What is the subgroup of $\mathrm{SL}(n,\mathbb{C})$ which preserves the discriminant?

Question

$\DeclareMathOperator{\SL}{\operatorname{SL}}$Let $\mathcal{P}_{n-1}$ be the space of complex polynomials in one variable, say $z$, of degree at most $n-1$. As a complex vector space, it is clearly $n$-dimensional. Consider the basis $1$, $z,\ldots,z^{n-1}$ of $\mathcal{P}_{n-1}$. This allows us to identify $\mathcal{P}_{n-1}$ with $\mathbb{C}^n$.

The complex Lie group $\SL(n,\mathbb{C})$ acts on $\mathbb{C}^n$, and thus, via our identification, it acts on $\mathcal{P}_{n-1}$. The discriminant is a homogeneous polynomial on $\mathcal{P}_{n-1}$ of degree $2n-2$.

My question can now be formulated. What is the subgroup of $\SL(n,\mathbb{C})$ which preserves the discriminant (using our identification of $\mathcal{P}_{n-1}$ with $\mathbb{C}^n$)? I have a feeling it is the image of $\SL(2,\mathbb{C})$ under a principal homomorphism from $\SL(2,\mathbb{C})$ to $\SL(n,\mathbb{C})$.

More specifically, $\SL(2,\mathbb{C})$ acts simultaneously on all linear factors of a polynomial of degree $n-1$, and this action preserves the discriminant. I think these are all "projective" transformations which preserve the discriminant, but I am not sure how to show that. Edit: I should really write that I think these are all transformations in $\SL(n,\mathbb{C})$ which preserve the discriminant (see the remark by @NoamD.Elkies below).

Edit 2: here is a conceptual proof for polynomials of degree at most $2$ (i.e. for $n=3$ using my notation). Let $L$ be the map on $\mathbb{C}P^2$ induced by some element $g \in \SL(3,\mathbb{C})$ which preserves the discriminant of quadratic polynomials. Note that we can think of $\mathbb{C}P^2$ as the set of unordered pairs of points on $\mathbb{C}P^1$ (namely the roots of the corresponding polynomial, viewed up to scaling). Note also that polynomials of vanishing discriminant correspond to an unordered pair of coinciding points on $\mathbb{C}P^1$.

Let $f$ be the holomorphic map from $\mathbb{C}P^1$ to $\mathbb{C}P^1$ obtained essentially by restricting $L$ to coinciding pairs of points on $\mathbb{C}P^1$, in turn corresponding to the vanishing locus of the discriminant. Note that $f$ induces a holomorphic map $\tilde{f}$ on the set of unordered pairs of points on $\mathbb{C}P^1$, and thus on $\mathbb{C}P^2$.

Moreover, $\tilde{f}$ and $L$ agree on the conic in $\mathbb{C}P^1$ coinciding to the vanishing locus of the discriminant, and thus they must be equal, since they are two linear isomorphism maps from $\mathbb{C}P^2$ to itself which agree on a conic in $\mathbb{C}P^2$.

We have thus shown the claim for $n=3$. I am hoping this can be generalized to higher $n$'s. I think it is doable.

Edit 3: the previous argument can be generalized using an additional hypothesis, that the element $g \in \SL(n,\mathbb{C})$ not only preserves the discriminant, but also maps any polynomial with a single root of multiplicity $n-1$ to a polynomial with a single root of multiplicity $n-1$. Under this additional hypothesis, essentially the same argument proves that $L$, which is the map on $\mathbb{C}P^{n-1}$ induced by $g$, is induced by some holomorphic automorphism of $\mathbb{C}P^1$, and thus by some element of $\SL(2,\mathbb{C})$. But is it necessary to make this assumption? Is my conclusion false without it? Or is it perhaps the case that this hypothesis can be proved?

Answer 1

At the request of the OP, I post my comment as an answer. View $\mathcal{P}_n$ as the space of homogeneous polynomials of degree $n$ in 2 variables. Let $\Delta _p\subset \mathcal{P}_n$ be the locus of polynomials with one linear factor of multiplicity $\geq p$. One can show that the singular locus of $\Delta _p$ is $\Delta _{p+1}$. Therefore the subgroup $G$ of $\operatorname{GL}(\mathcal{P}_n) $ preserving the discriminant hypersurface $\Delta _2$ preserves $\Delta _n$. Now the image of $\Delta _n$ in $\mathbb{P}(\mathcal{P}_n)\cong \mathbb{P}^n$ is a rational normal curve, that is, the image of the $n$-th Veronese embedding $V_n:\mathbb{P}^1\hookrightarrow \mathbb{P}^n$. Thus up to homotheties, $G$ is the group of automorphisms of $\mathbb{P}^n$ preserving $V_n(\mathbb{P}^1)$, which maps isomorphically to $\operatorname{Aut}(\mathbb{P}^1)=\operatorname{PGL}(2,\mathbb{C}) $.