Rocky Mountain Mathematics Consortium Summer School University of Wyoming, Laramie Large Subalgebras and the Structure of Crossed Products, Lecture 2: Large Subalgebras and their Basic 1–5 June 2015 Properties Lecture 1 (1 June 2015): Introduction, Motivation, and the Cuntz Semigroup. N. Christopher Phillips Lecture 2 (2 June 2015): Large Subalgebras and their Basic Properties. University of Oregon Lecture 3 (4 June 2015): Large Subalgebras and the Radius of 2 June 2015 Comparison. Lecture 4 (5 June 2015 [morning]): Large Subalgebras in Crossed Products by Z . Lecture 5 (5 June 2015 [afternoon]): Application to the Radius of Comparison of Crossed Products by Minimal Homeomorphisms. N. C. Phillips (U of Oregon) Large Subalgebras: Basics 2 June 2015 1 / 24 N. C. Phillips (U of Oregon) Large Subalgebras: Basics 2 June 2015 2 / 24 A rough outline of all five lectures Definition Let A be a C*-algebra, and let a , b ∈ ( K ⊗ A ) + . We say that a is Cuntz Introduction: what large subalgebras are good for. subequivalent to b over A , written a � A b , if there is a sequence ( v n ) ∞ Definition of a large subalgebra. n =1 in K ⊗ A such that lim n →∞ v n bv ∗ n = a . Statements of some theorems on large subalgebras. A very brief survey of the Cuntz semigroup. Open problems. Definition Basic properties of large subalgebras. Let A be an infinite dimensional simple unital C*-algebra. A unital A very brief survey of radius of comparison. subalgebra B ⊂ A is said to be large in A if for every m ∈ Z > 0 , Description of the proof that if B is a large subalgebra of A , then A a 1 , a 2 , . . . , a m ∈ A , ε > 0, x ∈ A + with � x � = 1, and y ∈ B + \ { 0 } , there and B have the same radius of comparison. are c 1 , c 2 , . . . , c m ∈ A and g ∈ B such that: A very brief survey of crossed products by Z . 1 0 ≤ g ≤ 1. Orbit breaking subalgebras of crossed products by minimal 2 For j = 1 , 2 , . . . , m we have � c j − a j � < ε . homeomorphisms. Sketch of the proof that suitable orbit breaking subalgebras are large. 3 For j = 1 , 2 , . . . , m we have (1 − g ) c j ∈ B . A very brief survey of mean dimension. 4 g � B y and g � A x . Description of the proof that for minimal homeomorphisms with Cantor factors, the radius of comparison is at most half the mean 5 � (1 − g ) x (1 − g ) � > 1 − ε . dimension. N. C. Phillips (U of Oregon) Large Subalgebras: Basics 2 June 2015 3 / 24 N. C. Phillips (U of Oregon) Large Subalgebras: Basics 2 June 2015 4 / 24
Dense subsets When A is finite B ⊂ A is large in A if for a 1 , a 2 , . . . , a m ∈ A , ε > 0, x ∈ A + with � x � = 1, and y ∈ B + \ { 0 } , there are c 1 , c 2 , . . . , c m ∈ A and g ∈ B such that: B ⊂ A is large in A if for a 1 , a 2 , . . . , a m ∈ A , ε > 0, x ∈ A + with � x � = 1, and y ∈ B + \ { 0 } , there are c 1 , c 2 , . . . , c m ∈ A and g ∈ B such that: 1 0 ≤ g ≤ 1. 1 0 ≤ g ≤ 1. 2 For j = 1 , 2 , . . . , m we have � c j − a j � < ε . 2 For j = 1 , 2 , . . . , m we have � c j − a j � < ε . 3 For j = 1 , 2 , . . . , m we have (1 − g ) c j ∈ B . 4 g � B y and g � A x . 3 For j = 1 , 2 , . . . , m we have (1 − g ) c j ∈ B . 5 � (1 − g ) x (1 − g ) � > 1 − ε . 4 g � B y and g � A x . Lemma 5 � (1 − g ) x (1 − g ) � > 1 − ε . In the definition, it suffices to let S ⊂ A be a subset whose linear span is Proposition dense in A , and verify the hypotheses only when a 1 , a 2 , . . . , a m ∈ S . Let A be a finite infinite dimensional simple unital C*-algebra, and let Unlike other approximation properties (such as tracial rank), it seems not B ⊂ A be a unital subalgebra satisfying the definition of a large subalgebra except for condition (5). Then B is large in A . to be possible to take S to be a generating subset, or even a selfadjoint generating subset. (We can do this for the definition of a centrally large subalgebra.) N. C. Phillips (U of Oregon) Large Subalgebras: Basics 2 June 2015 5 / 24 N. C. Phillips (U of Oregon) Large Subalgebras: Basics 2 June 2015 6 / 24 When A is finite (continued) When A is finite (continued) � � To show: x ∈ A + with � x � = 1, ε > 0. Then there is y ∈ xAx + \ { 0 } From the previous slide: such that whenever g ∈ A + satisfies 0 ≤ g ≤ 1 and g � A y , then Proposition � (1 − g ) x (1 − g ) � > 1 − ε . Let A be a finite infinite dimensional simple unital C*-algebra, and let Choose a sufficiently small number ε 0 > 0. Choose f : [0 , 1] → [0 , 1] such B ⊂ A be a unital subalgebra satisfying the definition of a large subalgebra that f = 0 on [0 , 1 − ε 0 ] and f (1) = 1. Construct a , b j , c j , d j ∈ f ( x ) Af ( x ) except for the condition � (1 − g ) x (1 − g ) � > 1 − ε . Then B is large in A . for j = 1 , 2 such that 0 ≤ d j ≤ c j ≤ b j ≤ a ≤ 1 , ab j = b j , b j c j = c j , c j d j = d j , and d j � = 0 , It suffices to prove: and b 1 b 2 = 0. Take x 0 = d 1 . Lemma If ε 0 is small enough, g � A d 1 , and � (1 − g ) x (1 − g ) � ≤ 1 − ε , use Let A be a finite simple infinite dimensional unital C*-algebra. Let x ∈ A + � � � = � ( b 1 + b 2 ) 1 / 2 (1 − g ) 2 ( b 1 + b 2 ) 1 / 2 � satisfy � x � = 1. Then for every ε > 0 there is x 0 ∈ xAx + \ { 0 } such � � � � (1 − g )( b 1 + b 2 )(1 − g ) � , that whenever g ∈ A + satisfies 0 ≤ g ≤ 1 and g � A x 0 , then ( b 1 + b 2 ) 1 / 2 x 1 / 2 ≈ x 1 / 2 � x 1 / 2 (1 − g ) 2 x 1 / 2 � � � (1 − g ) x (1 − g ) � = � , and � (1 − g ) x (1 − g ) � > 1 − ε . to get (details omitted) If we also require x 0 � A x , then we can use x 0 in place of x in the � > 1 − ε � � � (1 − g )( b 1 + b 2 )(1 − g ) 3 . definition. N. C. Phillips (U of Oregon) Large Subalgebras: Basics 2 June 2015 7 / 24 N. C. Phillips (U of Oregon) Large Subalgebras: Basics 2 June 2015 8 / 24
When A is finite (continued) When A is finite (continued) We assumed g � A d 1 and � (1 − g ) x (1 − g ) � ≤ 1 − ε , and we want a In search of a contradiction, we have gotten contradiction. We have c 1 + c 2 � A d 1 0 ≤ d j ≤ c j ≤ b j ≤ a ≤ 1 , ab j = b j , b j c j = c j , c j d j = d j , and d j � = 0 with for j = 1 , 2, and b 1 b 2 = 0. We also have c 1 d 1 = d 1 , c 1 c 2 = 0 , and c 2 � = 0 . � > 1 − ε � � � (1 − g )( b 1 + b 2 )(1 − g ) (1) 3 . This looks rather suspicious. Set r = (1 − c 1 − c 2 ) + d 1 . Use basic result (12) at the first step, From ( b 1 + b 2 )( c 1 + c 2 ) = c 1 + c 2 one gets, for any β ∈ [0 , 1), c 1 + c 2 � A d 1 at the second step, and basic result (13) and c 1 + c 2 � A [( b 1 + b 2 ) − β ] + . (2) d 1 (1 − c 1 − c 2 ) = 0 at the third step, to get (If we are in C ( X ), whenever ( c 1 + c 2 )( x ) � = 0, we have 1 � A (1 − c 1 − c 2 ) ⊕ ( c 1 + c 2 ) � A (1 − c 1 − c 2 ) ⊕ d 1 ∼ A (1 − c 1 − c 2 )+ d 1 = r . ( b 1 + b 2 )( x ) = 1 > β .) Take β = 1 − ε 3 . Combine (2) with the second Thus, there is v ∈ A such that � vrv ∗ − 1 � < 1 2 . It follows that vr 1 / 2 has a lemma on the list of basic results on Cuntz equivalence at the first step, (1) at the second step, and g � A d 1 at the last step, to get right inverse. Recall that c 2 d 2 = d 2 and d 2 � = 0. So rd 2 = 0, whence vr 1 / 2 d 2 = 0. Thus vr 1 / 2 is not invertible. We have contradicted finiteness � � c 1 + c 2 � A (1 − g )( b 1 + b 2 )(1 − g ) − β + ⊕ g = 0 ⊕ g � A d 1 . of A , and thus proved the lemma. N. C. Phillips (U of Oregon) Large Subalgebras: Basics 2 June 2015 9 / 24 N. C. Phillips (U of Oregon) Large Subalgebras: Basics 2 June 2015 10 / 24 Lemma Simplicity of a large subalgebra Let A be a finite infinite dimensional simple unital C*-algebra, and let Recall from Lecture 1: B ⊂ A be a large subalgebra. Let m , n ∈ Z ≥ 0 , let a 1 , a 2 , . . . , a m ∈ A , let Proposition b 1 , b 2 , . . . , b n ∈ A + , let ε > 0, let x ∈ A + satisfy � x � = 1, and let Let A be an infinite dimensional simple unital C*-algebra, and let B ⊂ A y ∈ B + \ { 0 } . Then there are c 1 , c 2 , . . . , c m ∈ A , d 1 , d 2 , . . . , d n ∈ A + , and be a large subalgebra. Then B is simple. g ∈ B such that: 1 0 ≤ g ≤ 1. (The result stated in Lecture 1 also included infinite dimensionality. Once 2 � c j − a j � < ε and � d j − b j � < ε . one has simplicity, infinite dimensionality is easy to prove, and we omit it.) 3 � c j � ≤ � a j � and � d j � ≤ � b j � . The proof of this proposition uses two preliminary lemmas. 4 (1 − g ) c j ∈ B and (1 − g ) d j (1 − g ) ∈ B . Lemma 5 g � B y and g � A x . Let A be a C*-algebra, let n ∈ Z > 0 , and let a 1 , a 2 , . . . , a n ∈ A . Set a = � n k =1 a k and x = � n k =1 a ∗ k a k . Then a ∗ a ∈ xAx . Sketch of proof. Lemma To get � c j � ≤ � a j � one takes ε > 0 to be a bit smaller, and scales down c j for any j for which � c j � is too big. To get d j , approximate b 1 / 2 Let A be a unital C*-algebra and let a ∈ A + . Suppose AaA = A . Then sufficiently j there exist n ∈ Z > 0 and x 1 , x 2 , . . . , x n ∈ A such that � n k =1 x ∗ k ax k = 1. well by r j (without increasing the norm), and take d j = r j r ∗ j . N. C. Phillips (U of Oregon) Large Subalgebras: Basics 2 June 2015 11 / 24 N. C. Phillips (U of Oregon) Large Subalgebras: Basics 2 June 2015 12 / 24
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