Blowing up a contracted line

Proof. Choose coordinates in which $$Q=\{x_0x_3+x_1x_4+x_2^2=0\},\qquad B=\{x_2=x_3=x_4=0\}.$$ Writing $[x_2:x_3:x_4]=[a:b:c]$ gives the equation $bx_0+cx_1+a^2t=0$ on the resolved projection. The three coefficients have no common zero, so its fibres are projective lines, including the fibre over $[1:0:0]$. With the quotient convention for projective bundles, we may write $Z=\mathbb P(\mathcal V)$, where $$0\longrightarrow\mathcal O(-2) \xrightarrow{(b,c,a^2)} \mathcal O(-1)^{\oplus2}\oplus\mathcal O \longrightarrow\mathcal V\longrightarrow0.$$ Here $c_1(\mathcal V)=0$, $c_2(\mathcal V)=1$; its tautological class $\xi$ satisfies $H_Q=\xi+A$ and $F_B=\xi$.

The linear forms through $B$ restrict to the complete degree-three linear series $\mathcal O_{C'}(1)(-B\cap C')$. Indeed they give three independent sections (otherwise $C'$ would lie in a hyperplane), and Riemann–Roch gives dimension three. A degree-three line bundle on a smooth elliptic curve is very ample. This proves the assertion about $C_Z$ and $T$, also when $B\cap C'$ is nonreduced.

The section $C_Z\subset\mathbb P(\mathcal V|_T)$ corresponds to a quotient $\mathcal V|_T\twoheadrightarrow\mathcal M$ with $$\deg\mathcal M=\xi\cdot C_Z =(H_Q-A)\cdot C_Z=5-3=2.$$ Define the elementary transform $$0\longrightarrow\mathcal V_1\longrightarrow\mathcal V \longrightarrow i_*\mathcal M\longrightarrow0, \qquad i:T\hookrightarrow\mathbb P^2.$$ It is locally free of rank two: locally the last map is the projection $\mathcal O^{\oplus2}\to\mathcal O/(t)$, whose kernel is generated by $(t,0)$ and $(0,1)$. Its Chern classes are $c_1(\mathcal V_1)=-3$ and $c_2(\mathcal V_1)=3$. For example $\operatorname{ch}(i_*\mathcal M)=3h+(2-9/2)h^2$, where $h=c_1(\mathcal O_{\mathbb P^2}(1))$.

Put $\mathcal W=\mathcal V_1(2)$; then $c_1(\mathcal W)=1$, $c_2(\mathcal W)=1$. We claim that $\mathcal W\simeq T_{\mathbb P^2}(-1)$. First $H^0(\mathcal W(-1))=H^0(\mathcal V_1(1))=0$: such a section would give a member of $|H_Q|$ containing $C_Z$, hence a hyperplane of $\mathbb P^4$ containing the nondegenerate curve $C'$. Thus $\mathcal W$ is slope stable. Riemann–Roch gives $\chi(\mathcal W)=3$, and Serre duality and stability give $H^2(\mathcal W)=0$. A nonzero section has no divisorial zeros by stability, and its zero scheme has length $c_2=1$. It therefore gives $$0\longrightarrow\mathcal O\longrightarrow\mathcal W \longrightarrow\mathcal I_z(1)\longrightarrow0$$ for one reduced point $z$. The extension space $\operatorname{Ext}^1(\mathcal I_z(1),\mathcal O)$ is one-dimensional; local freeness excludes its zero element. The Euler sequence supplies this same nonzero extension for $T_{\mathbb P^2}(-1)$, proving the claim.

The elementary transform is resolved by $Y'$, giving a morphism $g:Y'\to\mathbb P(\mathcal V_1)=\mathbb P(\mathcal W)$. Locally its rational map is $[u:v]\mapsto[tu:v]$, whose ideal of indeterminacy is $(t,v)$, the ideal of $C_Z$. It follows that $$g^*\mathcal O_{\mathbb P(\mathcal V_1)}(1)=f^*\xi-J, \qquad f^*(H_Q+2A)-J =g^*(\mathcal O_{\mathbb P(\mathcal W)}(1)+h).$$ On $\mathbb P(T_{\mathbb P^2}(-1))$, the tautological divisor is generated by global sections, is relatively ample, and its sum with the pullback of $h$ is ample. If its class is $\eta$, the projective-bundle relations give $\eta^3=c_1(\mathcal W)^2-c_2(\mathcal W)=0$, $\eta^2h=1$, and $\eta h^2=1$. Hence $(\eta+h)^3=6$. ◻

Proof. At an intersection point of length $k\in\{1,2\}$, choose an analytic parameter $z$ on $B$. In the tangent case it is also a parameter on $C$. For $k=1$ choose the parameter so that it is a parameter on both curves. Two transverse coordinates can then be chosen such that $$B=\{x=y=0\},\qquad C=\{x-z^k=y=0\}.$$ Indeed, first make $B$ the coordinate axis. One transverse coordinate of $C$ has order $k$ and can be made $z^k$, and the other has order at least $k$ and is removed by subtracting a multiple of the first.

On the chart of $\mathop{\mathrm{Bl}}_C$ where $y=t(x-z^k)$, the strict transform of $B$ is $(t,x)$. On the chart of its blow-up with $x=bt$, the coordinates are $(b,t,z)$ and $$x=bt,\qquad y=t(bt-z^k).$$ Conversely, on the chart of $\mathop{\mathrm{Bl}}_B$ where $y=vx$, the strict transform of $C$ is $(v,x-z^k)$. The corresponding chart after the second blow-up has coordinates $(b,v,z)$, with $$x=z^k+bv,\qquad y=v(z^k+bv).$$ These are the two small resolutions of $$W_k=\{ab=x(x-z^k)\},$$ given by the ideals $(b,x)$ and $(b,x-z^k)$; the displayed charts set $a=y$. The second chart of each small resolution is also smooth, and the exceptional fibre is $\mathbb P^1$. The morphisms are crepant because they are small resolutions of a Gorenstein hypersurface. Thus their birational exchange is a flop. For $k=1$ the hypersurface has an ordinary node; for $k=2$ it has a $cA_3$ singularity and the width-two local form above. The two distinct length-one intersection points give disjoint local flops.

In the first model, the exceptional surface of the blow-up of the strict transform of $B$ is $t=0$ in the displayed chart. On the second model, its strict transform is $$bv+z^k=0.$$ For $k=2$ this is an $A_1$ surface singularity. Its morphism to the smooth surface $\{t=0\}$, with coordinates $(b,z)$, is the blow-up of $(b,z^k)$. For $k=1$ this is the ordinary blow-up of a point. For $k=2$ its minimal resolution is obtained by first blowing up $(b,z)=(0,0)$ and then blowing up the point on the first exceptional curve corresponding to the tangent direction $b=0$. If $e_1,e_2$ denote the total transform classes of the exceptional curves of these two ordinary blow-ups, the pullback of the exceptional Cartier divisor of the blow-up of $(b,z^2)$ is $e_1+e_2$; its irreducible components on the resolution are $(e_1-e_2)+2e_2$. ◻

Proof. The map $F_B\to\mathbb P^2$ is the blow-up of $o$. Therefore its inverse image $R_B$ of $T$ is a smooth elliptic curve of class $3c+3f$ if $o\notin T$. If $o\in T$, it is the reduced union of $c$ and the smooth elliptic strict transform of $T$, of class $2c+3f$. These two components meet transversely at one point. In particular, $R_B$ has no ruling fibre.

First suppose $\Xi=\{p_1,p_2\}$ is reduced. The points lie on distinct ruling fibres, since their images are the two distinct points of $B\cap C'$. On the common resolution $\widetilde F=\operatorname{Bl}_{p_1,p_2}(F_B)$, write $e_1,e_2$ for the exceptional curves. Its map to $F$ contracts the disjoint curves $L_i=f-e_i$, and $$\rho^*s=c+f-e_1-e_2,\qquad \overline R=3c+3f-e_1-e_2 =\rho^*R-2L_1-2L_2.$$ At each $p_i$ the divisor $R_B$ is smooth. Indeed, if $R_B$ were singular there, the smooth surfaces $F_B$ and $r^{-1}(T)$ would have the same tangent plane; as $C_Z\subset r^{-1}(T)$, their intersection with $C_Z$ would have multiplicity at least two, contrary to the reducedness of $\Xi$. Thus $\overline R$ is just the strict transform of $R_B$. Pushing forward gives the asserted class and introduces no additional component. If $o\in T$, exactly one $p_i$, say $p_1$, lies on $c$. The image of $c-e_1$ is $s$, because $\rho^*s=(c-e_1)+L_2$. The other component gives $R_1$. If $o\notin T$, neither $p_i$ lies on $c$, and $s$ maps to a line in $\mathbb P^2$; it cannot be a component of the irreducible cubic transform.

Now suppose $\Xi$ is a curvilinear scheme of length two supported at $p$. Its tangent direction is transverse to the ruling fibre of $F_B$ through $p$. To see this, choose a parameter $z$ along $B$ at the tangency point of $C'$ and $B$. The smooth curve $C'$ is tangent to $B$, so $z$ is also a parameter on $C'$. Since the intersection length is exactly two, $\mathcal I_B\mathcal O_{C'}=(z^2)$. The length-two scheme $\Xi$ therefore maps isomorphically to $\operatorname{Spec}\mathbb C[z]/(z^2)$ on $B$.

Resolve the blow-up of $\Xi$ by blowing up $p$ and then the point in its specified tangent direction. Denote the total exceptional classes by $e_1,e_2$. The curve $K=e_1-e_2$ has self-intersection $-2$. The map to $F$ contracts first $L=f-e_1$ and then the image of $K$; its negative section has pullback $c+f-e_1-e_2$. The strict transform of $F_B$ in the threefold has an $A_1$ singularity, and its minimal resolution contracts $K$. The pullback of the intersection with the strict transform of $r^{-1}(T)$ has class $3c+3f-e_1-e_2$.

If $o\notin T$, the divisor $R_B$ is smooth and contains $\Xi$, so its successive multiplicities are $1,1$. Its strict transform is exactly this pullback. If $o\in T$, the unique point of $C_Z$ over $o$ lies on $c$, so $p\in c$. Since $C_Z$ is tangent to $F_B$, the restriction of $r$ to $F_B$ shows that $p$ is also the intersection of $c$ with the strict transform of $T$. Thus $R_B$ has a node at $p$. The tangent direction of $\Xi$ is neither branch of this node: otherwise the intersection of $C_Z$ with $F_B$ would have length at least three. The successive multiplicities of $R_B$ are therefore $2,0$. If $R_B^{\mathrm{str}}$ denotes its strict transform, the pulled-back intersection divisor is $R_B^{\mathrm{str}}+K$. The extra curve $K$ is contracted by the map to $F$ and contributes no component to $R$. The image of the strict transform $c-e_1$ is $s$. This proves reducedness and the absence of ruling fibres in all cases.

Finally, when $o\notin T$, the negative section on $F$ is the transform of the unique plane line with the prescribed first-order contact at $p$; it is disjoint from the contracted chain, so $A\notin s$. When $o\in T$, the strict transform $c-e_1$ meets $K$, so its image $s$ contains $A$. ◻

Proof. Suppose first that $o\notin T$. The curve $R$ is irreducible, has arithmetic genus three, and has elliptic normalization. For two reduced centres, each contracted $L_i$ meets its smooth strict transform in a scheme of length two; the resulting singularity has multiplicity two and delta-invariant one. It is a node or an ordinary cusp. The two singular points are distinct. For a nonreduced centre the contracted chain produces just one singular point, of multiplicity two and delta-invariant two. Thus it is $A_3$ or $A_4$. The thresholds of $A_n$, in the equation $y^2=x^{n+1}$, are $\min\{1,1/2+1/(n+1)\}$, proving the first column.

Suppose $o\in T$, so $R=s+R_1$. The curve $R_1$ has arithmetic genus two and elliptic normalization. For reduced centres, its unique singularity is a node or a cusp at the point $A_1\notin s$. The two intersections counted by $s\cdot R_1=2$ are either distinct transverse intersections, or one ordinary tangency. The latter gives an $A_3$ singularity of the union. This proves the upper right entry. For a nonreduced centre, the only singularity of $R_1$ is at $A\in s$, and both intersections with $s$ are concentrated there. If $R_1$ is nodal, $s$ is transverse to both branches, giving an ordinary triple point $D_4$. If $R_1$ is cuspidal, $s$ is transverse to its tangent, giving $D_5$. Their thresholds are $2/3$ and $5/8$, respectively. The latter can also be read directly from the weighted homogeneous equation $x(y^2-x^3)=0$, with weights $(2,3)$. ◻

Proof. Since $L$ embeds $\ell$ as a line, hyperplanes containing this line generate $\mathcal I_\ell\otimes L$. Thus $\sigma^*L-F$ is base-point-free, and $\sigma^*L-uF$ is nef for $0\leq u\leq1$. The blow-up intersection formulas give $\sigma^*L^2F=0$, $\sigma^*L\,F^2=-1$ and $F^3=-\deg N_{\ell/X}=1$. This proves the first cubic.

Let $J$ be the exceptional divisor of $Y'\to Z$. On this model the transform of $\sigma^*L-uF$ is $$M_u=(3-u)H_Q+uA-J.$$ The divisor $M_1$ is the transform of the base-point-free divisor above; it descends to the common small contraction in Lemma 62.2 and remains nef. Lemma 62.1 proves that $M_2=H_Q+2A-J$ is the pullback of an ample divisor on $\mathbb P(T_{\mathbb P^2}(-1))$. Hence every $M_u$ for $1\leq u\leq2$ is nef.

On $Z$ one has $$H_Q^3=2,\quad H_Q^2A=2,\quad H_QA^2=1,\quad A^3=0, \qquad H_QC_Z=5,\quad AC_Z=3.$$ Moreover $-K_Z=2H_Q+A$, so $\deg N_{C_Z/Z}=13$. If $K_u=(3-u)H_Q+uA$, the blow-up formula is $M_u^3=K_u^3-3K_uC_Z+13$. Expanding gives the second cubic in (62.2). This calculation applies equally to the width-two flop; no formula for two distinct ordinary flops is substituted in the tangent case.

Write $R_Y=3A-J$ for the strict transform of $r^{-1}(T)$ on $Y'$. It is the exceptional divisor of the elementary transform $g:Y'\to\mathbb P(T_{\mathbb P^2}(-1))$. A contracted ruling has intersection $-1$ with $R_Y$ and zero with $M_2$. For $2\leq u\leq3$, $$M_u=(3-u)M_2+(u-2)R_Y.$$ Intersecting an effective divisor with the moving rulings of $R_Y$ forces the coefficient $u-2$ of $R_Y$. Removing it leaves the nef, semiample divisor $(3-u)M_2$. This proves the last chamber and its volume. For $u>3$, intersect with a movable complete-intersection curve pulled back from the target of $g$ and avoiding its exceptional image. The intersection is negative, so $M_u$ is not pseudoeffective. Finally integrate the three cubics to obtain (62.3). ◻

Proof. The first row follows from (62.1). To obtain the second, restrict on $Y'$ and pull back to a smooth resolution of $F'$. For distinct points the restriction is the pullback of $js+(j+m)f$ minus $j-m$ times each exceptional curve. For a double point the same statement uses the total exceptional classes $e_1,e_2$ and the divisor $(j-m)(e_1+e_2)$. By the local calculation this is precisely the power of the ideal $(b,z^2)$, not two reduced conditions at the same point.

For the third row the fixed part in Lemma 62.6 forces $(j-2m)R$. On $F$ the curve $R$ has class $3s+4f$ and belongs to $\mathcal I^2$. Subtracting its class gives $$js+(j+m)f-(j-2m)(3s+4f) =2(3m-j)s+3(3m-j)f,$$ and subtracting its orders at the marked scheme leaves $(j-m)-2(j-2m)=3m-j$. We justify this division, rather than relying only on the membership $R\in\mathcal I^2$. At a reduced marked point $R$ has multiplicity exactly two. At a curvilinear point the successive multiplicities on the two ordinary point blow-ups are $(2,2)$ off $s$ and $(3,1)$ on $s$. In both cases its value for the valuation with weights $(2,1)$ is exactly four. The corresponding initial form has a nonzero coefficient after reduction to the residue field and is a nonzerodivisor in the polynomial associated graded ring of $\mathcal I=(b,z^2)$. Equivalently, this follows directly from the complete monomial ideals $(b,z^2)^n$, whose elements have weighted order at least $2n$. Consequently $$(\mathcal I^n:R^k)=\mathcal I^{n-2k}\qquad(n\geq2k).$$ The ordinary maximal-ideal version gives the same identity at the two reduced points. This proves all the inclusions. They suffice for the upper bounds below: a valuation-adapted basis can be extended to a basis of a containing space, adding an effective basis divisor with the original normalization. ◻