Prove $\operatorname{ sgn}(\sigma)=\frac{D(X_{\sigma(1)},\cdots,X_{\sigma(n)})}{D(X_1,\cdots,X_n)}$

Question

Q. The polynomial $D(X_1,\cdots,X_n)$ with n variables $X_1,\cdots,X_n$ is defined as follows: $$D(X_1,\cdots,X_n)\stackrel{\mathrm{def}}{=}\displaystyle\prod_{1\leq i<j\leq n}(X_i-X_j).$$

Then show that the following holds for any source of the $n$th order symmetry group $\mathfrak{S}_n$:

$$\displaystyle \operatorname{sgn}(\sigma)=\frac{D(X_{\sigma(1)},\cdots,X_{\sigma(n)})}{D(X_1,\cdots,X_n)}.$$

I know that when $\operatorname{sgn}(\sigma)=1$, there are an odd number of odd substitutions where the terms in the denominator and numerator do not match, but how do I prove this?

Answer 1

Outline of proof:

For $\sigma \in \mathfrak S_n$, denote $$ Q_\sigma = \frac{D(X_{\sigma(1)},\cdots,X_{\sigma(n)})}{D(X_1,\cdots,X_n)}. $$ Note that for $\sigma, \tau \in \mathfrak S_n$, we have $Q_\sigma Q_\tau = Q_{\sigma \tau}$. Because of this and the fact that the symmetric group is generated by transpositions, it suffices to show that $Q_\sigma = -1$ whenever $\sigma$ is a transposition.

Suppose that $Q_\sigma$ is the transposition that switches the elements $p,q$, where $1 \leq p < q \leq n$. If we compare the numerator of $Q_\sigma$ to its denominator, we can see that there are three kinds of terms that (might) appear in the numerator that do not appear in the denominator:

• $(X_p - X_j)$, where $p < j < q$,
• $(X_j - X_q)$, where $p < j < q$,
• $(X_p - X_q)$.

Argue that the total number of such terms (that appear) is odd.

Note: We could make this proof a bit quicker if we note that $\mathfrak S_n$ is generated by the set of transpositions that exchange $p$ and $p + 1$ for $1 \leq p \leq n-1$.