The refined threshold on the exceptional surface

Proof. Give $(r,s,a,b,z)$ the second weights $(1,0,2,0,1)$. Put $U=as^2$, $V=br^2$, and write $j=2h+\varepsilon$, where $\varepsilon\in\{0,1\}$. The primary weight is $2N+\varepsilon$, where $N=\alpha+\beta+h$; the second weight is $Q=i+j+2\alpha$. Put $$R=Q-\varepsilon-2N,\qquad S=m-Q,\qquad L=\max\left\{0,\left\lceil\frac{2N+\varepsilon-Q}{2}\right\rceil\right\},$$ $$K=\max\left\{0,Q-2m,\left\lceil\frac{L+Q-m}{2}\right\rceil\right\}, \qquad d=N-K-L.$$ The eigenspace is zero if $d<0$, and otherwise is exactly $$z^\varepsilon r^R s^S U^K V^L\operatorname{Sym}^d\langle U,V\rangle.$$ Indeed, a generator becomes, up to a nonzero constant, $z^\varepsilon r^R s^S U^\alpha V^\beta(U+V)^h$. The conditions on the original exponents are equivalent to $$\beta\ge L,\qquad \alpha\ge0,\qquad \alpha\ge Q-2m, \qquad 2\alpha-\beta\ge Q-m,\qquad \alpha+\beta\le N.$$ They imply $\alpha\ge K$. Conversely, taking $\beta=L$ and $\alpha=K,K+1,\ldots,N-L$ gives a basis of the binary space in the displayed formula. Negative values of $R$ or $S$ cause no pole: the factors $U^KV^L$ make every displayed section regular. This also accounts for all cancellations among the original generators.

Here are the integrations used below. Write $q=Q/m$, $n=N/m$ and $$l=\max\{0,n-q/2\},\qquad k=\max\{0,q-2,(l+q-1)/2\},\qquad d=n-k-l.$$ The domain is $d\ge0$. It is the union of the following eight polygons; in the first three rows $l=0$, and in the last five rows $l=n-q/2$. $$\begin{array}{c|c|c|c} q&n&k&d\\\hline {[0,1]}&[0,q/2]&0&n\\ {[1,3]}&[(q-1)/2,q/2]&(q-1)/2&n-(q-1)/2\\ {[3,4]}&[q-2,q/2]&q-2&n-q+2\\ {[0,1]}&[q/2,1-q/2]&0&q/2\\ {[0,1]}&[1-q/2,q/2+1]&(n+q/2-1)/2&q/4-n/2+1/2\\ {[1,3]}&[q/2,q/2+1]&(n+q/2-1)/2&q/4-n/2+1/2\\ {[3,4]}&[q/2,3q/2-3]&q-2&2-q/2\\ {[3,4]}&[3q/2-3,q/2+1]&(n+q/2-1)/2&q/4-n/2+1/2 \end{array}$$ Their total mass is $\int d\,dn\,dq=7/6$. For instance, the first polygon contributes $\int_0^1\int_0^{q/2}n\,dn\,dq=1/24$. Both parities have the same leading term. Thus normalized moments are $\frac67\int d(\cdots)\,dn\,dq$. Integrating on the eight polygons gives $$\begin{array}{c|c|c} \text{curve or weight}&\text{integrand after the factor }d&\text{moment}\\\hline r=0&q-2n+2l+d&11/28\\ s=0&1-q+2k+d&17/28\\ a=0&k+d/2&45/56\\ b=0&l+d/2&45/112\\ U+\kappa V=0\ (\kappa\ne0)&d/2&17/112\\ \text{primary weight}&2n&135/56\\ Q&q&2\\ m+2N+\varepsilon-Q&1+2n-q&79/56 \end{array}$$ These are maximal average orders along each prescribed invariant curve. For example, at $U+\kappa V=0$ the maximal total order of a basis of binary forms of degree $d$ is $d(d+1)/2$. The odd factor $z$ contributes $O(1/m)$ to its normalized coefficient. The same factor is included exactly in the two corner weights in the last two rows.

We explain why these numerical inequalities give log canonicity. A second-weight eigenvector factors into the four coordinate curves and curves $U+\kappa V=0$. The latter are smooth, meet only at $P=(r=a=0)$ and $P'=(s=b=0)$, and have contact order two there. Their intersections with the coordinate curves are also confined to these two points. At the other two coordinate corners the configuration has simple normal crossings. The limiting coefficients of the four coordinate curves are less than one. The coefficient of any other component is at most $17/112$, to which one must add $1/2$ when the component is $\Gamma$; this remains less than one.

At $P$ use the weights $(1,2)$ in $(r,a)$, and at $P'$ the weights $(1,2)$ in $(s,b)$. Their log discrepancies on $G$ are three, and their orders on $\Gamma$ are two. The required bounds on the basis divisor are therefore two. The last two rows give respectively two and $79/56$. These are moments of the whole divisor: adding the separate maxima in the preceding rows would not give the correct estimate. The weighted blow-ups separate the strict transforms of the curves with quadratic contact. On the smooth chart this is seen from $r=t$, $a=t^2u$, where the strict transforms have equations $u+\kappa=0$. The remaining chart is a quotient of a smooth surface by $\boldsymbol\mu_2$; its covering pair has two transverse coordinate curves, with coefficients at most one. Thus these inequalities are sufficient for log canonicity at both corners. Equivalently, if $c$ is the coefficient of the transverse coordinate curve, then the ordinary multiplicity at the corner equals one half of the weighted order plus $c/2$, and is at most $(3+1)/2=2$. Thus the ordinary first blow-up also has the required inequality.

There is no rounding error at the equality at $P$. Lemma 86.7 gives the exact weight sum $2mh_m$ at every level, including the odd factor $z$. All other inequalities have a fixed positive margin, so hold for all sufficiently large $m$.

Finally, in each primary graded piece choose a basis compatible with an arbitrary divisorial valuation and with the second weight filtration. Taking its second initial basis gives a divisor of the form just considered. Log canonicity of that divisor together with $\Gamma/2$, and semicontinuity of log canonical thresholds, imply the same bound for the original valuation. Passing to the asymptotic threshold proves the lemma. ◻

Proof. If $f$ is a square, take $f=s^2$. Coprimality implies that the coefficient of $r^2$ in $g$ is nonzero. Taking maximum initial weights $(r,s,a,b,z)=(1,0,2,0,1)$ gives $z^2+as^2+br^2$ after rescaling $b$. If $g$ is a square, take $g=r^2$; then $d\ne0$. This change of coordinates replaces $r$ by a nonzero multiple of $r+\lambda s$ and preserves the polynomial series: expanding a generator only increases its $s$-exponent, so preserves both bounds $\alpha\le m+k$ and $\beta\le k$. Taking minimum initial weights gives $z^2+ads^2+br^2$, since the omitted term $acrs$ has weight three. Rescale $a$.

The polynomial part of each primary graded piece, after removing $z^\varepsilon$, is the monomial space of Lemma 86.2. Hence it is preserved by the second torus action. We therefore degenerate the branch curve while keeping these polynomial spaces fixed. In the second case the family is explicitly $$\Gamma_\tau=(as^2+br^2+\tau acrs=0).$$ The first case is obtained by the opposite torus action. The odd factor contributes $\varepsilon \Gamma_\tau/2$ to the zero divisor of a section, and must move with the branch curve. The coefficient of the different is $1/2$ throughout. Choose bases compatible with both filtrations in every primary graded piece. Their initial divisors lie in the square model, for which the preceding lemma gives log canonicity. Semicontinuity of log canonical thresholds for this family, including the moving branch divisor, gives the same bound on the original pair. ◻