Define the discriminant by $$\Delta^2 = \prod_{i > j}(x_i -x_j)^2$$ Clearly $\Delta^2$ is a symmetric polynomial. My question is, why is the square root of of $\Delta^2$, i.e $$\Delta = \prod_{i > j}(x_i -x_j)$$ Not symmetric?
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2$\sqrt{a^2}$ is $|a|$, not $a$ as you seem to use. – Johan Löfberg Oct 12 '20 at 15:18
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Let's assume we only have $x,y$, so $\Delta^2 = (x-y)^2$, which is clearly symmetric since $(x-y)^2 = (y-x)^2$.
However, $\Delta = y-x = -(x-y) \ne x-y$, so $\Delta$ is not symmetric. That said, if you take the root carefully, you get $$ \Delta = \sqrt{\Delta^2} = \sqrt{(x-y)^2} = |x-y|, $$ that would be symmetric, as would be the more general $\prod_{i>j} |x_i-x_j|$. Of course, this would not really be a polynomial anymore.
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The transposition $(1 \ 2)$ changes $\Delta$ into $-\Delta$. More generally for any permutation $\sigma$ you have
$$\Delta(x_{\sigma(1)}, \dots , x_{\sigma(n)}) = \epsilon(\sigma)\Delta(x_1, \dots, x_n)$$ where $\epsilon(\sigma)$ is the sign of $\sigma$. Hence $\Delta(x_1, \dots, x_n)$ is not symmetric.
And for any number $a <0$, the square root of $a^2$ is not $a$ but $-a$.
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