5.43.6 Maps to a curve3110CB46064E

Proof. The critical set is a proper compact analytic subset of $X$. It has finitely many irreducible components, and $p$ is constant on each: its differential vanishes on the smooth locus of that component, and continuity extends the constant to the component. Thus the set $D\subset C$ of critical values is finite.

Let $U\in \mathrm{PSH}(X,p^*\beta )$. On a base coordinate disk choose a smooth $a$ with $\mathrm{dd}^{\mathrm{c}}a=\beta $. Over $C\setminus D$, the function $U+a\circ p$ restricts to a plurisubharmonic function, or the constant $-\infty $, on each compact connected fiber. The maximum principle makes it constant there. Local holomorphic sections then give a function $v$ with $U=v\circ p$ and $a+v$ subharmonic. This is the local argument of proposition 5.43.5; it uses compactness of the fibers, not compactness of the punctured total space.

The function $v$ is bounded above by $\sup _ XU$. The removable singularity theorem extends $a+v$ subharmonically across the finitely many punctures. The extensions glue, giving $v\in \mathrm{PSH}(C,\beta )$. Its pullback is $p^*\beta $-plurisubharmonic by lemma 5.43.2. It agrees with $U$ off the proper analytic set $p^{-1}(D)$, hence almost everywhere and everywhere by lemma 5.24.3.

Conversely pullback preserves the curvature inequality by the disk criterion. It is not identically $-\infty $, because $p$ is surjective. Surjectivity also proves uniqueness and preserves the supremum and every inequality $v\leq w+c$. $\square$

Proof. A smooth correction of $v$ pulls back to a smooth function, so we may assume that $v$ is subharmonic. The Taylor expansion of $f$ gives $|f(z)|\leq M|z|^ e$ on a small ball. Choose a unit vector $\xi $ on which the leading homogeneous term is nonzero. Then

For all small $r$, the remainder has modulus at most $|a|r^ e/4$ on $|\zeta |=r$. If $|t|<|a|r^ e/4$, Rouché's theorem shows that $f(\zeta \xi )-t$ has $e$ zeros in $|\zeta |<r$. Consequently, with $c=|a|/4$,

The suprema of $v$ on these three sets are ordered. Dividing by $\log r^2<0$, the two disk terms both tend to $e\nu (v,0)$ by the definition in section 5.34. The middle term therefore has the same limit, proving the formula. $\square$

Proof. On a smooth curve, local modifications are isomorphisms. By theorem 5.37.3, the base inclusions are therefore equivalent to $\nu (v,s)\geq \nu (w,s)$ for every $s\in C$. Let $\pi $ be any local modification of a neighborhood in $X$ with smooth source, and let $y$ be a point of its source. In a coordinate at $s=p(\pi (y))$, the germ $p\circ \pi -s$ is nonconstant. Indeed, a modification is an isomorphism on a dense open set, and a nonconstant holomorphic map cannot be constant on an open set. Its order $e_ y$ is positive and finite. The preceding lemma gives

The base inequalities thus imply all inequalities upstairs, and the same valuative theorem gives the ideal inclusions. Conversely the upstairs inclusions give the inequalities for the identity modification at any $x\in p^{-1}(s)$. Dividing by $\operatorname {ord}_ x(p-s)>0$ recovers the base inequalities. Apply this argument in both directions for equality. $\square$

Proof. Put $M_ j=\sup _ Cv_ j$ and $M=\sup _ Cv$. Hartogs' lemma applied at moving maximum points gives $\limsup M_ j\leq M$. A subsequence with limiting suprema below $M$ would give $v$ below that limit almost everywhere, hence everywhere, a contradiction. Thus $M_ j\to M$.

By surjectivity, $(v_ j-M_ j)\circ p$ has supremum zero. Normalized compactness, theorem 5.25.2, makes these pullbacks precompact in $L^1(X)$. On every relatively compact product chart over $C\setminus D$, where $D$ is the finite set of critical values, Fubini's theorem and bounded smooth volume densities give local $L^1$ convergence to $(v-M)\circ p$. Every subsequential limit therefore equals this pullback off $p^{-1}(D)$, and hence almost everywhere, because that analytic set has smooth volume zero. All subsequential limits agree, so the entire normalized sequence converges. Adding $M_ j\to M$ proves the assertion. $\square$

Proof. Composing a local analytic presentation with $p$ proves the forward implication. The pulled-back generators are not all identically zero, since a nonconstant map to a curve is open.

Conversely fix $s\in C$ and $x\in p^{-1}(s)$. In coordinates, choose a complex line through $x$ on which $p-s$ has its finite order $e$ and is nonconstant, as in the Lelong lemma. On that line write

where $h$ is a local biholomorphism fixing zero; take an analytic $e$th root of the nonvanishing factor of the germ. The restriction of $v\circ p$ to the line is not identically $-\infty $, since $v$ is finite almost everywhere and this map is open. By proposition 5.34.1, it has analytic singularities. In one variable its presentation is $r\log |\zeta |^2+O(1)$ with $r\in \mathbb Q_{\geq 0}$, obtained by taking the least order of its finitely many generators. Changing to $w=h(\zeta )$ gives

Every sufficiently small $t$ has an $e$th root $w$, so $v(s+t)=(r/e)\log |t|^2+O(1)$. This is an analytic presentation on the base, including a bounded one when $r=0$. $\square$