Difference between revisions of "Associators with Frozen Feet"

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{{Paperlets Navigation}}
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==The Goal==
 
==The Goal==
  
 
The purpose of the [[paperlets|paperlet]] is to find an explicit formula for an associator with frozen feet. As I'm starting to write, I don't know such a formula. My hope is that as I type up all the relevant equations, a solution will emerge. I'll be just as happy if it emerges in somebody else's mind, provided (s)he shares her/his thoughts.
 
The purpose of the [[paperlets|paperlet]] is to find an explicit formula for an associator with frozen feet. As I'm starting to write, I don't know such a formula. My hope is that as I type up all the relevant equations, a solution will emerge. I'll be just as happy if it emerges in somebody else's mind, provided (s)he shares her/his thoughts.
  
A '''horizontal associator''' is a solution <math>\Phi\in A_3</math> of the pentagon equation and the hexagon equations ({{ref|Drinfeld_90}}, {{ref|Drinfeld_91}}, {{ref|Bar-Natan_97}}):
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{| align=center width=100%
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|- align=center
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|[[Image:The Pentagon For Parenthesized Braids.jpg|350px]]
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|[[Image:The Hexagons For Parenthesized Braids.jpg|500px]]
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|- align=center
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|colspan=2|'''The Pentagon and the Hexagons for Parenthesized Braids'''
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|}
  
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A '''horizontal associator''' is a solution <math>\Phi\in A_3</math> of the pentagon equation and the hexagon equations ({{ref|Drinfeld_90}}, {{ref|Drinfeld_91}}, {{ref|Bar-Natan_97}}):
  
 
{{Equation|Pentagon|<math>\Phi^{123}\cdot(1\otimes\Delta\otimes 1)(\Phi)\cdot\Phi^{234}=(\Delta\otimes 1\otimes 1)(\Phi)\cdot(1\otimes 1\otimes\Delta)(\Phi)</math>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;in&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<math>A_4</math>,}}
 
{{Equation|Pentagon|<math>\Phi^{123}\cdot(1\otimes\Delta\otimes 1)(\Phi)\cdot\Phi^{234}=(\Delta\otimes 1\otimes 1)(\Phi)\cdot(1\otimes 1\otimes\Delta)(\Phi)</math>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;in&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<math>A_4</math>,}}
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==Why Bother?==
 
==Why Bother?==
  
This isn't the place to explain the need for an explicit associator. The need is there even if it is a bit under-appreciated. Why frozen feet? Because finding an honest explicit horizontal associator seems too hard, and as a warmup and perhaps a step, it seems worthwhile to look for associators in some quotient spaces. The frozen feet quotient is the simplest quotient I'm aware off for which the answer is unknown.
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This isn't the place to explain the need for an explicit associator. The need is there even if it is a bit under-appreciated. Why frozen feet? Because finding an honest explicit horizontal associator seems too hard, and as a warm up and perhaps a step, it seems worthwhile to look for associators in some quotient spaces. The frozen feet quotient is the simplest quotient I'm aware off for which the answer is unknown.
  
 
==Why might it be doable?==
 
==Why might it be doable?==
  
Because frozen feet are really molten bodies. It is known that there exists and associtor of the form <math>\Phi=1+F(t^{13},t^{23})</math>, where <math>F</math> is some non-commutative power series with no constant term. Once the body melts (but keeping the feet frozen), this becomes
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Because frozen feet are really molten bodies. It is known that there exists and associator of the form <math>\Phi=1+F(t^{13},t^{23})</math>, where <math>F</math> is some non-commutative power series with no constant term. Once the body melts (but keeping the feet frozen), this becomes
  
{{Equation|Phi|<math>\Phi=1+t^{13}F_{13}(t^{13},t^{23})+t^{23}F_{23}(t^{13},t^{23})</math>,}}
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<center><math>\Phi=1+t^{13}F_{13}(t^{13},t^{23})+t^{23}F_{23}(t^{13},t^{23})</math>,</center>
  
 
where <math>F_{13}</math> and <math>F_{23}</math> are ''commutative'' power series (i.e., "functions"). So finding an associator with frozen feet reduces to finding two, just two, functions of two, just two, variables, satisfying some algebraic equations. How hard can that be?
 
where <math>F_{13}</math> and <math>F_{23}</math> are ''commutative'' power series (i.e., "functions"). So finding an associator with frozen feet reduces to finding two, just two, functions of two, just two, variables, satisfying some algebraic equations. How hard can that be?
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Well, it's actually even easier. Taking <math>\Phi</math> to be the projection from <math>A_3</math> to <math>A^{ff}_3</math> of a group like horizontal associator we find (by means to be documented later) that me may actually take it to be of the form
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{{Equation|Phi|<math>\Phi=1+t^{13}t^{23}F(t^{13},t^{23})-t^{23}t^{13}F(t^{13},t^{23})</math>,}}
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where <math>F</math> is a single unknown function of two commuting variables.
  
 
==So is it doable?==
 
==So is it doable?==
  
Me? Why are you asking me? My forte is in daydreaming; not in solving math problems. Anyway, let's try. The hexagon equations are easier (they are written in <math>A_3</math> and not the "bigger" <math>A_4</math>) and in some sense more important (they force <math>\Phi</math> to be non-trivial; <math>\Phi=1</math> solves the pentagon but not the hexagons), so we will start with the hexagons.
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Me? Why are you asking me? My forte is in daydreaming; not in solving math problems. Anyway, let's try. The hexagon equations are easier (they are written in <math>A_3</math> and not the "bigger" <math>A_4</math>) and in some sense more important (they force <math>\Phi</math> to be non-trivial; <math>\Phi=1</math> solves the pentagon but not the hexagons), so we will start with the hexagons. First we simply the hexagon in <math>A^{ff}_3</math>, factor by factor:
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{| align=center border=1 cellspacing=0
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|-
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|Factor
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|Written in <math>A^{ff}_3</math>
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|Comments
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|-
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|<math>(\Delta\otimes 1)(R^{\pm 1})</math>
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|<math>1+t^{13}\frac{\exp(\pm t^{13}\pm t^{23})-1}{t^{13}+t^{23}}+t^{23}\frac{\exp(\pm t^{13}\pm t^{23})-1}{t^{13}+t^{23}}</math>
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|We use <math>e^{\pm x}=1+x\frac{e^{\pm x}-1}{x}</math> to write exponentials with their frozen parts made explicit
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|-
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|<math>\Phi^{123}</math>
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|<math>1+t^{13}t^{23}F(t^{13}, t^{23})-t^{23}t^{13}F(t^{13}, t^{23})</math>
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|
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|-
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|<math>(R^{\pm 1})^{23}</math>
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|<math>1+t^{23}\frac{\exp(\pm t^{23})-1}{t^{23}}</math>
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|
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|-
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|<math>(\Phi^{-1})^{132}</math>
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|
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|
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|-
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|<math>(R^{\pm 1})^{13}</math>
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|
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|-
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|<math>\Phi^{312}</math>
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|
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|
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|}
  
 
==References==
 
==References==

Latest revision as of 17:26, 12 October 2006

Contents

The Goal

The purpose of the paperlet is to find an explicit formula for an associator with frozen feet. As I'm starting to write, I don't know such a formula. My hope is that as I type up all the relevant equations, a solution will emerge. I'll be just as happy if it emerges in somebody else's mind, provided (s)he shares her/his thoughts.

The Pentagon For Parenthesized Braids.jpg The Hexagons For Parenthesized Braids.jpg
The Pentagon and the Hexagons for Parenthesized Braids

A horizontal associator is a solution \Phi\in A_3 of the pentagon equation and the hexagon equations ([Drinfeld_90], [Drinfeld_91], [Bar-Natan_97]):

[Pentagon]
\Phi^{123}\cdot(1\otimes\Delta\otimes 1)(\Phi)\cdot\Phi^{234}=(\Delta\otimes 1\otimes 1)(\Phi)\cdot(1\otimes 1\otimes\Delta)(\Phi)     in     A_4,


[Hexagons]
(\Delta\otimes 1)(R^{\pm 1}) = \Phi^{123}\cdot (R^{\pm 1})^{23}\cdot(\Phi^{-1})^{132}\cdot(R^{\pm 1})^{13}\cdot\Phi^{312}     in     A_3.

Here A_n is the algebra of horizontal chord diagrams on n vertical strands; i.e., A_n=\left.\left\langle t^{ij}=t^{ji}\right\rangle\right/[t^{ij},t^{kl}]=[t^{ij},t^{ik}+t^{jk}]=0, with i, j, k and l all different integers between 1 and n. Also, here R\in A_2 will be \exp\,t^{12} (slightly different than the normal convention of \exp\,t^{12}/2, just to save some denominators).

The frozen feet quotient of any associative algebra A is the quotient A^{ff}:=A\left/\left\langle xyz-xzy\right\rangle\right.. (In a vertical presentation of chord diagrams we may think of the first letter of a word as its foot. The frozen feet relation makes all letters of a word commute, except perhaps the first one. "Molten bodies" is perhaps more accurate, but it is definitely less catchy).

What we are looking for is an associator with frozen feet, a solution of exactly the same equations, except regarded with the frozen feet quotients A^{ff}_n of A_n.

Why Bother?

This isn't the place to explain the need for an explicit associator. The need is there even if it is a bit under-appreciated. Why frozen feet? Because finding an honest explicit horizontal associator seems too hard, and as a warm up and perhaps a step, it seems worthwhile to look for associators in some quotient spaces. The frozen feet quotient is the simplest quotient I'm aware off for which the answer is unknown.

Why might it be doable?

Because frozen feet are really molten bodies. It is known that there exists and associator of the form \Phi=1+F(t^{13},t^{23}), where F is some non-commutative power series with no constant term. Once the body melts (but keeping the feet frozen), this becomes

\Phi=1+t^{13}F_{13}(t^{13},t^{23})+t^{23}F_{23}(t^{13},t^{23}),

where F_{13} and F_{23} are commutative power series (i.e., "functions"). So finding an associator with frozen feet reduces to finding two, just two, functions of two, just two, variables, satisfying some algebraic equations. How hard can that be?

Well, it's actually even easier. Taking \Phi to be the projection from A_3 to A^{ff}_3 of a group like horizontal associator we find (by means to be documented later) that me may actually take it to be of the form

[Phi]
\Phi=1+t^{13}t^{23}F(t^{13},t^{23})-t^{23}t^{13}F(t^{13},t^{23}),

where F is a single unknown function of two commuting variables.

So is it doable?

Me? Why are you asking me? My forte is in daydreaming; not in solving math problems. Anyway, let's try. The hexagon equations are easier (they are written in A_3 and not the "bigger" A_4) and in some sense more important (they force \Phi to be non-trivial; \Phi=1 solves the pentagon but not the hexagons), so we will start with the hexagons. First we simply the hexagon in A^{ff}_3, factor by factor:

Factor Written in A^{ff}_3 Comments
(\Delta\otimes 1)(R^{\pm 1}) Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): 1+t^{13}\frac{\exp(\pm t^{13}\pm t^{23})-1}{t^{13}+t^{23}}+t^{23}\frac{\exp(\pm t^{13}\pm t^{23})-1}{t^{13}+t^{23}} We use e^{\pm x}=1+x\frac{e^{\pm x}-1}{x} to write exponentials with their frozen parts made explicit
\Phi^{123} 1+t^{13}t^{23}F(t^{13}, t^{23})-t^{23}t^{13}F(t^{13}, t^{23})
(R^{\pm 1})^{23} Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): 1+t^{23}\frac{\exp(\pm t^{23})-1}{t^{23}}
(\Phi^{-1})^{132}
(R^{\pm 1})^{13}
\Phi^{312}

References

[Bar-Natan_97] ^  D. Bar-Natan, Non-associative tangles, in Geometric topology (proceedings of the Georgia international topology conference), (W. H. Kazez, ed.), 139-183, Amer. Math. Soc. and International Press, Providence, 1997.

[Drinfeld_90] ^  V. G. Drinfel'd, Quasi-Hopf algebras, Leningrad Math. J. 1 (1990) 1419-1457.

[Drinfeld_91] ^  V. G. Drinfel'd, On quasitriangular Quasi-Hopf algebras and a group closely connected with Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \operatorname{Gal}(\bar{\mathbb Q}/{\mathbb Q}) , Leningrad Math. J. 2 (1991) 829-860.