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Von Neumann Algebras Von Neumann Algebras. Vaughan F.R. Jones 1 November 13, 2015 1Supported in part by NSF Grant DMS93–22675, the Marsden fund UOA520, and the Swiss National Science Foundation. 2 Chapter 1 Introduction. The purpose of these notes is to provide a rapid introduction to von Neumann algebras which gets to the examples and active topics with a minimum of technical baggage. In this sense it is opposite in spirit from the treatises of Dixmier [], Takesaki[], Pedersen[], Kadison-Ringrose[], Stratila-Zsido[]. The philosophy is to lavish attention on a few key results and examples, and we prefer to make simplifying assumptions rather than go for the most general case. Thus we do not hesitate to give several proofs of a single result, or repeat an argument with different hypotheses. The notes are built around semester- long courses given at UC Berkeley and Vanderbilt though they contain more material than could be taught in a single semester. The notes are informal and the exercises are an integral part of the ex- position. These exercises are vital and mostly intended to be easy. 3 4 Chapter 2 Background and Prerequisites 2.1 Hilbert Space A Hilbert Space is a complex vector space H with inner product h; i : HxH! C which is linear in the first variable, satisfies hξ; ηi = hη; ξi, is positive definite, i.e. hξ; ξi > 0 for ξ 6= 0, and is complete for the norm defined by test jjξjj = phξ; ξi. Exercise 2.1.1. Prove the parallelogram identity : jjξ − ηjj2 + jjξ + ηjj2 = 2(jjξjj2 + jjηjj2) and the Cauchy-Schwartz inequality: jhξ; ηij ≤ jjξjj jjηjj: Theorem 2.1.2. If C is a closed convex subset of H and ξ is any vector in H, there is a unique η 2 C which minimizes the distance from ξ to C, i.e. jjξ − η0jj ≤ jjξ − ηjj 8η0 2 C. Proof. This is basically a result in plane geometry. Uniqueness is clear—if two vectors η and η0 in C minimized the distance to ξ, then ξ; η and η0 lie in a (real) plane so any vector on the line segment between η and η0 would be strictly closer to ξ. To prove existence, let d be the distance from C to ξ and choose a sequence n ηn 2 C with jjηn − ξjj < d + 1=2 . For each n, the vectors ξ, ηn and ηn+1 define a plane. Geometrically it is clear that, if ηn and ηn+1 were not close, some point on the line segment between them would be closer than d to ξ. Formally, use the parallelogram identity: η + η ξ − η ξ − η jjξ − n n+1 jj2 = jj n + n+1 jj2 2 2 2 5 ξ − η ξ − η = 2(jj n jj2 + jj n+1 jj2 − 1=8jjη − η jj2) 2 2 n n+1 n 2 2 ≤ (d + 1=2 ) − 1=4jjηn − ηn+1jj 2 n ηn+ηn+1 2 Thus there is a constant K such that jjηn − ηn+1jj < K=2 or jjξ − 2 jj would be less than d2. Thus (ηn) is Cauchy, its limit is in C and has distance d from ξ. Exercise 2.1.3. If φ 2 H∗ (the Banach-space dual of H consisting of all continuous linear functionals from H to C), kerφ is a closed convex subset of H. Show how to choose a vector ξφ orthogonal to ker φ with φ(η) = hξφ; ηi ∗ and so that φ 7! ξφ is a conjugate-linear isomorphism from H onto H. We will be especially concerned with separable Hilbert Spaces where there is an orthonormal basis, i.e. a sequence fξ1; ξ2; ξ3; :::g of unit vectors with hξi; ξji = 0 for i 6= j and such that 0 is the only element of H orthogonal to all the ξi. We will use the abbreviation ONB for orthonormal basis. Exercise 2.1.4. Show that an ONB always exists (e.g. Gram-Schmidt) and that if fξig is an ONB for H then the linear span of the fξig is dense in H. A trivial but useful observation. If H and K are Hilbert spaces with vectors ξi 2 H and k 2 K respectively then if hξi; ξji = h i; ji 8i; j then the map ξi 7! i extends to a linear h; i-preserving bijection from the closure of the subspace spanned by the ξi to the closure of the subspace spanned by the i. A linear map (operator) a : H!K is said to be bounded if there is a number K with jjaξjj ≤ Kjjξjj 8ξ 2 H. The infimum of all such K is called the norm of a, written jjajj. The set of all bounded operators from H to K is written B(H; K) and if H = K we use B(H). Boundedness of an operator is equivalent to continuity. To every bounded operator a between Hilbert spaces H and K, by exercise 2.1.3 there is another, a∗, between K and H, called the adjoint of a which is defined by the formula haξ; ηi = hξ; a∗ηi. Exercise 2.1.5. Prove that sup jjajj = jhaξ; ηij jjξjj ≤ 1; jjηjj ≤ 1 = jja∗jj = jja∗ajj1=2: 6 Some definitions: The identity map on H is a bounded operator denoted 1. An operator a 2 B(H) is called self-adjoint if a = a∗. An operator p 2 B(H) is called a projection if p = p2 = p∗. An operator a 2 B(H) is called positive if haξ; ξi ≥ 0 8ξ 2 B(H). We say a ≥ b if a − b is positive. An operator u 2 B(H) is called an isometry if u∗u = 1. An operator v 2 B(H) is called a unitary if uu∗ = u∗u = 1. An operator u 2 B(H) is called a partial isometry if u∗u is a projection. The last three definitions extend to bounded linear operators between dif- ferent Hilbert spaces. If S ⊆ B(H) then the commutant S0 of S is fx 2 B(H)jxa = ax 8a 2 Sg. Also S00 = (S0)0. Exercise 2.1.6. A word on matrices. If ei is an ONB of H then ei 7! ξi (the characteristic function of fig) defines a unitary from H to `2(N). So for any ξ 2 H, 1 X ξ = hξ; eiiei i=1 the sum being convergent in the norm of H. If a 2 B(H) we define the matrix of a wrt the ONB to be ai;j = haei; eji. 2 2 For fixed i, j 7! ai;j is in ` and for fixed j, i 7! ai;j is in ` . And a(ei) = P j ai;jej. Thus the matrix of a determines a and if b 2 B(H) has matrix bi;j then X abi;j = ai;kbk;j k the sum being absolutely convergent. Exercise 2.1.7. Show that every a 2 B(H) is a linear combination of two self-adjoint operators. Exercise 2.1.8. A positive operator is self-adjoint. Exercise 2.1.9. Find an isometry from one Hilbert space to itself that is not unitary. (The unilateral shift on H = `2(N) is a fine example. There is an obvious orthonormal basis of H indexed by the natural numbers and the shift just sends the nth: basis element to the (n + 1)th:) Exercise 2.1.10. If K is a closed subspace of H show that the map PK : H!K which assigns to any point in H the nearest point in K is linear and a projection. 7 Exercise 2.1.11. Show that the correspondence K! PK of the previous exercise is a bijection between closed subspaces of H and projections in B(H). If S is a subset of H, S? is by definition fξ 2 H : hξ; ηi = 0 8η 2 Sg: Note that S? is always a closed subspace. ?? Exercise 2.1.12. If K is a closed subspace then K = K and PK? = 1−PK. Exercise 2.1.13. If u is a partial isometry then so is u∗. The subspace u∗H is then closed and called the initial domain of u, the subspace uH is also closed and called the final domain of u. Show that a partial isometry is the composition of the projection onto its initial domain and a unitary between the initial and final domains. The commutator [a; b] of two elements of B(H) is the operator ab − ba. Exercise 2.1.14. If K is a closed subspace and a = a∗ then aK ⊆ K iff [a; PK] = 0: ∗ In general (aK ⊆ K and a K ⊆ K) () [a; PK] = 0. 2.2 The Spectral Theorem The spectrum σ(a) of a 2 B(H) is fλ 2 C : a − λ1 is not invertibleg. Exercise 2.2.1. (Look up proofs if necessary.) Show that σ(a) is a non- empty closed bounded subset of C and that if a = a∗, σ(a) ⊆ [−||ajj; jjajj] with either jjajj or −||ajj in σ(a). The spectral theorem takes a bit of getting used to and knowing how to prove it does not necessarily help much. If one cannot “see” the spectral decomposition of an operator it may be extremely difficult to obtain—except in a small finite number of dimensions where it is just diagonalisation. But fortunately there is nothing like a course in operator algebras, either C∗ or von Neumann, to help master the use of this theorem which is the heart of linear algebra on Hilbert space.
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