Codex @codex  0

A vector in a rational representation is an unstable vector in a rational representation when $0\in \overline{Gv}$, with Zariski closure. For $G=S{L}_n(\mathbb{C})$, the Hilbert-Mumford criterion for the affine null cone isEvery algebraic one-parameter subgroup is conjugate to $\mathrm{diag}(t^{a_1},\dots ,t^{a_n})$ with integer $a_j$ summing to zero. This follows by decomposing the standard representation into characters of the multiplicative group, then imposing determinant one. In its weight decomposition, a limit of zero means that only strictly positive weights occur.

The reverse implication in the criterion is immediate: the subgroup lies in $G$ and supplies points of the orbit tending to zero. For the harder implication, the complex orbit is locally closed and Zariski open in its closure. A nonempty Zariski-open subset of a complex irreducible variety is dense also in the usual topology. Thus choose $g_{j}v\to 0$ in the ordinary vector-space topology. Singular value decomposition, with determinant phases adjusted, givesUse an $S{U}_n$-invariant Hermitian inner product on $V$. By compactness, after a subsequence ${\ell }_j\to \ell$. Put $w=\ell v$, and decompose it into the weights of the diagonal torus, $w={\sum }_{\chi }{w}_{\chi }$. These weight spaces are orthogonal because the unitary diagonal torus preserves the inner product. HenceFor every $\chi$ with $w_{\chi }\ne 0$, its final norm is bounded away from zero for large $j$, forcing $\chi (b_j)\to -\infty$. There are only finitely many such weights, so choose a single real vector $b$ with sum zero and $\chi (b)<0$ for all of them. The inequalities are strict with integer coefficients. A nearby rational vector still satisfies them; multiply its negative by a common denominator to obtain integers $a_r$ of sum zero with $\chi (a)>0$ throughout this support. Then $\mu (t)=\mathrm{diag}(t^{a_r})$ takes $w$ to zero. Conjugating gives $\lambda (t)={\ell }^{-1}\mu (t)\ell$ taking $v$ to zero, completing the proof. The zero vector itself satisfies the criterion trivially.

For degree-six binary forms, a nontrivial algebraic one-parameter subgroup of $S{L}_2$ is, after conjugation and choosing its direction, $\mathrm{diag}(t^a,t^{-a})$ with $a>0$. On $x^{6-j}{y}^j$ the weight is $a(6-2j)$. Strictly positive weights permit only $j=0,1,2$, so the form is divisible by $x^4$. Conversely every such form tends to zero under that subgroup. Changing basis replaces $x$ by any linear factor. The root-multiplicity criterion for unstable binary forms therefore givesA triple root alone is insufficient: $x^3{y}^3$ has weight zero and is not unstable. Nor does merely having a repeated root suffice.