06-1350/Class Notes for Tuesday October 24: Difference between revisions
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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]]". |
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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\ |
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>. |
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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>. |
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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>}} |
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==Scanned Notes== |
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Latest revision as of 14:37, 29 October 2006
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What Cyclic Permutations Can't See
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 denote the permutation group on letters and let denote its group ring. Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c:{\mathbb Q}S_n\to{\mathbb Q}} be the linear functional defined via its definition on generators by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c(\sigma)=1} if the permutation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma} is cyclic, and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c(\sigma)=0} otherwise. Turn into a (symmetric!) bilinear form (also called ) on by setting Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c(\tau,\sigma):=c(\tau\circ\sigma)} .
Question 1. Determine the kernel of the bilinear form . (Recall that the kernel of a bilinear form is .
Question 2. For , I know by a lengthy computation (see below) that is in , where
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H = [(12),[(13),(14)]]-(14)-(23)+(13)+(24)} |
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \ = 2341-2413-3142+4123-4231-1324+3214+1432} |
(here denotes the transposition of and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} , denotes the permutation for which , and the bracket is taken in the additive sense: ). Do you have quicker explanation?
Question 3. By another lengthy computation for , I also know that , where
Do you have quicker explanation?
Question 4. I suspect that in some sense, though I'm not sure in which, and generate the whole kernel or at least some easily definable special part of the kernel of for all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} . Can you make sense of that?
The Lengthy Computations
The lengthy computation for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H} (and likewise for ) involves multiplying 24 "test permutations" against a linear combination of 8 permutations and counting cycles in the resulting 192 permutations. Here's a Mathematica session that does that:
In[1]:=
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S[n_] := (P @@@ Permutations[Range[n]]);
c[p_P] := If[
Length[p] == Length[NestWhileList[p[[#]] &, p[[1]], # > 1 &]],
1, 0
];
c[x_] := x /. p_P :> c[p];
Unprotect[NonCommutativeMultiply];
p1_P ** p2_P := p1 /. Thread[Rule[Range[Length[p2]], List @@ p2]];
p1_ ** (p2_ + p3_) := p1 ** p2 + p1 ** p3;
(p1_ + p2_) ** p3_ := p1 ** p3 + p2 ** p3;
p1_ ** (c_*p2_P) := c p1 ** p2;
(c_*p1_P) ** p2_ := c p1 ** p2;
b[a_, b_] := a ** b - b ** a;
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In[2]:=
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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]
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Out[2]=
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-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]
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In[3]:=
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c[# ** H] & /@ S[4]
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Out[3]=
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{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
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In[4]:=
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Y4 = b[P[2, 1, 3, 4], P[1, 3, 2, 4]] - b[P[1, 3, 2, 4], P[1, 2, 4, 3]] +
b[P[1, 2, 4, 3], P[4, 2, 3, 1]] - b[P[4, 2, 3, 1], P[2, 1, 3, 4]]
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Out[4]=
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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]
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In[5]:=
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c[# ** Y4] & /@ S[4]
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Out[5]=
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{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
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