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In the following, we define infinite Hadamard matrices.

Let $\omega$ be the set of non-negative integers. If $f,g:\omega\to\{-1,1\}$ are maps, then we say $f,g$ are approximately orthogonal if $$\liminf_{n\to\infty}\sum_{k=0}^n \big(f(k)\cdot g(k)\big)=0.$$

An infinite Hadamard matrix is a map $M:\omega^2 \to \{-1,1\}$ such that whenever $i\neq j \in\omega$ then the "row vectors" $M[i,\cdot]$ and $M[j,\cdot]$ are approximately orthogonal. (Given $a\in\omega$, the map $M[a,\cdot]:\omega\to\{-1,1\}$ is defined by $n\mapsto M(a, n)$.)

Question. What is an example of an infinite Hadamard matrix?

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    $\begingroup$ Saying that lim.sup equals lim.inf is the same as saying that lim exists with that value. So your condition just says that $f(k)\cdot g(k)$ is summable with sum $0$. Which can't happen because $f(k)\cdot g(k)$ does not tend to $0$. What did you really mean to write? $\endgroup$
    – Gro-Tsen
    Commented Aug 29 at 10:26
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    $\begingroup$ LaTeX tip: \limsup and \liminf exist. $\endgroup$
    – Wojowu
    Commented Aug 29 at 11:43
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    $\begingroup$ Right, I have to remove $\lim\sup$. Thanks @Gro-Tsen $\endgroup$ Commented Aug 29 at 17:09

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One example would be the universal orthogonal array from the theory of factorial experimental design. The matrix is presented below, followed by its mathematical description.

$\begin{array}{cccccccc} —&{i=0}&1&2&3&4&5&...\\ j=0&1&1&1&1&1&1&...\\ j=1&1&-1&1&-1&1&-1&...\\ j=2&1&1&-1&-1&1&1&...\\ j=3&1&-1&-1&1&1&-1&...\\ j=4&1&1&1&1&-1&-1&...\\ j=5&1&-1&1&-1&-1&1&...\\ ...&...&...&...&...&...&...&...\end{array}$

Let the column number $i$ and row number $j$ be defined as indicated above. Express each as its binary representation according to

$i=\sum\limits_{k=0}^\infty b_{ik}2^k$

$j=\sum\limits_{k=0}^\infty b_{jk}2^k$

$b_{ik},b_{jk}\in\{0,1\}$

(Sums are written to infinity, but for any finite coordinates only finitely many terms are nonzero.)

Then the entry in column $i$, riw $j$ is given by

$\exp_{-1}[\sum\limits_{k=0}^\infty (b_{ik}b_{jk})]$

where $\exp_{-1}$ is use to mean exponential with base $-1$, this nomenclature being used to simplify subscripting. This matrix is exactly orthogonal if it is truncated to $2^n$ rows and $2^n$ columns for any whole number $n$, the resulting truncated arrays then representing the treatments for different factors in a two-level factorial experimental design matrix.

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  • $\begingroup$ Thanks a lot Oscar! Actually it occurred to me that the most natural definition for approximately orthogonal would be: there is a global constant $C_0\in\omega$ such that $\big|\sum_{k=0}^n \big(f(n)g(n)\big)\big| < C_0$ for all $n\in\omega$. I think your example works with this as well $\endgroup$ Commented Aug 29 at 19:23
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    $\begingroup$ It would not work for that more restrictive rule, nor could any IMO. You are pigeonholed into having indefinitely long parts of rows or columns all $1$ or all $-1$. But the fraction of such row or column pairs diminishes for larger truncations of this matrix. $\endgroup$ Commented Aug 29 at 19:26
  • $\begingroup$ For example, if we compute the "inner product" of columns $1,2$ we get the following partial sums: $1,0,-1,0,... $ (and then it repeats). Thus the $\liminf$ would be $-1$. $\endgroup$ Commented Aug 30 at 6:12
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    $\begingroup$ You nay wnt to pur absolute values aroubd the lim inf argument. $\endgroup$ Commented Aug 30 at 9:35
  • $\begingroup$ @OscarLanzi I am not sure your argument works - see the comment here : mathoverflow.net/questions/477848/… $\endgroup$ Commented Aug 30 at 11:17

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