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Smooth and K-finite vectors for SL2(R), and the (g,K)-module

Definition

Assume the Axiom of Choice. Let $G = {S L}_{2} (R)$ , $K = S O (2)$ with $k_{θ} = (\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array})$ , and $g = {s l}_{2} (R)$ with complexification $g_{C} = {s l}_{2} (C)$ , using the conventions of Iwasawa and minimal-parabolic data for SL2(R) and The special linear Lie algebra sl_2. Let $(π, H)$ be a strongly continuous unitary representation of $G$ (Strongly continuous unitary representations, invariant linear subspaces and intertwiners).

A vector $v \in H$ is smooth if its orbit map $g \mapsto π (g) v$ is $C^{\infty}$ in the norm topology of $H$ ; write $H^{\infty}$ for the smooth vectors. It is $K$ -finite if $span {π (k) v : k \in K}$ is finite-dimensional. Set $V : = H^{\infty, K}$ , the vector space of smooth, $K$ -finite vectors.

For $X \in g$ and $v \in H^{\infty}$ , define the derived operator $L_{X} v : = {\frac{d}{d t} ∣}_{t = 0} π (\exp_{G} (t X)) v,$ where $t \mapsto \exp_{G} (t X)$ is the one-parameter subgroup with tangent $X$ (One-parameter subgroup of a Lie group, One-parameter subgroups are exactly exponentials). Extend $X \mapsto L_{X}$ complex-linearly to $g_{C}$ . The maps $L_{X}$ preserve $H^{\infty}$ , give a Lie-algebra action there, and extend uniquely to a unital action of $U (g_{C})$ (The universal enveloping algebra as a tensor quotient, Lie algebra actions extend to unital actions of the enveloping algebra).

The $(g, K)$ -module associated to $π$ is $V$ , with the restricted $K$ -action $π ∣_{K}$ and derived $g_{C}$ -action $X \mapsto L_{X} ∣_{V}$ . These actions preserve $V$ and satisfy $π (k) L_{X} π (k)^{- 1} = L_{Ad (k) X} (k \in K, X \in g_{C}), Ad (k) X = k X k^{- 1} .$ The derivative of the $K$ -action agrees with the restriction of $L$ to $Lie (K) = R J$ , and every vector lies in a finite-dimensional smooth $K$ -invariant subspace. These are the compatibility and local finiteness conditions meant by a $(g, K)$ -module here; no finite-multiplicity assertion is included. The enveloping-algebra action restricts to $V$ .

For the fixed compact-adapted basis, put $J = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}), S = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}), D = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array}),$ and define $W : = - i J$ and $E_{\pm} : = \frac{1}{2} (D \pm i S)$ in $g_{C}$ . Direct multiplication gives $[W, E_{\pm}] = \pm 2 E_{\pm}$ and $[E_{+}, E_{-}] = W$ ; also $W = - i \frac{d}{d θ} ∣_{0} k_{θ}$ in the complexified Lie algebra. For $n \in Z$ , define $H_{n} : = {v \in H : π (k_{θ}) v = e^{i n θ} v for all θ \in R}, V_{n} : = V \cap H_{n} .$ If $v \in H_{n} \cap H^{\infty}$ , then $L_{J} v = {\frac{d}{d θ} ∣}_{0} π (k_{θ}) v = i n v$ , so $L_{W} v = - i L_{J} v = n v$ .

The quadratic Casimir element $Ω \in Z (U (g_{C}))$ is normalized by $Ω = \frac{1}{8} W^{2} - \frac{1}{4} W + \frac{1}{2} E_{+} E_{-} .$ In the ordered basis $(W, E_{+}, E_{-})$ , the displayed brackets give $B (W, W) = tr ((ad W)^{2}) = 8$ , $B (E_{+}, E_{-}) = B (E_{-}, E_{+}) = tr (ad E_{+} ad E_{-}) = 4$ , and zero for the remaining pairings. Thus the Killing-dual basis is $(W / 8, E_{-} / 4, E_{+} / 4)$ , so The quadratic Casimir element gives $W^{2} / 8 + (E_{+} E_{-} + E_{-} E_{+}) / 4$ ; the bracket relation above gives the displayed formula. Centrality follows from The quadratic Casimir element is central. All derived-action computations on this pair use this normalization.

Facts & Assumptions

Given: AC; the finite-dimensional real Lie group $G = {S L}_{2} (R)$ , a strongly continuous unitary representation $(π, H)$ , a smooth vector $v \in H^{\infty}$ , and real Lie-algebra elements $X, Y \in g$ .

[F1]

For every smooth vector $v$ , the orbit map $F_{v} : G \to H$ , $F_{v} (g) = π (g) v$ , is $C^{\infty}$ in norm by the definition above.

[F2]

For $X \in g$ , the curve $γ_{X} (t) = \exp_{G} (t X)$ is a smooth one-parameter subgroup with $γ_{X} (0) = e$ and $γ_{X}^{'} (0) = X$ (One-parameter subgroup of a Lie group, One-parameter subgroups are exactly exponentials).

[F3]

The left-invariant fields $X^{L}, Y^{L}$ satisfy $[X^{L}, Y^{L}] = [X, Y]^{L}$ for the fixed tangent Lie bracket (Lie bracket on the tangent space of a Lie group).

Proof

technique · direct

1.1F1F2algebra

The curve $t \mapsto F_{v} (γ_{X} (t))$ is smooth by [F1] and [F2], so its derivative at $0$ exists and is $L_{X} v$ . For every $g \in G$ , bounded linearity of $π (g)$ gives $π (g) L_{X} v = {\frac{d}{d t} ∣}_{0} F_{v} (g γ_{X} (t)) = (X^{L} F_{v}) (g)$ . In local coordinates $X^{L} = \sum_{i} a_{i} \partial_{i}$ with smooth coefficients, so $X^{L} F_{v} = \sum_{i} a_{i} \partial_{i} F_{v}$ is smooth as an $H$ -valued map. Hence the orbit map of $L_{X} v$ is smooth and $L_{X} v \in H^{\infty}$ .

2.1F1F2F3step 1.1

Applying the preceding identity twice gives $L_{X} L_{Y} v = (X^{L} (Y^{L} F_{v})) (e)$ and $L_{Y} L_{X} v = (Y^{L} (X^{L} F_{v})) (e)$ . For an $H$ -valued smooth map the commutator identity follows by applying every continuous linear functional on $H$ to it; these functionals separate points, and differentiation commutes with them. Therefore $L_{X} L_{Y} v - L_{Y} L_{X} v = ([X^{L}, Y^{L}] F_{v}) (e) = ([X, Y]^{L} F_{v}) (e) = L_{[X, Y]} v$ by [F3].

2.2F1F2step 1.1algebra

For $k \in K$ , the orbit map of $π (k) v$ is $g \mapsto F_{v} (g k)$ , so $K$ preserves $H^{\infty}$ . The curve $k \exp_{G} (t X) k^{- 1}$ is a one-parameter subgroup with tangent $k X k^{- 1}$ , and hence equals $\exp_{G} (t Ad (k) X)$ by [F2]. Differentiating gives $π (k) L_{X} v = L_{Ad (k) X} π (k) v$ for real $X$ , and complex-linearity gives it for complex $X$ . If $v \in V$ , let $E = span {π (k) v : k \in K}$ ; it is a finite-dimensional $K$ -invariant subspace of $H^{\infty}$ . For a basis $X_{1}, X_{2}, X_{3}$ of $g_{C}$ , the finite-dimensional span of all $L_{X_{i}} u$ , $u \in E$ , is $K$ -invariant by this covariance identity and contains every $L_{X} v$ . Thus $L_{X} v$ is $K$ -finite as well as smooth, and $L_{X}$ preserves $V$ .

3.1step 2.1algebra

Complex-linearity extends this bracket identity to $g_{C}$ , so $X \mapsto L_{X}$ is a Lie-algebra action on $H^{\infty}$ . The universal property of the enveloping algebra, Lie algebra actions extend to unital actions of the enveloping algebra, then gives the unique unital action of $U (g_{C})$ extending it.

4.1F1F2step 3.1step 2.2algebra∎

The $K$ -action preserves $V$ , since translating a finite-dimensional $K$ -orbit span leaves it unchanged. Its restriction to each such span is smooth: all its vectors are smooth and coordinate functionals on the finite-dimensional span recover smooth matrix entries from their orbit maps. Differentiating $π (k_{θ}) v$ gives $L_{J} v$ because $k_{θ} = \exp_{G} (θ J)$ ; this identity follows from [F2] since both curves have tangent $J$ . Hence the infinitesimal $K$ -action and the restricted Lie-algebra action agree. Together with step 2.2 this proves all stated $(g, K)$ compatibilities. Stability under each $L_{X}$ also makes $V$ stable under their finite products and sums, so the action from step 3.1 restricts to $V$ .

Choice. AC is inherited from the Iwasawa supplier's normalized Haar measure on $K$ ; its consequence ${A C}_{ω}$ is used by the one-parameter-subgroup and tangent-bracket suppliers. The smoothness, finiteness, basis, and K-type definitions here add no further choice (The Axiom of Choice).

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Sources

Matt Kerr, Notes on the Representation Theory of SL2(R) (CBMS workshop writeup) (standard reference, not scraped)
Pavel Etingof, Representations of Lie Groups (MIT 18.757 Lecture 9) (standard reference, not scraped)