Three refined surface calculations for Family 2.17

Proof. The complete polygon in (63.1) has area $a(b-a)+a^2/2$. In the second and third intervals the marked-point conditions remove area $r^2$. For two distinct points this is the sum of two triangles of area $r^2/2$. For the curvilinear ideal used below it is one triangle with side lengths $r$ and $2r$. The displayed formulas follow by substitution. The fixed coefficient is $$\frac14\int_2^3 (u-2)A(u)\,du =\frac34\int_2^3(u-2)(3-u)^2\,du=\frac1{16}.$$ If one has only an inclusion of refined spaces, equality of the leading total dimensions leaves $O(m^2)$ omitted sections. All the exponents are $O(m)$, so their omission changes a normalized first moment by $O(m^{-1})$. Thus the same limiting first moments give upper bounds for the actual refined series. ◻

Proof. In the first case an automorphism of $F$ places the points at $t\cap f_0$ and $t\cap f_\infty$. Indeed, after choosing coordinates on the base, translations by sections of $\mathcal O_{\mathbb P^1}(1)$ can prescribe the two fibre coordinates. The extra polygon inequalities are $$a-i+k\geq r,\qquad a-i+b-a+i-k\geq r.$$ In the second case we place the points at $s\cap f_0$ and $t\cap f_\infty$. The inequalities become $$i+k\geq r,\qquad a-i+b-a+i-k\geq r.$$ In the third case they are replaced by $2(a-i)+k\geq2r$.

For completeness, the moving first moments are (63.2)$$\label{f217:eq:surface-moment-table} \begin{array}{c|cccc} &s&t&f_0&f_\infty\\ \hline \text{two points off }s&35/48&2/3&5/6&5/6\\ \text{one point on }s&79/96&55/96&5/6&89/96\\ \text{curvilinear point off }s&35/48&2/3&15/16&35/48. \end{array}$$ Here is a direct check of every entry. Without the marked-point conditions the four moments are $23/24,17/24,49/48,49/48$. Put $$\mathcal J(\phi)=\frac14\int_1^2 \phi(u,u+1,u-1)\,du+ \frac14\int_2^3\phi(2c,3c,c)\,du.$$ The moments removed by the marked conditions are obtained by applying $\mathcal J$ to the entries of the following table: $$\begin{array}{c|cccc} &s&t&f_0&f_\infty\\ \hline \text{two off }s&r^2(a-r/3)&r^3/3&r^2(b-r/3)/2&r^2(b-r/3)/2\\ \text{one on }s&r^2a/2&r^2a/2&r^2(b-r/3)/2&r^2(b-a+r/3)/2\\ \text{curvilinear off }s&r^2(a-r/3)&r^3/3&2r^3/3&r^2(b-r). \end{array}$$ For example, in the first row the two removed triangles have centroids with $s$-coordinate $a-r/3$. Their $f_0$-coordinates are $r/3$ and $b-2r/3$. The other entries are obtained in exactly the same way. These are integrals of explicitly displayed polynomials; they give (63.2).

It remains to include the fixed divisor. Its monomials have the form (63.3)$$\label{f217:eq:R-monomial} x^i y^{3-i}z^k w^{1+i-k}.$$ In the first case the double-point conditions imply $k\geq i-1$ and $k\leq2$. Thus the four exponents are at most $3,3,2,2$, respectively. After adding the coefficient $1/16$ of $R$, all four coefficients are at most $23/24$.

In the second case the smallest $x$-exponent of $R$ is one. Among the terms with this exponent the double-point conditions give $1\leq k\leq2$. Choose a torus degeneration selecting one such term. Its initial divisor is $s+2t+kf_0+(2-k)f_\infty$. The four total coefficients are bounded by $85/96,67/96,23/24,95/96$, respectively.

In the third case first minimize the $z$-exponent. The hypothesis that $R$ does not contain $f_0$ makes this minimum zero. Since $R\in(y,z^2)^2$, a term with $k=0$ in (63.3) has $i\leq1$. The four total coefficients are consequently at most $19/24,41/48,15/16,41/48$.

We justify why these monomial estimates control every divisorial valuation. Fix such a valuation $v$ on $F$. In each graded piece choose a basis compatible with $v$ and with a one-parameter subgroup of the torus. The two-filtration lemma permits this choice. For each $m$ the subgroup can be chosen in the indicated open cone of weights and can distinguish all the finitely many monomials occurring in the moving spaces. The initial basis is then their monomial basis, and the fixed divisor has the initial monomial just specified. The limiting basis boundary is supported on $s,t,f_0,f_\infty$, a simple normal crossings divisor. For every positive number strictly less than $24/23$, $96/95$, or $16/15$, respectively, its multiple is log canonical for all sufficiently large $m$. Semicontinuity of log canonical thresholds gives the same assertion before degeneration. The bounds are independent of $v$. Letting $m$ tend to infinity proves the three assertions. ◻

Proof. Put $b=x+z$ and group the monomials in (63.1) by their ordinary degree $d$ at $A$. In the first two intervals their $x$-exponents run from $i_0=\max\{0,(d-1)/2\}$ to $i_1=\min\{u,d\}$. Set $n=i_1-i_0$ and $q=\max\{0,2r-d\}$. The corresponding binary space is $$(b-z)^{i_0}z^{d-i_1}\operatorname{Sym}^{n}\langle b,z\rangle.$$ Membership in $(b,z^2)^r$ forces divisibility by $b^q$. The other displayed factors are coprime to $b$, so the asymptotic dimension of the degree-$d$ slice is $n-q$ when $n\geq q$ and is zero otherwise. Its ordinary first moment is $d(n-q)$.

Here are the integration intervals and the values of $n,q$; thus the calculation can be checked without an implicit chamber choice: $$\begin{array}{c|c|c|c} u&d&n&q\\ \hline {[0,1]}&[0,u]&d&0\\ {[0,1]}&[u,1]&u&0\\ {[0,1]}&[1,1+2u]&u-(d-1)/2&0\\ \hline {[1,3/2]}&[u-1,2u-2]&d&2u-2-d\\ {[1,3/2]}&[2u-2,1]&d&0\\ {[1,3/2]}&[1,u]&(d+1)/2&0\\ {[1,3/2]}&[u,1+2u]&u-(d-1)/2&0\\ \hline {[3/2,2]}&[u-1,1]&d&2u-2-d\\ {[3/2,2]}&[1,2u-2]&(d+1)/2&2u-2-d\\ {[3/2,2]}&[2u-2,u]&(d+1)/2&0\\ {[3/2,2]}&[u,1+2u]&u-(d-1)/2&0. \end{array}$$ Multiplying by the normalization $1/4$ and integrating $d(n-q)$ over these regions gives $65/48$. In the final interval scale the calculation at $u=2$ by $c=3-u$. At $u=2$ the degree intervals are $1\leq d\leq2$, with $n=(d+1)/2,q=2-d$, and $2\leq d\leq5$, with $n=(5-d)/2,q=0$. Their unnormalized first moments sum to $8$. The last interval therefore contributes $\frac14\int_0^1 8c^3\,dc=1/2$. The moving total is $65/48+1/2=89/48$. Integrating the slice dimensions $(n-q)_+$ themselves gives $2/3$ in the first interval, $7/3$ in the second, and $1$ in the last, for a total of $4$. Thus these containing spaces have total dimension $4m^3+O(m^2)$, the same leading dimension as the actual refinement. The inclusion of Lemma 62.7 therefore omits only $O(m^2)$ dimensions. All orders under consideration are $O(m)$, so the normalized first moment differs by $O(m^{-1})$. This proves exactness for the actual series in this configuration. The fixed contribution is $\operatorname{mult}_A(R)/16=3/16$. The discrepancy of the ordinary point blow-up of a smooth surface is two, which proves the assertion. ◻

Proof. The restriction $q|_\ell$ is an isomorphism onto the line $B$, because $q^*\mathcal O_Q(1)\cdot\ell=1$. The ideal of $C'$ pulls back under $q$ to $\mathcal O_X(-E')$. Its restriction to $\ell$ therefore identifies the scheme $B\cap C'$ with the divisor $E'|_\ell$. The latter has degree two and contains no one-dimensional component. Thus $B\cap C'$ is nonreduced at $q(p)$ if and only if $E'|_\ell=2p$, if and only if $T_p\ell\subset T_pE'$. This also proves that the intersection multiplicity is exactly two, rather than merely at least two. The strict transform $C_Z$ in $\operatorname{Bl}_B(Q)$ is isomorphic to $C'$: the restricted centre ideal on the smooth curve $C'$ is invertible. Its intersection with the exceptional divisor is the same length-two Cartier divisor as $B\cap C'$. Thus nonreducedness here is exactly nonreducedness of the length-two centre scheme $\Xi$ in the surface calculation.

The tangent lines $T_p\ell$ and $T_p\ell'$ are distinct. Indeed, $dq$ is nonzero on the first line, since $q|_\ell$ is an isomorphism, whereas it vanishes on the second. In particular, $\ell'$ is transverse to $\ell$ as a pair of smooth curves in $X$. Its strict transform meets the exceptional surface $F$ in the point represented by the nonzero normal direction $T_p\ell'\bmod T_p\ell$ in $N_{\ell/X,p}$.

The normal sequence for $\ell\subset E\subset X$ is $$0\longrightarrow\mathcal O_\ell \longrightarrow N_{\ell/X} \longrightarrow\mathcal O_\ell(-1)\longrightarrow0.$$ It splits, since $H^1(\mathbb P^1,\mathcal O(1))=0$. The section of $\mathbb P(N_{\ell/X})$ corresponding, in the convention of lines, to the subbundle $N_{\ell/E}=\mathcal O_\ell$ has normal bundle $\operatorname{Hom}(\mathcal O_\ell,\mathcal O_\ell(-1))$ and hence self-intersection $-1$. It is therefore precisely $s$; equivalently it is the intersection of $F$ with the strict transform of $E$. Consequently $$A\in s\quad\Longleftrightarrow\quad T_p\ell'\bmod T_p\ell\in N_{\ell/E,p} \quad\Longleftrightarrow\quad T_p\ell'\subset T_pE.$$

Finally, $E$ and $E'$ are smooth divisors in the smooth threefold $X$. Their scheme-theoretic intersection is singular at $p$ precisely when their tangent planes coincide. Each plane already contains its own ruling tangent line. As these two lines are distinct, the two cross-tangency conditions $T_p\ell\subset T_pE'$ and $T_p\ell'\subset T_pE$ hold precisely when both planes equal their two-dimensional span. This proves all equivalences. ◻