Proof. By Lemma 2.7 we have the equality $K_{X}=K_{{\mathcal{F}}}+[N_{{\mathcal{F}}}^{\ast }]$ and by assumption $[N_{{\mathcal{F}}}^{\ast }]$ is a Cartier divisor. Thus $K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5}_{X}=(K_{X}+\unicode[STIX]{x1D6E5}_{X})-[N_{{\mathcal{F}}}^{\ast }]$ and so $K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5}_{X}$ is $\mathbb{R}$ -Cartier.

It suffices to prove the stated equality outside of a codimension 2 subset on $Y$ , and so we may assume that $Y$ is smooth.

By definition in case (i) $N_{{\mathcal{F}}_{Y}}^{\ast }$ is the saturation of $f^{\ast }N_{{\mathcal{F}}}^{\ast }$ in $\unicode[STIX]{x1D6FA}_{Y}^{1}$ (in case (ii) it is the saturation of $f^{\ast }((N_{{\mathcal{F}}}^{\ast })^{\ast \ast })$ in $\unicode[STIX]{x1D6FA}_{Y}^{1}$ ). And so $f^{\ast }N_{{\mathcal{F}}}^{\ast }=N_{{\mathcal{F}}_{Y}}^{\ast }-\unicode[STIX]{x1D6E9}$ (in case (ii) $f^{\ast }((N_{{\mathcal{F}}}^{\ast })^{\ast \ast })=N_{{\mathcal{F}}_{Y}}^{\ast }-\unicode[STIX]{x1D6E9}$ ). Apply Lemma 2.7 to finish. ◻

Proof. The only part which does not follow from Lemma 3.1 is the claim in item (1) that $\unicode[STIX]{x1D708}(\unicode[STIX]{x1D6E9})$ is either contained in $\text{sing}(D)$ or is tangent to ${\mathcal{F}}$ . This is a local problem and so may be checked in a neighborhood of a general point of $\unicode[STIX]{x1D708}(\unicode[STIX]{x1D6E9})$ where ${\mathcal{F}}$ is defined by a 1-form $\unicode[STIX]{x1D714}$ .

Write $\unicode[STIX]{x1D708}^{\ast }(K_{X}+D)=K_{D^{\unicode[STIX]{x1D708}}}+\unicode[STIX]{x1D6E5}$ where $\unicode[STIX]{x1D6E5}$ is the usual different. Observe that $\unicode[STIX]{x1D6E5}$ is supported on $\unicode[STIX]{x1D708}^{-1}(\text{sing}(D))$ .

Thus, we see that $\unicode[STIX]{x1D6E9}$ is supported on the union of the zero locus of $\unicode[STIX]{x1D708}^{\ast }\unicode[STIX]{x1D714}$ and $\text{supp}(\unicode[STIX]{x1D6E5})$ . Let $C\subset D^{\unicode[STIX]{x1D708}}$ be a component of the zero locus of $\unicode[STIX]{x1D708}^{\ast }\unicode[STIX]{x1D714}$ so that $\unicode[STIX]{x1D708}(C)$ is not contained in $\text{sing}(D)$ and let $i:\unicode[STIX]{x1D708}(C)\rightarrow X$ be the inclusion. Then observe that $i^{\ast }\unicode[STIX]{x1D714}=0$ and so $\unicode[STIX]{x1D708}(C)$ is tangent to  ${\mathcal{F}}$ .◻

Proof. We construct $\unicode[STIX]{x1D6E5}_{S^{\unicode[STIX]{x1D708}}}$ as follows: pass to a log resolution $\unicode[STIX]{x1D70B}:Y\rightarrow X$ of $(X,\unicode[STIX]{x1D6E5}+S)$ and write $K_{{\mathcal{F}}_{Y}}+S^{\prime }+\unicode[STIX]{x1D6E5}^{\prime }+E=\unicode[STIX]{x1D70B}^{\ast }(K_{{\mathcal{F}}}+S+\unicode[STIX]{x1D6E5})$ . Let $\unicode[STIX]{x1D70E}:S^{\prime }\rightarrow S^{\unicode[STIX]{x1D708}}$ factor through $S^{\prime }\rightarrow S$ . We may apply Corollary 3.3 item (i) to $({\mathcal{F}}_{Y},S^{\prime }+\unicode[STIX]{x1D6E5}^{\prime }+E)$ (since $Y$ is smooth) and push forward along $\unicode[STIX]{x1D70E}$ .

It remains to check that $\unicode[STIX]{x1D6E5}_{S^{\unicode[STIX]{x1D708}}}\geqslant 0$ . Using Corollary 2.26, the line of argument of the general adjunction statement in the $K_{X}$ case works equally well in the foliated situation; see for example [Reference AmbroAmb07, 9.2.1] or [Reference FujinoFuj11, 14.1]. For the reader’s convenience we explain the argument here.

By taking general hyperplane cuts we may reduce to the case where $X$ is a surface, ${\mathcal{F}}$ is a foliation by curves and $S$ is a curve. Let $\unicode[STIX]{x1D707}:Y\rightarrow X$ be a log resolution of $(X,\unicode[STIX]{x1D6E5}+S)$ , i.e. $Y$ is smooth and $\text{exc}(\unicode[STIX]{x1D707})\cup \unicode[STIX]{x1D707}_{\ast }^{-1}(\text{supp}(\unicode[STIX]{x1D6E5}+S))$ is a simple normal crossings (snc) divisor, and let ${\mathcal{G}}$ be the transformed foliation. Perhaps passing to a higher model we may assume that ${\mathcal{G}}$ has canonical singularities. Let $\unicode[STIX]{x1D6E5}^{\prime }$ denote the strict transform of $\unicode[STIX]{x1D6E5}_{Y}$ and let $S_{Y}$ denote the strict transform of $S$ .

By Corollary 2.26 we may run a $K_{{\mathcal{G}}}+S_{Y}$ -MMP over $X$ terminating in $\unicode[STIX]{x1D70B}:\widetilde{X}\rightarrow X$ . We claim that this MMP only contracts curves $E$ which are disjoint from the strict transform of $S$ . We prove this by induction on the number of steps in the MMP, and so let $Y\rightarrow W$ be some intermediate step of the MMP and let $S_{W}$ denote the strict transform of $S$ and $\unicode[STIX]{x1D6E5}_{W}$ the strict transform of $\unicode[STIX]{x1D6E5}$ . Observe that by assumption $W$ is smooth in a neighborhood of $S_{W}$ so if $E$ is any curve meeting $S_{W}$ we have that $S_{W}\cdot E\geqslant 1$ . On the other hand, the foliated MMP over $X$ only contracts curves with $0>K_{{\mathcal{F}}_{W}}\cdot E\geqslant -1$ . Thus if $(K_{{\mathcal{F}}_{W}}+\unicode[STIX]{x1D6E5}_{W}+S_{W})\cdot E<0$ we see that $E$ must be disjoint from $S_{W}$ .

Therefore we see that $\widetilde{X}$ is smooth in a neighborhood of $S_{\widetilde{X}}$ and moreover $S_{\widetilde{X}}$ is also smooth. Write

$$\begin{eqnarray}\displaystyle (K_{{\mathcal{F}}_{\widetilde{X}}}+\unicode[STIX]{x1D6E5}_{\widetilde{X}}+S_{\widetilde{X}})+\unicode[STIX]{x1D6E4}=\unicode[STIX]{x1D70B}^{\ast }(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5}+S) & & \displaystyle \nonumber\end{eqnarray}$$

where $\unicode[STIX]{x1D6E4}$ is $\unicode[STIX]{x1D70B}$ -exceptional. Observe that, since $K_{{\mathcal{F}}_{\widetilde{X}}}+S_{\widetilde{X}}$ is $\unicode[STIX]{x1D70B}$ -nef and since $\unicode[STIX]{x1D6E5}_{\widetilde{X}}$ is an effective divisor containing no $\unicode[STIX]{x1D70B}$ -exceptional divisor,

$$\begin{eqnarray}\displaystyle -\unicode[STIX]{x1D6E4}=(K_{{\mathcal{F}}_{\widetilde{X}}}+\unicode[STIX]{x1D6E5}_{\widetilde{X}}+S_{\widetilde{X}})-\unicode[STIX]{x1D70B}^{\ast }(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5}+S) & & \displaystyle \nonumber\end{eqnarray}$$

is $\unicode[STIX]{x1D70B}$ -nef and so the negativity lemma, [Reference Kollár and MoriKM98, Lemma 3.38], applies to show that $\unicode[STIX]{x1D6E4}\geqslant 0$ .

For ease of notation set $T=S_{\widetilde{X}}$ and observe that we have an isomorphism $T\rightarrow S^{\unicode[STIX]{x1D708}}$ .

Write $(K_{{\mathcal{F}}_{\widetilde{X}}}+\unicode[STIX]{x1D6E5}_{\widetilde{X}}+S_{\widetilde{X}}+\unicode[STIX]{x1D6E4})|_{T}=K_{{\mathcal{F}}_{T}}+\unicode[STIX]{x1D6E5}_{T}$ and by the fact that $\unicode[STIX]{x1D6E4}\geqslant 0$ and by Corollary 3.3 we see that $\unicode[STIX]{x1D6E5}_{T}\geqslant 0$ . On the other hand, by construction we see that $\unicode[STIX]{x1D6E5}_{T}=\unicode[STIX]{x1D6E5}_{S^{\unicode[STIX]{x1D708}}}$ and so we are done.◻

Proof. This is proven in [Reference BrunellaBru00, pp. 20–21] where $X,Y$ are surfaces, but the proof works equally well for any co-rank 1 foliation.◻

Proof. The statement can be checked locally on $X$ , so consider the following diagram.

Here $f:X^{\prime }\rightarrow X$ is the index 1 cover associated to $N_{{\mathcal{F}}}^{\ast }$ ; note $f$ is étale in codimension 2. Denote by ${\mathcal{F}}^{\prime }$ the foliation on $X^{\prime }$ and let $Y^{\prime }$ be the normalization of $X^{\prime }\times _{X}Y$ . Observe that $g$ is finite.

Next, note $f^{\ast }K_{X}=K_{X^{\prime }}$ and $f^{\ast }K_{{\mathcal{F}}}=K_{{\mathcal{F}}^{\prime }}$ . Write $\unicode[STIX]{x1D6E5}^{\prime }=f^{\ast }\unicode[STIX]{x1D6E5}$ .

Let $E$ be a divisor contracted by $\unicode[STIX]{x1D70B}$ and let $E^{\prime }$ be a divisor contracted by $\unicode[STIX]{x1D70B}^{\prime }$ such that $g(E^{\prime })=E$ ; let $r$ be the ramification index. Working around a general point of $E,E^{\prime }$ we may write

$$\begin{eqnarray}\displaystyle K_{Y}+\unicode[STIX]{x1D70B}_{\ast }^{-1}\unicode[STIX]{x1D6E5} & = & \displaystyle \unicode[STIX]{x1D70B}^{\ast }(K_{X}+\unicode[STIX]{x1D6E5})+aE,\nonumber\\ \displaystyle K_{{\mathcal{F}}_{Y}}+\unicode[STIX]{x1D70B}_{\ast }^{-1}\unicode[STIX]{x1D6E5} & = & \displaystyle \unicode[STIX]{x1D70B}^{\ast }(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5})+bE,\nonumber\\ \displaystyle K_{Y^{\prime }}+\unicode[STIX]{x1D70B}_{\ast }^{\prime -1}\unicode[STIX]{x1D6E5}^{\prime } & = & \displaystyle \unicode[STIX]{x1D70B}^{\prime \ast }(K_{X^{\prime }}+\unicode[STIX]{x1D6E5}^{\prime })+a^{\prime }E^{\prime },\nonumber\\ \displaystyle K_{{\mathcal{F}}_{Y^{\prime }}}+\unicode[STIX]{x1D70B}_{\ast }^{\prime -1}\unicode[STIX]{x1D6E5}^{\prime } & = & \displaystyle \unicode[STIX]{x1D70B}^{\prime \ast }(K_{{\mathcal{F}}^{\prime }}+\unicode[STIX]{x1D6E5}^{\prime })+b^{\prime }E^{\prime }.\nonumber\end{eqnarray}$$

We have $(N_{{\mathcal{F}}^{\prime }}^{\ast })^{\ast \ast }$ is a line bundle sub-sheaf of $\unicode[STIX]{x1D6FA}_{X^{\prime }}^{[1]}$ . Next, by [Reference Greb, Kebekus, Kovács and PeternellGKKP11, Theorem 4.3] we have a morphism

$$\begin{eqnarray}\displaystyle d\unicode[STIX]{x1D70B}^{\prime }:\unicode[STIX]{x1D70B}^{\prime [\ast ]}(N_{{\mathcal{F}}^{\prime }}^{\ast })^{\ast \ast }\rightarrow \unicode[STIX]{x1D6FA}_{Y^{\prime }}^{[1]}. & & \displaystyle \nonumber\end{eqnarray}$$

We apply Lemma 3.1 to see $b^{\prime }\leqslant a^{\prime }$ .

Next, by the Riemann–Hurwitz formula,

$$\begin{eqnarray}\displaystyle K_{Y^{\prime }}+\unicode[STIX]{x1D70B}_{\ast }^{\prime -1}\unicode[STIX]{x1D6E5}^{\prime } & = & \displaystyle g^{\ast }(K_{Y}+\unicode[STIX]{x1D70B}_{\ast }^{-1}\unicode[STIX]{x1D6E5})+(r-1)E^{\prime }\nonumber\\ \displaystyle & = & \displaystyle g^{\ast }(\unicode[STIX]{x1D70B}^{\ast }(K_{X}+\unicode[STIX]{x1D6E5})+aE)+(r-1)E^{\prime }.\nonumber\end{eqnarray}$$

Pulling back the other way around the diagram shows that

$$\begin{eqnarray}\displaystyle a^{\prime }=ra+(r-1). & & \displaystyle \nonumber\end{eqnarray}$$

Likewise, the foliated Riemann–Hurwitz formula, Proposition 3.7, tells us that

$$\begin{eqnarray}\displaystyle K_{{\mathcal{F}}_{Y^{\prime }}}+\unicode[STIX]{x1D70B}_{\ast }^{\prime -1}\unicode[STIX]{x1D6E5}^{\prime } & = & \displaystyle g^{\ast }(K_{{\mathcal{F}}_{Y}}+\unicode[STIX]{x1D70B}_{\ast }^{-1})+\unicode[STIX]{x1D716}(r-1)E^{\prime }\nonumber\\ \displaystyle & = & \displaystyle g^{\ast }(\unicode[STIX]{x1D70B}^{\ast }(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5})+bE)+\unicode[STIX]{x1D716}(r-1)E^{\prime }\nonumber\end{eqnarray}$$

where $\unicode[STIX]{x1D716}=0$ if $E^{\prime }$ is invariant and $=1$ otherwise. Again, pulling back the other way around the diagram gives

$$\begin{eqnarray}\displaystyle b^{\prime }=rb+\unicode[STIX]{x1D716}(E)(r-1). & & \displaystyle \nonumber\end{eqnarray}$$

Since $a\leqslant -1$ , we get that $a^{\prime }\leqslant -1$ . And so $rb+\unicode[STIX]{x1D716}(E)(r-1)=b^{\prime }\leqslant a^{\prime }\leqslant -1$ . This gives that $b\leqslant (-\unicode[STIX]{x1D716}(E)(r-1)-1)/r\leqslant -\unicode[STIX]{x1D716}(E)$ with strict inequality if $\unicode[STIX]{x1D716}(E)=0$ .◻

Proof. Perhaps passing to a higher resolution $\unicode[STIX]{x1D707}:X^{\prime \prime }\rightarrow X^{\prime }$ we may assume that $(X^{\prime \prime },\text{exc}(\unicode[STIX]{x1D70B}\circ \unicode[STIX]{x1D707}))$ is log smooth. Let ${\mathcal{G}}$ be the foliation pulled back to $X^{\prime \prime }$ . Perhaps shrinking around a general point of $Z$ we may assume that every exceptional divisor dominates $Z$ . Let $E$ be one such divisor.

By non-dicriticality of ${\mathcal{F}}$ we see that if ${\mathcal{H}}$ is the foliation restricted to $E$ then ${\mathcal{H}}$ is induced by the fibration $\unicode[STIX]{x1D70E}:E\rightarrow Z$ . Let $f$ be a general fibre of $\unicode[STIX]{x1D70E}$ . Notice that $K_{{\mathcal{H}}}\cdot f=K_{E}\cdot f$ and so

$$\begin{eqnarray}\displaystyle (K_{{\mathcal{G}}}+E)\cdot f=K_{{\mathcal{H}}}\cdot f=K_{E}\cdot f=(K_{X^{\prime \prime }}+E)\cdot f & & \displaystyle \nonumber\end{eqnarray}$$

where the first equality follows from foliation adjunction, Corollary 3.3, and the third equality follows from usual adjunction, hence $K_{{\mathcal{G}}}\cdot f=K_{X^{\prime \prime }}\cdot f$ .

Repeating this computation for every exceptional divisor $E$ shows (perhaps shrinking $X$ a bit more) that $K_{{\mathcal{G}}}$ and $K_{X^{\prime \prime }}$ are $f$ -numerically equivalent, and so their discrepancies agree.◻

Proof. The proof is standard; see for example [Reference Kollár and MoriKM98]. ◻

Proof. Let $\unicode[STIX]{x1D708}:W^{\unicode[STIX]{x1D708}}\rightarrow W$ be the normalization and let $G=\unicode[STIX]{x1D708}^{\ast }(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5})$ .

First, notice that since $W$ is transverse if $E$ is any divisor such that the center of $E$ on $X$ is $W$ then $E$ is transverse to the foliation and we have by Lemma 3.11 that $a(E,{\mathcal{F}},\unicode[STIX]{x1D6E5})=a(E,X,\unicode[STIX]{x1D6E5})$ . In particular, $W$ is a log canonical center of $(X,\unicode[STIX]{x1D6E5})$ .

Let $f:(Y,{\mathcal{H}})\rightarrow (X,{\mathcal{F}})$ be a dlt modification of $(X,\unicode[STIX]{x1D6E5})$ and write

$$\begin{eqnarray}\displaystyle f^{\ast }(K_{X}+\unicode[STIX]{x1D6E5})=K_{Y}+\unicode[STIX]{x1D6E4}^{\prime } & & \displaystyle \nonumber\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle f^{\ast }(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5})=K_{{\mathcal{H}}}+\unicode[STIX]{x1D6E4}. & & \displaystyle \nonumber\end{eqnarray}$$

Since $f$ only extracts divisors of usual discrepancy less than or equal to $-1$ , by Corollary 3.9 it only extracts divisors of foliation discrepancy less than or equal to $0$ , and so $\unicode[STIX]{x1D6E4}\geqslant 0$ .

Observe that if $E$ is a divisor then the coefficient of $E$ in $\unicode[STIX]{x1D6E4}$ (respectively $\unicode[STIX]{x1D6E4}^{\prime }$ ) is $-a(E,{\mathcal{F}},\unicode[STIX]{x1D6E5})$ (respectively $-a(E,X,\unicode[STIX]{x1D6E5})$ ) and so by Lemma 3.11 we know that $\unicode[STIX]{x1D6E4}$ and $\unicode[STIX]{x1D6E4}^{\prime }$ agree above the generic point of $W$ .

Let $E\rightarrow W$ be a divisor dominating $W$ which has coefficient 1 in $\unicode[STIX]{x1D6E4}$ . Notice that $E$ is transverse to ${\mathcal{H}}$ , and if we write ${\mathcal{H}}_{E}$ for the foliation restricted to $E$ we have that ${\mathcal{H}}_{E}$ is the foliation induced by $\unicode[STIX]{x1D70E}:E\rightarrow W^{\unicode[STIX]{x1D708}}$ .

Write $(K_{Y}+\unicode[STIX]{x1D6E4}^{\prime })|_{E}=K_{E}+\unicode[STIX]{x1D6E9}^{\prime }$ and $(K_{{\mathcal{H}}}+\unicode[STIX]{x1D6E4})|_{E}=K_{{\mathcal{H}}_{E}}+\unicode[STIX]{x1D6E9}$ . Note that $\unicode[STIX]{x1D6E9},\unicode[STIX]{x1D6E9}^{\prime }\geqslant 0$ and that if $D\subset E$ is a divisor dominating $W$ it is transverse to the foliation and therefore has the same coefficient in $\unicode[STIX]{x1D6E9}$ and $\unicode[STIX]{x1D6E9}^{\prime }$ (by the construction of the different, Proposition 3.4, and Lemma 3.11). Thus $({\mathcal{H}}_{E},\unicode[STIX]{x1D6E9})$ is log canonical (lc) above the generic point of $W^{\unicode[STIX]{x1D708}}$ .

Suppose for sake of contradiction that $\text{deg}(G)<0$ . Then $K_{{\mathcal{H}}_{E}}+\unicode[STIX]{x1D6E9}=\unicode[STIX]{x1D70E}^{\ast }G$ is not nef and we can apply the cone theorem for surface foliations, see Theorem 6.3 below, to conclude that there is some rational curve tangent to ${\mathcal{H}}_{E}$ which is $K_{{\mathcal{H}}_{E}}+\unicode[STIX]{x1D6E9}$ -negative. However, if $C$ is any curve tangent to ${\mathcal{H}}_{E}$ , then

$$\begin{eqnarray}\displaystyle (K_{{\mathcal{H}}_{E}}+\unicode[STIX]{x1D6E9})\cdot C=\unicode[STIX]{x1D70E}^{\ast }G\cdot C=0, & & \displaystyle \nonumber\end{eqnarray}$$

which contradicts the supposition. ◻

Proof. Let $\unicode[STIX]{x1D708}:S^{\unicode[STIX]{x1D708}}\rightarrow S$ be the normalization of $S$ . Write $\unicode[STIX]{x1D708}^{\ast }(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5}+S)=K_{{\mathcal{F}}_{S^{\unicode[STIX]{x1D708}}}}+\unicode[STIX]{x1D6E9}$ . A straightforward computation (see for example [Reference Kollár and MoriKM98, Propostion 5.46]) shows that the non-log canonical centers of $({\mathcal{F}}_{S^{\unicode[STIX]{x1D708}}},\unicode[STIX]{x1D6E9})$ are contained in non-log canonical centers of $({\mathcal{F}},\unicode[STIX]{x1D6E5}+S)$ .

Since $R\cdot S<0$ , there exists an extremal ray $R^{\prime }$ in $\overline{NE}(S)$ such that $\unicode[STIX]{x1D708}_{\ast }R^{\prime }=R$ in $\overline{NE}(X)$ .

By the cone theorem for surface foliations, see Theorem 6.3 below, $R^{\prime }$ is spanned by a curve $C$ and so $R$ is spanned by $\unicode[STIX]{x1D708}(C)$ . Furthermore:

1. (i) either $C$ is transverse to ${\mathcal{F}}_{S^{\unicode[STIX]{x1D708}}}$ and contained in the non-log canonical locus of $({\mathcal{F}}_{S^{\unicode[STIX]{x1D708}}},\unicode[STIX]{x1D6E9})$ ;

2. (ii) or $C$ is tangent to ${\mathcal{F}}_{S^{\unicode[STIX]{x1D708}}}$ .

In the first case observe that $\unicode[STIX]{x1D708}(C)$ is a non-log canonical center of $({\mathcal{F}},\unicode[STIX]{x1D6E5}+S)$ transverse to  ${\mathcal{F}}$ .

Thus we may find $0<\unicode[STIX]{x1D706}\leqslant 1$ so that $\unicode[STIX]{x1D708}(C)$ is a log canonical center of $K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5}+\unicode[STIX]{x1D706}S$ transverse to ${\mathcal{F}}$ . In fact, since $\unicode[STIX]{x1D708}(C)$ is not a log canonical center of $({\mathcal{F}},\unicode[STIX]{x1D6E5}+S)$ we see that $\unicode[STIX]{x1D706}<1$ . However, $(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5}+\unicode[STIX]{x1D706}S)\cdot C<0$ , a contradiction of Theorem 4.5. Thus $C$ is tangent to the foliation and so $\unicode[STIX]{x1D708}(C)$ is tangent to ${\mathcal{F}}$ .◻

Proof. See [Reference RockafellarRoc70, Corollary 18.7.1]. ◻

Proof. Let $W$ denote the closure of the right-hand side of the desired equality.

Assume that $W$ is strictly smaller than $\overline{NE}(X)$ . Then, by Lemma 6.2, if $H$ is a suitable ample divisor, if we choose $t$ so that $H_{R}=K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5}+tH$ is nef, it is zero precisely on one exposed extremal ray $R$ , not contained in $W$ .

We argue depending on $\unicode[STIX]{x1D708}(H_{R})$ . If $\unicode[STIX]{x1D708}(H_{R})\leqslant 1$ , then, we apply foliated bend and break lemma, Corollary 2.28, setting $D_{i}=H_{R}$ for $i\leqslant \unicode[STIX]{x1D708}(H_{R})+1$ and $D_{i}=H$ otherwise. Then, $D_{1}\cdot D_{2}=0$ and $(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5})\cdot D_{2}=-tH\cdot D_{2}<0$ . We can then conclude that through a general point of $S$ there is a rational curve $\unicode[STIX]{x1D6F4}$ with $D_{1}\cdot \unicode[STIX]{x1D6F4}=H_{R}\cdot \unicode[STIX]{x1D6F4}=0$ , with

$$\begin{eqnarray}\displaystyle M\cdot \unicode[STIX]{x1D6F4}\leqslant 4\frac{M\cdot D_{2}}{-K_{{\mathcal{F}}}\cdot D_{2}} & & \displaystyle \nonumber\end{eqnarray}$$

where $M$ is any nef divisor.

For $k\gg 0$ sufficiently large we see that $A:=kH_{R}-(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5})$ is ample and

$$\begin{eqnarray}\displaystyle A\cdot \unicode[STIX]{x1D6F4}\leqslant 4\frac{(kH_{R}-K_{{\mathcal{F}}}-\unicode[STIX]{x1D6E5})\cdot D_{2}}{-K_{{\mathcal{F}}}\cdot D_{2}}=4\frac{-(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5})\cdot D_{2}}{-K_{{\mathcal{F}}}\cdot D_{2}}\leqslant 4. & & \displaystyle \nonumber\end{eqnarray}$$

In particular, the extremal ray is spanned by the class $[\unicode[STIX]{x1D6F4}]$ and $A\cdot \unicode[STIX]{x1D6F4}=-(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5})\cdot \unicode[STIX]{x1D6F4}\leqslant 4$ .

If $\unicode[STIX]{x1D708}(H_{R})=2$ , then, writing $H_{R}=A+E$ where $A$ is ample and $E$ is effective we see that $E\cdot R<0$ , and hence $R$ is spanned by some component of $E$ ; call it $C$ .

Write $E=rC+E^{\prime }$ with $r>0$ . If $\unicode[STIX]{x1D6E5}$ is a boundary along $C$ , we see that there exists some $\unicode[STIX]{x1D6FC}\geqslant 0$ such that $K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5}+\unicode[STIX]{x1D6FC}E=K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5}^{\prime }+C$ where $\unicode[STIX]{x1D6E5}^{\prime }\geqslant 0$ and $C$ is not contained in $\text{supp}(\unicode[STIX]{x1D6E5}^{\prime })$ . However, we have $(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5}^{\prime }+C)\cdot C<0$ which is a contradiction of foliation adjunction, Proposition 3.4, if $C$ is not invariant. Thus $C$ must be invariant and so $R$ is spanned by an invariant curve, $C$ .

If $C$ is not contained in $\unicode[STIX]{x1D6E5}$ then $\unicode[STIX]{x1D708}^{\ast }(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5})=K_{C^{\unicode[STIX]{x1D708}}}+\unicode[STIX]{x1D6E9}\geqslant -2$ where $\unicode[STIX]{x1D6E9}\geqslant 0$ and where $\unicode[STIX]{x1D708}:C^{\unicode[STIX]{x1D708}}\rightarrow C$ is the normalization. Thus, either $C\subset \text{supp}(\unicode[STIX]{x1D6E5})$ or $C^{\unicode[STIX]{x1D708}}$ is rational and $(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5})\cdot C\geqslant -2$ . We see then that $W$ and $\overline{NE}(X)$ coincide.

Standard arguments then apply to show that the right-hand side of our equality is already closed, and that the extremal rays are locally discrete. ◻

Proof. Choose a suitable ample divisor $H$ so that $H_{R}=(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5})+H$ is nef, and such that $H_{R}$ is zero on precisely one exposed extremal ray, $R$ . We argue based on the numerical dimension of $H_{R}$ .

If $\unicode[STIX]{x1D708}=\unicode[STIX]{x1D708}(H_{R})<3$ , as in the proof of Theorem 6.3, we see that $R$ is spanned by a family of rational curves $\unicode[STIX]{x1D6F4}_{t}$ tangent to ${\mathcal{F}}$ , passing through a general point of $X$ and with $\unicode[STIX]{x1D6F4}_{t}\cdot (K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5})\geqslant -6$ by applying Corollary 2.28 with $D_{i}=H_{R}$ for $1\leqslant i\leqslant \unicode[STIX]{x1D708}(H_{R})+1$ and $D_{i}=H$ for $\unicode[STIX]{x1D708}(H_{R})+1<i\leqslant \text{dim}(X)=3$ .

So, suppose that $\unicode[STIX]{x1D708}(H_{R})=3$ , i.e. $H_{R}$ is big and nef. Taking $\unicode[STIX]{x1D716}>0$ sufficiently small, we may take $K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5}+(t-\unicode[STIX]{x1D716})H$ to be a big $\mathbb{Q}$ -divisor which is negative on $R$ and so there exists some effective prime divisor $D$ such that $D\cdot R<0$ . We have then that $R$ comes from an extremal ray $R^{\prime }\subset \overline{NE}(D^{\unicode[STIX]{x1D708}})$ where $\unicode[STIX]{x1D708}:D^{\unicode[STIX]{x1D708}}\rightarrow D$ is the normalization of $D$ .

Suppose first that $D$ is invariant. By Lemma 5.11 we may write $\unicode[STIX]{x1D708}^{\ast }(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5})=K_{D^{\unicode[STIX]{x1D708}}}+\unicode[STIX]{x1D6E5}_{D^{\unicode[STIX]{x1D708}}}$ where $\unicode[STIX]{x1D6E5}_{D^{\unicode[STIX]{x1D708}}}\geqslant 0$ . The (classical) cone theorem for surfaces shows that $R^{\prime }$ is spanned by a $K_{D^{\unicode[STIX]{x1D708}}}$ -negative rational curve or it is spanned by a curve contained in $\text{supp}(\unicode[STIX]{x1D6E5}_{D^{\unicode[STIX]{x1D708}}})$ . In either case, we see that $R$ is spanned by a curve tangent to ${\mathcal{F}}$ in which case may apply Lemma 5.14 to conclude.

Now suppose that $D$ is transverse to the foliation. In this case, since $R$ is $K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5}$ and $D$ -negative Lemma 4.7 applies to show that $R$ is spanned by the class of a curve tangent to the foliation, in which case Lemma 5.14 applies again.

In both cases, $R$ is spanned by the class of a rational curve $C$ tangent to the foliation with $(K_{{\mathcal{F}}}+\unicode[STIX]{x1D6E5})\cdot C\geqslant -4$ .

Our result then follows by standard arguments to show that the cone of curves indeed has the claimed structure. ◻

Proof. Let $H_{R}$ be a supporting hyperplane to $R$ .

Suppose that $\unicode[STIX]{x1D708}(H_{R})<3$ . As in the proof of the cone theorem, we see that $R$ is spanned by a family of rational curves tangent to ${\mathcal{F}}$ and passing through a general point of $X$ by applying Corollary 2.28 with $D_{i}=H_{R}$ for $1\leqslant i\leqslant \unicode[STIX]{x1D708}(H_{R})+1$ and $D_{i}=H$ for $\unicode[STIX]{x1D708}(H_{R})+1<i\leqslant \text{dim}(X)=3$ where $H$ is an ample divisor.

Let $B$ be the closed subscheme of the Hilbert scheme of $X$ parametrizing those subvarieties tangent to ${\mathcal{F}}$ ; see [Reference Loray, Pereira and TouzetLPT18, § 2.3]. We may consider the following diagram as in [Reference KollárKol91, § 4]: