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We say that a sequence of algebras $(\{1,\dots,2^{n}\},*_{n})_{n\in\omega}$ is an inverse system of fake Laver tables if for $x\in\{1,\dots,2^{n}\}$, we have

  1. $2^{n}*_{n}x=x$,

  2. $x*_{n}1=x+1\mod 2^{n}$,

  3. the mapping $\phi_{n}:\{1,\dots,2^{n}\}\rightarrow\{1,\dots,2^{n-1}\}$ where $\phi_{n}(x)=x\mod 2^{n-1}$ is a homomorphism, and

  4. for each $x\leq 2^{n}$, there is some $m\leq n$ where the sequence $(x*_{n}1,\dots,x*_{n}2^{m})$ is strictly increasing with $x*_{n}2^{m}=2^{n}$ and where $x*_{n}y=x*_{n}(y+2^{m})$ whenever $1\leq y<2^{n}-2^{m}$.

If $$\mathbf{x}=(x_{n})_{n\in\omega}\in(\varprojlim_{n}((\{1,...,2^{n}\},*_{n})_{n\in\omega}$$ and $x_{n}\neq 2^{n}$ for some $n$, then define $\mathrm{crit}(\mathbf{x})$ to be $\gcd(x_{n},2^{n})$ for sufficiently large $n$.

We say that a sequence $$((x_{n,1})_{n\in\omega},\dots,(x_{n,k})_{n\in\omega})\in(\varprojlim_{n}((\{1,...,2^{n}\},*_{n})_{n\in\omega},(\phi_{n})_{n\in\omega}))^{k}$$ is admissible if whenever $(a_{1},\dots,a_{s})\in\{1,\dots,k\}$, there is some $n$ where $$x_{n,a_{1}}*_{n}\cdots*_{n}x_{n,a_{s}}=2^{n-1}$$ but where $$x_{n,a_{1}}*_{n}\cdots*_{n}x_{n,a_{r}}<2^{n-1}$$ whenever $1\leq r<s$.

If $\mathbf{x}_{1}=((x_{n,1})_{n\in\omega},\dots,\mathbf{x}_{k}=(x_{n,k})_{n\in\omega})$ are admissible, then define non-commutative polynomials (by non-commutative, I mean $x_{i}x_{j}\neq x_{j}x_{i}$ for $i\neq j$) $p_{n,\mathbf{x}_{1},\dots,\mathbf{x}_{k}}(x_{1},\dots,x_{k})$ by letting $p_{n-1,\mathbf{x}_{1},\dots,\mathbf{x}_{k}}(x_{1},\dots,x_{k})$ $$=1+\sum\{x_{a_{1}}\dots x_{a_{s}}\mid a_{1},\dots,a_{s}\in\{1,\dots,k\},$$ $$x_{n,a_{1}}*_{n}\cdots*_{n}x_{n,a_{s}}=2^{n-1}, x_{n,a_{1}}*_{n}\cdots*_{n}x_{n,a_{s}}<2^{n-1}\}.$$

Then these polynomials satisfy the infinite product formula $$p_{n,\mathbf{x}_{1},\dots,\mathbf{x}_{k}}(x_{1},\dots,x_{k})\dots p_{0,\mathbf{x}_{1},\dots,\mathbf{x}_{k}}(x_{1},\dots,x_{k})=\frac{1}{1-(x_{1}+\dots+x_{k})}.$$

What are some examples of inverse systems of fake Laver tables, matrices $A_{1},...,A_{k}$ with algebraic coefficients, along with admissible sequences $\mathbf{x}_{1},\dots,\mathbf{x}_{k}$ where

  1. $1-A_{1}+\dots+A_{k}$ is non-singular

  2. $$\lim_{n\rightarrow\infty}p_{n}(A_{1},\dots,A_{k})\dots p_{0}(A_{1},\dots,A_{k})=\frac{1}{1-(A_{1}+\dots+A_{k})},$$

  3. we have $p_{n,\mathbf{x}_{1},\dots,\mathbf{x}_{k}}(A_{1},\dots,A_{k})\neq 1$ for infinitely many $n$,

  4. if $\alpha$ is a natural number, then $$\sum\{A_{i}\mid 1\leq i\leq k,\mathrm{crit}(\mathbf{x}_{i})=\alpha\}$$ is non-singular,

  5. if $B_{n}=p_{n}(A_{1},\dots,A_{k})\dots p_{0}(A_{1},\dots,A_{k})$ for all $n$, then there is a closed form expression for the sequence $(B_{n})_{n\in\omega}$, and

  6. the operations $*_{n}$, threads $\mathbf{x}_{1},\dots,\mathbf{x}_{k}$, and matrices $B_{n}$ are computable in polynomial time?

This question is a close cousin to this question, but now with no set theory (and this question is probably easier to answer since set theory is hard to control). Feel free to let $A_{1},\dots,A_{k}$ be $1\times 1$-matrices (i.e. complex numbers) if you want everything to commute.

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    $\begingroup$ Will someone care to explain the completely unjustified downvote for this question? $\endgroup$ Commented Mar 24, 2019 at 20:51
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    $\begingroup$ To get things started, I found that if we used the same fake Laver tables as in this question mathoverflow.net/questions/221344/… and where $\mathbf{x}_{1}=(1),\mathbf{x}_{2}=(2)$, then $p_{0}(x,-x^{2})=x+1,p_{1}(x,-x^{2})=1,p_{n}(x,-x^{2})=(-x)^{3\cdot 2^{n-2}}+1$ for $n\geq 2$ (As of 3/24/2019. I have not formally proven this fact yet but computer calculations indicate the veracity of this statement, and I suspect a proof would not be very difficult). $\endgroup$ Commented Mar 25, 2019 at 0:51
  • $\begingroup$ Let us obsere that if $\mathbf{x}_{1}=(1)_{n}$, then $p_{n}(x)=1+x^{2^{n}}$ for $n\in\omega$. $\endgroup$ Commented Mar 25, 2019 at 13:44

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