Proving Evenness of m and r in σ ∈ Sn

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Discussion Overview

The discussion revolves around proving the relationship between the evenness of two integers, m and r, in the context of permutations represented as products of transpositions. The focus is on the implications of this relationship for the Vandermonde polynomial and its behavior under permutations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes to prove that if σ ∈ Sn can be expressed as a product of m transpositions, then m is even if and only if r is even.
  • Another participant references a similar thread and expresses a desire for direct answers to the original question, indicating a need for clarity on the topic.
  • A participant suggests a strategy involving the action of a transposition on the Vandermonde polynomial, stating that an even number of transpositions leaves the polynomial unchanged, while an odd number changes its sign.
  • The same participant notes that if a permutation could be both even and odd, it would lead to a contradiction regarding the Vandermonde polynomial.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are multiple viewpoints on how to approach the proof and the implications of the relationship between m and r. Some participants express confusion and seek further clarification, indicating that the discussion remains unresolved.

Contextual Notes

The discussion highlights the complexity of the relationship between permutations and the Vandermonde polynomial, with participants acknowledging the need for more detailed insights and potentially missing assumptions in their arguments.

namzay300
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Suppose σ ∈ Sn and σ = τ1*τ2****τm = t1*t2****tr where each τi and tj is a transposition. Then, prove m is even iff r is even.

Note: τ(δ(r1,r2,...,rn) = -δ(r1,r2,...,rn)

Usually like to provide what I have done so far, but I've been racking my brain for awhile and can't come up with much. Any detailed insight would be much appreciated. Thanks!
 
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namzay300 said:
Suppose σ ∈ Sn and σ = τ1*τ2****τm = t1*t2****tr where each τi and tj is a transposition. Then, prove m is even iff r is even.

Note: τ(δ(r1,r2,...,rn) = -δ(r1,r2,...,rn)

Usually like to provide what I have done so far, but I've been racking my brain for awhile and can't come up with much. Any detailed insight would be much appreciated. Thanks!

http://mathhelpboards.com/linear-abstract-algebra-14/symmetric-polynomials-involving-discriminant-poly-17823.html

Are you two taking the same class? (Wink)
 
Deveno said:
http://mathhelpboards.com/linear-abstract-algebra-14/symmetric-polynomials-involving-discriminant-poly-17823.html

Are you two taking the same class? (Wink)

That would be quite a coincidence! I was hoping to receive an answer pertaining to my question though :( If anyone out there can help, that would be great! Thank you.
 
namzay300 said:
That would be quite a coincidence! I was hoping to receive an answer pertaining to my question though :( If anyone out there can help, that would be great! Thank you.

The link I posted was to another thread that essentially covers the same ground, although it may not seem like it.

The strategy is to show that if $\sigma = (i\ j)$ is a transposition, then the action of $\sigma$ on the Vandermonde polynomial $\delta(x_1,\dots,x_n) = \prod\limits_{k < m} (x_k - x_m)$ given by:
$\sigma(\delta(x_1,\dots,x_n)) = \delta(x_{\sigma(1)},\dots,x_{\sigma(n)})$ is equal to $-\delta(x_1,\dots,x_n)$.

So, by extension, if a permutation $\tau$ can be written as an even number of permutations, it leaves the Vandermonde polynomial unchanged, and if it can be written as an odd number of permutations, it changes the sign of the Vandermonde polynomial.

Now if a permutation could be written as both an even number AND an odd number of permutations, we would have:

$\delta(x_1,\dots,x_n) = -\delta(x_1,\dots,x_n)$, a contradiction.
 

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