Representations of the Lie algebra of vector fields on a sphere

TL;DR

This paper investigates modules over the Lie algebra of vector fields on a sphere, constructing simple modules and establishing an equivalence with finite-dimensional rational L_2-modules, revealing new algebraic structures.

Contribution

It introduces a new class of simple modules for the Lie algebra of vector fields on a sphere and proves their categorical equivalence to GL_2-modules.

Findings

Constructed finitely generated simple modules over the algebra of functions on the sphere.

Proved the monoidal category generated by these modules is equivalent to finite-dimensional rational GL_2-modules.

Established a categorical equivalence linking vector field modules to GL_2-representations.

Abstract

For an affine algebraic variety $X$ we study a category of modules that admit compatible actions of both the algebra of functions on $X$ and the Lie algebra of vector fields on $X$. In particular, for the case when $X$ is the sphere $\mathbb{S}^2$, we construct a set of simple modules that are finitely generated over $A$. In addition, we prove that the monoidal category that these modules generate is equivalent to the category of finite-dimensional rational $\mathrm{GL}_2$-modules.

Equations

$$\eta \cdot (a\cdot m)=\eta (a)\cdot m+a\cdot (\eta \cdot m).$$

$$a\cdot (m\otimes n)=(a\cdot m)\otimes n,$$

$$\eta (m\otimes n)=(\eta \cdot m)\otimes n+m\otimes (\eta \cdot n),$$

$$\eta \cdot (a\cdot (m\otimes n))$$

$$=(\eta (a)\cdot m)\otimes n+(a\cdot (\eta \cdot m))\otimes n+a\cdot (m\otimes (\eta \cdot n))$$

$$=\eta (a)\cdot (m\otimes n)+a\cdot (\eta \cdot (m\otimes n)).$$

$$M^{\circ }:={\mathrm{Hom}}_A(M,A).$$

$$(\eta \cdot \phi )(m)=-\phi (\eta \cdot m)+\eta (\phi (m))$$

$$\mu (f)=-\frac{\mu (p_n)f^n+\cdots +\mu (p_1)f+\mu (p_0)}{n{p}_n{f}^{n-1}+\cdots +p_1},$$

$$J=\left(\frac{\partial g_i}{\partial x_j}\right)\in {\mathrm{Mat}}_{n\times m}(A).$$

$$\{N(\det J^{\alpha,\beta})\;\big{|}\;|\alpha|=|\beta|=r,\;\det(J^{\alpha,\beta})\neq 0\}.$$

$$\sum _{i\notin \beta }{f}_i\frac{\partial }{\partial x_i}+\sum _{j\in \beta }{f}_j\frac{\partial }{\partial x_j}$$

$$\sum _{i=1}^s{f}_i\frac{\partial x_j}{\partial t_i}=\sum _{i=1}^s{f}_i{\tau }_i(x_j).$$

$$\mathrm{Vect}(X)\simeq \mathrm{Der}(A)\subset \mathrm{Der}(A_{(h)})=\underset{p=1}{\overset{n}{⨁}}{A}_{(h)}\frac{\partial }{\partial t_p},$$

$$\nu_{P}(f)=\min\left\{\sum_{i=s}^{n}\alpha_{i}\;\Bigg{|}\left(\prod_{i=1}^{s}\left(\tfrac{\partial}{\partial t_{i}}\right)^{\alpha_{i}}f\right)(P)\neq 0\;\right\}.$$

$$d(a_0)=-\left(\sum _{i=1}^{N}d(a_i)f^i+a_{i}i{f}^{i-1}d(f)\right).$$

$$\left(\sum _{i=1}^s{f}_i\frac{\partial }{\partial t_i}\right)\cdot (g\otimes u):=\sum _{i=1}^s{f}_i\frac{\partial g}{\partial t_i}\otimes u+\sum _{p=1}^s\sum _{i=1}^{s}g\frac{\partial f_i}{\partial t_p}\otimes (E_{p,i}\cdot u).$$

$$\cdots \subset h^{k+1}A\subset h^{k}A\subset h^{k-1}A\subset \cdots .$$

$$\mathrm{deg}a:=\max \{k\in \mathbb{Z}|a\in h^{k}A\}.$$

$$\mathrm{deg}m:=\max \{k\in \mathbb{Z}|m\in h^{k}A\otimes U\}.$$

$$\mathrm{deg}(M):=\inf \{\mathrm{deg}m|m\in M,m\ne 0\}.$$

$$h^{K}A\otimes U\subset M\subset h^{k}A\otimes U.$$

$$[{\Delta }_{12},{\Delta }_{23}]={\Delta }_{31},[{\Delta }_{23},{\Delta }_{31}]={\Delta }_{12},[{\Delta }_{31},{\Delta }_{12}]={\Delta }_{23}.$$

$${\Delta }_{12}=y\frac{\partial }{\partial x}-x\frac{\partial }{\partial y},{\Delta }_{23}=z\frac{\partial }{\partial y},{\Delta }_{31}=-z\frac{\partial }{\partial x}.$$

$$E_{1,2}\cdot f=X\frac{\partial f}{\partial Y},E_{2,1}\cdot f=Y\frac{\partial f}{\partial X},(E_{1,1}-E_{2,2})\cdot f=X\frac{\partial f}{\partial X}-Y\frac{\partial f}{\partial Y}.$$

$$E_{1,1}\cdot v_i=\frac{1}{2}(\alpha +m-2i)v_i,E_{1,2}\cdot v_i=(m-i+1)v_{i-1},$$

$$E_{2,1}\cdot v_i=(i+1)v_{i+1},E_{2,2}\cdot v_i=\frac{1}{2}(\alpha -m+2i)v_i.$$

$$(E_{1,1}-E_{2,2})\cdot v_i=(m-2i)v_i,(E_{1,1}+E_{2,2})\cdot v_i=\alpha v_i.$$

$${\Delta }_{2,3}(z^k{a}_k\otimes v_0)$$

$$=z\left(z^k\frac{\partial a_k}{\partial y}-k{a}_k{z}^{k-1}\frac{y}{z}\right)\otimes v_0-z^k{a}_k\frac{y}{z}\otimes \frac{\alpha }{2}{v}_0,$$

$$-(k+\frac{\alpha }{2})a_{k}y{z}^{k-1}\otimes v_0.$$

$$\left(\sum _{i=1}^s{f}_i\frac{\partial }{\partial t_i}\right)\cdot \sum _{j=1}^s(g_{j}d{t}_j)=\sum _{i,j=1}^s{f}_i\frac{\partial g_j}{\partial t_i}d{t}_j+g_{j}d\left(f_i\frac{\partial }{\partial t_i}{t}_j\right)$$

Full text

1 Introduction

In 1986 David Jordan proved simplicity of Lie algebras of polynomial vector fields on smooth irreducible affine algebraic varieties ([8], see also [9] and [4]). Yet, representation theory for this important class of Lie algebras remains largely undeveloped. The goal of the present paper is to investigate as a test case, representation theory of the Lie algebra of polynomial vector fields on a sphere $\mathbb{S}^{2}$. Previously, representation theory was developed only for the classical Lie algebras of Cartan type – polynomial vector fields on an affine space and on a torus. Representations of the Lie algebras of vector fields on an affine space were studied by Rudakov in 1974 [15].

An important classification result on representations of the Lie algebra of vector fields on a circle was established by Mathieu [12]. The case of an $N$-dimensional torus was studied by Larsson [11], Eswara Rao [6], [7], Billig [1], Mazorchuk-Zhao [13] and Billig-Futorny [2]. The culmination of this work was the proof in [3] of Rao’s conjecture on classification of irreducible weight modules for the Lie algebra of vector fields on a torus with finite-dimensional weight spaces. According to this classification, every such module is either of the highest weight type, or a quotient of a tensor module.

When we move from the torus to general affine algebraic varieties, the first difficulty that arises is the absence of a Cartan subalgebra. It was shown in [4] that even in the case of an affine elliptic curve, the Lie algebra of vector fields does not contain non-zero semisimple or nilpotent elements. This demonstrates that the theory of this class of simple Lie algebras is very different from the classical theory of simple finite-dimensional Lie algebras, where roots and weights play a fundamental role. When studying representation theory of a simple Lie algebra, one has to impose some reasonable restrictions on the class of modules, since the description of simple modules in full generality is only known for $\mathfrak{sl}_{2}$ [5].

In case of vector fields on a torus, a natural restriction is the existence of a weight decomposition with finite-dimensional weight spaces. A theorem proved in [3] states that a simple weight module $M$ for the Lie algebra of vector fields on a torus admits a cover $\widehat{M}\rightarrow M$, where $\widehat{M}$ is a module for both the Lie algebra of vector fields and the commutative algebra of functions on a torus. This suggests that a reasonable category of modules for the Lie algebra $\mathcal{V}$ of vector fields on an affine variety $X$ will be those that admit a compatible action of the algebra $A$ of polynomial functions on $X$. We refer to such modules as $A\mathcal{V}$-modules. A finiteness condition will be a requirement that the module is finitely generated over $A$.

We begin by discussing the general theory of $A\mathcal{V}$-modules, defining dual modules and tensor product in this category. Then the focus of the paper shifts to the study of the case of a sphere $\mathbb{S}^{2}$. A new feature here compared to the torus and circle is that Lie algebra of vector fields on $\mathbb{S}^{2}$, as well as its $A\mathcal{V}$-modules, are not free as modules over $A$. We construct a class of tensor modules (geometrically these are modules of tensor fields on a sphere), and prove their simplicity in the category of $A\mathcal{V}$-modules. We also show that the monoidal category generated by simple tensor modules on $\mathbb{S}^{2}$ is semisimple and is equivalent to the category of finite-dimensional rational $\mathrm{GL}_{2}$-modules.

The methods that we employ are a combination of Lie theory and commutative algebra. Hilbert’s Nullstellensatz is an essential ingredient in the proof of simplicity of tensor modules.

2 Generalities

Let $X\subset\mathbb{A}^{n}$ be an algebraic variety over an algebraically closed field ${\bf k}$ of characteristic zero. Write $A_{X}$ for the algebra of polynomial functions on $X$, and let $\mathcal{V}_{X}=\mathrm{Der}_{\bf k}(A_{X})$ be the Lie algebra of polynomial vector fields on $X$. When the variety $X$ is understood by the context we shall drop it as a subscript. Note that $\mathcal{V}$ is an $A$-module and that $A$ is a $\mathcal{V}$-module. For $a\in A$ and $\eta\in\mathcal{V}$ we shall write the latter action as $\eta(a)$.

Consider a vector space $M$ equipped with a module structure for both the associative commutative unital algebra $A$ and for the Lie algebra $\mathcal{V}$, such that the two structures are compatible in the following sense: For every $a\in A$, $\eta\in\mathcal{V}$, and $m\in M$ we have

[TABLE]

This is equivalent to saying that $M$ is a module over the smash product algebra $A\#U(\mathcal{V})$, see [14] for details. For simplicity we shall write just$A\mathcal{V}$ for $A\#U(\mathcal{V})$. A morphism of $A\mathcal{V}$-modules is a map that preserves both the $A$- and the $\mathcal{V}$-structures. The category of $A\mathcal{V}$-modules is clearly abelian since $A$-Mod and $\mathcal{V}$-Mod are.

For two $A\mathcal{V}$-modules $M$ and $N$ we may form the tensor product $M\otimes_{A}N$ which makes sense since $A$ is commutative. This is equipped with the natural $A$-module structure

[TABLE]

where the right side also equals $m\otimes(a\cdot n)$.

The Lie algebra of vector fields acts on the tensor product by

[TABLE]

as usual. We verify that these two structures are compatible in the above sense. We have

[TABLE]

This shows that $M\otimes_{A}N$ is an $A\mathcal{V}$-module, and that $A\mathcal{V}$-Mod is a monoidal category.

For any $A\mathcal{V}$-module $M$ we define

[TABLE]

The algebra $A$ acts naturally on $M^{\circ}$ by $(a\cdot\varphi)(m)=a\varphi(m)$, and we define an action of $\mathcal{V}$ on $M^{\circ}$ by

[TABLE]

for all $\eta\in\mathcal{V}$, $\varphi\in M^{\circ}$, and $m\in M$. These two actions are compatible in the sense defined above, so $M^{\circ}$ is indeed an $A\mathcal{V}$-module. The contravariant functor $M\mapsto M^{\circ}$ provides a duality on $A\mathcal{V}$-Mod.

Chart Parameters

Let $h\in A$ be a function and consider the corresponding chart for $X$ consisting of points where $h$ does not vanish: $N(h)=\{p\in X|h(p)\neq 0\}$.

Definition 1**.**

We shall say that $t_{1},\ldots,t_{s}\in A$ are chart parameters in the chart $N(h)$ provided that the following conditions are satisfied:

$t_{1},\ldots,t_{s}$* are algebraically independent over ${\bf k}$, so ${\bf k}[t_{1},\ldots,t_{s}]\subset A$.* 2. 2.

Each element of $A$ is algebraic over ${\bf k}[t_{1},\ldots,t_{s}]$. 3. 3.

For each index $p$, the derivation $\frac{\partial}{\partial t_{p}}\in\mathrm{Der}({\bf k}[t_{1},\ldots,t_{s}])$ extends to a derivation of the localized algebra $A_{(h)}$.

Note that part $2$ also implies that each element of $A_{(h)}$ is algebraic over ${\bf k}[t_{1},\ldots,t_{s}]$ since algebraic elements are closed under taking inverses. Some further consequences of the definition are given below.

Lemma 2**.**

Let $t_{1},\ldots,t_{s}\in A$ be chart parameters in the chart $N(h)$. Then

The extension of the derivation $\frac{\partial}{\partial t_{p}}\in\mathrm{Der}({\bf k}[t_{1},\ldots,t_{s}])$ to the localized algebra $A_{(h)}$ is unique. 2. 2.

$\mathrm{Der}(A_{(h)})=\displaystyle\bigoplus_{p=1}^{s}A_{(h)}\frac{\partial}{\partial t_{p}}$.

Proof.

The uniqueness in the first claim follows from part $2$ of Definition 1. Indeed let $f$ be a non-zero element of $A_{(h)}$ and let $p_{n}T^{n}+\cdots+p_{1}T+p_{0}$ be its minimal polynomial with $p_{i}\in{\bf k}[t_{1},\ldots,t_{s}]$. Let $\mu\in\mathrm{Der}(A_{(h)})$. Then

[TABLE]

hence every derivation of $A_{(h)}$ is uniquely determined by its values on ${\bf k}[t_{1},\ldots,t_{s}]$.

The second claim follows from part $3$ of Definition 1: For any $d\in\mathrm{Der}(A_{(h)})$, let $d^{\prime}=\sum_{i=1}^{s}d(t_{i})\frac{\partial}{\partial t_{i}}$. Then $d$ and $d^{\prime}$ are both derivations of $A_{(h)}$ which are equal on ${\bf k}[t_{1},\ldots,t_{s}]\subset A_{(h)}$, hence $d=d^{\prime}$. Moreover, the expression of a derivation as an $A_{(h)}$-combination of $\{\frac{\partial}{\partial t_{1}},\ldots,\frac{\partial}{\partial t_{s}}\}$ is unique; this is seen when applying such a combination to $t_{1},\ldots,t_{s}$. ∎

Our prototypical example is the following. Let $X=\mathbb{S}^{2}$ and let $h=z$. Then $x,y$ are chart parameters outside the equator $z=0$: First ${\bf k}[x,y]\subset A$. Next we have $q(z)=0$ for $q(T)=T^{2}+(x^{2}+y^{2}-1)$ so $z$ is algebraic over ${\bf k}[x,y]$. Since $x,y,z$ generate the ring $A$, every element of $A$ is algebraic over ${\bf k}[x,y]$. For the third part we note that $0=\frac{\partial}{\partial x}(x^{2}+y^{2}+z^{2}-1)=2x+2z\frac{\partial z}{\partial x}$, so for the extension of $\frac{\partial}{\partial x}$ to $A_{(z)}$ we have $\frac{\partial z}{\partial x}=-\frac{x}{z}$ which uniquely determines the derivation $\frac{\partial}{\partial x}$ on $A_{(z)}$. By symmetry the same goes for $\frac{\partial}{\partial y}$.

An Atlas for $X$

Let $X$ be the zero locus of polynomials $g_{1},\ldots,g_{n}\in{\bf k}[x_{1},\ldots,x_{m}]$ such that $A={\bf k}[x_{1},\ldots,x_{m}]/\langle g_{1},\ldots,g_{n}\rangle$, and define

[TABLE]

Let $F$ be the field of fractions of $A$ and define $r:=\operatorname{rank}_{F}J$. This means that there exists some $r\times r$-minor which is a nonzero element of $A$. By the Nullstellensatz we have $\operatorname{rank}_{F}J=\max_{P\in X}\operatorname{rank}_{\bf k}J(P)$, and when $X$ is smooth, $\operatorname{rank}_{\bf k}J(P)$ is independent of the point $P$ (see [16] Section 2.1.4).

We now consider $r\times r$-minors of $J$. For $\alpha\subset\{1,\ldots,n\}$ and $\beta\subset\{1,\ldots,m\}$ we write $J^{\alpha,\beta}$ for the corresponding $r\times r$-minor of $J$. For $h\in A$, let $N(h)=\{P\in X\;|\;h(P)\neq 0\}$.

Lemma 3**.**

Let $X$ be smooth and let $r=\operatorname{rank}_{F}J$. Then the following set of charts forms an atlas for $X$:

[TABLE]

Proof.

Let $P\in X$. Since $X$ is smooth, $\operatorname{rank}_{\bf k}J(P)=\operatorname{rank}_{F}J=r$. Thus there exists a nonzero minor in $J(P)$: $\det(J^{\alpha,\beta}(P))\neq 0$ for some $\alpha$ and $\beta$ of order $r$, so $P\in N(\det J^{\alpha,\beta})$. Thus the above atlas covers $X$. ∎

From here on we shall fix this atlas for the variety $X$.

For the sphere $X=\mathbb{S}^{2}$, we shall sometimes write $(x,y,z)$ for $(x_{1},x_{2},x_{3})$. We have $A=k[x,y,z]/\langle g\rangle$ with $g=x^{2}+y^{2}+z^{2}-1$, so $J=(2x\;\;2y\;\;2z)$. Here $\operatorname{rank}_{F}J=1$ and there are three nonzero minors: $2x,2y$, and $2z$. So in this case our charts are $N(x)$, $N(y)$, and $N(z)$ - each obtained by removing a great circle from the sphere.

As described in [4], the matrix $J$ also grants an explicit description of the vector fields on $X$: for $f_{i}\in A$, the combination $\sum_{i=1}^{m}f_{i}\frac{\partial}{\partial x_{i}}$ is a vector field on $X$ if and only if the vector $(f_{1},\ldots,f_{m})$ belongs to the kernel of $J$.

Lemma 4**.**

Let $r=\operatorname{rank}_{F}J$ and let $J^{\alpha,\beta}$ be a minor of size $r$ with $h=\det J^{\alpha,\beta}\neq 0$. Then $\{x_{i}\;|\;i\not\in\beta\}$ are chart parameters in the chart $N(h)$.

Proof.

Consider $\mathrm{Der}_{\bf k}F$. It is easy to see that for any $s\in F$ and $\eta\in\mathrm{Der}(A)$ we have $s\eta\in\mathrm{Der}_{\bf k}F$ and conversely, for any $\mu\in\mathrm{Der}_{\bf k}F$ there exists $q\in A$ such that $q\mu\in\mathrm{Der}_{\bf k}(A)$. Then derivations of $F$ can be written as $\sum_{i=1}^{m}f_{i}\frac{\partial}{\partial x_{i}}$ where $(f_{1},\ldots,f_{m})$ are solutions over $F$ of a system of linear homogeneous equations with matrix $J$. Since $\operatorname{rank}_{F}J=r$, we can keep only the rows $\alpha$ and choose variables $\{f_{i}\;|\;i\in\beta\}$ as leading, and $\{f_{i}\;|\;i\not\in\beta\}$ as free. Writing down the fundamental solutions, we get derivations $\tau_{j}=\frac{\partial}{\partial x_{j}}+\sum_{i\in\beta}f_{i,j}\frac{\partial}{\partial x_{i}}\in\mathrm{Der}_{\bf k}F$ for each $j\not\in\beta$. Note that only $h$ may appear in the denominators of $f_{i,j}$, hence $h\tau_{j}\in\mathrm{Der}_{\bf k}(A)$ for all $j\not\in\beta$. This implies that $\{x_{i}\;|\;i\not\in\beta\}$ are algebraically independent. Indeed if $p$ is a polynomial in $\{x_{i}\;|\;i\not\in\beta\}$ which vanishes in $F$, we can apply a sequence of derivations $\tau_{j}$ which brings $p$ to $1$ and obtain a contradiction $1=0$. If all free variables have zero values, the solution of a homogeneous system is trivial. This implies that every derivation of $F$ which is zero on ${\bf k}[x_{i}\;|\;i\not\in\beta]$, is zero on $F$. By [10] Chapter VIII, Prop. 5.2, $F$ is algebraic over ${\bf k}(x_{i}\;|\;i\not\in\beta)$. ∎

Remark 5**.**

In the above lemma, we may assume without loss of generality that $\{i\;|\;i\not\in\beta\}=\{1,\ldots,s\}$ and $\{i\;|\;i\in\beta\}=\{s+1,\ldots,n\}$.

If we treat the chart parameters $t_{i}=x_{1},\ldots,t_{s}=x_{s}$ as independent variables and $x_{s+1},\ldots,x_{n}$ as dependent we can write a derivation

[TABLE]

simply as $\sum_{i=1}^{s}f_{i}\frac{\partial}{\partial t_{i}}$ with understanding that for $j\in\beta$ we have

[TABLE]

Embedding of Vector Fields

The embedding $A\subset A_{(h)}$ gives a corresponding embedding of polynomial vector fields:

[TABLE]

In other words, a polynomial vector field on $X$ can be written as $\sum_{i=1}^{n}f_{i}\frac{\partial}{\partial t_{i}}\in\mathrm{Der}(A_{(h)})$ for some unique $f_{i}\in A_{(h)}$.

Valuation

For each point $P\in N(h)$, define $\nu_{P}:A_{(h)}\setminus\{0\}\rightarrow\mathbb{N}$ by

[TABLE]

Then $\nu_{P}$ is well-defined by the following lemma.

Lemma 6**.**

We have $\nu_{P}(f)<\infty$ for each nonzero $f\in A_{(h)}$.

Proof.

First of all, $\nu_{P}$ is finite on ${\bf k}[t_{1},\ldots,t_{s}]\subset A_{(h)}$ as it is bounded by the degree function. Now any $f\in A_{(h)}$ is by definition algebraic over ${\bf k}[t_{1},\ldots,t_{s}]$ so we can consider its minimal polynomial $m(X)=\sum_{i=0}^{N}a_{i}X^{k}$, where $a_{i}\in{\bf k}[t_{1},\ldots,t_{s}]$ and $m(f)=0$ and $a_{0}\neq 0$. We claim that $\nu_{P}(f)\leq\nu_{P}(a_{0})$. To prove this, suppose for contradiction that $\left(\prod_{i=1}^{s}(\tfrac{\partial}{\partial t_{i}})^{\alpha_{i}}f\right)(P)=0$ for all $\alpha_{1}+\cdots+\alpha_{s}\leq\nu_{P}(a_{0})$. Pick $\alpha_{i}$’s such that $\left(\prod_{i=1}^{s}(\tfrac{\partial}{\partial t_{i}})^{\alpha_{i}}a_{0}\right)(P)\neq 0$. Then for any derivation $d=\frac{\partial}{\partial t_{p}}$ we have

[TABLE]

But by assumption, both $f(P)=0$ and $d(f)(P)=0$ so the right hand side is zero at $P$, hence also $d(a_{0})(P)=0$. Iterating this $\nu_{P}(a_{0})$ times we get $\left(\prod_{i=1}^{s}(\tfrac{\partial}{\partial t_{i}})^{\alpha_{i}}a_{0}\right)(P)=0$, a contradiction. ∎

Note that the above proof uses our assumption that $\mathrm{char}\;{\bf k}=0$.

3 A class of $A\mathcal{V}$-modules

Let $s:=\dim X$. Let $\{t_{1},\ldots,t_{s}\}$ be chart parameters in a chart $N(h)$ and let $A_{(h)}$ be the localization of the algebra $A$ at $h$. For any $\mathfrak{gl}_{s}$-module $U$ we consider the space $A_{(h)}\otimes_{\bf k}U$. The algebra $A$ acts on this space by multiplication on the left factor.

The proof of the following lemma is straightforward and we leave it to the reader.

Lemma 7**.**

Let $U$ be a finite-dimensional $\mathfrak{gl}_{s}$-module. Consider the vector fields $\mathcal{V}$ as embedded in $\mathrm{Der}(A_{(h)})=\bigoplus_{i=1}^{s}A_{(h)}\frac{\partial}{\partial t_{i}}$. Define an action of $\mathcal{V}$ on $A_{(h)}\otimes U$ by

[TABLE]

Here $E_{p,i}$ is a standard basis element of $\mathfrak{gl}_{s}$, $g\in A_{(h)}$, and $u\in U$. This equips $A_{(h)}\otimes U$ with the structure of an $A\mathcal{V}$-module.

The algebra $A_{(h)}$ has a natural doubly infinite filtration with respect to powers of $h$:

[TABLE]

For each $a\in A_{(h)}$ we define its degree by

[TABLE]

We extend our notion of degree to elements of $A_{(h)}\otimes U$ in the natural way: for nonzero $m\in A_{(h)}\otimes U$ we define

[TABLE]

If $M\subset A_{(h)}\otimes U$ is a nonzero submodule, we define

[TABLE]

Let $M\subset A_{(h)}\otimes U$ be a submodule. We shall call $M$ bounded if $\deg(M)$ is finite. It is easy to see that if $M$ is finitely generated over $A$ then it is bounded. Conversely, every bounded submodule in $A_{(h)}\otimes U$ is finitely generated over $A$. Indeed, as an $A$-module $h^{k}A\otimes U$ is isomorphic to $A\otimes U$ and since $A$ is noetherian, every submodule in a finitely generated $A$-module is finitely generated.

On the other hand, we shall call $M$ dense if $M\supset h^{k}A\otimes U$ for some $k$. Note that $M$ is both dense and bounded when there exist two integers $K\geq k$ such that

[TABLE]

4 The Sphere

From here on we shall focus on the case when $X$ is the sphere $\mathbb{S}^{2}$. Some results still hold in a more general setting.

Let $X=\mathbb{S}^{2}\subset\mathbb{A}^{3}$. With notation as above we have $A={\bf k}[x_{1},x_{2},x_{3}]/\langle x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-1\rangle$. However, we shall sometimes write $x,y,z$ for $x_{1},x_{2},x_{3}$. Let $\Delta_{ij}=x_{j}\frac{\partial}{\partial x_{i}}-x_{i}\frac{\partial}{\partial x_{j}}$. Then it is easy to check that $\mathcal{V}$ is generated by $\Delta_{12}$, $\Delta_{23}$, and $\Delta_{31}$ as an $A$-module, and that these generators satisfy

[TABLE]

However, $\mathcal{V}$ is not a free $A$-module since we have the relation $x_{1}\Delta_{23}+x_{2}\Delta_{31}+x_{3}\Delta_{12}=0$.

In the chart $N(z)$ with chart parameters $\{x,y\}$ these generating vector fields are expressed as

[TABLE]

5 Explicit construction of modules

The Lie algebra $\mathfrak{sl}_{2}$ acts naturally on ${\bf k}[X,Y]$ by

[TABLE]

For any $\alpha\in{\bf k}$ we may extend this to a $\mathfrak{gl}_{2}$-module ${\bf k}_{\alpha}[X,Y]$ by requiring $(E_{1,1}+E_{2,2})\cdot f=\alpha f$.

Homogeneous components are preserved by this action, and ${\bf k}_{\alpha}[X,Y]=\bigoplus_{m\geq 0}U_{m}^{\alpha}$ where $U_{m}^{\alpha}$ is the homogeneous component of degree $m$. Explicitly, for $m\in\mathbb{N}$ and for $0\leq i\leq m$, we define $v_{i}^{m}:={m\choose i}X^{i}Y^{m-i}$ (we shall drop the upper $m$ when it is understood by the context). Then $U_{m}^{\alpha}=\mathrm{span}\{v_{0}^{m},\ldots,v_{m}^{m}\}$ is the homogeneous component of degree $m$, and the action on these basis elements is given by

[TABLE]

[TABLE]

In particular this implies that

[TABLE]

Here $v_{-1}$ and $v_{m+1}$ are [math] by definition. When restricted to $\mathfrak{sl}_{2}$, $U_{m}^{\alpha}$ is the unique simple module of dimension $m+1$. In particular, any finite-dimensional simple $\mathfrak{gl}_{2}$-module is isomorphic to $U_{m}^{\alpha}$ for a unique $\alpha$ and $m$.

Modules of rank $1$

We first consider the case $m=0$. Here $U_{0}^{\alpha}$ is one dimensional and the identity acts by $\alpha$. We consider the chart $N(z)$ on $\mathbb{S}^{2}$ with chart parameters $\{x,y\}$.

Proposition 8**.**

The module $A_{(z)}\otimes U_{0}^{\alpha}$ contains a bounded $A\mathcal{V}$-submodule if and only if $\alpha\in 2\mathbb{Z}$.

Proof.

Let $M$ be a bounded submodule of $A_{(z)}\otimes U_{0}^{\alpha}$. Let $w$ be a non-zero element of $M$ of lowest possible degree $k$. Then $w$ can be expressed as $w=\sum_{i\geq k}z^{i}a_{i}\otimes v_{0}$ where $a_{k}\neq 0$. We compute

[TABLE]

which modulo the space $z^{k}A\otimes U_{0}^{\alpha}$ equals

[TABLE]

So by the minimality of $k$ we must have $\alpha=-2k$. On the other hand it is easy to check that for $\alpha=2k$ the space $A^{\alpha}=z^{k}\otimes U_{0}^{\alpha}$ is an $A\mathcal{V}$-submodule in $A_{(z)}\otimes U_{0}^{\alpha}$. ∎

Higher rank

Module of $1$-forms on $X$

In this section we consider for a moment an arbitrary $s$-dimensional variety $X$. The space of $1$-forms $\Omega$ is an $A\mathcal{V}$-module where $A$ acts by left multiplication and vector fields act as follows:

[TABLE]

[TABLE]

By identifying $e_{i}\leftrightarrow dt_{i}$ we see that $\Omega\subset A_{(h)}\otimes V$, where the action on $V$ now is $E_{p,i}e_{j}=\delta_{i,j}e_{p}$ which shows that $V$ is the natural $\mathfrak{gl}_{s}$-module.

For $X=\mathbb{S}^{2}$ this means that $\Omega\subset A_{(z)}\otimes U_{1}^{1}$. In this case, the submodule $\Omega$ is generated by $dz=-z^{-1}xdx-z^{-1}ydy$.

Module of vector fields on $X$

Similarly, the Lie algebra $\mathcal{V}$ of vector fields themselves forms an $A\mathcal{V}$-module in a natural way: $A$ acts by left multiplication, and $\mathcal{V}$ acts adjointly. We may rewrite this $\mathcal{V}$-action in the following way:

[TABLE]

[TABLE]

Comparing with the definition $(1)$, we see that $\mathcal{V}$ is isomorphic to a submodule of $A_{(h)}\otimes U$ with $v_{i}\leftrightarrow\frac{\partial}{\partial t_{i}}$. The action on this module $\mathfrak{gl}_{s}$-module $U$ is seen to be given by $E_{p,i}\cdot v_{j}=-\delta_{p,j}v_{i}$, which shows that $U$ is the dual of the natural $\mathfrak{gl}_{s}$-module.

For $X=\mathbb{S}^{2}$ this means that $\mathcal{V}\subset A_{(z)}\otimes U_{1}^{-1}$. This $A\mathcal{V}$-submodule is generated by $\Delta_{1,2}=y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}$.

Proposition 9**.**

Let $M$ be an $A\mathcal{V}_{\mathbb{S}^{2}}$-submodule of $A_{(z)}\otimes U$, where $U$ is a finite-dimensional $\mathfrak{gl}_{2}$-module. Then for $\sum_{k}g_{k}\otimes u_{k}\in M$ we also have $\sum_{k}(zg_{k}\otimes E_{i,j}\cdot u_{k})\in M$ for all $1\leq i,j\leq 2$. In other words, $M$ is closed under the operators $z\otimes E_{i,j}$.

Proof.

It suffices to prove the statement for a single term $g\otimes u$. For each vector field $\mu\in\mathcal{V}$ and for each function $f\in A$ we have

[TABLE]

Taking $\mu=\Delta_{2,3}$ we obtain the following element in $M$:

[TABLE]

[TABLE]

[TABLE]

Taking $f=x$ we obtain $zg\otimes E_{1,2}u\in M$. If we instead take $f=y$ we obtain $zg\otimes E_{2,2}u$. Analogously, by taking $\mu=\Delta_{3,1}$ and $f=x,y$ we obtain $zg\otimes E_{1,1}u\in M$ and $zg\otimes E_{2,1}u\in M$. ∎

In what follows we shall use the following version of Hilbert’s Nullstellensatz, see [16, Section 1.2.2].

Lemma 10**.**

Let $I\lhd A$ be an ideal. Suppose that $g\in A$ satisfies $g(P)=0$ at all points $P\in X$ for which $f(P)=0$ for all $f\in I$. Then $g^{k}\in I$.

Proposition 11**.**

Let $U$ be a finite-dimensional irreducible $\mathfrak{gl}_{2}$-module. Then every nonzero $A\mathcal{V}_{\mathbb{S}^{2}}$-submodule of $A_{(z)}\otimes U$ is dense.

Proof.

Since $U$ is simple and finite-dimensional it has form $U_{m}^{\alpha}$ as above. Let $M\subset A_{(z)}\otimes U$ be a submodule and define

[TABLE]

Then $I$ is an ideal of $A$. To show that $M$ is dense we need to show that $z^{N}\in I$ for some $N$.

Let $v\in M$ and express this element in the form $v=\sum_{i=0}^{m}f_{i}\otimes v_{i}$, with $f_{i}\in A_{(z)}$. In fact we may assume that $f_{i}\in A$ (otherwise just multiply by a power of $z$). By Lemma 9, $M$ is closed under the operator $z\otimes E_{1,2}$, so we obtain $z^{k}f_{0}\otimes v_{0}\in M$ for some $k$ and for some nonzero $f_{0}$. Acting by $z\otimes E_{2,1}$ repeatedly on this element we get $z^{k+i}f_{0}\otimes v_{i}$, so in particular we have $z^{N}f_{0}\otimes U\subset M$, which shows that $z^{N}f_{0}\in I$ so $I$ is non-zero.

We now aim to apply Hilbert’s Nullstellensatz to the function $g=z$. Fix $P\in\mathbb{S}^{2}$ with nonzero $z$-coordinate. We need to show that there exists $f\in I$ with $f(P)\neq 0$. We had already found $z^{N}f_{0}\in I$ so if $f_{0}(P)\neq 0$ we are done. Otherwise, consider the element $z\frac{\partial}{\partial x}(z^{N+1}f_{0}\otimes v_{0})\in M$. This expands as

[TABLE]

and since the second term lies in $z^{N}f_{0}(A\otimes U)\subset M$, we also get $z^{N+2}\frac{\partial f_{0}}{\partial x}\otimes v_{0}\in M$. This shows that $z^{N^{\prime}}\frac{\partial f_{0}}{\partial x}\in I$ for some $N^{\prime}$, and by symmetry we also get $z^{N^{\prime}}\frac{\partial f_{0}}{\partial y}\in I$. Since $\nu_{P}(f_{0})$ is finite there is some product $d$ of derivations with $d(f_{0})(P)\neq 0$. So acting repeatedly with elements as above we eventually obtain $z^{K}d(f_{0})\in M$ and $z^{K}d(f_{0})$ is nonzero at $P$. By the Nullstellensatz this means that $z^{K}\in I$ for some $K$, which in turn means that $M$ is dense. ∎

Corollary 12**.**

When $U$ is an irreducible $\mathfrak{gl}_{s}$-module there exists at most one simple $A\mathcal{V}_{\mathbb{S}^{2}}$-submodule of $A_{(z)}\otimes U$.

Proof.

Let $M$ and $M^{\prime}$ be simple submodules in $A_{(z)}\otimes U$. By Proposition 11 both modules are dense, so they both contain $z^{N}A\otimes U$ for large enough $N$. Thus $M\cap M^{\prime}$ is a nonzero submodule of both $M$ and $M^{\prime}$ so by simplicity we must have $M=M^{\prime}$. ∎

6 Tensor modules

Consider two charts $N(h)$ and $N(\tilde{h})$ in our atlas for $X$. Let $t_{1},\ldots,t_{s}$ be chart parameters for $N(h)$ and let $\tilde{t}_{1},\ldots,\tilde{t}_{s}$ be chart parameters for $N(\tilde{h})$.

Let $V$ be the natural $\mathfrak{gl}_{s}$-module with basis $\{e_{1},\ldots,e_{s}\}$. Define an $A_{(\tilde{h})}$-linear map $C:A_{(h,\tilde{h})}\otimes V\rightarrow A_{(h,\tilde{h})}\otimes V$ by

[TABLE]

Note that $C$ is invertible.

From now on we shall understand all $\mathrm{GL}_{s}$-modules (resp. $\mathfrak{gl}_{s}$-modules) as $\mathrm{GL}(V)$-modules (resp. $\mathfrak{gl}(V)$-modules).

For a finite-dimensional rational $\mathrm{GL}_{s}$-module $U$, write $\rho:\mathrm{GL}_{s}\rightarrow\mathrm{GL}(U,k)$ for the corresponding representation. Then we may consider $\rho(C)$ as an element of $\mathrm{GL}(U,A_{(h)})$. We can also consider it as a map $A_{(h,\tilde{h})}\otimes U\rightarrow A_{(h,\tilde{h})}\otimes U$.

Denote by $\mathfrak{T}$ the full subcategory of $A\mathcal{V}$-Mod consisting of those objects $M$ that satisfy

• •

$M$ is finitely generated as an $A$-module,

• •

For each chart $N(h)$ in our fixed atlas there exists an injective $A\mathcal{V}$-module homomorphisms $\varphi_{h}:M\rightarrow A_{(h)}\otimes U$,

• •

The following diagram commutes for each pair of charts $(N(h),N(\tilde{h}))$:

[TABLE]

The objects of the category $\mathfrak{T}$ will be called tensor modules.

We point out that the modules $\Omega$ and $\mathcal{V}$ are in fact tensor modules. For $\Omega$, the natural $\mathfrak{gl}_{s}$-module appearing in its construction is identified with $V$ via $e_{i}\leftrightarrow dt_{i}$. Then the transformation $C$ corresponds to the usual change of variables formula for the differentials. Clearly all compatibility conditions are satisfied and $\Omega$ is a tensor module.

The $\mathfrak{gl}_{s}$-module corresponding to $\mathcal{V}$ is the dual of the natural module $(V^{*},\rho^{*})$. It can easily be checked that $\rho^{*}(C)$ corresponds to the change of variables formula for partial derivatives.

The category of tensor modules is also closed under taking tensor products and duals, in particular we have $\Omega\simeq\mathcal{V}^{\circ}$.

Finally, for the sphere the rank $1$ modules $A^{\alpha}$ (for $\alpha\in 2\mathbb{Z}$) are also tensor modules with $\rho(C)=\det(C)^{\frac{\alpha}{2}}$.

Theorem 13**.**

Let $M\in\mathfrak{T}$ be a tensor module on $\mathbb{S}^{2}$ corresponding to a simple rational $\mathrm{GL}_{2}$-module $U$. Then $M$ is a simple $A\mathcal{V}$-module.

Proof.

Let $M^{\prime}$ be a nonzero submodule of $M$ and define

[TABLE]

Then $I$ is an ideal and it does not depend on the chart we use. Since $M$ is bounded, Proposition 11 implies that $z^{N}\in I$. By symmetry we also have $x^{N}\in I$ and $y^{N}\in I$ for some large enough $N$. But then the set of common zeros $V(I)\subset X$ is empty, and Hilbert’s weak Nullstellensatz gives $1\in I$. In view of the definition of $I$ this says that $M=M^{\prime}$. ∎

Theorem 14**.**

When $\frac{m-\alpha}{2}\in\mathbb{Z}$ the vector

[TABLE]

generates a bounded $A\mathcal{V}$-submodule inside $A_{(z)}\otimes U_{m}^{\alpha}$. On the other hand, when $\frac{m-\alpha}{2}\not\in\mathbb{Z}$, the module $A_{(z)}\otimes U_{m}^{\alpha}$ contains no bounded submodules.

Proof.

We first prove the second part. Let $M$ be a nonzero bounded submodule of $A_{(z)}\otimes U^{\alpha}_{m}$. Since $M$ is dense, it contains a vector of form $z^{N}\otimes v_{0}$ for large enough $N$. We may pick such a vector with minimal $N$.

We now compute

[TABLE]

By the minimality of $N$ we must have $N+\frac{1}{2}(\alpha-m)=0$ which since $N$ is an integer means that $\frac{m-\alpha}{2}\in\mathbb{Z}$.

To prove the first statement, assume that $\frac{m-\alpha}{2}\in\mathbb{Z}$. First note that $w_{m}$ gives a correct generator for the cases discussed above: vector fields $\mathcal{V}$, $1$-forms $\Omega$, and all rank $1$-modules $A^{\alpha}$. We now proceed by induction on $m$. Dropping the tensor signs we may write $w_{m}$ as

[TABLE]

Consider three $\mathfrak{sl}_{2}$-submodules $U_{m}$, $U_{n}$, and $U_{m+n}$ of ${\bf k}[X,Y]$. The map $\varphi:U_{m}\otimes U_{n}\rightarrow U_{m+n}$ given by multiplication in ${\bf k}[X,Y]$, ($f\otimes g\mapsto fg$), is a surjective homomorphism of $\mathfrak{sl}_{2}$-modules. Introducing an action of the identity gives a corresponding surjective homomorphism of $\mathfrak{gl}_{2}$-modules: $U_{m}^{\alpha}\otimes U_{n}^{\beta}\rightarrow U_{m+n}^{\alpha+\beta}$. This morphism can now be further extended to a surjective homomorphism of $A\mathcal{V}$-modules:

[TABLE]

By the inductive assumption, $w_{m}$ and $w_{n}$ generate bounded submodules in $A_{(z)}\otimes U_{m}^{\alpha}$ and $A_{(z)}\otimes U_{n}^{\beta}$ respectively, so $w_{m}\otimes w_{n}$ generates a bounded submodule in $(A_{(z)}\otimes U_{m}^{\alpha})\otimes(A_{(z)}\otimes U_{n}^{\beta})$.

Applying our multiplication map we conclude that $\varphi(w_{m}\otimes w_{n})$ generates a bounded submodule in $A_{(z)}\otimes U_{m+n}^{\alpha+\beta}$. But

[TABLE]

This concludes the proof. ∎

7 Tensor product decomposition

Note in $A_{(h)}\otimes_{A}A_{(h)}$ we have

[TABLE]

This shows that $A_{(h)}\otimes_{A}A_{(h)}\simeq A_{(h)}$. Therefore, from the classical decomposition of $\mathfrak{gl}_{2}$-modules it follows that

[TABLE]

We shall show that our $A\mathcal{V}$-modules which appear as submodules in $A_{(h)}\otimes U$ respect this decomposition.

Theorem 15**.**

In the category of tensor modules on the sphere, tensor products of simple modules decompose as a direct sums of simple tensor modules.

Proof.

Let $M$ and $N$ be simple tensor modules embedded as $M\subset A_{(z)}\otimes U_{m}^{\alpha}$ and $N\subset A_{(z)}\otimes U_{n}^{\beta}$ with $m\geq n$. Then

[TABLE]

Write $\pi_{k}$ for the projection onto the $k$-th direct summand:

[TABLE]

Then $\mathrm{id}_{A_{(z)}\otimes(U_{m}^{\alpha}\otimes U_{n}^{\beta})}=\bigoplus_{k=0}^{n}\pi_{k}$. We shall show that this decomposition still holds when restricted to the subspace $M\otimes_{A}N$. For this it suffices to check that $\pi_{k}(M\otimes N)\subset M\otimes N$. Let $v\in M\otimes_{A}N$. By the density of $M$ and $N$, we see that $M\otimes N$ is dense in $\bigoplus_{i=0}^{n}A_{(z)}\otimes U^{\alpha+\beta}_{m+n-2i}$ as well. This implies that $\pi_{k}(M\otimes N)$ is nonzero for each $k$, so $\pi_{k}(M\otimes N)$ is a nonzero bounded submodule of $A_{(z)}\otimes U_{m+n-2k}^{\alpha+\beta}$, and by previous results $\pi_{k}(M\otimes N)$ is dense and simple.

Now for arbitrary $j\geq 0$ we have $z^{j}v=\sum_{k}\pi_{k}(z^{j}v)$, and for $j$ large enough we get $\pi_{k}(z^{j}v)\in M\otimes N$ by the density of $M\otimes N$.

Next, by the simplicity of $\pi_{k}(M\otimes N)$, we have

[TABLE]

Thus we have shown that $id_{M\otimes N}=\bigoplus\pi_{k}|_{M\otimes N}$ and hence

[TABLE]

∎

Corollary 16**.**

The category of finite-dimensional rational $\mathrm{GL}_{2}$-modules is equivalent to a full subcategory of the category $\mathfrak{T}$ of tensor modules on the sphere. Moreover, this subcategory is generated by $A^{-2}$ and $\Omega$ as a monoidal abelian category.

Proof.

The equivalence is provided by the functor $F:U\mapsto\text{soc}(A_{(z)}\otimes U)$. This is a bijection between simple finite-dimensional rational $\mathrm{GL}_{2}$-modules and simple tensor modules.

For the second statement we note that the tensor products of $A\mathcal{V}$-modules of rank $1$ is given by

[TABLE]

In particular, we get $A^{2k}=(A^{2})^{\otimes k}$ and $A^{-2k}=(A^{-2})^{\otimes k}$ for any $k\in\mathbb{N}$. We also have

[TABLE]

Let $M$ be a bounded submodule of $A_{(z)}\otimes U_{m}^{\alpha}$. Then $\frac{m-\alpha}{2}$ is an integer by Lemma 14. Now let $V=U_{1}^{1}$ be the natural $\mathfrak{gl}_{2}$-module. It is well known that $U_{m}\subset V^{\otimes m}$ as $\mathfrak{sl_{2}}$-modules, so we get $U_{m}^{\alpha}\subset A^{\alpha-m}\otimes V^{\otimes n}$. This in turn implies that $M\subset A^{\alpha-m}\otimes_{A}\Omega^{\otimes m}$.

Thus the category $\mathfrak{T}$ is generated by $A^{2}$, $A^{-2}$, and $\Omega$ as a monoidal abelian category. However, $A^{2}$ is a direct summand of $\Omega\otimes\Omega$ so it may be dropped as a generator. ∎