06-1350/Class Notes for Tuesday October 24

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

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

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

$H=[(12),[(13),(14)]]-(14)-(23)+(13)+(24)$

$\ \ =[2134,[3214,4231]]-4231-1324+3214+1432$

$\ \ =2341-2413-3142+4123-4231-1324+3214+1432$

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

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

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

Do you have quicker explanation?

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

The Lengthy Computations

The lengthy computation for $H$ (and likewise for $Y_{4}$) 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}