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  Can one classify irreducible unitary representations of the Weyl algebra?

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I saw in this MO post: Is there a machinery describing all the irreducible representations ? that classifying irreducible representations of the Weyl algebra is essentially intractable. My question is if it becomes more reasonable if one restricts to unitary irreps.

To be completely precise: By the Weyl algebra I mean the complex algebra generated by $1$, $a_{+}$, and $a_{-}$ subject to the relations $[a_{+},a_{-}] = 1$. A unitary representation is a complex vector space $V$ with a Hermitian form, on which the Weyl algebra acts in such a way that $a_{+}$ and $a_{-}$ are adjoint pairs (and $1$ acts as $1$). There is no topological component: $V$ need not be complete, the operators must be everywhere defined, and irreducible means no subrepresentations (not just 'no closed subrepresentations').

I know that the Fock space (generated by a vector $v$ with $a_{-}v = 0$) is one such irrep, and one can conjugate by automorphisms to the algebra to obtain other irreps: $a_{+} \mapsto \alpha a_{+} + \beta a_{-}$; $a_{-} \mapsto \overline{\beta}a_{+} + \overline{\alpha} a_{-}$; where $|\alpha|^2 - |\beta|^2 = 1$. The automorphisms where $\beta = 0$ do not give distinct irreps, so one gets a two-dimensional space of irreps (unless there are other redundancies here).

Any futher explanations or references would be great, thanks!

This post imported from StackExchange MathOverflow at 2014-11-18 15:07 (UTC), posted by SE-user Alex Zorn
asked Nov 17, 2014 in Mathematics by Alex Zorn (25 points) [ no revision ]
retagged Nov 21, 2014 by dimension10
en.wikipedia.org/wiki/Stone%E2%80%93von_Neumann_theorem

This post imported from StackExchange MathOverflow at 2014-11-18 15:07 (UTC), posted by SE-user Qiaochu Yuan
But those are only the ones that come from representations of the Heisenberg group (see, for example, "the Weyl form of the CCR" on that page).

This post imported from StackExchange MathOverflow at 2014-11-18 15:07 (UTC), posted by SE-user Alex Zorn

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