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The affine null cone is the set of vectors whose orbit closure contains the origin. For a reductive algebraic group it is also the common zero locus of all positive-degree homogeneous invariants. A nonzero vector in this set represents an unstable projective point. The Hilbert-Mumford criterion for the affine null cone detects membership by a algebraic one-parameter subgroup with only positive active weights.

For a complex reductive algebraic group acting linearly on $V$, decompose a nonzero vector into weights for each algebraic one-parameter subgroup. With the displayed sign convention, semistability is equivalent to $\mu \ge 0$ for every such subgroup, and stability is equivalent to strict positivity for every nontrivial subgroup. One must test conjugates of the subgroups as well as those diagonal in a single fixed coordinate system.

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 conjugation of $n$ by $n$ complex matrices by the special linear group, the affine null cone consists exactly of the nilpotent matrices. If a conjugate tends to zero along an algebraic one-parameter subgroup, its constant characteristic polynomial must be $T^n$. Conversely, put a nilpotent matrix in strictly upper triangular form and conjugate it by $\mathrm{diag}(t^{n-1},t^{n-3},\dots ,t^{1-n})$. Every active matrix entry has positive weight and tends to zero. The Hilbert-Mumford criterion for projective stability therefore identifies the semistable locus of the projective conjugation representation as the complement of the projectivized nilpotent cone.

Use the natural linearization of a line bundle on ${\mathcal{O}}_X(1)$ induced by the conjugation representation on $M_n\mathbb{C}$. The notation involving the dual does not change this matrix description: the nondegenerate pairing $(A,B)\mapsto \mathrm{tr}(AB)$ identifies the representation with its dual equivariantly. Let $[A]$ be a projective point, so $A\ne 0$.

The Hilbert-Mumford criterion for projective stability says that $[A]$ is not a semistable projective point in geometric invariant theory precisely when some algebraic one-parameter subgroup $\lambda :{\mathbb{G}}_m\to S{L}_n$ sends its affine representative to zero:Equivalently, all active weights for that subgroup are strictly positive. With the convention $\mu ([A],\lambda )=-\min \{\text{active weights}\}$, semistability requires $\mu \ge 0$ for every subgroup, including conjugates of diagonal ones.

Suppose such a zero limit exists. The coefficients $c_j(A)$ inare invariant under conjugation. Continuity along the zero limit gives $c_j(A)=c_j(0)=0$ for every $j$. By the Cayley-Hamilton theorem, $A^n=0$: the matrix is nilpotent.

Conversely, if $A$ is a nonzero nilpotent matrix, choose a basis in which it is strictly upper triangular, for example a Jordan normal form. The change of basis can be chosen in $S{L}_n$: multiply any invertible change-of-basis matrix by a scalar with the required $n$th power, which does not affect its conjugation. In this basis takeThe exponents sum to zero, so this is an algebraic one-parameter subgroup of the special linear group. The $(i,j)$ matrix entry has weightEvery potentially nonzero entry has $i<j$ and therefore positive weight. Thus the conjugated matrix tends to zero. Conjugate this subgroup back to the original basis to obtain the required destabilizing subgroup for $A$.

It follows that the nilpotent cone for conjugation of matrices is exactly the affine null cone, andEach $c_j$ is a homogeneous invariant of degree $j$, so the second description also gives invariant open sets witnessing semistability. Equivalently, a representative has at least one nonzero eigenvalue. Nonzero singular matrices can be semistable; invertibility is not the criterion. For example, a diagonal matrix with diagonal entries $1,0,\dots ,0$ is semistable, while a nonzero strictly upper triangular matrix is unstable.

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.

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.

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.