While studying a larger problem I came across the question of which binary quartic forms have finite stabilizer. More precisely. Let's denote five-dimensional irreducible representations of $SL_2(\Bbbk)$ by $U_4$ ($\Bbbk$ is algebraically closed). $U_4$ can be associated with a space of binary forms of degree 4. And a $PSL_2(\Bbbk)$-action comes from natural action of $SL_2(\Bbbk)$. I would like to find points with a finite stabilizer. And probably to consider points not only from $U_4$, but also from $\mathbb{P}(U_4)$.
I want to learn about this, the purpose of this question is not to ask for an answer to the specific question. A reference/link to any resource (e.g. a textbook) which contains a systematic discussion on the subject from which I can start learning would be most appreciated. Thanks in advance!
I don't know a reference offhand.
A binary form $A \in U_n$ (that is, of degree $n$) is determined up to scale by the roots of $[A]$ in the projective line $\Bbb P^1(\Bbb K)$ and their multiplicities. If $A$ is nonzero, there are exactly $n$ roots up to multiplicity, and the possible multiplicities of the roots of $[A] \in \Bbb P(U_n)$ are given just by the (unordered) partitions of $n$.
Herein, let $(X, Y)$ denote the standard basis of $\Bbb K^2$ and $(x, y)$ its dual basis.
Example: Binary quartic forms In the case $n = 4$ the possible multiplicities are those that occur in the Petrov classification of Weyl tensors in general relativity, where they have been assigned special names. \begin{array}{cc} \textrm{partition} & \textrm{type} \\ \hline \ast & \textrm{O} \\ (4) & \textrm{N} \\ (3, 1) & \textrm{III} \\ (2, 2) & \textrm{D} \\ (2, 1, 1) & \textrm{II} \\ (1, 1, 1, 1) & \textrm{I} \end{array} Here, type $O$ refers to the zero form.
If for example, if $[A]$ has a quadruple root (type N), we can pick $g \in G := PSL(2, \Bbb k)$ that maps the sole root of $[A]$ to $[X]$, so that $$[g \cdot A] = [y^4]$$ and so $g \cdot A = \alpha y^4$ for some $\alpha \in \Bbb k - \{ 0 \}$. Since the stabilizers in $G$ of $A$ and $g \cdot A$ are conjugate by $g$, they have the same cardinality.
Now, if $h \in G$ stabilizes $A$, then it must map $X$ to $\beta X$, where $\beta^4 = 1$. By definition of $G$, $h$ must send $Y$ to $b X + \beta^{-1} Y$, where $b$ is arbitrary, and so $$\operatorname{Stab}_G(g \cdot A) = \left\{\left[\pmatrix{\beta & b \\ & \beta^{-1}}\right] : \beta^4 = 1, b \in \Bbb k\right\} ,$$ which is infinite. Since $\operatorname{Stab}_G(g \cdot A) \subseteq \operatorname{Stab}_G([g \cdot A])$, this latter stabilizer is infinite, too. (It turns out to be same as $\operatorname{Stab}_G(g \cdot A)$, except that $\beta$ is free to take on any nonzero value in $\Bbb k$.)
Likewise, for all other possible types except type I, there are at most three roots (ignoring multiplicity), so we find a $g \in G$ that maps $[A]$ to a chosen element of $\Bbb P(U_4)$ with the same multiplicities.
In the case where all four roots of $A$ are distinct (type I), we can use the action of $PSL(2, \Bbb k)$ to produce a normal form (this time with a parameter) by normalizing three of the roots to $[Y], [X + Y], [X]$. We can write the fourth root as $[\mu X + Y]$ for some $\mu \in \Bbb k - \{0, 1\}$, so $$[g \cdot A] = [x (x - y) y (x - \mu y)] .$$ (There is generically a threefold redundancy in $\mu$, but this still leaves infinitely many $G$-orbits.)
Remark The parameter $\mu$ is closely related to the cross ratio of projective geometry.