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\def\dsec#1#2{{{\partial^2 #1 } \over {\partial #2^2}}}
\def\dpar#1#2{{{\partial #1 } \over {\partial #2}}}
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\def\norma#1{{\left\| #1 \right\|}}
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\def\references{\immediate\closeout\fileref
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\null\msk
\centerline{\bf References}
\bsk
\input ref.tmp}
%Fine delle def. generali
\def \K {{\cal K}}
\def \V {{\cal V}}
\def \vp {{\varphi}}
\def \A {{\cal A}}
\def \D {{\cal D}}
\def \Dp {{\D^\prime}}
\def \F {{\cal F}}
\def \L {{\cal L}}
\def \Z {{\cal Z}}
\def \Ldue {L^2 ([0, 2\pi)}
\def \endproof {$q.e.d.$}
\def \intdp {{\int_{0}^{2 \pi}}}
\def \intinf {{\int_{-\infty}^{\infty}}}
\def \suminf {{\sum_{n= -\infty}^{\infty}}}
\def \sumN {{\sum_{n= -N}^{N}}}
\def \eps {\varepsilon}
\def \He {Heisenberg \ }
\def \op {operator \ }
\def \al {algebra \ }
\def \h {{\cal H}}
\def \hx {{{\cal H}_{ext}}}
\def \psz {{\psi_0}}
\def \psth {{\psi_\theta}}
\def \psm {{\psi}_{-1}}
\def \psn {{\psi}_{n}}
\def \psne {{\psi}_{n}^{(\varepsilon)}}
\def \psl {{\psi}_{\lambda}}
\def \udp {{U(2\pi)}}
\def \norm {{1 \over 2\pi}}
\def \norme {{\eps \over 2\pi}}
\def \normi {{i \over 2\pi}}
\def \normim {{-i \over 2\pi}}
\def \ens {e^{i n s}}
\def \ent {e^{i n t}}
\def \emt {e^{i m t}}
\def \ekt {e^{i k t}}
\def \ensm {e^{- i n s}}
\def \as {a^{*}}
\def \asb {\bar a^{*}}
\def \ab {\bar a}
\def \nb {{\bar n}}
\def \psnb {{\psi}_{\nb}}
\def \udpn {U(2 \pi n)}
\def \udpm {U(2 \pi m)}
\def \udpk {U(2 \pi k)}
%Fine delle definizioni
\hfill IFUP--TH 70/95
\bsk\bsk\bsk
\centerline{\titfnt Irreducible representations of the Heisenberg algebra}
\ssk
\centerline{\titfnt in Krein spaces}
\bsk
\centerline{M. Mnatsakanova$ ^{*}$ }
\centerline{\it Institute for Nuclear Physics
of Moskow State University, Moskow, Russia }
\msk
\centerline{G. Morchio}
\centerline{\it Dipartimento di Fisica dell'Universit\`a and INFN,
Pisa, Italy}
\msk
\centerline{F. Strocchi}
\centerline{\it Scuola Normale Superiore and INFN,
Pisa, Italy}
\bsk
\centerline{Yu. Vernov$ ^{*}$ }
\centerline{\it Institute for Nuclear Research
of Russian Academy of Sciences, Moskow, Russia }
\bsk\msk
\ni {\bf Abstract}.
The representations of the \He algebra in Krein spaces,
more generally in weakly complete inner product spaces,
are classified under general regularity and irreducibility
conditions. Besides the Fock representation, two other types
appear, one with negative, the other with two sided discrete
spectrum of the number operator.
\bsk\bsk\bsk\bsk
\ni $ ^{*} $ Supported by
the Russian Foundation for Basic Research
\vfill
\eject
\ni
{\bf 1. Introduction.}
\bsk
The \He \al is the algebraic structure at the basis of
Quantum Mechanics (QM), and it is of great interest not only from
a physical point of view, but also as a basic mathematical structure.
It is the Lie \al generated by the so--called canonical variables,
$q_i$, $p_i$,
$i = 1 \ldots n$, satisfying the following
commutation relations (CCR in unbounded form)
\citaref{W.Heisenberg, {\it The Physical Principles
of the Quantum Theory}, Dover Publ. 1930.}
\citaref{P.A.M.Dirac, {\it The Principles of Quantum Mechanics},
Oxford University Press, Oxford 1986.}
\citaref{J.Von Neumann, {\it Mathematical Foundations of
Quantum Mechanics}, Princeton University Press, 1955.}
$$ [q_i , p_j ] = i \, \delta_{i,j} \ \ \ , \ \ \ \
[q_i , q_j ] = 0 = [p_i , p_j ] \eqno(1.1) $$
The foundations of QM rely on the analysis of the representations
of such an \al and great clarification of the mathematical
structures of QM has been obtained by this analysis
\citaref{J.M.Chaiken, Ann. Phys. {\bf 42}, 23 (1967).}
\citaref{O.Bratteli and D.W.Robinson, {\it Operator Algebras
and Quantum Statistical Mechanics}, Vol.II, Springer--Verlag
1967.}
\citaref{C.R.Putnam, {\it Commutation Properties of Hilbert
Space Operators and Related Topics}, Springer--Verlag 1967.}
>From a technical point of view, non trivial problems arise
because the commutation relations (1.1) imply that $q_i$
and $p_i$ cannot be represented by bounded operators
(in fact, the corresponding group is not compact);
as a consequence, domain questions become important, the
concept of irreducibility is delicate (see below), etc.
For these reasons, one usually considers the algebra $\A$
generated by the Weyl exponentials $\exp i \alpha q$,
$\exp i \beta p$, $\alpha, \beta \in \reali^n$
(briefly, the Weyl algebra).
This amounts to consider, instead of the Lie \al (1.1),
the associated group (the \He group), and to analyse
its representations. A systematic classification of such
representations in a Hilbert space, under general regularity
conditions, is available [5][6]. It should however
be remarked that the choice of the Weyl algebra, generated by
rather special functions of the canonical variables
$q_i$, $p_i$, gives rise to problems, e.g. for the algebraic
setting of the dynamics, since in general the time evolution
does not leave the Weyl algebra stable
\citaref{M.Fannes and A.Verbeure, Comm. Math. Phys. {\bf 35},
257 (1974).}.
Actually, most of the standard treatments of QM models deal with the
canonical variables directly, and therefore it is of some
interest to consider the problem of analysing the representations
of the \He \al directly.
This strategy is actually the most convenient one if one wants to
consider the case of the representations of the CCR's in
indefinite metric spaces, in particular in Krein spaces, since
in this case $q_i$ and $p_i$ are not selfadjoint operators,
the construction of the Weyl exponentials is in general problematic
and requires domain and regularity assumptions,
and the advantage is not significant since the Weyl exponentials
need not to be bounded operators.
A notable case of an analysis of the Hilbert space representations
of the CCR's based on the \He \al is the theorem by Rellich
and Dixmier \citaref{F.Rellich, Nach. Akad. Wiss. G\"ott.
Math--Phys. Klasse, {\bf 110} (1946).}
\citaref{J.Dixmier, Comp. Math. {\bf 13}, 263, (1958).}
\citaref{C.R.Putnam, loc. cit., Ch.IV, 4.6.},
which overcomes the difficulties mentioned above by assuming that
$N = 1/2 \, (p^2 + q^2 - 1)$ is essentially selfadjoint
on a dense domain invariant under the closure of
$p$ and $q$. The result of this note is a generalization
of the theorem by Rellich and Dixmier, which covers the case
of representations of the \He \al in Krein spaces,
toghether with a discussion of irreducibility.
The main motivation of this note is that of a preparatory
result for the classification of the representation of the CCR's
in Krein space, when the number of degrees of freedom
is infinite, a problem which is relevant for
realistic gauge quantum field theories
\citaref{F.Strocchi and A.S.Wightman, J. Math. Phys. {\bf 15},
2198 (1974).}
\citaref{F.Strocchi, {\it Selected Topics on the General Properties
of Quantum Field Theory}, World Scientific 1993.}.
Crucial technical points of our approach are:
i) a proper position of the irreducibility problem in
a situation in which the algebra consists of unbounded operators,
ii) regularity assumptions in the same spirit of Rellich
and Dixmier.
The result is that the spectrum of
$N = 1/2 \, (p^2 + q^2 - 1)$ is discrete (but not semibounded,
in general), and in terms of it one has a classification
of the irreducible representations of the \He \al in Krein
spaces (more generally, in weakly complete non--degenerate
inner product spaces).
Up to isometries between indefinite spaces, such
representations are completely described by the action of the algebra
on the (weakly) dense domain generated by the eigenvector of $N$.
\bsk\bsk\goodbreak\ni
{\bf 2. General setting}
\bsk
As mentioned in the Introduction, in order to analyse the
representations of the \He \al $\h$ in Krein spaces
\citaref{J.Bognar, {\it Indefinite Inner--Product Spaces},
Springer--Verlag, Berlin 1974.}
\citaref{T.Ya.Azizov, I.S.Iokhvidov, {\it Linear Operators
in Spaces with an Indefinite Metric}, J.Wiley and Sons 1989.},
we have to specify domain and regularity conditions and a proper
definition of irreducibility.
For simplicity, in the following we restrict our attention to the \He
algebra for one degree of freedom, the extension to a finite
number of degrees of freedom being straightforward.
We briefly recall that a Krein space $\K$ is an indefinite,
non--degenerate, inner product space, with the properties
\ni
i) it is weakly complete
\ni
ii) there is an operator $J$, self--adjoint
with respect to the inner product,
$J^\dagger = J$ and with $J^2 =1$, such that
$< \cdot , J \cdot>$ is a positive inner product
(giving a Hilbert space structure to $\K$)
Only property i) is needed in the following analysis,
which remains therefore valid for any (sequentially) weakly
complete non--degenerate inner product space $\V$
(see Proposition 11 below). We recall that (sequential)
weak completeness means that if
$\forall x \in \K$
$$ is a convergent sequence, then there exists $y \in \K$
such that $$ converges to $$,
$\forall x \in \K$. The uniqueness of the weak limit is
guaranteed by the non--degeneracy of the inner product
($ = 0$
$\forall x \in \K \Rightarrow y = 0$).
\bsk\ni
{\bf Definition 1}. {\it A representation of a ${*}$ algebra
$\A$ in a Krein space $\K$ is a ${*}$--morphism
$\rho$ of $\A$ into the algebra of operators acting
on a dense invariant domain $\D \subset \K$, namely
$$ \rho (\A) \, \D \subset \D $$
and, $\forall x,y \in \D$, $A \in \A$}
$$ = < \rho (A^{*}) x , y > \eqno(2.1) $$
\bsk
Eq.(2.1) implies that all the operators $\rho(A)$,
$A \in \A$, have a densely defined adjoint $\rho(A)^\dagger$,
with respect to the indefinite inner product,
and $\rho(A^{*}) \subset \rho(A)^\dagger$.
All the operators $\rho(A)$ are therefore closable.
We shall denote by $\D_c$ the common domain of the
closure $\overline{\rho(A)}$ of $\rho(A)$, $A \in \A$,
$$ \D_c = \bigcap_{A \in \A} \D_{\overline{\rho(A)}} \supset \D
\eqno(2.2) $$
In the following, for simplicity, we will denote $\rho(A)$ by $A$.
In general, the domain $\D_c$ is not invariant under
the closures $\bar A$, $A \in \A$.
Any invariant (non--trivial)
subspace $\D^\prime \subset \D_c$, ${\bar A} \D^\prime
\subset \D^\prime$ $\forall A \in \A$, provides a representation
of the algebra in the sense of Definition 1. In fact,
eq.(2.1) implies $\bar A \subset (A^{*})^\dagger $,
and the invariance of $\D^\prime$ implies,
$\forall x \in \D^\prime $,
$\bar B x \in \D_{\bar A} \subset \D_{(A^{*})^\dagger} $,
so that
$\forall y \in \D$,
$x_n \in \D^\prime $ with $x_n \to x$ and $B x_n$ convergent,
$$ \ = \ \ \to $$
$$ \ \ = \
< y, (A^{*})^\dagger \bar B x > \ = \ < y, \bar A \bar B x > $$
By Definition 1, the representation of $\h$ in $\K$
is also generated by the representative of the
so--called creation and annihilation
operators
$$ a^{*} \equiv 1/ \sqrt 2 \, (q - ip) \ \ \ \ \ \ , \ \ \ \ \ \ \ \
a \equiv 1/ \sqrt 2 \, (q + ip) $$
leaving $\D$ invariant and satisfying $[a,a^{*}] =1$ on $\D$.
By the \He relations, $N = \as a$ on $\D$.
\msk \ni
{\bf Remark 1}.
Domain conditions are crucial because of the
unboundedness of the operators $a, a^{*}$; moreover,
for the analysis of the representations it is essential
to consider the closures of the operators,
since otherwise the restrictions of the same representation
to different domains would appear as different, in general
inequivalent representations.
\msk\ni
{\bf Remark 2}.
In a separable Krein space the (strong, Hilbert space)
closure of an
operator $A$ (with densely defined adjoint) coincides with its
weak closure (the weak topologies defined by the indefinite
and by the positive inner product coincide).
We recall that an operator $A$ on a (separable) Krein space
is weakly closable if $x_n \to x$ weakly, and $A x_n$ weakly
convergent implies that the limit of $A x_n$ only depends on $x$;
equivalently, $x_n \to 0$ weakly and $A x_n $ weakly
convergent imply $A x_n \to 0$ weakly.
Operators with densely defined adjoints are weakly
closable, and, by taking subsequences and means, it is easy
to show that closure and weak closure coincide.
By the same argument, in non--separable Krein spaces
the weak closure by sequences is
always contained in the strong closure.
The notion of weak closure remains valid for all sequentially
weakly complete inner product spaces, and is actually all
what is needed in the following. In such spaces, the
extension of the algebraic relations to invariant domains
$\D^\prime \subset \D_c$ follows by the same argument as above.
\bsk
According to the discussion given above, we will consider
representations of the \He \al which satisfy an
additional {\it regularity condition}.
\bsk\ni
{\bf Definition 2}. {\it A representation of the \He \al
on a domain $\D$ will be called regular iff it satisfies
the following condition}:
\msk\ni
A. ({\it Regularity condition}). {\it The operator
$N = 1/2 \, (q^2 + p^2 -1)$ is the weak derivative, on $\D$,
of a one parameter group $U(s)$, $s \in \reali$}:
$$ {d\over ds} |_{s=0} = i \ \ \ \ \ \ \ \ \
\forall y \in \K, \; x\in \D \eqno(2.3) $$
{\it with}
$$ U(s) \D \subset \D \eqno(2.4) $$
$$ U(s) U(t) = U(s+t) \ \ \ \ \ \ \ \ \ \ \
{\rm on} \ \D \eqno(2.5) $$
\msk
Condition A implies
$$ __ \ = \ \ \ \ \ \ \ \ \ \
\forall x,y \in D \eqno(2.6) $$
and therefore
$ U(-s) \subset U(s)^\dagger $,
so that $U(s)$, $s \in \reali$, are closable operators.
Moreover, weakly on $\D$,
$$ {d \over ds} (U(s) \, a \, U(-s)) =
i U(s) \, [ a^{*} a \, , a ] \, U(-s) =
-i \, U(s) \, a \, U(-s) $$
so that
$$ U(s) \, a \, U(-s) = e^{-is} \, a \ \ \ ,
\ \ \ \ U(s) \, a^{*} \, U(-s) =
e^{is} \, a^{*} \ \ \ \ \ \ {\rm on} \ \D \eqno(2.7)$$
Therefore $U(s)$ implements the one--parameter group
of gauge automorphisms of $\h$
$$ \gamma^s (a) = e^{-is} \, a \ \ \ \ \ \ , \ \ \ \ \ \ \
\gamma^s (a^{*}) = e^{is} \, a^{*} \eqno(2.8)$$
Condition A, eq.(2.3), also implies that $U(s)$
is weakly continuous at the origin, and then, by the group law,
for all $s \in \reali$. Eqs.(2.3)--(2.5)
immediately imply
$$ {d\over ds} U(s) = i N U(s) \ \ \ \ \ \ \forall s \in \reali $$
as a weak derivative, on $\D$.
\msk
We denote by $\hx$ the ${*}$ algebra generated by $\h$ and the group
$U(s)$, $s \in \reali$, $U(s)^{*} = U(-s)$,
with the relations (2.7). By Def.2, a regular representation
of $\h$ on a domain $\D \subset \K$ defines a representation
of $\hx$ on $\D$.
In the following, we will denote by $\D_c$ the common domain
of the closure of the operators of $\hx$ (see eq.(2.2));
we will omit the bar which denotes the closure,
when no ambiguity may arise.
\bsk\ni
{\bf Remark 3}.
The regularity condition plays the r\^ole of the Rellich--Dixmier
condition (essential self--adjointness of $N$ on $\D$),
and in fact, in the ordinary Hilbert space case, the stability
of $\D$ under $U(s)$ implies the essential self--adjointness
of its generator $N$ on $\D$. A stronger form of the Rellich--Dixmier
condition is here justified because, without positivity of the
inner product, the $U(s)$ need not to be continuous
everywhere defined operators.
\bsk
The next issue is a proper notion of irreducibility.
As it is well known, the concept of irreducibility is delicate
in the case of algebras of unbounded operators. Algebraic
reducibility is a very weak property
for algebras of {\it unbounded\/} operators, but also the
existence of a proper closed subspace
$\K_1 \subset \K$ containing a common linear domain $\D_1$ invariant
under $\h$ ({\it topological reducibility})
\citaref{A.O.Barut, R.Rackza, \ {\it Theory of Group Representations and
Applications\/}, \ World Scientific 1986.} \ is too weak,
in particular for the \He algebra;
in fact, in the ordinary
irreducible (Schr\"odinger) representation of the Weyl
algebra in $L^2 (\reali)$, it is satisfied by the \He algebra, taken
on the dense domain of $C^\infty$ functions
$f(x)$ of fast decrease,
with all derivatives vanishing at $x =0$;
the subdomain of function with positive support is in fact invariant
under the \He algebra, and
its closure is a proper subspace of $L^2(\reali)$
(see the Appendix for more details).
A more relevant notion, which uses the regularity
properties of the representation, is that of topological
irreducibility {\it for the extended algebra\/} $\hx$:
\bsk\ni
{\bf Definition 3}. {\it A regular representation of $\h$
is called domain irreducible if there is no non--trivial
domain $\D_1 \subset \D_c$ (the common dense domain of the
weak closures of the elements of $\hx$) such that
\ni
i) $\D_1$ is invariant under $\hx$
\ni
ii) $\D_1$ is not dense in $\K$}
\bsk
The above definition recognizes the crucial r\^ole played by
the \lq\lq integrated\rq\rq\ elements $U(s)$,
very close to the r\^ole of the one--parameter
group generated by the Nelson elliptic operator $\Delta$
in the classification of the Hilbert space representations
of unbounded Lie algebras [15]
\citaref{E.Nelson, W.F.Stinespring, Amer. J. Math.
{\bf 81}, 547 (1959).}.
It is immediate to see that domain irreducibility
in the above sense is equivalent to
cyclicity of all vectors $x$ belonging to subspaces of
$\D_c$ invariant under $\hx$, i.e.
$\hx \, x$ is dense in $\K$.
In fact, if $x$ belongs to an invariant domain, by applying
polynominals of elements of $\hx$ to $x$ one gets an
invariant subspace of $\D_c$, which must be dense, by Def.3.
An alternative definition of irreducibility is the one related
to the triviality of the commutant, and it is often used
in the analysis of unbounded operator algebras [15].
Given the representation of $\h$ on the domain $\D$,
one could formulate commutant irreducibility
by asking that any (weakly closable)
operator $B$ which leaves $\D$ invariant and commutes
on $\D$ with $\h$
and $U(s)$, is a multiple of the identity.
The relation between this notion and Definition 3 does not appear
as trivial as in the bounded positive case, because a spectral
theory of unbounded operators in Krein spaces is not available.
One of the points is that reduction may take place only
after the operators have been extended by closure,
and in fact a more relevant notion of
commutant irreducibility
must be given in terms of the extended domain $\D_c$
\bsk\ni
{\bf Definition 4}. {\it A regular representation of $\h$ is
said to be commutant irreducible if whenever a (densely defined)
weakly closable operator $B$ has the properties
\ni
i) $B$ leaves invariant a non--trivial domain $\D_1 \subset \D_c$,
invariant under $\hx$
\ni
ii) $B$ commutes with $\hx$ on $\D_1$
\ni
then $B$ is a multiple of the identity}.
It is easy to see that commutant irreducibility in the sense of
Definition 4 implies domain irreducibility (Definition 3).
In sections 3 and 4 we analyze irreducible representations
in the sense of Def.4.
In Sect.5 the same classification will be obtained (Proposition 12)
assuming only domain irreducibility (Def.3), but using additional
regularity assumptions on $U(s)$.
\bsk\bsk\goodbreak\ni
{\bf 3. Regular irreducible representations of the \He \al
in Krein spaces}
\bsk
The general setting discussed in the previous section allows
for the following characterization of the regular (Def.2)
irreducible (Def.4) representations (Def.1) of the \He algebra $\h$.
\bsk\ni
{\bf Theorem 5}. {\it The regular irreducible representations
of the \He algebra $\h$ in a Krein space $\K$ fall in the following
classes, up to isometries defined on dense subspaces:
\msk\ni
1) $\pi_{\interi^{+}}$: it is characterized by the existence of
a vector $\psz \in \D_c$ (Fock vector) satisfying
$$ a \, \psz = 0 \eqno(3.1) $$
The operators $U(s)$, $s \in \reali$ have a common complete
set of eigenvectors $\psn$, $n \in \naturali$ with
$$ U(s) \, \psn = e^{i n s} \, \psn \eqno(3.2) $$
\msk\ni
2) $\pi_{\interi^{-}}$: it is characterized by the existence of
a vector $\psm \in \D_c$ (anti--Fock vector) satisfying
$$ a^{*} \psm = 0 \eqno(3.3) $$
The operators $U(s)$ have a common complete
set of eigenvectors $\psn$, $n = -1 , -2 , \ldots$ satisfying
eq.(3.2).
\msk\ni
3) $\pi_{\interi}^\theta $, $\theta \in (0,1)$:
it is characterized by the existence of
a vector $\psth \in \D_c$ satisfying
$$ U(s) \, \psth = e^{i \theta s} \, \psth \eqno(3.4) $$
The operators $U(s)$ have a common complete
set of eigenvectors $\psi_n$, $n \in \interi$, with
$$ U(s) \, \psi_n = e^{i (\theta + n) s} \,
\psi_n \eqno(3.5) $$
(\lq\lq Dirac sea states\rq\rq).
\msk\ni
In all the cases, the common eigenvectors of the (closure of)
operators $U(s)$, $s \in \reali$, are unique (i.e there are
no multiplicities), and there are no other eigenvalues}.
\bsk\ni
The proof follows from a few lemmas:
\bsk\ni
{\bf Lemma 6}. {\it The closures of $U(s)$, $s\in \reali$, have
common eigenvectors $\psn$, $n \in \interi$,
$$ U(s) \psn = e^{i (n+\theta)s} \, \psn \eqno(3.6)$$
with $\theta$ fixed in $[0,1)$; the linear span of such
vectors is dense in $\K$}.
\msk\ni
{\bf Proof}. By eq.(2.4), the operator $U(2\pi)$ leaves
$\D$ invariant, and by eq.(2.8) it commutes with $\h$
(and obvously with $U(s)$, $\forall s \in \reali$).
Therefore, by the irreducibility of the representation,
$\udp$ is a multiple of the identity, and actually a phase,
by eq.(2.6):
$$ \udp = e^{2 \pi i \theta} \ \ \ , \ \ \ \ \theta \in [0,1)
\eqno(3.7) $$
Thus, the one--parameter group
$$ V(s) \equiv U(s) \, e^{- i s \theta} \ \ \ ,
\ \ \ \ s \in \reali \eqno(3.8) $$
is a $U(1)$ group:
$$ V(2\pi) = 1 \eqno(3.9) $$
and, $\forall \psi \in \D$, the vector valued function
$V(s) \psi$ is periodic and weakly continuous, by
condition A. Hence, the Riemann sums
$$ \psne \equiv \norme
\sum_{s_k = k \eps, \ 0 \leq k < 2 \pi / \varepsilon} e^{-i n s_k}
\, V(s_k) \, \psi \eqno(3.10) $$
converge weakly, as $\eps \to 0$, to the (weak) integral
$$ \psn \equiv \norm \, \intdp e^{-i n s} \, V(s) \, \psi \, ds
\eqno(3.11) $$
which defines a vector in $\K$, by
weak completeness. Moreover, $\forall t \in \reali$,
$V(t) \, \psne$ also converges weakly as $\eps \to 0$:
$$ V(t) \, \psne = \norme
\sum_{s_k = k \eps, \ 0 \leq k < 2 \pi / \varepsilon} e^{-i n s_k} \,
V(t + s_k ) \, \psi = $$
$$ = \norme
\sum_{s_k} e^{-i n s_k} \, e^{i n t} \,
V(s_k) \, \psi \, \to \, e^{int} \, \psn \eqno(3.12) $$
(where the periodicity of $V(s)$ has been used).
Thus, all the vectors $\psn$ are in the domain of the closures
of $V(t)$, and
$$ V(s) \, \psn = \ens \psn \eqno(3.13) $$
Furthermore, the vector valued weakly continuous periodic
functions $V(s) \, \psi$ can be expressed (pointwise in $s$) by
their (weakly convergent) Fourier series,
$$V(s) \, \psi = \sum_{n \in \interi} \ens \, \psn \eqno(3.14) $$
For $s = 0$, eq.(3.14) expresses all $\psi \in \D$ as weak
limits of linear combinations of the vectors $\psn$, so
that their linear span is dense in $\K$.
\bsk\ni
{\bf Lemma 7}. {\it The vectors
$\psn$ constructed in Lemma 6 are eigenvectors
of the closure of $a^{*} a$}:
$$ \as a \, \psn = (n + \theta) \, \psn \eqno(3.15) $$
\msk\ni
{\bf Proof}. With the notations of the proof of Lemma 5,
by condition A, in the sense of weak derivatives:
$$ (\as a - \theta) \, \psne =
\normim \, \eps \sum_{s_k} e^{-i n s_k} \, {d\over dt} V(t) \,
V(s_k) \, \psi \, |_{t=0} = $$
$$ = \normim \, \eps \sum_{s_k} e^{-i n s_k} \, V(s_k) \,
{d\over dt} V(t) \, \psi \, |_{t=0} $$
$$ \to \normim \, \intdp ds \, e^{-i n s} \,
{d\over ds} V(s) \, \psi \, =
n \, \norm \, \intdp ds \, \ensm \, V(s) \, \psi $$
where an integration by part has been done in the last
step.
\bsk\ni
{\bf Lemma 8}. {\it The linear span $\D_1$ of all the vectors
$\psn$ constructed in Lemma 6, as $\psi$ varies on $\D$,
is contained in $\D_c$ (the common domain of the weak closures
of the elements of $\hx$), and it is left invariant by $\hx$.
The \He relations hold on $\D_1$, and each vector in $\D_1$
is cyclic for $\hx$.
For any $\psi \in \D$ there is an integer $\nb$ such that the
vectors $(\as)^k \psnb$, $a^l \psnb$ generate a dense
subspace of $\D_1$}.
\msk\ni
{\bf Proof}. By eq.(2.7),
$$ a \, \psne = \norme
\sum_{s_k} e^{-i n s_k} \, a \, V(k \eps) \, \psi = $$
$$ = \norme
\sum_{s_k} e^{-i n s_k} \, e^{i s_k} V(s_k) \, a \, \psi =
(a \psi)^{(\eps)}_{n-1} \eqno(3.16) $$
and similarly
$$ \as \psne = (\as \psi)^{(\eps)}_{n+1} \eqno(3.17) $$
Since the r.h.s. of eqs.(3.16),(3.17) converge by Lemma 6,
the vectors $\psn$ belong to the domain of the weak closure
of $a$ and $\as$, and
$$ \ab \, \psn = (a \, \psi)_{n-1} \ \ \ \ \ , \ \ \ \ \ \
\asb \, \psn = (\as \psi)_{n+1} \eqno(3.18) $$
The same treatment applies to all the products of $a$ and $\as$,
so that $\D_1 \subset \D_c$ and $\hx D_1 \subset \D_1 $.
By irreducibility, $\hx \chi$ is dense in $\K$,
$\forall \chi \in \D_1$.
The \He commutation relations hold on $\D_1$, as a consequence
of their validity on $\D$, by eq.(3.18):
$$ (a \as - \as a) \, \psn = a\, (\as \psi)_{n+1} -
\as (a \psi)_{n-1} =
((a \as - \as a) \, \psi)_n = \psn \eqno(3.19) $$
>From eqs.(3.19) and (3.15) one easily gets
$$ a \, \as \, \psn = (n+1+\theta) \, \psn \eqno(3.20) $$
(an equation which could have been derived directly
as in Lemma 7).
For each $\psi$ in $\D$, by eq.(3.14) there is at least one
$\psnb$ which does not vanish, and is then cyclic.
Given any product of operators $a$, $\as$ applied to $\psnb$,
by using eqs.(3.18),(3.15),(3.20), one can write it as a
constant times $(\as)^k \, \psnb$, if the number of $\as$ operators
exceeds, by $k$, that of the $a$'s, or, otherwise, as $a^k \, \psnb$.
The ciclicity of $\psnb$ implies therefore the density in $\K$ of
the linear span of the vectors
$\{ (\as )^k \, \psnb \, , \, a^l \psnb$, $k,l \in \naturali \}$.
\bsk\ni
{\bf Proof of Theorem 5}. By Lemma 8, the analysis of regular
irreducible representations of $\h$ can be done on the domain
$\D_1$. Given a vector $\psnb \in \D_1$, we consider first
the case in which at least one of the sequences $a^k \, \psnb$
terminates. If, for some $ k > 0 $,
$$ a^k \, \psnb = 0 \ \ \ \ \ \ \ \ \ a^{k-1} \psnb \neq 0 \eqno(3.21)$$
then, by eqs.(3.15),(3.18) one has
$$ 0 = \as a \, a^{k-1} \, \psnb =
\as a \, (a^{k-1} \, \psi)_{\nb -k +1} =
(\nb -k +1 +\theta) \, a^{k-1} \, \psnb $$
i.e.
$$ \nb -k +1 +\theta = 0 \eqno(3.22) $$
which implies $\theta = 0$, since $\theta \in [0,1)$.
Similarly,
$$ (\as)^l \, \psnb = 0 \ \ \ \ \ \ \ \ \ (\as)^{l-1} \, \psnb \neq 0
\eqno(3.23)$$
implies, by eqs.(3.18),(3.20),
$$ \nb +l +\theta = 0 \eqno(3.24) $$
and therefore again $ \theta = 0 $
Eqs.(3.22) and (3.24) are incompatible, and therefore only one
sequence can terminate.
\ssk\ni
{\bf Case 1}. Eq.(3.21) holds. Then, putting
$$ \psz \equiv a^{k-1} \, \psnb = (a^{k-1} \, \psi)_0 \eqno(3.26) $$
one has
$$ a \, \psz = 0 \eqno(3.25) $$
and $\D_1$ is generated by the vectors $(\as)^l \, \psz$, $l > 0$.
({\it Fock representation}). By using eqs.(3.18),(3.20),(3.25)
one gets
$$ < (\as)^l \psz , (\as)^m \psz > \, =
\cases{& $0$ \ \ \ \ for \ $l \neq m$
\cr & $m! \, <\psz, \psz>$ \ \ \ \ for \ $l = m$ } \eqno(3.27) $$
Since the linear span of the vectors $(\as)^m \,
\psz$ is dense in $\K$ (Lemma 8), the non--degeneracy
of the inner product implies $<\psz,\psz> \neq 0$, and eq.(3.27)
fully determines the inner product in $\K$, which turns out to be
definite. Without loss of generality
we can normalize $< \psz , \psz> \equiv \eta_0 = \pm 1$,
and the Krein metric operator is then
$$ J = \eta_0 1 $$
Apart from the sign of the inner product,
the representation is the usual Fock representation in a Hilbert
space.
\msk\ni
{\bf Case 2}. Eq.(3.23) holds. Then, putting
$$ \psm \equiv (\as)^{l-1} \, \psnb = ((\as)^{l-1} \, \psi)_{-1}
\eqno(3.28) $$
one has
$$ \as \, \psm = 0 \eqno(3.29) $$
and $\D_1$ is generated by the vectors $a^l \, \psz$ , $l > 0$.
({\it \lq\lq Anti--Fock representation\rq\rq }).
By using eqs.(3.15),(3.18),(3.28) one gets
$$ < a^l \psm , a^m \psm > = \delta_{l,m} \, m! \,
(-1)^m \, <\psm, \psm> \eqno(3.30) $$
The non--degeneracy
of the inner product implies $<\psm,\psm> \neq 0$, and
the inner product in $\K$ is fully determined by eq.(3.30),
and it is not definite.
The metric operator which defines the Hilbert space structure of
$\K$ is not uniquely determined by eq.(3.30).
There is however a distinguished Krein structure
associated to eq.(3.30), defined by the metric operator
$$ J = \eta_1 \, (-1)^N \eqno(3.31) $$
where $ \eta_1 \equiv < \psm , \psm> = \pm 1$.
In general, the representation defined (on $\D_1$) by eq.(3.30) may
live in a Krein space obtained by closing $\D$ with
a (weak) topology different from
that defined by the above metric operator.
However, all such Krein spaces contain dense domains $\D_1$,
on which the corresponding representations are related
by (indefinite space) isometries. Thus, any representation
of type 2 is isometrically equivalent to the representation
defined by eq.(3.30) on the Krein space given by
the metric (3.31).
\msk\ni
{\bf Case 3}. None of the sequences $a^k \, \psnb$, $(\as)^l \, \psnb$
terminates. Then one of the sequences contains a vector $\psth$ with
$$ \as a \, \psth = \theta \, \psth \eqno(3.32) $$
and $\D_1$ is generated by the vectors
$ a^l \, \psth$, $(\as)^k \, \psth$, $ l,k \in \naturali $.
({\it \lq\lq Dirac sea states\rq\rq }).
Different eigenvalues of $U(s)$ define orthogonal eigenspaces
(with the indefinite inner product), and,
using the commutation relations of Lemma 8, one has
$$ < (\as)^l \psth , (\as)^m \psth > \, = \delta_{l,m} \,
(\theta +1) (\theta + 2) \ldots (\theta + m) \,
<\psth, \psth> \eqno(3.33) $$
$$ < a^l \psth , a^m \psth > \, = \delta_{l,m} \,
\theta (\theta -1) \ldots (\theta - m) \, <\psth, \psth> = $$
$$ = \delta_{l,m} \, (-1)^{m-1} \,
\theta (1 - \theta) \ldots (m-1-\theta) \, <\psth, \psth>
\eqno(3.34) $$
Again $<\psth,\psth> \neq 0$ follows from non--degeneracy of
the inner product, which is fully determined
by eqs.(3.33),(3.34), and is not definite.
Again, the metric operator is not uniquely determined, and
a distinguished Krein structure is defined by
$$ J \, (\as)^m \, \psth = \eta_0 \, (\as)^m \, \psth \ \ \ \ \
\ \ \ \ \ J \, a^l \, \psth =
\eta_0 \, (-1)^{l-1} \, a^l \, \psth \eqno(3.35) $$
with $\eta_0 \equiv < \psth , \psth> = \pm 1$.
The same arguments as in case 2 then apply to the equivalence
of representations in different Krein spaces.
In order to control all the common eigenvectors of $U(s)$,
$s \in \reali$,
assume that a vector $\chi$, in the domain of the closure
of all the operators $U(s)$, satisfies
$$ U(s) \,\chi = \lambda (s) \chi $$
By weak continuity, and eq.(2.5),
$$\lambda (s) = e^{i \mu s} $$
>From eq.(2.6) (extended to the closure of the operators),
it follows that $\chi$ is orthogonal to all the eigenvectors
of $U(s)$ with different eigenvalues, and therefore, by
the density of the domains $\D_1$ constructed above,
$\exp i \mu s$ is, in the three different cases, one
of the eigenvalues classified above,
$$ e^{i \mu s} =
e^{i (n + \lambda) s} $$
For each $\psi \in \D$ the
set of eigenvectors constructed in Lemma 8
generates a dense subspace; since, by the argument given above,
$\chi$ is orthogonal to
all the such eigenvectors, except for the eigenvalue
$ \exp i (\nb + \theta) s $, it follows that
$\chi$ is a multiple of the corresponding eigenvector,
which is therefore unique.
It also follows that the all the subspaces of $\D_1$
introduced in Lemma 8 actually coincide with $\D_1$.
\bsk\bsk\goodbreak\ni
{\bf 4. Properties of the regular irreducible representations
of the \He algebra}.
\bsk
We list some properties of the regular irreducible
representation of $\h$ in Krein spaces, and show how the result
generalize to weakly complete inner product space.
We recall that the characterization of Theorem 5 amounts to
the construction of a dense subspace of $\D_1$ of $\K$, where the
representation is given explicitly. The indefinite inner product
is completely determined on $\D_1$, and
the representations are therefore classified up to isometries
of such domains.
Different positive scalar product, and therefore different
weak topologies and different completions,
are compatible with the representations on
$\D_1$, but the simple choice of Hilbert product given
in the proof of Theorem 5 is always possible and gives rise to
very simple spectral properties of the operators $U(s)$:
\bsk\ni
{\bf Proposition 9}. {\it In the regular irreducible
representations of the \He algebra, with the
Krein structures introduced in the proof of Theorem 5,
the operators
$U(s)$ which implement the gauge transformations (2.8)
are unitary operators with simple pure point spectrum}.
\msk\ni
{\bf Proof}. All the Krein metric operators $J$ introduced
in the proof of Theorem 5 commute with $U(s)$;
eq.(2.6) then implies that $U(s)$ are unitary (with
respect to the positive scalar product). From
Theorem 5 it follows immediately that the spectrum is
pure point, without multiplicities.
\msk
The logic of Theorem 5 is that, {\it if\/} regular irreducible
representations exist, they are of the types 1)--3); to
complete the classification, it is necessary to show that
indeed all the representations constructed in Theorem 5
fulfill the regularity and irreducibility conditions.
Algebraic relations and regularity are obvious, on suitable
domains (e.g., the linear span of the vectors $\psn$),
and for the irreducibilty we have:
\bsk\ni
{\bf Proposition 10}. {\it All the representations characterized
in Theorem 5 are commutant irreducible (Def.4),
and therefore also domain irreducible (Def.3)}.
\msk\ni
{\bf Proof}. The proof follows from the absence of
multiplicities for the common eigenvalues of $U(s)$, $s \in \reali$.
If a closable operator $B$ leaves a domain
$\D^\prime \subset \D_c$ invariant, with
$\hx \, \D^\prime \subset \D^\prime $, then for any
$\psi \in \Dp$ one can construct vectors $\psne$ as in
Lemma 6, eq.(3.10), which converge weakly to vectors
$\psn$, eq.(3.12), still belonging to an invariant domain
contained in $\D_c$;
if $B$ commutes with $\hx$, then
$$ B \, \psne \to (B \, \psi)_n \ \ \ \ \ \ \ {\rm weakly}
\eqno(4.1) $$
We can fix $n$ so that $\psn$ is non--vanishing;
since $B$ is (weakly) closable, $\psn$
belongs to the domain of its weak closure $\bar B$, and
$$ \bar B \, \psn = (B \psi)_n \eqno(4.2) $$
\ni
Both $\psn$ and $(B \psi)_n$ are eigenvectors of $U(s)$,
$\forall s \in \reali$, as in Lemma 6, eq.(3.6). Since
there are no multiplicities, eq.(4.2) implies
$$ B \, \psn = \lambda \, \psn \eqno(4.3) $$
Since $B$ commutes with $\h$,
eq.(4.3) also holds, with the same $\lambda$,
for the vectors $ \{ a^k \psnb , (a^{*})^l \psnb \} $,
generating a dense domain.
$B$ is closable, and therefore $B = \lambda 1$
\bsk
The same analysis as in Theorem 5 can be done under the weaker
assumption that $\D$ is a dense subspace of a sequentially
weakly complete
non--degenerate inner product space $\V$, rather than a Krein space.
The assumption is indeed weaker, since there exist weakly complete
inner product space which are not Krein spaces.
\bsk\ni
{\bf Proposition 11}. {\it The regular, commutant irreducible
representations of the \He \al in sequentially weakly complete
spaces are classified by Theorem 5, up to isometries
defined on dense subspaces}.
\msk\ni
{\bf Proof}.
The proof of Theorem 5 only uses the representation of
$\h$ on a weakly dense domain $\D$ and the regularity condition
A on $\D$, and constructs the common eigenvectors of $U(s)$
by taking weak limits and extending weakly closable operators.
\bsk
Absence of multiplicities for the eigenvalues of $U(s)$ and
irreducibility of the representations follow in the same way.
On the dense domain $\D_1$ constructed in the proof of Theorem 5,
the representations are clearly identical to those in
Krein spaces; their classification has therefore nothing to
do with the completion of $\D_1$, which can always
be embedded in the distinguished Krein spaces constructed
in the proof of Theorem 5.
Weak completeness is however
necessary, in general, for the construction of the
domain $\D_1$ starting from the domain $\D$ of a given representation;
the domain $\D$ is in fact not unique, and may be any dense
subspace, with the only constraints of invariance
and validity of the regularity condition.
\bsk\bsk\goodbreak\ni
{\bf 5. An alternative derivation}.
\bsk
The same results as in Theorem 5 can be obtained,
in any complete non degenerate inner product space $\V$,
under the following assumptions:
\ni
i) Regularity (Def.2)
\ni
ii) Domain Irreducibility (Def.3)
\ni
iii) Temperedness of $U(s)$, $s \in \reali$: $\forall
x \in \D$, there exists a polynomial $P_{x} (n)$ such that
$\forall y \in \V$ there exists a constant $C_y$, such that
$$ | | \leq C_y \, P_{x} (n)
\ \ \ \ \ \ \ \ \ \ \forall n \in \interi \eqno(5.1) $$
In fact we have
\msk\ni
{\bf Proposition 12}. {\it Under the assumptions i), ii),
iii), the operator $ \udpn $ is a multiple of the identity},
$$ \udpn = e^{2 \pi i \theta} \, 1 \eqno(5.2) $$
\msk\ni
{\bf Proof}.
By eq.(2.7), $\udpn$ commutes with the algebra
$\h$ on $\D$, but eq.(5.2) does not immediately follow, as in the
standard case (bounded operators in Hilbert spaces),
and a crucial r\^ole is played by the regularity and
temperedness conditions (a counterexample, if regularity
does not hold, was given in Sect.2).
A substitute for the spectral theory of the group $\udpn$
is necessary, and can be constructed if condition iii)
is satisfied. The Fourier transform of $\udpn$,
$$ T(t) = \suminf \udpn \, e^{i n t} $$ is in fact well defined as
an operator valued distribution on $\D$ by the following
argument: define
$$ T_N(t) \equiv \sumN \udpn \, e^{i n t} $$
and, for all $f(t)$ infinitely differentiable on the circle,
$t \in [0,2\pi)$
$$ T_N(f) \equiv \intdp T_N(t) \, f(t) \, dt =
\sumN \, c_f^{(n)} \, \udpn \eqno(5.3) $$
with $c_f^{(n)}$ the Fourier coefficients of $f$. Since
such coefficients decrease faster than any polynomial,
the regularity condition iii) immediately implies that
$T_N(f)$ converges weakly on $\D$, as $N \to \infty$.
By the weak completness of $\V$, the limit defines an operator
which maps $\D$ in $\V$. Furthermore, since all the elements
$A \in \h$ are closable,
$$ A \, T_N(f) \, x = T_N(f) \, A \, x \to T(f) \, A \, x
\ \ \ \ \ \ \ \forall x \in \D$$
implies
$$ A \, T_N(f) \, x \to A \, T(f) \, x \eqno(5.4)$$
and therefore
$$ [A , T(f)] = 0 \ \ \ \ \ \ {\rm on} \ \D \eqno(5.5) $$
The operator valued distribution $T(f)$ satisfies
on $\D$
$$ T(1) = \suminf \udpn \, \intdp \ent \, dt = U(0) = 1 $$
If $f_1(t) $ and $f_2(t)$ have disjoint support,
$f_1 (t) \, f_2(t) = 0$, then
$$ T(f_1) \, T(f_2) = 0 \eqno(5.6) $$
In fact, on $\D$,
$$ T(f_1) \, T(f_2) = \sum_{n,m} \udpn \, \udpm \,
c_{f_1}^{(n)} \, c_{f_2}^{(m)} = $$
$$ = \sum_{k,m} \udpk \,
c_{f_1}^{(k-m)} \, c_{f_2}^{(m)} \equiv
\sum_{k} \udpk \, c^{(k)} $$
with
$$ c^{(k)} = \sum_m c_{f_1}^{(k-m)} \, c_{f_2}^{(m)} =
\sum_m \intdp \ekt \, e^{-i m t} \, f_1(t) \, dt \, \intdp
e^{i m t^\prime} \, f_2(t^\prime) \, dt^\prime = $$
$$ \sum_m \intdp \ekt \, e^{-i m (t- t^\prime)} \,
f_1(t) \, f_2(t^\prime) \, dt \, dt^\prime =
\intdp \ekt \, f_1(t) \, f_2(t) \, dt = 0 $$
\ni
In order to prove that $\udpn$ is a multiple of the identity
we consider two possibilities:
\ni
i) The support of $T$ consists of only one point,
denoted by $t_0$. Let us fix $x \in \D$;
the bound (5.1) and the support property of $T$ implie,
$\forall y \in \D$,
$$ =
\sum_{k=0}^N a_k (y) \, n^k \eqno(5.7) $$
Then, by putting
$$ x_k \equiv \sum_{n=0}^N \, A_{kn} \, \udpn \, e^{-i n t_0} x
\ \ \ \ , \ \ \ \ \ \ k = 0, 1 \ldots N \eqno(5.8) $$
with $A_{kn}$ the inverse matrix of $B_{nj} \equiv n^j$,
$$ \sum_{n=0}^N \, A_{kn} \, n^j = \delta_{kj} $$
one has
$$ a_k(y) = $$
and therefore
$$ \udpn \, x = \sum_{j=0}^N \, n^j \, e^{i n t_0} \, x_j \eqno(5.9) $$
The lowest integer $N$ for which eq.(5.7) holds will be called
the degree of $x$.
The subspace $\D_M \subset \D$ of vectors of degree less or
equal to some $M$ is
invariant under $\hx$, and non--empty for sufficiently large $M$;
we have therefore constructed operators $O_j$, $j = 0 \ldots M$,
defined on $\D_M$ by $O_j \, x = x_j$; by eq.(5.8), the operators
$O_j$ are given by finite linear combinations of
operators $\udpn$, so that they are closable
and leave $\D_M$ invariant. The operator
$O_M$ does not vanish, if $M$ is chosen small enough, and,
if $M>0$, it satisfies, by the group law for $U(s)$,
$$ O_M ^2 = 0 \eqno(5.10) $$
$O_M \, \D_M$ is invariant under $\hx$ and is therefore dense,
by domain irreducibility;
this is incompatible with eq.(5.10); all vectors
in $\D$ are therefore of degree $0$, i.e.
$$ \udpn x = e^{i n t_0} x \ \ \ \ \ \ \
\forall x \in \D \eqno(5.11) $$
\ni
ii) The support contains at least two points.
There are then two functions, $f_1$, $f_2$, with
$T(f_i) \neq 0$, $i = 1,2$, and
$f_1 f_2 = 0$. Then, $T(f_1) \D$, by eq.(5.4) a subspace of
$\D_c$, must be dense, since it is invariant under $\h$
and the representation is domain irreducible.
But then $T(f_2) T(f_1) = 0 $ is incompatible with
$T(f_2) \neq 0$, since all $T(f)$ are closable.
\msk\ni
The proof of Theorem 5 then proceeds as in Sect.3
\bsk\bsk\goodbreak\ni
{\bf Appendix}
\bsk\ni
We briefly discuss here:
\ni
1. a class of \lq\lq counterexamples\rq\rq\
concerning the notion of irreducibility, and its relation with
regularity, for the \He algebra in a Hilbert space;
\ni
2. an example of a weakly complete inner product space which
is {\it not\/} a Krein space.
\bsk\ni
1. The (ordinary) regular representation of
the Weyl algebra in the Hilbert space $L^2(\reali)$,
$$ e^{ i \alpha q} \; e^{ i \beta p} \; \psi (x) =
e^{i \alpha x} \; \psi (x + \beta) $$
defines (by taking strong derivatives) representations
of the \He algebra with reducibility and regularity properties
which depend substantially on the domains which are considered.
In fact, regularity and (topological) irreducibility hold if an invariant
domain of analytic vectors for $q$ and $p$ is chosen,
e.g. the domain generated by the eigenvectors of $N$, but
very different situations arise for different domains.
Consider, e.g., the dense domain $\D_0$ of $C^\infty$ functions
$f(x)$ of fast decrease, vanishing toghether with
all their derivatives at $x =0$. $\D_0$ is clearly
invariant under the \He algebra,
but the representation of the \He algebra on $\D_0$ does
{\it not} satisfy the regularity condition, since the operators
$U(s)$ exist but do not leave $\D_0$ invariant.
The subdomains $\D_0^{+}$
and $\D_0^{-}$, consisting of function with positive, resp. negative,
support are invariant, and their closure are
proper orthogonal subspaces of $L^2(\reali)$.
The representation is therefore {\it domain reducible\/}. It is
also {\it commutant reducible\/}, since the projector on
$L^2(0,\infty)$ commutes with the \He algebra on $\D_0$
(not only in the weak sense,
since the domain is left invariant by the projector).
If the representation of $\h$ is defined,
by taking strong derivatives, on a domain $\D \supset \D_0$
invariant under $\h$ and $U(s)$, $s \in \reali$, then the
regularity condition is satisfied.
However, domain irreducibility does {\it not} hold for
{\it the \He algebra} on $\D$, since the previously
considered domains are obviously still invariant under $\h$.
Domain irreducibility {\it holds
for the extended algebra\/} $\hx$; in
fact, invariance of a domain $\D'$ under $U(s)$ implies,
using the construction
given in the proof of Theorem 5, that one, and therefore every,
eigenvector of the operators $U(s)$ belongs to the closure of $\D'$,
and $\D'$ is therefore dense.
Identical conclusions hold for commutant reducibility, since
one may still consider the same projectors as above, as
operators defined on $\D_0$, commuting there with $\h$.
On the other hand, commutativity
with $\hx$ on an invariant domain (Def.4) implies triviality as in the
proof of Proposition 10.
It may also be remarked that the operator $N$ is
{\it not\/} essentially self--adjoint on $\D_0$, and that
the extended algebra (with the validity of
relations (2.7)) can be constructed only on the basis
of {\it one} choice for its self--adjoint extension
(namely by taking functions continuous,
toghether with their derivative, at $x=0$).
All other choices lead to different extensions of $\h$ (with
algebraic relations different from (2.7)), and in general
to different spectra of $N$.
Very similar examples can be constructed by using any sequence of
intervals (rather than $(-\infty,0)$ and $(0,\infty)$)
to define invariant subspaces, with very similar results.
Our conclusion is that both the {\it regularity\/} property and
the use of the {\it extended algebra\/} are essential in order to
avoid pathologies in the discussion of the
representations of the \He algebra, even in positive (Hilbert) spaces.
In indefinite spaces, domain questions become even more critical.
E.g., on different domains in the same
Krein space one may define different, and in general inequivalent,
positive scalar products, so that the same domains can also be
seen as dense subspaces of {\it different Krein spaces}.
On the other hand, since a sufficiently general
spectral theory is not available in indefinite
spaces, all {\it regularity} properties must be expressed in terms
of domains.
Roughly speaking, in order to fix the representation one
must start with a domain which is large enough for
the control of the regularity condition;
no extension is then needed, and the main r\^ole of the Hilbert
structure is that of giving rise to complete spaces.
\bsk\ni
2. In order to construct a weakly complete inner product
space which is not a Krein space, we start with
the space $V_0$ of {\it finite\/} sequences $a_n$, $n\in \interi$,
with the inner product
$$ < a_n \, , \, b_m > \equiv \sum_n \bar a_{-n} b_n $$
On the space $V_0$ it is easy to define positive scalar
products which majorize the indefinite inner product (see [13]),
and therefore $V_0$ can be completed to a Krein space
(inequivalent scalar products defining different Krein completions).
However, not all weakly complete inner product spaces
which contain $V_0$ as a dense subspace are Krein spaces.
Consider in fact the space $V \supset V_0$ consisting of {\it all\/}
sequences which terminate on the right, $a_n = 0$ for $n$ large enough.
The space $V$ is weakly complete, because i)
weak convergence implies convergence of each term of the
sequences, ii) by completeness of the $l^2$ spaces,
the limit sequences $x_n$ belong to all the spaces
$l^2$ with {\it any} weight function $p_n > 0$
on $n \in \naturali$,
$$ \sum_{n=0}^\infty |x_n|^2 \, p_n < \infty $$
and therefore $x_n$ vanishes for $n$ large enough.
Moreover, $V$ is not a Krein space, since in any Krein space
containing $V_0$ there are always
sequences which are infinite on the right, and have
weakly convergent truncations;
one such sequence is immediately constructed,
e.g., as $ 1 / (c_k \, k^2) $, $k > 0$, with $c_k$ the Hilbert
norm of the sequence $ \delta_{n,k}$, in the given Krein space.
The argument also shows that $V$ is not a Banach space, so that it
cannot be obtained from $V_0$ by completion with respect to a norm.
\vfill\eject\references
\bye
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