The weighted point extraction and its actual section ring

Proof. Serre’s construction gives a rank-two bundle $\mathcal N$ fitting in $$0\longrightarrow\mathcal O\longrightarrow\mathcal N(2) \longrightarrow\mathcal I_C(4)\longrightarrow0.$$ It has $c_1(\mathcal N)=0,c_2(\mathcal N)=1$. No quadric contains a smooth elliptic quintic in $\mathbb P^3$. Indeed, on a smooth quadric a curve of bidegree $(a,b)$ has degree $a+b$ and genus $(a-1)(b-1)$, which cannot be five and one simultaneously. On a quadric cone its strict transform on $\mathbb F_2$ is $as+5f$ and has genus $-a^2+5a-4$, again never one for integral $a$. The strict transform is isomorphic to the original smooth curve, since its map to that curve is finite and birational. A reducible quadric would force the curve into a plane, which is also impossible. Consequently $H^0(\mathcal N)=0$, and the normalized rank-two bundle $\mathcal N$ is stable. By the classification of stable rank-two bundles with these Chern classes [41, Chapter II, Lemma 4.3.2], it is a null-correlation bundle. Its projectivization is $Y$ above. The zero locus of the induced relative hyperplane section is $\operatorname{Bl}_C\mathbb P^3$. For clarity, let $\mathcal S=\mathcal O_{\mathbb P^3}(-1)$. The null-correlation bundle is $\mathcal S^\perp/\mathcal S$. On the flag variety let $\mathcal T=\Lambda/\mathcal S$ be the tautological line in this quotient, and put $\xi=c_1(\mathcal T^*)$. Since $\det(\Lambda)=\mathcal S\otimes\mathcal T$, we have $B=H+\xi$. The wedge of the section of $\mathcal N(2)$ with $\mathcal T$ defines a section of $\mathcal T^*\otimes\mathcal O(2H)$ (the determinant of $\mathcal N$ is trivial). Its zero divisor therefore has class $\xi+2H=B+H$. Since $-K_Y=2\xi+4H=2(H+B)$, adjunction gives $-K_X=H+B$. ◻

Proof. The displayed formula is obtained by taking the isotropic plane spanned by $(1,x,y,z)$ and $(0,1,t,y-xt)$, and putting $a=y-xt$, $b=xa-z$. The bilinear sections of $H+B$ have weight-zero space spanned by $1$, weight-one space spanned by $x,t$, weight-two space spanned by $a,xt$, and weight-three space spanned by $b,xa,at+xt^2$. Both projection fibres pass through the point and lie on $X$, so its constant and linear $x,t$ terms vanish. Write the equation as $F=F_0(t,a,b)+xF_1(t,a,b)$. The exceptional divisor of the second projection is locally $F_1=0$. The double intersection with the first projection fibre says $(F_1)_t(0)=0$. The triple intersection of the second fibre with the first exceptional divisor is read from $$Q|_{t=a=b=0}=(F_0)_t+x((F_1)_t-(F_0)_a) -x^2((F_1)_a+(F_0)_b)-x^3(F_1)_b,$$ where all coefficients are evaluated at zero. It forces $(F_0)_a(0)=(F_1)_t(0)=0$ and $(F_1)_a(0)=-(F_0)_b(0)$. Smoothness of $X$ now implies $(F_0)_b(0)\ne0$. Thus all weights below three vanish, and if the weight-three form is $\alpha b+\beta xa+\gamma(at+xt^2)$ then $\beta=-\alpha$ and $\alpha\ne0$. The first exceptional divisor is cut out by $$Q=F_t-xF_a-x^2F_b.$$ Moreover $F_1(t,0,0)=\gamma t^2$: the constant and linear terms vanish, and the restriction has degree at most two. Its zero divisor is $E'|_\ell=2p$, so $\gamma\ne0$. Equivalently, the linear term of $Q|_X$ is $\gamma a$, which is nonzero because $E$ is smooth. Division by $\alpha$ proves the assertion. ◻

Proof. Retain the nonzero coefficient $\nu$. Write homogeneous coordinates $h_0,\ldots,h_3$ and $b_0,\ldots,b_4$. The flag incidence equations are $$-h_0b_2-h_1b_1+h_2b_0=0,\quad h_0b_3-h_1b_2+h_3b_0=0,$$ $$h_0b_4-h_2b_2+h_3b_1=0,\quad h_1b_4-h_2b_3-h_3b_2=0, \qquad b_0b_4=b_2^2+b_1b_3.$$ The extra equation is $h_3b_0-\nu h_2b_1=0$. On $h_0b_0\ne0$ it solves uniquely for $b$, giving affine three-space. There is no component in the boundary. Indeed, on $h_0=0,h_1\ne0$ we have $b_1=(h_2/h_1)b_0$, $b_2=(h_3/h_1)b_0$, so the extra equation is $(h_3-\nu h_2^2/h_1)b_0=0$. The first branch has a one-dimensional base and one-dimensional fibres; the second has a two-dimensional base and a unique fibre point. Both therefore have dimension two. On $h_0=h_1=0$ the base is a line and fibres have dimension at most one, again giving dimension at most two. On $b_0=0,h_0\ne0$, the extra equation forces $h_2=0$: otherwise $b_1=0$ and the incidence equations force all $b_i=0$. The remaining base is a plane, and the fibre point with $b_0=0$ is unique. Thus this boundary also has dimension at most two. As a Cartier divisor on smooth $Y$, $X_0$ is Cohen–Macaulay; its dense smooth affine three-space and absence of boundary components show that it is integral.

The weights on the fundamental coordinates are $$(0,1,2,3)\quad\text{and}\quad(0,1,2,3,4).$$ They come from $\operatorname{diag}(1,\lambda,\lambda^2,\lambda^3)$ in the general symplectic group, of similitude multiplier $\lambda^3$, with a shift of the Plücker linearization. Its projective action preserves $Y$. Hence they give a grading on the section ring of $Y$. Let $R_Y$ be the diagonal section ring of $Y$. For every nonzero $g\in R_Y$, the smallest-weight form of $gF$ is $\operatorname{in}(g)f_0$: the ambient graded ring is a domain. Hence the initial ideal of $(F)$ is exactly $(f_0)$. This gives a flat degeneration of $R_Y/(F)$ to $R_Y/(f_0)$, with the same Hilbert function. Restriction and the divisor exact sequence identify the latter quotient with the section ring of the integral divisor $X_0$. It is therefore a domain, so localization at $h_0b_0$ is injective. On that chart its weights are precisely the weighted order in the regular parameters $x,t,a$. Therefore there are no further cancellations, and the associated graded ring is the claimed quotient. Restriction identifies its diagonal pieces with $H^0(X,m(H+B))$, by vanishing of the higher cohomology of $\mathcal O_Y((m-1)(H+B))$. ◻

Proof. The Weyl dimension formula gives $$\dim H^0(Y,\mathcal O_Y(aH+bB)) =\frac{(a+1)(b+1)(a+b+2)(a+2b+3)}6.$$ On the diagonal this is $(m+1)^4$. Weyl symmetry gives average weight $3a/2+2b$ under the shifted grading above. The defining equation has weight three. Hence, with $N_m=(m+1)^4-m^4$, the sum of the weights on the quotient diagonal piece is $$\frac{7m}{2}(m+1)^4- \left(3+\frac{7(m-1)}2\right)m^4 =\frac{7m}{2}N_m+\frac12m^4.$$ Divide by $mN_m$ and let $m$ tend to infinity. This gives $S_X(G)=7/2+1/8=29/8$. The discrepancy is $1+1+2=4$. ◻

Proof. On the initial divisor, the equation gives $b=xa-\nu t(a+xt)$. Therefore $h_3=\nu tA$ and $$b_2=A-xt,\qquad b_3=(x-\nu t)A-x^2t, \qquad b_4=A(A-xt-\nu t^2).$$ Multiplication of sections of dominant line bundles on the flag variety is surjective: by the Borel–Weil description it is a nonzero equivariant map onto the irreducible highest-weight representation of the sum of the weights. Thus the degree-$(m,m)$ sections are spanned by degree-$m$ products of each set of fundamental coordinates. Restriction to $X$ is surjective by the exact sequence for its divisor of class $H+B$ and vanishing for $(m-1)(H+B)$. Lemma 66.3 identifies its associated graded quotient with the integral initial divisor. Substituting the displayed formulas, and dividing $h_3$ by the nonzero constant $\nu$, proves the assertion. The dimension was computed in Lemma 66.4. ◻

Proof. The largest weights of the generators in $\mathcal H$ and $\mathcal B$ are three and four, respectively. Hence $j\le7m$. There is a nonzero degree-one section of every weight from zero to seven; their products show that every integral $j$ in this interval occurs at level $m$. If $i+h+2k=j\le m$, the monomial $x^i t^h A^k$ uses $i+k\le m$ factors from $\mathcal H$ and $h\le m$ factors from $\mathcal B$. Pad each product with the constant generator. Thus every polynomial of weight $j$ occurs. On the ray $(m,j)=(2k,2k)$ these are the complete spaces of the ample Cartier bundles $\mathcal O(2k)$. The ray is interior to $0<j<7m$, as required for refinement. ◻