The Heisenberg group

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

The Heisenberg group

The Heisenberg group $G = \{(\begin{matrix} 1 & x & z \\ 1 & y \\ 1 \end{matrix}) | x, y, z \in ℝ\}$ has the Lie algebra $𝔤 = \{(\begin{matrix} 0 & x & z \\ 0 & y \\ 0 \end{matrix}) | x, y, z \in ℝ\},$ where we exhibit $𝔤$ as a Lie subalgebra of $M_{n} (ℝ)$ since $H$ is a subgroup of ${GL}_{n} (ℝ) .$ Setting $X = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 \\ 0 \end{matrix}), Y = (\begin{matrix} 0 & 0 & 0 \\ 0 & 1 \\ 0 \end{matrix}), Z = (\begin{matrix} 0 & 0 & 1 \\ 0 & 0 \\ 0 \end{matrix})$ and $𝔥 = span - \{X, Y, Z\}$ with bracket determined by $[X, Y] = Z, [X, Z] = 0, [Y, Z] = 0 .$

Since $[𝔤, [𝔤, 𝔤]] = 0,$ the lower central series of $𝔤$ is $𝔤 \supseteq [𝔤, 𝔤] \supseteq 0, where [𝔤, 𝔤] = \{(\begin{matrix} 0 & 0 & z \\ 0 & 0 \\ 0 \end{matrix}) | z \in ℝ\},$ is the center of $𝔤 .$ The exponential map is given by $\begin{array}{rcl} 𝔤 & \to & G \\ (\begin{matrix} 0 & x & z \\ 0 & y \\ 0 \end{matrix}) & \mapsto & (\begin{matrix} 1 & x & z + x y / 2 \\ 1 & y \\ 1 \end{matrix}) \end{array},$ since $\exp (\begin{matrix} 0 & x & z \\ 0 & y \\ 0 \end{matrix}) = 1 + (\begin{matrix} 0 & x & z \\ 0 & y \\ 0 \end{matrix}) + \frac{1}{2} = (\begin{matrix} 1 & x & z + x y / 2 \\ 1 & y \\ 1 \end{matrix}) .$ Let $X = (\begin{matrix} 0 & x & z \\ 0 & y \\ 0 \end{matrix}) \in 𝔤, g = e^{X} = (\begin{matrix} 1 & x & z + x y / 2 \\ 1 & y \\ 1 \end{matrix}) \in G, and Y = (\begin{matrix} 0 & a & c \\ 0 & b \\ 0 \end{matrix}) \in 𝔤 .$ Then ${Ad}_{g} Y = g Y g^{- 1} = (\begin{matrix} 0 & a & c + b x \\ 0 & b \\ 0 \end{matrix}) (\begin{matrix} 1 & - x & - z + x y / 2 \\ 1 & - y \\ 1 \end{matrix}) = (\begin{matrix} 0 & a & - a y + b x + c \\ 0 & b \\ 0 \end{matrix})$ describes the adjoint representation of $G .$ (Alternatively one can use the formula $({Ad}_{e^{X}}) (Y) = (e^{{ad}_{X}}) (Y)$ to complete this calculation.)

To compute the coadjoint representation we need a good way of looking at $𝔤 * = Hom (𝔤 ℝ) .$ Use the trace on $M_{n} (ℝ)$ $\begin{array}{rcl} Tr : & M_{n} (ℝ) & \to & ℝ \\ X & \mapsto & Tr (X) \end{array}$ to define a symmetric bilinear form $⟨,⟩ : M_{n} (ℝ) \times M_{n} (ℝ)$ by $⟨X, Y⟩ = Tr (X Y), for X, Y \in 𝔤 .$ The form $⟨,⟩$ provides a vector space isomorphism $\begin{array}{rcl} M_{n} (ℝ) & \to & M_{n} (ℝ) \\ X & \mapsto & ⟨X .⟩ \end{array} where \begin{array}{rcl} ⟨X .⟩ : & M_{n} (ℝ) & \to & ℝ \\ Y & \mapsto & ⟨X, Y⟩ \end{array}, for X \in M_{n} (ℝ) .$ Use this isomorphism to identify $M_{n} (ℝ)$ and ${M_{n} (ℝ)}^{*} .$ Our favourite basis of $M_{n} (ℝ)$ is $\{E_{i j} | 1 \leq i, j \leq n\},$ where $E_{i j}$ denotes the matrix with a 1 in the $(i, j)$ -th entry and 0 everywhere else. Then $\{E_{i j}^{*} = E_{j i} | 1 \leq i, j \leq n\}$ is the dual basis with respect to $⟨,⟩ .$ Since $𝔤 = span - \{E_{12}, E_{13}, E_{23}\},$ $𝔤^{*} = span - \{E_{12}^{*}, E_{13}^{*}, E_{23}^{*}\} = \{(\begin{matrix} 0 & 0 & 0 \\ α & 0 & 0 \\ γ & β & 0 \end{matrix}) | α, β, γ \in ℝ\} .$ Let $g = e^{X} = (\begin{matrix} 1 & x & z + x y / 2 \\ 1 & y \\ 1 \end{matrix}) \in G, φ = (\begin{matrix} 0 & 0 & 0 \\ α & 0 & 0 \\ γ & β & 0 \end{matrix}) \in 𝔤^{*}, Y = (\begin{matrix} 0 & a & c \\ 0 & b \\ 0 \end{matrix}) \in 𝔤 .$ Then $\begin{array}{rcl} ({Ad}_{g}^{*}) (Y) & = & φ ({Ad}_{g^{- 1}} Y) = φ (g^{- 1} Y g) = ⟨φ, g^{- 1} Y g⟩ \\ = & ⟨(\begin{matrix} 0 & 0 & 0 \\ α & 0 & 0 \\ γ & β & 0 \end{matrix}), (\begin{matrix} 0 & a & - a y + b x + c \\ 0 & b \\ 0 \end{matrix})⟩ \\ = & α a + β b + γ a y - γ b x + c \\ = & ⟨(\begin{matrix} 0 \\ α + γ y & 0 \\ γ & β - γ x \end{matrix}), (\begin{matrix} 0 & a & c \\ 0 & b \\ 0 \end{matrix})⟩ . \end{array}$ Thus ${Ad}_{g}^{*} (α, β, γ) = (α + γ y, β - γ x, γ),$ where we use the more concise notation $(α, β, γ)$ for the matrix $(\begin{matrix} 0 \\ α & 0 \\ γ & β & 0 \end{matrix}) \in 𝔤^{*} .$ To compute the coadjoint orbits note that

If $γ \neq 0$ then we can choose $x, y \in ℝ$ so that ${Ad}_{g}^{*} (α, β, γ) = (0, 0, γ) .$ So $(0, 0, γ)$ is in the orbit of $(α, β, γ) .$
If $γ = 0$ then ${Ad}_{g}^{*} (α, β, γ) = (0, 0, γ),$ for all $g \in G .$ So $(α, β, 0)$ is in an orbit all by itself.

Thus there are two kinds of coadjoint orbits

Ω_{γ} = \{(α, β, γ) | α, β \in ℝ, γ \in ℝ \ \{0\}\} and Ω_{α, β} = \{(α, β, 0) | α, β \in ℝ\},

with

\dim Ω_{γ} = 2

and

\dim Ω_{α, β} = 0 .

Note that $\dim 𝔤 = 3$ and $𝔥 = \{(\begin{matrix} 0 & 0 & c \\ 0 & b \\ 0 \end{matrix}) | b, c \in ℝ\}$ is a two dimensional subalgebra of $𝔤$ such that $, [𝔥, 𝔥] = 0 .$ So $𝔥$ is a subalgebra of $𝔤$ which is subordinate to any $φ \in 𝔤^{*},$ $φ ([x, y]) = φ (0) = 0, for all x, y \in 𝔥 .$ So, if $φ \in 𝔤^{*},$ then either a) there is a 3 dimensional subalgebra subordinate to $φ$ (which must be all of $𝔤$ ) or b) $𝔥$ is a maximal subordinate subalgebra to $φ .$

If $φ = (α, β, 0) = (\begin{matrix} 0 \\ α & 0 \\ 0 & β & 0 \end{matrix}) \in Ω_{α, β}$ then $𝔤$ is subordinate to $φ$ and $\begin{array}{rcl} G & \to & ℂ^{*} \\ e^{X} & \mapsto & e^{2 π i ⟨φ, X⟩} \end{array}$ is the representation $V^{Ω_{α, β}}$ associated with the orbit $Ω_{α, β} .$ More precisely, $V^{Ω_{α, β}} = span - \{v\}$ and $(\begin{matrix} 1 & x & z + x y / 2 \\ 1 & y \\ 1 \end{matrix}) v = e^{2 π i (α x + β y) v} v, for all x, y, z \in ℝ,$ since $⟨φ, X⟩ = ⟨(\begin{matrix} 0 \\ α & 0 \\ 0 & β & 0 \end{matrix}), (\begin{matrix} 0 & x & y \\ 0 & y \\ 0 \end{matrix})⟩ = α x + β y .$
If $φ = (α, β, γ) = (\begin{matrix} 0 \\ α & 0 \\ 0 & β & 0 \end{matrix}) \in Ω_{γ}, γ \neq 0,$ then $𝔤$ is not subordinate to $φ$ and so $𝔥$ is a maximal subordinate subalgebra for $φ .$ The one dimensional representation of $H = \exp (𝔥)$ is given by $\begin{array}{rcl} W^{φ} : & H & \to & ℂ^{*} \\ (\begin{matrix} 1 & 0 & z \\ 1 & y \\ 1 \end{matrix}) & \mapsto & e^{2 π i (y β + t z)} \end{array}$ since $\exp ((\begin{matrix} 0 & 0 & z \\ 0 & y \\ 0 \end{matrix})) = (\begin{matrix} 1 & 0 & z \\ 1 & y \\ 1 \end{matrix}) and ⟨(\begin{matrix} 0 & 0 & z \\ 0 & y \\ 0 \end{matrix}), (\begin{matrix} 0 \\ α & 0 \\ 0 & β & 0 \end{matrix})⟩ = y β + γ z .$ The representation of $G$ corresponding to the orbit $Ω_{γ}$ is $V^{Ω_{γ}} = {Ind}_{H}^{G} (W^{φ}),$ and we can view elements of ${Ind}_{H}^{G} (W^{φ})$ as functions on $H \ G .$ Since $H = \{(\begin{matrix} 1 & 0 & z \\ 1 & y \\ 1 \end{matrix}) | y, z \in ℝ\} and G = \{(\begin{matrix} 1 & x & z \\ 1 & y \\ 1 \end{matrix}) | y, z \in ℝ\},$ the elements $(\begin{matrix} 1 & t & 0 \\ 1 & 0 \\ 1 \end{matrix}), t \in ℝ,$ are coset representatives of the cosets in $H \ G .$ So we may view elements of $V^{Ω_{γ}}$ as functions $\begin{array}{rcl} f : & ℝ & \to & ℂ = W^{φ} \\ t & \mapsto & f (t) \end{array} .$ If $X = (\begin{matrix} 0 & x & z \\ 0 & y \\ 0 \end{matrix})$ then $(e^{X} f) (t) = f (t e^{X}) = f ((\begin{matrix} 1 & t & 0 \\ 1 & 0 \\ 1 \end{matrix}) (\begin{matrix} 1 & x & z + x y / 2 \\ 1 & y \\ 1 \end{matrix})) = f ((\begin{matrix} 1 & 0 & a_{2} \\ 1 & a_{1} \\ 1 \end{matrix}) (\begin{matrix} 1 & s & 0 \\ 1 & 0 \\ 1 \end{matrix})),$ where $a_{1} = y, s = t + x$ and $a_{2} = z + (x y) / 2 + t y .$ So $\begin{array}{rcl} (e^{X} f) (t) & = & f ((\begin{matrix} 1 & x & z + x y / 2 + t y \\ 1 & y \\ 1 \end{matrix}) (\begin{matrix} 1 & t + x & 0 \\ 1 & 0 \\ 1 \end{matrix})) \\ = & (\begin{matrix} 1 & x & z + x y / 2 + t y \\ 1 & y \\ 1 \end{matrix}) f ((\begin{matrix} 1 & t + x & 0 \\ 1 & 0 \\ 1 \end{matrix})) \\ = & e^{2 π i (y β + γ z + γ (x y) / 2 + γ t y)} f (t + x), \end{array}$ and this formula describes the action of $G$ on $V^{Ω_{γ}} .$

Summary: For each orbit $Ω_{α, β}, α, β \in ℝ,$ we get a one dimensional representation $V^{Ω_{α, β}} = ℂ v$ with action $e^{X} v = e^{2 π i (α x + β y)} v, if X = (\begin{matrix} 0 & x & z \\ 0 & y \\ 0 \end{matrix}) \in 𝔤,$ and for each orbit $Ω_{γ}, γ \neq 0$ we get an action of $G$ on functions $W^{Ω_{γ}} = \{f : ℝ \to ℂ\} given by (e^{X} f) (t) = e^{2 π i (y β + γ z + γ (x y) / 2 + γ t y)} f (t + x) .$

Let us sho that the representations are irreducible by computing ${Hom}_{G} (V^{Ω_{g}}, V^{Ω_{γ}}) = {Hom}_{G} ({Ind}_{H}^{G} (W^{φ}), {Ind}_{H}^{G} (W^{φ})) ≅ \{T : G \to Hom (W^{φ}, W^{φ}) | T (h_{1} g h_{2}) = T (g)\} .$

Then $\begin{array}{rcl} T (\begin{matrix} 1 & x & z \\ 1 & y \\ 1 \end{matrix}) & = & T ((\begin{matrix} 1 & 0 & z \\ 1 & y \\ 1 \end{matrix}) (\begin{matrix} 1 & x & 0 \\ 1 & 0 \\ 1 \end{matrix})) = e^{2 π i (β y + γ z)} T (x) \\ = & T ((\begin{matrix} 1 & x & 0 \\ 1 & 0 \\ 1 \end{matrix}) (\begin{matrix} 1 & 0 & z - x y \\ 1 & y \\ 1 \end{matrix})) = e^{2 π i (β y + γ z - γ x y)} T (x) . \end{array}$ So $T (x) = 0$ unless $e^{2 π i (- γ x y)} = 1$ for all $y \in ℝ .$ So $T (x) = 0$ unless $x = 0 .$ So $T$ is determined by its value at thw identity matrix. So $\dim {Hom}_{G} (V^{Ω_{g}}, V^{Ω_{g}}) = 1 .$ So $V^{Ω_{g}}$ is irreducible.

The following is an attempt to make this same argument work in general. (????? correct)

Let $φ \in 𝔤^{*}$ and let $𝔥$ be a maximal subalgebra of $G$ subordinate to $φ .$ Then (if $V^{φ}$ is unitary) $V^{φ} = {Ind}_{H}^{G} (W^{φ})$ is irreducible.

		Proof.
	Let $x$ be a coset representative for a coset in $H \ G .$ Let $t : G \to Hom (W^{φ}, W^{φ})$ be such that $T (h_{1} x h_{2}) = h_{1} T (x) h_{2}, for all h_{1}, h_{2} \in H .$ Let $η \in 𝔥 .$ Then $e^{2 π i φ (η)} T (x) = T (e^{η} x) = T (x x^{- 1} e^{η} x) = T (x e^{{Ad}_{x^{- 1}} η}) = T (x) e^{2 π i φ ({Ad}_{x^{- 1}} η)} .$ So $T (x) = 0$ unless $φ ({Ad}_{x^{- 1}} η) = φ (η)$ for all $η \in 𝔥 .$ So $T (x) = 0$ unless $φ (e^{ad (- X)} η) = φ (η) .$ So $T (x) = 0$ unless $φ ((e^{ad (- X)} - 1) η) = 0 .$ Now $ad (- X) = \log (1 + (e^{ad (- X)} - 1)) = (e^{ad (- X)} - 1) - \frac{{(e^{ad (- X)} - 1)}^{2}}{2} + \dots$ So $φ (ad (- X) η) = 0$ for all $η \in 𝔥 .$ So $φ (- [X, η]) = 0$ for all $η \in 𝔥 .$ Let $\tilde{𝔥} = ℂ X \oplus 𝔥 .$ We know $X \notin 𝔥$ since $x$ is a coset representative of $H \ G$ (not 1). Then $\tilde{h}$ is a subalgebra of $𝔤$ subordinate to $φ .$ This is a contradiction to the maximality of $𝔥 .$ So $T (x) = 0 .$ So $T (x) = 0$ unless $x = 1 .$ So $T$ is determined by its value at $e^{0} = 1 .$ So $\dim {Hom}_{G} (V^{Ω}, V^{Ω}) = 1 . □$

Remark 1. Let $ℤ$ be the subgroup of $G$ given by $ℤ = \{(\begin{matrix} 1 & 0 & n \\ 1 & 0 \\ 1 \end{matrix}) | n \in ℤ\} .$ The group $G / ℤ$ is often called the Heisenberg group. This group has a subgroup $S^{1} = \{(\begin{matrix} 1 & 0 & t \\ 1 & 0 \\ 1 \end{matrix}) | t \in ℝ\} ≅ \{z \in ℂ | |z| = 1\} ≅ ℝ / ℤ,$ which is contained in $Z (G / ℤ) = [G / ℤ, G / ℤ],$ and this can be used to show that $G / ℤ$ does not have a fithful finite dimensional representation. The Lie group $G$ is the simply connected cover of $G / ℤ .$ There is an exact sequence $1 \to S^{1} \to G / ℤ \to 1 and if ⟨ξ, η⟩ = ξ_{1} η_{2} - ξ_{2} η_{1},$ then $G / ℤ = \{u \exp (ξ) | u \in S^{1}, ξ \in ℝ^{2}\}, with u \exp (ξ) . v \exp (η) = u v e^{i ⟨ξ, η⟩} \exp (ξ + η) .$

Let $P, Q$ be such that $P Q - Q P = - i .$ Then $e^{i a P} e^{i b Q} = - e^{i a b} e^{i b Q} e^{i a P}$ and $G / ℤ$ is the group generated by $e^{i a P}$ and $e^{i b Q} .$ Let $T_{a}, M_{b}$ and $U_{c}$ be the operators on $L^{2} (ℝ)$ given by $(T_{a} f) (x) = f (x - a), M_{b} f = e^{2 π i λ x} f, U_{c} f = e^{2 π i λ c} f, for fixed λ \in ℝ^{*} .$ Then this is a unitary representation of $G / ℤ$ for each $λ$ and $G / ℤ ≅ \{T_{a} M_{b} U_{c} | a, b, c \in ℝ\} .$

Remark 2. If $P, Q$ are such that $P Q - Q P = - i$ and if $a = \frac{1}{\sqrt{2}} (P + i Q) and a^{*} = \frac{1}{\sqrt{2}} (P - i Q), then [a, a^{*}] = 1 .$ Then, on $L^{2} (ℝ),$ $a^{*} acts by \frac{- i}{\sqrt{2}} (\frac{d}{d x} + x) and a^{*} Ω = 0 for Ω = e^{- \frac{1}{2} x^{2}} .$ So $Ω$ is a lowest weight vector and ${\{a^{n} Ω\}}_{n \geq 0}$ is a basis of $L^{2} (ℝ) .$ $𝔰 𝔩_{2} (ℝ) ≅ (Lie algebra generated by \{(i \ 2) P^{2}, (i \ 2) (P Q + Q P), (i \ 2) Q^{2}\}) \subseteq 𝔤 .$

Remark 3. If $P = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 \\ 0 \end{matrix}), Q = (\begin{matrix} 0 & 0 & 0 \\ 0 & 1 \\ 0 \end{matrix}), Z = (\begin{matrix} 0 & 0 & 1 \\ 0 & 0 \\ 0 \end{matrix}),$ then $[P, Q] = Z$ and these act on functions $V^{Ω_{γ}} = \{f : ℝ \to ℂ\},$ via the differential of the representation $V^{Ω_{γ}}$ of $G .$ Then $(Z f) (t) = 2 π i γ f (t) .$ In particular, by considering the action of $P, Q$ on $V^{Ω_{γ}}$ when $γ = - 1 / 2 π,$ we have operators $P, Q$ on functions $f : ℝ \to ℂ$ which satisfy $[P, Q] = P Q - Q P = - i .$ This solves a basic problem in QM, see Dirac and Heisenberg.

Remark 4. Define an action of ${SL}_{2} (ℝ) = \{(\begin{array}{rcl} a & b \\ c & d \end{array}) | a d - b c = 1\}$ on $𝔤$ by $(\begin{array}{rcl} a & b \\ c & d \end{array}) P = a P + c Q, (\begin{array}{rcl} a & b \\ c & d \end{array}) Q = b P + d Q, (\begin{array}{rcl} a & b \\ c & d \end{array}) Z = Z .$ Then $[(\begin{array}{rcl} a & b \\ c & d \end{array}) P, (\begin{array}{rcl} a & b \\ c & d \end{array}) Q] = [a P + c Q, b P + d Q] = a d [P, Q] + c b [Q, P] = (a d - b c) [P, Q] = [P, Q],$ and so ${SL}_{2} (ℝ)$ acts on $𝔤$ by automorphisms. Let $m \in {SL}_{2} (ℝ)$ and let $φ \in 𝔤^{*} .$ Let $𝔥$ be subordinate to $φ .$ The $m 𝔥$ is subordinate to $φ$ since $m 𝔥$ is, after all, abelian. Then let us make the isomorphism $\begin{array}{rcl} {Ind}_{H}^{G} (W^{φ}) & \to & {Ind}_{\tilde{H}}^{G} (W^{φ}) \\ f & \mapsto & \tilde{f} \end{array}$ where $\tilde{H} = \exp (m 𝔥) \dots$

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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