Derived Heisenberg representation on Schwartz space

In symmetric coordinates $(a,b,z)$ on the Heisenberg group, the Schrödinger action is

(\pi(a,b,z)f)(x)=e^{i(z+bx+ab/2)}f(x+a).

On its smooth domain $\mathcal S(\mathbb R)$ , the derived representation for the coordinate Lie-algebra basis $A,B,C$ is

d\pi(A)f=f',\qquad d\pi(B)f=ixf,\qquad d\pi(C)f=if.

It realizes $[A,B]=C$ and $[A,C]=[B,C]=0$ .

Bracket calculation

All three operators preserve Schwartz space, and

[d\pi(A),d\pi(B)]f=(ixf)'-ixf'=if=d\pi(C)f.

Thus the commutator is an identity on a specified common domain. These operators are continuous for the Schwartz topology but the first two are unbounded in the $L^2$ norm.

Unsymmetrized coordinates

Putting $c=z+ab/2$ changes the group law to

(a,b,c)(a',b',c')=(a+a',b+b',c+c'+ab')

and the action to $\pi(a,b,c)f(x)=e^{i(c+bx)}f(x+a)$ . This is the same representation in different coordinates, not a change of central sign. In terms of the existing Weyl convention, $\pi(a,b,z)=e^{iz}V(b,a)$ .

Unitary versus algebraic irreducibility

The representation on $L^2$ is unitarily irreducible, but its derived action on Schwartz space is not algebraically irreducible. The nonzero proper subspace $C_c^\infty(\mathbb R)$ is preserved by differentiation and multiplication by $x$ . It is dense in $L^2$ , so it does not contradict the absence of proper closed invariant Hilbert subspaces.

References

Rahul Garg and Sundaram Thangavelu, On the structure of analytic vectors for the Schrödinger representation. §2, group law and Schrödinger formula; the coordinate conversion and operator identities are calculated directly above.