9.11 Potentials and equations on unibranch spaces77EFAFB55325

Proof. In a local embedding, choose an ambient smooth strictly plurisubharmonic function $r$ with $\mathrm{dd}^{\mathrm{c}}r\geq \theta $ on a smaller neighborhood. For a disk $f$ and a local potential $a$ of $f^*\theta $, the function $r\circ f-a$ is subharmonic. Hence $(u+r)\circ f$ is subharmonic or identically $-\infty $. The Fornaess–Narasimhan criterion, together with the assumption that $u$ is not identically $-\infty $ on any open set, shows that $u+r$ is plurisubharmonic. The smooth case now follows from lemma 5.43.2. $\square$

Proof. Pullback preserves the disk inequalities. Every nonempty open subset of $Y$ meets the locus where $\pi $ is biholomorphic, so a pullback is not identically $-\infty $ on any nonempty open set. Injectivity follows from surjectivity of $\pi $.

For the converse, take $v\in \mathrm{PSH}(Y,\pi ^*\theta )$. Near a fiber, choose the local correction $r$ on $X$ used above. The function $v+r\circ \pi $ is plurisubharmonic on the inverse image of that neighborhood, since

Its restriction to each compact irreducible component of the fiber is constant, possibly $-\infty $, by the maximum principle on a resolution of that component. Connectedness of the fiber makes these constants equal: intersecting components have the same value, and the finite intersection graph of the components is connected. Since $r\circ \pi $ is constant on the fiber, so is $v$.

Thus $v=u\circ \pi $ for a unique function $u$ on $X$. This function is upper semicontinuous: for each real $c$, the closed set $\{ v\geq c\} $ has closed image under the proper map $\pi $, and that image is $\{ u\geq c\} $. If $u$ were identically $-\infty $ on a nonempty open set, the same would hold for $v$ on its inverse image. Hence $u$ is not identically $-\infty $ on any nonempty open set.

It remains to verify the disk condition for $u$ with the background $\theta $ itself. Let $f:\Delta \to X$ be a holomorphic disk with $u\circ f\not\equiv -\infty $. Near a given point of $\Delta $, take a curve in the fiber product $\Delta \times _ XY$ dominating the disk. Its normalization gives, after shrinking, a finite cover $\rho (t)=t^ k$ and a holomorphic lift $f':\Delta \to Y$ with $\pi \circ f'=f\circ \rho $. Choose a smooth local function $a$ on the original disk with $\mathrm{dd}^{\mathrm{c}}a=f^*\theta $. The function

is subharmonic by lemma 5.43.2. Subharmonicity descends through $\rho $ by the finite-disk argument following that lemma. Thus $u\circ f+a$ is subharmonic near every point, which proves that pullback is a bijection.

Since $\pi $ is surjective, pullback preserves suprema, uniform bounds and inequalities up to additive constants; this proves the remaining assertions. $\square$

Proof. The preceding theorem identifies the competitor families. Let $s$ be their pointwise supremum on $X$; its pullback is the pointwise supremum upstairs. The regularization of $s\circ \pi $ on $Y$ is $\pi ^*\theta $-plurisubharmonic and not identically $-\infty $, since there is a competitor below it. By the theorem, it descends to a $\theta $-plurisubharmonic function $q$. Since $q\geq s$ and $q$ is upper semicontinuous, $q\geq s^*$.

For the reverse inequality, fix $x\in X$ and $y\in \pi ^{-1}(x)$. For each neighborhood $U$ of $x$, the unregularized supremum upstairs is bounded by $\sup _ U s$ on the open neighborhood $\pi ^{-1}(U)$ of $y$. Therefore $q(x)\leq \sup _ U s$, and taking the infimum over $U$ gives $q(x)\leq s^*(x)$. This proves the identity and shows that $P_\theta (f)=s^*$ is $\theta $-plurisubharmonic. The proof for $P_\theta [u]$ is identical, because pullback preserves the ordinary singularity order together with its constants. $\square$

Proof. For a fixed resolution, the right-hand side exists by theorem 18.4.2 and descends by theorem 9.11.5. Under a further smooth modification, the competitor families of the ideal envelopes correspond by the descent theorem and lemma 5.43.24, and the regularized suprema commute by the argument of proposition 9.11.7. Thus two resolutions give the same pullback on a common refinement, and hence the same function on $X$. Idempotence, preservation of the class and the normalization follow from the corresponding assertions on $Y$ and surjectivity of $\pi $. $\square$

Proof. Split the quadrant along $q=Ap/B$. On the part $q\geq Ap/B$, set $p_0=\max \{ T,BT/A\} $. After discarding the strip $T<p<p_0$, integration in $q$ gives

The discarded strip has finite integral because $\beta >0$. On the other part, integration in $p$ gives the corresponding tail with coefficient $\beta -B+\alpha B/A$; its discarded strip is finite because $\alpha >0$. Both coefficients are positive exactly when $\alpha /A+\beta /B>1$. If that sum equals one, either tail is an integral of a positive constant over a half-line. $\square$

Proof. The function is the restriction of a maximum of two plurisubharmonic functions on $\mathbb C^3$. On the axes parametrized by $(z_1,z_2,z_3)=(\zeta ,0,0)$ and $(0,\zeta ,0)$, its logarithmic coefficients are $\sqrt2$ and $2$. An analytic singularity $c\log \sum _ j|f_ j|^2+O(1)$ restricts to a logarithmic coefficient equal to $c$ times an integer on either axis, provided that it is finite off the origin there. The ratio of the two coefficients would therefore be rational, which it is not.

Pullback by the finite degree-two map reduces local integrability to the cover. Neighborhoods of the origin contain and are contained in images of sufficiently small bidisks, so it suffices to integrate over a bidisk in $s,t$. The pullback weight is

For $s^ at^ b$, $a,b\geq 0$, set $p=-\log |s|^2$ and $q=-\log |t|^2$. By (9.11.11), angular integration gives, up to a positive constant,

for the boundary $D$. For zero boundary, replace $a+\tfrac 12$ by $a+1$. Thus lemma 9.11.13 gives the respective conditions

These monomial tests determine the entire ideal. Indeed, the measure and the weight on the cover are invariant under separate rotations of $s,t$. Parseval's identity and Tonelli's theorem show that every nonzero coefficient of a weighted-integrable holomorphic series must satisfy its monomial test. Only exponents with $a+b$ even occur.

In the first test, the minimal allowed exponents, under addition of nonnegative even-total exponents, are

To see this, every term with $a\geq 2$ is divisible by $s^2$, every term with $a=1$ has odd $b\geq 1$ and is divisible by $st$, and a term with $a=0$ has even $b$ and passes the test exactly when $b\geq 4$. Each of the three displayed exponents passes the test. Splitting a convergent invariant series into these three groups and dividing gives convergent invariant coefficient series on a smaller bidisk. They descend to holomorphic germs on $X$, bounded on a smaller neighborhood. Since the three generators are weighted-integrable, every such combination is weighted-integrable. This proves the first ideal at the origin. For the second test, every nonconstant even-total monomial is allowed, with minimal exponents $(2,0),(1,1),(0,2)$, and the same argument gives the second ideal.

Away from the origin, $u$ is locally bounded and $X$ is smooth. The only additional volume singularity lies along $E$: with $z_2\neq 0$, $z_3$ is a normal coordinate and $z_1=z_3^2/z_2$, so $\mathrm{d}V_ D$ is a smooth positive factor times $|z_3|^{-1}$ times smooth volume. This is locally integrable. Both ideals therefore equal $\mathcal O_ X$ there, as do the ideals generated by the displayed coordinates, which proves the sheaf identities. $\square$

Proof. Set $F=g-\ell $. Suppose its recession is positive on $C\setminus \{ 0\} $. Then $F(x)\geq \gamma |x|-A$ on $C$ for some $\gamma >0$ and $A$. Otherwise there are $x_ j\in C$ with $|x_ j|\to \infty $ and $F(x_ j)/|x_ j|\leq 1/j$. After taking a subsequence, $x_ j/|x_ j|\to v\in C$ with $|v|=1$. For fixed $t>0$, convexity gives

Continuity gives $F(tv)\leq F(0)$ for every $t$, contradicting positive recession. A bound on the remaining compact set gives the asserted $A$. Exponential decay proves integrability. The same bound persists when $\ell $ is changed by a sufficiently small linear form, so $\ell $ lies in the stated interior.

Conversely, if $(\operatorname {rec}F)(v)\leq 0$ for some nonzero $v\in C$, then $F(tv)\leq F(0)$ for $t\geq 0$. Choose a compact ball $K$ in the interior of $C$. On the half-tube $\{ tv+x/2:t\geq 1,\ x\in K\} $, convexity gives

Thus $\mathrm e^{-2F}$ has a positive lower bound on a set of infinite volume. If instead $\ell $ is in the stated interior, for each nonzero $v\in C$ one may perturb $\ell $ within that interior by a linear form positive on $v$. Dividing its defining upper bound along $tv$ by $t$ gives the strict recession inequality for $\ell $. Finally, translating $C$ replaces $g(x)$ by $g(x+b)$ and multiplies the exponential by a constant; recession is unchanged. $\square$

Proof. On the torus, polar coordinates turn the weighted monomial integral into a positive constant times

where $Q$ is the logarithmic image of the domain. The omitted toric divisors have measure zero on the regular locus. A basis of neighborhoods of $o$ is given by $|\chi ^{b_ j}|<\epsilon $ for every $j$, with $0<\epsilon <1$. Their logarithmic domains satisfy

for $b\in \operatorname {int}\sigma $ sufficiently far along an interior ray. Indeed, every $\langle b_ j,b\rangle $ is strictly positive, and the $b_ j$ generate the dual cone. The convex-cone lemma gives necessity and sufficiency at $o$, including divergence when a recession inequality is an equality.

For a compact subset of $X_\sigma $, the embedding coordinates have absolute value at most some $R>1$. Its logarithmic image is contained in $\{ x:\langle b_ j,x\rangle \leq \log R\} $, which is contained in $b_ R-\sigma $ for a sufficiently large $b_ R\in \operatorname {int}\sigma $. Under the same strict condition the integral over this translated cone is finite. This proves local integrability everywhere. The converse follows by testing at $o$, and the interior formulation follows from the lemma. $\square$

Proof. For a finite sum and $c>0$, its $c$-th power is comparable to the sum of the $c$-th powers of its terms. Thus the integrand is comparable to

Each term gives the preceding logarithmic integral with exponent $m+w_ D+ca_ j$. The cone criterion applies to real exponents, so no rationality of $c$ is needed for this integrability assertion. $\square$

Proof. On the torus use $x,y,z$ as characters, so $w=yz/x$. The character lattice is $\mathbb Z^3$, with semigroup generators

The dual cone has primitive rays $(0,1,0)$, $(0,0,1)$, $(1,1,0)$ and $(1,0,1)$. It is full-dimensional with four rays. The residue form $\Omega =\mathrm dx\wedge \mathrm dy\wedge \mathrm dz/x$ generates the canonical sheaf of this hypersurface; it equals $yz$ times the logarithmic torus form. Hence $w_0=(0,1,1)$, which pairs to one with each primitive ray.

Write $d(m)=m_1+m_2+m_3$. Each semigroup generator has degree one. The logarithmic function $g$ differs by a bounded term from $\lambda $ times the maximum of the four linear forms given by those generators. Their convex hull is the section of $\sigma ^\vee $ by $d=1$, so

Since $w_0\in \operatorname {int}\sigma ^\vee $ and $d(w_0)=2$, the character condition is $d(m)+2>\lambda $. The smallest allowed integer degree is $k=\max \{ 0,\lfloor \lambda \rfloor -1\} $.

This also gives the ideal for whole holomorphic germs. Lift a germ to a convergent power series in $x,y,z,w$. On a smaller invariant neighborhood, angular Parseval and Tonelli force every nonzero character coefficient to satisfy the same test. All homogeneous terms of degree less than $k$ therefore vanish in the coordinate ring. They are multiples of the homogeneous relation $xw-yz$, and subtracting those finitely many terms leaves a lift of order at least $k$. Thus the germ belongs to $\mathfrak m_ o^ k$. Conversely, every degree-$k$ monomial is integrable by the character test, and bounded holomorphic coefficient germs preserve integrability. This proves the identity and its strict endpoint. $\square$

Proof. Given two resolutions, choose a proper analytic subset $S\subset X$ containing $X_{\mathrm{sing}}$ and the images of both exceptional sets. Each map is an isomorphism over $X\setminus S$, and there the two measures are given by the same bounded local products, followed by truncation and exhaustion. On each resolution, the inverse image of $S$ is a proper analytic subset and has zero extended mass by theorem 13.14.3. Hence both pushforwards give $S$ zero mass. This proves independence, including for resolutions whose exceptional sets lie over regular points.

Positivity follows from positivity on $Y$. On $X\setminus S$ either resolution is a Borel isomorphism, so a countable exhaustion of $Y$ by sets of finite measure pushes forward to one on $X$; adjoining the null set $S$ proves $\sigma $-finiteness. A locally pluripolar subset of $X$ pulls back to a pluripolar set off the exceptional locus, and the exceptional locus itself has zero mass. This gives the asserted vanishing.

If $u$ is bounded, its pullback is bounded on the compact manifold $Y$, and local bounded-product estimates give finite total mass. For general nonpositive $u$, apply theorem 13.15.23 on $Y$ to the smooth background $\pi ^*\theta $. Pushforward and the identity $u\circ \pi $ identify the weighted integrals exactly, and the constants come from $\pi ^*\theta $ and a Hermitian metric on $Y$. $\square$

Proof. We first establish the divisor formula

Indeed, $\log x=\log |t|^2+\log S-\log (1+|t|^2S)$ and $\mathrm{dd}^{\mathrm{c}}\log |t|^2=[t=0]$. Along $E$ the kernels of $\beta $ and $\sigma $ are complementary, and $\beta >0$ off $E$. Thus $\beta _ a$ is a smooth Kähler form. Fiber integration also gives

By (9.11.31), $P_ a(0)=a$, $P_ a(1)=1$ and

The numerator defining $Q_ a'$ vanishes at both ends, so $Q_ a$ is smooth. On $0<x<1$, a local potential for the full positive current is $F_ a(\rho _\psi )$, where

Direct differentiation gives $F_ a'=P_ a(s)$ and $F_ a''=c_ a k(s)s^ n(1-s)/P_ a(s)^{n-1}$. Consequently

The coefficients are smooth since $P_ a\geq a>0$. On $E$ the restriction is $a\sigma _\psi $; at infinity the horizontal coefficient is one and the transverse coefficient relative to $B_\psi -\sigma _\psi $ is $c_ a k(1)>0$. This proves all the asserted positivity properties, including along both sections.

The functions $\varphi \circ \pi $ and $\varphi \circ \pi +q_ a$ are quasi-plurisubharmonic on $Y$. By (9.11.36), their curvature currents in the background $\beta $ are $a[E]+\beta _ a$ and $a[E]+A_ a$, respectively, so both are positive. The descent theorem theorem 9.11.5 gives $\varphi \in \mathrm{PSH}(X,\theta _0)$ and a potential $u_0$ with $u_0\circ \pi =\varphi \circ \pi +q_ a$. The bounded residual $q_ a$ puts $u_0$ in the ordinary class of $\varphi $ and gives the uniform logarithmic asymptotics at $p$. Its finite supremum is the constant $K$ of the theorem.

Let $H$ be admissible. Self-adjointness makes $\exp H$ positive, and $\det \exp H=\exp (\operatorname {tr}H)=1$. Hence

The function $u=u_0-K$ does not depend on $H$. On the regular locus its $\theta _ H$-curvature form is the positive form in (9.11.39); near $p$ the background and the potential are unchanged. Thus $u$ is $\theta _ H$-plurisubharmonic. Formula (9.11.36), together with the deformation, gives $\pi ^*\theta _ H+\mathrm{dd}^{\mathrm{c}}(u\circ \pi )=a[E]+A_ a+\pi ^*\eta _ H$; removing the divisor proves the equation. The mass of the solution is $c_ aM=V$. This calculation uses the preservation of the determinant, rather than a mass identity for the class of a nonclosed form.

Bounded residuals and uniqueness. Let $w\in \mathcal C_{a,H}$. On $Y\setminus E$, its residual $g=w\circ \pi -\varphi \circ \pi $ is bounded and $\beta _{a,H}$-plurisubharmonic. By lemma 5.24.7, it extends across $E$ with the same background. Conversely, for any bounded $\beta _{a,H}$-plurisubharmonic $g$, the function $\varphi \circ \pi +g$ has the positive curvature current $a[E]+\beta _{a,H}+\mathrm{dd}^{\mathrm{c}}g$ in the pulled-back background. It descends by theorem 9.11.5, and its uniform bounds relative to $\varphi \circ \pi $ put the descended potential in $\mathcal C_{a,H}$. Thus bounded residuals correspond exactly to members of the prescribed ordinary class.

The bounded product is finite on the compact manifold $Y$ and gives $E$ zero mass, by theorem 5.27.1, proposition 5.28.2. Thus

For $H=0$, closedness of $\beta _ a$ makes every term of positive degree in $\mathrm{dd}^{\mathrm{c}}g$ in the bounded expansion integrate to zero, so (9.11.37) gives mass $V$ for every member of the class. In the nonclosed case only finiteness is used here.

If $w$ solves the equation with a scalar $c'>0$, its lifted measure is $c'\nu $, since both sides give $E$ zero mass. Both residuals, $g$ and $q_ a-K$, are continuous by corollary 14.19.29, and the uniqueness of the scalar in proposition 14.19.34 gives $c'=c_ a$. The density $\nu /\beta _{a,H}^ n$ is smooth and nonnegative, and its positive locus is exactly $Y\setminus E$. This locus is connected and dense: it is a bundle over the connected base $B$ whose fibers, $\mathbb P^1$ minus a point, are connected. Theorem 14.24.1 shows that $g-(q_ a-K)$ is constant. The normalization fixes this constant, and both full functions equal $-\infty $ at $p$. This proves uniqueness in the ordinary class.

The uniform logarithmic asymptotics give the continuity of the full solution with values in $[-\infty ,+\infty )$. Its residual restricts to $E$ as $a\psi +Q_ a(0)-K$. If $\psi $ is nonconstant, this restriction is nonconstant, so the residual is not a pullback from $X$, since a pullback is constant on $E$. This concerns only the residual; the full potential is $-\infty $ on the whole fiber $E$ and does descend.

Nonclosed deformations. Fix $z_0\in O$ and a smooth $A_ a$-unitary frame on a smaller ball. Let $K_0=\operatorname {diag}(1,-1,0,\ldots ,0)$ in this frame. Choose a real compactly supported function $\chi $ with $\chi (z_0)=0$ such that $\mathrm{d}\chi (z_0)$ is a nonzero covector in the first two complex directions, and, for $\epsilon \in \mathbb R$, put $H=\epsilon \chi K_0$. At $z_0$,

Terms containing derivatives of the frame and of $A_ a$ vanish at $z_0$ because $\chi (z_0)=0$. The restriction of $\gamma $ to these four real directions is nondegenerate, so wedge multiplication by $\gamma $ sends a nonzero one-form to a nonzero three-form. Thus $\mathrm{d}\theta _ H(z_0)\ne 0$ for $\epsilon \ne 0$. On the fixed compact support, both strict inequalities (9.11.33) persist for sufficiently small $|\epsilon |$, and $H\to 0$ smoothly as $\epsilon \to 0$. This proves the existence of arbitrarily small nonclosed admissible choices; admissibility itself does not force nonclosedness.

For any resolution $\tau :Z\to X$, theorem 9.11.26 identifies the pushforward of its non-pluripolar product with (9.11.40), which proves (9.11.35). Its total mass equals the finite mass on $X$. Since the ordinary class, the equation and the normalization are intrinsic to $X$, uniqueness gives the same solution and scalar on every resolution and common refinement. $\square$

Proof. Set $P(s)=sR(s)$ with the data in (9.11.42). Then

Smoothness and positivity of $R$ follow from its integral formula. Since $R(1)=1$, $Q$ is smooth at $s=1$, and its derivative is also smooth at zero. Put $U=\ell +Q(s)$. It is smooth on $Y$ and equals $Q(0)$ on $E$. On $0<x<1$, a local potential for $\beta +\mathrm{dd}^{\mathrm{c}}U$ is $F(\rho _\psi )$, where $F(r)=h(r)+Q(h'(r))$. Differentiation gives $F'=P(s)$ and $F''=c_ bk(s)s^ n(1-s)/P(s)^{n-1}$.

For $A=\beta +\mathrm{dd}^{\mathrm{c}}U$, it follows that

This formula shows smoothness at zero directly. At infinity the horizontal coefficient is one and the transverse coefficient is positive. Thus $A$ is semipositive on $Y$ and strictly positive on $Y\setminus E$. By theorem 9.11.5, $U=u_0\circ \pi $ for some $u_0\in \mathrm{PSH}(X,\theta _0)$. The proper surjection $\pi $ is a quotient map, so continuity of $U$ makes $u_0$ continuous. Subtracting $K_ b=\sup _ Xu_0$ gives the required solution.

Let $w$ be a bounded $\theta _0$-plurisubharmonic function. Its pullback $g=w\circ \pi $ is bounded and $\beta $-plurisubharmonic. Its bounded Monge–Ampère measure gives analytic sets zero mass and pushes forward to $\mathrm{MA}^{\mathrm{ext}}_{X,\theta _0}(w)$. Integrating the bounded expansion and using that $\beta $ is closed gives $\int _ Y(\beta +\mathrm{dd}^{\mathrm{c}}g)^ n=\int _ Y\beta ^ n=d$; this forces the scalar of any solution to be $c_ b=d/M$. The resolution identity follows from theorem 9.11.26. Now suppose that $w$ solves the equation, and put $q=U-K_ b$, $D=\beta +\mathrm{dd}^{\mathrm{c}}g$ and $v=g-q$. The bounded integration-by-parts calculus of section 5.27 gives

Each integral in the sum is nonnegative, so all of them vanish; in particular, so does the one with the smooth factor $A^{n-1}$. In the weak-gradient interpretation, strict positivity of $A$ on each relatively compact chart of $Y\setminus E$ forces $\mathrm{d} v=0$ there. Since $Y\setminus E$ is connected, $v$ is constant almost everywhere on it. The divisor $E$ has volume zero, so lemma 5.24.3 gives equality everywhere on $Y$. Surjectivity of $\pi $ and the normalization give $w=u$ on $X$.

If $\Psi =0$ and the solution is constant, then $U=Q(x)$ is constant, so $Q'=0$ and $R=1$. Differentiating (9.11.44) then gives $c_ bk=1$, so $k$ is constant. If $\psi $ is nonconstant, the restriction of $U$ to the infinity section is $\psi $, so the solution is nonconstant for every $k$. $\square$

Proof. With the normalization of $\mathrm{dd}^{\mathrm{c}}$ fixed in the Conventions, $\mathrm{dd}^{\mathrm{c}}v=(\Delta v)/(4\pi )\, dA$ and $\mathrm{dd}^{\mathrm{c}}\log |z|^2=\delta _0$, as in lemma 5.37.18. Cutting off this logarithm in a coordinate disk, with a cutoff equal to one near zero, gives a function $L$ on $E$ with $\mathrm{dd}^{\mathrm{c}}L=\delta _0+q\beta _0$, where $q$ is smooth and $\int _ Eq\beta _0=-1$ by Stokes' theorem.

Let $e_\xi $ be the characters of the torus, with $-\Delta e_\xi =\lambda _\xi e_\xi $. The nonconstant characters have $\lambda _\xi >0$, growing quadratically on the dual lattice. Define the correction $S$, of mean zero, by its Fourier coefficients

where $(q+1)_\xi $ is the Fourier coefficient of $q+1$. Integration by parts shows that these coefficients decay faster than any power of $\lambda _\xi $. Hence the series and all its derivatives converge uniformly, and $\mathrm{dd}^{\mathrm{c}}S=-(q+1)\beta _0$. Conjugate symmetry of the coefficients makes $S$ real. Subtracting the mean of $L+S$ from $L+S$ proves (9.11.52). In particular, $G$ belongs to $L^ r$ for every finite $r$.

The convolution uses the group law and the Haar probability $\beta _0$. Write $K=G*(r\beta _0)$. Testing against smooth functions and using Fubini's theorem proves its distributional equation. Smooth input gives smooth output by the same Fourier argument. For $p'=p/(p-1)$, Hölder's inequality gives

Approximating $G$ by smooth functions shows that translation is continuous in $L^{p'}$, and hence that $K$ is continuous. $\square$

Proof. Set $c=M/F$ and

The lemma gives a bounded continuous $v$ and the positive current

In each disk, adding a smooth local primitive of $\beta $ gives a function with positive distributional Laplacian, and hence a subharmonic representative. Continuity off the poles and the explicit logarithmic values identify this representative with $U$ as defined above. The same argument applies to $\Psi $. The supremum $S=\sup _ EU$ is finite, and we put $u=(U-S)\circ \pi ^{-1}$.

Every holomorphic disk $g:\Delta \to X$ lifts holomorphically to $\widetilde g=\pi ^{-1}\circ g$. Indeed, for a nonconstant disk the inverse image of the finite singular set is discrete, and off this set the lift is holomorphic. Since $\pi ^{-1}$ is a homeomorphism, $\widetilde g$ is continuous on the whole disk. Near an omitted point its image therefore stays in a bounded coordinate neighborhood in $E$, and the removable singularity theorem in one variable extends it holomorphically. A constant disk lifts to a constant map, and a nonconstant disk cannot have image in a finite set.

Choose a smooth local primitive $\eta $ of $\beta $ near the lifted image, that is, $\mathrm{dd}^{\mathrm{c}}\eta =\beta $. Then $\eta \circ \widetilde g$ is a primitive of $g^*h$, and

is subharmonic or identically $-\infty $. That the composition is subharmonic follows from decreasing smooth local regularizations and the holomorphic chain rule for the Laplacian. Any other primitive on the disk differs by a harmonic function. These are exactly the disk tests of definition 9.11.2, and the same argument shows that $\psi \in \mathrm{PSH}(X,h)$. A constant disk at a pole has the permitted restriction $\equiv -\infty $. A nonconstant lift meeting $q_ j$ with order $\ell $ gives the pole coefficient $\ell a_ j$.

The homeomorphism $\pi $ gives continuity off the poles. Near a pole, the smooth logarithmic remainder of $\Psi $ and the bounded continuous function $v$ give continuity with extended values. The residual is the descent of $v-S$, including at the poles, so no difference of two infinite values is involved. The solution belongs to the stated ordinary type.

For the mass and uniqueness assertions, let $w$ be any member of the type and put $W=w\circ \pi $. Pulling back along coordinate disks of the normalization shows that $W$ is $\beta $-plurisubharmonic and not identically $-\infty $. The two type inequalities make $W$ locally bounded away from $Q=\{ q_ j\} $, and at each pole they give

with value $-\infty $ at $q_ j$. The positive Radon measure $T_ W=\beta +\mathrm{dd}^{\mathrm{c}}W$ has total mass $H$ by Stokes' theorem. Its atom at $q_ j$ has weight exactly $a_ j$. This follows from the planar atom identity of lemma 5.37.18. Explicitly, choose a smooth local primitive $\eta $ of $\beta $ near $q_ j$, and let $m(r)$ be the circular mean of $W+\eta $ around $q_ j$. Since $\mathrm{dd}^{\mathrm{c}}(W+\eta )=T_ W$, it satisfies

for almost every $r$, by the divergence theorem and regularization. If $\tau =T_ W(\{ q_ j\} )$, integration gives $m(r)/\log r^2\to \tau $, while the bounded logarithmic remainder gives the limit $a_ j$.

Hence the non-pluripolar product of $W$, defined in dimension one by truncation, is

Indeed, on each relatively compact set avoiding $Q$, a sufficiently low truncation equals $W$, and the indicator removes the poles. Its mass is $H-A=M$ for every member of the type. For (9.11.55), removing the atoms leaves $cf\beta _0$. The definition (9.11.25) gives the asserted equation on $X$. A locally pluripolar set pulls back to a set on which some quasi-subharmonic function, not identically $-\infty $, equals $-\infty $; such sets have zero flat area. Hence $\nu $ is non-pluripolar.

Suppose that $w$ solves the equation with a constant $c'>0$. Comparing total masses gives $c'F=M$, so $c'=c$. Since $\pi $ is a Borel homeomorphism and the atoms are fixed, $T_ W=T_ U$. Thus $\mathrm{dd}^{\mathrm{c}}(W-U)=0$, where $W-U$ is an integrable function. Every nonconstant Fourier coefficient of $W-U$ vanishes, so the difference is constant almost everywhere. Subharmonic functions are determined by their averages over shrinking disks, so the values of $W$ and $U$ differ by the same constant everywhere, and the same holds on $X$. This includes members of the type with no regularity assumed.

Finally, a resolution $\rho :Y\to X$ factors through $E$ by the disk-lifting argument. The resulting proper holomorphic map $Y\to E$ has generic degree one. For smooth compact curves, the multiplicities in a fiber sum to the degree, so every fiber is a single point of multiplicity one, and the map is biholomorphic. The same holds for every common refinement, which proves the independence assertion directly. $\square$

Proof. For the square lattice define

The summands are $O(|\lambda |^{-3})$ on compact sets away from the lattice, so the series converges normally. Pairing opposite lattice points shows that $x$ is even. The differentiated series is absolutely and normally convergent, and reindexing shows that $y$ is periodic. Thus $x(z+\omega )-x(z)$ is constant for $\omega =1,\mathrm i$; evaluating at $-\omega /2$ and using evenness shows that this constant is zero. Both functions are meromorphic on $E$, with their only poles at zero and expansions

Cancelling the Laurent expansions leaves an elliptic function with no poles and zero constant term, which vanishes by the maximum principle. This gives

The argument principle on a translated fundamental parallelogram shows that $x-c$ has two zeros, counted with multiplicity, for every finite $c$. Its fibers are therefore of the form $\{ z,-z\} $. If $y(z)=0$, the local multiplicity is at least two. Distinct points $z,-z$ would then give at least four zeros, so $z=-z$ on $E$, the multiplicity is exactly two, and $x''(z)\ne 0$. Since $y$ is odd, the map $(x,y)$ is injective and immersive away from zero: use $dx$ where $y\ne 0$, and $dy=x''dz$ otherwise.

Define

In the last affine chart its holomorphic coordinates are

Equation (9.11.58) becomes

Dividing $t$ by an analytic square root of the nonvanishing factor gives the ordinary cusp $t'^2=s^3$. Choose the local parameter $\zeta $ so that $s=\zeta ^2$ and $t'=\zeta ^3$. The local ring of the cusp is $\mathbb C\{ \zeta ^2,\zeta ^3\} $, with normalization $\mathbb C\{ \zeta \} $. To see this directly, note that division by $t'^2-s^3$ writes every germ as $A(s)+t'B(s)$, and the even and odd parts identify the image ring. Its fraction field contains $\zeta $, which is integral since $\zeta ^2=s$, and the larger ring is a finite module generated by $1$ and $\zeta $. An integral meromorphic germ cannot have negative order: in a monic equation, its highest power would have strictly smallest order. Hence $\mathbb C\{ \zeta \} $ is the full integral closure.

The image is the reduced projective locus

On $Z_0\ne 0$ these equations describe the graph of the affine cubic, and since the fibers of $x$ have degree two, both square-root values of $y$ occur. On $Z_0=0$ they force $Z_1=Z_3=Z_2=0$, leaving only $\pi (0)$. In the last chart they give exactly the cusp displayed above. The finite local module at the cusp and the biholomorphic charts elsewhere show that $\pi $ is finite. Hence $\pi $ is the bijective normalization of a connected curve that is smooth away from this one cusp.

For the restricted Fubini–Study form, $\beta =b\beta _0$ is positive away from zero and

near zero. Indeed, the derivatives of the coordinates in the last chart vanish at zero, and $ds=2z\, dz+O(z^5)dz$ gives the leading term. On the complement of zero, the Fubini–Study potential is

Globally $\mathrm{dd}^{\mathrm{c}}L=\beta -6\delta _0$, so integrating gives $H=6$. Since the normalization has generic degree one, the degree two of $x$ contributes no factor to the measure.

For the pole of weight one, the reference and the continuous correction are

Direct differentiation gives $\beta +\mathrm{dd}^{\mathrm{c}}\Psi =5\beta _0+\delta _0$ and $\mathrm{dd}^{\mathrm{c}}v=(5f-5)\beta _0$. Since $F=1$, theorem 9.11.54 gives $c=5$, the mass, and uniqueness in the whole type. The residual is the descent of $v-\sup _ E(\Psi +v)$. Finally, $f/b$ is comparable to $|z|^{-2}$ near zero, and the integral of its $r$th power against $\beta =b\beta _0$ is comparable to

This is finite exactly when $r<2$, with logarithmic divergence at $r=2$. $\square$

Proof. The coefficient below differs from that in (10.23.2) because here $\beta _0=dA/V$. Define on the punctured coordinate disk

Direct differentiation gives

off $q$. The flux of $P_\gamma $ through the circle of radius $\mathrm{e}^{-t}$, which computes the mass of $\mathrm{dd}^{\mathrm{c}}P_\gamma $ inside this circle, is

as $t\to \infty $. Thus the distributional extension across $q$ adds no atom. Also $P_\gamma $ is locally integrable and $|P_\gamma (t)|/t\to 0$, so its Lelong number at $q$ is zero. These are the same flux and growth mechanisms as in theorem 10.23.1, with the present normalization of area.

Take a smooth cutoff $\chi $ supported in the coordinate disk and equal to one near $q$, and extend $L=c_\gamma \chi P_\gamma $ by zero. The signed measure

has a smooth density $\rho $ on all of $E$: near $q$ the singular densities cancel, leaving $-M\beta _0$. Its integral is zero by Stokes' theorem and $c_\gamma F_\gamma =M$. Hence lemma 9.11.51 gives a smooth correction $R=G*(\rho \beta _0)$. Set $v_\gamma =L+R$ and $U_\gamma =\Psi +v_\gamma $. Then

As before, the positive local Laplacians give subharmonic representatives. The disk-lifting argument in the proof of theorem 9.11.54 descends $U_\gamma -\sup _ EU_\gamma $. It is smooth away from the original poles and from $q$. Its non-pluripolar product removes the original atoms, and a possible new value $-\infty $ at $q$ carries no atom. The mass is therefore $M$, and (9.11.25) gives the equation on $X$.

For $\gamma >2$, $P_\gamma \to 0$, so $v_\gamma $ extends continuously and boundedly, and the solution belongs to the prescribed ordinary type. The difference of $v_\gamma $ from its value at $q$ has a nonzero leading term proportional to $t^{2-\gamma }$, with a smooth remainder of order $O(r)$. This decays more slowly than every power $r^\alpha $, $\alpha >0$, so the residual is not Hölder continuous for any positive exponent. Uniqueness follows from the mass of the whole type, the restoration of the fixed pole atoms and the harmonic-difference argument used in the proof of theorem 9.11.54; none of these steps requires $p>1$ once the bounded correction exists.

For $1<\gamma \leq 2$, $P_\gamma \to -\infty $. The correction is bounded above but unbounded below near $q$, so the inequality $\Psi \leq U_\gamma +C$ fails off the pole set. Since its Lelong number is zero, the point atoms of $\beta +\mathrm{dd}^{\mathrm{c}}U_\gamma $ are those of $\beta +\mathrm{dd}^{\mathrm{c}}\Psi $. Applying lemma 5.37.18 to each positive multiple, after a smooth local correction, shows that all multiplier ideals of $U_\gamma $ and $\Psi $ on $E$ coincide, also when $q$ is not one of the original poles. On $X$, this is resolution-ideal equivalence in the sense defined before proposition 9.11.9.

If another solution belonged to the ordinary type, its mass would be $M$, which forces the constant $c_\gamma $. Restoring its prescribed atoms in the extended equation would give exactly (9.11.62). Its difference from $U_\gamma $ would then be harmonic on the torus, hence constant, with equality of the values everywhere. This contradicts the unboundedness of the correction near $q$, so no positive constant admits a solution in the ordinary type.

Finally, the integral of the density is a constant times $\int ^\infty t^{-\gamma }dt$, whereas the integral of its $p$th power, $p>1$, is comparable to

This proves both integrability assertions and completes the proof. $\square$

Proof. Choose real smooth functions $A$ on $C$, supported away from $q$, and $B$ on $E_2$. Let

The form $\alpha $ is real and satisfies $\mathrm{dd}^{\mathrm{c}}\alpha =0$. At a point where $\partial A=0$, $\partial \bar\partial A\ne 0$ and $\partial B\ne 0$,

Thus $\mathrm d\alpha \ne 0$. The form $\alpha $ vanishes near the singular locus. Its operator norm relative to $H_0$ is bounded on its support, and its norm relative to $K_0$ is bounded everywhere. Choose $|\varepsilon |$ so small that both norms, multiplied by $|\varepsilon |$, are less than $1/2$. This proves positivity of $h$ in the ambient sense, semipositivity of $H$, and the inequality $K_0+\varepsilon \pi ^*\alpha \geq \tfrac 12K_0$.

The equation $\partial _ z\partial _{\bar z}v=\lambda -\rho $ on $E_1$, whose right-hand side has mean zero, has a smooth real periodic solution, obtained by dividing its nonzero Fourier coefficients. Its pullback $v_0$ satisfies $\mathrm{dd}^{\mathrm{c}}v_0=(\lambda -\rho )\kappa _1$, which proves the formula for $k$. The forms $H_0$ and $K_0$ are closed. Hence $\mathrm dk$ and $\mathrm d^{\mathrm c}k$ come from the complex surface $E_1\times E_2$, so that $\mathrm dk\wedge \mathrm d^{\mathrm c}k=0$ for degree reasons, and $\mathrm{dd}^{\mathrm{c}}k=0$. The same identities hold for $H$. The Leibniz rule now gives $\mathrm{dd}^{\mathrm{c}}(k^ r)=\mathrm{dd}^{\mathrm{c}}(H^ r)=0$ for all $r$; this uses both torsion identities, not only pluriclosedness. Expanding $(H+\mathrm{dd}^{\mathrm{c}}v_0)^ n$ and integrating by parts shows that every term containing $\mathrm{dd}^{\mathrm{c}}v_0$ integrates to zero. Hence the masses agree, and positivity of $k$ makes them positive. $\square$

Proof. Set $b_1=\lambda $ and $b_ j=1$ for $j\geq 2$. The operator

has Fourier eigenvalues $\ell (m)=-\pi ^2\sum _ jb_ j^{-1}(m_{2j-1}^2+m_{2j}^2)$. It has an inverse $H^ s_0\to H^{s+2}_0$ of norm

Let $L_\varepsilon w=nk^{n-1}\wedge \mathrm{dd}^{\mathrm{c}}w/\Omega $. It preserves zero mean because $\mathrm{dd}^{\mathrm{c}}(k^{n-1})=0$. Fix $\varepsilon _0\leq 1$ within the positivity range. The finite expansion of $k^{n-1}$ gives

Choose bounds $e_ j$ for the norms $\| P_ j\| _{H^{s+2}\to H^ s}$, which are finite since multiplication by smooth coefficients is bounded on $H^ s$, and set $B=\sum _ j\varepsilon _0^{j-1}e_ j$. If $|\varepsilon |\leq \varepsilon _0$ and $|\varepsilon |A_0B\leq 1/2$, the Neumann series gives $\| L_\varepsilon ^{-1}\| \leq A:=2A_0$.

The remainder is

For $s>n$, Fourier convolution and the Cauchy–Schwarz inequality, together with $\sum _ m(1+|m|^2)^{-s}<\infty $, make $H^ s$ an algebra. Multiplication by smooth coefficients and the finite expansion therefore give a constant $C_ Q$ such that, on the unit ball of $H^{s+2}$,

Each term of $Q(w)$ has zero integral against $\Omega $: for smooth $w$, its integral is $\int w\, \mathrm{dd}^{\mathrm{c}}(k^{n-r})\wedge (\mathrm{dd}^{\mathrm{c}}w)^{r-1}=0$, and Sobolev approximation gives the general case. The Sobolev embedding also gives a constant $D$ such that the pointwise operator norm of $\mathrm{dd}^{\mathrm{c}}w$ with respect to $K_0$ is at most $D\| w\| _{H^{s+2}}$. Choose

The equation is equivalent to $L_\varepsilon w+Q(w)=g$. The map $w\mapsto L_\varepsilon ^{-1}(g-Q(w))$ takes the closed ball of radius $R$ in $H^{s+2}_0$ into the ball of radius $3R/4$ and has Lipschitz constant at most $1/2$. Its fixed point gives existence and uniqueness in that ball. Since $k\geq K_0/2$ and the operator norm of $\mathrm{dd}^{\mathrm{c}}w$ is at most $1/4$, the form $k+\mathrm{dd}^{\mathrm{c}}w$ is at least $K_0/4$. The embedding $H^{s+2}\subset C^2$ follows from $s\geq n+3$. $\square$

Proof. The products are the bounded products of proposition 5.28.2. For $r\geq 1$, the bounded local recursion and Stokes' theorem on the compact manifold $Y$ give

Alternatively, one may convolve $U$ on the torus, add a fixed local positive quadratic function, and pass to the limit in the decreasing bounded expansion; every smooth integral vanishes by the same identity. Expanding the top power proves the assertion. The proof for $H$ is identical, since its powers are also $\mathrm{dd}^{\mathrm{c}}$-closed and its mass is $M$. $\square$

Proof. Convolve $U$, $V$ and $k$ with the same nonnegative radial kernel $\varrho _ j$ on the torus, of radius $r_ j\downarrow 0$, and write

Translations preserve positivity, so $A_ j\geq 0$ and $B_ j\geq \beta K_0$ with $\beta =1/4$. The forms $k_ j$ converge smoothly, and $V_ j$ converges in $C^2$. On lifted coordinate balls, choose a common positive quadratic function $q$ with $\mathrm{dd}^{\mathrm{c}}q\geq k$. Convolution gives $q*\varrho _ j=q+b r_ j^2$ for a fixed $b>0$. The radial averages of the bounded plurisubharmonic functions $U+q$ and $V+q$ decrease as the radii decrease. Hence $U_ j+b r_ j^2+q$ and $V_ j+b r_ j^2+q$ decrease to $U+q$ and $V+q$; the identical constants cancel in the difference.

Put $\phi _ j=U_ j-V_ j$ and $T_ j=\sum _{r=0}^{n-1}A_ j^ r\wedge B_ j^{n-1-r}$. The identities $\mathrm{dd}^{\mathrm{c}}k_ j=0$ and $\mathrm dk_ j\wedge \mathrm d^{\mathrm c}k_ j=0$ persist under convolution, since the torsion is still pulled back from $E_1\times E_2$. Differentiating the monomials therefore gives $\mathrm{dd}^{\mathrm{c}}T_ j=0$. Also $T_ j\geq B_ j^{n-1}\geq \beta ^{n-1}K_0^{n-1}$. Stokes' theorem applied to $\mathrm{dd}^{\mathrm{c}}(\phi _ j^2)$ yields

Only the $\mathrm{dd}^{\mathrm{c}}$-closedness of $T_ j$ is needed here.

We treat the four terms on the left separately. Expand $A_ j^ n$ with respect to the local plurisubharmonic potential $U_ j+b r_ j^2+q$ and the smoothly converging coefficient $k_ j-\mathrm{dd}^{\mathrm{c}}q$. The weighted limit lemma 5.27.2 gives $\int U_ jA_ j^ n\to \int U(k+\mathrm{dd}^{\mathrm{c}}U)^ n$. Decreasing convergence and the uniform convergence of $V_ j$ give the corresponding limit for $\int V_ jA_ j^ n$. Since $B_ j^ n$ has uniformly converging continuous densities, bounded pointwise convergence and dominated convergence give the limit of $\int U_ jB_ j^ n$. The term $\int V_ jB_ j^ n$ converges by $C^2$ convergence. A finite partition of unity gives the global statements. Since the two limiting measures are equal, the left-hand side of (9.11.70) tends to zero.

The flat Fourier Poincaré inequality now shows that the $L^2$ norm of $\phi _ j$ minus its average tends to zero. Bounded convergence gives $\phi _ j\to U-V$ in $L^2$, so $U-V$ is constant almost everywhere. Local smooth corrections and the canonical plurisubharmonic representatives extend this equality to every point. $\square$

Proof. The function $g=cf-f_\varepsilon $ has zero mean. Let $w$ be the solution given by theorem 9.11.66. The potential $v_0+w$ is bounded and $H$-plurisubharmonic. Descent through the resolution $\pi $ (theorem 9.11.5) gives

which satisfies the disk condition with the background $h$ itself. Since the normalization is a homeomorphism, $u$ is continuous. The definition of the extended measure and theorem 9.11.26 give the equation. Any other bounded solution pulls back to a bounded $H$-plurisubharmonic potential. Since $\pi $ is a Borel homeomorphism, equality of the pushforward measures means equality on $Y$. Subtracting $v_0$ and applying proposition 9.11.69 gives uniqueness. The mass identity of lemma 9.11.68 forces the scalar of every bounded solution to be $M/\nu (X)=c$.

The singular locus pulls back to the divisor $\{ p\} \times E_2\times \cdots \times E_ n$, which has zero smooth volume. If a set is locally pluripolar, a local plurisubharmonic function equal to $-\infty $ on it pulls back to a plurisubharmonic function, not identically $-\infty $, away from that divisor; its pole set has zero smooth volume. Thus $\nu $ is non-pluripolar. On the regular locus, the change of variables formula gives the density $f/J$. Near the cusp $\alpha =0$, so

The ratio gives the density $\lambda f/\rho $; the generic degree of $\pi $ and both factorials have been accounted for.

For the Hölder assertion, put $d=z-z'$ and $a=z+z'$. Then

If $|a|\geq |d|/4$, the first modulus is at least $|d|^2/4$, which controls $|d|^3$ on a bounded chart. Otherwise the second modulus is at least $13|d|^3/64$. Hence $|z-z'|\leq C(|z^2-z'^2|+|z^3-z'^3|)^{1/3}$. The $C^1$ regularity of $v_0+w$ in $z$ and in the other product coordinates gives the asserted ambient estimate.

Every resolution $\sigma :Z\to X$ factors through $Y$: in the cusp chart, the quotient $y/x$ is bounded off the exceptional set and extends holomorphically to the smooth source, giving the normalization coordinate. Write $\sigma =\pi \circ \mu $. The map $\mu $ is a proper modification. The bounded pullback products agree off its exceptional locus, and they give this analytic set zero mass by proposition 5.28.2. Hence their pushforwards agree on all of $Y$. This proves the equation on every resolution, and the same argument applies to further common refinements. The full potential is exactly $u\circ \sigma $, which is constant on every fiber, and surjectivity preserves all ordinary comparisons. $\square$

Proof. The torus $Y$ is Kähler, so it has bounded mass by lemma 19.9.12. Since $k>0$, zero is an ordinary model for $k$: $P_ k[0]=0$. Its lower volume is $M>0$ by lemma 9.11.68, because the potentials of the same ordinary type as zero are exactly the bounded ones. Hence theorem 10.31.3, applied on $Y$ with background $k$ and reference zero, gives a bounded $k$-plurisubharmonic solution for the measure $f\Omega $, and its scalar is $M/\int f\Omega $. Adding $v_0$ and using theorem 9.11.5, theorem 9.11.26 gives the normalized equation on $X$. The datum is non-pluripolar as in theorem 9.11.71. Neither smallness nor strict positivity of $f$ is needed for this existence statement. $\square$