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Let $P=\{(i, j)|1\leq iFor $\sigma\in S_n$, define $\operatorname{sgn}\colon S_{n}\rightarrow \{\pm 1\}$ by $$ \operatorname{sgn}(\sigma)=\prod_{(i, j)\in P}\operatorname{sgn}(\sigma(j)-\sigma(i)).$$ How to prove the following?

  1. $\displaystyle{\operatorname{sgn}(\sigma)=\prod_{1\leq i.

  2. Consider the polynomial $\displaystyle{p(x_{1},...,x_{n})=\prod_{1\leq i.

    For $\sigma\in S_n$ prove that $\displaystyle{\operatorname{sgn}(\sigma)=\frac{p(x_{\sigma(1)},\ldots,x_{\sigma(n)}) }{p(x_{1},\ldots,x_{n})}}$.

  3. Let $N_{\sigma}$ be the number of orbits of $\sigma$ acting on $\{1,\ldots,n\}$. Show $\operatorname{sgn}(\sigma)=(-1)^{n-N_{\sigma}}$.

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    Surely you know how to solve *some* of these! Please show what you tried and where you are stuck.2012-05-24
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    You have a conflicting notation: you use sgn for two different functions.2012-05-24
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    In fact, once you fix this, you should be able to solve the first question *immediately*.2012-05-24
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    Hi Didier not I do not take any part :(2012-05-24
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    in your definition in the right hand side by $\mbox{sgn}(\sigma(j)-\sigma(i))$ you mean $$\mbox{sgn}(\sigma(j)-\sigma(i))=\cases{+1&; $\sigma(j)-\sigma(i)>0$ \\ -1&; $\sigma(j)-\sigma(i)<0$}.$$ Is that right? If yes, does that help for part 1?2012-05-24
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    Part 2 is the immediate consequence of part 1.2012-05-24
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    Even part 1.? What are the possible values of the product in the RHS?2012-05-24
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    Keivan my problem is that I can't write the proves of 1,2,3 formally. Could you help me?2012-05-24
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    Maybe you should first write them informally?2012-05-24
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    I write for particular cases but the true I cant not see the prove in general. Sorry my ignorance.2012-05-24
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    @Andres If $x\in\mathbb{R}$ is non zero, we have $sgn(x)=\frac{|x|}{x}$ (**by definition**). If this doesn't help, I don't know what will.2012-05-25
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    Then write what you have for particular cases, but write something. Yourself. Mathematics is not [a spectator sport](http://www.cut-the-knot.org/books/Reviews/NotSpectatorSport.shtml).2012-05-25

1 Answers 1

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Pay attention to the fact that in the formula in $\,(1)\,$ , for any $\,1\leq i, the denominator $\,j-i\,$ is always positive , so what "decides" the sign there is merely the numerator $\,\sigma(j)-\sigma(i)\,$ , thus: both "definitions" you wrote down of sgn are exactly the same.

Now, let us write down the given information: $$\frac{p\left(x_{\sigma(1)},...,x_{\sigma(n)}\right)}{p\left(x_1,...,x_n\right)}=\frac{\prod_{1\leq iand as before: the factors in the denominator are the very same factors in the numerator up to sign, and there's a change of sign, say for some pairs $\,(r,s)\,,\,(i,j)\in P\,$: $$\frac{x_{\sigma(j)}-x_{\sigma(i)}}{x_r-x_s}=\frac{-(x_r-x_s)}{x_r-x_s}=-1$$iff exactly the same change exists in $$\frac{\sigma(j)-\sigma(i)}{r-s}=\frac{-(r-s)}{r-s}=-1$$ so from here clearly we have that $\,\,(**)=sgn(\sigma)\,$

Finally, we can try to attack $\,(3)\,$ with the following:

Lemma: For every permutation $\,\sigma\in S_n\,\,,\,n-N_\sigma=T_\sigma $ , with $\,T_\sigma:=\,$number of transpositions $\,(i \,j)\,,\,1\leq i\neq j\leq n\,$ , needed to express $\,\sigma\,$ as a product of transpositions.

Proof: It is enough to prove for cycles, as any permutation can be written as a product of (disjoint) cycles: let $\,\sigma = (i_1\,i_2\,,...,i_r)\,$ be an $\,r-\,$cycle , then $$\sigma=(i_1\,i_r)(i_1\,i_{r-1})\cdot...\cdot (i_1\,i_3)(i_1\,i_2)$$and we can see that the number of transpositions in the right hand side is $$\,r-1=n-N_\sigma\Longleftrightarrow N_\sigma=n-r+1$$ which is clearly the number of orbits of $\,\sigma\,$ , as it fixes $\,n-r\,$ points in $\,\{1,2,...,n\}\,$ Q.E.D.

Well, then we already are there since any transposition has sign $\,-1\,$, obviously, so that $$(-1)^{n-N_\sigma}=(-1)^{r-1}$$and let us not forget that the number of orbits of $\,\sigma\,$ is just another name for "the number of different (disjoint) cycles in the decomposition of $\,\sigma\,$"

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    @Andres Did you see my answer above?2012-06-10