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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.

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    It looks like $C\subset \overline{\Sigma}$. A vector is linearly independent from most "nearby" vectors, so the limit $w\to v$ (usual topology) forces $C$ into $\overline{\Sigma}$. The fact that we use Zariski topology does not really change that. The points of $Sym^4V$ can be viewed as homogeneous quartic polynomials in two unknowns (=the basis vectors of $V$). Such a polynomial is in $\Sigma$, iff it has a triple zero. But a polynomial with a quadruple zero will pass any polynomial equation checking the presence of a triple zero. I need to think about the details, so this is not an answer.2011-11-20
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    Thanks; $C\subset \overline{\Sigma}$ is indeed true according to the book (Harris, A First Course), also $C\subset \overline{\Phi}$ holds. They are both contained in the closure of $\Psi$, though I obviously did not check on that :)2011-11-20

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