Four open problems in frame theory Dustin G. Mixon Frames and Algebraic & Combinatorial Geometry July 27 – 31, 2015 1/34
Part I The Paulsen Problem 1/34
A quick recap of unit norm tight frames Definition of M × N UNTF: � (i) Rows have norm N / M (ii) Rows are orthogonal (iii) Columns have norm 1 UNTFs form an algebraic variety of known dimension Singular points of the variety are the orthodecomposable frames UNTF variety is a manifold when M and N are relatively prime Strawn, J. Fourier Anal. Appl., 2011 2/34
A quick recap of unit norm tight frames Topology of the real 2 × 4 UNTFs modulo rotation: Dykema, Strawn, Internat. J. Pure Appl. Math., 2006 3/34
A quick recap of unit norm tight frames To find a UNTF, pick a matrix and optimize for UNTF-ness When can we be sure that a UNTF is close by? The Paulsen Problem If Φ is close to being a UNTF, that is, ΦΦ ∗ ≈ N diag(Φ ∗ Φ) ≈ 1 , M I , how far is the closest UNTF? (One might pose the analogous question for ETFs...) Bodmann, Casazza, J. Funct. Anal., 2010 4/34
Progress on the Paulsen problem Easy solution: Apply the � Lojasiewicz inequality to the function N 2 � 2 � � � ΦΦ ∗ − N � � ϕ i � 2 − 1 � f (Φ) = F + M I � � � i =1 Observe UNTF = { Φ : f (Φ) = 0 } For every ǫ > 0, there exist α = α ( ǫ ) and C = C ( ǫ ) such that dist(Φ , UNTF) α ≤ C · f (Φ) f (Φ) ≤ ǫ ⇒ We want the smallest possible α in terms of M and N Fernando, Gamboa, Math. Res. Lett., 2010 5/34
Progress on the Paulsen problem Goal: Find nearby point in UNTF = TF ∩ UNF 6/34
Progress on the Paulsen problem Goal: Find nearby point in UNTF = TF ∩ UNF Method 1: 1. Project onto TF: Φ �→ ( M N ΦΦ ∗ ) − 1 / 2 Φ 2. Follow an ODE to flow through TF towards UNTF Result: α = 2 when gcd( M , N ) = 1, no estimate otherwise Bodmann, Casazza, J. Funct. Anal., 2010 7/34
Progress on the Paulsen problem Goal: Find nearby point in UNTF = TF ∩ UNF Method 2: 1. Project onto UNF: Φ �→ Φ diag(Φ ∗ Φ) − 1 / 2 2. Locally minimize the frame potential in UNF 3. “Jump” when nearly orthodecomposable Result: α = 2 when gcd( M , N ) = 1, otherwise α ≤ 2 · 7 M Casazza, Fickus, M., Appl. Comput. Harmon. Anal., 2012 8/34
Better results with new techniques? Definition of M × ( N + 1) eigensteps matrix ( λ ij ) M N j =0 : i =1 , (i) 0th column is all zeros (ii) Adjacent columns interlace: λ M , j ≤ λ M , j +1 ≤ λ M − 1 , j ≤ · · · ≤ λ 2 , j +1 ≤ λ 1 , j ≤ λ 1 , j +1 Facts: ◮ E M , N = { M × ( N + 1) eigensteps matrices } is a convex polytope ◮ Λ: C M × N → E M , N , Λ ij (Φ) = λ i (Φ j Φ ∗ j ) is onto, continuous ◮ Λ(TF), Λ(UNF), Λ(UNTF) are convex subpolytopes of E M , N Open problem: How does distance in C M × N relate to distance in E M , N ? Cahill, Fickus, M., Poteet, Strawn, Appl. Comput. Harmon. Anal., 2013 9/34
Part II The Fickus Conjecture 10/34
A quick recap of equiangular tight frames Goal: Find optimal packings of lines through the origin Solution: prove uniform bound, then achieve equality in bound � N − M 1. Welch bound: max i � = j |� ϕ i , ϕ j �| ≥ M ( N − 1) 2. Φ achieves equality in Welch bound iff Φ is an ETF Strohmer, Heath, Appl. Comput. Harmon. Anal., 2003 11/34
A quick recap of equiangular tight frames Definition of M × N ETF: � (i) Rows have norm N / M (ii) Rows are orthogonal (iii) Columns have norm 1 � N − M (iv) Columns satisfy |� ϕ i , ϕ j �| = M ( N − 1) whenever i � = j In the real case, we know a lot: ◮ Φ ∗ Φ ← → adjacency matrix of strongly regular graph ◮ ETF exists only if f ( M , N ) is integer/nonnegative for several f Open problem: Necessary/sufficient conditions for complex ETFs Sustik, Tropp, Dhillon, Heath, Linear Algebra Appl., 2007 Waldron, Linear Algebra Appl., 2009 12/34
A quick recap of equiangular tight frames After investigating all known ETFs, Matt posed a conjecture: The Fickus Conjecture Consider the three quantities: M , N − M , N − 1 . An M × N ETF exists only if one of these quantities divides the product of the other two. ◮ Prove it, then Matt owes you US$200 ◮ Disprove it, then Matt owes you US$100 Fickus, M., arXiv:1504.00253 M., dustingmixon.wordpress.com/2015/07/08/conjectures-from-sampta/ 13/34
Towards a proof of the Fickus Conjecture Our knowledge to date: A complex ETF exists only if √ M ) , M 2 � ◮ N ∈ { M , M + 1 } ∪ � M + Ω( ◮ ( M , N ) � = (3 , 8) How to prove the second condition: 1. Characterize ETFs with 667 polynomials in 12 variables 2. Take about an hour to compute a Gr¨ obner basis 3. Find 1 in the ideal generated by the Gr¨ obner basis 4. Conclude that no solutions exist What’s the next thing to try? Strohmer, Heath, Appl. Comput. Harmon. Anal., 2003 Sz¨ oll˝ osi, arXiv:1402.6429 14/34
Towards a proof of the Fickus Conjecture Proposed program to prove that no M × N ETFs exist: 1. Find a nonnegative polynomial p ∈ R [ x 1 , . . . , x 2 MN ] whose roots are the M × N ETFs 2. Show that min x p ( x ) > 0 (no roots means no ETFs) For step 1, here’s a choice for p : 2 � � | Φ ∗ Φ | 2 − W � N − M p (Φ) = F , W ii = 1 , W ij = � � M ( N − 1) � For step 2, exploit duality: min x p ( x ) = max p − ǫ ≥ 0 ǫ Unfortunately, testing for nonnegativity is NP-hard in general Parrilo, Math. Program., Ser. B, 2003 15/34
Towards a proof of the Fickus Conjecture A sum-of-squares polynomial f ∈ R [ x 1 , . . . , x n ] has the form k � g i ( x ) 2 , f ( x ) = g 1 , . . . , g k ∈ R [ x 1 , . . . , x n ] i =1 ◮ SOS is a convex subcone of nonnegative polynomials, and so p − ǫ ≥ 0 ǫ ≥ max p − ǫ ∈ SOS ǫ max ◮ Bound is often tight, e.g., p = F ◮ RHS solved in polynomial time using semidefinite programming 10 x 4 + 1 3 x 6 + xy − 4 y 2 +4 y 4 F ( x , y ) = 4 x 2 − 21 Parrilo, Math. Program., Ser. B, 2003 16/34
Towards a proof of the Fickus Conjecture Proposal: Find ǫ > 0 such that p − ǫ is SOS � | Φ ∗ Φ | 2 − W � 2 � � Good news: p (Φ) = F is SOS ıve SDP has O (( MN ) 16 ) matrix entries Bad news: Na¨ Ought to exploit p ’s structure: ◮ p is sparse ⇒ exponents in each g i lie in a small known set ◮ p enjoys symmetries: p ( U Φ) = p (Φ), p (Φ P ) = p (Φ) Can we recover ( M , N ) � = (3 , 8)? Is ( M , N ) � = (4 , 8) necessary? Reznick, Duke Math. J., 1978 Gatermann, Parrilo, J. Pure Appl. Algebra, 2004 17/34
Part III Zauner’s Conjecture 18/34
Maximal ETFs An M × N ETF exists only if N ≤ M 2 An M × M 2 ETF is called maximal or a SIC-POVM Maximal ETF constructions are known for each M ∈ { 1 , 2 , . . . , 17 , 19 , 24 , 28 , 35 , 48 } Numerical evidence suggests existence whenever M ≤ 67 Zauner, gerhardzauner.at/sicfiducialsd.html Chein, Ph.D. thesis, U. Auckland, 2015 Scott, Grassl, J. Math. Physics, 2010 19/34
Maximal ETFs Zauner’s Conjecture For each M ≥ 1, there exists an M × M 2 ETF (with very specific structure) . How to construct a maximal ETF: 1. Pick a function ϕ : Z / M Z → C (the fiducial vector) 2. Take the Gabor frame Φ = { T a E b ϕ } a , b ∈ Z / M Z , where ( M b ψ )( x ) = e 2 π ibx / M ψ ( x ) ( T a ψ )( x ) = ψ ( x − a ) , 3. Pray that Φ is an ETF To date, if we have an M × M 2 ETF, we have one that’s Gabor Zauner, Ph.D. thesis, U. Vienna, 1999 20/34
Recent progress on Zauner’s conjecture Chein’s program to find explicit maximal ETFs: 1. Take a numerically approximated ETF fiducial vector 2. Locally optimize to obtain many (say, 2000) digits of precision 3. Apply field structure conjectures to guess analytic expression 4. Verify ETF properties by symbolic computation This program recently produced the first explicit 17 × 17 2 ETF Conditionally finite-time algorithm! But step 3 is slow... Chein, Ph.D. thesis, U. Auckland, 2015 21/34
Recent progress on Zauner’s conjecture Claim: Every method that reports explicit fiducial vectors is slow ◮ Explicit fiducial vectors are available on Zauner’s webpage ◮ Count characters in each fiducial vector’s description and plot: Exponential description length! We need an alternative... M., dustingmixon.wordpress.com/2015/03/10/zauners-conjecture-is-true-in-dimension-17/ 22/34
How to avoid being explicit? Shor, mathoverflow.net/questions/30894/fixed-point-theorems-and-equiangular-lines 23/34
How to avoid being explicit? Perhaps fiducial vectors are simpler in lifted space For M odd, define the discrete Wigner transform by 1 � f ( t + τ 2 ) f ( t − τ 2 ) e − 2 π i τω/ M ( Wf )( t , ω ) = √ M τ ∈ Z / M Z Useful properties: ◮ ( Wf )( t , ω ) ∈ R for every t , ω ∈ Z / M Z ◮ � Wf , Wg � = |� f , g �| 2 ◮ W ( T a E b ϕ ) = T ( a , b ) ( W ϕ ) Goal: Find F ∈ im( W ) such that translates of F are equiangular 24/34
Part IV Vinzant’s Conjecture 25/34
A quick intro to coherent diffractive imaging Disclaimer: I am not a physicist. What does the diffraction pattern say about the object? 26/34
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