$ -module. This requires some explicit computations, for which the following three formulæ in $\mathfrak{g}$ will prove useful and will be used frequently without comment: for all $1\leqslant i,j\leqslant n$ we have $[\unicode[STIX]{x1D6E5}_{1},\,\unicode[STIX]{x2202}_{j}]=[\unicode[STIX]{x1D6E5}_{1},\,D_{ij}]=0$ , while

$$\begin{eqnarray}[\unicode[STIX]{x1D6E5}_{1},X_{j}]=\biggl[{\displaystyle \frac{1}{2}}\unicode[STIX]{x2202}_{j}^{2},\,X_{j}\biggr]=\unicode[STIX]{x2202}_{j}\quad \text{and}\quad [\mathtt{E}_{1},\,\unicode[STIX]{x1D6E5}_{1}]={\displaystyle \frac{1}{2}}\mathop{\sum }_{j}[\mathtt{E}_{1},\,\unicode[STIX]{x2202}_{j}^{2}]=-2\unicode[STIX]{x1D6E5}_{1}.\end{eqnarray}$$

First, computing in ${\mathcal{D}}(R_{r},\,R_{r-1})$ we have

$$\begin{eqnarray}\displaystyle \widetilde{\unicode[STIX]{x1D713}}(\widetilde{P}_{j})(\overline{\unicode[STIX]{x1D6E5}}_{1}) & = & \displaystyle P_{j}\biggl({\displaystyle \frac{n}{2}}-r+1\biggr)\overline{\unicode[STIX]{x1D6E5}}_{1}-\overline{\unicode[STIX]{x1D6E5}}_{1}P_{j}\biggl({\displaystyle \frac{n}{2}}-r\biggr)\nonumber\\ \displaystyle & = & \displaystyle (X_{j}\unicode[STIX]{x1D6E5}_{1}\overline{\unicode[STIX]{x1D6E5}}_{1}-\overline{\unicode[STIX]{x1D6E5}}_{1}X_{j}\unicode[STIX]{x1D6E5}_{1})+(-\mathtt{E}_{1}\unicode[STIX]{x2202}_{j}\overline{\unicode[STIX]{x1D6E5}}_{1}+\overline{\unicode[STIX]{x1D6E5}}_{1}\mathtt{E}_{1}\unicode[STIX]{x2202}_{j})\nonumber\\ \displaystyle & & \displaystyle +\,\biggl(-\biggl({\displaystyle \frac{n}{2}}-r+1\biggr)\unicode[STIX]{x2202}_{j}\overline{\unicode[STIX]{x1D6E5}}_{1}+\biggl({\displaystyle \frac{n}{2}}-r\biggr)\overline{\unicode[STIX]{x1D6E5}}_{1}\unicode[STIX]{x2202}_{j}\biggr)\nonumber\\ \displaystyle & = & \displaystyle -\unicode[STIX]{x2202}_{j}\overline{\unicode[STIX]{x1D6E5}}_{1}+2\unicode[STIX]{x2202}_{j}\overline{\unicode[STIX]{x1D6E5}}_{1}-\unicode[STIX]{x2202}_{j}\overline{\unicode[STIX]{x1D6E5}}_{1}=0.\nonumber\end{eqnarray}$$

Next,

$$\begin{eqnarray}\widetilde{\unicode[STIX]{x1D713}}(\widetilde{\mathtt{E}}_{1})(\overline{\unicode[STIX]{x1D6E5}}_{1})=\biggl(\mathtt{E}_{1}+{\displaystyle \frac{n}{2}}-r+1\biggr)\overline{\unicode[STIX]{x1D6E5}}_{1}-\overline{\unicode[STIX]{x1D6E5}}_{1}\biggl(\mathtt{E}_{1}+{\displaystyle \frac{n}{2}}-r\biggr)=-\overline{\unicode[STIX]{x1D6E5}}_{1}+2\overline{\unicode[STIX]{x1D6E5}}_{1}=-\overline{\unicode[STIX]{x1D6E5}}_{1}.\end{eqnarray}$$

Similarly, $\widetilde{\unicode[STIX]{x1D713}}(D_{ij})(\overline{\unicode[STIX]{x1D6E5}}_{1})=[D_{ij},\overline{\unicode[STIX]{x1D6E5}}_{1}]=0$ and $\widetilde{\unicode[STIX]{x1D713}}(X_{j})(\overline{\unicode[STIX]{x1D6E5}}_{1})=[D_{ij},\overline{\unicode[STIX]{x1D6E5}}_{1}]=0$ . Therefore, $\widetilde{\unicode[STIX]{x1D713}}(\mathfrak{g})(\overline{\unicode[STIX]{x1D6E5}}_{1})\subseteq V$ .

Now we compute that

$$\begin{eqnarray}\displaystyle \widetilde{\unicode[STIX]{x1D713}}(\widetilde{P}_{i})(\overline{\unicode[STIX]{x2202}}_{i}) & = & \displaystyle P_{i}\biggl({\displaystyle \frac{n}{2}}-r+1\biggr)\overline{\unicode[STIX]{x2202}}_{i}-\overline{\unicode[STIX]{x2202}}_{i}P_{i}\biggl({\displaystyle \frac{n}{2}}-r\biggr)\nonumber\\ \displaystyle & = & \displaystyle (X_{i}\unicode[STIX]{x1D6E5}_{1}\overline{\unicode[STIX]{x2202}}_{i}-\overline{\unicode[STIX]{x2202}}_{i}X_{i}\unicode[STIX]{x1D6E5}_{1})+(-\mathtt{E}_{1}\unicode[STIX]{x2202}_{i}\overline{\unicode[STIX]{x2202}}_{i}+\overline{\unicode[STIX]{x2202}}_{i}\mathtt{E}_{1}\unicode[STIX]{x2202}_{i})+\biggl(\!-\biggl({\displaystyle \frac{n}{2}}-r+1\biggr)\unicode[STIX]{x2202}_{i}\overline{\unicode[STIX]{x2202}}_{i}+\biggl({\displaystyle \frac{n}{2}}-r\biggr)\overline{\unicode[STIX]{x2202}}_{i}\unicode[STIX]{x2202}_{i}\!\biggr)\nonumber\\ \displaystyle & = & \displaystyle -\overline{\unicode[STIX]{x1D6E5}}_{1}-\unicode[STIX]{x2202}_{i}\overline{\unicode[STIX]{x2202}}_{i}+\overline{\unicode[STIX]{x2202}}_{i}\unicode[STIX]{x2202}_{i}=-\overline{\unicode[STIX]{x1D6E5}}_{1}.\nonumber\end{eqnarray}$$

If $i\not =j$ we obtain

$$\begin{eqnarray}\displaystyle \widetilde{\unicode[STIX]{x1D713}}(\widetilde{P}_{j})(\overline{\unicode[STIX]{x2202}}_{i}) & = & \displaystyle (X_{j}\unicode[STIX]{x1D6E5}_{1}\overline{\unicode[STIX]{x2202}}_{i}-\overline{\unicode[STIX]{x2202}}_{i}X_{j}\unicode[STIX]{x1D6E5}_{1})+(-\mathtt{E}_{1}\unicode[STIX]{x2202}_{j}\overline{\unicode[STIX]{x2202}}_{i}+\overline{\unicode[STIX]{x2202}}_{i}\mathtt{E}_{1}\unicode[STIX]{x2202}_{j})+\biggl(\!-\biggl({\displaystyle \frac{n}{2}}-r+1\biggr)\unicode[STIX]{x2202}_{j}\overline{\unicode[STIX]{x2202}}_{i}+\biggl({\displaystyle \frac{n}{2}}-r\biggr)\overline{\unicode[STIX]{x2202}}_{i}\unicode[STIX]{x2202}_{j}\!\biggr)\nonumber\\ \displaystyle & = & \displaystyle -\unicode[STIX]{x2202}_{j}\overline{\unicode[STIX]{x2202}}_{i}+\overline{\unicode[STIX]{x2202}}_{i}\unicode[STIX]{x2202}_{j}=0.\nonumber\end{eqnarray}$$

On the other hand, $\widetilde{\unicode[STIX]{x1D713}}(\widetilde{\mathtt{E}}_{1})(\overline{\unicode[STIX]{x2202}}_{i})=(\mathtt{E}_{1}+n/2-r+1)\overline{\unicode[STIX]{x2202}}_{i}-\overline{\unicode[STIX]{x2202}}_{i}(\mathtt{E}_{1}+n/2-r)=\overline{\unicode[STIX]{x2202}}_{i}-\overline{\unicode[STIX]{x2202}}_{i}~=0$ , while

$$\begin{eqnarray}\widetilde{\unicode[STIX]{x1D713}}(D_{ab})\overline{\unicode[STIX]{x2202}}_{i}=(X_{a}\unicode[STIX]{x2202}_{b}-X_{b}\unicode[STIX]{x2202}_{a})\overline{\unicode[STIX]{x2202}}_{i}-\overline{\unicode[STIX]{x2202}}_{i}(X_{a}\unicode[STIX]{x2202}_{b}-X_{b}\unicode[STIX]{x2202}_{a})=\mathop{\sum }_{u}\unicode[STIX]{x1D6FC}_{u}\overline{\unicode[STIX]{x2202}}_{u}\end{eqnarray}$$

for any $1\leqslant a,b,i\leqslant n$ and appropriate scalars $\unicode[STIX]{x1D6FC}_{u}\in \mathbb{C}$ . Since $\widetilde{\unicode[STIX]{x1D713}}(X_{k})\overline{\unicode[STIX]{x2202}}_{i}=[X_{k},\overline{\unicode[STIX]{x2202}}_{i}]~=-\unicode[STIX]{x1D6FF}_{k,i}\overline{1}$ , it follows that $\widetilde{\unicode[STIX]{x1D713}}(\mathfrak{g})(\overline{\unicode[STIX]{x2202}}_{i})\subseteq V$ . The easy fact that $\widetilde{\unicode[STIX]{x1D713}}(\mathfrak{g})(\overline{1})\subseteq V$ is left to the reader and completes the proof.◻

Proof. (1) If $\unicode[STIX]{x1D703}\in {\mathcal{L}}(A,B)$ and $\unicode[STIX]{x1D719}\in {\mathcal{L}}(B,C)$ for $\mathfrak{g}$ -modules $A,B$ and $C$ , then $\unicode[STIX]{x1D719}\unicode[STIX]{x1D703}\in {\mathcal{L}}(A,C)$ ; see [Reference JantzenJan83, 6.8(3)]. Thus the result follows from Proposition 6.3 and induction on $r-s$ .

(2) By Theorem 4.13 each $R_{s}/Z_{s}$ is finite-dimensional (or zero), say with $\mathfrak{a}_{s}=\operatorname{ann}_{U(\mathfrak{g})}(R_{s}/Z_{s})$ . Since $Z_{r}$ is infinite-dimensional, the projection map $\unicode[STIX]{x1D70B}_{rs}\in {\mathcal{L}}(R_{r},\,R_{s})$ has an infinite-dimensional image, as does its restriction $\unicode[STIX]{x1D70B}_{rs}^{\prime }:Z_{r}\rightarrow R_{s}$ . Hence $\mathfrak{a}_{s}\unicode[STIX]{x1D70B}_{rs}^{\prime }\not =0$ , and so for some $a\in \mathfrak{a}_{s}$ one has $0\not =a\circ \unicode[STIX]{x1D70B}_{rs}^{\prime }\in {\mathcal{L}}(Z_{r},Z_{s})$ . Thus ${\mathcal{L}}(Z_{r},Z_{s})\not =0$ . By [Reference JantzenJan83, §6.9(5)], this also implies that ${\mathcal{L}}(Z_{s},\,Z_{r})\not =0$ .◻

Proof. By [Reference Joseph and StaffordJS84, Corollary 2.10], ${\mathcal{L}}(M,M)$ is a maximal order and it is noetherian since it is finitely generated as both a left and a right $U$ -module. So it remains to prove uniqueness. Suppose that $U\subseteq T$ for some other order $T$ equivalent to $U$ . Then $aTb\subseteq U$ for some regular elements $a,b\in U$ . Let $T^{\prime }=UaT+U$ ; this is an overring of $U$ with $T^{\prime }b\subseteq U$ . So, since $b$ is regular, $T^{\prime }$ is certainly finitely generated as a left $U$ -module. Therefore, by [Reference Joseph and StaffordJS84, Theorem 2.9], $T^{\prime }\subseteq {\mathcal{L}}(M,M)$ . In particular, $T^{\prime }$ is also finitely generated as a right $U$ -module. Since $aT\subseteq T^{\prime }$ , it follows that $T$ is also a finitely generated right $U$ -module. Hence [Reference Joseph and StaffordJS84, Theorem 2.9] implies that $T\subseteq {\mathcal{L}}(M,M)$ .◻

Proof. (i) By Theorem 4.13, Hypotheses 5.1 are satisfied and so the result follows from Theorem 5.7.

(ii) By [Reference JantzenJan83, Lemma 8.14], $\operatorname{ann}_{U(\mathfrak{g})}(R_{b})=\operatorname{ann}_{U(\mathfrak{g})}(Z_{b})=J_{b}$ and so $R_{b}$ is ann-homogeneous in the sense of [Reference Joseph and StaffordJS84, §2.7]. Thus, by [Reference Joseph and StaffordJS84, Lemma 2.8], ${\mathcal{L}}(R_{a},R_{b})$ is GK-homogeneous in the sense that any nonzero left or right $\mathfrak{g}$ -submodule $N$ of ${\mathcal{L}}(R_{a},R_{b})$ has $\operatorname{GKdim}(N)=\operatorname{GKdim}({\mathcal{L}}(R_{a},R_{b}))$ . Since each $U(\mathfrak{g})/J_{c}$ is a prime ring by (i), this is equivalent to ${\mathcal{L}}(R_{a},R_{b})$ being torsion-free on both sides. Exactly the same argument works for ${\mathcal{L}}(Z_{a},Z_{b})$ .

It therefore remains to prove that these modules are nonzero. For ${\mathcal{L}}(Z_{a},Z_{b})$ this is Corollary 6.4. In particular, ${\mathcal{L}}(Z_{a},Z_{b})$ is infinite-dimensional, and hence ${\mathcal{L}}(Z_{a},Z_{b})\mathfrak{a}_{a}\not =0$ , where $\mathfrak{a}_{a}$ is the annihilator of the finite-dimensional $\overline{U}_{a}$ -module $R_{a}/Z_{a}$ . Thus

$$\begin{eqnarray}0\not ={\mathcal{L}}(Z_{a},Z_{b})\mathfrak{a}_{a}\subseteq {\mathcal{L}}(R_{a},Z_{b})\subseteq {\mathcal{L}}(R_{a},R_{b}).\end{eqnarray}$$

(iii) This is automatic from (ii). ◻

Proof. (i) Clearly $R_{a}$ has a local artinian ring of fractions $(R_{a})_{(\mathtt{F})}$ . Now apply [Reference MussonMus89, Corollary 2.6].

(ii) This follows from [Reference MussonMus89, Lemma 2.4 and (2.5)].

(iii) The projection map $\unicode[STIX]{x1D70B}:R_{r}\rightarrow R_{1}$ is an $A$ -module map and so belongs to ${\mathcal{D}}(R_{r},R_{1})$ . Similarly, the $A$ -module isomorphism $\unicode[STIX]{x1D719}:R_{1}\rightarrow \mathtt{F}^{r-1}R_{r}$ belongs to ${\mathcal{D}}(R_{1},R_{r})$ . Thus, ${\mathcal{D}}(R_{r},R_{1})\not =0\not ={\mathcal{D}}(R_{1},R_{r})$ .

As right ${\mathcal{D}}(R_{r})$ -modules, ${\mathcal{D}}(R_{r},R_{1})\cong {\mathcal{D}}(R_{r},\mathtt{F}^{r-1}R_{r}){\hookrightarrow}{\mathcal{D}}(R_{r},R_{r})$ . Therefore, by (i), it is a torsion-free right ${\mathcal{D}}(R_{r})$ -module. For the left action, consider $0\not =\unicode[STIX]{x1D719}\in {\mathcal{D}}(R_{r},R_{1})$ and pick $s$ maximal such that $\unicode[STIX]{x1D719}(\mathtt{F}^{s}R_{r})\not =0$ . If $\bar{r}\in R_{1}$ , write $\bar{r}=\unicode[STIX]{x1D70B}(r)$ for some $r\in R_{r}$ and define $\widetilde{\unicode[STIX]{x1D719}}(\bar{r})=\unicode[STIX]{x1D719}(\mathtt{F}^{s}r)$ . Since $\unicode[STIX]{x1D719}(\mathtt{F}^{s+1}R_{r})=0$ , this is a well-defined morphism and hence a differential operator; thus $\widetilde{\unicode[STIX]{x1D719}}\in {\mathcal{D}}(R_{1},R_{1})$ . Moreover, $\widetilde{\unicode[STIX]{x1D719}}\not =0$ by the choice of $s$ , and so $\widetilde{\unicode[STIX]{x1D719}}$ is not torsion as an element of the left ${\mathcal{D}}(R_{1})$ -module ${\mathcal{D}}(R_{1})$ . Consequently, $\unicode[STIX]{x1D719}$ is not torsion under the left ${\mathcal{D}}(R_{1})$ action.