§ 1. The Lie algebra sl(2, k) and its representations

§ 1. The Lie algebra $\mathfrak{s}\mathfrak{l}$ (2 $\boldsymbol{, k}$ ) and its representations

1. Canonical basis of $\mathfrak{s}\mathfrak{l}$ (2 $\boldsymbol{, k}$ )

Lemma 1 00XS

Let A be an associative algebra over $k,H$ and $X$ elements of A such that $[H, X] = 2X$ .

(i) $[H, X^n] = 2nX^n$ for any integer $n\geq 0$ .

(ii) If $Z$ is an element of A such that $[Z, X] =H$ , then, for any integer $n >0$ ,

[Z, X^n] =nX^{n-1}(H+n-1) =n(H-n+ 1)X^{n-1}

The map $T \rightarrow [H, T]$ from A to A is a derivation, which implies (i). With the assumptions in (ii),

[Z, X^n] =\sum_{i+j=n-1}X^iHX^j

=\sum_{i+j=n-1}(X^iX^jH+X^i2jX^j)

=nX^{n-1}H+ 2X^{n-1}\frac{n(n- 1)}{2}

=nX^{n-1}(H+n-1)

On the other hand, $X^{n-1}(H+n-1) = (H-n+ 1)X^{n-1}$ by (i). Q.E.D.

Recall that we denote by $\mathfrak{s}\mathfrak{l}(2, k)$ the Lie algebra consisting of the square matrices of order 2, trace zero, and with entries in $k$ . This Lie algebra is simple of dimension 3 (Chap. I, §6, no. 7, Example). The canonical basis of $\mathfrak{s}\mathfrak{l}(2, k)$ is the basis $(X_+, X_-, H)$ , where

$(01)($ 0 $0)(1$ 0 $)$

$X_+=$ 0 0 $X_-=-1$ 0 $H=$ 0 $-1$ .

We have

[H, X_+] = 2X_+[H, X_-] =-2X_-[X_+, X_-] =-H \tag{1}

Since the identity representation of $\mathfrak{s}\mathfrak{l}(2, k)$ is injective, $H$ is a semi-simple element of $\mathfrak{s}\mathfrak{l}(2, k)$ and $X_+, X_-$ are nilpotent elements of $\mathfrak{s}\mathfrak{l}(2, k)$ (Chap. I, §6, no. 3, Th. 3). By Chap. VII, §2, no. 1, Example $4,kH$ is a Cartan subalgebra of $\mathfrak{s}\mathfrak{l}(2, k)$ . The map $U \rightarrow -^tU$ is an involutive automorphism of the Lie algebra $\mathfrak{s}\mathfrak{l}(2, k)$ , called the canonical involution of $\mathfrak{s}\mathfrak{l}(2, k)$ ; it transforms $(X_+, X_-, H)$ into $(X_-, X_+,-H)$ .

Lemma 2 00XT

In the enveloping algebra of $\mathfrak{s}\mathfrak{l}(2, k)$ ,

[H, X_+^n] = 2nX_+^n[H, X_-^n] =-2nX_-^n

for any integer $n\geq 0$ , and

[X_-, X_+^n] =nX_+^{n-1}(H+n-1) =n(H-n+ 1)X_+^{n-1}

[X_+, X_-^n] =nX_-^{n-1}(-H+n-1) =n(-H-n+ 1)X_-^{n-1}

if $n >0$ .

The first and third relations follow from Lemma 1. The others can be deduced from them by using the canonical involution of $\mathfrak{s}\mathfrak{l}(2, k)$ .

2. Primitive elements of $\mathfrak{s}\mathfrak{l}$ (2 $\boldsymbol{, k}$ )-modules

Let E be an $\mathfrak{s}\mathfrak{l}(2, k$ )-module. If $A\in \mathfrak{s}\mathfrak{l}(2, k)$ and $x\in E$ , we shall often write $Ax$ instead of $A_Ex$ . Let $\lambda \in k$ . If $Hx=\lambda x$ we say, by abuse of language, that $x$ is an element of E of weight $\lambda$ , or that $\lambda$ is the weight of $x$ . If E is finite dimensional, $H_E$ is semi-simple, so the set of elements of weight $\lambda$ is the primary subspace of E relative to $H_E$ and $\lambda$ (cf. Chap. VII, §1, no. 1).

Lemma 3 00XU

If $x$ is an element of weight $\lambda$ , then $X_+x$ is an element of weight $\lambda + 2$ and $X_-x$ is an element of weight $\lambda -2$ .

Indeed, $HX_+x= [H, X_+]x+X_+Hx= 2X_+x+X_+\lambda x= (\lambda + 2)X_+x$ , and similarly $HX_-x= (\lambda -2)X_-x$ (cf. also Chap. VII, §1, no. 3, Prop. 10 (ii)).

Definition 1 00XV

Let E be an $\mathfrak{s}\mathfrak{l}(2, k)$ -module. An element of E is said to be primitive if it is a non-zero eigenvector of $H_E$ and belongs to the kernel of $X_{+E}$ .

A non-zero element $e$ of E is primitive if and only if $ke$ is stable under the operation of $kH+kX_+$ ; this follows for example from Lemma 3.

Examples The element $X_+$ is primitive of weight 2 for the adjoint representation of $\mathfrak{s}\mathfrak{l}(2, k)$ . The element $(1,0)$ of $k^2$ is primitive of weight 1 for the identity representation of $\mathfrak{s}\mathfrak{l}(2, k)$ on $k^2$ .

Lemma 4 00XW

Let E be a non-zero finite dimensional $\mathfrak{s}\mathfrak{l}(2, k)$ -module. Then E has primitive elements.

Since $X_+$ is a nilpotent element of $\mathfrak{s}\mathfrak{l}(2, k),X_{+E}$ is nilpotent. Assume that $X_{+E}^{m-1}\not= 0$ and $X_{+E}^m$ = 0. By Lemma 2,

m(H_E-m+ 1)X_{+E}^{m-1}= [X_{-E}, X_{+E}^m] = 0

and hence the elements of $X_+^{m-1}(E)$ -- $\{0\}$ are primitive.

Proposition 1 00XX

Let E be an $\mathfrak{s}\mathfrak{l}(2, k)$ -module, and $e$ a primitive element of E of weight $\lambda$ . Put $e_n=\frac{(-1)^n}{n}X_-^ne$ for $n\geq 0$ , and $e_{-1}= 0$ . Then

He_n= (\lambda -2n)e_n

X_-e_n=-(n+ 1)e_{n+1} \tag{2}

X_+e_n= (\lambda -n+ 1)e_{n-1}

The first formula follows from Lemma 3, and the second from the definition of the $e_n$ . We prove the third by induction on $n$ . It is satisfied for $n= 0$ since $e_{-1}= 0$ . If $n >0$ ,

nX_+e_n=-X_+X_-e_{n-1}=-[X_+, X_-]e_{n-1}-X_-X_+e_{n-1}

=He_{n-1}-X_-(\lambda -n+ 2)e_{n-2}

= (\lambda -2n+ 2 + (n-1)(\lambda -n+ 2))e_{n-1}

=n(\lambda -n+ 1)e_{n-1}

Corollary 00XY

The submodule of E generated by $e$ is the vector subspace generated by the $e_n$ .

This follows from the formulas (2).

The integers $n\geq 0$ such that $e_n\not= 0$ constitute an interval in $\mathbf{N}$ , and the corresponding elements $e_n$ form a basis over $k$ of the submodule generated by $e$ (indeed, they are linearly independent because they are non-zero elements of distinct weights). This basis will be said to be associated to the primitive element $e$ .

Proposition 2 00XZ

If the submodule V of E generated by the primitive element $e$ is finite dimensional, then:

(i) the weight $\lambda$ of $e$ is integral and equal to dim $V-1$ ;

(ii) $(e_0, e_1, . . . , e_{\lambda})$ is a basis of V, and $e_n= 0$ for $n > \lambda$ ;

(iii) the eigenvalues of $H_V$ are $\lambda , \lambda -2, \lambda -4, . . . ,-\lambda$ ; they are all of multiplicity 1;

(iv) every primitive element of V is proportional to $e$ ;

(v) the commutant of the module V is reduced to the scalars; in particular, V is absolutely simple.

Let $m$ be the largest integer such that $e_m\not= 0$ . Then $0 =X_+e_{m+1}=$ $(\lambda -m)e_m$ , so $\lambda =m$ ; since $(e_0, e_1, . . . , e_m)$ is a basis of V, this proves (i) and (ii). Assertion (iii) follows from the equality $He_n= (\lambda -2n)e_n$ . We have $X_+e_n\not= 0$ for $1\leq n\leq m$ , hence (iv). Let $c$ be an element of the commutant of the module V. Then $Hc(e) =cH(e) =\lambda c(e)$ , so there exists $\mu\in k$ such that $c(e) =\mu e$ ; then

cX_-^qe=X_-^qce=\mu X_-^qe

for all $q\geq 0$ , so $c=\mu.1$ , proving (v).

Corollary 00Y0

Let E be a finite dimensional $\mathfrak{s}\mathfrak{l}(2, k)$ -module.

(i) The endomorphism $H_E$ is diagonalizable and its eigenvalues are rational integers.

(ii) For any $p\in \mathbf{Z}$ , let $E_p$ be the eigenspace of $H_E$ corresponding to the eigenvalue $p$ . Let $i$ be an integer $\geq 0$ . The map $X_{-E}^i|E_p: E_p\rightarrow E_{p-2i}$ is injective for $i\leq p$ , bijective for $i=p$ , and surjective for $i\geq p$ . The map $X_{+E}^i|E_{-p}: E_{-p}\rightarrow E_{-p+2i}$ is injective for $i\leq p$ , bijective for $i=p$ , and surjective for $i\geq p$ .

(iii) The length of E is equal to dim Ker $X_{+E}$ and to dim Ker $X_{-E}$ .

(iv) Let $E'$ (resp. $E''$ ) be the sum of the $E_p$ for $p$ even (resp. odd). Then $E'$ (resp. $E''$ ) is the sum of the simple submodules of E of odd (resp. even) dimension; and $E = E'\oplus E''$ . The length of $E'$ is dim $E_0$ , and that of $E''$ is dim $E_1$ .

(v) Ker $X_{+E}\cap$ Im $X_{+E}\subset \sum_{p>0}E_p$ and Ker $X_{-E}\cap$ Im $X_{-E}\subset \sum_{p<0}E_p$ .

If E is simple, E is generated by a primitive element (Lemma 4), and it suffices to apply Propositions 1 and 2. The general case follows since every finite dimensional $\mathfrak{s}\mathfrak{l}(2, k$ )-module is semi-simple.

3. The simple modules v( $\boldsymbol{m}$ )

Let $(u, v)$ be the canonical basis of $k^2$ . For the identity representation of $\mathfrak{s}\mathfrak{l}(2, k)$ ,

X_+u= 0Hu=uX_-u=-v

X_+v=uHv=-vX_-v= 0

Consider the symmetric algebra $\mathbf{S}(k^2)$ of $k^2($ Algebra, Chap. III, §6, no. 1, Def. 1). The elements of $\mathfrak{s}\mathfrak{l}(2, k)$ extend uniquely to derivations of $\mathbf{S}(k^2)$ , giving $\mathbf{S}(k^2)$ the structure of an $\mathfrak{s}\mathfrak{l}(2, k$ )-module (Chap. I, §3, no. 2). Let $V(m)$ be the set of homogeneous elements of $\mathbf{S}(k^2)$ of degree $m$ . Then $V(m)$ is an $\mathfrak{s}\mathfrak{l}(2, k$ )-submodule of $\mathbf{S}(k^2)$ of dimension $m+ 1$ , the $m$ th symmetric power of $V(1) =k^2$ (Chap. III, Appendix). If $m, n$ are integers such that $0\leq n\leq m$ , put

e^{(m)}_n=(m)u^{m-n}v^n\in V(m)

n

Proposition 3 00Y1

For any integer $m\geq 0$ , $V(m)$ is an absolutely simple $\mathfrak{s}\mathfrak{l}(2, k)$ -module. In this module, $e^{(m)}_0=u^m$ is primitive of weight $m$ .

We have $X_+u^m= 0$ and $Hu^m=mu^m$ , so $u^m$ is primitive of weight $m$ . The submodule of $V(m)$ generated by $u^m$ is of dimension $m+ 1$ (Prop. 2 (i)) and so is equal to $V(m)$ . By Prop. 2 (v), $V(m)$ is absolutely simple.

Theorem 1 00Y2

Every simple $\mathfrak{s}\mathfrak{l}(2, k)$ -module of finite dimension $n$ is isomorphic to $V(n-1)$ . Every finite dimensional $\mathfrak{s}\mathfrak{l}(2, k)$ -module is a direct sum of submodules isomorphic to the modules $V(m)$ .

This follows from Lemma 4 and Prop. 1, 2 and 3.

Remark 1 00Y3

The adjoint representation of $\mathfrak{s}\mathfrak{l}(2, k)$ defines on $\mathfrak{s}\mathfrak{l}(2, k)$ the structure of a simple $\mathfrak{s}\mathfrak{l}(2, k$ )-module. This module is isomorphic to V(2) by an isomorphism that takes $u^2$ to $X_+, 2uv$ to $-H$ , and $v^2$ to $X_-$ .

Remark 2 00Y4

For $n\geq 0$ and $m > n$ ,

X_-e^{(m)}_n=-(m-n)(m)u^{m-n-1}v^{n+1}=-(n+ 1)e^{(m)}_{n+1}

n

Hence, $(e_0^{(m)}, e_1^{(m)}, . . . , e_m^{(m)})$ is the basis of $V(m)$ associated to the primitive element $e_0^{(m)}$ .

Remark 3 00Y5

Let $\Phi$ be the bilinear form on $V(m)$ such that

$\Phi (e^{(m)}_n, e^{(m)}_{n'}) = 0$ if $n+n'\not=m$

\Phi (e^{(m)}_n, e^{(m)}_{m-n}) = (-1)^n(m)

n

If $x=au+bv$ and $y=cu+dv$ , then $\Phi (x^m, y^m) = (ad-bc)^m$ . It is now easy to check that $\Phi$ is invariant, and that $\Phi$ is symmetric for $m$ even, and alternating for $m$ odd.

Proposition 4 00Y6

Let E be a finite dimensional $\mathfrak{s}\mathfrak{l}(2, k)$ -module, $m$ an integer $\geq 0$ , $P_m$ the set of primitive elements of weight $m$ . Let L be the vector space of homomorphisms from the $\mathfrak{s}\mathfrak{l}(2, k)$ -module $V(m)$ to the $\mathfrak{s}\mathfrak{l}(2, k)$ -module E. The map $f \rightarrow f(u^m)$ from L to E is linear, injective, and its image is $P_m\cup \{0\}$ .

This map is clearly linear, and it is injective because $u^m$ generates the $\mathfrak{s}\mathfrak{l}(2, k$ )-module $V(m)$ . If $f\in L$ ,

X_+(f(u^m)) =f(X_+u^m) = 0,H(f(u^m)) =f(Hu^m) =mf(u^m)

so $f(u^m)\in P_m\cup \{0\}$ . Let $e\in P_m$ , and V the submodule of E generated by $e$ . By Prop. 1, there exists an isomorphism from the module $V(m)$ to the module V that takes $u^m$ to $e$ . Then $L(u^m) = P_m\cup \{0\}$ .

Corollary 00Y7

The isotypical component of E of type $V(m)$ has length

dim(P $_m\cup \{0\}$ .

4. Linear representations of the group sl(2 $\boldsymbol{, k}$ )

Recall (Algebra, Chap. III, §8, no. 9) that we denote by $\mathbf{S}\mathbf{L}(2, k)$ the group of square matrices of order 2 with coefficients in $k$ whose determinant is equal to 1. If $x\in \mathfrak{s}\mathfrak{l}(2, k)$ is nilpotent, then $x^2= 0 ($ Algebra, Chap. VII, §5, Cor. 3 of Prop. 5) and $e^x= 1 +x\in \mathbf{S}\mathbf{L}(2, k)$ . If E is a finite dimensional vector space and $\rho$ is a linear representation of $\mathfrak{s}\mathfrak{l}(2, k)$ on E, then $\rho (x)$ is nilpotent and so $e^{\rho(x)}$ is defined (Chap. I, §6, no. 3).

Definition 2 00Y8

Let E be a finite dimensional vector space, and $\rho$ (resp. $\pi$ ) a linear representation of $\mathfrak{s}\mathfrak{l}(2, k)$ (resp. $\mathbf{S}\mathbf{L}(2, k)$ ) on E. Then $\rho$ and $\pi$ are said to be compatible if, for every nilpotent element $x$ of $\mathfrak{s}\mathfrak{l}(2, k),\pi (e^x) =e^{\rho(x)}$ .

In other words, $\rho$ and $\pi$ are compatible if, for every nilpotent element $x$ of $\mathfrak{s}\mathfrak{l}(2, k)$ , the restriction of $\rho$ to $kx$ is compatible with the restriction of $\pi$ to the group $1 +kx$ (Chap. VII, §3, no. 1).

If $\rho$ and $\pi$ are compatible, so are the dual representations, the $m$ th tensor powers, and the $m$ th symmetric powers of $\rho$ and $\pi$ , respectively (Chap. VII, §5, no. 4, Lemma 1 (i) and (ii)). Similarly for the representations induced by $\rho$ and $\pi$ on a vector subspace stable under $\rho$ and $\pi ($ loc. cit.).

In particular, the representation $\rho_m$ of $\mathfrak{s}\mathfrak{l}(2, k)$ on $V(m)$ (no. 3) is compatible with the $m$ th symmetric power $\pi_m$ of the identity representation $\pi_1$ of $\mathbf{S}\mathbf{L}(2, k)$ . Putting $e^{(m)}_n=(^m_n)u^{m-n}v^n$ as above, we have

(m)(m)m-nn

$\pi_m(s)e_n$ = $(su)(sv)$ (3)

n

for $s\in \mathbf{S}\mathbf{L}(2, k)$ and $0\leq n\leq m$ .

Theorem 2 00Y9

Let $\rho$ be a linear representation of $\mathfrak{s}\mathfrak{l}(2, k)$ on a finite dimensional vector space E.

(i) There exists a unique linear representation $\pi$ of $\mathbf{S}\mathbf{L}(2, k)$ on E that is compatible with $\rho$ .

(ii) A vector subspace F of E is stable under $\pi$ if and only if it is stable under $\rho$ .

(iii) Let $x\in E$ . Then $\pi (s)x=x$ for all $s\in \mathbf{S}\mathbf{L}(2, k)$ if and only if $x$ is invariant under $\rho ($ that is, $\rho (a)x= 0$ for all $a\in \mathfrak{s}\mathfrak{l}(2, k))$ .

The existence of $\pi$ follows from the preceding and Th. 1. On the other hand, we know that the group $\mathbf{S}\mathbf{L}(2, k)$ is generated by the elements of the form

_{tX}(1t)_{-tX_-}(10)

$e^+=e$ =

0 1 $t$ 1

where $t\in k($ Algebra, Chap. III, §8, no. 9, Prop. 17). This proves the uniqueness of $\pi$ .

Assertions (ii) and (iii) follow from what we have said, together with Chap. VII, §3, no. 1, Lemma 1 (i). Q.E.D.

Every finite dimensional $\mathfrak{s}\mathfrak{l}(2, k$ )-module therefore has a unique $\mathbf{S}\mathbf{L}(2, k)$ -module structure, which is said to be associated to its $\mathfrak{s}\mathfrak{l}(2, k$ )-module structure.

Remark 00YA

When $k$ is $\mathbf{R}$ or $\mathbf{C}$ or a complete non-discrete ultrametric field, $\mathfrak{s}\mathfrak{l}(2, k)$ is the Lie algebra of $\mathbf{S}\mathbf{L}(2, k)$ . Let $\rho$ and $\pi$ be as in Th. 2. The homomorphism $\pi$ is a homomorphism of Lie groups from $\mathbf{S}\mathbf{L}(2, k)$ to $\mathbf{G}\mathbf{L}(E):$ this is clear when $E = V(m)$ , and the general case follows, in view of Th. 1. By Chap. VII, §3, no. $1,\rho (X_+) = L(\pi )(X_+),\rho (X_-) = L(\pi )(X_-)$ . Hence $\rho = L(\pi )$ (for a converse, see Exerc. 18).

Proposition 5 00YB

Let E, F be finite dimensional $\mathfrak{s}\mathfrak{l}(2, k)$ -modules, and let $f\in$ Hom $_k(E,F)$ . The following conditions are equivalent:

(i) $f$ is a homomorphism of $\mathfrak{s}\mathfrak{l}(2, k)$ -modules;

(ii) $f$ is a homomorphism of $\mathbf{S}\mathbf{L}(2, k)$ -modules.

Condition (i) means that $f$ is an invariant element of the $\mathfrak{s}\mathfrak{l}(2, k$ )-module Hom $_k(E,F)$ , and condition (ii) means that $f$ is an invariant element of the $\mathbf{S}\mathbf{L}(2, k$ )-module Hom $_k(E,F)$ . Since these module structures are associated by Chap. VII, §5, no. 4, Lemma 1 (iii), the proposition follows from Th. 2 (iii).

Definition 3 00YC

The adjoint representation of the group $\mathbf{S}\mathbf{L}(2, k)$ is the linear representation Ad of $\mathbf{S}\mathbf{L}(2, k)$ on $\mathfrak{s}\mathfrak{l}(2, k)$ defined by

Ad( $s$ ) $.a=sas^{-1}$

for all $a\in \mathfrak{s}\mathfrak{l}(2, k)$ and all $s\in \mathbf{S}\mathbf{L}(2, k)$ . When $k$ is $\mathbf{R}$ or $\mathbf{C}$ or a complete non-discrete ultrametric field, we recover Def. 7 of Chap. III, §3, no. 12 (cf. loc. cit., Prop. 49).

By Chap. VII, §5, no. 4, Lemma 1 (i) and (ii), the adjoint representations of $\mathfrak{s}\mathfrak{l}(2, k)$ and $\mathbf{S}\mathbf{L}(2, k)$ are compatible. By Chap. VII, §3, no. 1, Remark 2, Ad( $\mathbf{S}\mathbf{L}(2, k)$ ) $=$ Aut $_e(\mathfrak{s}\mathfrak{l}(2, k))$ .

5. Some elements of sl(2 $\boldsymbol{, k}$ )

For any $t\in k^*$ , put

\theta (t) =e^{tX_+}e^{t^{-1}X_-}e^{tX_+}

$(1t)(($ 1 $0)((1t)$

= 0 1 $-t^{-1}$ 1 0 1

$($ 0 t)

= $-t^{-1}$ 0

=e^{t^{-1}X_-}e^{tX_+}e^{t^{-1}X_-}

With the notations of no. 3,

\theta (t)u=-t^{-1}v\theta (t)v=tu

\theta (t)e^{(m)}_n= (-1)^{m-n}t^{2n-m}e^{(m)}_{m-n} \tag{4}

Hence, the element $\theta (t)^2=(-1$ 0 $)$ operates by $(-1)^m$ on $V(m)$ . If

0 $-1$

E is an odd-dimensional simple $\mathfrak{s}\mathfrak{l}(2, k$ )-module, $\theta (t)_E$ is thus an involutive automorphism of the vector space E. In particular, taking E to be the adjoint representation:

\theta (t)_EX_+=t^{-2}X_-\theta (t)_EX_-=t^2X_+\theta (t)_EH=-H \tag{5}

so that $\theta (1)_E=\theta (-1)_E$ is the canonical involution of $\mathfrak{s}\mathfrak{l}(2, k)$ .

For any $t\in k^*$ , put

h(t) =(0tt^-0_1)=\theta (t)\theta (-1)

Then $h(t)u=tu,h(t)v=t^{-1}v$ , so

h(t)e^{(m)}_n=t^{m-2n}e^{(m)}_n \tag{6}

Proposition 6 01L4

Let E be a finite dimensional $\mathfrak{s}\mathfrak{l}(2, k)$ -module, and $t\in k^*$ . Let $E_p$ be the set of elements of E of weight $p$ .

(i) $\theta (t)_E|E_p$ is a bijection from $E_p$ to $E_{-p}$ .

(ii) $h(t)_E|E_p$ is the homothety with ratio $t^p$ on $E_p$ .

If $E = V(n)$ , the proposition follows from formulas (4) and (6). The general case follows from Th. 1.

Corollary 00YD

Let $E = E'\oplus E''$ be the decomposition of E defined in the Cor. of Prop. 2. The element $(-10)$ of $\mathbf{S}\mathbf{L}(2, k)$ operates by +1 on $E'$