Simple $\mathfrak{sl}(V)$-modules which are free over an abelian subalgebra

TL;DR

This paper classifies certain simple modules over rak{sl}(V) that are free of rank 1 over a nilradical of a maximal parabolic subalgebra, parametrized by polynomials, and describes their structure and simplicity.

Contribution

It provides a complete classification of rak{sl}(V)-modules that are free of rank 1 over a nilradical, including their parametrization and submodule structure.

Findings

Modules are parametrized by polynomials in im V - 1 variables.

Most of these modules are simple.

Submodule structures are explicitly determined.

Abstract

Let $\mathfrak{p}$ be a parabolic subalgebra of $\mathfrak{sl}(V)$ of maximal dimension and let $\mathfrak{n} \subset \mathfrak{p}$ be the corresponding nilradical. In this paper we classify the set of $\mathfrak{sl}(V)$-modules whose restriction to $U(\mathfrak{n})$ is free of rank $1$. It turns out that isomorphism classes of such modules are parametrized by polynomials in $\dim V-1$ variables. We determine the submodule structure for these modules and we show that they generically are simple.

Equations

$$p_{ij}:=x_i\frac{\partial p}{\partial x_j}+{\delta }_{ij}p(0)/n\text{ and }{q}_i:=-\frac{1}{x_i}{\int }_0^{x_i}\sum _{r=1}^n(p_{ii}^r{p}_{ri}+x_r{p}_{ii}^{ir}+p_{ii}^i{p}_{rr})d{x}_i,$$

$$e_{i,n+1}\cdot f$$

$$h_i\cdot f$$

$$e_{ij}\cdot f$$

$$e_{n+1,i}\cdot f$$

$$e_{13}\cdot f$$

$$h_1\cdot f$$

$$e_{12}\cdot f$$

$$e_{31}\cdot f$$

$$e_{32}\cdot f$$

$$\mathrm{Hom}_{U(\mathfrak{a})}(\mathrm{Res}_{U(\mathfrak{a})}^{U(\mathfrak{g})}M,N)\simeq\mathrm{Hom}_{U(\mathfrak{g})}\big{(}M,\mathrm{Hom}_{U(\mathfrak{a})}(U(\mathfrak{g}),N)\big{)}.$$

$${\mathrm{Hom}}_{U(\mathfrak{a})}({\mathrm{Res}}_{U(\mathfrak{a})}^{U(\mathfrak{g})}M,N)\simeq {\mathrm{Hom}}_{U(\mathfrak{a})}(U(\mathfrak{a}),N)\simeq N,$$

$$\dim\mathrm{Hom}_{U(\mathfrak{g})}\big{(}M,\mathrm{Hom}_{U(\mathfrak{a})}(U(\mathfrak{g}),N)\big{)}=\dim N.$$

$$f\in \mathrm{Ker}({\overline{\phi }}_{\alpha })\iff {\overline{\phi }}_{\alpha }(f(z))=0\iff {\overline{\phi }}_{\alpha }(f(z))(x)=0\forall x\in U(\mathfrak{g})$$

$$\iff {\phi }_{\alpha }(x\cdot f(z))=0\forall x\in U(\mathfrak{g})\iff (x\cdot f)(\alpha )=0\forall x\in U(\mathfrak{g}).$$

$$h\cdot f(h)=hf(h),x\cdot f(h)=(h+b)f(h-1),y\cdot f(h)=-(h-b)f(h+1).$$

$${\mathfrak{p}}_{\Delta }:=\mathrm{Stab}(\Delta )=\{f\in \mathfrak{sl}(V)|f(\Delta )\subset \Delta \},$$

$${\mathfrak{n}}_{\Delta }:=\{f\in \mathfrak{sl}(V)|f(V)\subset \Delta \}.$$

$${\mathfrak{p}}_{\Delta }=\left[\begin{array}{ccccc}\ast & \ast & \cdots & \ast & 0\\ \ast & \ast & \cdots & \ast & 0\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ \ast & \ast & \cdots & \ast & 0\\ \ast & \ast & \cdots & \ast & \ast \end{array}\right]{\mathfrak{p}}_{{\Delta }^{\perp }}=\left[\begin{array}{ccccc}\ast & \ast & \cdots & \ast & \ast \\ \ast & \ast & \cdots & \ast & \ast \\ ⋮& ⋮& \ddots & ⋮& ⋮\\ \ast & \ast & \cdots & \ast & \ast \\ 0& 0& \cdots & 0& \ast \end{array}\right]$$

$${\mathfrak{n}}_{\Delta }=\left[\begin{array}{ccccc}0& 0& \cdots & 0& 0\\ 0& 0& \cdots & 0& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& \cdots & 0& 0\\ \ast & \ast & \cdots & \ast & 0\end{array}\right]{\mathfrak{n}}_{{\Delta }^{\perp }}=\left[\begin{array}{ccccc}0& 0& \cdots & 0& \ast \\ 0& 0& \cdots & 0& \ast \\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& \cdots & 0& \ast \\ 0& 0& \cdots & 0& 0\end{array}\right].$$

$$q(x):=-\frac{1}{2x}\int_{0}^{x}\big{(}p(t)p^{\prime}(t)+tp^{\prime\prime}(t)\big{)}dt.$$

$$x\cdot f(x)$$

$$h\cdot f(x)$$

$$y\cdot f(x)$$

$$x\cdot y\cdot f-$$

$$=$$

$$h\cdot x\cdot f-$$

$$=$$

$$h\cdot y\cdot f-$$

$$=$$

$$-q(pf+2x{f}^{\prime })+p(p^{\prime }f+p{f}^{\prime }+2f^{\prime }+2x{f}^{\prime \prime })$$

$$+x(p^{\prime \prime }f+p^{\prime }{f}^{\prime }+p^{\prime }{f}^{\prime }+p{f}^{\prime \prime }+4f^{\prime \prime }+2x{f}^{\prime \prime \prime })$$

$$=$$

$$h\cdot x^{k+1}=h\cdot x\cdot x^k=x\cdot h\cdot x^k+[h,x]\cdot x^k=x(p(x)+2k)x^k+2x\cdot x^k$$

$$=(p(x)+2k)x^{k+1}+2x^{k+1}=(p(x)+2(k+1))x^{k+1}.$$

$$y\cdot x^{k+1}=y\cdot x\cdot x^k=x\cdot y\cdot x^k+[y,x]\cdot x^k=x(y\cdot x^k)-h\cdot x^k$$

Full text

1. Introduction

Lie algebras and their representations appear throughout multiple areas of mathematics, and the elemental objects of representation theory are simple modules. Unfortunately, a complete classification of simple modules for a Lie algebra $\mathfrak{g}$ is too broad a project, only for the Lie algebra $\mathfrak{sl}_{2}$ does a version of such a classification exist, see [B, Maz1]. Nevertheless many classes of $\mathfrak{g}$-modules are well studied. For example, when $\mathfrak{g}$ is a simple finite-dimensional complex Lie algebra, all simple finite-dimensional modules were classified early, see [Ca, Di]. More generally, simple highest weight modules (see [Di, Hu, BGG]) and simple weight modules with finite-dimensional weight spaces (see [BL, Fu, Fe, Mat]) are also completely classified.

Several classes of non-weight modules have also been studied. These include Whittaker modules (see [Kos, BM]), Gelfand-Zetlin modules (see [DFO]), and various others (see for example  [FOS]).

Recently several authors have studied $\mathfrak{g}$-modules whose restriction to certain $\mathfrak{g}$-subalgebras are free. For example, when $\mathfrak{g}$ is a simple complex finite-dimensional Lie algebra, the set of modules which are free of rank $1$ when restricted to the universal enveloping algebra of a Cartan subalgebra were classified in [N1, N2]. Corresponding and related results were also obtained for a multitude of other Lie algebras such as the Witt- and Virasoro-algebras, the Heisenberg-Virasoro algebra, Schrödinger algebras, and for basic Lie-super algebras, see [CC, CG, CLNZ, CZ, CTZ, HCS, LZ, MP, N3, TZ1, TZ2] and references therein. A common theme for many of the modules in the papers listed above are that they are free when restricted to some commutative $\mathfrak{g}$-subalgebra, often involving a Cartan subalgebra and central elements of $\mathfrak{g}$.

In the present paper we study $\mathfrak{sl}(V)$-modules which are free over another maximal commutative subalgebra: the nilradical of a parabolic subalgebra of maximal dimension. A concrete example of such a module is given in the following result which is a restatement of Theorem 16 in Section 4.2.

Let $\mathfrak{g}=\mathfrak{sl}_{n+1}$ and let $\mathfrak{n}=\mathrm{span}(e_{1,n+1},\ldots,e_{n,n+1})\subset\mathfrak{g}$ (where as usual $e_{i,j}$ are the standard basis elements for $\mathfrak{gl}_{n+1}$). Then $\mathfrak{n}$ is the nilradical of the parabolic subalgebra corresponding to removing the simple root whose root space is spanned by $e_{n+1,n}$.

Theorem 1**.**

Fix a polynomial $p({\bf x})\in\mathbb{K}[x_{1},\ldots,x_{n}]$ and for $1\leq i,j\leq n$ define polynomials

[TABLE]

where upper indices indicate derivatives: $p^{k}=\tfrac{\partial p}{\partial x_{k}}$.

Then the following action equips the space $M(p)=\mathbb{K}[x_{1},\ldots,x_{n}]$ with an $\mathfrak{sl}_{n+1}$-module structure:

[TABLE]

for $f\in\mathbb{K}[x_{1},\ldots,x_{n}]$, $1\leq i,j\leq n$, and $h_{i}:=e_{ii}-\frac{1}{n+1}I$.

Moreover, any $\mathfrak{sl}_{n+1}$-module $M$ for which $\mathrm{Res}_{U(\mathfrak{n})}^{U(\mathfrak{sl}_{n+1})}M$ is free of rank $1$ is isomorphic to $M(p)$ for a unique $p\in\mathbb{K}[x_{1},\ldots,x_{n}]$.

The easiest case is when $p$ is constant as in the following example:

• Example

Taking $n=2$ and $p=-\tfrac{3}{2}\lambda\in\mathbb{K}$ we obtain the following $\mathfrak{sl}_{3}$-module structure on $\mathbb{K}[x_{1},x_{2}]$:

[TABLE]

where we have written $d$ for the degree operator $x_{1}\tfrac{\partial}{\partial x_{1}}+x_{2}\tfrac{\partial}{\partial x_{2}}$. In this case our module $M(-\tfrac{3\lambda}{2})$ is actually parabolically induced: Let $\mathfrak{p}=\mathrm{span}(h_{1},h_{2},e_{12},e_{21},e_{31},e_{32})$, and let $\mathbb{K}_{\lambda}$ be the $1$-dimensional $\mathfrak{p}$-module where $h_{1}$ and $h_{2}$ both act by $\lambda$, and the other basis elements of $\mathfrak{p}$ act trivially. Then $\mathrm{Ind}_{U(\mathfrak{p})}^{U(\mathfrak{sl}_{3})}\mathbb{K}_{\lambda}=U(\mathfrak{sl}_{3})\otimes_{U(\mathfrak{p})}\mathbb{K}_{\lambda}\simeq M(-\tfrac{3\lambda}{2})$. However, when $p$ is nonconstant $M(p)$ is not parabolically induced.

The layout of this paper is as follows. In Section 2 we discuss parabolic subalgebras of $\mathfrak{sl}(V)$ and their nilradicals, and we look at some general theory for modules free over subalgebras. In Section 3 we focus on $\mathfrak{g}=\mathfrak{sl}_{2}$, in this case we get somewhat nicer formulas for our module structure. We determine the Jordan-Hölder components of the modules we construct, and in Section 3.1 we give a Glebsch-Gordan style decomposition theorem for tensor products of $U(\mathfrak{n})$-free modules and finite-dimensional modules. In Section 4 we generalize most of these results to $\mathfrak{sl}_{n+1}$. In Section 4.2 we obtain the classification of $U(\mathfrak{n})$-free modules of rank $1$ for $\mathfrak{sl}(V)$ and in Section 4.3 we prove that our modules are irreducible in general, and determine the submodule structure for the exceptional cases.

2. Preliminaries

Denote the nonnegative integers by $\mathbb{N}$, and let $\mathbb{K}$ be an algebraically closed field of characteristic zero.

2.1. Modules which are free over a subalgebra

We first discuss some general results relating the modules we study to previously known modules.

In this section let $\mathfrak{g}$ be an arbitrary Lie algebra over $\mathbb{K}$ and let $\mathfrak{a}\subset\mathfrak{g}$ be a subalgebra.

First let us briefly recall that we have an adjunction between the functors $\mathrm{Res}_{U(\mathfrak{a})}^{U(\mathfrak{g})}:U(\mathfrak{g})\text{-Mod}\rightarrow U(\mathfrak{a})\text{-Mod}$ and $\mathrm{Hom}_{U(\mathfrak{a})}(U(\mathfrak{g}),-):U(\mathfrak{a})\text{-Mod}\rightarrow U(\mathfrak{g})\text{-Mod}$. In particular this means that for every $\mathfrak{g}$-module $M$ and any $\mathfrak{a}$-module $N$ we have a natural vector space isomorphism

[TABLE]

Here the $\mathfrak{g}$-action on $\mathrm{Hom}_{U(\mathfrak{a})}(U(\mathfrak{g}),N)$ is given by $(x\cdot f)(y):=f(yx)$, and the correspondence above maps $\varphi\in\mathrm{Hom}_{U(\mathfrak{a})}(\mathrm{Res}_{U(\mathfrak{a})}^{U(\mathfrak{g})}M,N)$ to \overline{\varphi}\in\mathrm{Hom}_{U(\mathfrak{g})}\big{(}M,\mathrm{Hom}_{U(\mathfrak{a})}(U(\mathfrak{g}),N)\big{)}, where $\overline{\varphi}(m)\in\mathrm{Hom}_{U(\mathfrak{a})}(U(\mathfrak{g}),N)$ is defined by $\overline{\varphi}(m)(x):=\varphi(x\cdot m)$.

Now let $\mathfrak{g}$ be a finite-dimensional complex simple Lie algebra, and let $\mathfrak{a}\subset\mathfrak{g}$ be a subalgebra. Let $M$ be a $\mathfrak{g}$-module such that $\mathrm{Res}_{U(\mathfrak{a})}^{U(\mathfrak{g})}M$ is a free module of rank $1$. This just means that $M\simeq U(\mathfrak{a})$ as an $\mathfrak{a}$-module. As vector spaces we then have

[TABLE]

so the $\mathrm{Res}-\mathrm{Hom}$ adjunction above gives

[TABLE]

For example, take $\mathfrak{a}=\mathrm{span}(z)$ for some $z\in\mathfrak{g}$, and take and take $N=\mathbb{K}_{\alpha}$ to be the one-dimensional $\mathfrak{a}$-module where $z$ acts by the scalar $\alpha$. Then the space $\mathrm{Hom}_{U(\mathfrak{a})}(U(\mathfrak{a}),\mathbb{K}_{\alpha})=\mathrm{Hom}_{\mathbb{K}[z]}(\mathbb{K}[z],\mathbb{K}_{\alpha})$ is one-dimensional and spanned by the evaluation map $\varphi_{\alpha}$ where $\varphi_{\alpha}(f(z))=f(\alpha)$. Corresponding to $\varphi_{\alpha}$ we get a $\mathfrak{g}$-submodule $\mathrm{Ker}(\overline{\varphi}_{\alpha})\subset M$. We can describe this kernel explicitly: Let $f(z)\in\mathbb{K}[z]=U(\mathfrak{a})$. Then

[TABLE]

[TABLE]

As an example we may take $\mathfrak{a}$ to be a Cartan-subalgebra of $\mathfrak{g}$. Then the situation becomes as in the papers [N1, N2], where we studied simple module structures on $U(\mathfrak{h})$. In particular, if we take $\mathfrak{g}=\mathfrak{sl}_{2}$ and pick a basis $\{x,y,h\}$ satisfying $[h,x]=x$, $[h,y]=-y$ and $[x,y]=2h$, for any scalar $b$ we have a module structure on $M_{b}=U(\mathfrak{h})$ in which

[TABLE]

As above, $\mathrm{Ker}(\overline{\varphi}_{\alpha})$ is a submodule for each $\alpha$, and the conditions from [N1] for $M_{b}$ to be simple correspond precisely to our derived condition $\mathrm{Ker}\overline{\varphi}_{\alpha}=M_{b}$ as above.

If we stick with $\mathfrak{sl}_{2}$, another option is to instead take $\mathfrak{a}=\mathrm{span}(x)$ and study modules free over $U(\mathfrak{a})$. This is what we do in Section 3 below.

We can generalize these results to construct a new type of modules for $\mathfrak{sl}_{n}$. Here instead of taking $\mathfrak{a}$ as the Cartan subalgebra, we pick $\mathfrak{a}$ as another abelian subalgebra of dimension $n-1$, namely the nilradical of a parabolic subalgebra of maximal dimension. This is discussed starting from Section 4.

2.2. Nilradicals of maximal parabolics

We first describe the set of parabolic subalgebras of $\mathfrak{sl}(V)$ of maximal dimension. Given a proper nontrivial subspace $\Delta\subset V$ we define subalgebras of $\mathfrak{sl}(V)$ as follows:

[TABLE]

[TABLE]

We summarize some classical results on such subalgebras, see [Kob, Lemma 7.3.1] for details.

Lemma 2**.**

We have

1. (1)

$\mathfrak{p}_{\Delta}$* is a parabolic subalgebra of $\mathfrak{sl}(V)$.* 2. (2)

$\mathfrak{p}_{\Delta}$* is maximal with respect to inclusion: it is not contained in any other parabolic subalgebra.* 3. (3)

$\mathfrak{n}_{\Delta}$* is the nilradical of $\mathfrak{p}_{\Delta}$.* 4. (4)

$\mathfrak{n}_{\Delta}$* is an abelian subalgebra.* 5. (5)

$(\mathfrak{p}_{\Delta})^{\perp}=\mathfrak{n}_{\Delta}$* with respect to the Killing form on $\mathfrak{sl}(V)$.* 6. (6)

$\mathfrak{n}_{\Delta}$* is an ideal of $\mathfrak{p}_{\Delta}$ and $\mathfrak{p}_{\Delta}/\mathfrak{n}_{\Delta}$ is semi-simple.*

Lemma 3**.**

The following statements are equivalent:

1. (1)

$\mathfrak{p}_{\Delta}$* is a parabolic subalgebra of maximal dimension.* 2. (2)

$\dim\mathfrak{n}_{\Delta}=\dim V-1$. 3. (3)

$\dim\Delta=1$* or $\mathrm{codim}\>\Delta=1$.*

For any subspace $\Delta\subset V$, denote by $\mathfrak{C}_{\Delta}$ the full subcategory $\mathfrak{sl}(V)$-Mod consisting of modules which are free of rank $1$ when restricted to $U(\mathfrak{n}_{\Delta})$.

When $\Delta\subset V$ is a one-dimensional subspace, we may fix a basis $v_{1}\ldots v_{n}$ of $V$ such that $v_{n}\in\Delta$. We write $\Delta^{\perp}$ for the subspace spanned by $v_{1},\ldots,v_{n-1}$. This choice of basis lets us identify $\mathfrak{sl}(V)=\mathfrak{sl}_{n}$ which gives

[TABLE]

[TABLE]

Lemma 4**.**

There are equivalences of categories $\mathfrak{C}_{\Delta}\simeq\mathfrak{C}_{\Delta^{\prime}}$ for any pairs of subspaces $\Delta$ and $\Delta^{\prime}$ of either dimension or codimension $1$.

Proof.

First assume that $\dim\Delta=\dim\Delta^{\prime}$. Then we may pick an invertible $S$ that bijectively maps $\Delta$ to $\Delta^{\prime}$. But then the automorphism $\varphi:x\mapsto SxS^{-1}$ maps $\mathfrak{n}_{\Delta}$ to $\mathfrak{n}_{\Delta^{\prime}}$. And therefore, if $M$ is a module free over $\mathfrak{n}_{\Delta^{\prime}}$, then the twisted module ${}^{\varphi}M$ (in which the action is $x\bullet m:=\varphi(x)\cdot m$) will be free over $\mathfrak{n}_{\Delta}$. Similarly, we note that if $\dim\Delta=1$ the category $\mathfrak{C}_{\Delta}$ is equivalent to $\mathfrak{C}_{\Delta^{\perp}}$ by twisting by the outer isomorphism $x\mapsto-x^{T}$ (minus transpose). Finally, by combining the above statements we see that if $\dim\Delta=1$ and $\mathrm{codim}\>\Delta^{\prime}=1$ we have $\mathfrak{C}_{\Delta}\simeq\mathfrak{C}_{\Delta^{\perp}}\simeq\mathfrak{C}_{\Delta^{\prime}}$. ∎

Thus we shall restrict our focus to modules which are free of rank $1$ over the universal enveloping algebra of the fixed subalgebra $\mathfrak{n}_{\Delta^{\perp}}$ of $\mathfrak{sl}_{n}$ as described above. All other $\mathfrak{sl}(V)$-modules free over the nilradical of a maximal-dimensional parabolic can be obtained from these by twisting by automorphisms.

3. $\mathfrak{sl}_{2}$-modules

We treat the case $\dim V=2$ separately, because we obtain more extensive results and nicer formulas in this setting. We fix the standard basis $\{x,y,h\}$ for $\mathfrak{sl}_{2}(\mathbb{K})$. These elements satisfy $[h,x]=2x,[h,y]=-2y,\;[x,y]=h$.

When $\dim V=2$ the only parabolic subalgebras of $\mathfrak{sl}(V)$ are Borel-subalgebras $\mathfrak{p}=\mathfrak{h}\oplus\mathfrak{n}$, where the corresponding Cartan and nilradical subalgebras both are $1$-dimensional. By picking a basis $(v_{1},v_{2})$ for $V$ where $v_{1}\in\mathfrak{h}$ and $v_{2}\in\mathfrak{n}$, we obtain an identification $\mathfrak{sl}(V)=\mathfrak{sl}_{2}$, $\mathfrak{p}=\mathrm{span}(h,x)$, and $\mathfrak{n}=\mathrm{span}(x)$. Since $U(\mathfrak{n})=\mathbb{K}[x]$, our classification problem reduces to describing all possible $\mathfrak{sl}_{2}$-module structures on $\mathbb{K}[x]$ in which we have $x\cdot f(x)=xf(x)$.

We start by defining some such modules.

Proposition 5**.**

Fix a polynomial $p(x)\in\mathbb{K}[x]$ and define a second polynomial

[TABLE]

Then the following $\mathfrak{sl}_{2}$-action defines an $\mathfrak{sl}_{2}$-module structure on $\mathbb{K}[x]$:

[TABLE]

We denote this module by $V(p)$.

Proof.

We verify that the above action respects the $\mathfrak{sl}_{2}$ structure. We have

[TABLE]

and

[TABLE]

and

[TABLE]

∎

It turns out that the modules $V(p)$ defined above are pairwise non-isomorphic and exhaust all modules whose restriction to $\mathbb{K}[x]$ is free of rank $1$.

Proposition 6**.**

$V(p)\simeq V(\overline{p})$* if and only if $p=\overline{p}$.*

Proof.

Let $\varphi:V(p)\rightarrow V(\overline{p})$ be an isomorphism. Then $\varphi(f)=f\varphi(1)$ so $\varphi(1)$ is a nonzero constant. The relation $\varphi(h\cdot f)=h\cdot\varphi(f)$ is equivalent to the condition $\varphi(1)(p-\overline{p})=0$ so $p=\overline{p}$. ∎

Proposition 7**.**

Any $M\in\mathfrak{sl}_{2}$-Mod such that $\mathrm{Res}_{\mathbb{K}[x]}^{U(\mathfrak{sl}_{2})}M$ is free of rank $1$ is isomorphic to $V(p)$ for some polynomial $p$.

Proof.

Let $M=\mathbb{K}[x]$ with a given $\mathfrak{sl}_{2}$-module structure such that $x\cdot f(x)=xf(x)$. Define $p(x):=h\cdot 1$. We claim that this implies that $h\cdot x^{k}=(p(x)+2k)x^{k}$. Indeed, it holds for $k=0$, and by induction we have

[TABLE]

[TABLE]

Note that $h\cdot x^{k}=(p(x)+2k)x^{k}$ for all $k$ can be written more compactly as $h\cdot f=pf+2xf^{\prime}$ as in the definition of $V(p)$ above.

Next, we define $q(x):=y\cdot 1$ and claim that this implies that $y\cdot f=qf-pf^{\prime}-xf^{\prime\prime}$ as in the above definition. This equality is equivalent to $y\cdot x^{k}=qx^{k}-kpx^{k-1}-k(k-1)x^{k-1}=(qx-kp-k(k-1))x^{k-1}$ for all $k$. The latter statement can again be proved by induction: it holds trivially for $k=0$ and we have

[TABLE]

[TABLE]

[TABLE]

This proves that the $h$- and $y$-action are completely determined by $p$ and $q$. It remains only to verify that $q$ is uniquely determined by $p$ as in the above definition. For this we expand the equality $[h,y]\cdot f-(h\cdot y\cdot f-y\cdot h\cdot f)=0$. Our previous considerations show that the left side expands as follows.

[TABLE]

This should hold for all $f$, which implies that $2q+2xq^{\prime}+pp^{\prime}+xp^{\prime\prime}=0$. This can be rewritten $\frac{\partial}{\partial x}(xq)=-\frac{1}{2}(pp^{\prime}+xp^{\prime\prime})$, which has the unique polynomial solution $q=-\frac{1}{2x}\int(pp^{\prime}+xp^{\prime\prime})dx$ (where the integration constant is forced to be zero). Thus $q$ is determined by $p$ and the module structure is just as in the above definition. ∎

Next we investigate the simplicity of the modules $V(p)$.

Proposition 8**.**

The module $V(p)$ is simple if and only if $p(0)\not\in-\mathbb{N}$. Otherwise $V(p(x))$ has length $2$ and we have a short exact sequence

[TABLE]

where $L(-p(0))$ is the simple highest weight module of highest weight $-p(0)\in\mathbb{N}$.

Proof.

Let $S\subset V(p(x))$ be a proper nonzero submodule. We first claim that $S$ is a homogeneous ideal of $\mathbb{K}[x]$. This follows because $d:=\frac{1}{2}(h-p(x))\in U(\mathfrak{sl}_{2})$ is the degree operator, which acts by $f\mapsto xf^{\prime}(x)$, so by repeatedly acting by $(k-d)$ for different $k\in\mathbb{N}$ we can reduce any element $f$ to its lowest degree homogeneous component. Thus $S=x^{k}\mathbb{K}[x]$ for some $k>0$. Then $S\ni(q(x)-y)\cdot x^{k}=k(p(x)+(k-1))x^{k-1}$, so $x^{k}|k(p(x)+(k-1))x^{k-1}$ and $x|(p(x)+(k-1))$, which in turn means that $p(0)=1-k$. Thus if $p(0)\not\in-\mathbb{N}$, this is a contradiction so $V(p(x))$ is simple. On the other hand, if $p(0)=1-k\in-{\mathbb{N}}$ then $V(p(x))$ has a unique proper nontrivial submodule $x^{k}\mathbb{K}[x]$, so $V(p(x))$ has length $2$. Finally we analyze the quotient $V(p(x))/\langle x^{k}\rangle$ for $p(0)=1-k\in-\mathbb{N}$. This quotient is finite-dimensional and we have $x\cdot x^{k-1}=0$ and

[TABLE]

so $x^{k-1}$ is a highest weight vector, and the quotient is the simple highest weight module $L(k-1)$, and we recall that we had $k-1=-p(0)$.

It remains only to verify that the submodule $x^{k}\mathbb{K}[x]$ is isomorphic to $V(p(x)-2p(0)+2)$. The submodule $x^{k}\mathbb{K}[x]$ is free of rank $1$ over $\mathbb{K}[x]$ so by Proposition 7 we have $x^{k}\mathbb{K}[x]\simeq V(\overline{p}(x))$ for some polynomial $\overline{p}$. Any isomorphism $\varphi:x^{k}\mathbb{K}[x]\rightarrow V(\overline{p}(x))$ must be a multiple of $\varphi:V(\overline{p}(x))\rightarrow V(p(x))$ defined by $\varphi(f)=x^{k}f$ since both modules are free over $\mathbb{K}[x]$ and $\varphi$ needs to be bijective map between $V(\overline{p}(x))$ and the submodule $x^{k}\mathbb{K}[x]\subset V(p(x))$. Since $\varphi$ is an isomorphism we have

[TABLE]

from which it follows that $\overline{p}(x)=p(x)+2k=p(x)+2(1-p(0))$. ∎

Actually, the family of modules $V(p(x))$ includes the lowest weight Verma-modules as seen below. Take $p=\lambda\in\mathbb{K}$. Then $q=0$ and the action on the basis $\{x^{k}\}$ of the module $V(\lambda)$ is given by

[TABLE]

Thus each $x^{k}$ is a weight vector of weight $\lambda+2k$, and $V(\lambda)$ is a weight module of lowest weight $\lambda$. As in the proposition, $V(\lambda)$ is reducible precisely when $p(0)=\lambda\in-\mathbb{N}$. The quotient is the unique simple highest weight module of highest weight $-\lambda$.

3.1. Tensor product decomposition

In this section we shall give a formula for decomposing $V(p(x))\otimes E$ when $E$ is a finite-dimensional and $V(p(x))$ is simple.

Let $L(k)$ be the unique simple $\mathfrak{sl}_{2}$ module of dimension $k+1$. For natural numbers $k\geq m$ we then have

[TABLE]

which is known as the Clebsch-Gordan formula (see for example [Maz1]).

Recall that $L(1)$ is isomorphic to the natural module; it has a basis $\{e_{1},e_{2}\}$ on which $\mathfrak{sl}_{2}$ acts by $e_{ij}\cdot e_{k}=\delta_{jk}e_{i}$. Then

[TABLE]

and using Proposition 5 we see that the $\mathfrak{sl}_{2}$ action on the tensor product is given by

[TABLE]

Lemma 9**.**

For any polynomial $p$ we have

[TABLE]

Proof.

The action of $\mathbb{K}[x]$ on the tensor product can be written $r(x)\cdot(f,g)=(rf+r^{\prime}g,rg)$, so any submodule isomorphic to $\mathbb{K}[x]$ can be generated by a single element $(f,g)$. By taking $(f,g):=(\tfrac{1}{2x}(p-p(0)),1)$ we get a $\mathbb{K}[x]$-submodule as follows: Define $\varphi:\mathbb{K}[x]\rightarrow V(p)\otimes L(1)$ by

[TABLE]

Then

[TABLE]

is a $\mathbb{K}[X]$-submodule in which $x\cdot\varphi(f)=\varphi(xf)$. We claim that $\overline{V}$ in fact is an $\mathfrak{sl}_{2}$-submodule. Since $\varphi(f)=f\cdot\varphi(1)$ it suffices to verify that $h\cdot\varphi(1)\in\overline{V}$ and $h\cdot\varphi(1)\in\overline{V}$. We have

[TABLE]

Next we claim that $y\cdot\varphi(1)=\varphi(q+\tfrac{1}{2x}(p-p(0)))$. To see this first note that

[TABLE]

We now calculate

[TABLE]

In this difference the second component is indeed zero and in the first component we obtain the following after multiplying by $2x$:

[TABLE]

and we need to show that this is zero too. Inserting our expression above for $-2xq^{\prime}$ and multiplying again by $x$ we get

[TABLE]

which is zero when $x=0$. The derivative is

[TABLE]

Hence we have shown that $y\cdot\varphi(1)=\varphi(q+\tfrac{1}{2x}(p-p(0)))\in\overline{V}$, and it follows that $\overline{V}$ is an $\mathfrak{sl}_{2}$-submodule.

Next we define $\psi:\mathbb{K}[x]\rightarrow V(p)\otimes L(1)$ by

[TABLE]

so that

[TABLE]

We claim that $\tilde{V}$ is a $\mathfrak{sl}_{2}$ submodule complementary to $\overline{V}$. Verification of this is analogous to the calculations above and we omit it here. We note however that $x\cdot\psi(f)=\psi(xf)$, $h\cdot\psi(1)=\psi(p+1)$ and $y\cdot\psi(1)=\psi(q-\tfrac{1}{2x}(p-p(0)))$.

By Propositions 6 and 7 it follows from the facts that $h\cdot\varphi(1)=\varphi(p-1)$ and $h\cdot\psi(1)=\psi(p+1)$ that $\varphi$ is an isomorphism $V(p-1)\rightarrow\overline{V}$ and that $\psi$ is an isomorphism $V(p+1)\rightarrow\tilde{V}$.

Finally we verify that $V(p)=\overline{V}\oplus\tilde{V}$. Since

[TABLE]

and since $p(0)\neq 1$ we have $(\mathbb{K}[x],0)\subset\overline{V}+\tilde{V}$ and then clearly also $(0,\mathbb{K}[x])\subset\overline{V}+\tilde{V}$ since we may form $\varphi(g)-(\tfrac{1}{2x}(p-p(0))g+g^{\prime},0)=(0,g)$. Thus $V(p)\otimes L(1)=\overline{V}+\tilde{V}$. Next, assume that $\varphi(g)=\psi(f)$, then $g=xf$ and $(\tfrac{1}{2}(p-p(0))f+(xf)^{\prime},xf)=(\tfrac{1}{2}(p+p(0))f+xf^{\prime},xf)$ so $\tfrac{1}{2}(p-p(0))f+xf^{\prime}+f=\tfrac{1}{2}(p+p(0))f+xf^{\prime}$ implying $(1-p(0))f=0$. Since $p(0)\neq 1$ we get $f=g=0$ showing that $\overline{V}\cap\tilde{V}=\{0\}$. Thus $V(p)\otimes L(1)=\overline{V}\oplus\tilde{V}$. ∎

Since each finite-dimensional module $E$ is a direct sum of modules $L(k)$, the following proposition determines the tensor product decomposition of $V(p)\otimes E$ completely.

Proposition 10**.**

We have

[TABLE]

Proof.

We proceed by induction. The statement holds trivially for $k=0$, and also for $k=1$ by Lemma 9. Using the inductive assumption and Lemma 9 we find that

[TABLE]

But on the other hand we can use the Clebsch-Gordan formula to obtain

[TABLE]

By cancelling the isomorphic summands \bigoplus_{i=0}^{k-1}V\big{(}p+(k-1)-2i\big{)} in the two above expressions we finally get

[TABLE]

and the statement of the proposition follows by induction. ∎

4. $\mathfrak{sl}_{n+1}$-modules

In this section we generalize most of the results of the previous section from $\mathfrak{sl}_{2}$ to $\mathfrak{sl}_{n+1}$.

4.1. Preliminaries

For $1\leq i,j\leq n+1$, let $e_{ij}$ be the standard basis for $\mathfrak{gl}_{n+1}$, and recall that these satisfy $[e_{ij},e_{kl}]=\delta_{jk}e_{il}-\delta_{li}e_{jk}$. Let $\mathfrak{n}\subset\mathfrak{sl}_{n+1}$ be the subalgebra with basis $\{e_{i,n+1}\;|\;1\leq i\leq n\}$. Since $\mathfrak{n}$ is abelian we have $U(\mathfrak{n})\simeq\mathbb{K}[x_{1},\ldots,x_{n}]$.

For $f\in\mathbb{K}[x_{1},\ldots,x_{n}]$ write $f^{i}:=\frac{\partial}{\partial x_{i}}f$ for the partial derivative, and define degree operators

[TABLE]

Note that $d_{i}$ and $d$ are $\mathbb{K}$-linear derivations of $\mathbb{K}[x_{1},\ldots,x_{n}]$. We also note that $d$ is invertible on the space of polynomials with zero constant term. We define maps $d^{\prime},d_{i}^{\prime}:\mathbb{K}[x_{1},\ldots,x_{n}]\rightarrow\mathbb{K}[x_{1},\ldots,x_{n}]$ by

[TABLE]

and

[TABLE]

Then the following lemma is easy to verify.

Lemma 11**.**

For all $1\leq i,j\leq n$ we have

[TABLE]

For $1\leq i\leq n$ define $x_{i}:=e_{i,n+1}$. Let $I:=\sum_{i=1}^{n+1}e_{ii}$ and for $1\leq i\leq n+1$ define $h_{i}:=e_{ii}-\frac{1}{n+1}I$ and $\overline{h}=\sum_{i=1}^{n}h_{i}$. Note that $h_{n+1}=-\sum_{i=1}^{n}h_{i}=-\overline{h}$.

We fix the basis $\{e_{ij}\;|\;1\leq i,j\leq n+1;i\neq j\}\cup\{h_{1},\ldots,h_{n}\}$ for $\mathfrak{sl}_{n+1}$ and note that for $1\leq i,j\leq n$ and $i\neq j$ we have

[TABLE]

The following lemma tells us how to commute elements of $\mathfrak{sl}_{n+1}$ with the $x_{i}$.

Lemma 12**.**

The following relations hold in $U(\mathfrak{sl}_{n+1})$ for all $1\leq i,j,k\leq n$ with $i\neq j$ and for all $m\in\mathbb{N}$.

[TABLE]

Proof.

These relations can be easily proved by induction on $m$. We verify only Equation (5). It holds trivially for $m=0$ and also for $m=1$. Assuming it holds for a fixed $m$, we have

[TABLE]

where in the second equality we used Equations (2) and (5) for $m=1$. Thus Equation (5) holds for all $m\in\mathbb{N}$ by induction. ∎

Corollary 13**.**

Let $f\in\mathbb{K}[x_{1},\ldots,x_{n}]=U(\mathfrak{n})$. The following relations hold in $U(\mathfrak{sl}_{n+1})$ for all $1\leq i,j\leq n$ with $i\neq j$.

[TABLE]

Proof.

Since the above formulas are linear in $f$, it suffices to prove them when $f$ is a monomial. We do by using Lemma 12 repeatedly. Equations (6)-(7) follows easily from Equations (1)-(2). For (8) we take $f=\prod x_{k}^{a_{k}}$ and compute

[TABLE]

where we used that $e_{ij}$ commutes with $x_{k}$ for $k\neq j$. For (9) we instead proceed by induction. The equation holds for $f=1$ and assuming that the equation holds for a fixed monomial $f$ it suffices to prove that it holds when we replace $f$ by $x_{j}f$. We divide this into cases: for $j\neq i$ we have

[TABLE]

And similarly, if we instead take $j=i$ we have

[TABLE]

Therefore Equation (9) holds by induction. ∎

4.2. Classification

In this section we shall study all possible $\mathfrak{sl}_{n+1}$-modules $M$ such that $\mathrm{Res}^{U(\mathfrak{sl}_{n+1})}_{U(\mathfrak{n})}M$ is free of rank $1$.

Let $M$ be a given $\mathfrak{sl}_{n+1}$-modules $M$ such that $\mathrm{Res}^{U(\mathfrak{sl}_{n+1})}_{U(\mathfrak{n})}$ is free of rank $1$. Without loss of generality we may assume that $M=\mathbb{K}[x_{1},\ldots,x_{n}]$ as a vector space and that $e_{i,n+1}\cdot f=x_{i}f$ for $f\in M$. For $1\leq i,j\leq n$ and $i\neq j$, define $p_{ij}:=e_{ij}\cdot 1$, $p_{ii}:=h_{i}\cdot 1$ and $q_{i}:=e_{n+1,i}\cdot 1$. Also let $\overline{p}:=\sum_{i=1}^{n}p_{ii}$ and $p:=d^{\prime}(\overline{p})$.

Lemma 14**.**

The elements $p_{ij},q_{i}\in\mathbb{K}[x_{1},\ldots,x_{n}]$ uniquely determines the module structure on $M$. Explicitly, for $i\neq j$ we have

[TABLE]

where $\overline{p}=\sum_{i=1}^{n}p_{ii}$.

Proof.

This immediately follows by acting on $1\in\mathbb{K}[x_{1},\ldots,x_{n}]$ in both sides of each equation of Corollary 13. ∎

Next we determine what relations are required between polynomials $p_{ij}$ and $q_{i}$ in order that the action in Lemma 14 should define an $\mathfrak{sl}_{n+1}$-module structure on $\mathbb{K}[x_{1},\ldots,x_{n}]$.

First, let $\mathfrak{p}\subset\mathfrak{sl}_{n+1}$ be the parabolic subalgebra spanned by $\{h_{1},\ldots,h_{n}\}$ and all $e_{i,j}$ for $1\leq i\leq n$ and $1\leq j\leq n+1$ where $i\neq j$.

First we note that

[TABLE]

so the conditions that $h_{i}\cdot h_{j}\cdot f-h_{j}\cdot h_{i}\cdot f=[h_{i},h_{j}]\cdot f=0$ for all $f$ reduces to the conditions

[TABLE]

Summing over $j$ we obtain $x_{i}\overline{p}^{i}=d(p_{ii})$ and applying $d^{\prime}$ we have

[TABLE]

Thus each $p_{ii}$ is determined up to addition of a constant by $\overline{p}$. We conclude that

[TABLE]

for some constants $c_{i}$.

Next, for $1\leq i,j,k\leq n$ with $i\neq j$ we have

[TABLE]

On the other hand,

[TABLE]

So the conditions $[h_{k},e_{ij}]\cdot f=h_{k}\cdot e_{ij}\cdot f-e_{ij}\cdot h_{k}\cdot f$ translates to the conditions

[TABLE]

Note that by Equation (10), the above equality also holds when $i=j$.

Now for $1\leq i,j,k,l\leq n$ where $i\neq j$ and $k\neq l$ we have

[TABLE]

while

[TABLE]

So the condition $[e_{ij},e_{kl}]\cdot f=e_{ij}\cdot e_{kl}\cdot f-e_{kl}\cdot e_{ij}\cdot f$ for all $f$ is equivalent to

[TABLE]

Lemma 15**.**

We have

[TABLE]

Proof.

Since $[h_{k},e_{n+1,i}]\cdot f=h_{k}\cdot e_{n+1,i}\cdot f-e_{n+1,i}\cdot h_{k}\cdot f$ for all $f$, we have

[TABLE]

where the last equality followed by using equations (10) and (11). Since the above equality holds for all $f$ we have

[TABLE]

Taking $k=i$ we obtain

[TABLE]

Note that since $q_{i}$ is a polynomial the integral above is well-defined since it needs to be divisible by $x_{i}$. ∎

Theorem 16**.**

Fix a polynomial $p\in\mathbb{K}[x_{1},\ldots,x_{n}]$ and for $1\leq i,j\leq n$ define polynomials $p_{ij}:=x_{i}\frac{\partial p}{\partial x_{j}}+\delta_{ij}p(0)/n$ and $q_{i}:=-\frac{1}{x_{i}}\int\sum_{r=1}^{n}(p_{ii}^{r}p_{ri}+x_{r}p_{ii}^{ir}+p_{ii}^{i}p_{rr})dx_{i}$.

Then the following action defines an $\mathfrak{sl}_{n+1}$-module structure on the space $M(p)=\mathbb{K}[x_{1},\ldots,x_{n}]$:

[TABLE]

where $h_{i}=e_{ii}-\frac{1}{n+1}\sum_{j=1}^{n+1}e_{jj}$.

Moreover, any $\mathfrak{sl}_{n+1}$-module $M$ for which $\mathrm{Res}_{U(\mathfrak{n})}^{U(\mathfrak{sl}_{n+1})}M$ is free of rank $1$ is isomorphic to $M(p)$ for a unique $p\in\mathbb{K}[x_{1},\ldots,x_{n}]$.

Proof.

We first verify that the definition in the theorem indeed gives an $\mathfrak{sl}_{n+1}$-module structure. Lemma 13 guarantees that for any $y\in\mathfrak{sl}_{n+1}$ we have $[y,x_{k}]\cdot f=y\cdot x_{k}\cdot f-x_{k}\cdot y\cdot f$.

By equations (10), (11), (12), and (15) we see that the relations

[TABLE]

holds for all $f$ if and only if

[TABLE]

for all $1\leq i,j,k,l\leq n$. We verify these relations for $p_{ij}=x_{i}p^{j}+\delta_{ij}c$:

[TABLE]

It remains only to show that $[e_{n+1,i},y]\cdot f=e_{n+1,i}\cdot y\cdot f-y\cdot e_{n+1,i}\cdot f$ for each $y\in\mathfrak{sl}_{n+1}$. We verify only the relation for $y=h_{k}$ here, the remaining relations are similar.

By Lemma 15, the condition $[h_{k},e_{n+1,i}]\cdot f=h_{k}\cdot e_{n+1,i}\cdot f-e_{n+1,i}\cdot h_{k}\cdot f$ for all $f$ is equivalent to

[TABLE]

For $k=i$, this equation holds by the definition of $q_{i}$, so assume that $k\neq i$. Then we have

[TABLE]

Substituting $p_{ij}=x_{i}p^{j}+\delta_{ij}p(0)/n$, the verification of the above equality reduces to a simple but long calculation which we omit here.

Finally we note that the conditions

[TABLE]

[TABLE]

reduces to the two conditions

[TABLE]

[TABLE]

which can be verified similarly.

Next we adress the uniqueness claim of the theorem. Suppose that we are given an $\mathfrak{sl}_{n+1}$-module structure on $M=\mathbb{K}[x_{1},\ldots,x_{n}]$ such that $e_{i,n+1}\cdot f=x_{i}f$ for $1\leq i\leq n$. Define $p_{ij}:=e_{i,j}\cdot 1$ and $q_{i}:=e_{n+1,i}\cdot 1$ as before. By Lemma 14, $M$ is determined by these polynomials up to isomorphism. We need to show that $M\simeq M(p)$ for some $p\in\mathbb{K}[x_{1},\ldots,x_{n}]$.

Equation (10) says that $x_{i}p_{jj}^{i}=x_{j}p_{ii}^{j}$ for all $1\leq i,j\leq n$. Summing over $j$ we obtain $x_{i}\overline{p}^{i}=d(p_{ii})$ and applying $d^{\prime}$ we have

[TABLE]

Thus each $p_{ii}$ is determined up to addition of a constant by $\overline{p}$. We conclude that

[TABLE]

for some constants $c_{i}$.

Next recall that by Equation 12 we had $x_{i}p_{kl}^{j}-x_{k}p_{ij}^{l}=\delta_{kj}p_{il}-\delta_{il}p_{kj}$. Taking $k=i$ and $j=l$ here we obtain

[TABLE]

Substituting $p_{ii}=d^{\prime}(x_{i}\overline{p}^{i})+c_{i}$ in (15) and considering the constant term it follows that $c_{i}=c_{j}$ for all $i,j$ and

[TABLE]

Define $p:=d^{\prime}(\overline{p})$. Then $p_{ii}=x_{i}p^{i}+c$, and $\overline{p}(0)=\sum_{i=1}^{n}p_{ii}(0)=nc$, so

[TABLE]

Finally, recall that by (11) we had $x_{k}p_{ij}^{k}-x_{i}p_{kk}^{j}=(\delta_{ki}-\delta_{kj})p_{ij}$. Summing over $1\leq k\leq n$ in we get

[TABLE]

Thus $p_{ij}-p_{ij}(0)=d^{\prime}\circ d(p_{ij})=d^{\prime}(x_{i}\overline{p}^{j})$. Taking $k=i$ in (11) we get $x_{i}p_{ij}^{i}-x_{i}p_{ii}^{j}=p_{ij}$ which shows that $p_{ij}(0)=0$, and we conclude that

[TABLE]

Thus we have shown that the $p_{ij}$ defined above are the same as those in $M(p)$. Finally, by Lemma 15 each $q_{i}$ is uniquely determined by the $p_{ij}$. We have therefore proved that $M\simeq M(p)$, which shows that the $\mathfrak{sl}_{n+1}$ modules of form $M(p)$ exhaust the set of modules which are free when restricted to $U(\mathfrak{n})$. ∎

4.3. Submodule structure

Proposition 17**.**

We have $M(p)\simeq M(\tilde{p})$ if and only if $p=\tilde{p}$.

Proof.

Let $\varphi:M(p)\rightarrow M(\tilde{p})$ be an isomorphism. Since $\varphi(f)=\varphi(f\cdot 1)=f\cdot\varphi(1)$, $\varphi(1)$ is a nonzero constant. For $i\neq j$ we have $\varphi(e_{ij}\cdot 1)=e_{ij}\cdot\varphi(1)$, which implies $x_{i}p^{j}=x_{i}\tilde{p}^{j}$ and $p^{j}=\tilde{p}^{j}$. Since this holds for each $j$, $p-\tilde{p}$ is a constant. Finally, $\varphi(h_{j}\cdot 1)=h_{j}\cdot\varphi(1)$ implies $x_{j}p^{j}+p(0)/n=x_{j}\tilde{p}^{j}+\tilde{p}(0)/n$ and in turn $p(0)=\tilde{p}(0)$. Thus $p=\tilde{p}$. ∎

Lemma 18**.**

Every submodule $N$ of $M(p)$ is a homogeneous ideal of form

[TABLE]

for some $k\in\mathbb{N}$.

Proof.

Let $N$ be a submodule of $M(p)$. Freeness of $M(p)$ over $\mathbb{K}[x_{1},\ldots,x_{n}]$ implies that any submodule is an ideal. Note that $D_{i}:=(h_{i}-x_{i}p^{i}-p(0)/n)\in U(\mathfrak{sl}_{n+1})$ acts as the $i$-degree operator on $M(p)$, and $n-D_{i}$ acts on $f$ by killing all terms of $f$ which has $i$-degree $n$. Thus any $f\in N$ can be reduced to each of its monomial terms. Next, let $f\in N$. Then for $i\neq j$, $(e_{ij}-x_{i}p^{j})\cdot f=x_{i}f^{j}$ which shows that $f$ can be mapped to any polynomial of the same degree by a product of such elements in $U(\mathfrak{sl}_{n+1})$. Thus $N\supset\{g\>|\>d(g)=d(f)\}$, and since $\mathbb{K}[x]$ acts freely on $N$, we have $N\supset\{g\>|\>d(g)\geq d(f)\}$. Therefore if $f\in N$ is taken to have minimal degree we see that $N$ has the form stated in the lemma. ∎

Theorem 19**.**

$M(p)$* is a simple $\mathfrak{sl}_{n+1}$-module if and only if $k:=-\frac{n+1}{n}p(0,\ldots,0)\not\in\mathbb{N}_{+}$. Otherwise, if $k\in\mathbb{N}_{+}$, the module $M(p)$ has length $2$ and the top of $M$ is a simple highest weight module.*

Proof.

Let $N\subset M(p)$ be a nonzero proper submodule, and $f\in N$ be a monomial with $\deg f$ minimal. Then

[TABLE]

Subtracting terms of degree $\geq\deg(f)$ we obtain

[TABLE]

But by the minimality of $\deg(f)$, the coefficient of $f^{i}$ must be zero and we have $\deg(f^{i})=-\frac{n+1}{n}p(0)$. This is a contradiction if $k\not\in\mathbb{N}_{+}$, so $M(p)$ is simple in this case. Conversely, assume $k\in\mathbb{N}_{+}$ and let

[TABLE]

Then $W_{k}$ is invariant under the action of each $e_{n+1,i}$ by the above calculation. $W_{k}$ is also invariant under the remaining basis elements of $\mathfrak{sl}_{n+1}$ because $\deg(x_{i}f^{j})=\deg(f)$ for all $1\leq i,j\leq n$. Thus $W_{k}$ is the unique submodule of $M(p)$. This means that $M(p)/W_{k}$ is simple finite-dimensional module, and therefore a weight module. We note that the image of $x_{1}^{k}$ in the quotient is a highest weight vector since it is annihilated by all $e_{ij}$ for $i<j$. Since

[TABLE]

we have $M(p)/W_{k}\simeq L(\lambda)$ where $\lambda\in\mathfrak{h}^{*}$ is given by $\lambda(h_{i})=\delta_{ik}\frac{n+2}{n}p(0)$. We note that $\dim M(p)/W_{k}={k+n-2\choose k-1}$, the dimension of the space of polynomials of degree less than $k$. ∎