Pome

11.9 Convex norms and Gauduchon degrees93399AABE3EC

93399AABE3EC

For holomorphic line bundles, the trace identity of proposition 11.5.1 becomes a statement about curvature. Let $F$ be a holomorphic line bundle on $X$ and $k$ a smooth Hermitian metric on $F$. In a local holomorphic frame $e$ of $F$ we identify $k$ with the positive function $|e|_ k^2$, and the curvature form of $k$ is $R_ k=-\mathrm{dd}^{\mathrm{c}}\log k$. Let $\Theta $ be a smooth strictly positive real $(n-1,n-1)$-form with $\mathrm{dd}^{\mathrm{c}}\Theta =0$; by the root calculation preceding the proof of theorem 11.6.1, $\Theta =\gamma ^{n-1}$ for a unique Gauduchon metric $\gamma $. The degree of $F$ with respect to $\Theta $ is

\[ \deg _\Theta F=\int _ XR_ k\wedge \Theta . \]

Two metrics on $F$ differ by a factor $\mathrm e^{-u}$ with $u$ smooth, which adds $\mathrm{dd}^{\mathrm{c}}u$ to the curvature, so the degree is independent of $k$ by lemma 11.1.2. It is the Gauduchon degree of the curvature form with respect to $\gamma $; the form $\Theta $ need not be $\mathrm d$-closed.

The squared norm $v\mapsto |v|_ k^2$ is plurisubharmonic on the total space of $F$ off the zero section exactly when $\log k$ is plurisubharmonic, that is, when $R_ k\leq 0$; this follows from the Levi form computation in the proof of lemma 11.11.1 below. A line bundle of positive degree therefore admits no metric with plurisubharmonic squared norm. This section extends the obstruction to direct sums of line bundles of positive degree, to all holomorphic structures sufficiently close to such a direct sum, and to norms that need not be Hermitian. The mechanism is again integration against $\Theta $: averaging the horizontal complex Hessians of a norm near the line summands produces a negative multiple of the degrees, and this negativity is stable under small changes of the holomorphic structure.

For a smooth complex vector bundle $F$, a smooth complex norm is a continuous function $p:F\to [0,+\infty )$, positive away from the zero section and smooth there, such that

\[ p(\lambda v)=|\lambda |p(v),\qquad p(v+w)\leq p(v)+p(w) \]

for $\lambda \in \mathbb C$ and vectors $v,w$ in the same fiber. Its square $G=p^2$ satisfies $G(\lambda v)=|\lambda |^2G(v)$. Near each line summand of a direct sum, the triangle inequality controls the fiber derivatives of $G$, and these estimates control the averaged horizontal complex Hessians.

Lemma 11.9.1.A592FEE54E20 Fix a smooth positive volume form $\mu $ on $X$. Every holomorphic line bundle $F$ has a Hermitian metric $k$ satisfying

\[ R_ k\wedge \Theta =\lambda \mu , \qquad \lambda =\frac{\deg _\Theta F}{\int _ X\mu }. \]

Proof. Start with any metric $k^0$, and put $\mathcal Lu=(\mathrm{dd}^{\mathrm{c}}u\wedge \Theta )/\mu $. This scalar operator is uniformly elliptic, has no zeroth-order term, and its kernel consists of the constants by the maximum principle. On Hölder spaces, the homotopy to a Laplace operator used in the proof of theorem 11.4.2 shows that its index is zero, so its cokernel is one-dimensional. Since $\mathrm{dd}^{\mathrm{c}}\Theta =0$, integration by parts gives $\int _ X\mathcal Lu\, \mu =0$, and therefore its image consists exactly of the smooth functions of zero $\mu $-mean. Hence we can solve

\[ \mathrm{dd}^{\mathrm{c}}u\wedge \Theta =\lambda \mu -R_{k^0}\wedge \Theta , \]

whose right-hand side has integral zero, and elliptic regularity makes $u$ smooth. Then $k=k^0\mathrm e^{-u}$ has the required curvature. $\square$

Let $F_0=\bigoplus _{s=1}^ rF_ s$ be a direct sum of holomorphic line bundles, with $r\geq 2$. Choose Hermitian metrics $k_ s$ on $F_ s$ and write $k=\bigoplus _ sk_ s$. Let $P_ s$ be the orthogonal projection onto $F_ s$, and let

\[ \Sigma =\{ v\in F_0:|v|_ k=1\} \]

be the unit sphere bundle, with fibers $\Sigma _ x$. On each fiber sphere $\Sigma _ x$ we use the rotation-invariant probability measure $\mathrm d\sigma _ x$. Fix a nonnegative smooth function $\chi $ supported in $(-1,1)$ and positive near zero. For $0<\varepsilon <1/2$, define weights $\rho _ s$ on $\Sigma $ by

\[ \rho _ s(v)=c_\varepsilon \chi \left(\frac{|(I-P_ s)v|_ k^2}{\varepsilon ^2}\right), \qquad \int _{\Sigma _ x}\rho _ s\, \mathrm d\sigma _ x=1. \]

In unitary frames respecting the splitting, these weights and the normalizing constant $c_\varepsilon $ are independent of the base point and of the index $s$. Independent phase rotations of the summands preserve them, so they are globally defined. The support of $\rho _ s$ lies within distance $\varepsilon $ of the unit circle of $F_ s$.

Lemma 11.9.2.FD645EB1208B For a smooth complex norm on $F_0$ with square $G$, put

\[ a_ s(x)=G(e_ s),\qquad N(x)=\sum _ sa_ s(x),\qquad D_ jG(v)=\mathrm d_ vG(P_ jv), \]

where $e_ s$ is any $k_ s$-unit vector in the fiber of $F_ s$ at $x$; complex homogeneity makes $a_ s$ well defined. Then

\[ G(v)\leq N|v|_ k^2,\qquad |\mathrm d_ vG(w)|\leq 2N|v|_ k|w|_ k. \]

On the support of $\rho _ s$,

\[ |D_ jG|\leq 2\varepsilon N\quad (j\ne s),\qquad |D_ sG-2a_ s|\leq 10\varepsilon N. \]

Proof. The triangle inequality and the Cauchy–Schwarz inequality give $p(v)\leq \sum _ s|v_ s|_ k\sqrt{a_ s}\leq \sqrt N|v|_ k$. Differentiating $|p(v+tw)-p(v)|\leq |t|p(w)$ in the real variable $t$ gives $|\mathrm d_ vp(w)|\leq p(w)$, and both global bounds follow.

For $v$ in the support of $\rho _ s$, set $w=(I-P_ s)v$ and $e_ s=P_ sv/|P_ sv|_ k$. Then $|w|_ k<\varepsilon $ and $|v-e_ s|_ k\leq |w|_ k+1-|P_ sv|_ k<2\varepsilon $. The segment from $e_ s$ to $v$ stays in the unit ball and avoids zero, so integrating the derivative bound along it gives $|G(v)-a_ s|\leq 4\varepsilon N$. For $j\ne s$, we have $|P_ jv|_ k<\varepsilon $, which gives the first axis bound. Euler's identity $\mathrm d_ vG(v)=2G(v)$ yields

\[ D_ sG-2a_ s=2(G(v)-a_ s)-\sum _{j\ne s}D_ jG. \]

The sum on the right is $\mathrm d_ vG(w)$, whose absolute value is at most $2\varepsilon N$. This proves the second axis bound. $\square$

Write $\pi :F_0\to X$ for the bundle projection. For $v\in F_0$ with $\pi (v)=x$, let $H_{0,v}:T_ xX\to T_ vF_0$ be the horizontal lift for the Chern connection of $k$. It is complex linear, and at unit vectors it is tangent to $\Sigma $, since the Chern connection preserves $k$. For a smooth real function $G$ on $F_0\setminus \{ 0\} $, define a $(1,1)$-form $\beta _ s(G)$ and functions $U_ s$, $V_{sj}$ on $X$ by

\[ \begin{aligned} \beta _ s(G)& =\int _{\Sigma _ x}\rho _ s H_{0,v}^*(\mathrm{dd}^{\mathrm{c}}G)\, \mathrm d\sigma _ x,\\ U_ s& =\int _{\Sigma _ x}G\rho _ s\, \mathrm d\sigma _ x,\\ V_{sj}& =\int _{\Sigma _ x}D_ jG\rho _ s\, \mathrm d\sigma _ x. \end{aligned} \]

Lemma 11.9.3.EE774D339C2C With $R_ j$ the curvature of $k_ j$,

\[ \beta _ s(G)=\mathrm{dd}^{\mathrm{c}}U_ s-\frac12\sum _ jR_ jV_{sj}. \]

Proof. At a fixed base point, choose holomorphic frames of the line bundles with $k_ j=\mathrm e^{-\varphi _ j}$ and $\mathrm d\varphi _ j=0$ at that point. There the horizontal lift has zero vertical component. Write the smooth unitary fiber coordinates as $\zeta _ j=\mathrm e^{-\varphi _ j/2}v_ j$ and $G(z,v)=g(z,\zeta )$. At the base point, the first base derivatives of $\zeta _ j$ vanish, and its mixed second base derivatives are $-\frac12(\varphi _ j)_{\alpha \bar\beta }\zeta _ j$ and their conjugates. The chain rule therefore gives

\[ G_{\alpha \bar\beta } =g_{\alpha \bar\beta }-\frac12\sum _ j(\varphi _ j)_{\alpha \bar\beta } (\zeta _ jg_{\zeta _ j}+\bar\zeta _ jg_{\bar\zeta _ j}). \]

The expression in parentheses is $D_ jG$. Both the spherical measure and $\rho _ s$ are independent of $z$ in unitary coordinates, so averaging the identity gives the result, since $R_ j=\mathrm{dd}^{\mathrm{c}}\varphi _ j$. $\square$

Theorem 11.9.4.C222D1BB3C65 Suppose that for each $s$ there is a smooth strictly positive real $(n-1,n-1)$-form $\Theta _ s$ satisfying

\[ \mathrm{dd}^{\mathrm{c}}\Theta _ s=0,\qquad \deg _{\Theta _ s}F_ s>0. \]

Then there is a $C^4$ neighborhood of the split Dolbeault operator on the underlying smooth bundle of $F_0$ such that no integrable Dolbeault operator in this neighborhood admits a smooth complex norm whose square is plurisubharmonic off the zero section.

Proof. Step 1: the split structure. Fix $\mu $ and use lemma 11.9.1 to choose $k_ s$ with $R_ s\wedge \Theta _ s=\lambda _ s\mu $, where $\lambda _ s>0$. Put

\[ \lambda _* =\min _ s\lambda _ s,\qquad C=5\sum _ s\lambda _ s+ \sum _ s\sum _{j\ne s}\sup _ X\left|\frac{R_ j\wedge \Theta _ s}{\mu }\right|. \]

For a squared norm $G$, define $I_0(G)=\sum _ s\int _ X\beta _ s(G)\wedge \Theta _ s$. By lemma 11.9.3, the term $\mathrm{dd}^{\mathrm{c}}U_ s$ integrates to zero against $\Theta _ s$, and the axis estimates of lemma 11.9.2 then give

\[ I_0(G)=-\frac12\sum _{s,j}\int _ XV_{sj}R_ j\wedge \Theta _ s \leq -(\lambda _* -\varepsilon C)\int _ XN\mu . \]

Choose $\varepsilon $ so that $\varepsilon C<\lambda _*/2$, and keep all weights fixed.

Step 2: nearby holomorphic structures. Write a nearby operator as $\bar\partial _ A=\bar\partial _0+A$, with $\bar\partial _ A^2=0$. Its Chern connection for $k$ is $D_ A=D_0+A-A^{*k}$, where the adjoint changes the type from $(0,1)$ to $(1,0)$. Let $J_ A$ be the corresponding complex structure on the total space and $H_{A,v}$ the horizontal lift of $D_ A$. These depend smoothly on $A$, and $H_{A,v}$ is complex linear and tangent to the fixed sphere bundle $\Sigma $.

For $g\in C^\infty (\Sigma ,\mathbb R)$, define its homogeneous extension by

\[ (\mathcal Eg)(v)=|v|_ k^2g(v/|v|_ k), \]

and define a real differential operator on $\Sigma $ by

\[ (\mathcal P_ Ag)(v)=\sum _ s\rho _ s(v) \frac{H_{A,v}^*(\mathrm{dd}^{\mathrm{c}}_{J_ A}\mathcal Eg)_ v\wedge \Theta _ s(x)}{\mu (x)}, \qquad x=\pi (v). \]

It has order at most two. By the identity $\mathrm d^{\mathrm c}_ Jf=-(4\pi )^{-1}\mathrm df\circ J$, its coefficients depend on $A$ and its first derivatives, as well as on the fixed smooth data. Consequently they converge in $C^2$ as $A\to 0$ in $C^4$.

Step 3: the adjoint operator. Take the formal adjoint $\mathcal P_ A^*$ for the fixed density $\mathrm d\nu =\mathrm d\sigma _ x\, \mu (x)$ on the compact sphere bundle, and put $b_ A=\mathcal P_ A^*1$. Locally, if $\mathrm d\nu =m(y)\, \mathrm dy$ and $\mathcal P_ A=\sum _{|\alpha |\leq 2}c_{\alpha ,A}\partial ^\alpha $, then

\[ b_ A=m^{-1}\sum _{|\alpha |\leq 2}(-1)^{|\alpha |} \partial ^\alpha (mc_{\alpha ,A}). \]

Thus $\| b_ A-b_0\| _{C^0}\to 0$ as $A\to 0$ in $C^4$. For $g=G|_\Sigma $, homogeneity gives $\mathcal Eg=G$ and

\[ I_ A(G):=\int _\Sigma \mathcal P_ Ag\, \mathrm d\nu =\int _\Sigma gb_ A\, \mathrm d\nu . \]

Since $0<g\leq N\circ \pi $ and the spherical measures have mass one,

\[ |I_ A(G)-I_0(G)| \leq \| b_ A-b_0\| _{C^0}\int _ XN\mu . \]

Choose the neighborhood so that $\| b_ A-b_0\| _{C^0}<\lambda _*/4$ in it. Then $I_ A(G)\leq -\frac14\lambda _*\int _ XN\mu <0$ for every smooth squared complex norm $G$. If $G$ were plurisubharmonic for $J_ A$, all horizontal pullbacks of $\mathrm{dd}^{\mathrm{c}}_{J_ A}G$ would be semipositive, and their pairings with $\Theta _ s$, weighted by $\rho _ s\geq 0$, would give $I_ A(G)\geq 0$, a contradiction. $\square$

The neighborhood depends on the bundle and on the test forms $\Theta _ s$, and different line summands may require different tests. Its size is independent of all derivatives of the candidate norms: integration over the fixed sphere bundle transfers those derivatives to the coefficients of the adjoint operator.

To apply theorem 11.9.4 to norms obtained by duality, we first need to regularize continuous norms while preserving a strict convexity margin. Polar duality then turns the obstruction into a failure of semipositive Kobayashi curvature on the dual bundle. Stable ample bundles on Hirzebruch surfaces provide concrete examples, and a fixed Stein hypersurface detects the failure for every Finsler metric on such a bundle.

Dual support norms also connect this obstruction with the geometry of neighborhoods of the zero section. This leads to bounds for the convexity of a projective complement. Curvature supplies a second way to study these complements, through explicit exhaustions. A uniformly negative bundle over a Kähler base with a proper psh exhaustion instead gives logarithmic weights and a complete metric on its punctured total space (11.19). The distinction between these spaces matters: the final flat and Hopf examples use restrictions and topology to obtain further bounds for the projective complement.

Cite this page

Comments

Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.