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Topological Properties of the Unitary Group 3

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Topological Properties of the Unitary Group 3 TOPOLOGICAL PROPERTIES OF THE UNITARY GROUP JESUS´ ESPINOZA AND BERNARDO URIBE Abstract. We show that the strong operator topology, the weak operator topology and the compact-open topology agree on the space of unitary opera- tors of a infinite dimensional separable Hilbert space. Moreover, we show that the unitary group endowed with any of these topologies is a Polish group. Introduction The purpose of this short note is to settle some topological properties of the unitary group U( ) of an infinite dimensional separable Hilbert space , whenever the group is endowedH with the compact open topology. H When dealing with equivariant Hilbert bundles and its relation with its associ- ated unitary principal equivariant bundles (see [1]), one is obliged to consider the compact-open topology on the structural group U( ). In one of the foundational papers for twisted equivariantH K-theory, Atiyah and Segal claimed that the unitary group endowed with the compact-open topology was not a topological group [1, Page 40], based on the fact that the inverse map on GL( ) is not continuous when GL( ) is endowed with the strong operator topol- ogy.H This unfortunate claim obligedH Atiyah and Segal to device a set of ingenious constructions in order to make U( ) into a topological group with the desired topo- logical properties suited for the classificationH of Fredholm bundles. Nevertheless, these ingenious constructions of Atiyah and Segal added difficulties on the quest of finding local cross sections for equivariant projective bundles, and therefore a clarification on the veracity of the claim was due. The purpose of this note is to show that the unitary group U( ) endowed with the compact-open topology is indeed a topological group, moreoveHr a Polish group, and that this topology agrees with the strong operator topology, as the weak op- arXiv:1407.1869v1 [math.AT] 7 Jul 2014 erator topology; this is the content of Theorem 1.2 which is the main result of this note. 1. Operator topologies on the unitary group Let be a separable Hilbert space with inner product , inducing the norm H h i x := x, x for x . Let L( , ) denote the vector space of linear operators from| | toh iand denote∈ H by B( H) theH vector space of bounded linear operators Hp H H B( ) := T L( , ): M > 0 x T x M x . H { ∈ H H ∃ ∀ ∈ H | |≤ | |} 2010 Mathematics Subject Classification. (primary) 47D03 (secondary) 57S20. Key words and phrases. Unitary group, strong operator topology, compact-open topology. The first author acknowledges the financial support of CONACyT and of the Universidad de Sonora. The second author acknowledges the financial support of the Alexander von Humboldt Foundation. 1 2 JESUS´ ESPINOZA AND BERNARDO URIBE The space of bounded linear operators B( ) endowed with the norm H T = sup T x | | |x|≤1 | | becomes a Banach space. Denote the adjoint operator T ∗ as the operator defined by the equality Tx,y = x, T ∗y for all x, y . The space B(h ) cani beh endowedi with several∈ H operator topologies and we will list the ones that interestH us in this work. They all can be defined by specifying which are the convergent sequences. Then let Tk be a sequence of linear operators on { } B( ) and denote by Tk ? T the statement that Tk converges to T in the topology ?, andH denote B( ) the→ space of bounded operators endowed with the topology H ? ?. In what follows the notation xk x denotes that the sequence xk k in converges to x. → { } H Norm Topology (Uniform convergence): • Tk n T if sup Tkx T x 0. → |x|≤1{| − |} → Strong Operator Topology (Pointwise convergence): • Tk s T if for all x , Tkx T x. Weak→ Operator Topology:∈ H → • Tk w T if for all x, y , Tkx, y Tx,y Compact→ Open Topology∈ H h (Uniformi → h convergencei on compact sets): • Tk co T if for all compact subsets C , the restricted sequence Tk C k → ⊂ H { | } converges uniformly to T C. Strong* Operator Topology:| • ∗ ∗ ∗ Tk s T if both Tk s T and Tk s T . Weak*→ Operator Topology:→ → • ∗ ∗ ∗ Tk w T if both Tk w T and Tk w T . Compact→ Open* Topology:→ → • ∗ ∗ Tk co∗ T if both Tk co T and T co T . → → k → The identity map on B( ) induces the following commutative diagram of con- tinuous maps H / / / (1.1) B( ) / B( ) ∗ / B( ) ∗ / B( ) ∗ H n H co H s H w B( ) / B( ) / B( ) H co H s H w with the property that none of the maps is a homeomorphism. Let U( ) denote the group of unitary operators on , i.e. H H U( )= U B( ): UU ∗ = U ∗U = Id , H { ∈ H H} and note that U −1 = U ∗ and that x, y = Ux,Uy for all x, y . Since U( ) B( ), the group U( ) can beh endowedi h with anyi of the operator∈ H topolo- H ⊂ H H gies previously defined. The group U( )n endowed with the norm topology is a topological group and it is the prototypicalH example of what is known as a Banach Lie group, see [3]. The fact that U( ) endowed with any of the other topologies defined above is also a topological groupH is the main result of this section Theorem 1.2. The operator topologies: compact open, strong, weak and their * counterparts, all agree on the group U( ), i.e. H U( ) ∗ = U( ) ∗ = U( ) ∗ = U( ) = U( ) = U( ) . H co H s H w H co H s H w TOPOLOGICAL PROPERTIES OF THE UNITARY GROUP 3 Moreover, the group U( ) endowed with any of these topologies is a Polish group, i.e. a completely metrizableH topological group. The proof of the theorem is built out from various Lemmas: Lemma 1.3. The map U( ) U( ) , T T ∗ is continuous. In particular H w → H w 7→ U( ) ∗ = U( ) . H w H w Proof. Consider Tk w T and x, y . Then → ∈ H ∗ ∗ x, T y = Tkx, y Tx,y = x, T y , h k i h i → h i h i ∗ ∗ ∗ and therefore Tk w T . This proves that T T is continuous. Hence we conclude that the weak→ and the weak* topologies7→ agree on U( ). H Lemma 1.4. The composition of operators U( )s U( )s U( )s, (T,S) TS, is continuous in the strong operator topology.H × H → H 7→ Proof. Consider a convergent sequence (Tk,Sk) s (T,S) in U( )s U( )s. For x we have → H × H ∈ H TkSkx T Sx = TkSkx TkSx + TkSx T Sx | − | | − − | TkSkx TkSx + TkSx T Sx ≤ | − | | − | = Tk(Skx Sx) + (Tk T )Sx | − | | − | = Skx Sx + (Tk T )Sx , | − | | − | and since Skx Sx and Tk(Sx) T (Sx), then we have that (TkSk)x (TS)x. Therefore the composition→ of operators→ is continuous in the strong operator→ topol- ogy. Lemma 1.5. The weak and the strong operator topologies agree on U( ). There- H fore we have that U( )s is a topological group and moreover that U( )s∗ = U( )s = U( ) . H H H H w Proof. Consider a convergent sequence Tk w T in U( ) . To prove that Tk s T → H w → it is enough to show the convergence Tkx T x for x a unit vector in . We have → H then that x = 1 and therefore Tkx =1= T x . Now we compute | | | | | | 2 2 2 Tkx T x = Tkx 2Re Tkx, T x + T x =2 2Re Tkx, T x | − | | | − h i | | − h i and since Tk w T , we have that Tkx, T x T x, T x = 1. Therefore Tkx 2 → h i → h i | − T x 0 and hence we conclude that Tkx T x. This shows that the identity map| U→( ) U( ) is continuous, and therefore→ U( ) = U( ) . H w → H s H w H s Now, by Lemma 1.3 we know that the inverse map T T ∗ = T −1 is continuous, and by Lemma 1.4 we know that the composition is continuous.7→ Then we have that U( ) with the strong (or weak) operator topology is a topological group; the fact thatH the strong* topology agrees with the strong topology follows from the continuity of the inverse map. The proofs of the previous lemmas follow the proofs that appear in [2, Cor. 9.4]. Lemma 1.6. The topological group U( ) is metrizable. H s 4 JESUS´ ESPINOZA AND BERNARDO URIBE Proof. Consider an orthonormal basis ej j N of . Consider the map { } ∈ H N Ψ: U( ) , T (Tej)j N H s → H 7→ ∈ N where is endowed with the product topology. For a convergent sequence Tk s H → T we have that Tkej Tej for all j N, this implies that the maps U( ) → ∈ H s → , T Tej are all continuous and therefore by the universal properties of the H 7→ product topology we obtain that the map Ψ is continuous. Since the basis ej j∈N generates a dense subset of , we have that any two operators which agree{ on} the H basis ej j∈N must be equal; hence we have that the map Ψ is injective. Now{ let} us show that the map Ψ is an embedding. For this purpose let us take a convergent sequence Ψ(Tk) Ψ(T ) in the image of Ψ. Since Tkej Tej for → → all basis vectors, we have that the sequence Tk k converges pointwise to T in { } the dense subset of generated by the basis ej j N, and since the operators are H { } ∈ unitary we can conclude that Tk s T . Let us see in more detail this argument: → take x and let xj j N with each xj belonging to the dense subset generated ∈ H { } ∈ by ej j N and such that xj x.
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