14

I recently noticed (while playing around) that the product of a Laver matrix with a Hadamard matrix gives a very sparse matrix. In particular, all but logarithmically few rows are all zero. The nonzero rows all seem to occur near powers of 2, or multiples of large powers of 2. For instance, multiplying the 256x256 Laver table by the 256x256 Hadamard matrix, the only nonzero rows are

{1, 2, 4, 8, 9, 16, 32, 33, 64, 65, 96, 97, 128, 129, 160, 161, 192, 193, 224, 225, 256}

Is this a known phenomenon? Is there an explanation for it, other than the high periodicity of the Laver table?

In particular since all current proofs of the unbounded growth of the period of the first row of a Laver table rely on strong cardinal axioms, I wonder if understanding this phenomenon could help us understand it.

Pedro Sánchez Terraf
  • 954
Alex Meiburg
  • 1,193
  • 3
    Can you provide a definition (or a link) of what you mean by Laver table and Hadamard matrix? – Ilya Bogdanov Sep 11 '16 at 15:21
  • 1
    Edited, please check if the links point where they should. – Pedro Sánchez Terraf Sep 11 '16 at 19:42
  • 2
    Laver tables are broken by row and column permutations whereas Hadamard matrices are not. Are you asserting this for an arbitrary Hadamard matrix, or just those in a standard format? Gerhard "Permutations Affect Sparsity Of Products" Paseman, 2016.09.11. – Gerhard Paseman Sep 11 '16 at 20:07
  • Just to add to what Gerhard said: You may also multiply columns (and rows, but this does not affect anything) by $\pm 1$; also, I'm not sure whether a Hadamard matrix of order $2^n$ is unique even up to these transformations. – Ilya Bogdanov Sep 11 '16 at 21:32
  • For small orders the equivalence classes are not many. Once you get past order 24, there are a lot of such classes of inequivalent Hadamard matrices. Gerhard "Check Out Hadamard Matrix Websites" Paseman, 2016.09.11. – Gerhard Paseman Sep 11 '16 at 22:12
  • Sorry, yes. The Laver table link is correct. For the Hadamard matrix I meant form described on that page as the "Sylvester construction". – Alex Meiburg Sep 11 '16 at 23:31
  • I thought so. I encourage you to edit your post to specify that you are using Sylvester's matrix. I suspect that sparsity of the product is nonexistent for most other Hadamard matrices, and that the phenomenon you see is (as Joseph Van Name remarks) due to the periodicity more than due to it being a Laver table. Gerhard "Is Identity A Sparse Concept?" Paseman, 2016.09.11. – Gerhard Paseman Sep 12 '16 at 00:12
  • The unbounded growth of the periods in the Laver tables implies that the Laver tables behave wildly and that many patterns that appear to exist in the classical Laver tables must eventually end. The unbounded growth in the periods of the Laver tables implies that every white point at the image http://mathoverflow.net/a/257366/22277 has infinitely complex behavior if you zoom in far enough. – Joseph Van Name Jan 16 '17 at 17:53

1 Answers1

2

The sparsity of the product of a Laver table and a Hadamard matrix only follows from the periodicity of the Laver tables since the non-distributive but still periodic fake Laver tables (see this question for a more general construction) still exhibit this phenomenon. I have used computer calculations to test this phenomenon with the fake Laver tables. My computer calculations indicate that the product with a fake Laver table matrix and a Hadamard matrix is just as sparse as the product of a real classical Laver table and a Hadamard matrix.

Community
  • 1
Joseph Van Name
  • 27,791
  • 1
    I'm guessing you are using Sylvester's construction S for a Hadamard matrix . What happens if you use PS instead of S for P a random or clever choice of signed permutation matrix? Gerhard "Irregularity May Lead To Density" Paseman, 2016.09.11. – Gerhard Paseman Sep 11 '16 at 23:51
  • 1
    Yes. I have used the Sylvester construction. When I use PS instead of S, the phenomenon does not hold for either the real or fake Laver tables. – Joseph Van Name Sep 12 '16 at 20:04