Unstable vector in a rational representation ($0\in \overline{Gv}$)

A vector is unstable if zero lies in the Zariski closure of its orbit. For complex $S{L}_n$, this is equivalent to an algebraic one-parameter subgroup taking it to zero as $t\to 0$. In the subgroup's weight decomposition, exactly the positive-weight components can occur in such a vector.

For complex $S{L}_n$, choose a sequence taking an unstable vector to zero and apply singular value decomposition to each group element. Compactness gives a limiting right-unitary factor. Every nonzero torus-weight component of the transformed vector forces a strict linear inequality on the diagonal exponents. A rational solution of these finitely many strict inequalities scales to integral exponents with sum zero, producing an algebraic one-parameter subgroup that takes the vector to zero. The converse follows immediately from the subgroup limit.

For $\mathrm{diag}(t,t^{-1})$, the monomial $x^{d-i}{y}^i$ has weight $d-2i$. Positive weights force divisibility by $x^{\lfloor d/2\rfloor +1}$. Conjugating the subgroup replaces $x$ by any nonzero linear factor. Thus a nonzero binary form is unstable exactly when a projective root has multiplicity greater than half its degree. Equality gives zero weight and does not imply instability.