Associators with Frozen Feet - Revision history

Drorbn at 22:26, 12 October 2006

2006-10-12T22:26:18Z

← Older revision		Revision as of 18:26, 12 October 2006
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			{{Paperlets Navigation}}

	==The Goal==		==The Goal==

Drorbn: /* The Goal */

2006-07-20T14:56:40Z

The Goal

← Older revision		Revision as of 10:56, 20 July 2006
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	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.

			{\| align=center width=100%
⚫	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}}):
			\|- align=center
			\|[[Image:The Pentagon For Parenthesized Braids.jpg\|350px]]
			\|[[Image:The Hexagons For Parenthesized Braids.jpg\|500px]]
			\|- align=center
			\|colspan=2\|'''The Pentagon and the Hexagons for Parenthesized Braids'''
			\|}

		⚫	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>     in     <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>     in     <math>A_4</math>,}}

Drorbn: /* So is it doable? */

2006-07-17T03:40:01Z

So is it doable?

← Older revision		Revision as of 23:40, 16 July 2006
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	\|-		\|-
	\|<math>\Phi^{123}</math>		\|<math>\Phi^{123}</math>
	\|<math>1+t^{13}~~F_1~~(t^{13}, t^{23})+t^{23}~~F_2~~(t^{13}, t^{23})</math>		\|<math>1+t^{13}t^{23}F(t^{13}, t^{23})-t^{23}t^{13}F(t^{13}, t^{23})</math>
	\|		\|
	\|-		\|-

Drorbn: /* Why might it be doable? */

2006-07-17T03:39:14Z

Why might it be doable?

← Older revision		Revision as of 23:39, 16 July 2006
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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		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>,}}		<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?

			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

			{{Equation\|Phi\|<math>\Phi=1+t^{13}t^{23}F(t^{13},t^{23})-t^{23}t^{13}F(t^{13},t^{23})</math>,}}

			where <math>F</math> is a single unknown function of two commuting variables.

	==So is it doable?==		==So is it doable?==

Drorbn: /* So is it doable? */

2006-07-16T17:16:40Z

So is it doable?

← Older revision		Revision as of 13:16, 16 July 2006
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	\|-		\|-
	\|Factor		\|Factor
	\|~~Reduction~~ in <math>A^{ff}_3</math>		\|Written in <math>A^{ff}_3</math>
			\|Comments
	\|-		\|-
	\|<math>(\Delta\otimes 1)(R^{\pm 1})</math>		\|<math>(\Delta\otimes 1)(R^{\pm 1})</math>
			\|<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>
⚫	\|
			\|We use <math>e^{\pm x}=1+x\frac{e^{\pm x}-1}{x}</math> to write exponentials with their frozen parts made explicit
	\|-		\|-
	\|<math>\Phi^{123}</math>		\|<math>\Phi^{123}</math>
			\|<math>1+t^{13}F_1(t^{13}, t^{23})+t^{23}F_2(t^{13}, t^{23})</math>
	\|		\|
	\|-		\|-
	\|<math>(R^{\pm 1})^{23}</math>		\|<math>(R^{\pm 1})^{23}</math>
			\|<math>1+t^{23}\frac{\exp(\pm t^{23})-1}{t^{23}}</math>
	\|		\|
	\|-		\|-
	\|<math>(\Phi^{-1})^{132}</math>		\|<math>(\Phi^{-1})^{132}</math>
		⚫	\|
	\|		\|
	\|-		\|-
	\|<math>(R^{\pm 1})^{13}</math>		\|<math>(R^{\pm 1})^{13}</math>
			\|
	\|		\|
	\|-		\|-
	\|<math>\Phi^{312}</math>		\|<math>\Phi^{312}</math>
			\|
	\|		\|
	\|}		\|}

Drorbn at 16:51, 16 July 2006

2006-07-16T16:51:29Z

@@ Line 31: / Line 31: @@
 ==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.
+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:
 ==References==

Drorbn: Spell checking

2006-07-16T15:16:08Z

Spell checking

← Older revision		Revision as of 11:16, 16 July 2006
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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.		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		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>,}}		{{Equation\|Phi\|<math>\Phi=1+t^{13}F_{13}(t^{13},t^{23})+t^{23}F_{23}(t^{13},t^{23})</math>,}}

Drorbn: /* So is it doable? */

2006-07-13T14:43:53Z

So is it doable?

← Older revision		Revision as of 10:43, 13 July 2006
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	==So is it doable?==		==So is it doable?==

	Me? Why are you asking me? My forte is to ~~daydream~~; not to ~~solve~~ 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.		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.

	==References==		==References==

Drorbn at 13:44, 13 July 2006

2006-07-13T13:44:41Z

← Older revision		Revision as of 09:44, 13 July 2006
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	==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^{12},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		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

	{{Equation\|Phi\|<math>\Phi=1+t^{12}F_{12}(t^{12},t^{23})+t^{23}F_{23}(t^{12},t^{23})</math>,}}		{{Equation\|Phi\|<math>\Phi=1+t^{13}F_{13}(t^{13},t^{23})+t^{23}F_{23}(t^{13},t^{23})</math>,}}

	where <math>F_{12}</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?

			==So is it doable?==

			Me? Why are you asking me? My forte is to daydream; not to solve 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.

	==References==		==References==

Drorbn: /* The Goal */

2006-07-13T13:27:38Z

The Goal

← Older revision		Revision as of 09:27, 13 July 2006
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	{{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)~~\quad\mbox{~~in~~}\quad~~ 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>     in     <math>A_4</math>,}}


	{{Equation\|Hexagons\|<math>(\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}~~\quad\mbox{~~in~~}\quad~~ A_3</math>.}}		{{Equation\|Hexagons\|<math>(\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}</math>     in     <math>A_3</math>.}}

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

← Older revision		Revision as of 10:56, 20 July 2006
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	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.

			{\| align=center width=100%
⚫	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}}):
			\|- align=center
			\|[[Image:The Pentagon For Parenthesized Braids.jpg\|350px]]
			\|[[Image:The Hexagons For Parenthesized Braids.jpg\|500px]]
			\|- align=center
			\|colspan=2\|'''The Pentagon and the Hexagons for Parenthesized Braids'''
			\|}

		⚫	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>     in     <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>     in     <math>A_4</math>,}}