06-1350/Class Notes for Tuesday October 24 - Revision history

128.100.219.73: /* What Cyclic Permutations Can't See */

2006-10-29T19:37:08Z

What Cyclic Permutations Can't See

← Older revision		Revision as of 15:37, 29 October 2006
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	Believe or not, but the following questions are directly related to class material - specifically, to the determination of "[[The Envelope of The Alexander Polynomial]]".		Believe or not, but the following questions are directly related to class material - specifically, to the determination of "[[The Envelope of The Alexander Polynomial]]".

	Let <math>S_n</math> denote the permutation group on <math>n</math> letters and let <math>{\mathbb Q}S_n</math> denote its group ring. Let <math>c:{\mathbb Q}S_n\to{\mathbb Q}</math> be the linear functional defined via its definition on generators by <math>c(\sigma)=1</math> if the permutation <math>\sigma</math> is cyclic, and <math>c(\sigma)=0</math> otherwise. Turn <math>c</math> into a (symmetric!) bilinear form (also called <math>c</math>) on <math>{\mathbb Q}S_n\~~otimes~~{\mathbb Q}S_n</math> by setting <math>c(\tau,\sigma):=c(\tau\circ\sigma)</math>.		Let <math>S_n</math> denote the permutation group on <math>n</math> letters and let <math>{\mathbb Q}S_n</math> denote its group ring. Let <math>c:{\mathbb Q}S_n\to{\mathbb Q}</math> be the linear functional defined via its definition on generators by <math>c(\sigma)=1</math> if the permutation <math>\sigma</math> is cyclic, and <math>c(\sigma)=0</math> otherwise. Turn <math>c</math> into a (symmetric!) bilinear form (also called <math>c</math>) on <math>{\mathbb Q}S_n\times{\mathbb Q}S_n</math> by setting <math>c(\tau,\sigma):=c(\tau\circ\sigma)</math>.

	'''Question 1.''' Determine the kernel <math>\ker c</math> of the bilinear form <math>c</math>. (Recall that the kernel of a bilinear form <math>\gamma</math> is <math>\{w:\forall v,\ \gamma(v,w)=0\}</math>.		'''Question 1.''' Determine the kernel <math>\ker c</math> of the bilinear form <math>c</math>. (Recall that the kernel of a bilinear form <math>\gamma</math> is <math>\{w:\forall v,\ \gamma(v,w)=0\}</math>.

Andy: Added scanned notes

2006-10-25T14:36:31Z

Added scanned notes

← Older revision		Revision as of 10:36, 25 October 2006
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	{{InOut\|n=5\|in=<nowiki>c[# ** Y4] & /@ S[4]</nowiki>\|out=<nowiki>{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}</nowiki>}}		{{InOut\|n=5\|in=<nowiki>c[# ** Y4] & /@ S[4]</nowiki>\|out=<nowiki>{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}</nowiki>}}

			==Scanned Notes==

			{\| align=left
			\|[[Image:06-1350-scan1024-0001.jpg\|thumb]]
			\|[[Image:06-1350-scan1024-0002.jpg\|thumb]]
			\|[[Image:06-1350-scan1024-0003.jpg\|thumb]]
			\|}

Drorbn at 19:20, 24 October 2006

2006-10-24T19:20:30Z

← Older revision		Revision as of 15:20, 24 October 2006
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	Believe or not, but the following questions are directly related to class material - specifically, to the determination of "[[The Envelope of The Alexander Polynomial]]".		Believe or not, but the following questions are directly related to class material - specifically, to the determination of "[[The Envelope of The Alexander Polynomial]]".

	Let <math>S_n</math> denote the permutation group on <math>n</math> letters and let <math>{\mathbb Q}S_n</math> denote its group ring. Let <math>c:{\mathbb Q}S_n\to{\mathbb Q}</math> be the linear functional defined via its definition on generators by <math>c(\sigma)=1</math> if the permutation <math>\sigma</math> is cyclic, and <math>c(\sigma)=0</math> otherwise. Turn <math>c</math> into a (symmetric!) bilinear form (also called <math>c</math>) on <math>{\mathbb Q}\otimes{\mathbb Q}</math> by setting <math>c(\tau,\sigma):=c(\tau\circ\sigma)</math>.		Let <math>S_n</math> denote the permutation group on <math>n</math> letters and let <math>{\mathbb Q}S_n</math> denote its group ring. Let <math>c:{\mathbb Q}S_n\to{\mathbb Q}</math> be the linear functional defined via its definition on generators by <math>c(\sigma)=1</math> if the permutation <math>\sigma</math> is cyclic, and <math>c(\sigma)=0</math> otherwise. Turn <math>c</math> into a (symmetric!) bilinear form (also called <math>c</math>) on <math>{\mathbb Q}S_n\otimes{\mathbb Q}S_n</math> by setting <math>c(\tau,\sigma):=c(\tau\circ\sigma)</math>.

	'''Question 1.''' Determine the kernel <math>\ker c</math> of the bilinear form <math>c</math>. (Recall that the kernel of a bilinear form <math>\gamma</math> is <math>\{w:\forall v,\ \gamma(v,w)=0\}</math>.		'''Question 1.''' Determine the kernel <math>\ker c</math> of the bilinear form <math>c</math>. (Recall that the kernel of a bilinear form <math>\gamma</math> is <math>\{w:\forall v,\ \gamma(v,w)=0\}</math>.

Drorbn at 12:30, 24 October 2006

2006-10-24T12:30:47Z

← Older revision		Revision as of 08:30, 24 October 2006
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	'''Question 1.''' Determine the kernel <math>\ker c</math> of the bilinear form <math>c</math>. (Recall that the kernel of a bilinear form <math>\gamma</math> is <math>\{w:\forall v,\ \gamma(v,w)=0\}</math>.		'''Question 1.''' Determine the kernel <math>\ker c</math> of the bilinear form <math>c</math>. (Recall that the kernel of a bilinear form <math>\gamma</math> is <math>\{w:\forall v,\ \gamma(v,w)=0\}</math>.

	[[Image:TheHRelation.svg\|thumb\|~~600px~~\|center\|The H Relation]]		[[Image:TheHRelation.svg\|thumb\|550px\|center\|The H Relation]]

	'''Question 2.''' For <math>n=4</math>, I know by a lengthy computation (see below) that <math>H</math> is in <math>\ker c</math>, where		'''Question 2.''' For <math>n=4</math>, I know by a lengthy computation (see below) that <math>H</math> is in <math>\ker c</math>, where
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	(here <math>(jk)</math> denotes the transposition of <math>j</math> and <math>k</math>, <math>k_1k_2k_3k_4</math> denotes the permutation for which <math>i\mapsto k_i</math>, and the bracket is taken in the additive sense: <math>[\tau,\sigma]:=\tau\sigma-\sigma\tau</math>). Do you have quicker explanation?		(here <math>(jk)</math> denotes the transposition of <math>j</math> and <math>k</math>, <math>k_1k_2k_3k_4</math> denotes the permutation for which <math>i\mapsto k_i</math>, and the bracket is taken in the additive sense: <math>[\tau,\sigma]:=\tau\sigma-\sigma\tau</math>). Do you have quicker explanation?

	[[Image:The4YRelation.svg\|thumb\|~~480px~~\|center\|The 4Y Relation]]		[[Image:The4YRelation.svg\|thumb\|440px\|center\|The 4Y Relation]]

	'''Question 3.''' By another lengthy computation for <math>n=4</math>, I also know that <math>4Y\in\ker c</math>, where		'''Question 3.''' By another lengthy computation for <math>n=4</math>, I also know that <math>4Y\in\ker c</math>, where

Drorbn at 12:29, 24 October 2006

2006-10-24T12:29:47Z

← Older revision		Revision as of 08:29, 24 October 2006
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	'''Question 1.''' Determine the kernel <math>\ker c</math> of the bilinear form <math>c</math>. (Recall that the kernel of a bilinear form <math>\gamma</math> is <math>\{w:\forall v,\ \gamma(v,w)=0\}</math>.		'''Question 1.''' Determine the kernel <math>\ker c</math> of the bilinear form <math>c</math>. (Recall that the kernel of a bilinear form <math>\gamma</math> is <math>\{w:\forall v,\ \gamma(v,w)=0\}</math>.

	[[Image:TheHRelation.svg\|thumb\|~~480px~~\|center\|The H Relation]]		[[Image:TheHRelation.svg\|thumb\|600px\|center\|The H Relation]]

	'''Question 2.''' For <math>n=4</math>, I know by a lengthy computation (see below) that <math>H</math> is in <math>\ker c</math>, where		'''Question 2.''' For <math>n=4</math>, I know by a lengthy computation (see below) that <math>H</math> is in <math>\ker c</math>, where
Line 22:		Line 22:
	(here <math>(jk)</math> denotes the transposition of <math>j</math> and <math>k</math>, <math>k_1k_2k_3k_4</math> denotes the permutation for which <math>i\mapsto k_i</math>, and the bracket is taken in the additive sense: <math>[\tau,\sigma]:=\tau\sigma-\sigma\tau</math>). Do you have quicker explanation?		(here <math>(jk)</math> denotes the transposition of <math>j</math> and <math>k</math>, <math>k_1k_2k_3k_4</math> denotes the permutation for which <math>i\mapsto k_i</math>, and the bracket is taken in the additive sense: <math>[\tau,\sigma]:=\tau\sigma-\sigma\tau</math>). Do you have quicker explanation?

			[[Image:The4YRelation.svg\|thumb\|480px\|center\|The 4Y Relation]]
⚫	'''Question 3.''' By another lengthy computation for <math>n=4</math>, I also know that <math>~~Y_4~~\in\ker c</math>, where

⚫	<center><math>~~Y_4~~=[(12),(23)]-[(23),(34)]+[(34),(41)]-[(41),(12)]</math>.</center>
		⚫	'''Question 3.''' By another lengthy computation for <math>n=4</math>, I also know that <math>4Y\in\ker c</math>, where
		⚫	<center><math>4Y=[(12),(23)]-[(23),(34)]+[(34),(41)]-[(41),(12)]</math>.</center>
	Do you have quicker explanation?		Do you have quicker explanation?

	'''Question 4.''' I suspect that in some sense, though I'm not sure in which, <math>H</math> and <math>~~Y_4~~</math> generate the whole kernel or at least some easily definable special part of the kernel of <math>c</math> for ''all'' <math>n</math>. Can you make sense of that?		'''Question 4.''' I suspect that in some sense, though I'm not sure in which, <math>H</math> and <math>4Y</math> generate the whole kernel or at least some easily definable special part of the kernel of <math>c</math> for ''all'' <math>n</math>. Can you make sense of that?

	==The Lengthy Computations==		==The Lengthy Computations==

Drorbn: /* What Cyclic Permutations Can't See */

2006-10-24T11:37:37Z

What Cyclic Permutations Can't See

← Older revision		Revision as of 07:37, 24 October 2006
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	'''Question 1.''' Determine the kernel <math>\ker c</math> of the bilinear form <math>c</math>. (Recall that the kernel of a bilinear form <math>\gamma</math> is <math>\{w:\forall v,\ \gamma(v,w)=0\}</math>.		'''Question 1.''' Determine the kernel <math>\ker c</math> of the bilinear form <math>c</math>. (Recall that the kernel of a bilinear form <math>\gamma</math> is <math>\{w:\forall v,\ \gamma(v,w)=0\}</math>.

			[[Image:TheHRelation.svg\|thumb\|480px\|center\|The H Relation]]

	'''Question 2.''' For <math>n=4</math>, I know by a lengthy computation (see below) that <math>H</math> is in <math>\ker c</math>, where		'''Question 2.''' For <math>n=4</math>, I know by a lengthy computation (see below) that <math>H</math> is in <math>\ker c</math>, where

Drorbn: /* What Cyclic Permutations Can't See */

2006-10-24T10:49:35Z

What Cyclic Permutations Can't See

← Older revision		Revision as of 06:49, 24 October 2006
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	Believe or not, but the following questions are directly related to class material - specifically, to the determination of "[[The Envelope of The Alexander Polynomial]]".		Believe or not, but the following questions are directly related to class material - specifically, to the determination of "[[The Envelope of The Alexander Polynomial]]".

	Let <math>S_n</math> denote the permutation group on <math>n</math> letters and let <math>{\mathbb Q}S_n</math> denote its group ring. Let <math>c:{\mathbb Q}S_n\to{\mathbb Q}</math> be the linear functional defined via its definition on generators by <math>c(\sigma)=1</math> if the permutation <math>\sigma</math> is cyclic, and <math>c(\sigma)=0</math> otherwise. Turn <math>c</math> into a (symmetric!) bilinear form (also called <math>c</math>) on <math>{\mathbb Q}\otimes{\mathbb Q}</math> by setting <math>c(\tau,\sigma):=c(\tau~~^{-1}~~\sigma)</math>.		Let <math>S_n</math> denote the permutation group on <math>n</math> letters and let <math>{\mathbb Q}S_n</math> denote its group ring. Let <math>c:{\mathbb Q}S_n\to{\mathbb Q}</math> be the linear functional defined via its definition on generators by <math>c(\sigma)=1</math> if the permutation <math>\sigma</math> is cyclic, and <math>c(\sigma)=0</math> otherwise. Turn <math>c</math> into a (symmetric!) bilinear form (also called <math>c</math>) on <math>{\mathbb Q}\otimes{\mathbb Q}</math> by setting <math>c(\tau,\sigma):=c(\tau\circ\sigma)</math>.

	'''Question 1.''' Determine the kernel <math>\ker c</math> of the bilinear form <math>c</math>. (Recall that the kernel of a bilinear form <math>\gamma</math> is <math>\{w:\forall v,\ \gamma(v,w)=0\}</math>.		'''Question 1.''' Determine the kernel <math>\ker c</math> of the bilinear form <math>c</math>. (Recall that the kernel of a bilinear form <math>\gamma</math> is <math>\{w:\forall v,\ \gamma(v,w)=0\}</math>.

Drorbn: /* What Cyclic Permutations Can't See */

2006-10-24T10:43:31Z

What Cyclic Permutations Can't See

← Older revision		Revision as of 06:43, 24 October 2006
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	\|<math>\ \ = 2341-2413-3142+4123-4231-1324+3214+1432</math>		\|<math>\ \ = 2341-2413-3142+4123-4231-1324+3214+1432</math>
	\|}		\|}
	(here <math>(jk)</math> denotes the transposition of <math>j</math> and <math>k</math>, <math>k_1k_2k_3k_4</math> denotes the permutation for which <math>i\mapsto k_i</math>, and the bracket is taken in the additive sense: <math>[\tau,\sigma]:=\tau\sigma-\sigma\tau)</math>. Do you have quicker explanation?		(here <math>(jk)</math> denotes the transposition of <math>j</math> and <math>k</math>, <math>k_1k_2k_3k_4</math> denotes the permutation for which <math>i\mapsto k_i</math>, and the bracket is taken in the additive sense: <math>[\tau,\sigma]:=\tau\sigma-\sigma\tau</math>). Do you have quicker explanation?

	'''Question 3.''' By another lengthy computation for <math>n=4</math>, I also know that <math>Y_4\in\ker c</math>, where		'''Question 3.''' By another lengthy computation for <math>n=4</math>, I also know that <math>Y_4\in\ker c</math>, where

Drorbn: /* The Lengthy Computations */

2006-10-24T01:39:24Z

The Lengthy Computations

← Older revision		Revision as of 21:39, 23 October 2006
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	{{InOut\|n=3\|in=<nowiki>c[# ** H] & /@ S[4]</nowiki>\|out=<nowiki>{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}</nowiki>}}		{{InOut\|n=3\|in=<nowiki>c[# ** H] & /@ S[4]</nowiki>\|out=<nowiki>{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}</nowiki>}}


	{{InOut\|n=4\|in=<nowiki>Y4 = b[P[2, 1, 3, 4], P[1, 3, 2, 4]] - b[P[1, 3, 2, 4], P[1, 2, 4, 3]] +		{{InOut\|n=4\|in=<nowiki>Y4 = b[P[2, 1, 3, 4], P[1, 3, 2, 4]] - b[P[1, 3, 2, 4], P[1, 2, 4, 3]] +

Drorbn: /* The Lengthy Computations */

2006-10-24T01:38:46Z

The Lengthy Computations

← Older revision		Revision as of 21:38, 23 October 2006
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	(c_p1_P) * p2_ := c p1 ** p2;		(c_p1_P) * p2_ := c p1 ** p2;
	b[a_, b_] := a b - b a;</nowiki>}}		b[a_, b_] := a b - b a;</nowiki>}}


	{{InOut\|n=2\|in=<nowiki>H = b[P[2, 1, 3, 4], b[P[3, 2, 1, 4], P[4, 2, 3, 1]]]		{{InOut\|n=2\|in=<nowiki>H = b[P[2, 1, 3, 4], b[P[3, 2, 1, 4], P[4, 2, 3, 1]]]
	- P[1, 3, 2, 4] + P[1, 4, 3, 2] + P[3, 2, 1, 4] - P[4, 2, 3, 1]</nowiki>\|out=<nowiki>-P[1, 3, 2, 4] + P[1, 4, 3, 2] + P[2, 3, 4, 1] - P[2, 4, 1, 3] -		- P[1, 3, 2, 4] + P[1, 4, 3, 2] + P[3, 2, 1, 4] - P[4, 2, 3, 1]</nowiki>\|out=<nowiki>-P[1, 3, 2, 4] + P[1, 4, 3, 2] + P[2, 3, 4, 1] - P[2, 4, 1, 3] -
	P[3, 1, 4, 2] + P[3, 2, 1, 4] + P[4, 1, 2, 3] - P[4, 2, 3, 1]</nowiki>}}		P[3, 1, 4, 2] + P[3, 2, 1, 4] + P[4, 1, 2, 3] - P[4, 2, 3, 1]</nowiki>}}


	{{InOut\|n=3\|in=<nowiki>c[# ** H] & /@ S[4]</nowiki>\|out=<nowiki>{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}</nowiki>}}		{{InOut\|n=3\|in=<nowiki>c[# ** H] & /@ S[4]</nowiki>\|out=<nowiki>{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}</nowiki>}}
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	b[P[1, 2, 4, 3], P[4, 2, 3, 1]] - b[P[4, 2, 3, 1], P[2, 1, 3, 4]]</nowiki>\|out=<nowiki>P[1, 3, 4, 2] - P[1, 4, 2, 3] - P[2, 3, 1, 4] + P[2, 4, 3, 1] +		b[P[1, 2, 4, 3], P[4, 2, 3, 1]] - b[P[4, 2, 3, 1], P[2, 1, 3, 4]]</nowiki>\|out=<nowiki>P[1, 3, 4, 2] - P[1, 4, 2, 3] - P[2, 3, 1, 4] + P[2, 4, 3, 1] +
	P[3, 1, 2, 4] - P[3, 2, 4, 1] - P[4, 1, 3, 2] + P[4, 2, 1, 3]</nowiki>}}		P[3, 1, 2, 4] - P[3, 2, 4, 1] - P[4, 1, 3, 2] + P[4, 2, 1, 3]</nowiki>}}


	{{InOut\|n=5\|in=<nowiki>c[# ** Y4] & /@ S[4]</nowiki>\|out=<nowiki>{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}</nowiki>}}		{{InOut\|n=5\|in=<nowiki>c[# ** Y4] & /@ S[4]</nowiki>\|out=<nowiki>{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}</nowiki>}}

← Older revision		Revision as of 21:38, 23 October 2006
Line 43:		Line 43:
	(c_p1_P) * p2_ := c p1 ** p2;		(c_p1_P) * p2_ := c p1 ** p2;
	b[a_, b_] := a b - b a;</nowiki>}}		b[a_, b_] := a b - b a;</nowiki>}}


	{{InOut\|n=2\|in=<nowiki>H = b[P[2, 1, 3, 4], b[P[3, 2, 1, 4], P[4, 2, 3, 1]]]		{{InOut\|n=2\|in=<nowiki>H = b[P[2, 1, 3, 4], b[P[3, 2, 1, 4], P[4, 2, 3, 1]]]
	- P[1, 3, 2, 4] + P[1, 4, 3, 2] + P[3, 2, 1, 4] - P[4, 2, 3, 1]</nowiki>\|out=<nowiki>-P[1, 3, 2, 4] + P[1, 4, 3, 2] + P[2, 3, 4, 1] - P[2, 4, 1, 3] -		- P[1, 3, 2, 4] + P[1, 4, 3, 2] + P[3, 2, 1, 4] - P[4, 2, 3, 1]</nowiki>\|out=<nowiki>-P[1, 3, 2, 4] + P[1, 4, 3, 2] + P[2, 3, 4, 1] - P[2, 4, 1, 3] -
	P[3, 1, 4, 2] + P[3, 2, 1, 4] + P[4, 1, 2, 3] - P[4, 2, 3, 1]</nowiki>}}		P[3, 1, 4, 2] + P[3, 2, 1, 4] + P[4, 1, 2, 3] - P[4, 2, 3, 1]</nowiki>}}


	{{InOut\|n=3\|in=<nowiki>c[# ** H] & /@ S[4]</nowiki>\|out=<nowiki>{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}</nowiki>}}		{{InOut\|n=3\|in=<nowiki>c[# ** H] & /@ S[4]</nowiki>\|out=<nowiki>{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}</nowiki>}}
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	b[P[1, 2, 4, 3], P[4, 2, 3, 1]] - b[P[4, 2, 3, 1], P[2, 1, 3, 4]]</nowiki>\|out=<nowiki>P[1, 3, 4, 2] - P[1, 4, 2, 3] - P[2, 3, 1, 4] + P[2, 4, 3, 1] +		b[P[1, 2, 4, 3], P[4, 2, 3, 1]] - b[P[4, 2, 3, 1], P[2, 1, 3, 4]]</nowiki>\|out=<nowiki>P[1, 3, 4, 2] - P[1, 4, 2, 3] - P[2, 3, 1, 4] + P[2, 4, 3, 1] +
	P[3, 1, 2, 4] - P[3, 2, 4, 1] - P[4, 1, 3, 2] + P[4, 2, 1, 3]</nowiki>}}		P[3, 1, 2, 4] - P[3, 2, 4, 1] - P[4, 1, 3, 2] + P[4, 2, 1, 3]</nowiki>}}


	{{InOut\|n=5\|in=<nowiki>c[# ** Y4] & /@ S[4]</nowiki>\|out=<nowiki>{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}</nowiki>}}		{{InOut\|n=5\|in=<nowiki>c[# ** Y4] & /@ S[4]</nowiki>\|out=<nowiki>{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}</nowiki>}}