Chapter 2 C*-algebras, states and representations 2.1 C*-algebras Let be an associative algebra over C. is a normed algebra if there is a norm x x A A A3 7! k k2 R+ such that xy x y . A complete normed algebra is a Banach algebra. A mapping k kk kk k x x of into itself is an involution if 7! ⇤ A (x⇤)⇤ = x; (x + y)⇤ = x⇤ + y⇤; (xy)⇤ = y⇤x⇤; (λx)⇤ = λx⇤. An algebra with an involution is a *-algebra. Definition 1. A Banach *-algebra is called a C*-algebra if A 2 x⇤x = x ,x. k k k k 2A Proposition 1. Let be a C*-algebra. A 1. x = x ; k ⇤k k k 2. If does not have an identity, let be the algebra obtained from by adjoining an A A A identity 1. Then is a C*-algebra with norm defined by A k·k e λy + xy e λ1+x =supk k,λC. k k y=0 y 2 6 k k Proof. Exercise. In the following, will always denote a C*-algebras with an identity if not specified other- A wise. Definition 2. The spectrum Sp(x) of x is the set 2A Sp(x):= λ C : x λ1 is not invertible in . { 2 − A} 3 1 n 1 If λ > x ,thentheseriesλ− n N (x/λ) is norm convergent and sums to (λ1 x)− . | | k k 2 − Hence, Sp(x) B x (0). Assume now that x is a self-adjoint element and that a +ib ⇢ k k P 2A 2 Sp(x), a, b R.Thena+i(b+t) Sp(x+it1). Since x+it1 2 = x+it1 x it1 = x2+t21 2 2 k k k kk − k k k x 2 + t2, and by the remark above, a +i(b + t) 2 x 2 + t2, and further 2bt x a2 b2 k k | | k k nk k − for all t R, so that b = 0. For any polynomial P over C, P (µ) λ = A i=1(µ zi), and 2 n − − P (x) λ1=A (x zi) for any x . Hence, λ Sp(P (x)) i↵ zj Sp(x) for a − i=1 − 2A 2A 2 Q 2 1 j n.SinceP (zj)=λ, we have that λ Sp(P (x)) i↵ λ P (Sp(x)). We have proved Q 2 2 Proposition 2. Let x . 2A 1. Sp(x) B x (0); ⇢ k k 2. if x = x , then Sp(x) [ x , x ]; ⇤ ⇢ k k k k 3. if xx = x x, i.e. x is normal, then x =sup λ : λ Sp(x) ; ⇤ ⇤ k k {| | 2 } 4. for any polynomial P , Sp(P (x)) = P (Sp(x)); The proof of 3. is left as an exercise. Note that the condition holds in particular for x = x⇤ An element x is positive if it is self-adjoint and Sp(x) R+. 2A ⇢ Proposition 3. Let x , x =0. The following are equivalent: 2A 6 1. x is positive; 2. there is a self-adjoint z such that x = z2; 2A 3. there is y such that x = y y; 2A ⇤ Proof. (3) (2) by choosing y = z. (3) (1) since z2 is self-adjoint and since, by Proposition 2, ) ) Sp(z2) [0, z 2]. To show (1) (3), we note that1 for any µ>0, ⇢ k k ) 1 1 1 2 µ = 1 pλ dλ (2.1) ⇡ λ − λ + µ Z0 ✓ ◆ Since x is positive, (x+λ1) is invertible for all λ>0 so that z := ⇡ 1 pλ λ 1 (x + λ1) 1 dλ 0 − − − is well defined as a norm convergent integral, and x = z2. Using again (2.1) with µ = 1, we R have that 1/2 1/2 x 1 pλ 1 1 x 1 z = k k (ˆx + λ1)− (ˆx 1)dλ, xˆ = x x − k k − ⇡ λ +1 − k k Z0 Butx ˆ positive implies Sp(ˆx) [0, 1], hence Sp(1 xˆ) [0, 1] and 1 xˆ 1. Moreover, since ⇢ − ⇢ k − k Sp((ˆx + λ1) 1)=(Sp(ˆx + λ1)) 1 [(1 + λ) 1,λ 1], we have (ˆx + λ1) 1 λ 1 for λ>0. − − ⇢ − − k − k − Hence, 1 z x 1/2 1 so that Sp(z) [0, 2 x 1/2] and finally z is positive. k − k k− k ⇢ k k It remains to prove (2) (1). Since x = y y is self-adjoint, x2 is positive and we denote by ) ⇤ x its positive square root defined by the integral above. Then x := ( x x)/2 is positive and | | ± | |± x x+ = x+x = 0. Decomposing yx = s +it, with self-adjoint s, t,wehave(yx )⇤(yx )+ − − 2 2 − 3 − − (yx )(yx )⇤ = 2(s +t ) 0. But (yx )⇤(yx )= x ( x +x+)x = x is positive, so that − − ≥ − − − − − − − − − (yx )(yx )⇤ is positive. On the other hand, Sp((yx )(yx )⇤) 0 =Sp((yx )⇤(yx )) 0 − − − − [{ } − − [{ }⇢ R ,hence(yx )⇤(yx ) = 0 so that x = 0, and finally x = x+ is positive. − − − − 1 1/2 3/2 Convergence follows from the asymptotics O(s− )ass 0andO(s− )ass ,whilethechangeof ! !1 variables λ = µ⇠ yields immediately that the integral is pµ,uptoaconstant. 4 Definition 3. A *-morphism between two *-algebras and is a linear map ⇡ : A B A!B such that ⇡(A A )=⇡(A )⇡(A ) and ⇡(A )=⇡(A) ,forallA, A ,A .Itiscalleda 1 2 1 2 ⇤ ⇤ 1 2 2A *-isomorphism if it is bijective. A *-isomorphism is an automorphism. A!A Proposition 4. Let , be two C*-algebras and ⇡ : a *-morphism. Then ⇡(x) A B A!B k kB x , and the range ⇡(A):A is a *-subalgebra of . k kA { 2A} B Proof. If x is self-adjoint, so is ⇡(x) and ⇡(x) =sup λ : λ Sp(⇡(x)) .Ifx λ1isinvertible, k k {| | 2 } − then 1 = ⇡((x λ1) 1(x λ1)) = ⇡((x λ1) 1)⇡(x λ1) so that ⇡(x) λ1isinvertible,whence − − − − − − − Sp(⇡(x)) Sp(x), we have that ⇡(x) sup λ : λ Sp(x) = x . The general case follows ⇢ k k {| | 2 } k k from ⇡(x) 2 = ⇡(x x) x x = x 2. k k k ⇤ kk ⇤ k k k Let Γbe a locally compact Hausdor↵space, and let C0(Γ) be the algebra, under pointwise multiplication, of all complex valued continuous functions that vanish at infinity. Theorem 5. If is a commutative C*-algebra, then there is a locally compact Hausdor↵space A Γ such that is *-isomorphic to C (Γ). A 0 Proof. See Robert’s lectures. In classical mechanics, the space Γis usually referred to as the phase space. If is a commutative C*-algebra with an identity, then is isomorphic to C(K), the A A algebra of continuous functions on a compact Hausdor↵space K. Let be a *-subalgebra of ( ). The commutant is the subset of ( ) of operators U L H U 0 L H that commute with every element of , and so forth with := ( ) . In particular, , U U 00 U 0 0 U⇢U00 and further = . U 0 U 000 Definition 4. A von Neumann algebra or W*-algebra on is a *-subalgebra of ( ) such H U L H that 00 = .Itscenter is ( ):= 0,and is a factor if ( )=C 1. U U Z U U\U U Z U · Theorem 6. Let be a *-subalgebra of ( ) such that = . Then is a von Neumann U L H UH H U algebra i↵ is weakly closed. U Note that = is automatically satisfied if 1 . Furthermore, for any *-subalgebra UH H 2U U of ( ), let be its weak closure for which 00 = by the theorem. Since ,wehave L H U U U U⇢U 0 . Furthermore, if x , and y ,with y *y,thenx commutes with y for all U ⇢U0 2U0 2 U U3 n n n and hence with y, so that 0. It follows that 0 = ,whence 00 = and so: U 0 ⇢ U U U 0 U U 00 Corollary 7. is weakly dense in , namely = . U U 00 U U 00 2.2 Representations and states Definition 5. Let be a C*-algebra and a Hilbert space. A representation of in is a A H A H *-morphism ⇡ : ( ). Moreover, A!LH 1. Two representations ⇡,⇡ in , are equivalent if there is a unitary map U : 0 H H0 H!H0 such that U⇡(x)=⇡0(x)U; 2. A representation ⇡ is topologically irreducible if the only closed subspaces that are invari- ant under ⇡( ) are 0 and ; A { } H 3. A representation ⇡ is faithful if it is an isomorphism, namely Ker⇡ = 0 . { } 5 Note that in general ⇡(x) x , with equality if and only if ⇡ is faithful. One can further k kk k show that for any C*-algebra, there exists a faithful representation. n n If ( ,⇡) is a representation of and n N,thenn⇡(A)( i):= ⇡(A) i defines a H A 2 ⊕i=1 ⊕i=1 representation n⇡ on n . ⊕i=1H Proposition 8. Let ⇡ be a representation of in .T.f.a.e A H 1. ⇡ is topologically irreducible; 2. ⇡( )0 := B ( ):[B,⇡(x)] = 0, for all x = C 1; A { 2LH 2A} · 3. Any ⇠ , ⇠ =0is cyclic: ⇡(x)⇠ = ,or⇡ =0. 2H 6 H Proof. (1) (3) : If ⇡( )⇠ is not dense, then ⇡( )⇠ = 0 . It follows that C⇠ is an invariant ) A A { } subspace, and hence = C⇠ and ⇡ =0 H (3) (1) : Let = 0 be a closed invariant subspace. For any ⇠ , ⇡( )⇠ and since ⇠ ) K6 { } 2K A ⇢K is cyclic, ⇡( )⇠ is dense in A H (2) (1) : Let = 0 be a closed invariant subspace, and let P be the orthogonal projection ) K6 { } K on .ThenP ⇡( )0, since for ⇠ ,⌘ ?, ⇠,⇡(x)⌘ = ⇡(x⇤)⇠,⌘ = 0 so that K K 2 A 2K 2K h i h i ⇡(x)⌘ ? for any x .