Laver table: Difference between revisions

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There is no known [[closed-form expression]] to calculate the entries of a Laver table directly,<ref>{{citation|contribution=Laver Tables: from Set Theory to Braid Theory|title=Annual Topology Symposium, Tohoku University, Japan|year=2014|first=Victoria|last=Lebed|url=http://www.maths.tcd.ie/~lebed/Lebed_ATS14_beamer.pdf}}. See slide 8/33.</ref> but Patrick Dehornoy provides a simple [[algorithm]] for filling out Laver tables.<ref name="Dehornoy">Dehornoy, Patrick. [https://dehornoy.lmno.cnrs.fr/Talks/Dyz.pdf Laver Tables] (starting on slide 26). Retrieved 2025-05-06.</ref>
There is no known [[closed-form expression]] to calculate the entries of a Laver table directly,<ref>{{citation|contribution=Laver Tables: from Set Theory to Braid Theory|title=Annual Topology Symposium, Tohoku University, Japan|year=2014|first=Victoria|last=Lebed|url=http://www.maths.tcd.ie/~lebed/Lebed_ATS14_beamer.pdf}}. See slide 8/33.</ref> but [[Patrick Dehornoy]] provides a simple [[algorithm]] for filling out Laver tables.<ref name="Dehornoy">Dehornoy, Patrick. [https://dehornoy.lmno.cnrs.fr/Talks/Dyz.pdf Laver Tables] (starting on slide 26). Retrieved 2025-05-06.</ref>


== Properties ==
== Properties ==


# For all ''p'', ''q'' in {1,...,2<sup>''n''</sup>}: <math>\ \ 2^n \star_n q = q;\ \ p \star_n 2^n = 2^n;\ \ (2^n-1)\star_n q = 2^n;\ \ p\star_n 2^{n-1}=2^n\text{ if }p\ne 2^n</math>.
# For all ''p'', ''q'' in {1,...,2<sup>''n''</sup>}: <math>\ \ 2^n \star_n q = q;\ \ p \star_n 2^n = 2^n;\ \ (2^n-1)\star_n q = 2^n;\ \ p\star_n 2^{n-1}=2^n\text{ if }p\ne 2^n</math>.
# For all ''p'' in {1,...,2<sup>''n''</sup>}: <math>\ \ (p \star_n q)_{q=1,2,3,...}</math> is periodic with period π<sub>n</sub>(p) equal to a power of two.
# For all ''p'' in {1,...,2<sup>''n''</sup>}: <math>\ \ (p \star_n q)_{q=1,2,3,...}</math> is periodic with period π<sub>n</sub>(p) equal to a [[power of two]].
# For all ''p'' in {1,...,2<sup>''n''</sup>}: <math>\ \ (p \star_n q)_{q=1,2,3,...,\pi_n(p)}</math> is strictly increasing from <math>p \star_n 1 = p+1\ </math> to <math>\ p \star_n \pi_n(p) = 2^n</math>.
# For all ''p'' in {1,...,2<sup>''n''</sup>}: <math>\ \ (p \star_n q)_{q=1,2,3,...,\pi_n(p)}</math> is strictly increasing from <math>p \star_n 1 = p+1\ </math> to <math>\ p \star_n \pi_n(p) = 2^n</math>.
# For all ''p'',''q'': <math>\ p \star_n q = (p+1)^{(q)}, \text{ where } x^{(1)}=x,\ x^{(k+1)}=x^{(k)} \star_n x.</math><ref name="Biane19" />
# For all ''p'',''q'': <math>\ p \star_n q = (p+1)^{(q)}, \text{ where } x^{(1)}=x,\ x^{(k+1)}=x^{(k)} \star_n x.</math><ref name="Biane19" />


== Are the first-row periods unbounded? ==
== Are the first-row periods unbounded? ==
{{unsolved|mathematics|Is ZFC set theory able to prove that the periods of the first rows of Laver tables are unbounded?}}


Looking at just the first row in the ''n''-th Laver table, for ''n'' = 0, 1, 2,&nbsp;..., the entries in each first row are seen to be periodic with a period that's always a power of two, as mentioned in Property&nbsp;2 above. The first few periods are 1, 1, 2, 4, 4, 8, 8, 8, 8, 16, 16,&nbsp;... {{OEIS|A098820}}. This sequence is nondecreasing, and in 1995 Richard Laver [[mathematical proof|proved]], ''under the assumption that there exists a [[rank-into-rank]] (a [[large cardinal]] property)'', that it actually increases without bound. (It is not known whether this is also provable in [[ZFC]] without the additional large-cardinal axiom.)<ref>{{citation
Looking at just the first row in the ''n''-th Laver table, for ''n'' = 0, 1, 2,&nbsp;..., the entries in each first row are seen to be periodic with a period that's always a power of two, as mentioned in Property&nbsp;2 above. The first few periods are 1, 1, 2, 4, 4, 8, 8, 8, 8, 16, 16,&nbsp;... {{OEIS|A098820}}. This sequence is nondecreasing, and in 1995 Richard Laver [[mathematical proof|proved]], ''under the assumption that there exists a [[rank-into-rank]] (a [[large cardinal]] property)'', that it actually increases without bound. (It is not known whether this is also provable in [[ZFC]] without the additional large-cardinal axiom.)<ref>{{citation
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  | hdl = 10338.dmlcz/127328
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  }}.</ref> In any case, it grows extremely slowly; Randall Dougherty showed that 32 cannot appear in this sequence (if it ever does) until ''n'' > A(9,&nbsp;A(8,&nbsp;A(8,&nbsp;254))), where A denotes the [[Ackermann function|Ackermann–Péter function]].<ref>{{citation
  }}.</ref> In any case, it grows extremely slowly; [[Randall Dougherty]] showed that 32 cannot appear in this sequence (if it ever does) until ''n'' > A(9,&nbsp;A(8,&nbsp;A(8,&nbsp;254))), where A denotes the [[Ackermann function|Ackermann–Péter function]].<ref>{{citation
  | last = Dougherty | first = Randall |author-link=Randall Dougherty  
  | last = Dougherty | first = Randall |author-link=Randall Dougherty  
  | arxiv = math.LO/9205202
  | arxiv = math.LO/9205202

Latest revision as of 23:09, 24 November 2025

Template:Short description In mathematics, Laver tables (named after Richard Laver, who discovered them towards the end of the 1980s in connection with his works on set theory) are tables of numbers that have certain properties of algebraic and combinatorial interest. They occur in the study of racks and quandles.

Definition

For any nonnegative integer n, the n-th Laver table is the 2n × 2n table whose entry in the cell at row p and column q (1 ≤ p,q ≤ 2n) is defined as[1]

Ln(p,q):=pnq

where n is the unique binary operation on {1,...,2n} that satisfies the following two equations for all p, q:

Template:NumBlk

and

Template:NumBlk

Note: Equation (1) uses the notation xmod2n to mean the unique member of {1,...,2n} congruent to x modulo 2n.

Equation (2) is known as the (left) self-distributive law, and a set endowed with any binary operation satisfying this law is called a shelf. Thus, the n-th Laver table is just the multiplication table for the unique shelf ({1,...,2n}, n) that satisfies Equation (1).

Examples: Following are the first five Laver tables,[2] i.e. the multiplication tables for the shelves ({1,...,2n}, n), n = 0, 1, 2, 3, 4:

0 1
1 1
1 1 2
1 2 2
2 1 2
2 1 2 3 4
1 2 4 2 4
2 3 4 3 4
3 4 4 4 4
4 1 2 3 4
3 1 2 3 4 5 6 7 8
1 2 4 6 8 2 4 6 8
2 3 4 7 8 3 4 7 8
3 4 8 4 8 4 8 4 8
4 5 6 7 8 5 6 7 8
5 6 8 6 8 6 8 6 8
6 7 8 7 8 7 8 7 8
7 8 8 8 8 8 8 8 8
8 1 2 3 4 5 6 7 8
4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 12 14 16 2 12 14 16 2 12 14 16 2 12 14 16
2 3 12 15 16 3 12 15 16 3 12 15 16 3 12 15 16
3 4 8 12 16 4 8 12 16 4 8 12 16 4 8 12 16
4 5 6 7 8 13 14 15 16 5 6 7 8 13 14 15 16
5 6 8 14 16 6 8 14 16 6 8 14 16 6 8 14 16
6 7 8 15 16 7 8 15 16 7 8 15 16 7 8 15 16
7 8 16 8 16 8 16 8 16 8 16 8 16 8 16 8 16
8 9 10 11 12 13 14 15 16 9 10 11 12 13 14 15 16
9 10 12 14 16 10 12 14 16 10 12 14 16 10 12 14 16
10 11 12 15 16 11 12 15 16 11 12 15 16 11 12 15 16
11 12 16 12 16 12 16 12 16 12 16 12 16 12 16 12 16
12 13 14 15 16 13 14 15 16 13 14 15 16 13 14 15 16
13 14 16 14 16 14 16 14 16 14 16 14 16 14 16 14 16
14 15 16 15 16 15 16 15 16 15 16 15 16 15 16 15 16
15 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16
16 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

There is no known closed-form expression to calculate the entries of a Laver table directly,[3] but Patrick Dehornoy provides a simple algorithm for filling out Laver tables.[4]

Properties

  1. For all p, q in {1,...,2n}:   2nnq=q;  pn2n=2n;  (2n1)nq=2n;  pn2n1=2n if p2n.
  2. For all p in {1,...,2n}:   (pnq)q=1,2,3,... is periodic with period πn(p) equal to a power of two.
  3. For all p in {1,...,2n}:   (pnq)q=1,2,3,...,πn(p) is strictly increasing from pn1=p+1  to  pnπn(p)=2n.
  4. For all p,q:  pnq=(p+1)(q), where x(1)=x, x(k+1)=x(k)nx.[1]

Are the first-row periods unbounded?

<templatestyles src="Unsolved/styles.css" />

Unsolved problem in mathematics
Is ZFC set theory able to prove that the periods of the first rows of Laver tables are unbounded?

Looking at just the first row in the n-th Laver table, for n = 0, 1, 2, ..., the entries in each first row are seen to be periodic with a period that's always a power of two, as mentioned in Property 2 above. The first few periods are 1, 1, 2, 4, 4, 8, 8, 8, 8, 16, 16, ... (sequence A098820 in the OEIS). This sequence is nondecreasing, and in 1995 Richard Laver proved, under the assumption that there exists a rank-into-rank (a large cardinal property), that it actually increases without bound. (It is not known whether this is also provable in ZFC without the additional large-cardinal axiom.)[5] In any case, it grows extremely slowly; Randall Dougherty showed that 32 cannot appear in this sequence (if it ever does) until n > A(9, A(8, A(8, 254))), where A denotes the Ackermann–Péter function.[6]

References

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  4. Dehornoy, Patrick. Laver Tables (starting on slide 26). Retrieved 2025-05-06.
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Further reading