Norms of idempotent Schur multipliers Rupert Levene University College Dublin Banach Algebras and Applications 29 July 2013 Rupert Levene (Dublin) Norms of idempotent Schur multipliers 1 / 14
Outline Schur multipliers 1 Some norm calculations 2 Gaps in the set of norms 3 Rupert Levene (Dublin) Norms of idempotent Schur multipliers 2 / 14
Schur multipliers The Schur product of A , B ∈ M m × n is A • B = [ a ij b ij ] (a.k.a. the Hadamard product) The Schur multiplier corresponding to B is the linear map S B : M m × n → M m × n , S B ( A ) = A • B . We also consider “ m = n = ∞ ”: change M m × n to B ( ℓ 2 ) and take infinite matrices B that give bounded maps S B . These form a commutative semisimple Banach algebra. Rupert Levene (Dublin) Norms of idempotent Schur multipliers 1. Schur multipliers 3 / 14
Idempotent Schur multipliers In this talk B will be a matrix of 0s and 1s. Then B • B = B = ⇒ S B ◦ S B = S B , so S B is idempotent Motivating question What are the possible values of � S B : M m × n → M m × n � ? Trivially, 0 and 1 are possible values, but nothing in between: B � ≤ � S B � 2 = � S B � = � S 2 ⇒ � S B � ∈ { 0 } ∪ [ 1 , ∞ ) . We can have � S B � > 1: Example � √ � 1 1 � � 4 1 2 1 � B = has � S B � ≥ � U • B � = for U = . √ √ 0 1 3 3 − 1 2 C = B ⊗ k has � S C � = � S B � k → ∞ as k → ∞ Rupert Levene (Dublin) Norms of idempotent Schur multipliers 1. Schur multipliers 4 / 14
� 4 Showing that � S � 1 1 � � = 3 0 1 Theorem (Grothendieck) ⇒ ∃ v i , w j ∈ ball ( ℓ 2 ): B = [ � v i , w j � ] � S B � ≤ 1 ⇐ v2 w2 � 1 1 � 3 � [ � v i , w j � ] = 4 0 1 v1 � � 4 4 So � S � 1 1 � � = 3 � S � � 1 1 � � ≤ 3 . 3 w1 0 1 0 1 4 � Have unitary U with � S � 1 1 4 � ( U ) � = 3 , so we have equality. 0 1 v2 � w 3 w2 4 In fact, � S � 1 1 0 � � = 3 too: 0 1 1 v1 w1 Rupert Levene (Dublin) Norms of idempotent Schur multipliers 2. Some norm calculations 5 / 14
“Diagonal + superdiagonal” idempotents 1 1 1 1 1 1 1 1 ... ... ... ... Let B = ∈ M n × n and C = ∈ M n × ( n + 1 ) . 1 1 1 1 1 1 1 Theorem (L., 2012) 2 π � S B � = � S C � = n + 1 cot 2 ( n + 1 ) . Question Given a matrix Y , which submatrices X have � S X � = � S Y � ? Example [Davidson–Donsig 2007] 1 1 1 1 ... ... 2 For n odd, D = ∈ M ( n + 1 ) × ( n + 1 ) has � S D � = n + 1 cot 2 ( n + 1 ) . π 1 1 1 1 Note that B and C are both submatrices. Rupert Levene (Dublin) Norms of idempotent Schur multipliers 2. Some norm calculations 6 / 14
Livshits’ two gaps theorem Theorem (Livshits, 1995) �� � 4 For any 0 – 1 matrix B, we have � S B � ∈ { 0 , 1 } ∪ 3 , ∞ . We can say more: there are at least six gaps. Rupert Levene (Dublin) Norms of idempotent Schur multipliers 3. Gaps in the set of norms 7 / 14
Bipartite graphs { 0–1 matrices in M m × n } ← → { (m,n) bipartite graphs } rows and columns ← → vertices entries equal to 1 ← → edges Examples � 1 1 � B = ← → G B = 0 1 � 1 1 0 0 0 � C = ← → G C = 0 1 1 1 0 0 0 0 1 1 � B 0 � B ⊕ C = ← → G B ⊕ C = 0 C Rupert Levene (Dublin) Norms of idempotent Schur multipliers 3. Gaps in the set of norms 8 / 14
Dictionary of operations norm � S B � 0 – 1 matrix B bipartite graph G B shuffle rows bipartite graph equal or columns isomorphism duplicate rows duplicate vertices equal or columns and their edges submatrix induced subgraph decreases direct sum disjoint union max Rupert Levene (Dublin) Norms of idempotent Schur multipliers 3. Gaps in the set of norms 9 / 14
A proof of Livshits’ theorem in this language Theorem (Livshits, 1995) �� � 4 For any 0 – 1 matrix B, we have � S B � ∈ { 0 , 1 } ∪ 3 , ∞ . Proof. Let G = G B . WLOG: G is connected. If G is complete bipartite then � S B � ∈ { 0 , 1 } . Otherwise, take vertices c and r so that: ◮ c and r are in different parts of the bipartition; ◮ ( r , c ) is not an edge of G ; and ◮ the distance from c to r is as small as possible A minimal path joining c to r starts with the subgraph ⇒ c and r are the ends of this path Minimality = ( c , r ) not an edge of G = ⇒ is an induced subgraph of G � 4 = ⇒ � S B � ≥ 3 . Rupert Levene (Dublin) Norms of idempotent Schur multipliers 3. Gaps in the set of norms 10 / 14
Small idempotent Schur multipliers Define η k , E k , F k for 1 ≤ k ≤ 6 as follows: k 0 1 2 3 4 5 6 √ √ √ � � � � 4 1 + 2 1 3 2 η k 0 1 169 + 38 19 5 + 2 5 3 2 15 2 5 E k F k Theorem (L., 2012) If G = G B is a connected, duplicate-free bipartite graph and 1 ≤ k ≤ 6 , then the following are equivalent: � S B � = η k 1 η k − 1 < � S B � ≤ η k 2 E k ≤ G ≤ F k 3 Rupert Levene (Dublin) Norms of idempotent Schur multipliers 3. Gaps in the set of norms 11 / 14
Small idempotent Schur multipliers Define η k , E k , F k for 1 ≤ k ≤ 6 as follows: k 0 1 2 3 4 5 6 √ √ √ � � � � 4 1 + 2 1 3 2 η k 0 1 169 + 38 19 5 + 2 5 3 2 15 2 5 E k F k Corollary If G = G B is any bipartite graph and 1 ≤ k ≤ 6 , then the following are equivalent: � S B � = η k 1 η k − 1 < � S B � ≤ η k 2 (i) Some component H of df ( G ) has E k ≤ H ≤ F k , and 3 (ii) Every component H of df ( G ) has E j ≤ H ≤ F j for some j ≤ k. Rupert Levene (Dublin) Norms of idempotent Schur multipliers 3. Gaps in the set of norms 11 / 14
Six gaps Corollary If S B is any idempotent Schur multiplier, then � S B � ∈ { η 0 , η 1 , η 2 , η 3 , η 4 , η 5 } ∪ [ η 6 , ∞ ) . Using tools of Katavolos–Paulsen (2005), this generalises: Theorem The same is true if we replace S B with any idempotent normal masa-bimodule map S : B ( H ) → B ( H ) where H is a separable Hilbert space. Rupert Levene (Dublin) Norms of idempotent Schur multipliers 3. Gaps in the set of norms 12 / 14
Some natural questions N = {� S B � : B ∈ { 0 , 1 } m × n , m , n ∈ N ∪ {∞} } contains left 2 accumulation points, such as 4 /π = lim n →∞ n + 1 cot ( 2 ( n + 1 ) ) . π Is 4 /π the smallest accumulation point in N ? Is N countable? Does N contain an open interval? Which graphs give Schur idempotents of the same norm? Find a combinatorial characterisation of the idempotent Schur multipliers on B ( ℓ 2 ) . Rupert Levene (Dublin) Norms of idempotent Schur multipliers 3. Gaps in the set of norms 13 / 14
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