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One way to define sign of permutation is to consider the polynomial $P(x_1,x_2,\cdots, x_n)=\prod_{1\leq i<j\leq n} (x_i-x_j)$ and define $$Sign(\sigma)= \frac{P(x_{\sigma(1)},x_{\sigma(2)},\cdots, x_{\sigma(n)})}{P(x_1,x_2,\cdots, x_n)}.$$

I want to see the proof of $Sign(\sigma\tau)=Sign(\sigma)Sign(\tau)$. For this, $$ Sign(\sigma\tau)=\frac{P(x_{\sigma\tau(1)},x_{\sigma\tau(2)},\cdots, x_{\sigma\tau(n)})}{P(x_1,x_2,\cdots, x_n)}$$

$$=\frac{P(x_{\sigma\tau(1)},x_{\sigma\tau(2)},\cdots, x_{\sigma\tau(n)})}{P(x_{\tau(1)},x_{\tau(2)},\cdots, x_{\tau(n)})}\times\frac{P(x_{\tau(1)},x_{\tau(2)},\cdots, x_{\tau(n)})}{P(x_1,x_2,\cdots, x_n)}.$$

I didn't get why the first term on RHS in last step is $Sign(\sigma)$. Can anybody help me?

(This could be elementary, but I am not justified.)

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$\newcommand{\Q}[0]{\mathbb{Q}}\newcommand{\Sign}[0]{\mathrm{Sign}}$The main point here is that you can define an action of the symmetric group $S_{n}$ on the set $\Q(x_{1}, \dots, x_{n})$ of rational functions with rational coefficients in the indeterminates $x_{1}, \dots, x_{n}$.

This action is given by declaring it on the indeterminates $$ \sigma(x_{i}) = x_{\sigma(i)}. $$ This is immediately verified to be an action on the set $\{ x_{1}, \dots, x_{n} \}$.

Then you extend it to the whole of $\Q[x_{1}, \dots, x_{n}]$ via the universal property of polynomials, and then also on rational functions $\Q(x_{1}, \dots, x_{n})$ - you have to prove it is well defined here, but that's easy enough. Here you let every constant be fixed by every permutation, so that in the end the action is just $$ \sigma\left(R(x_{1}, x_{2}, \dots, x_{n})\right) = R(x_{\sigma(1)}, x_{\sigma(2)}, \dots, x_{\sigma(n)}) $$ if $R(x_{1}, x_{2}, \dots, x_{n}) \in \Q(x_{1}, \dots, x_{n})$.

Once you have that, your answer is easily obtained as $$ \Sign(\sigma\tau) = \frac{P(x_{\sigma\tau(1)},x_{\sigma\tau(2)},\cdots, x_{\sigma\tau(n)})}{P(x_1,x_2,\cdots, x_n)} =\\= \frac{P(x_{\sigma\tau(1)},x_{\sigma\tau(2)},\cdots, x_{\sigma\tau(n)})}{P(x_{\sigma(1)},x_{\sigma(2)},\cdots, x_{\sigma(n)})} \cdot \frac{P(x_{\sigma(1)},x_{\sigma(2)},\cdots, x_{\sigma(n)})}{P(x_1,x_2,\cdots, x_n)} =\\= \sigma\left( \frac{P(x_{\tau(1)},x_{\tau(2)},\cdots, x_{\tau(n)})}{P(x_{1},x_{2},\cdots, x_{n})} \right) \cdot \frac{P(x_{\sigma(1)},x_{\sigma(2)},\cdots, x_{\sigma(n)})}{P(x_{1},x_{2},\cdots, x_{n})} =\\= \sigma(\Sign(\tau)) \cdot \Sign(\sigma) = \Sign(\tau) \cdot \Sign(\sigma),$$ as $\Sign(\tau) \in \{ 1, -1 \}$, and a constant is fixed by definition by any permutation.

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  • $\begingroup$ Very nice argument. I didn't think the action of $S_n$ on constant coefficients is trivial; but your answer shows how it is interesting. Thanks. $\endgroup$ – Groups Jan 9 '15 at 6:00
  • $\begingroup$ @Groups, thank, and you're welcome. $\endgroup$ – Andreas Caranti Jan 9 '15 at 13:59
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Hint:

The definition

$$ Sign(\sigma){}={}\dfrac{P(x_{\sigma(1)},x_{\sigma(2)},\cdots, x_{\sigma(n)})}{P(x_1,x_2,\cdots, x_n)} $$

is valid for any initial order of $x_1,\ldots,x_n\,$.

For example,

$$ \dfrac{(x_1-x_2)}{(x_2-x_1)}{}={}\dfrac{(x_2-x_1)}{(x_1-x_2)}{}={}-1\,. $$

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  • $\begingroup$ Nice answer. Thanks for the help. $\endgroup$ – Groups Jan 9 '15 at 6:04

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