Liu’s method: the CM line, GIT, and the reconstruction problem

Proof. Diagonalize a nontrivial one-parameter subgroup. Interchanging the eigenvectors in each factor if necessary, we may take the weights on $u,v,x,y$ to be $-a,a,b,-b$, where $a,b\ge0$ are integers and $(a,b)\ne(0,0)$. The weights of the two sorts of monomials are $$-a+(2i-5)b,\qquad a+(2i-5)b.$$ If $f(0,1)\ne0$, the monomial $uy^5$ has negative weight. Otherwise $g(0,1)\ne0$. Write $r=\operatorname{ord}_{(0:1)}f$; then $1\le r\le5$, since the two quintics are linearly independent. The support contains monomials of weights $$-a+(2r-5)b,\qquad a-5b.$$ Since $r\le5$, their minimum is at most $\min\{-a+5b,a-5b\}=-|a-5b|\le0$. This proves semistability. If $b=0$, the minimum is negative. If $b>0$, it can be zero only when $a=5b$ and $r=5$. Conversely, when $f=cx^5$ and $g(0,1)\ne0$, the subgroup with $a=5$, $b=1$ has minimum weight zero. This proves the assertion about stability.

In the equality case its limit is $cux^5+g(0,1)vy^5$. Thus every zero-weight limit of a smooth graph is in the displayed monomial orbit. For the monomial graph itself, such a limit is again in its orbit; the polystable form of the Hilbert–Mumford criterion proves that this orbit is closed. A graph with one totally ramified point is not equivalent to the monomial graph, which has two. If there are two, their ramification contributions exhaust eight, and the map is equivalent to $[x:y]\mapsto[x^5:y^5]$. This proves all the remaining assertions. ◻

Proof. The universal divisor $\mathcal C\subset S\times P$ is smooth: evaluation at a point of $S$ is a nonzero linear functional on $V$, so the projection $\mathcal C\to S$ is a projective bundle. It is a codimension-two regular embedding in $M\times P$. Moreover, it is a relative Cartier divisor on $S\times P$, flat over $P$. For its ideal $I$, each $I^j/I^{j+1}$ is $\operatorname{Sym}^j(I/I^2)$ and is flat over $P$. Induction gives flatness of $\mathcal O/I^j$, hence of $I^j$, and base change for these ideals. Taking the relative Proj of the Rees algebra proves flatness and the fibre assertion. The total blow-up is smooth because its centre and ambient space are smooth. Over $U$ every centre is a smooth graph and every fibre is the smooth Fano threefold described above. ◻

Proof. We compute on $S\times P$. Let $A,B$ be the two ruling classes of $S$ and let $h=c_1(\mathcal O_P(1))$. Thus $A^2=B^2=0$, $\int_S AB=1$, and $H|_S=2B$. The universal centre has class $F=A+5B+h$. Its normal bundle $N$ in $M\times P$ fits into $$0\longrightarrow\mathcal O_{\mathcal C}(F) \longrightarrow N \longrightarrow\mathcal O_{\mathcal C}(4B) \longrightarrow0.$$ In particular, $c_1(N)=A+9B+h$ and $c_2(N)=4B(A+5B+h)$. The class $K=-K_M$ restricts to $2A+6B$.

For the inclusion $i:\mathcal C\hookrightarrow M\times P$, the blow-up pushforward identities are $$\rho_*\mathcal E=0,\quad \rho_*\mathcal E^2=-i_*1,\quad \rho_*\mathcal E^3=-i_*c_1(N),\quad \rho_*\mathcal E^4=-i_*(c_1(N)^2-c_2(N)).$$ These follow by restricting to the projective bundle $\mathcal E\to\mathcal C$; in particular the last two signs are negative. Since $K^4=0$ on $M\times P$, expansion of $(\rho^*K-\mathcal E)^4$ gives $$-p_*(\mathcal L^4) =\operatorname{pr}_{P*} \bigl((6K^2-4Kc_1(N)+c_1(N)^2-c_2(N))F\bigr).$$ Using $A^2=B^2=0$, the factor in parentheses is $$62AB-6Ah-10Bh+h^2.$$ Only the coefficient of $ABh$ survives the pushforward. It is $62-30-10=22$. Hence $c_1(\lambda_\mathrm{ac})=22h$. Since $\operatorname{Pic}(P)=\mathbb Z[\mathcal O_P(1)]$, the underlying line is $\mathcal O_P(22)$. Finally, $G$ has no characters, so two $G$-linearizations on the same line differ trivially. This proves the equivariant statement. ◻

Proof. The CM weight has the same sign as the Donaldson–Futaki invariant. By Proposition 102.4, it is $22\mu(F,\lambda)$. Apply Theorem 102.1. The ampleness and normality assumptions ensure that this is a test configuration of the required kind; a blow-up of an arbitrary boundary graph is not asserted to have these properties. ◻

Proof. These are the central models in Lemma 93.2. The subgroup having weights $-r,r,1,-1$ on $u,v,x,y$ fixes $[C_r]$; both monomials have weight $r-5$. Its weight on the CM line is $22(5-r)\ne0$. Consequently the product test configurations associated to this subgroup and its inverse have nonzero opposite Donaldson–Futaki invariants. One of them is negative.

There is also a direct check using the earlier first moment. In the primitive local normalization, the two Futaki values have absolute value $$(r+1)-S_{X_r}(G)=\frac{11(5-r)}{80}.$$ The subgroup just used doubles the local weights, giving the same nonvanishing test. The common factor $y^{5-r}$ in the centre must be retained throughout this calculation. ◻

Proof. The volume is constant in the smoothing and equals $20$. Liu’s local-to-global inequality [38] gives $\widehat{\operatorname{vol}}(z,Z)\ge20(3/4)^3=135/16$. The remaining assertions follow from [37, Theorem 4.2], whose volume assumption is exactly $(-K_Z)^3\ge20$. ◻

Proof. Intersect the threefold classes with $L=-K_X$. The identities in Section 89 give $$T^2L=0,\quad THL=3,\quad TDL=1,\quad H^2L=2,\quad HDL=2,\quad D^2L=-2.$$ This proves the matrix and its nonzero determinant. Put $F_1=T$, $F_2=T+D$, $R=H-8T-3D$. Then $F_1^2=F_2^2=0$, $F_1F_2=1$, $RF_1=RF_2=0$, and $R^2=-28$. The change of basis is integral with determinant one up to sign. Finally, $L=2T+H+D=6F_1+4F_2+R$, whose square is $48-28=20$. No assertion that this lattice is the entire Picard group of every anticanonical surface is needed. ◻

Proof. In the dependent case, a change of target coordinates gives $F=uf$. A target one-parameter subgroup acts on this vector with strictly positive weight, proving instability.

Suppose that the pencil has rank two. Diagonalize a one-parameter subgroup and use weights $-a,a,b,-b$ on $u,v,x,y$, where $b\ge0$. The case $b=0$ is strictly positive in the Hilbert–Mumford convention used above, since both target coefficients are nonzero. Suppose $b>0$, and put $p=(0:1)$. If the orders of the chosen coefficients are $k,l$, the minimum weight is $$\min\{-a+(2k-5)b,\ a+(2l-5)b\}.$$ For any basis of the pencil, its two orders are either $s_p,r_p$ or $s_p,s_p$. In particular, $k+l\le s_p+r_p$. The displayed minimum is at most $(k+l-5)b$. It is therefore nonpositive, respectively negative, under the stated semistability, respectively stability, assumption.

Conversely, choose a basis with orders $r_p,s_p$, and take $a=r_p-s_p$, $b=1$. The minimum weight is $r_p+s_p-5$. This proves the two criteria. If $r_p+s_p=5$, its zero-weight limit is, after rescaling, $$x^{s_p}y^{s_p}(ux^{5-2s_p}+vy^{5-2s_p}).$$ Here $s_p=0,1,2$. In each of these three examples, equality in the criterion occurs only at zero and infinity, and every zero-weight limit is in its own orbit. The polystable Hilbert–Mumford criterion proves that these orbits are closed in the semistable locus. Every nonstable semistable pencil has one of them in its orbit closure, so these are all the nonstable polystable orbits. ◻

Proof. Write $uf+vg=c(x,y)(uf_0+vg_0)$, where $f_0,g_0$ are relatively prime. Near a root of $c$, choose a local source coordinate $x$. The residual graph has an equation $h=t-\eta_0(x)$ after choosing an affine target coordinate. Together with a local equation $z$ of $S$, the ideal of $C$ is $(z,x^s h)$. One chart of its blow-up is smooth, and the other is the displayed hypersurface. The same description with $h$ a unit applies away from the residual graph. The hypersurface is normal: its singular locus has codimension at least two, and it is Cohen–Macaulay. It is Gorenstein and toric. Its toric Gorenstein height function has value one on every primitive ray generator, so every nonzero lattice point in the cone has positive integral height. Thus the singularity is canonical. Where the centre is reduced and smooth the blow-up is smooth. This proves the assertions about singularities.

The ideal of $C$ twisted by $\mathcal O_M(1,3)$ is globally generated. Indeed, use the equation of $S$ multiplied by the sections of $\mathcal O_M(1,1)$ and the two cubic lifts of $x(uf+vg)$ and $y(uf+vg)$ from $S$. Consequently $B=T+3H-E$ is globally generated and $-K_{X_F}=T+B$.

The centre is finite flat of degree five over the target $\mathbb P^1$: on each target fibre the quintic $uf+vg$ is nonzero, because $f,g$ are independent. Since $C$ is a relative complete intersection, the powers of its ideal commute with this base change. Thus a fibre $V$ of $X_F\to\mathbb P^1$ is the blow-up of $\mathbb P^2$ in a length-five Cartier subscheme $Z\subset Q$. Its exceptional Cartier divisor is $E_V$, and $B|_V=3h-E_V$. This is ample. To see this, its square is four. If a curve is contracted to a point in $\mathbb P^2$, positivity follows from the relative ampleness of $-E_V$. The strict transform of $Q$ is isomorphic to $Q$ and has intersection $6-5=1$ with $3h-E_V$. For the strict transform $R$ of any other irreducible plane curve of degree $d$, the effective Cartier divisor $Q'=2h-E_V$ does not contain $R$. Therefore $$(3h-E_V)\cdot R=d+Q'\cdot R\ge d>0.$$ Nakai–Moishezon gives ampleness on $V$.

The globally generated line bundles $T$ and $B$ define a morphism to a product of projective spaces. It has no contracted curve: such a curve would be in a target fibre and have zero intersection with $B|_V$. The morphism is finite, and $T+B$ is its pullback of an ample line bundle. Thus $-K_{X_F}$ is ample. Its cube is twenty by flatness of the universal blow-up, or by the same intersection calculation as for a smooth graph. Finally, the GIT criterion gives $2s+1\le s+r\le5$, hence $s\le2$. ◻

Proof. The strict transform $D$ is smooth, the threefold is smooth along $D$, and $D|_D=\mathcal O(-1,-1)$. Let $Z$ be the source ruling through $p$, and let $s$ be its multiplicity as a component of $C$. The GIT criterion gives $s\le2$. The nef and fixed parts of $L-uD$ have the same classes as in Lemma 90.1. In the second interval the fixed restriction is $(u-1)C$. Consequently the first moment along $Z$ is the old moving moment plus the contribution of its fixed multiplicity: $$S(W^D;Z)=\frac{61}{80} +\frac{3s}{20}\int_1^{3/2}4(6-4u)(u-1)\,du =\frac{61+4s}{80}.$$ After removing $sZ$, the remaining divisor $C-sZ$ meets $Z$ transversely at one point: the other base rulings are disjoint from $Z$, and the residual graph has degree one over the source. Thus the secondary moving moment is $69/80$ and its fixed contribution is at most $1/20$, exactly as in Lemma 90.3. The flag inequality gives $$\delta_p(X_F)\ge \min\left\{\frac{40}{31},\frac{80}{61+4s},\frac{80}{73}\right\} =\frac{80}{73}.$$ ◻

Proof. Put $d=5-2b$. The torus $$([u:v],[x:y])\longmapsto ([\lambda^{-d}u:v],[\lambda x:y])$$ and the involution $([u:v],[x:y])\mapsto([v:u],[y:x])$ preserve the centre, including its multiplicities. They act on $X_b$ and generate a reductive group $G=\mathbb G_m\rtimes\boldsymbol\mu_2$. There is no point of the target $\mathbb P^1$ fixed by all of $G$.

The surface $D$ is still $\mathbb P^1\times\mathbb P^1$, with normal bundle $\mathcal O(-1,-1)$. The nef and fixed parts in Lemma 90.1 hold on $X_b$ with the same proof: $B$ is globally generated; restriction to $D$ forces the stated multiples of $D$; and the fibres of every component of $E$ force the stated multiples of $E$. Over a base component of multiplicity $s$, locally $E=sE_\mathrm{red}$ and $E_\mathrm{red}\cdot f=-1/s$ on an exceptional fibre; hence the forced fixed part is still $(u-1)E$. Thus the expected orders and the moving series on a target fibre are unchanged. Vanishing for the nef and big positive parts applies because $X_b$ is canonical.

The preceding lemma gives $\delta_p(X_b)\ge80/73$ at all points of $D$, including points on the multiple components of $C_b$.

For $b=1$, a general target fibre is a smooth quartic del Pezzo surface: the length-five scheme consists of the two simple base points and the three distinct points of a general residual fibre. At its points outside $D$, the proof of Lemma 90.4 gives $\delta_p(X_b)\ge10/9$. For $b=2$, a general fibre has cluster type $(2,2,1)$ and two nodes. At its smooth points outside $D$, the polarized-surface computation in [14, Appendix A], used in Lemma 96.1, gives the bound $100/99$. The restricted classes and the marked conic are exactly the same. The pair $(X_b,T)$ is plt by inversion of adjunction, since $T$ is Cartier and has Du Val singularities; thus the same refinement applies. We need this estimate only at points where the ambient threefold is smooth. The two singular curves of $X_2$ are disjoint and exchanged by the involution, so no irreducible $G$-invariant centre lies in their union.

Let $F$ be a $G$-invariant prime divisor over $X_b$, and put $Z=c_{X_b}(F)$. Its image in the target is irreducible, closed and $G$-invariant, so it is the whole target. Choose a general point $p$ of $Z$ on a general target fibre, outside the singular locus when $b=2$. If $Z$ lies in $D$, use the bound $80/73$; otherwise choose $p$ outside $D$ and use the smooth or nodal fibre bound. In all cases $A_{X_b}(F)>S_{X_b}(F)$. The equivariant valuative criterion for K-polystability, [3, Theorem 1.22], proves the assertion. ◻

Proof. The zero-weight one-parameter subgroup in the GIT lemma gives an isotrivial degeneration to one of the three displayed closed orbits. Its universal blow-up is flat, and all its fibres are canonical Fano threefolds by the rank-two pencil lemma. Thus it is a Q-Gorenstein Fano family. Its central fibre is K-polystable by Lemma 91.1 or the preceding proposition. Openness of K-semistability, followed by the isomorphisms of the nonzero fibres, proves the assertion. ◻

Proof. The ideal of $C$ restricts to the principal ideal of its graph on each $S_i$. Consequently its strict transform $D_i$ is isomorphic to $S_i$, and the exceptional Cartier divisor restricts to $\mathcal O(1,a_i)$. Since $-K_M|_{S_i}=\mathcal O(2,3)$, one has $$-K_Y|_{D_i}=\mathcal O(1,3-a_i).$$ Ampleness would require $a_1,a_2\le2$, contrary to their sum. Nefness requires $a_1,a_2\le3$, leaving only the stated pair. ◻