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Using the printed convention that the projective points represent lines of binary quartics, the locus of GIT stable points isThe root at infinity is included. We determine stability by calculating every relevant algebraic one-parameter subgroup weight.

Use the natural action $(g\cdot f)(X,Y)=f(g^{-1}(X,Y))$. Every nontrivial algebraic one-parameter subgroup of $S{L}_2$ is conjugate to ${\lambda }_m(t)=\mathrm{diag}(t^m,t^{-m})$ for some integer $m>0$: its two integral weights of a representation sum to zero, and a basis of weight vectors can be rescaled to have determinant one. WriteIf $r=\min \{j:a_j\ne 0\}$, then $r$ is the multiplicity of a root at the projective point $[1:0]$: in the local coordinate $Y/X$, the polynomial begins with a nonzero multiple of $(Y/X{)}^r$.

The Hilbert-Mumford criterion for projective stability, with $\mu =-\min$(active weights), now givesBeing a GIT stable point requires this number to be strictly positive for every nontrivial subgroup. Conjugating the subgroup lets $[1:0]$ range over the entire projective line. Thus the root-multiplicity criterion for stable binary forms becomes $r_p(f)<2$ at every point $p$. A nonzero degree-four form over $\mathbb{C}$ has four projective roots counted with multiplicity, so this means precisely four distinct roots. This is equivalently nonvanishing of its discriminant.

The boundary cases show why a weak inequality would give the wrong answer. A triple or quadruple root can be moved to $[1:0]$, leaving only strictly positive weights; its affine representative tends to zero under that subgroup, so it is unstable. If all root multiplicities are at most two, all Hilbert-Mumford values are nonnegative and the point is a GIT semistable point. A double root gives a zero value for a suitable subgroup, so never represents a GIT stable point. More concretely, after putting that double root at $[1:0]$ one hasand ${\lambda }_1(t)\cdot f\to a{X}^2{Y}^2$. When there are three distinct roots, this limiting form, representing a GIT semistable point with two distinct roots, is outside the original group orbit, so that group orbit is not closed in the semistable locus. When there are two double roots, a change of coordinates gives $X^2{Y}^2$, which has the positive-dimensional diagonal torus as stabilizer. Neither case represents a GIT stable point.

For a form with four distinct roots, the projective stabilizer is finite: it permutes the four roots, and a Möbius transformation fixing three distinct points is the identity. Thus its image in $PG{L}_2$ injects into the finite permutation group on the roots. The kernel of a group homomorphism for $S{L}_2\to PG{L}_2$ is $\{\pm I\}$, also finite. This agrees with the finite-stabilizer condition in geometric invariant theory. The weight calculation already supplies the required strict stability condition for every conjugate subgroup.