Closed orbit of $\operatorname{PGL_2} K$ acting on $\mathbb{P}(\operatorname{Sym}^4 V)$, $\operatorname{dim} V = 2$ The standard action of $\operatorname{GL_2} K$ on $V$ induces an action of $\operatorname{PGL_2} K$ on $\mathbb{P}(\operatorname{Sym}^4 V)$. So far, I understood how all the orbits can be obtained via the $j$-function, but I'm struggling to see some algebraic properties of the more obvious orbits.
Specifically, define
$C = \{ [v^4] | v \in V\}$
$\Sigma = \{ [v^3\cdot w] | v, w \operatorname{independent} \in V\}$
$\Phi = \{ [v^2\cdot w^2] | v, w \operatorname{independent} \in V\}$
$\Psi = \{ [v^2\cdot w \cdot u] | v, w,u \operatorname{pairwise independent} \in V\}$
Now these are orbits because three pairwise independent vectors are projectively equivalent in $\mathbb{P}^1$, and C is a closed orbit because it can be described as a rational normal curve of degree 4.
Supposedly, this is the only closed orbit of all the above. How, for instance, can I see that $\Sigma$ is not closed in the Zariski topology? (So far, I have never computed why some set is NOT a variety)
I'd appreciate any help.
Edit: K is assumed to be algebraically closed.
 A: I can show every orbit other than $C$ is not Zariski closed. Actually, for an arbitrary point $p\in\mathbb{P}(\operatorname{Sym}^4 V)$, if $\Omega_p$ denotes its orbit then I'm going to prove the stronger statement that $\overline{\Omega_p}\cap C\neq\varnothing$.
Let $\{x,y\}$ be a basis for $V$. Then \begin{equation*} p=(a_1x+b_1y)\cdots(a_4x+b_4y)=c_4x^4+c_3x^3y+c_2x^2y^2+c_1xy^3+c_0y^4\end{equation*} for some $a_i,b_i,c_i\in K$, with the $c_i$ not all zero. Fix $t\in K$.  Consider the map \begin{align*} \sigma_t:\mathbb{P}^1&\to\mathbb{P}(\operatorname{Sym}^4 V)\\ [a:b]&\mapsto c_4a^4x^4+c_3a^3x^3(tx+by)+c_2a^2x^2(tx+by)^2+c_1ax(tx+by)^3+c_0(tx+by)^4\end{align*} which is rational end everywhere regular.  We have \begin{equation*}
\sigma_t[1:0]=(c_4+c_3t+c_2t^2+c_1t^3+c_0t^4)x^4.
\end{equation*} Since $K$ is infinite, we can take $t\in K$ such that $c_4+c_3t+c_2t^2+c_1t^3+c_0t^4\neq 0$, and then $\sigma_t[1:0]\in C$. For all $[a:b]\in\mathbb{P}^1\setminus\{[1:0],[0:1]\}$ the matrix \begin{equation*}A_t=\begin{pmatrix} a &t\\ 0 &b\end{pmatrix}\in\operatorname{PGL_2} K\end{equation*} satisfies $A_t(p)=\sigma_t[a:b]$. Therefore $\sigma_t[a:b]\in\Omega_p$ for all $[a:b]\neq [1:0],[0:1]$. Since $\sigma_t:\mathbb{P}^1\to\sigma_t(\mathbb{P}^1)$ is an isomorphism (in particular a homeomorphism), we have \begin{equation*} \sigma_t[1:0]\in\overline{\Omega_p}\cap C.\end{equation*} This proves that $C$ is the only closed orbit. I think a refinement of this argument might show that $C$ is contained in the closure of every orbit. EDIT: indeed, just take $[v^4]\in C$ and do the same for a basis $\{v,w\}$ instead of $\{x,y\}$. Therefore $C$ is contained in the closure of every orbit.
