Tutorial on Von Neumann Algebras

Tutorial on von Neumann algebras

Adrian Ioana

University of California, San Diego

Model Theory and Operator Algebras UC Irvine September 20-22, 2017

1/174 the adjoint T ⇤ B(H) given by T ⇠,⌘ = ⇠,T ⇤⌘ ,forall⇠,⌘ H. 2 h i h i 2 the spectrum of T is (T )= C T I not invertible . { 2 | } Fact: (T ) is a compact non-empty subset of C.

Deﬁnition An operator T B(H)iscalled 2 1 self-adjoint if T = T ⇤. 2 positive if T ⇠,⇠ 0, for all ⇠ H. (positive self-adjoint) h i 2 ) 3 a projection if T = T = T 2 T is the orthogonal projection onto ⇤ , a closed subspace K H. ⇢ 4 a unitary if TT ⇤ = T ⇤T = I .

The algebra of bounded linear operators

H separable Hilbert space, e.g. `2(N), L2([0, 1], Leb). B(H) algebra of bounded linear operators T : H H, ! i.e. such that T =sup T ⇠ ⇠ =1 < . k k {k k|k k } 1

2/174 the spectrum of T is (T )= C T I not invertible . { 2 | } Fact: (T ) is a compact non-empty subset of C.

The algebra of bounded linear operators

H separable Hilbert space, e.g. `2(N), L2([0, 1], Leb). B(H) algebra of bounded linear operators T : H H, ! i.e. such that T =sup T ⇠ ⇠ =1 < . k k {k k|k k } 1 the adjoint T ⇤ B(H) given by T ⇠,⌘ = ⇠,T ⇤⌘ ,forall⇠,⌘ H. 2 h i h i 2

3/174 Deﬁnition An operator T B(H)iscalled 2 1 self-adjoint if T = T ⇤. 2 positive if T ⇠,⇠ 0, for all ⇠ H. (positive self-adjoint) h i 2 ) 3 a projection if T = T = T 2 T is the orthogonal projection onto ⇤ , a closed subspace K H. ⇢ 4 a unitary if TT ⇤ = T ⇤T = I .

The algebra of bounded linear operators

4/174 3 a projection if T = T = T 2 T is the orthogonal projection onto ⇤ , a closed subspace K H. ⇢ 4 a unitary if TT ⇤ = T ⇤T = I .

The algebra of bounded linear operators

Deﬁnition An operator T B(H)iscalled 2 1 self-adjoint if T = T ⇤. 2 positive if T ⇠,⇠ 0, for all ⇠ H. (positive self-adjoint) h i 2 )

5/174 4 a unitary if TT ⇤ = T ⇤T = I .

The algebra of bounded linear operators

6/174 The algebra of bounded linear operators

7/174 strong operator topology (SOT): T T T ⇠ T ⇠ 0, i ! () k i k! for all ⇠ H. 2 weak operator topology (WOT): T T T ⇠,⌘ T ⇠,⌘ , i ! () h i i!h i for all ⇠,⌘ H. 2 Exercise 1: Suppose that T T (WOT) and T , T are all projections i ! i (or all unitaries). Prove that T T (SOT). i ! Deﬁnition A subalgebra A B(H)iscalleda -algebra if T ⇤ A,forallT A. ⇢ ⇤ 2 2 A -subalgebra A B(H)iscalleda ⇤ ⇢ 1 C⇤-algebra if it is closed in the norm topology. 2 von Neumann algebra if it is WOT-closed.

Terminology. A homomorphism ⇡ : A B between two -algebras is ! ⇤ called a -homomorphism if ⇡(T )=⇡(T ) ,forallT A. ⇤ ⇤ ⇤ 2 Abijective -homomorphism ⇡ : A B is called a -isomorphism. ⇤ ! ⇤

Topologies on B(H)

norm topology: T T T T 0. i ! () k i k!

8/174 Exercise 1: Suppose that T T (WOT) and T , T are all projections i ! i (or all unitaries). Prove that T T (SOT). i ! Deﬁnition A subalgebra A B(H)iscalleda -algebra if T ⇤ A,forallT A. ⇢ ⇤ 2 2 A -subalgebra A B(H)iscalleda ⇤ ⇢ 1 C⇤-algebra if it is closed in the norm topology. 2 von Neumann algebra if it is WOT-closed.

Topologies on B(H)

norm topology: T T T T 0. i ! () k i k! strong operator topology (SOT): T T T ⇠ T ⇠ 0, i ! () k i k! for all ⇠ H. 2 weak operator topology (WOT): T T T ⇠,⌘ T ⇠,⌘ , i ! () h i i!h i for all ⇠,⌘ H. 2

9/174 Deﬁnition A subalgebra A B(H)iscalleda -algebra if T ⇤ A,forallT A. ⇢ ⇤ 2 2 A -subalgebra A B(H)iscalleda ⇤ ⇢ 1 C⇤-algebra if it is closed in the norm topology. 2 von Neumann algebra if it is WOT-closed.

Topologies on B(H)

10 / 174 A -subalgebra A B(H)iscalleda ⇤ ⇢ 1 C⇤-algebra if it is closed in the norm topology. 2 von Neumann algebra if it is WOT-closed.

Topologies on B(H)

norm topology: T T T T 0. i ! () k i k! strong operator topology (SOT): T T T ⇠ T ⇠ 0, i ! () k i k! for all ⇠ H. 2 weak operator topology (WOT): T T T ⇠,⌘ T ⇠,⌘ , i ! () h i i!h i for all ⇠,⌘ H. 2 Exercise 1: Suppose that T T (WOT) and T , T are all projections i ! i (or all unitaries). Prove that T T (SOT). i ! Deﬁnition A subalgebra A B(H)iscalleda -algebra if T ⇤ A,forallT A. ⇢ ⇤ 2 2

11 / 174 2 von Neumann algebra if it is WOT-closed.

Topologies on B(H)

12 / 174 Terminology. A homomorphism ⇡ : A B between two -algebras is ! ⇤ called a -homomorphism if ⇡(T )=⇡(T ) ,forallT A. ⇤ ⇤ ⇤ 2 Abijective -homomorphism ⇡ : A B is called a -isomorphism. ⇤ ! ⇤

Topologies on B(H)

13 / 174 Topologies on B(H)

Terminology. A homomorphism ⇡ : A B between two -algebras is ! ⇤ called a -homomorphism if ⇡(T )=⇡(T ) ,forallT A. ⇤ ⇤ ⇤ 2 Abijective -homomorphism ⇡ : A B is called a -isomorphism. ⇤ ! ⇤ 14 / 174 Let B((T )) denote the -algebra of bounded Borel functions ⇤ f : (T ) C,withthe -operation given by f ⇤(z)=f (z). ! ⇤ Theorem: a -homomorphism ⇡ : B((T )) B(H)suchthat 9 ⇤ ! ⇡(z)=T (and thus ⇡(¯z)=T ⇤). ⇡(f ) f ,forallf B((T )), with equality if f C((T )). k kk k1 2 2 If fi B((T )) is a net such that supi fi < and fi (z) f (z), 2 k k1 1 ! for all z (T ), then ⇡(f ) ⇡(f )(SOT). 2 i ! Notation. f (T ):=⇡(f ), for all f B((T )). 2 Consequences. Assume M is a von Neumann algebra containing T . 1) f (T ) M,forallf B((T )). 2 2 2) 1 (T ) M is a projection, for any Borel set (T ). 2 ⇢ 3) T is a norm limit of linear combinations of projections in M. 4) If T is positive, then there exists S M such that T = S S. 2 ⇤ (since (T ) [0, ), we can take S = T 1/2). ⇢ 1

Borel functional calculus

Let T B(H) be an operator which is normal: T ⇤T = TT ⇤. 2

15 / 174 Theorem: a -homomorphism ⇡ : B((T )) B(H)suchthat 9 ⇤ ! ⇡(z)=T (and thus ⇡(¯z)=T ⇤). ⇡(f ) f ,forallf B((T )), with equality if f C((T )). k kk k1 2 2 If fi B((T )) is a net such that supi fi < and fi (z) f (z), 2 k k1 1 ! for all z (T ), then ⇡(f ) ⇡(f )(SOT). 2 i ! Notation. f (T ):=⇡(f ), for all f B((T )). 2 Consequences. Assume M is a von Neumann algebra containing T . 1) f (T ) M,forallf B((T )). 2 2 2) 1 (T ) M is a projection, for any Borel set (T ). 2 ⇢ 3) T is a norm limit of linear combinations of projections in M. 4) If T is positive, then there exists S M such that T = S S. 2 ⇤ (since (T ) [0, ), we can take S = T 1/2). ⇢ 1

Borel functional calculus

Let T B(H) be an operator which is normal: T ⇤T = TT ⇤. 2 Let B((T )) denote the -algebra of bounded Borel functions ⇤ f : (T ) C,withthe -operation given by f ⇤(z)=f (z). ! ⇤

16 / 174 ⇡(z)=T (and thus ⇡(¯z)=T ⇤). ⇡(f ) f ,forallf B((T )), with equality if f C((T )). k kk k1 2 2 If fi B((T )) is a net such that supi fi < and fi (z) f (z), 2 k k1 1 ! for all z (T ), then ⇡(f ) ⇡(f )(SOT). 2 i ! Notation. f (T ):=⇡(f ), for all f B((T )). 2 Consequences. Assume M is a von Neumann algebra containing T . 1) f (T ) M,forallf B((T )). 2 2 2) 1 (T ) M is a projection, for any Borel set (T ). 2 ⇢ 3) T is a norm limit of linear combinations of projections in M. 4) If T is positive, then there exists S M such that T = S S. 2 ⇤ (since (T ) [0, ), we can take S = T 1/2). ⇢ 1

Borel functional calculus

17 / 174 ⇡(f ) f ,forallf B((T )), with equality if f C((T )). k kk k1 2 2 If fi B((T )) is a net such that supi fi < and fi (z) f (z), 2 k k1 1 ! for all z (T ), then ⇡(f ) ⇡(f )(SOT). 2 i ! Notation. f (T ):=⇡(f ), for all f B((T )). 2 Consequences. Assume M is a von Neumann algebra containing T . 1) f (T ) M,forallf B((T )). 2 2 2) 1 (T ) M is a projection, for any Borel set (T ). 2 ⇢ 3) T is a norm limit of linear combinations of projections in M. 4) If T is positive, then there exists S M such that T = S S. 2 ⇤ (since (T ) [0, ), we can take S = T 1/2). ⇢ 1

Borel functional calculus

18 / 174 If fi B((T )) is a net such that supi fi < and fi (z) f (z), 2 k k1 1 ! for all z (T ), then ⇡(f ) ⇡(f )(SOT). 2 i ! Notation. f (T ):=⇡(f ), for all f B((T )). 2 Consequences. Assume M is a von Neumann algebra containing T . 1) f (T ) M,forallf B((T )). 2 2 2) 1 (T ) M is a projection, for any Borel set (T ). 2 ⇢ 3) T is a norm limit of linear combinations of projections in M. 4) If T is positive, then there exists S M such that T = S S. 2 ⇤ (since (T ) [0, ), we can take S = T 1/2). ⇢ 1

Borel functional calculus

19 / 174 Notation. f (T ):=⇡(f ), for all f B((T )). 2 Consequences. Assume M is a von Neumann algebra containing T . 1) f (T ) M,forallf B((T )). 2 2 2) 1 (T ) M is a projection, for any Borel set (T ). 2 ⇢ 3) T is a norm limit of linear combinations of projections in M. 4) If T is positive, then there exists S M such that T = S S. 2 ⇤ (since (T ) [0, ), we can take S = T 1/2). ⇢ 1

Borel functional calculus

20 / 174 Consequences. Assume M is a von Neumann algebra containing T . 1) f (T ) M,forallf B((T )). 2 2 2) 1 (T ) M is a projection, for any Borel set (T ). 2 ⇢ 3) T is a norm limit of linear combinations of projections in M. 4) If T is positive, then there exists S M such that T = S S. 2 ⇤ (since (T ) [0, ), we can take S = T 1/2). ⇢ 1

Borel functional calculus

21 / 174 1) f (T ) M,forallf B((T )). 2 2 2) 1 (T ) M is a projection, for any Borel set (T ). 2 ⇢ 3) T is a norm limit of linear combinations of projections in M. 4) If T is positive, then there exists S M such that T = S S. 2 ⇤ (since (T ) [0, ), we can take S = T 1/2). ⇢ 1

Borel functional calculus

22 / 174 2) 1 (T ) M is a projection, for any Borel set (T ). 2 ⇢ 3) T is a norm limit of linear combinations of projections in M. 4) If T is positive, then there exists S M such that T = S S. 2 ⇤ (since (T ) [0, ), we can take S = T 1/2). ⇢ 1

Borel functional calculus

23 / 174 3) T is a norm limit of linear combinations of projections in M. 4) If T is positive, then there exists S M such that T = S S. 2 ⇤ (since (T ) [0, ), we can take S = T 1/2). ⇢ 1

Borel functional calculus

24 / 174 4) If T is positive, then there exists S M such that T = S S. 2 ⇤ (since (T ) [0, ), we can take S = T 1/2). ⇢ 1

Borel functional calculus

25 / 174 Borel functional calculus

Let T B(H) be an operator which is normal: T ⇤T = TT ⇤. 2 Let B((T )) denote the -algebra of bounded Borel functions ⇤ f : (T ) C,withthe -operation given by f ⇤(z)=f (z). ! ⇤ Theorem: a -homomorphism ⇡ : B((T )) B(H)suchthat 9 ⇤ ! ⇡(z)=T (and thus ⇡(¯z)=T ⇤). ⇡(f ) f ,forallf B((T )), with equality if f C((T )). k kk k1 2 2 If fi B((T )) is a net such that supi fi < and fi (z) f (z), 2 k k1 1 ! for all z (T ), then ⇡(f ) ⇡(f )(SOT). 2 i ! Notation. f (T ):=⇡(f ), for all f B((T )). 2 Consequences. Assume M is a von Neumann algebra containing T . 1) f (T ) M,forallf B((T )). 2 2 2) 1 (T ) M is a projection, for any Borel set (T ). 2 ⇢ 3) T is a norm limit of linear combinations of projections in M. 4) If T is positive, then there exists S M such that T = S S. 2 ⇤ (since (T ) [0, ), we can take S = T 1/2). ⇢ 1 26 / 174 Exercise 2. Show that K(H) is not a von Neumann algebra if dim H = . 1 Exercise 3. Let S B(H)beasetsuchthatx⇤ S,forallx S. ⇢ 2 2 Show that S0 = y B(H) xy = yx, x S (the commutant of S) { 2 | 8 2 } is a von Neumann algebra. Notation.IfM is a von Neumann algebra, its center is (M)=M M . Z \ 0 Theorem (von Neumann) If M B(H)isaunital(1 M) -algebra, then TFAE: ⇢ 2 ⇤ 1 M is WOT-closed. 2 M is SOT-closed.

3 M = M00 := (M0)0.

(3) (1) (2), so it remains to prove that (2) (3). ) ) )

von Neumann’s bicommutant theorem

Examples. 1) B(H) is a von Neumann algebra. 2) The algebra K(H) of all compact operators on H is a C⇤-algebra.

27 / 174 Exercise 3. Let S B(H)beasetsuchthatx⇤ S,forallx S. ⇢ 2 2 Show that S0 = y B(H) xy = yx, x S (the commutant of S) { 2 | 8 2 } is a von Neumann algebra. Notation.IfM is a von Neumann algebra, its center is (M)=M M . Z \ 0 Theorem (von Neumann) If M B(H)isaunital(1 M) -algebra, then TFAE: ⇢ 2 ⇤ 1 M is WOT-closed. 2 M is SOT-closed.

3 M = M00 := (M0)0.

(3) (1) (2), so it remains to prove that (2) (3). ) ) )

von Neumann’s bicommutant theorem

Examples. 1) B(H) is a von Neumann algebra. 2) The algebra K(H) of all compact operators on H is a C⇤-algebra. Exercise 2. Show that K(H) is not a von Neumann algebra if dim H = . 1

28 / 174 Show that S0 = y B(H) xy = yx, x S (the commutant of S) { 2 | 8 2 } is a von Neumann algebra. Notation.IfM is a von Neumann algebra, its center is (M)=M M . Z \ 0 Theorem (von Neumann) If M B(H)isaunital(1 M) -algebra, then TFAE: ⇢ 2 ⇤ 1 M is WOT-closed. 2 M is SOT-closed.

3 M = M00 := (M0)0.

(3) (1) (2), so it remains to prove that (2) (3). ) ) )

von Neumann’s bicommutant theorem

29 / 174 Notation.IfM is a von Neumann algebra, its center is (M)=M M . Z \ 0 Theorem (von Neumann) If M B(H)isaunital(1 M) -algebra, then TFAE: ⇢ 2 ⇤ 1 M is WOT-closed. 2 M is SOT-closed.

3 M = M00 := (M0)0.

(3) (1) (2), so it remains to prove that (2) (3). ) ) )

von Neumann’s bicommutant theorem

30 / 174 Theorem (von Neumann) If M B(H)isaunital(1 M) -algebra, then TFAE: ⇢ 2 ⇤ 1 M is WOT-closed. 2 M is SOT-closed.

3 M = M00 := (M0)0.

(3) (1) (2), so it remains to prove that (2) (3). ) ) )

von Neumann’s bicommutant theorem

Examples. 1) B(H) is a von Neumann algebra. 2) The algebra K(H) of all compact operators on H is a C⇤-algebra. Exercise 2. Show that K(H) is not a von Neumann algebra if dim H = . 1 Exercise 3. Let S B(H)beasetsuchthatx⇤ S,forallx S. ⇢ 2 2 Show that S0 = y B(H) xy = yx, x S (the commutant of S) { 2 | 8 2 } is a von Neumann algebra. Notation.IfM is a von Neumann algebra, its center is (M)=M M . Z \ 0

31 / 174 (3) (1) (2), so it remains to prove that (2) (3). ) ) )

von Neumann’s bicommutant theorem

3 M = M00 := (M0)0.

32 / 174 von Neumann’s bicommutant theorem

3 M = M00 := (M0)0.

(3) (1) (2), so it remains to prove that (2) (3). ) ) ) 33 / 174 SOT In order to show that x M, it suces to argue that x M : 2 2 ⇠ ,...,⇠ H and ">0, y M s.t. x⇠ y⇠ <", i =1,...,n. 8 1 n 2 9 2 k i i k 8 First,weassumethatn = 1 and prove the following: Claim Let K H be an M-invariant closed subspace. (zK K,forallz M) ⇢ ⇢ 2 If p is the orthogonal projection onto K,thenp M . 2 0 Proof of the claim. If z M,thenzpH = zK K. 2 ⇢ Hence (1 p)zpH (1 p)K = 0 and thus (1 p)zp =0(a). ⇢ { } Since z M (1 p)z p =0 pz(1 p)=0(b). ⇤ 2 ) ⇤ ) (a) and (b) pz zp = pz(1 p) (1 p)zp = 0, proving the claim. )

Proof of the bicommutant theorem, I

Assumption: M B(H) is an SOT-closed unital -algebra. ⇢ ⇤ Goal: M = M .Tothisend,ﬁxx M . 00 2 00

34 / 174 First,weassumethatn = 1 and prove the following: Claim Let K H be an M-invariant closed subspace. (zK K,forallz M) ⇢ ⇢ 2 If p is the orthogonal projection onto K,thenp M . 2 0 Proof of the claim. If z M,thenzpH = zK K. 2 ⇢ Hence (1 p)zpH (1 p)K = 0 and thus (1 p)zp =0(a). ⇢ { } Since z M (1 p)z p =0 pz(1 p)=0(b). ⇤ 2 ) ⇤ ) (a) and (b) pz zp = pz(1 p) (1 p)zp = 0, proving the claim. )

Proof of the bicommutant theorem, I

35 / 174 Claim Let K H be an M-invariant closed subspace. (zK K,forallz M) ⇢ ⇢ 2 If p is the orthogonal projection onto K,thenp M . 2 0 Proof of the claim. If z M,thenzpH = zK K. 2 ⇢ Hence (1 p)zpH (1 p)K = 0 and thus (1 p)zp =0(a). ⇢ { } Since z M (1 p)z p =0 pz(1 p)=0(b). ⇤ 2 ) ⇤ ) (a) and (b) pz zp = pz(1 p) (1 p)zp = 0, proving the claim. )

Proof of the bicommutant theorem, I

36 / 174 If p is the orthogonal projection onto K,thenp M . 2 0 Proof of the claim. If z M,thenzpH = zK K. 2 ⇢ Hence (1 p)zpH (1 p)K = 0 and thus (1 p)zp =0(a). ⇢ { } Since z M (1 p)z p =0 pz(1 p)=0(b). ⇤ 2 ) ⇤ ) (a) and (b) pz zp = pz(1 p) (1 p)zp = 0, proving the claim. )

Proof of the bicommutant theorem, I

37 / 174 Proof of the claim. If z M,thenzpH = zK K. 2 ⇢ Hence (1 p)zpH (1 p)K = 0 and thus (1 p)zp =0(a). ⇢ { } Since z M (1 p)z p =0 pz(1 p)=0(b). ⇤ 2 ) ⇤ ) (a) and (b) pz zp = pz(1 p) (1 p)zp = 0, proving the claim. )

Proof of the bicommutant theorem, I

38 / 174 Hence (1 p)zpH (1 p)K = 0 and thus (1 p)zp =0(a). ⇢ { } Since z M (1 p)z p =0 pz(1 p)=0(b). ⇤ 2 ) ⇤ ) (a) and (b) pz zp = pz(1 p) (1 p)zp = 0, proving the claim. )

Proof of the bicommutant theorem, I

Assumption: M B(H) is an SOT-closed unital -algebra. ⇢ ⇤ Goal: M = M00.Tothisend,ﬁxx M00. 2 SOT In order to show that x M, it suces to argue that x M : 2 2 ⇠ ,...,⇠ H and ">0, y M s.t. x⇠ y⇠ <", i =1,...,n. 8 1 n 2 9 2 k i i k 8 First,weassumethatn = 1 and prove the following: Claim Let K H be an M-invariant closed subspace. (zK K,forallz M) ⇢ ⇢ 2 If p is the orthogonal projection onto K,thenp M . 2 0 Proof of the claim. If z M,thenzpH = zK K. 2 ⇢

39 / 174 Since z M (1 p)z p =0 pz(1 p)=0(b). ⇤ 2 ) ⇤ ) (a) and (b) pz zp = pz(1 p) (1 p)zp = 0, proving the claim. )

Proof of the bicommutant theorem, I

40 / 174 (a) and (b) pz zp = pz(1 p) (1 p)zp = 0, proving the claim. )

Proof of the bicommutant theorem, I

41 / 174 Proof of the bicommutant theorem, I

42 / 174 In the present case n = 1, let p be the orthogonal projection onto

M⇠ := y⇠ y M . 1 { 1 | 2 } Since M⇠ H is a closed M-invariant subspace, the claim implies p M . 1 ⇢ 2 0 Since x (M ) ,wehavexp = px,andthusx⇠ = xp⇠ = p(x⇠ ) M⇠ . 2 0 0 1 1 1 2 1 This implies that x⇠ y⇠ <", for some y M, and ﬁnishes the proof k 1 1k 2 in the case n = 1.

Proof of the bicommutant theorem, II

Goal:IfM B(H)isaunital -algebra and x M00,then ⇢ ⇤ 2 ⇠ ,...,⇠ H and ">0, y M s.t. x⇠ y⇠ <", i =1,...,n. 8 1 n 2 9 2 k i i k 8

43 / 174 Since M⇠ H is a closed M-invariant subspace, the claim implies p M . 1 ⇢ 2 0 Since x (M ) ,wehavexp = px,andthusx⇠ = xp⇠ = p(x⇠ ) M⇠ . 2 0 0 1 1 1 2 1 This implies that x⇠ y⇠ <", for some y M, and ﬁnishes the proof k 1 1k 2 in the case n = 1.

Proof of the bicommutant theorem, II

Goal:IfM B(H)isaunital -algebra and x M00,then ⇢ ⇤ 2 ⇠ ,...,⇠ H and ">0, y M s.t. x⇠ y⇠ <", i =1,...,n. 8 1 n 2 9 2 k i i k 8 In the present case n = 1, let p be the orthogonal projection onto

M⇠ := y⇠ y M . 1 { 1 | 2 }

44 / 174 Since x (M ) ,wehavexp = px,andthusx⇠ = xp⇠ = p(x⇠ ) M⇠ . 2 0 0 1 1 1 2 1 This implies that x⇠ y⇠ <", for some y M, and ﬁnishes the proof k 1 1k 2 in the case n = 1.

Proof of the bicommutant theorem, II

M⇠ := y⇠ y M . 1 { 1 | 2 } Since M⇠ H is a closed M-invariant subspace, the claim implies p M . 1 ⇢ 2 0

45 / 174 This implies that x⇠ y⇠ <", for some y M, and ﬁnishes the proof k 1 1k 2 in the case n = 1.

Proof of the bicommutant theorem, II

M⇠ := y⇠ y M . 1 { 1 | 2 } Since M⇠ H is a closed M-invariant subspace, the claim implies p M . 1 ⇢ 2 0 Since x (M ) ,wehavexp = px,andthusx⇠ = xp⇠ = p(x⇠ ) M⇠ . 2 0 0 1 1 1 2 1

46 / 174 Proof of the bicommutant theorem, II

47 / 174 We will use a “matrix trick”. We deﬁne a diagonal -homomorphism n ⇤ ⇡ : B(H) B(H ) Mn(B(H)) by letting ! ⌘ y 0 .. 0 0 y ... 0 ⇡(y)=0 1 ...... B 00... y C B C @ A Equivalently: ⇡(y)(⌘ ... ⌘ )=y⌘ ... y⌘ . 1 n 1 n Exercise 4. Show that ⇡(M00) Mn(M0)0 and ⇡(M)0 Mn(M0). ⇢ ⇢ This implies that ⇡(M ) (⇡(M) ) ,hence⇡(x) ⇡(M) . 00 ⇢ 0 0 2 00 If ⇠ = ⇠ ... ⇠ Hn,thecasen =1 y M s.t. 1 n 2 )9 2 ⇡(x)⇠ ⇡(y)⇠ <",orequivalently n x⇠ y⇠ 2 1/2 <". k k i=1 k i i k P

Proof of the bicommutant theorem, III

Goal:IfM B(H)isaunital -algebra and x M00,then ⇢ ⇤ 2 ⇠ ,...,⇠ H and ">0, y M s.t. x⇠ y⇠ <", i =1,...,n. 8 1 n 2 9 2 k i i k 8 Since we have veriﬁed the case n = 1, we may assume n 2.

48 / 174 Equivalently: ⇡(y)(⌘ ... ⌘ )=y⌘ ... y⌘ . 1 n 1 n Exercise 4. Show that ⇡(M00) Mn(M0)0 and ⇡(M)0 Mn(M0). ⇢ ⇢ This implies that ⇡(M ) (⇡(M) ) ,hence⇡(x) ⇡(M) . 00 ⇢ 0 0 2 00 If ⇠ = ⇠ ... ⇠ Hn,thecasen =1 y M s.t. 1 n 2 )9 2 ⇡(x)⇠ ⇡(y)⇠ <",orequivalently n x⇠ y⇠ 2 1/2 <". k k i=1 k i i k P

Proof of the bicommutant theorem, III

49 / 174 Exercise 4. Show that ⇡(M00) Mn(M0)0 and ⇡(M)0 Mn(M0). ⇢ ⇢ This implies that ⇡(M ) (⇡(M) ) ,hence⇡(x) ⇡(M) . 00 ⇢ 0 0 2 00 If ⇠ = ⇠ ... ⇠ Hn,thecasen =1 y M s.t. 1 n 2 )9 2 ⇡(x)⇠ ⇡(y)⇠ <",orequivalently n x⇠ y⇠ 2 1/2 <". k k i=1 k i i k P

Proof of the bicommutant theorem, III

50 / 174 This implies that ⇡(M ) (⇡(M) ) ,hence⇡(x) ⇡(M) . 00 ⇢ 0 0 2 00 If ⇠ = ⇠ ... ⇠ Hn,thecasen =1 y M s.t. 1 n 2 )9 2 ⇡(x)⇠ ⇡(y)⇠ <",orequivalently n x⇠ y⇠ 2 1/2 <". k k i=1 k i i k P

Proof of the bicommutant theorem, III

51 / 174 If ⇠ = ⇠ ... ⇠ Hn,thecasen =1 y M s.t. 1 n 2 )9 2 ⇡(x)⇠ ⇡(y)⇠ <",orequivalently n x⇠ y⇠ 2 1/2 <". k k i=1 k i i k P

Proof of the bicommutant theorem, III

52 / 174 Proof of the bicommutant theorem, III

Goal:IfM B(H)isaunital -algebra and x M00,then ⇢ ⇤ 2 ⇠ ,...,⇠ H and ">0, y M s.t. x⇠ y⇠ <", i =1,...,n. 8 1 n 2 9 2 k i i k 8 Since we have verified the case n = 1, we may assume n 2. We will use a “matrix trick”. We define a diagonal -homomorphism n ⇤ ⇡ : B(H) B(H ) Mn(B(H)) by letting ! ⌘ y 0 .. 0 0 y ... 0 ⇡(y)=0 1 ...... B 00... y C B C @ A Equivalently: ⇡(y)(⌘ ... ⌘ )=y⌘ ... y⌘ . 1 n 1 n Exercise 4. Show that ⇡(M00) Mn(M0)0 and ⇡(M)0 Mn(M0). ⇢ ⇢ This implies that ⇡(M ) (⇡(M) ) ,hence⇡(x) ⇡(M) . 00 ⇢ 0 0 2 00 If ⇠ = ⇠ ... ⇠ Hn,thecasen =1 y M s.t. 1 n 2 )9 2 ⇡(x)⇠ ⇡(y)⇠ <",orequivalently n x⇠ y⇠ 2 1/2 <". k k i=1 k i i k 53 / 174 P 2 Define a -homomorphism ⇡ : L1(X ,µ) B(L (X ,µ)) by ⇡(f )(⇠)=f ⇠, ⇤ ! for all f L (X )and⇠ L2(X ). 2 1 2 2 Then ⇡(L1(X ))0 = ⇡(L1(X )).Therefore, ⇡(L1(X )) B(L (X )) ⇢ is a maximal abelian von Neumann subalgebra.

Proof. Let T ⇡(L (X )) and put g = T (1 ) L2(X ). 2 1 0 X 2 Then fg = ⇡(f )T (1)=T ⇡(f )(1)=T (f ), for all f L (X ). 2 1 Since L (X ) L2(X )isdense,itsucestoshowthatg L (X ). 1 ⇢ 2 1 For n 1, let fn = 1 x X g(x) T +1/n .Then fng ( T +1/n) fn , { 2 || |k k } | | k k | | hence T f T (f ) = f g ( T +1/n) f ,sof = 0, a.e. k kk nk2 k n k2 k n k2 k k k nk2 n Thus, µ( x X g(x) T +1/n )=0foralln 1 g T . { 2 || |k k } )k k1 k k

L1-algebras as von Neumann algebras

Proposition Let (X ,µ)beastandard probability space (X is a Polish space, µ is a Borel measure on X with µ(X ) = 1).

54 / 174 2 Then ⇡(L1(X ))0 = ⇡(L1(X )).Therefore, ⇡(L1(X )) B(L (X )) ⇢ is a maximal abelian von Neumann subalgebra.

L1-algebras as von Neumann algebras

55 / 174 2 Therefore, ⇡(L1(X )) B(L (X )) ⇢ is a maximal abelian von Neumann subalgebra.

L1-algebras as von Neumann algebras

Proposition Let (X ,µ)beastandard probability space (X is a Polish space, µ is a Borel measure on X with µ(X ) = 1). 2 Deﬁne a -homomorphism ⇡ : L1(X ,µ) B(L (X ,µ)) by ⇡(f )(⇠)=f ⇠, ⇤ ! for all f L (X )and⇠ L2(X ). 2 1 2 Then ⇡(L1(X ))0 = ⇡(L1(X )).

56 / 174 Proof. Let T ⇡(L (X )) and put g = T (1 ) L2(X ). 2 1 0 X 2 Then fg = ⇡(f )T (1)=T ⇡(f )(1)=T (f ), for all f L (X ). 2 1 Since L (X ) L2(X )isdense,itsucestoshowthatg L (X ). 1 ⇢ 2 1 For n 1, let fn = 1 x X g(x) T +1/n .Then fng ( T +1/n) fn , { 2 || |k k } | | k k | | hence T f T (f ) = f g ( T +1/n) f ,sof = 0, a.e. k kk nk2 k n k2 k n k2 k k k nk2 n Thus, µ( x X g(x) T +1/n )=0foralln 1 g T . { 2 || |k k } )k k1 k k

L1-algebras as von Neumann algebras

57 / 174 Then fg = ⇡(f )T (1)=T ⇡(f )(1)=T (f ), for all f L (X ). 2 1 Since L (X ) L2(X )isdense,itsucestoshowthatg L (X ). 1 ⇢ 2 1 For n 1, let fn = 1 x X g(x) T +1/n .Then fng ( T +1/n) fn , { 2 || |k k } | | k k | | hence T f T (f ) = f g ( T +1/n) f ,sof = 0, a.e. k kk nk2 k n k2 k n k2 k k k nk2 n Thus, µ( x X g(x) T +1/n )=0foralln 1 g T . { 2 || |k k } )k k1 k k

L1-algebras as von Neumann algebras

Proof. Let T ⇡(L (X )) and put g = T (1 ) L2(X ). 2 1 0 X 2

58 / 174 Since L (X ) L2(X )isdense,itsucestoshowthatg L (X ). 1 ⇢ 2 1 For n 1, let fn = 1 x X g(x) T +1/n .Then fng ( T +1/n) fn , { 2 || |k k } | | k k | | hence T f T (f ) = f g ( T +1/n) f ,sof = 0, a.e. k kk nk2 k n k2 k n k2 k k k nk2 n Thus, µ( x X g(x) T +1/n )=0foralln 1 g T . { 2 || |k k } )k k1 k k

L1-algebras as von Neumann algebras

Proof. Let T ⇡(L (X )) and put g = T (1 ) L2(X ). 2 1 0 X 2 Then fg = ⇡(f )T (1)=T ⇡(f )(1)=T (f ), for all f L (X ). 2 1

59 / 174 For n 1, let fn = 1 x X g(x) T +1/n .Then fng ( T +1/n) fn , { 2 || |k k } | | k k | | hence T f T (f ) = f g ( T +1/n) f ,sof = 0, a.e. k kk nk2 k n k2 k n k2 k k k nk2 n Thus, µ( x X g(x) T +1/n )=0foralln 1 g T . { 2 || |k k } )k k1 k k

L1-algebras as von Neumann algebras

Proof. Let T ⇡(L (X )) and put g = T (1 ) L2(X ). 2 1 0 X 2 Then fg = ⇡(f )T (1)=T ⇡(f )(1)=T (f ), for all f L (X ). 2 1 Since L (X ) L2(X )isdense,itsucestoshowthatg L (X ). 1 ⇢ 2 1

60 / 174 hence T f T (f ) = f g ( T +1/n) f ,sof = 0, a.e. k kk nk2 k n k2 k n k2 k k k nk2 n Thus, µ( x X g(x) T +1/n )=0foralln 1 g T . { 2 || |k k } )k k1 k k

L1-algebras as von Neumann algebras

61 / 174 Thus, µ( x X g(x) T +1/n )=0foralln 1 g T . { 2 || |k k } )k k1 k k

L1-algebras as von Neumann algebras

62 / 174 L1-algebras as von Neumann algebras

63 / 174 Example. '(T )= T (⇠),⇠ is a state on B(H), for ⇠ H, ⇠ = 1. h i 2 k k Exercise 5. Show that a) '(y x) 2 '(x x)'(y y), for all x, y A. | ⇤ |  ⇤ ⇤ 2 b) ' is bounded and ' = '(1). k k Deﬁnition A von Neumann algebra M B(H)iscalledtracial if there exists a state ⇢ ' : M C (called a trace)whichis: ! 1 faithful if x 0and'(x) = 0, then x = 0. 2 normal (or completely additive) '( i I pi )= i I '(pi ), for any 2 2 collection of mutually orthogonal projections pi i I in M. P { P} 2 3 tracial '(xy)='(yx), for all x, y M. 2

Examples. 1) L1(X ,µ)with'(f )= X f dµ. 2) ( )withtheusualnormalizedtrace'([a ]) = (a + .... + a )/n. Mn C R i,j 1,1 n,n 3) B(H) is not tracial if dim H = . 1

States and traces

Deﬁnition.LetA be a unital C⇤-algebra. A unital linear functional ' : A C is called a state if is positive: '(x) 0, for all x A, x 0. ! 2

64 / 174 Exercise 5. Show that a) '(y x) 2 '(x x)'(y y), for all x, y A. | ⇤ |  ⇤ ⇤ 2 b) ' is bounded and ' = '(1). k k Deﬁnition A von Neumann algebra M B(H)iscalledtracial if there exists a state ⇢ ' : M C (called a trace)whichis: ! 1 faithful if x 0and'(x) = 0, then x = 0. 2 normal (or completely additive) '( i I pi )= i I '(pi ), for any 2 2 collection of mutually orthogonal projections pi i I in M. P { P} 2 3 tracial '(xy)='(yx), for all x, y M. 2

Examples. 1) L1(X ,µ)with'(f )= X f dµ. 2) ( )withtheusualnormalizedtrace'([a ]) = (a + .... + a )/n. Mn C R i,j 1,1 n,n 3) B(H) is not tracial if dim H = . 1

States and traces

65 / 174 Deﬁnition A von Neumann algebra M B(H)iscalledtracial if there exists a state ⇢ ' : M C (called a trace)whichis: ! 1 faithful if x 0and'(x) = 0, then x = 0. 2 normal (or completely additive) '( i I pi )= i I '(pi ), for any 2 2 collection of mutually orthogonal projections pi i I in M. P { P} 2 3 tracial '(xy)='(yx), for all x, y M. 2

Examples. 1) L1(X ,µ)with'(f )= X f dµ. 2) ( )withtheusualnormalizedtrace'([a ]) = (a + .... + a )/n. Mn C R i,j 1,1 n,n 3) B(H) is not tracial if dim H = . 1

States and traces

66 / 174 2 normal (or completely additive) '( i I pi )= i I '(pi ), for any 2 2 collection of mutually orthogonal projections pi i I in M. P { P} 2 3 tracial '(xy)='(yx), for all x, y M. 2

Examples. 1) L1(X ,µ)with'(f )= X f dµ. 2) ( )withtheusualnormalizedtrace'([a ]) = (a + .... + a )/n. Mn C R i,j 1,1 n,n 3) B(H) is not tracial if dim H = . 1

States and traces

Deﬁnition.LetA be a unital C⇤-algebra. A unital linear functional ' : A C is called a state if is positive: '(x) 0, for all x A, x 0. ! 2 Example. '(T )= T (⇠),⇠ is a state on B(H), for ⇠ H, ⇠ = 1. h i 2 k k Exercise 5. Show that a) '(y x) 2 '(x x)'(y y), for all x, y A. | ⇤ |  ⇤ ⇤ 2 b) ' is bounded and ' = '(1). k k Deﬁnition A von Neumann algebra M B(H)iscalledtracial if there exists a state ⇢ ' : M C (called a trace)whichis: ! 1 faithful if x 0and'(x) = 0, then x = 0.

67 / 174 3 tracial '(xy)='(yx), for all x, y M. 2

Examples. 1) L1(X ,µ)with'(f )= X f dµ. 2) ( )withtheusualnormalizedtrace'([a ]) = (a + .... + a )/n. Mn C R i,j 1,1 n,n 3) B(H) is not tracial if dim H = . 1

States and traces

68 / 174 Examples. 1) L1(X ,µ)with'(f )= X f dµ. 2) ( )withtheusualnormalizedtrace'([a ]) = (a + .... + a )/n. Mn C R i,j 1,1 n,n 3) B(H) is not tracial if dim H = . 1

States and traces

69 / 174 States and traces

Examples. 1) L1(X ,µ)with'(f )= X f dµ. 2) ( )withtheusualnormalizedtrace'([a ]) = (a + .... + a )/n. Mn C R i,j 1,1 n,n 3) B(H) is not tracial if dim H = . 1 70 / 174 p is dominated by q (p q)if q M such that q q and p q 9 0 2 0  ⇠ Lemma 1. If pMq = 0 p , q M s.t. p p, q q,andp q . 6 { })9 0 0 2 0  0  0 ⇠ 0 [If x M satisﬁes pxq = 0, then polar decomposition gives a partial 2 6 isometry v M s.t. x = v x .Onechecksthatvv p and v v q.] 2 | | ⇤  ⇤  Lemma 2. a projection z (M)s.t. pz qz and q(1 z) p(1 z). 9 2Z Proof sketch. Let (p0, q0) be a maximal pair of projections such that p , q M, p p, q q and p q . Since p p and q q have no 0 0 2 0  0  0 ⇠ 0 0 0 equivalent subprojections, Lemma 1 (p p )M(q q )= 0 . ) 0 0 { } Let z be the ortho. proj. onto M(q q )H.Thenz (M). 0 2Z Since (p p )z =0 pz = p z q z qz pz qz. 0 ) 0 ⇠ 0  ) Since (q q ) z (q q )(1 z)=0 q(1 z) p(1 z). 0  ) 0 )

Equivalence of projections

71 / 174 Lemma 1. If pMq = 0 p , q M s.t. p p, q q,andp q . 6 { })9 0 0 2 0  0  0 ⇠ 0 [If x M satisﬁes pxq = 0, then polar decomposition gives a partial 2 6 isometry v M s.t. x = v x .Onechecksthatvv p and v v q.] 2 | | ⇤  ⇤  Lemma 2. a projection z (M)s.t. pz qz and q(1 z) p(1 z). 9 2Z Proof sketch. Let (p0, q0) be a maximal pair of projections such that p , q M, p p, q q and p q . Since p p and q q have no 0 0 2 0  0  0 ⇠ 0 0 0 equivalent subprojections, Lemma 1 (p p )M(q q )= 0 . ) 0 0 { } Let z be the ortho. proj. onto M(q q )H.Thenz (M). 0 2Z Since (p p )z =0 pz = p z q z qz pz qz. 0 ) 0 ⇠ 0  ) Since (q q ) z (q q )(1 z)=0 q(1 z) p(1 z). 0  ) 0 )

Equivalence of projections

72 / 174 [If x M satisﬁes pxq = 0, then polar decomposition gives a partial 2 6 isometry v M s.t. x = v x .Onechecksthatvv p and v v q.] 2 | | ⇤  ⇤  Lemma 2. a projection z (M)s.t. pz qz and q(1 z) p(1 z). 9 2Z Proof sketch. Let (p0, q0) be a maximal pair of projections such that p , q M, p p, q q and p q . Since p p and q q have no 0 0 2 0  0  0 ⇠ 0 0 0 equivalent subprojections, Lemma 1 (p p )M(q q )= 0 . ) 0 0 { } Let z be the ortho. proj. onto M(q q )H.Thenz (M). 0 2Z Since (p p )z =0 pz = p z q z qz pz qz. 0 ) 0 ⇠ 0  ) Since (q q ) z (q q )(1 z)=0 q(1 z) p(1 z). 0  ) 0 )

Equivalence of projections

73 / 174 Lemma 2. a projection z (M)s.t. pz qz and q(1 z) p(1 z). 9 2Z Proof sketch. Let (p0, q0) be a maximal pair of projections such that p , q M, p p, q q and p q . Since p p and q q have no 0 0 2 0  0  0 ⇠ 0 0 0 equivalent subprojections, Lemma 1 (p p )M(q q )= 0 . ) 0 0 { } Let z be the ortho. proj. onto M(q q )H.Thenz (M). 0 2Z Since (p p )z =0 pz = p z q z qz pz qz. 0 ) 0 ⇠ 0  ) Since (q q ) z (q q )(1 z)=0 q(1 z) p(1 z). 0  ) 0 )

Equivalence of projections

74 / 174 Proof sketch. Let (p0, q0) be a maximal pair of projections such that p , q M, p p, q q and p q . Since p p and q q have no 0 0 2 0  0  0 ⇠ 0 0 0 equivalent subprojections, Lemma 1 (p p )M(q q )= 0 . ) 0 0 { } Let z be the ortho. proj. onto M(q q )H.Thenz (M). 0 2Z Since (p p )z =0 pz = p z q z qz pz qz. 0 ) 0 ⇠ 0  ) Since (q q ) z (q q )(1 z)=0 q(1 z) p(1 z). 0  ) 0 )

Equivalence of projections

75 / 174 Since p p and q q have no 0 0 equivalent subprojections, Lemma 1 (p p )M(q q )= 0 . ) 0 0 { } Let z be the ortho. proj. onto M(q q )H.Thenz (M). 0 2Z Since (p p )z =0 pz = p z q z qz pz qz. 0 ) 0 ⇠ 0  ) Since (q q ) z (q q )(1 z)=0 q(1 z) p(1 z). 0  ) 0 )

Equivalence of projections

76 / 174 Let z be the ortho. proj. onto M(q q )H.Thenz (M). 0 2Z Since (p p )z =0 pz = p z q z qz pz qz. 0 ) 0 ⇠ 0  ) Since (q q ) z (q q )(1 z)=0 q(1 z) p(1 z). 0  ) 0 )

Equivalence of projections

Deﬁnition Let M B(H) be a von Neumann algebra and p, q M be projections. ⇢ 2 p is equivalent to q (p q)if v M such that v v = p, vv = q. ⇠ 9 2 ⇤ ⇤ (v is an isometry between pH and qH,calledapartial isometry). p is dominated by q (p q)if q M such that q q and p q 9 0 2 0  ⇠ Lemma 1. If pMq = 0 p , q M s.t. p p, q q,andp q . 6 { })9 0 0 2 0  0  0 ⇠ 0 [If x M satisﬁes pxq = 0, then polar decomposition gives a partial 2 6 isometry v M s.t. x = v x .Onechecksthatvv p and v v q.] 2 | | ⇤  ⇤  Lemma 2. a projection z (M)s.t. pz qz and q(1 z) p(1 z). 9 2Z Proof sketch. Let (p0, q0) be a maximal pair of projections such that p , q M, p p, q q and p q . Since p p and q q have no 0 0 2 0  0  0 ⇠ 0 0 0 equivalent subprojections, Lemma 1 (p p )M(q q )= 0 . ) 0 0 { }

77 / 174 Since (p p )z =0 pz = p z q z qz pz qz. 0 ) 0 ⇠ 0  ) Since (q q ) z (q q )(1 z)=0 q(1 z) p(1 z). 0  ) 0 )

Equivalence of projections

78 / 174 Since (q q ) z (q q )(1 z)=0 q(1 z) p(1 z). 0  ) 0 )

Equivalence of projections

Note: If H is separable, any vNa M is an integral M = X Mx dµ(x) of factors Mx x X “indexed” over a probability space (X ,µ). { } 2 R Definition. A projection p in a von Neumann algebra M is minimal if q M, q p q 0, p . 2  ) 2{ } finite if q M, q p, q p q = p. 2  ⇠ ) Remark. Any minimal projection is finite. If M = B(H), then a projection p is minimal i↵dim(pH) 0, 1 and finite i↵dim(pH) < . 2{ } 1 AfactorM B(H)iscalledof ⇢ 1 type I if it has a non-zero minimal projection. 2 type II if it has a non-zero finite projection, but not minimal proj. If 1 M is finite, M is of type II1; otherwise, M is of type II . 2 1 3 type III if it contains no non-zero finite projection.

Classiﬁcation into types for factors

A von Neumann algebra M is called a factor if (M)=M M0 = C1. Z \

81 / 174 Note: If H is separable, any vNa M is an integral M = X Mx dµ(x) of factors Mx x X “indexed” over a probability space (X ,µ). { } 2 R Definition. A projection p in a von Neumann algebra M is minimal if q M, q p q 0, p . 2  ) 2{ } finite if q M, q p, q p q = p. 2  ⇠ ) Remark. Any minimal projection is finite. If M = B(H), then a projection p is minimal i↵dim(pH) 0, 1 and finite i↵dim(pH) < . 2{ } 1 AfactorM B(H)iscalledof ⇢ 1 type I if it has a non-zero minimal projection. 2 type II if it has a non-zero finite projection, but not minimal proj. If 1 M is finite, M is of type II1; otherwise, M is of type II . 2 1 3 type III if it contains no non-zero finite projection.

Classiﬁcation into types for factors

A von Neumann algebra M is called a factor if (M)=M M0 = C1. Z \ Corollary. If M is a factor, then p, q projections, either p q or q p. 8

82 / 174 Remark. Any minimal projection is finite. If M = B(H), then a projection p is minimal i↵dim(pH) 0, 1 and finite i↵dim(pH) < . 2{ } 1 AfactorM B(H)iscalledof ⇢ 1 type I if it has a non-zero minimal projection. 2 type II if it has a non-zero finite projection, but not minimal proj. If 1 M is finite, M is of type II1; otherwise, M is of type II . 2 1 3 type III if it contains no non-zero finite projection.

Classiﬁcation into types for factors

A von Neumann algebra M is called a factor if (M)=M M0 = C1. Z \ Corollary. If M is a factor, then p, q projections, either p q or q p. 8

83 / 174 AfactorM B(H)iscalledof ⇢ 1 type I if it has a non-zero minimal projection. 2 type II if it has a non-zero finite projection, but not minimal proj. If 1 M is finite, M is of type II1; otherwise, M is of type II . 2 1 3 type III if it contains no non-zero finite projection.

Classiﬁcation into types for factors

A von Neumann algebra M is called a factor if (M)=M M0 = C1. Z \ Corollary. If M is a factor, then p, q projections, either p q or q p. 8

84 / 174 Classiﬁcation into types for factors

A von Neumann algebra M is called a factor if (M)=M M0 = C1. Z \ Corollary. If M is a factor, then p, q projections, either p q or q p. 8

85 / 174 Proof. Let p M be a minimal projection. Let pi i I be a maximal 2 { } 2 family of mutually orthogonal proj. in M which are equivalent to p. Put q =1 p .Thenp q or q p. Maximality q p. i I i ) Since p is minimal2 q =0orq = p (impossible by maximality). P ) Thus, q =0andhence i I pi =1. (a) 2 Since p is minimal pMp = Cp. (b) ) P Combining (a) and (b) M = B(`2I ). ) ⇠ Let an inﬁnite countable group, (X ,µ) non-atomic measure space.

inﬁnite conjugacy class II1 factor L(). y (X ,µ) non-singular free ergodic action factor L1(X ) o of 1 type II i↵ -invariant probability measure on X equivalent to µ. 1 9 2 type II i↵ -invariant inﬁnite measure on X equivalent to ⌫. 1 9 3 type III i↵no measure on X equivalent to µ is -invariant.

Examples of factors

B(K) is a factor of type I, for any Hilbert space K. Any factor M of type I is isomorphic to B(K), for a Hilbert space K.

86 / 174 Put q =1 p .Thenp q or q p. Maximality q p. i I i ) Since p is minimal2 q =0orq = p (impossible by maximality). P ) Thus, q =0andhence i I pi =1. (a) 2 Since p is minimal pMp = Cp. (b) ) P Combining (a) and (b) M = B(`2I ). ) ⇠ Let an inﬁnite countable group, (X ,µ) non-atomic measure space.

Examples of factors

87 / 174 Since p is minimal q =0orq = p (impossible by maximality). ) Thus, q =0andhence i I pi =1. (a) 2 Since p is minimal pMp = Cp. (b) ) P Combining (a) and (b) M = B(`2I ). ) ⇠ Let an inﬁnite countable group, (X ,µ) non-atomic measure space.

Examples of factors

Put q =1 i I pi .Thenp q or q p. Maximality q p. 2 ) P

88 / 174 Thus, q =0andhence i I pi =1. (a) 2 Since p is minimal pMp = Cp. (b) ) P Combining (a) and (b) M = B(`2I ). ) ⇠ Let an inﬁnite countable group, (X ,µ) non-atomic measure space.

Examples of factors

89 / 174 Since p is minimal pMp = Cp. (b) ) Combining (a) and (b) M = B(`2I ). ) ⇠ Let an inﬁnite countable group, (X ,µ) non-atomic measure space.

Examples of factors

B(K) is a factor of type I, for any Hilbert space K. Any factor M of type I is isomorphic to B(K), for a Hilbert space K. Proof. Let p M be a minimal projection. Let pi i I be a maximal 2 { } 2 family of mutually orthogonal proj. in M which are equivalent to p. Put q =1 p .Thenp q or q p. Maximality q p. i I i ) Since p is minimal2 q =0orq = p (impossible by maximality). P ) Thus, q =0andhence i I pi =1. (a) 2 P

90 / 174 Combining (a) and (b) M = B(`2I ). ) ⇠ Let an inﬁnite countable group, (X ,µ) non-atomic measure space.

Examples of factors

91 / 174 Let an inﬁnite countable group, (X ,µ) non-atomic measure space.

Examples of factors

92 / 174 inﬁnite conjugacy class II1 factor L(). y (X ,µ) non-singular free ergodic action factor L1(X ) o of 1 type II i↵ -invariant probability measure on X equivalent to µ. 1 9 2 type II i↵ -invariant inﬁnite measure on X equivalent to ⌫. 1 9 3 type III i↵no measure on X equivalent to µ is -invariant.

Examples of factors

93 / 174 y (X ,µ) non-singular free ergodic action factor L1(X ) o of 1 type II i↵ -invariant probability measure on X equivalent to µ. 1 9 2 type II i↵ -invariant inﬁnite measure on X equivalent to ⌫. 1 9 3 type III i↵no measure on X equivalent to µ is -invariant.

Examples of factors

inﬁnite conjugacy class II1 factor L().

94 / 174 1 type II i↵ -invariant probability measure on X equivalent to µ. 1 9 2 type II i↵ -invariant inﬁnite measure on X equivalent to ⌫. 1 9 3 type III i↵no measure on X equivalent to µ is -invariant.

Examples of factors

inﬁnite conjugacy class II1 factor L(). y (X ,µ) non-singular free ergodic action factor L1(X ) o of

95 / 174 2 type II i↵ -invariant inﬁnite measure on X equivalent to ⌫. 1 9 3 type III i↵no measure on X equivalent to µ is -invariant.

Examples of factors

inﬁnite conjugacy class II1 factor L(). y (X ,µ) non-singular free ergodic action factor L1(X ) o of 1 type II i↵ -invariant probability measure on X equivalent to µ. 1 9

96 / 174 3 type III i↵no measure on X equivalent to µ is -invariant.

Examples of factors

97 / 174 Examples of factors

98 / 174 Moreover, if M is a II factor, then !tracialfaithfulnormalstate 1 9 ⌧ : M C (called the trace of M). ! Proof of . Assume that atrace⌧ : M C. ( 9 ! Goal. M is type II : (a) 1 M is a finite projection and (b) M not type I. 1 2 (a) Let p M such that p 1 v M with v v =1, vv = p. 2 ⇠ )9 2 ⇤ ⇤ Thus, ⌧(1 p)=⌧(v v vv )=⌧(v v) ⌧(vv ) = 0. ⇤ ⇤ ⇤ ⇤ Since 1 p 0and⌧ is faithful p =1 1isfinite.(b) If M is type I, ) ) since dim M = M = B(K), where dim K = .But1 B(K), with 1) ⇠ 1 2 dim K = , is not a finite projection, contradiction. 1 Notation. If M is a II factor and ⌧ its trace, we put x := ⌧(x x)1/2. 1 k k2 ⇤ L2(M) is the closure of M w.r.t. . . k k2 We have an injective SOT-cont. -homomorphism ⇡ : M B(L2(M)) ⇤ ! given by ⇡(x)(y)=xy,forx, y M.(the standard representation) 2

II1 factors are tracial

Theorem AfactorM is of type II it is an inﬁnite dimensional tracial factor. 1 ()

99 / 174 Proof of . Assume that atrace⌧ : M C. ( 9 ! Goal. M is type II : (a) 1 M is a finite projection and (b) M not type I. 1 2 (a) Let p M such that p 1 v M with v v =1, vv = p. 2 ⇠ )9 2 ⇤ ⇤ Thus, ⌧(1 p)=⌧(v v vv )=⌧(v v) ⌧(vv ) = 0. ⇤ ⇤ ⇤ ⇤ Since 1 p 0and⌧ is faithful p =1 1isfinite.(b) If M is type I, ) ) since dim M = M = B(K), where dim K = .But1 B(K), with 1) ⇠ 1 2 dim K = , is not a finite projection, contradiction. 1 Notation. If M is a II factor and ⌧ its trace, we put x := ⌧(x x)1/2. 1 k k2 ⇤ L2(M) is the closure of M w.r.t. . . k k2 We have an injective SOT-cont. -homomorphism ⇡ : M B(L2(M)) ⇤ ! given by ⇡(x)(y)=xy,forx, y M.(the standard representation) 2

II1 factors are tracial

Theorem AfactorM is of type II it is an inﬁnite dimensional tracial factor. 1 () Moreover, if M is a II factor, then !tracialfaithfulnormalstate 1 9 ⌧ : M C (called the trace of M). !

100 / 174 Goal. M is type II : (a) 1 M is a finite projection and (b) M not type I. 1 2 (a) Let p M such that p 1 v M with v v =1, vv = p. 2 ⇠ )9 2 ⇤ ⇤ Thus, ⌧(1 p)=⌧(v v vv )=⌧(v v) ⌧(vv ) = 0. ⇤ ⇤ ⇤ ⇤ Since 1 p 0and⌧ is faithful p =1 1isfinite.(b) If M is type I, ) ) since dim M = M = B(K), where dim K = .But1 B(K), with 1) ⇠ 1 2 dim K = , is not a finite projection, contradiction. 1 Notation. If M is a II factor and ⌧ its trace, we put x := ⌧(x x)1/2. 1 k k2 ⇤ L2(M) is the closure of M w.r.t. . . k k2 We have an injective SOT-cont. -homomorphism ⇡ : M B(L2(M)) ⇤ ! given by ⇡(x)(y)=xy,forx, y M.(the standard representation) 2

II1 factors are tracial

101 / 174 (a) Let p M such that p 1 v M with v v =1, vv = p. 2 ⇠ )9 2 ⇤ ⇤ Thus, ⌧(1 p)=⌧(v v vv )=⌧(v v) ⌧(vv ) = 0. ⇤ ⇤ ⇤ ⇤ Since 1 p 0and⌧ is faithful p =1 1isﬁnite.(b) If M is type I, ) ) since dim M = M = B(K), where dim K = .But1 B(K), with 1) ⇠ 1 2 dim K = , is not a ﬁnite projection, contradiction. 1 Notation. If M is a II factor and ⌧ its trace, we put x := ⌧(x x)1/2. 1 k k2 ⇤ L2(M) is the closure of M w.r.t. . . k k2 We have an injective SOT-cont. -homomorphism ⇡ : M B(L2(M)) ⇤ ! given by ⇡(x)(y)=xy,forx, y M.(the standard representation) 2

II1 factors are tracial

102 / 174 Thus, ⌧(1 p)=⌧(v v vv )=⌧(v v) ⌧(vv ) = 0. ⇤ ⇤ ⇤ ⇤ Since 1 p 0and⌧ is faithful p =1 1isﬁnite.(b) If M is type I, ) ) since dim M = M = B(K), where dim K = .But1 B(K), with 1) ⇠ 1 2 dim K = , is not a ﬁnite projection, contradiction. 1 Notation. If M is a II factor and ⌧ its trace, we put x := ⌧(x x)1/2. 1 k k2 ⇤ L2(M) is the closure of M w.r.t. . . k k2 We have an injective SOT-cont. -homomorphism ⇡ : M B(L2(M)) ⇤ ! given by ⇡(x)(y)=xy,forx, y M.(the standard representation) 2

II1 factors are tracial

103 / 174 Since 1 p 0and⌧ is faithful p =1 1isﬁnite.(b) If M is type I, ) ) since dim M = M = B(K), where dim K = .But1 B(K), with 1) ⇠ 1 2 dim K = , is not a ﬁnite projection, contradiction. 1 Notation. If M is a II factor and ⌧ its trace, we put x := ⌧(x x)1/2. 1 k k2 ⇤ L2(M) is the closure of M w.r.t. . . k k2 We have an injective SOT-cont. -homomorphism ⇡ : M B(L2(M)) ⇤ ! given by ⇡(x)(y)=xy,forx, y M.(the standard representation) 2

II1 factors are tracial

104 / 174 (b) If M is type I, since dim M = M = B(K), where dim K = .But1 B(K), with 1) ⇠ 1 2 dim K = , is not a ﬁnite projection, contradiction. 1 Notation. If M is a II factor and ⌧ its trace, we put x := ⌧(x x)1/2. 1 k k2 ⇤ L2(M) is the closure of M w.r.t. . . k k2 We have an injective SOT-cont. -homomorphism ⇡ : M B(L2(M)) ⇤ ! given by ⇡(x)(y)=xy,forx, y M.(the standard representation) 2

II1 factors are tracial

105 / 174 Notation. If M is a II factor and ⌧ its trace, we put x := ⌧(x x)1/2. 1 k k2 ⇤ L2(M) is the closure of M w.r.t. . . k k2 We have an injective SOT-cont. -homomorphism ⇡ : M B(L2(M)) ⇤ ! given by ⇡(x)(y)=xy,forx, y M.(the standard representation) 2

II1 factors are tracial

106 / 174 We have an injective SOT-cont. -homomorphism ⇡ : M B(L2(M)) ⇤ ! given by ⇡(x)(y)=xy,forx, y M.(the standard representation) 2

II1 factors are tracial

107 / 174 II1 factors are tracial

Theorem AfactorM is of type II it is an infinite dimensional tracial factor. 1 () Moreover, if M is a II factor, then !tracialfaithfulnormalstate 1 9 ⌧ : M C (called the trace of M). ! Proof of . Assume that atrace⌧ : M C. ( 9 ! Goal. M is type II : (a) 1 M is a finite projection and (b) M not type I. 1 2 (a) Let p M such that p 1 v M with v v =1, vv = p. 2 ⇠ )9 2 ⇤ ⇤ Thus, ⌧(1 p)=⌧(v v vv )=⌧(v v) ⌧(vv ) = 0. ⇤ ⇤ ⇤ ⇤ Since 1 p 0and⌧ is faithful p =1 1isfinite.(b) If M is type I, ) ) since dim M = M = B(K), where dim K = .But1 B(K), with 1) ⇠ 1 2 dim K = , is not a finite projection, contradiction. 1 Notation. If M is a II factor and ⌧ its trace, we put x := ⌧(x x)1/2. 1 k k2 ⇤ L2(M) is the closure of M w.r.t. . . k k2 We have an injective SOT-cont. -homomorphism ⇡ : M B(L2(M)) ⇤ ! given by ⇡(x)(y)=xy,forx, y M.(the standard representation) 2 108 / 174 2 2 1/2 Put ` = f : C f 2 := ( g f (g) ) < . { ! |k k2 2 | | 1} An orthonormal basis for ` is g g . {P} 2 The left regular representation : (`2) is given by !U (g)( )= ,forallg, h . Note. is a unitary representation: h gh 2 (g) is a unitary operator and (gh)=(g)(h), for all g, h . 2 Define := g F ag (g) F finite, ag C, g F . A { 2 | ⇢ 2 8 2 } Then B(`2) is a -subalgebra. A⇢ P ⇤ Remark. is isomorphic to the complex group algebra C[]. A Definition WOT L() := B(`2) is called the group von Neumann algebra of . A ⇢

Group von Neumann algebras, I

Let be a countable group.

109 / 174 2 An orthonormal basis for ` is g g . { } 2 The left regular representation : (`2) is given by !U (g)( )= ,forallg, h . Note. is a unitary representation: h gh 2 (g) is a unitary operator and (gh)=(g)(h), for all g, h . 2 Define := g F ag (g) F finite, ag C, g F . A { 2 | ⇢ 2 8 2 } Then B(`2) is a -subalgebra. A⇢ P ⇤ Remark. is isomorphic to the complex group algebra C[]. A Definition WOT L() := B(`2) is called the group von Neumann algebra of . A ⇢

Group von Neumann algebras, I

Let be a countable group. 2 2 1/2 Put ` = f : C f 2 := ( g f (g) ) < . { ! |k k 2 | | 1} P

110 / 174 The left regular representation : (`2) is given by !U (g)( )= ,forallg, h . Note. is a unitary representation: h gh 2 (g) is a unitary operator and (gh)=(g)(h), for all g, h . 2 Define := g F ag (g) F finite, ag C, g F . A { 2 | ⇢ 2 8 2 } Then B(`2) is a -subalgebra. A⇢ P ⇤ Remark. is isomorphic to the complex group algebra C[]. A Definition WOT L() := B(`2) is called the group von Neumann algebra of . A ⇢

Group von Neumann algebras, I

Let be a countable group. 2 2 1/2 Put ` = f : C f 2 := ( g f (g) ) < . { ! |k k2 2 | | 1} An orthonormal basis for ` is g g . {P} 2

111 / 174 Note. is a unitary representation: (g) is a unitary operator and (gh)=(g)(h), for all g, h . 2 Define := g F ag (g) F finite, ag C, g F . A { 2 | ⇢ 2 8 2 } Then B(`2) is a -subalgebra. A⇢ P ⇤ Remark. is isomorphic to the complex group algebra C[]. A Definition WOT L() := B(`2) is called the group von Neumann algebra of . A ⇢

Group von Neumann algebras, I

112 / 174 Define := g F ag (g) F finite, ag C, g F . A { 2 | ⇢ 2 8 2 } Then B(`2) is a -subalgebra. A⇢ P ⇤ Remark. is isomorphic to the complex group algebra C[]. A Definition WOT L() := B(`2) is called the group von Neumann algebra of . A ⇢

Group von Neumann algebras, I

113 / 174 Then B(`2) is a -subalgebra. A⇢ ⇤ Remark. is isomorphic to the complex group algebra C[]. A Deﬁnition WOT L() := B(`2) is called the group von Neumann algebra of . A ⇢

Group von Neumann algebras, I

114 / 174 Deﬁnition WOT L() := B(`2) is called the group von Neumann algebra of . A ⇢

Group von Neumann algebras, I

Let be a countable group. 2 2 1/2 Put ` = f : C f 2 := ( g f (g) ) < . { ! |k k2 2 | | 1} An orthonormal basis for ` is g g . {P} 2 The left regular representation : (`2) is given by !U (g)( )= ,forallg, h . Note. is a unitary representation: h gh 2 (g) is a unitary operator and (gh)=(g)(h), for all g, h . 2 Deﬁne := g F ag (g) F ﬁnite, ag C, g F . A { 2 | ⇢ 2 8 2 } Then B(`2) is a -subalgebra. A⇢ P ⇤ Remark. is isomorphic to the complex group algebra C[]. A

115 / 174 Group von Neumann algebras, I

116 / 174 ⌧ is faithful:letx L(), x 0s.t.⌧(x) = 0. Let y L() s.t. x = y y. 2 2 ⇤ Then 0 = ⌧(x)= x , = y y , = y 2 y =0 x = 0. h e e i h ⇤ e e i k e k ) e ) e 2 Deﬁne the right regular rep. ⇢ : (` ) by ⇢(k)(h)= 1 . !U hk Then ⇢(k) commutes with (g), for all g ⇢(k) L() . 2 ) 2 0 Hence x( )=x(⇢(k 1) )=⇢(k 1)(x ) = 0, for all k x = 0. k e e 2 ) ⌧ is tracial:ifg , then ⌧((g)) = (g) , = , = . 2 h e e i h g e i g,e Thus, if g, h , we have ⌧((g)(h)) = = = ⌧((h)(g)). 2 gh,e hg,e Since ⌧ is WOT-continuous ⌧ is tracial. ) Since ⌧ is normal, we are done.

Group von Neumann algebras, II

Proposition 1

⌧ : L() C given by ⌧(x)= x(e ),e is a trace. ! h i Proof. ⌧ is a state,since = 1. k e k

117 / 174 Then 0 = ⌧(x)= x , = y y , = y 2 y =0 x = 0. h e e i h ⇤ e e i k e k ) e ) e 2 Deﬁne the right regular rep. ⇢ : (` ) by ⇢(k)(h)= 1 . !U hk Then ⇢(k) commutes with (g), for all g ⇢(k) L() . 2 ) 2 0 Hence x( )=x(⇢(k 1) )=⇢(k 1)(x ) = 0, for all k x = 0. k e e 2 ) ⌧ is tracial:ifg , then ⌧((g)) = (g) , = , = . 2 h e e i h g e i g,e Thus, if g, h , we have ⌧((g)(h)) = = = ⌧((h)(g)). 2 gh,e hg,e Since ⌧ is WOT-continuous ⌧ is tracial. ) Since ⌧ is normal, we are done.

Group von Neumann algebras, II

Proposition 1

⌧ : L() C given by ⌧(x)= x(e ),e is a trace. ! h i Proof. ⌧ is a state,since = 1. k e k ⌧ is faithful:letx L(), x 0s.t.⌧(x) = 0. Let y L() s.t. x = y y. 2 2 ⇤

118 / 174 2 Deﬁne the right regular rep. ⇢ : (` ) by ⇢(k)(h)= 1 . !U hk Then ⇢(k) commutes with (g), for all g ⇢(k) L() . 2 ) 2 0 Hence x( )=x(⇢(k 1) )=⇢(k 1)(x ) = 0, for all k x = 0. k e e 2 ) ⌧ is tracial:ifg , then ⌧((g)) = (g) , = , = . 2 h e e i h g e i g,e Thus, if g, h , we have ⌧((g)(h)) = = = ⌧((h)(g)). 2 gh,e hg,e Since ⌧ is WOT-continuous ⌧ is tracial. ) Since ⌧ is normal, we are done.

Group von Neumann algebras, II

Proposition 1

119 / 174 Hence x( )=x(⇢(k 1) )=⇢(k 1)(x ) = 0, for all k x = 0. k e e 2 ) ⌧ is tracial:ifg , then ⌧((g)) = (g) , = , = . 2 h e e i h g e i g,e Thus, if g, h , we have ⌧((g)(h)) = = = ⌧((h)(g)). 2 gh,e hg,e Since ⌧ is WOT-continuous ⌧ is tracial. ) Since ⌧ is normal, we are done.

Group von Neumann algebras, II

Proposition 1

120 / 174 ⌧ is tracial:ifg , then ⌧((g)) = (g) , = , = . 2 h e e i h g e i g,e Thus, if g, h , we have ⌧((g)(h)) = = = ⌧((h)(g)). 2 gh,e hg,e Since ⌧ is WOT-continuous ⌧ is tracial. ) Since ⌧ is normal, we are done.

Group von Neumann algebras, II

Proposition 1

121 / 174 Thus, if g, h , we have ⌧((g)(h)) = = = ⌧((h)(g)). 2 gh,e hg,e Since ⌧ is WOT-continuous ⌧ is tracial. ) Since ⌧ is normal, we are done.

Group von Neumann algebras, II

Proposition 1

⌧ : L() C given by ⌧(x)= x(e ),e is a trace. ! h i Proof. ⌧ is a state,since = 1. k e k ⌧ is faithful:letx L(), x 0s.t.⌧(x) = 0. Let y L() s.t. x = y y. 2 2 ⇤ Then 0 = ⌧(x)= x , = y y , = y 2 y =0 x = 0. h e e i h ⇤ e e i k e k ) e ) e 2 Deﬁne the right regular rep. ⇢ : (` ) by ⇢(k)(h)= 1 . !U hk Then ⇢(k) commutes with (g), for all g ⇢(k) L() . 2 ) 2 0 Hence x( )=x(⇢(k 1) )=⇢(k 1)(x ) = 0, for all k x = 0. k e e 2 ) ⌧ is tracial:ifg , then ⌧((g)) = (g) , = , = . 2 h e e i h g e i g,e

122 / 174 Since ⌧ is normal, we are done.

Group von Neumann algebras, II

Proposition 1

123 / 174 Group von Neumann algebras, II

Proposition 1

124 / 174 (Note. ˆ is a compact metrizable group.) Then L() is -isomorphic to L (ˆ,µ), where µ is the Haar measure of . ⇤ 1 Sketch of proof. For g , deﬁneg ˆ C(ˆ) byg ˆ(h):=h(g). 2 2 Claim. If g e ,then h ˆ suchthath(g) = 1. 2 \{ } 9 2 6 Claim + Stone-Weierstrass span gˆ g is dense in C(ˆ) ) { | 2 } ) span gˆ g is dense in L2(ˆ,µ) gˆ g is an O.B. for L2(ˆ,µ). { | 2 } ){ | 2 } Thus, U( )=ˆg deﬁnes a unitary U : `2 L2(ˆ,µ). g ! Since U(g)U =ˆg,forg , we have C(ˆ) UL() U L (ˆ,µ). ⇤ 2 ⇢ ⇤ ⇢ 1 Since C(ˆ) is WOT-dense in L1(ˆ,µ), we get UL() U⇤ = L1(ˆ,µ)

Group von Neumann algebras, III

Proposition 2 Let be a countable abelian group. Let ˆ bethePontryagin dual of , i.e. ˆ = h : C h homomorphism such that h(g) =1, g . { ! | | | 8 2 }

125 / 174 Then L() is -isomorphic to L (ˆ,µ), where µ is the Haar measure of . ⇤ 1 Sketch of proof. For g , deﬁneg ˆ C(ˆ) byg ˆ(h):=h(g). 2 2 Claim. If g e ,then h ˆ suchthath(g) = 1. 2 \{ } 9 2 6 Claim + Stone-Weierstrass span gˆ g is dense in C(ˆ) ) { | 2 } ) span gˆ g is dense in L2(ˆ,µ) gˆ g is an O.B. for L2(ˆ,µ). { | 2 } ){ | 2 } Thus, U( )=ˆg deﬁnes a unitary U : `2 L2(ˆ,µ). g ! Since U(g)U =ˆg,forg , we have C(ˆ) UL() U L (ˆ,µ). ⇤ 2 ⇢ ⇤ ⇢ 1 Since C(ˆ) is WOT-dense in L1(ˆ,µ), we get UL() U⇤ = L1(ˆ,µ)

Group von Neumann algebras, III

Proposition 2 Let be a countable abelian group. Let ˆ bethePontryagin dual of , i.e. ˆ = h : C h homomorphism such that h(g) =1, g . { ! | | | 8 2 } (Note. ˆ is a compact metrizable group.)

126 / 174 Sketch of proof. For g , deﬁneg ˆ C(ˆ) byg ˆ(h):=h(g). 2 2 Claim. If g e ,then h ˆ suchthath(g) = 1. 2 \{ } 9 2 6 Claim + Stone-Weierstrass span gˆ g is dense in C(ˆ) ) { | 2 } ) span gˆ g is dense in L2(ˆ,µ) gˆ g is an O.B. for L2(ˆ,µ). { | 2 } ){ | 2 } Thus, U( )=ˆg deﬁnes a unitary U : `2 L2(ˆ,µ). g ! Since U(g)U =ˆg,forg , we have C(ˆ) UL() U L (ˆ,µ). ⇤ 2 ⇢ ⇤ ⇢ 1 Since C(ˆ) is WOT-dense in L1(ˆ,µ), we get UL() U⇤ = L1(ˆ,µ)

Group von Neumann algebras, III

127 / 174 Claim. If g e ,then h ˆ suchthath(g) = 1. 2 \{ } 9 2 6 Claim + Stone-Weierstrass span gˆ g is dense in C(ˆ) ) { | 2 } ) span gˆ g is dense in L2(ˆ,µ) gˆ g is an O.B. for L2(ˆ,µ). { | 2 } ){ | 2 } Thus, U( )=ˆg deﬁnes a unitary U : `2 L2(ˆ,µ). g ! Since U(g)U =ˆg,forg , we have C(ˆ) UL() U L (ˆ,µ). ⇤ 2 ⇢ ⇤ ⇢ 1 Since C(ˆ) is WOT-dense in L1(ˆ,µ), we get UL() U⇤ = L1(ˆ,µ)

Group von Neumann algebras, III

128 / 174 Claim + Stone-Weierstrass span gˆ g is dense in C(ˆ) ) { | 2 } ) span gˆ g is dense in L2(ˆ,µ) gˆ g is an O.B. for L2(ˆ,µ). { | 2 } ){ | 2 } Thus, U( )=ˆg deﬁnes a unitary U : `2 L2(ˆ,µ). g ! Since U(g)U =ˆg,forg , we have C(ˆ) UL() U L (ˆ,µ). ⇤ 2 ⇢ ⇤ ⇢ 1 Since C(ˆ) is WOT-dense in L1(ˆ,µ), we get UL() U⇤ = L1(ˆ,µ)

Group von Neumann algebras, III

129 / 174 ) span gˆ g is dense in L2(ˆ,µ) gˆ g is an O.B. for L2(ˆ,µ). { | 2 } ){ | 2 } Thus, U( )=ˆg deﬁnes a unitary U : `2 L2(ˆ,µ). g ! Since U(g)U =ˆg,forg , we have C(ˆ) UL() U L (ˆ,µ). ⇤ 2 ⇢ ⇤ ⇢ 1 Since C(ˆ) is WOT-dense in L1(ˆ,µ), we get UL() U⇤ = L1(ˆ,µ)

Group von Neumann algebras, III

130 / 174 Thus, U( )=ˆg deﬁnes a unitary U : `2 L2(ˆ,µ). g ! Since U(g)U =ˆg,forg , we have C(ˆ) UL() U L (ˆ,µ). ⇤ 2 ⇢ ⇤ ⇢ 1 Since C(ˆ) is WOT-dense in L1(ˆ,µ), we get UL() U⇤ = L1(ˆ,µ)

Group von Neumann algebras, III

Proposition 2 Let be a countable abelian group. Let ˆ bethePontryagin dual of , i.e. ˆ = h : C h homomorphism such that h(g) =1, g . { ! | | | 8 2 } (Note. ˆ is a compact metrizable group.) Then L() is -isomorphic to L (ˆ,µ), where µ is the Haar measure of . ⇤ 1 Sketch of proof. For g , deﬁneg ˆ C(ˆ) byg ˆ(h):=h(g). 2 2 Claim. If g e ,then h ˆ suchthath(g) = 1. 2 \{ } 9 2 6 Claim + Stone-Weierstrass span gˆ g is dense in C(ˆ) ) { | 2 } ) span gˆ g is dense in L2(ˆ,µ) gˆ g is an O.B. for L2(ˆ,µ). { | 2 } ){ | 2 }

131 / 174 Since U(g)U =ˆg,forg , we have C(ˆ) UL() U L (ˆ,µ). ⇤ 2 ⇢ ⇤ ⇢ 1 Since C(ˆ) is WOT-dense in L1(ˆ,µ), we get UL() U⇤ = L1(ˆ,µ)

Group von Neumann algebras, III

132 / 174 Since C(ˆ) is WOT-dense in L1(ˆ,µ), we get UL() U⇤ = L1(ˆ,µ)

Group von Neumann algebras, III

133 / 174 Group von Neumann algebras, III

134 / 174 Remark. Since x = ⌧(x x)1/2 = x and x =(x (g)) , k k2 ⇤ k e k2 g g g e we have that x = g xg (g), w.r.t. convergence in . 2. 2 k k Exercise 7. Let x, y L() and x = g xg (g), y = h yh(h)be P 2 2 2 their Fourier expansions. Show that xy = (⌃ x y 1 )(g). g h h h g P 2 2 P Proposition 3 P L() is a II factor is infinite conjugacy class (icc): 1 , hgh 1 h is infinite, for all g e . { | 2 } 2 \{ } Proof of .IfC = hgh 1 h is finite, for some g e , ) { | 2 } 2 \{ } then x = k C (k) (L()). 2 2Z Since L() is a factor x C1 x = ⌧(x)1 = 0, a contradiction. P ) 2 )

Group von Neumann algebras, IV

2 Deﬁnition. Let x L() and write xe = g xg g ` . 2 2 2 Then xg g are called the Fourier coecients of x. { } 2 P

135 / 174 Exercise 7. Let x, y L() and x = g xg (g), y = h yh(h)be 2 2 2 their Fourier expansions. Show that xy = (⌃ x y 1 )(g). g h h h g P 2 2 P Proposition 3 P L() is a II factor is infinite conjugacy class (icc): 1 , hgh 1 h is infinite, for all g e . { | 2 } 2 \{ } Proof of .IfC = hgh 1 h is finite, for some g e , ) { | 2 } 2 \{ } then x = k C (k) (L()). 2 2Z Since L() is a factor x C1 x = ⌧(x)1 = 0, a contradiction. P ) 2 )

Group von Neumann algebras, IV

136 / 174 Proposition 3 L() is a II factor is infinite conjugacy class (icc): 1 , hgh 1 h is infinite, for all g e . { | 2 } 2 \{ } Proof of .IfC = hgh 1 h is finite, for some g e , ) { | 2 } 2 \{ } then x = k C (k) (L()). 2 2Z Since L() is a factor x C1 x = ⌧(x)1 = 0, a contradiction. P ) 2 )

Group von Neumann algebras, IV

137 / 174 Since L() is a factor x C1 x = ⌧(x)1 = 0, a contradiction. ) 2 )

Group von Neumann algebras, IV

2 Definition. Let x L() and write xe = g xg g ` . 2 2 2 Then xg g are called the Fourier coecients of x. { } 2 P Remark. Since x = ⌧(x x)1/2 = x and x =(x (g)) , k k2 ⇤ k e k2 g g g e we have that x = g xg (g), w.r.t. convergence in . 2. 2 k k Exercise 7. Let x, y L() and x = g xg (g), y = h yh(h)be P 2 2 2 their Fourier expansions. Show that xy = (⌃ x y 1 )(g). g h h h g P 2 2 P Proposition 3 P L() is a II factor is infinite conjugacy class (icc): 1 , hgh 1 h is infinite, for all g e . { | 2 } 2 \{ } Proof of .IfC = hgh 1 h is finite, for some g e , ) { | 2 } 2 \{ } then x = k C (k) (L()). 2 2Z P

138 / 174 Group von Neumann algebras, IV

139 / 174 Let x = g xg (g) be the Fourier expansion of x. 2 1 If h , then x = (h)x(h )= x 1 (g) P g h gh 2 2 Uniqueness of the Fourier expansion x 1 = xg ,forallg, h . P) h gh 2 2 2 2 Since g xg = g xg g = xe < and is icc 2 | | k 2 k k k 1 we get that xg = 0, for all g e .Hencex = xe 1 C1. P P 2 \{ } 2 Exercise 8. Show that the following groups are icc: S = ⇡ permutation of N ⇡(n) = n, for ﬁnitely many n . 1 { | 6 } the free group Fn = a1,...,an on n 2 generators. h i SL3(Z); more generally, PSLn(Z), for n 2.

Group von Neumann algebras, V

Proof of . Assume is icc and let x (L()). ( 2Z

140 / 174 1 If h , then x = (h)x(h )= x 1 (g) g h gh 2 2 Uniqueness of the Fourier expansion x 1 = xg ,forallg, h . P) h gh 2 2 2 2 Since g xg = g xg g = xe < and is icc 2 | | k 2 k k k 1 we get that xg = 0, for all g e .Hencex = xe 1 C1. P P 2 \{ } 2 Exercise 8. Show that the following groups are icc: S = ⇡ permutation of N ⇡(n) = n, for ﬁnitely many n . 1 { | 6 } the free group Fn = a1,...,an on n 2 generators. h i SL3(Z); more generally, PSLn(Z), for n 2.

Group von Neumann algebras, V

Proof of . Assume is icc and let x (L()). ( 2Z Let x = g xg (g) be the Fourier expansion of x. 2 P

141 / 174 Uniqueness of the Fourier expansion x 1 = xg ,forallg, h . ) h gh 2 2 2 2 Since g xg = g xg g = xe < and is icc 2 | | k 2 k k k 1 we get that xg = 0, for all g e .Hencex = xe 1 C1. P P 2 \{ } 2 Exercise 8. Show that the following groups are icc: S = ⇡ permutation of N ⇡(n) = n, for ﬁnitely many n . 1 { | 6 } the free group Fn = a1,...,an on n 2 generators. h i SL3(Z); more generally, PSLn(Z), for n 2.

Group von Neumann algebras, V

Proof of . Assume is icc and let x (L()). ( 2Z Let x = g xg (g) be the Fourier expansion of x. 2 1 If h , then x = (h)x(h )= x 1 (g) P g h gh 2 2 P

142 / 174 2 2 2 Since g xg = g xg g = xe < and is icc 2 | | k 2 k k k 1 we get that xg = 0, for all g e .Hencex = xe 1 C1. P P 2 \{ } 2 Exercise 8. Show that the following groups are icc: S = ⇡ permutation of N ⇡(n) = n, for ﬁnitely many n . 1 { | 6 } the free group Fn = a1,...,an on n 2 generators. h i SL3(Z); more generally, PSLn(Z), for n 2.

Group von Neumann algebras, V

143 / 174 Exercise 8. Show that the following groups are icc: S = ⇡ permutation of N ⇡(n) = n, for ﬁnitely many n . 1 { | 6 } the free group Fn = a1,...,an on n 2 generators. h i SL3(Z); more generally, PSLn(Z), for n 2.

Group von Neumann algebras, V

144 / 174 the free group Fn = a1,...,an on n 2 generators. h i SL3(Z); more generally, PSLn(Z), for n 2.

Group von Neumann algebras, V

Proof of . Assume is icc and let x (L()). ( 2Z Let x = g xg (g) be the Fourier expansion of x. 2 1 If h , then x = (h)x(h )= x 1 (g) P g h gh 2 2 Uniqueness of the Fourier expansion x 1 = xg ,forallg, h . P) h gh 2 2 2 2 Since g xg = g xg g = xe < and is icc 2 | | k 2 k k k 1 we get that xg = 0, for all g e .Hencex = xe 1 C1. P P 2 \{ } 2 Exercise 8. Show that the following groups are icc: S = ⇡ permutation of N ⇡(n) = n, for ﬁnitely many n . 1 { | 6 }

145 / 174 SL3(Z); more generally, PSLn(Z), for n 2.

Group von Neumann algebras, V

146 / 174 Group von Neumann algebras, V

147 / 174 Define the Hilbert space H = L2(X ,µ) `2 ⌦ ⇠ ,⇠ = ⇠ ,⇠ , = ⇠ ⇠¯ dµ h 1 ⌦ g1 2 ⌦ g2 i h 1 2ih g1 g2 i g1,g2 1 2 ZX a -homomorphism ⇡ : L1(X ) B(H)andaunitaryrep.u : (H) ⇤ ! !U ⇡(f )(⇠ )=f ⇠ and u (⇠ )= (⇠) . ⌦ h ⌦ h g ⌦ h g ⌦ gh Fact. View L1(X ) B(H). Then ug fu⇤ = g (f ), for f L1(X ), g . ⇢ g 2 2 Let := g F ag ug F finite, ag L1(X ), g . A { 2 | ⇢ 2 8 2 } Then B(H)isa -subalgebra. A⇢P ⇤ Definition WOT L1(X ) o := B(H)iscalledthegroup measure space von A ⇢ Neumann algebra (or crossed product vNa) of y (X ,µ).

Group measure space von Neumann algebras, I

Let y (X ,µ) be a measure preserving action of a countable group on a probability space (X ,µ). For f L2(X ), g , let (f )(x)=f (g 1x). 2 2 g

148 / 174 a -homomorphism ⇡ : L1(X ) B(H)andaunitaryrep.u : (H) ⇤ ! !U ⇡(f )(⇠ )=f ⇠ and u (⇠ )= (⇠) . ⌦ h ⌦ h g ⌦ h g ⌦ gh Fact. View L1(X ) B(H). Then ug fu⇤ = g (f ), for f L1(X ), g . ⇢ g 2 2 Let := g F ag ug F ﬁnite, ag L1(X ), g . A { 2 | ⇢ 2 8 2 } Then B(H)isa -subalgebra. A⇢P ⇤ Deﬁnition WOT L1(X ) o := B(H)iscalledthegroup measure space von A ⇢ Neumann algebra (or crossed product vNa) of y (X ,µ).

Group measure space von Neumann algebras, I

149 / 174 Fact. View L1(X ) B(H). Then ug fu⇤ = g (f ), for f L1(X ), g . ⇢ g 2 2 Let := g F ag ug F ﬁnite, ag L1(X ), g . A { 2 | ⇢ 2 8 2 } Then B(H)isa -subalgebra. A⇢P ⇤ Deﬁnition WOT L1(X ) o := B(H)iscalledthegroup measure space von A ⇢ Neumann algebra (or crossed product vNa) of y (X ,µ).

Group measure space von Neumann algebras, I

150 / 174 Let := g F ag ug F ﬁnite, ag L1(X ), g . A { 2 | ⇢ 2 8 2 } Then B(H)isa -subalgebra. A⇢P ⇤ Deﬁnition WOT L1(X ) o := B(H)iscalledthegroup measure space von A ⇢ Neumann algebra (or crossed product vNa) of y (X ,µ).

Group measure space von Neumann algebras, I

151 / 174 Then B(H)isa -subalgebra. A⇢ ⇤ Deﬁnition WOT L1(X ) o := B(H)iscalledthegroup measure space von A ⇢ Neumann algebra (or crossed product vNa) of y (X ,µ).

Group measure space von Neumann algebras, I

Let y (X ,µ) be a measure preserving action of a countable group on a probability space (X ,µ). For f L2(X ), g , let (f )(x)=f (g 1x). 2 2 g Deﬁne the Hilbert space H = L2(X ,µ) `2 ⌦ ⇠ ,⇠ = ⇠ ,⇠ , = ⇠ ⇠¯ dµ h 1 ⌦ g1 2 ⌦ g2 i h 1 2ih g1 g2 i g1,g2 1 2 ZX a -homomorphism ⇡ : L1(X ) B(H)andaunitaryrep.u : (H) ⇤ ! !U ⇡(f )(⇠ )=f ⇠ and u (⇠ )= (⇠) . ⌦ h ⌦ h g ⌦ h g ⌦ gh Fact. View L1(X ) B(H). Then ug fu⇤ = g (f ), for f L1(X ), g . ⇢ g 2 2 Let := g F ag ug F ﬁnite, ag L1(X ), g . A { 2 | ⇢ 2 8 2 } P

152 / 174 Deﬁnition WOT L1(X ) o := B(H)iscalledthegroup measure space von A ⇢ Neumann algebra (or crossed product vNa) of y (X ,µ).

Group measure space von Neumann algebras, I

153 / 174 Group measure space von Neumann algebras, I

154 / 174 2 2 Deﬁnition. Let a L1(X ) o . Since a(1 e ) L (X ) ` , we can 2 ⌦ 2 2 ⌦ write a(1 e )= g ag g ,whereag L (X ). ⌦ 2 ⌦ 2 Remark.Wehavea = g ag ug ,w.r.t. . 2-convergence. P 2 k k Exercise 9. If a = g ag ug , b = g bg ug L1(X ) o , then 2P 2 2 ab = ( a (b 1 )). g h h h h g 2 2 P P DeﬁnitionP P A p.m.p. (probability measure preserving) action y (X ,µ)is free if x X gx = x is a µ-null set, for all g e . { 2 | } 2 \{ } ergodic if any -invariant mess. set A X has µ(A) 0, 1 . ⇢ 2{ }

Group measure space von Neumann algebras, II

Proposition 4

⌧ : L1(X ) o C given by ⌧(x)= x(1 e ), 1 e is a trace. ! h ⌦ ⌦ i

155 / 174 Remark.Wehavea = g ag ug ,w.r.t. . 2-convergence. 2 k k Exercise 9. If a = g ag ug , b = g bg ug L1(X ) o , then 2P 2 2 ab = ( a (b 1 )). g h h h h g 2 2 P P DeﬁnitionP P A p.m.p. (probability measure preserving) action y (X ,µ)is free if x X gx = x is a µ-null set, for all g e . { 2 | } 2 \{ } ergodic if any -invariant mess. set A X has µ(A) 0, 1 . ⇢ 2{ }

Group measure space von Neumann algebras, II

Proposition 4

⌧ : L1(X ) o C given by ⌧(x)= x(1 e ), 1 e is a trace. ! h ⌦ ⌦ i 2 2 Deﬁnition. Let a L1(X ) o . Since a(1 e ) L (X ) ` , we can 2 ⌦ 2 2 ⌦ write a(1 e )= g ag g ,whereag L (X ). ⌦ 2 ⌦ 2 P

156 / 174 Exercise 9. If a = g ag ug , b = g bg ug L1(X ) o , then 2 2 2 ab = ( a (b 1 )). g h h h h g 2 2 P P DeﬁnitionP P A p.m.p. (probability measure preserving) action y (X ,µ)is free if x X gx = x is a µ-null set, for all g e . { 2 | } 2 \{ } ergodic if any -invariant mess. set A X has µ(A) 0, 1 . ⇢ 2{ }

Group measure space von Neumann algebras, II

Proposition 4

157 / 174 Deﬁnition A p.m.p. (probability measure preserving) action y (X ,µ)is free if x X gx = x is a µ-null set, for all g e . { 2 | } 2 \{ } ergodic if any -invariant mess. set A X has µ(A) 0, 1 . ⇢ 2{ }

Group measure space von Neumann algebras, II

Proposition 4

158 / 174 Group measure space von Neumann algebras, II

Proposition 4

⌧ : L1(X ) o C given by ⌧(x)= x(1 e ), 1 e is a trace. ! h ⌦ ⌦ i 2 2 Deﬁnition. Let a L1(X ) o . Since a(1 e ) L (X ) ` , we can 2 ⌦ 2 2 ⌦ write a(1 e )= g ag g ,whereag L (X ). ⌦ 2 ⌦ 2 Remark.Wehavea = g ag ug ,w.r.t. . 2-convergence. P 2 k k Exercise 9. If a = g ag ug , b = g bg ug L1(X ) o , then 2P 2 2 ab = ( a (b 1 )). g h h h h g 2 2 P P DeﬁnitionP P A p.m.p. (probability measure preserving) action y (X ,µ)is free if x X gx = x is a µ-null set, for all g e . { 2 | } 2 \{ } ergodic if any -invariant mess. set A X has µ(A) 0, 1 . ⇢ 2{ }

159 / 174 Claim 1. a L (X ). 2 1 If b L1(X ), then g bag ug = ba = ab = g ag g (b)ug . 2 2 2 Identifying coecients ag b = ag g (b). Let Yg = x X ag (x) =0 . P ) P { 2 | 6 } For all b L (X )wehaveb(x)=b(g 1x), for almost every x Y . 2 1 2 g The action is free µ(Y )=0hencea = 0, for all g e . ) g g 2 \{ } Thus, a = a 1 L (X ). e 2 1 Claim 2. a is a constant function, i.e. a C1. 2 For all g , (a)=u au = a,hencea(g 1x)=a(x), a.e. x X . 2 g g g⇤ 2 If a has two essential values v = v ,put = v v /2. 1 6 2 | 1 2| Then X = x f (x) v < , i =1, 2, is non-negligible and -invariant. i { || i | } Since X X = , this contradicts that the action is ergodic. 1 \ 2 ;

Group measure space von Neumann algebras, III