# 06-1350/Class Notes for Tuesday October 24

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

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

The H Relation

Question 2. For ${\displaystyle n=4}$, I know by a lengthy computation (see below) that ${\displaystyle H}$ is in ${\displaystyle \ker c}$, where

 ${\displaystyle H=[(12),[(13),(14)]]-(14)-(23)+(13)+(24)}$ ${\displaystyle \ \ =[2134,[3214,4231]]-4231-1324+3214+1432}$ ${\displaystyle \ \ =2341-2413-3142+4123-4231-1324+3214+1432}$

(here ${\displaystyle (jk)}$ denotes the transposition of ${\displaystyle j}$ and ${\displaystyle k}$, ${\displaystyle k_{1}k_{2}k_{3}k_{4}}$ denotes the permutation for which ${\displaystyle i\mapsto k_{i}}$, and the bracket is taken in the additive sense: ${\displaystyle [\tau ,\sigma ]:=\tau \sigma -\sigma \tau }$). Do you have quicker explanation?

The 4Y Relation

Question 3. By another lengthy computation for ${\displaystyle n=4}$, I also know that ${\displaystyle 4Y\in \ker c}$, where

${\displaystyle 4Y=[(12),(23)]-[(23),(34)]+[(34),(41)]-[(41),(12)]}$.

Do you have quicker explanation?

Question 4. I suspect that in some sense, though I'm not sure in which, ${\displaystyle H}$ and ${\displaystyle 4Y}$ generate the whole kernel or at least some easily definable special part of the kernel of ${\displaystyle c}$ for all ${\displaystyle n}$. Can you make sense of that?

## The Lengthy Computations

The lengthy computation for ${\displaystyle H}$ (and likewise for ${\displaystyle Y_{4}}$) 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]:= 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;

 In[2]:= 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] Out[2]= -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]

 In[3]:= c[# ** H] & /@ S[4] Out[3]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

 In[4]:= 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]] Out[4]= 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]

 In[5]:= c[# ** Y4] & /@ S[4] Out[5]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}