Kirillov's classification

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 23 May 2010

Kirillov's classification

Let $G$ be a simply connected nilpotent Lie group. The coadjoint representation of $G$ is the action of $G$ on $𝔤^{*} = Hom (𝔤 ℝ)$ given by $(g φ) (x) = φ ({Ad}_{g^{- 1}} (x)), for all g \in G and φ \in 𝔤^{*} .$ A coadjoint orbit is a set $G φ, φ \in 𝔤^{*} .$ Then $\{coadjoint orbits\} \leftrightarrow \{irreducible unitary representations of G\} .$

Let $Ω$ be a coadjoint orbit $(Ω \subseteq 𝔤^{*}) .$ Let $φ \in Ω$ so that $Ω = G φ .$ A Lie subalgebra $𝔥 \subseteq 𝔤$ is subordinate to $φ$ if $φ |_{[𝔥, 𝔥]} = 0, or, equivalently, φ ([x, y]) = 0 for all x, y \in 𝔥 .$ A subalgebra $𝔥$ is subordinate to $𝔤$ iff $\begin{array}{rcl} H & \to & ℂ \\ e^{X} & \mapsto & e^{2 π i φ (X)} \end{array}$ defines a representation of $H = \exp (𝔥) .$

Choose a maximal dimensional subalgebra $𝔥 \subseteq 𝔤$ which i subordinate to $φ .$ (Choosing a maximal subalgebra $𝔥$ with respect to inclusion turns out to be the same thing.) Let $H = \exp (𝔥)$ and define a one dimensional representation of $H$ by $\begin{array}{rcl} 1_{φ} : & H & \to & ℂ^{*} \\ e^{X} & \mapsto & e^{2 π i φ (X)} \end{array}$ for $x \in 𝔥 .$ Then $V^{Ω} = {Ind}_{G}^{H} (1_{φ})$ is the irreducible representation of $G$ associated with $Ω .$ This says that every irreducible unitary representation of $G$ is obtained by inducing from a one dimensional representation of some subgroup $H .$ We will want to show that

$V$ is an irreducible representation of $G,$
$V$ does not depend on the choice of $φ$ and $𝔥$ (only on the orbit $Ω$ ).

and answer the questions

Why is it sufficient to define $W^{φ}$ bu its values on $e^{X}$ for $X \in 𝔥$ ?
What is ${Ind}_{H}^{G} (W)$ ?

Let $𝔥 \subseteq \tilde{𝔥}$ be Lie subalgebras of $𝔤$ and let $φ \in 𝔤^{*}$ be such that $𝔥$ and $\tilde{𝔥}$ are both subordinate to $φ .$ Then, since $H \supseteq \tilde{H},$ there are fewer functions in ${Ind}_{\tilde{H}}^{G} (W^{φ})$ than in ${Ind}_{H}^{G} (W^{φ}) .$ The $G$ -actions on the two spaces are defined in the same way so that ${Ind}_{\tilde{H}}^{G} (W^{φ})$ is a submodule of ${Ind}_{H}^{G} (W^{φ}) .$ So $V^{Ω}$ is not irreducible unless $𝔥$ is maximal.

Suppose that $Ω$ is a coadjoint orbit, $φ \in Ω$ is not irreducible unless $𝔥$ is maximal.

Suppose that $Ω$ is a coadjoint orbit and that $φ \in Ω$ and $𝔥$ is a maximal dimensional subalgebra of $𝔤$ subordinate to $φ .$ Suppose that $\tilde{φ}$ is another element of $Ω .$ Then there is a $g \in G$ such that $\tilde{φ} = g φ - {Ad}_{g}^{*} φ .$ Let $\tilde{𝔥} = g 𝔥 = {Ad}_{g} 𝔥 .$ Then, for $g X \in g 𝔥,$ $\begin{array}{rcl} \tilde{φ} (g X) & = & (g φ) (g X) = φ (g^{- 1} g X) = φ (X), \end{array}$ and since $\begin{array}{rcl} g [X, Y] & = & {Ad}_{g} ([X, Y]) = R_{g} [X, Y] R_{g^{- 1}} \end{array} = & R_{g} (X Y - Y X) R_{g^{- 1}} = & (R_{g} X R_{g^{- 1}}) (R_{g} Y R_{g^{- 1}}) - (R_{g} Y R_{g^{- 1}}) (R_{g} X R_{g^{- 1}}) = & [g X, g Y],$ it follows that $\tilde{𝔥}$ is subordinate to $\tilde{φ}$ iff $𝔥$ is subordinate to $φ .$ If $H = \exp (𝔥)$ and $\tilde{H} = \exp (\tilde{𝔥})$ then $\overset{`}{H} = g H g^{- 1},$ since ${Ad}_{g}$ is the differential of ${Int}_{g} .$ So we get one dimensional representations $\begin{array}{rcl} W^{φ} : & H & \to & ℂ \\ h & \mapsto & e^{2 π i φ (X)} \end{array} and \begin{array}{rcl} W^{\tilde{φ}} & g H g^{- 1} & \to & ℂ \\ g h g^{- 1} & \mapsto & e^{2 π i \tilde{φ} (g X)} = e^{2 π i φ (X)} \end{array}$ if $h = e^{X} .$ Define a map $\begin{array}{rcl} {Ind}_{H}^{G} (W^{φ}) & \overset{Φ}{\to} & {Ind}_{\tilde{H}}^{G} (W^{\tilde{φ}}) \\ f & \mapsto & \tilde{f} \end{array} by \tilde{f} (t) = [f (g^{- 1} t)]^~ .$ Then $\tilde{f} (g h g^{- 1} t) = [f (h g^{- 1} t)]^~ = [h f (g^{- 1} t)]^~ = \tilde{h} \tilde{f} (t)$ and $[x f]^~ (t) = [x f (g^{- 1} t)]^~ = [f (g^{- 1} t x)]^~ = \tilde{f} (t x) = (x \tilde{f}) (t) .$ So $Φ$ is a well defined $G$ -module isomorphism.

In general, how do we construct maximal subordinate subalgebras?

(Vergne's lemma) Let $V$ be a vector space and let $⟨,⟩$ be a skew symmetric bilinear form on $V .$ $f = (0 \subseteq V \subseteq V_{2} \subseteq \dots \subseteq V_{n} = V), \dim V_{i} = i .$ Let $W (f, ⟨,⟩) = \sum_{i = 0}^{n} V_{i} \cap V_{i}^{⊥} .$ Then

$W (f, ⟨,⟩)$ is an isotropic subspace of $V,$
If $⟨x, y⟩ = ⟨φ, [x, y]⟩$ for some $φ \in 𝔤^{*}$ then $𝔥$ is a subalgebra of $𝔤 .$

Let $𝔤$ be a nilpotent Lie algebra such that $\dim (Z (𝔤)) = 1 .$ Then $𝔤 = ℝ x \oplus ℝ y \oplus ℝ z \oplus W,$ where $ℝ z = Z (𝔤), [x, y] = z,$ and $[y, w] = 0$ for $w \in W .$

		Proof.
	Let $y \in 𝔤$ such that the image of $y$ in $𝔤 / Z (𝔤)$ is a nonzero element of $Z (𝔤 / Z (𝔤)) .$ Then $\begin{array}{rcl} {ad}_{y} : & 𝔤 & \to & Z (𝔤) \\ t & \mapsto & [y, t], \end{array}$ has $im ({ad}_{y}) = Z (𝔤),$ and $\ker ({ad}_{y}) = Z_{𝔤} (y) .$ Thus $\dim (\ker ({ad}_{y})) = \dim (𝔤) - 1$ and $\ker {ad}_{y}$ is an ideal of $𝔤$ since ${ad}_{y} ([x, t]) = [y, [x, t]] = - [t, [y, x]] - [x, [t, y]] = 0, for t \in \ker {ad}_{y} and x \in 𝔤 .$ So $Z_{𝔤} (y) = ℝ y \oplus ℝ z \oplus W, where [y, w] = 0 for w \in W,$ is an ideal of $𝔤 .$ Let $x \in 𝔤$ such that $[x, y] \neq 0$ and then normalise $x$ so that $[x, y] = z . □$

Note that, as in the previous lemma, $ℝ x + ℝ y + ℝ z$ is a subalgebra isomorphic to a Heisenberg algebra: $𝔤 = \{(\begin{matrix} 0 & x & z \\ 0 & y \\ 0 \end{matrix}) | x, y, z \in ℝ\}$ and let $x = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 \\ 0 \end{matrix}), y = (\begin{matrix} 0 & 0 & 0 \\ 0 & 1 \\ 0 \end{matrix}), z = (\begin{matrix} 0 & 0 & 1 \\ 0 & 0 \\ 0 \end{matrix}) .$ Then $𝔤 = ℝ x + ℝ y + ℝ z, [x, y] = z,$ and $ℝ z = Z (𝔤) .$ So, if $G$ is a nilpotent Lie group with one dimensional center then $G_{0} = \exp (Z_{𝔤} (y)), where Z_{𝔤} (y) = ℝ y \oplus ℝ z \oplus W,$ and $G_{0}$ is normal since $e^{t x} e^{h} e^{- t x} = e^{t [x, h] + \dots},$ where $t [h, x] + \dots \in Z_{𝔤} (y)$ since only involves brackets of $x$ and $h$ and $Z_{𝔤} (y)$ is an ideal. So $G = \{g_{0} e^{t x} | g_{0} \in G_{0}, t \in ℝ\} .$ Since $Z_{𝔤} (y)$ is an ideal of $𝔤,$ $G_{0} = \exp (Z_{𝔤} (y)) is a normal subgroup of G .$ In fact, every element of $G$ can be written uniquely in the form $g_{0} e^{t x}, g_{0} \in G_{0}, t \in ℝ .$ Since $G_{0}$ is normal in $G,$ the group $G$ acts on $G_{0}$ by automorphisms. Let $V_{0}$ be an irreducible representation of $G_{0} .$ Define $g : V_{0}$ to be the $G_{0}$ -module with the same vector space $g : V_{0} = V$ and with $G_{0}$ action $g_{0} . v = (g^{- 1} g_{0} g) v, for g_{0} \in G_{0} and v \in V_{0} .$ Then $g : V_{0}$ is a new $G_{0}$ representation and $g : V_{0}$ is irreducible iff $V_{0}$ is (since, if $P \subseteq V_{0}$ is a submodule of $V_{0}$ the $g : V_{0}$ then $g : P$ is a submodule of $g : V_{0}) .$ So $G$ acts on the irreducible representations of $G_{0} .$

Why consider Lie algebras with one dimensional center? Suppose we are trying to prove that nilpotent Lie groups are monomial:
Let $M$ be an irreducible representation. Then $Z (G)$ acts on $M$ by scalars. Let $z_{1}, \dots, z_{k}$ be a basis of $Z (𝔤) .$ Let $φ : Z (𝔤) \to ℂ$ be a map such that $e^{2 π i φ (z)} = ρ (e^{z}), for z \in Z (𝔤), and let Z_{0} = \{z \in Z (𝔤) | φ (z) = 0\} .$ So $z \in Z_{0}$ if $\sum_{i = 1}^{k} c_{i} φ (b, z, i) = 0 .$ So $\dim (Z_{0}) \geq \dim Z - 1$ and $\exp (Z_{0})$ acts trivially on $M .$ Let $ρ : G \to GL (M)$ has a nontrivial kernel unless $\dim Z (𝔤) = 1 .$ In the case when $\dim Z (𝔤) = 1$ we can use Kirillov's lemma to obtain the normal subgroup $G_{0} = \exp (Z_{𝔤} (y)) .$

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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