Topological Properties of the Unitary Group 3

Home , Operator topologies, Weak operator topology

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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 continuous 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→ topology.

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. We compute { } ∈ → Tkx T x Tkx Tkxj + Tkxj T xj + T xj T x | − | ≤ | − | | − | | − | = Tk(x xj ) + Tkxj T xj + T (xj x) | − | | − | | − | =2 x xj + Tkxj T xj , | − | | − | and since Tkxj T xj, we have that Tkx T x and therefore Tk s T . This implies that the→ map Ψ induces a homeomorphism→ with its image, and→ hence it is an embedding. The Hilbert space is a metric space, and the product N of countable copies H NH of can be endowed with a metric. Since Ψ : U( )s is an embedding, then U(H) inherits the induced metric and hence it isH metrizable.→ H H s The previous argument follows the proof that appears in [3, Prop II.1]. In [3, Prop II.1] it is also shown that the the map Ψ also provides an embedding N ∗ Iso( ) where Iso( ) := T B( ): T T = IdH is the monoid of all isometriesH → of H , which isH moreover{ complete∈ H with respect to} the induced metric. H ∗ It is furthermore shown that U( ) = T Iso( ): T Iso( ) is a Gδ set H { ∈ H ∈ H } in Iso( ), and since Gδ sets on a complete metrizable space are complete in the H induced metric [4, Thm 1, p. 93] this implies that U( )s is a complete metrizable space. H Lemma 1.7 ([3, Prop II.1]). The unitary group U( ) is a Polish group. H s So far we have shown that U( ) is a Polish group and we have that U( ) = H s H s U( )s∗ = U( )w = U( )w∗ . Note in particular that U( )s is compactly generated sinceH it is a metrizableH space.H H Let us now see the relation with the compact open topology on U( ). H Lemma 1.8. The compact open topology and the strong operator topology agree on U( ), i.e. U( ) = U( ) . H H co H s Proof. Let us start by recalling the retraction functor k defined in [5, Def. 3.1]. For X a Hausdorff topological space the associated compactly generated space k(X) is the set X whose topology is defined as follows: a set in k(X) is closed, if its intersection with every compact set in X is itself closed. By [5, Thm 3.2] we know that the identity map k(X) X is a continuous map, that k(X) is compactly → TOPOLOGICAL PROPERTIES OF THE UNITARY GROUP 5 generated, that k(X) and X have the same compact sets and that X = k(X) whenever X is compactly generated. Let C be a compact subset in U( ) . Take a sequence of operators Tn n in H s { } C, since C is compact in U( )s there exists a convergent subsequence Tnk k, and since U( ) is completely metrizableH and hence C is furthermore closed,{ this} con- H s vergent subsequence Tnk s T converges to an operator T C. By the Bannach- Steinhaus Theorem (see [6,→ Cor. of Thm 33.1, p. 348]), we have∈ that the sequence

Tnk k converges uniformly on every compact set of , and therefore we have that { } H Tnk co T . Thus the space C is also compact in U( ) , and by the same argument → H co as before, the induced topology of C in U( )co agrees with the induced topology of C in U( ) . H H s We conclude that the spaces U( )co and U( )s have the same compact sets with the same induced topologies. ThisH implies thatH the retraction functor k applied on the map U( ) U( ) induces a homeomorphism H co → H s ∼ k(U( ) ) = k(U( ) ). H co → H s Then we have the commutative diagram k(U( ) ) / U( ) H co H co ∼= ∼ k(U( ) ) = / U( ) H s H s which implies that U( ) = U( ) . H co H s proof of Theorem 1.2. By Lemma 1.5 we know that U( )s∗ = U( )s = U( )w, so the vertical arrow in the middle and the bottom right horizontalH H arrow of diagramH (1.9) are homeomorphisms; by Lemma 1.8 we know that U( )co = U( )s, and then the bottom left horizontal arrow is also a homeomorphism.H The proofH of Lemma 1.8 can also be used to show that U( )co∗ = U( )s∗ . Finally, the right vertical arrow is also a homeomorphism becauseH of LemmaH 1.3. Hence diagram (1.1) restricted to U( ) becomes H

/ ∼= / / (1.9) U( ) / U( ) ∗ / U( ) ∗ / U( ) ∗ H n H co H s H w ∼= =∼ ∼ ∼ U( ) = / U( ) = / U( ) H co H s H w Therefore, besides U( ) , every topological group in diagram (1.9) can be con- H n nected using a homeomorphism with the Polish group U( )s. This ﬁnishes the proof of Theorem 1.2. H

The norm topology on U( ) has strictly more open sets than the strong operator topology as can be easily checkedH with the following sequence of operators. 2 Consider := l (N) and take the sequence of operators Tk : such that H H → H for all x = (ξn) the n-th coordinate of Tkx is deﬁned by ∈ H ξn n = k (Tkx)n = 6 √ 1ξn n = k. ( − 6 JESUS´ ESPINOZA AND BERNARDO URIBE

∗ −1 It follows that T = T and therefore Tk k U( ). In the strong operator k k { } ⊂ H topology the sequence Tk k converges to the identity operator Id : , since we have that { } H → H Tkx Idx = iξk ξk = ξk i 1 0 | − | | − | | ||(k−) | → for all x = (ξn) . On the other hand, if xk = (ξn ) is defined by ∈ H ∈ H (k) 0 n = k ξn = 6 (i n = k, then Tkxk Idxk = 1 i = √2 and it follows that sup Tkx Idx 0 | − | |− − | |x|≤1{| − |} 6→ whenever k . Hence we have that Tk s Id but Tk n Id. Therefore→ we ∞ conclude that the operator→ topologies defined6→ at the beginning of the chapter reduced to only two once restricted to the unitary group. The unitary group with the norm topology, making U( )n into a Banach Lie group, and the strong operator topology, making U( ) intoH a Polish group. H s References [1] Michael Atiyah and Graeme Segal. Twisted K-theory. Ukr. Mat. Visn., 1(3):287–330, 2004. [2] Joachim Hilgert and Karl-Hermann Neeb. Lie semigroups and their applications, volume 1552 of Lecture Notes in Mathematics. Springer-Verlag, Berlin, 1993. [3] Karl-Hermann Neeb. On a theorem of S. Banach. J. Lie Theory, 7(2):293–300, 1997. [4] Laurent Schwartz. Radon measures on arbitrary topological spaces and cylindrical measures. Published for the Tata Institute of Fundamental Research, Bombay by Oxford University Press, London, 1973. Tata Institute of Fundamental Research Studies in Mathematics, No. 6. [5] N. E. Steenrod. A convenient category of topological spaces. Michigan Math. J., 14:133–152, 1967. [6] Fran¸cois Trèves. Topological vector spaces, distributions and kernels. Academic Press, New York, 1967.

Departamento de matematicas,´ Universidad de Sonora, Blvd Lu´ıs Encinas y Rosales S/N, Colonia Centro. Edificio 3K-1. C.P. 83000. Hermosillo, Sonora, Mexico´ E-mail address: [email protected]

Departamento de Matematicas´ y Estad´ıstica, Universidad del Norte, Km 5 V´ıa Puerto Colombia, Barranquilla, Colombia E-mail address: [email protected]