Hilbert–Mumford inequalities for smooth pairs

Proof. Bertini’s theorem gives a nonsingular member $A$: outside the base locus apply Bertini, while on the base locus every nonzero linear combination of the two gradients is nonzero. Write a second member as $B$, and let $T=A^{-1}B$, which is self-adjoint for the nondegenerate symmetric form $A$. A repeated eigenvalue would give an eigenvector $v\ne0$ with $A(v,v)=0$. If its eigenspace has dimension at least two, choose an isotropic vector in that eigenspace. If it has dimension one, a Jordan chain gives $v=(T-c)w$ and hence $A(v,v)=A(w,(T-c)v)=0$. Then $B(v,v)=0$ and $Bv=cAv$, producing a singular point of $V$. Thus $T$ has six distinct eigenvalues. This proves the assertions about the determinant and ranks.

Conversely, a pencil with six distinct determinant zeros has a nonsingular member $A$ and an $A$-orthogonal eigenbasis for $A^{-1}B$. In this basis its equations are $\sum y_i^2=\sum c_i y_i^2=0$ with the $c_i$ distinct. Dependent gradients would force a point to be supported in a single coordinate, which cannot satisfy the first equation. Thus its intersection is smooth as well.

A projective transformation preserving the pencil permutes its six singular members. The kernel of this action fixes every member of the pencil. After rescaling a linear representative, it preserves $A$ and $B$ and hence commutes with $T$. In an $A$-orthogonal eigenbasis of $T$ it is a diagonal sign matrix. Thus the kernel is finite, and the action on the pencil has finite image. ◻

Proof. Being adapted means that the initial forms of the basis give a basis of the associated graded pencil. Thus the initial Plücker coordinate has weight $d_1+d_2$, proving the last equality.

Suppose first that $d_1\ge0$. Every monomial of both quadrics then has nonnegative weight. Every term in their determinant polynomial has total weight $2\sum a_i=0$. Consequently that determinant is unchanged if both quadrics are replaced by their weight-zero parts. The latter therefore form a pencil with six distinct determinant zeros. By the same elementary linear-algebra criterion used in Lemma 58.1, its intersection is smooth. The one-parameter subgroup fixes both weight-zero quadrics. This contradicts the finiteness of its projective stabilizer. Hence $d_1<0$.

If $Q_2$ is nonsingular, a nonzero determinant term gives $6d_2\le0$, which suffices. If $Q_2$ is singular, its zero in the determinant sextic is simple. The coefficient of $s$ in $\det(Q_2+sQ_1)$ is therefore nonzero. Each of its terms uses one matrix entry of $Q_1$ and five of $Q_2$, and again has total coordinate weight zero. Thus $d_1+5d_2\le0$. ◻

Proof. Choose an eigenbasis within each weight space. Row reduction relative to the weight filtration gives a unipotent transformation $u$ such that $$u\Pi=\lim_{t\to0}\lambda(t)\Pi, \qquad \lim_{t\to0}\lambda(t)u\lambda(t)^{-1}=1.$$ Concretely, a basis of the plane consists of pivot vectors plus terms of higher vector weight, and $u$ removes those terms. Such a transformation preserves initial vectors in every representation. Equivalently, $u$ belongs to the unipotent radical of the parabolic associated to $\lambda$. It therefore preserves the limits and fibre weights of the three linearized bundles. Replace $(V,C)$ by $(uV,uC)$. The limiting plane is invariant; a change of basis inside the weight spaces makes it a coordinate plane. ◻

Proof. The plane is a subspace of $E$, whose vector weights are the negatives of the coordinate weights. Thus $\det\mathcal S^*$ has weight $\sigma$. The weight of $\mathcal O_{\mathcal H}(1)$ is $-b$ by its definition. The matrix of $q$ is nonsingular. A nonzero determinant term uses three entries, each with weight at least $b$, and its total coordinate weight is $2\sigma$. This proves $3b\le2\sigma$.

The plane is not contained in $V$. Indeed, on such a plane the normal derivatives of two defining quadrics would give an everywhere surjective map $\mathcal O_{\mathbb P^2}^{\oplus3}\to\mathcal O_{\mathbb P^2}(1)^{\oplus2}$ if $V$ were smooth. Its kernel would be $\mathcal O_{\mathbb P^2}(-2)$, contradicting the equality of second Chern classes in this exact sequence. Therefore some member of $P$ restricts to a nonzero scalar multiple of $q$. Every element of $P$ has least weight at least $d_1$. Restricting to an invariant coordinate plane only deletes monomials, so $b\ge d_1$. ◻

Proof. Apply the preceding lemmas. From $3b\le2\sigma$ and $b\ge d_1$ we get $$20\sigma-6b\ge24b\ge24d_1.$$ Moreover, $d_1+d_2\le4d_1/5$. Substitution gives $$\mu_{\mathcal L}\ge24d_1-\frac{264}{5}d_1 =-\frac{144}{5}d_1>0.$$ The line bundle is ample by Lemma 57.5, so the Hilbert–Mumford criterion applies. ◻