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Nov 15, 2020 at 20:47 comment added Abdelmalek Abdesselam For the result about duality, for coincident root loci, see Corollary 2.7 of this article by Lee and Sturmfels arxiv.org/abs/1508.00202
Nov 15, 2020 at 20:44 comment added Abdelmalek Abdesselam Just a short complement: the description of singular loci for coincident root loci and in particular the discriminant is given in Thm 5.4 of the article by Chipalkatti arxiv.org/abs/math/0110224
Nov 15, 2020 at 19:10 comment added Malkoun This answer is great. I just wanted to point out an even shorter argument, along similar lines, provided to me by I.D. (I am not sure if he wants to be named or not). It is based on the statement that the discriminant hypersurface is the dual hypersurface of the Veronese curve. So a projective automorphism which leaves the discriminant hypersurface invariant also must preserve the Veronese curve. This is really neat.
Nov 15, 2020 at 19:06 vote accept Malkoun
Nov 15, 2020 at 16:10 comment added abx @YCor: I am using the projective set-up because I think it is more classical, you just have to pull back to $\operatorname{GL}(n)$. What you get is the standard representation of $\operatorname{SL}(2,\mathbb{C}) $ in $\operatorname{Sym}^n \mathbb{C}^2$. Indeed if $n$ is even the corresponding homomorphism $\operatorname{SL}(2)\rightarrow \operatorname{SL}(n) $ factors through $\operatorname{PGL}(2)$.
Nov 15, 2020 at 15:55 comment added YCor In $\mathrm{SL}(n,\mathbf{C})$ there are many copies of $\mathrm{(P)SL}(2,\mathbf{C})$ (finitely many up to conjugation), and in particular the irreducible one, which is unique up to conjugacy. Is this the irreducible copy? If $n$ is even, this is a copy of $\mathrm{SL}(2,\mathbf{C})$ rather than $\mathrm{PGL}(2,\mathbf{C})$.
Nov 15, 2020 at 15:43 history answered abx CC BY-SA 4.0