The D5 point and the cubic section ring

Proof. The blow-up resolves the pencil generated by the two equations of $\Pi$. Its graph is a closed subvariety of $V\times\mathbb P^1$, and $H+A$ is the restriction of $\mathcal O_V(1)\boxtimes\mathcal O_{\mathbb P^1}(1)$, hence is ample. The blow-up formula gives $-K_X=2H-E_0=H+A$. Since $N_{C/V}=\mathcal O_C(1)^{\oplus2}$, the intersection numbers are $H^3=3$, $H^2E_0=0$, $HE_0^2=-3$, and $E_0^3=-6$. Expanding $(2H-E_0)^3$ gives $12$.

For normality, the Gauss map of $V$ is defined by its five partial derivatives, which have no common zero on $V$. Its pullback of the hyperplane class is $\mathcal O_V(2)$, an ample class. It is therefore finite: a positive-dimensional fibre would contain a curve on which this class has degree zero. The singular locus of a hyperplane section is contained in a fibre of the Gauss map, so it is finite. A reducible or nonreduced cubic surface in $\mathbb P^3$ has positive-dimensional singular locus; hence our section is integral. As a hypersurface with only isolated singularities, it is normal. The curve $C$ is Cartier on each such section, so its blow-up is the same surface; this identifies it with the fibre of $\varphi$. ◻

Proof. We derive the surface equations directly. At a Du Val point of type $D_5$ or $E_6$, the local analytic equations can be written $x^2+yz^2+y^4=0$ and $x^2+y^3+z^4=0$, respectively. In particular, the quadratic tangent cone has rank one. Put $P=[1:0:0:0]$ on its hyperplane and write its cubic equation as $x_0x^2+f_3(x,y,z)=0$. A linear change of $x_0$ removes the terms of $f_3$ divisible by $x^2$. Thus the equation is $$x_0x^2+x(\alpha y^2+\beta yz+\gamma z^2)+c_3(y,z)=0.$$ The restriction of the cubic jet to the kernel of the quadratic tangent cone is invariant, up to a linear change and a scalar, under analytic coordinate changes: the contribution of a quadratic coordinate change to the cubic jet is divisible by $x$. The analytic equations just displayed therefore allow us to take $c_3=yz^2$ in type $D_5$, and $c_3=y^3$ in type $E_6$.

In the first case take new coordinates $Y=y+\gamma x$, $Z=z+(\beta/2)x$, removing the resulting $x^2$-multiples by changes of $x_0$. After renaming $Y,Z$ as $y,z$, the equation becomes $x_0x^2+\alpha xy^2+yz^2=0$. If $\alpha=0$, the line $x=z=0$ is singular, contrary to the isolated singularity at $P$. Rescaling gives the asserted $D_5$ form.

In the second case $\gamma\ne0$, since otherwise $x=y=0$ is a singular line. Taking the new coordinate $Z=z+\beta y/(2\gamma)$ diagonalizes the binary quadratic form. Taking a further new coordinate $Y=y+\alpha' x/3$, where $\alpha'$ is its remaining $y^2$ coefficient, removes the $xy^2$ term; the new $x^2$-multiples are again absorbed in $x_0$. Rescaling $z$ gives $x_0x^2+xz^2+y^3=0$. These equations agree with [7] and with entries $17^\mathrm{o}$ and $14^\mathrm{o}$ of [9, Table 3].

Extend these coordinates by the fibre equation $t=0$. Since $\Pi$ avoids $P$, its second equation has nonzero $x_0$ coefficient and has the stated form $\ell$. Replacing $x_0$ by $\ell$ adds only $x^2$ times a linear form in $x,y,z$. All derivatives along $t=0$ vanish at $P$, so smoothness of $V$ forces the $t\ell^2$ coefficient to be nonzero; rescale $t$ to make it one. ◻

Proof. The ideal $\mathcal I_C=(\ell,t)$ is a complete-intersection ideal on $V$. For $m\ge1$ its power has the resolution $$0\longrightarrow\mathcal O_V(-m-1)^{\oplus m} \longrightarrow\mathcal O_V(-m)^{\oplus(m+1)} \longrightarrow\mathcal I_C^m\longrightarrow0.$$ After twisting by $2m$, the first cohomology of the left term vanishes: this follows from the hypersurface sequence for $V\subset\mathbb P^4$ and the vanishing of intermediate cohomology of line bundles on $\mathbb P^4$. Hence $H^0(V,\mathcal I_C^m(2m))$ is generated by $\ell^{m-q}t^q$ times homogeneous degree-$m$ polynomials on $V$.

Adjoining $a=\ell/e$, $b=t/e$, and $e$ to the homogeneous coordinate ring and eliminating $\ell,t$ gives (8.1). There are no further relations: after inverting $e$, the quotient becomes the coordinate ring of $V$ with $e^{\pm1}$ adjoined. The displayed relation is irreducible after localization and is not divisible by $e$, so its principal ideal is already saturated with respect to $e$. The generated degree piece is therefore $H^0(V,\mathcal I_C^m(2m))=H^0(X,mL)$. The case $m=0$ is immediate. Solving the two degree equations gives the exponents of $a,e$ and proves (8.2). ◻

Proof. Each $\Psi$ is primitive and linear in $b$: its coefficient $e^3a^2$ is coprime to its constant term. Hence $R_0$ is a domain. For a principal ideal in a polynomial ring, the initial ideal for a weight filtration is generated by the initial form of the generator, because initial forms multiply. Consequently $R_0$ is the weight-associated graded ring of $R$. In bidegree $(2m,-m)$ the polynomial filtration is finite: (8.2) bounds every monomial weight by $2dm$. Thus passage from the graded quotient to a basis of the original section space involves no completion or infinite reduction.

It remains to identify this algebraic filtration with the geometric valuation. The completed local ring at $P$ is $\mathbb C[[x/\ell,y/\ell,z/\ell]]$. The formal implicit-function theorem, with coefficient one at $t/\ell$, gives $t/\ell=-g+\text{terms of weight greater than }d$; this also follows recursively in the positive weighted filtration. Hence the weights assigned to all generators agree with their actual valuation orders. Localize at $ae$ and set $x'=x/(ea)$, $y'=y/(ea)$, $z'=z/(ea)$ and $b'=b/a$. Then $$R_0[(ae)^{-1}]= \frac{\mathbb C[x',y',z',b']}{(b'+g(x',y',z'))}[a^{\pm1},e^{\pm1}],$$ where $g=x'^2+x'y'^2+y'z'^2$ or $g=x'^2+x'z'^2+y'^3$. The degrees of $a,e$ form a basis of $\mathbb Z^2$, so a fixed bidegree has a common Laurent factor in them. After removing it, setting $a=e=1$ is injective on that homogeneous component. Localization itself is injective because $R_0$ is a domain. Finally, eliminating $b'$ gives $\mathbb C[x',y',z']$ with the stated native weights, which is the associated graded ring of the monomial valuation in the regular parameters. Thus no nonzero global initial form disappears after passage to the local valuation. This proves the assertion, including possible cancellation among global sections. ◻

Proof. The quotient $R_0/(b)$ is a domain: its relation is primitive and linear in $a$, with coefficient $ex^2$ coprime to $xy^2+yz^2$ or $xz^2+y^3$. Thus $(b)$ is prime and contains neither $a$ nor $e$. For every $q\ge0$, (8.5)$$\label{f25:eq:saturation} (b^q):(ae)^\infty=(b^q).$$ In fact, if $(ae)^Nh\in(b^q)$, reduction modulo $b$ gives $h=bh_1$; cancellation in the domain and induction give $h\in(b^q)$.

Now we prove that for $s\in W_{m,j}$, $\mathop{\mathrm{ord}}_B(s)\ge q$ if and only if the corresponding element of $R_0$ is in the ideal $(b^q)$. After localizing at $ae$, we have $b'=b/a=-g(x',y',z')$. Viewing $g$ as a monic quadratic in $x'$, the calculation in Example 8.3 shows that its discriminant is not a square. Thus $g$ is irreducible, and the cone $(g=0)$ is reduced and irreducible. This cone is not contained in any coordinate hyperplane, so all three coordinates are nonzero on a dense open subset. Since the original weight triple given in (8.3), either $(4,2,3)$ or $(6,4,3)$, has greatest common divisor one, the weighted projective quotient has trivial stabilizer on this open subset. It follows that the vanishing order along $B$ equals the vanishing order along $(g=0)$ on the cone.

Fix $s\in W_{m,j}$, and let $f\in R_0$ be its homogeneous representative under Lemma 8.5. After localizing at $ae$ and removing the common Laurent factor in $a,e$, write its representative as $\bar f(x',y',z')$. The preceding geometric identification gives $$\mathop{\mathrm{ord}}_B(s)\ge q\quad\Longleftrightarrow\quad g^q\mid\bar f\quad\Longleftrightarrow\quad f\in(b^q)R_0[(ae)^{-1}].$$ For the last equivalence, we use $b=-ag$ and the invertibility of $a$. Clearing denominators, the last condition means that $(ae)^Nf\in(b^q)$ for some $N\ge0$. Equation (8.5) then gives $f\in(b^q)$. Consequently, $$\{s\in W_{m,j}:\mathop{\mathrm{ord}}_B(s)\ge q\}\simeq(b^q)\cap(R_0)_{(2m,-m),j},$$ where the subscripts on the right specify the bidegree and weight. Thus the geometric filtration by vanishing along $B$ agrees with the $b$-adic filtration on each homogeneous component. ◻

Proof. In type $D_5$ take lexicographic order $z>y>x>b>a>e$, so $\Psi_D$ has leading term $yz^2$. In type $E_6$ take $y>x>z>b>a>e$, so $\Psi_E$ has leading term $y^3$. Standard monomials are the monomials in $\mathcal P$ that are not divisible by this leading coefficient. They respect both bidegree and valuation weight. Since the leading term of $\Psi$ is monic and contains no $b$, the quotient $R_0=\mathcal P/(\Psi)$ is a free $\mathbb C[b]$-module on the $b$-free standard monomials. The resulting basis is adapted to the filtration by powers of $b$, and hence, by Lemma 8.6, to the filtration by vanishing order along $B$. ◻

Proof. For the polynomial ring $\mathcal P$, let $D_m(s)=\dim\mathcal P_{(s,-m)}$ and let $T_m(s)$ be the sum of the weights of all monomials of bidegree $(s,-m)$ in $\mathcal P$. If $h$ is the total exponent of $a,b$, then the exponent of $e$ is $h-m$, and the total exponent of $x,y,z$ is $s-h$. For fixed $h$, there are $h+1$ choices in $a,b$ and $\binom{s-h+2}{2}$ choices in $x,y,z$. Among the monomials in $\mathcal P_{(s,-m)}$ whose exponents of $a$ and $b$ sum to $h$, the mean weight is $w(s-h)/3+dh/2$, by symmetry. Thus (8.7)$$\begin{aligned} D_m(s)&=\sum_{h=m}^s(h+1)\binom{s-h+2}{2},\label{f25:eq:Dsum}\end{aligned}$$

(8.8)$$\begin{aligned} T_m(s)&=\sum_{h=m}^s(h+1)\binom{s-h+2}{2} \left(\frac w3(s-h)+\frac d2h\right).\label{f25:eq:Tsum} \end{aligned}$$ In the above, empty sums are zero. The relation $\Psi$ has bidegree $(3,0)$ and weight $d$, and multiplication by $\Psi$ is injective on $\mathcal P$, so the homogeneous exact sequence $$0\to\mathcal P_{(s-3,-m)}\xrightarrow{\cdot\Psi}\mathcal P_{(s,-m)}\to(R_0)_{(s,-m)}\to0$$ for $s=2m$ and the isomorphism $(R_0)_{(2m,-m)}\cong\bigoplus_j W_{m,j}$ of Lemma 8.5 give (8.9)$$\label{f25:eq:weight-subtraction} \begin{split} N_m&=D_m(2m)-D_m(2m-3),\\ M_m(G)&=T_m(2m)-T_m(2m-3)-dD_m(2m-3). \end{split}$$ For the second equality above, multiplication by $\Psi$ sends the monomial basis of $\mathcal P_{(2m-3,-m)}$ to a basis of its image, increasing the weight of each basis element by $d$. The total weight of this image is therefore $T_m(2m-3)+dD_m(2m-3)$. Subtracting this quantity from $T_m(2m)$ gives the total weight of the quotient, which is $M_m(G)$ by Lemma 8.5.

To compute $N_m$ and $M_m(G)$, we evaluate the terms on the right-hand side of (8.9). For $r=s-m\ge0$, the binomial convolution identities give $$\begin{aligned} D_m(s)&=m\binom{r+3}{3}+\binom{r+4}{4},\\ T_m(s)&=w\left(m\binom{r+3}{4}+\binom{r+4}{5}\right)\\ &\quad+d\left(\binom{m+1}{2}\binom{r+3}{3} +(m+1)\binom{r+3}{4}+\binom{r+3}{5}\right). \end{aligned}$$ For example, the $w$-coefficient follows from $\frac{s-h}{3}\binom{s-h+2}{2}=\binom{s-h+2}{3}$; the $d$-coefficient follows by writing $h=m+k$ and expanding $\binom{m+k+1}{2}=\binom{m+1}{2}+(m+1)k+\binom{k}{2}$. These identities also follow by multiplying the elementary generating functions $(1-z)^{-j}$.

To obtain the formulas for $N_m$ and $M_m(G)$, we substitute the above expressions in (8.9) and expand; for $m<3$ use the empty-sum convention directly. For (8.6), by Lemma 2.1 and the formulas above, $$S_X(G)=\lim_{m\to\infty}\frac{M_m(G)}{mN_m}=\frac12\left(\frac{5w}{8}+\frac{7d}{6}\right)=\frac{5w}{16}+\frac{7d}{12},$$ where we use $mN_m=2m^4+O(m^3)$. Substituting $(w,d)=(9,8)$ and $(w,d)=(13,12)$ gives the expressions for $S_X(G_D)$ and $S_X(G_E)$.

More explicitly, the two cases of $M_m(G)$ in Lemma 8.9 are $$\begin{aligned} M_m(G_D)&=\frac{359m^4+502m^3+541m^2+182m}{24},\\ M_m(G_E)&=\frac{177m^4+250m^3+267m^2+90m}{8}. \end{aligned}$$ ◻

Proof. By Lemmas 8.6 and 8.7, the standard monomials give a basis compatible with vanishing along $B$, and for each the exponent of $b$ is equal to its order along $B$. Thus $K_m$ is the sum of these exponents.

Let $B_m(s)$ denote the sum of the $b$-exponents of all monomials in $\mathcal P_{(s,-m)}$. Grouping them by their total exponent $h$ in $a,b$ gives $$B_m(s)=\frac12\sum_{h=m}^s h(h+1)\binom{s-h+2}{2},$$ with empty sums understood to be zero. Indeed, for each choice of the exponents of $x,y,z$, the exponent of $b$ ranges from $0$ to $h$.

The monomials excluded from the standard basis are precisely the multiples of the leading monomial of $\Psi$. This leading monomial has bidegree $(3,0)$ and contains no $b$, so multiplication by it preserves the $b$-exponent. So $K_m=B_m(2m)-B_m(2m-3)$, and using the above expression for $B_m(s)$ this gives the formula for $K_m$ in the lemma statement. Finally, since $N_m=2m^3+O(m^2)$ by Lemma 8.9, the above expression for $K_m$ gives $S(W;B)=\lim_{m\to\infty}\frac{K_m}{mN_m}=\frac{11}{16}$. ◻

Proof. On the chart indexed by $i$, use the finite cover with coordinates $(u,v_j,v_k)$ and map $x_i=u^{w_i}$, $x_j=u^{w_j}v_j$, $x_k=u^{w_k}v_k$. The cyclic group $\mu_{w_i}$ acts with weights $(1,-w_j,-w_k)$, and $G$ is the image of $u=0$. Primitivity implies that this cover has no divisorial ramification: a nontrivial element fixing the generic point of $u=0$ would have order dividing all three weights. No other divisor has generic stabilizer either. Thus the smooth pair with boundary $u=0$ descends to a plt pair, by the discrepancy formula for finite maps. Relative ampleness of $-G$ follows from the weighted Proj construction.

The pulled-back volume form has factor $u^{w_1+w_2+w_3-1}$ times a nonzero constant, so its discrepancy along $G$ is $w_1+w_2+w_3-1$. This proves the formula for the log discrepancy. At a general point of $G\cap(x_k=0)$, the stabilizer has order $r_k$, and on the transverse cover it acts with weights $(1,-w_k)$. Since $\gcd(r_k,w_k)=1$, restriction to $u=0$ has ramification index $r_k$ along the coordinate curve. Adjunction consequently contributes $1-1/r_k$ to the different. The calculation is unchanged after an étale pullback from regular local coordinates. ◻

Proof. Lemma 8.12 gives the plt property and log discrepancy $4+2+3=9$. For the description of $G_D$, consider $\mathbb C[x,y,z]$ with weights $(4,2,3)$ as given in (8.3); its second Veronese subring is $\mathbb C[x,y,z^2]$, with regraded weights $(2,1,3)$. To compute $-G_D|_{G_D}$, write $D_y$ for the strict transform of $(y=0)$; then $\sigma^*(y=0)=D_y+2G_D$. Restriction to $G_D$ gives $-2G_D|_{G_D}\sim_{\mathbb Q}D_y|_{G_D}=(V=0)\sim_{\mathbb Q}H_G$, so $-G_D|_{G_D}\sim_{\mathbb Q}\frac12H_G$.

At the generic point of $z=0$, the quotient index is $\gcd(4,2)=2$; the other two coordinate curves have index one. Adjunction therefore gives $\Delta_G=\frac12(W=0)$. Substituting the coarse coordinates in the initial form of $b$ gives the equation of $B$. The formulas satisfy $-(K_{G_D}+\Delta_G)=\frac92H_G=9(-G_D|_{G_D})$. Finally, Lemma 3.2 gives $\alpha(G_D,\Delta_G;H_G)=\min\{2,1,3/2\}=1$. ◻

Proof. On $V=1$, the equation solves for $W$ and identifies this open set of $B$ with the affine $r$-line. The only remaining point is $[0:0:1]$. Its index-three cover has coordinates $(u,v)$ with $\mu_3$-weights $(2,1)$, and the curve is $u^2+uv^2+v=0$. It is smooth, with parameter $u$ and $v=-u^2+O(u^5)$. The quotient has parameter $u^3$, so it too is smooth. Moreover, $r=u/v^2$ has a simple pole on the quotient. This proves $B\cong\mathbb P^1$.

The boundary $(W=0)$ meets $B$ transversely at $Q_0,Q_{-1}$ and misses $Q_\infty$. The smooth transverse charts and the smooth index-three cover prove plt. Adjunction contributes $1/2$ at the first two points and $1-1/3$ at the last. On the cover, $V$ has order two and $U$ has order one at infinity; division by three gives their stated rational orders. Their remaining orders follow on the affine chart, where $U/V^2=r$ and $W/V^3=-r(r+1)$. ◻

Proof. A section in $W_{m,j}$ has class $jH_G/2$, and removing $B^q$ subtracts $4qH_G$. Consequently $$R_D=\tfrac12S_X(G_D)-4S(W;B) =\frac{359}{96}-4\cdot\frac{11}{16}=\frac{95}{96},$$ by Lemmas 8.9 and 8.10. Both averages use the same denominator $mN_m$. ◻

Proof. For $k=0$, the allowed powers are $r^i$ for $a\le i\le\lfloor n/2\rfloor$. For $j=0$, one has $i=(n-3k)/2$ and $$0\le k\le\min\{\lfloor n/3\rfloor,2T-n\},\qquad k\equiv n\pmod2.$$ The polynomial is $r^{(n-k)/2}(r+1)^k$. It is divisible by $r^a$ and has leading degree $(n+k)/2$. The positive allowed values of $k$ give exactly the consecutive leading degrees from $\lfloor n/2\rfloor+1$ through $D$. Together with the first group, they give a triangular basis of the claimed interval.

The common factor $r^a$ gives the fixed order at zero. A polynomial of degree $d$ represents $V^nf(r)$ and has order $2n/3-d$ at infinity, by Lemma 8.14. Taking the largest possible degree gives the fixed order there. After removing these two factors, the interval is a complete series of degree $D-a$, so it has no further fixed points. ◻

Proof. Apply Lemma 3.1 to the fixed factors and the complete moving interval in Lemma 8.16. The residue factor $z^\eta$ is interpreted by Lemma 2.14. Since its square is $W^\eta$, it adds $\eta/2$ at $Q_0,Q_{-1}$ and nothing at infinity. Its class is $3\eta H_G/2$, proving the last assertion. As a degree check, the fixed orders sum to $a+(2n/3-D)+\eta$ and the moving degree is $D-a$. Their total is $2n/3+\eta=(n+3\eta/2)\deg(H_G|_B)$, since $\deg(H_G|_B)=2/3$. ◻

Proof. We verify all the positivity conditions, including eventual nonvanishing on every integral interior ray. Write $J$ for the native weight and put $M=\min\{m,2m-q\}$. The support cone of the second refinement is (8.11)$$\label{f25:eq:D-support} \mathcal C_D=\{(m,J,q):m\ge0,\ 0\le q\le2m,\ 8q\le J\le8q+4M\}.$$ Every standard monomial satisfies these inequalities. Conversely, fix an integral point $(m_0,J_0,q_0)$ in the interior of this cone, and set $M_0=\min\{m_0,2m_0-q_0\}$. For its $k$th multiple choose $\eta_k\in\{0,1\}$ with $\eta_k\equiv kJ_0\pmod2$, and put $$n_k=\frac{kJ_0-8kq_0-3\eta_k}{2},\qquad T_k=kM_0-\eta_k.$$ The strict inequalities $0<J_0-8q_0<4M_0$ imply that both $n_k$ and $2T_k-n_k$ grow linearly. In Lemma 8.16, $D-a$ therefore grows linearly as well. The summand is nonzero for every sufficiently large $k$, with no divisibility restriction on $k$. The same argument applies to rational interior rays after rescaling. Such rays are dense in the cone, proving equality in (8.11) and the required nonvanishing assertion.

The projection to $(m,J)$ is $0\le J\le16m$. If $0<J_0<16m_0$, choose a rational $q_0$ just below $J_0/8$, so that $(m_0,J_0,q_0)$ is strictly inside (8.11). For each $k$ approximate $kq_0$ by an integer. The resulting bounded errors, together with the parity correction just used, do not affect eventual nonvanishing. This proves the condition on every integral interior ray for the first refinement too. Both support slices are bounded and have nonempty interior.

For the complete ample subseries on $G_D$, take $(m,J)=(12k,12k)$. Every native weighted polynomial of weight $12k$ has ordinary degree at most $6k\le m$, so it comes from a section with no $b$ factor in its expression. These give all of $H^0(G_D,\mathcal O_{G_D}(6k))$, and $6H_G$ is ample Cartier. On $B$, take $(m,J,q)=(12k,14k,k)$: then $\eta=0$, $n=3k$, $a=0$, $D=2k$, and the fixed order at infinity is zero. The series is exactly $H^0(B,\mathcal O_B(2k))$. Both rays are interior. Finally, Lemmas 8.13 and 8.14 verify the plt conditions. This proves every hypothesis of Proposition 2.7 directly, in agreement with the general refinement statement of [4, Example 2.15]. ◻