Lemma 11.29.1.E495F7FE3CE6 The function $G_ h$ is strictly plurisubharmonic off the zero section if and only if $h$ is a smooth positive Hermitian metric with strictly Griffiths-negative curvature. Equivalently, its dual metric on $E$ is strictly Griffiths-positive.
11.29 Projective first jets and positive bundle metricsA1CB99CAE47B
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RC-positivity asks for one positive tangent direction for each fiber vector, whereas Griffiths positivity asks for positivity on every pair of nonzero tangent and fiber vectors. The Lamari criterion of theorem 11.6.1 characterizes positivity of a $(1,1)$-form by a duality with Gauduchon tests, and Griffiths positivity on a projective manifold admits an analogous duality. A Griffiths-positive metric on $E$ is the same as a strictly plurisubharmonic squared norm on the dual bundle, which is a linear condition on the dual Hermitian form. Twisting by powers of a very ample line bundle gives finite-dimensional spaces of candidate forms. At each level, either some candidate is strictly plurisubharmonic or a finite atomic positive measure annihilates the Levi forms of all candidates, and if every level fails, a weak limit of these measures annihilates the Levi forms of all smooth Hermitian forms. This measure lives over the dual bundle, whereas the Gauduchon currents in theorem 11.6.1 live on the base.
Throughout this section, $X$ is a projective manifold of dimension $n\geq 2$, $E\to X$ is a holomorphic vector bundle of positive rank, and $F=E^*$. Fix a very ample line bundle $A$, a basis $t_1,\ldots ,t_ d$ of $H^0(X,A)$, and the Fubini–Study metric $a$ on $A$ determined by this basis. Fix also an integer $m_0\geq 0$ for which $E\otimes A^{m_0}$ is globally generated; this integer is part of the data. In a local frame of $A$, put
We use the curvature conventions of 11.18. A Hermitian metric $g$ on a holomorphic vector bundle over $X$ is strictly Griffiths-positive if $g(R_ g(u,\bar u)v,v)>0$ for every $x\in X$, every nonzero $u\in T_ x^{1,0}X$ and every nonzero vector $v$ in the fiber over $x$; it is strictly Griffiths-negative if the reverse strict inequality holds.
For a smooth Hermitian form $h$ on $F$, not initially assumed positive, let $G_ h(v)=h(v,v)$ on its total space. Write $\mathcal L G_ h(\xi )$ for the complex Hessian evaluated on $\xi \in T^{1,0}F$; in local coordinates it is $\sum (G_ h)_{j\bar k}\xi _ j\bar\xi _ k$, the coefficient form of $\mathrm i\partial \bar\partial G_ h$, without the factor $1/(2\pi )$.
Proof. On vertical tangent vectors the Levi form is $h$ itself, so strict positivity forces $h>0$. At a fixed base point, choose a holomorphic frame in which the matrix $H$ of $h$ is the identity and has vanishing first derivatives. Then $R_ h(u,\bar u)=-H_{u\bar u}$ at this point. A tangent vector to $F$ has horizontal and vertical components $(u,w)$, and
For $v\ne 0$, this is positive for every nonzero $(u,w)$ precisely when the curvature evaluation is negative for every nonzero $u$. The curvature of the dual metric is the negative transpose of the original curvature, with respect to the dual Hermitian pairing, which gives the last assertion. $\square$
The Levi forms will be tested on a compact set. Choose any smooth Hermitian metric $k$ on $F$, let $S=\{ v:k(v,v)=1\} $ be its unit sphere bundle, and choose a smooth positive Hermitian form $b$ on the complex vector bundle $W=T^{1,0}F|_ S$ over $S$. The set
is compact. Define the real-linear operator $D$ by $D h=\mathcal L G_ h|_ S$, viewed as a Hermitian form on $W$. It is continuous from the $C^2$ norm of Hermitian forms on $F$ to the uniform norm on $W$, since its coefficients involve at most two base derivatives of $h$ and the fiber vectors range over a compact set. Fiber dilation $v\mapsto cv$ gives
Thus strict positivity of $\mathcal LG_ h$ on $K$ is equivalent to strict positivity off the zero section.
11.29.2 The finite-dimensional alternative6243CAB7B202
For $m\geq m_0$, choose a basis of $V_ m=H^0(X,E\otimes A^ m)$ and write $N_ m=\dim V_ m$. Pairing these sections with $v\in F_ x$ and using a local frame of $A^ m$, we obtain a column $p_ m(v)$ of local holomorphic functions on the total space of $F$. For any Hermitian $N_ m\times N_ m$ matrix $B$, set
This function is globally defined and quadratic on each fiber, so $G_{m,B}=G_{h_{m,B}}$ for a smooth Hermitian form $h_{m,B}$ on $F$. The matrix $B$ is allowed to have negative eigenvalues.
For $(v,\xi )\in K$, let $u$ be the projection of $\xi $ to $T_ x^{1,0}X$, write $\phi _ u=\partial \phi (u)$, and put
Differentiating the weighted quadratic function gives
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11.29.3
All entries involve only the first jets of the sections: if $t=(t_\alpha )$ is the column of the sections of $A$, then
Changing the frame of $A$ by a nowhere-zero holomorphic factor $\lambda $ changes $p_ m$ to $\lambda ^{-m}p_ m$ and $\phi $ to $\phi -\log |\lambda |^2$. It changes $r_ m$ by the same factor and leaves $\kappa $ unchanged, so $\Psi _ m$ is frame independent. A change of the basis of $V_ m$ replaces $\Psi _ m$ by a congruent matrix, so whether the convex hull of the values of $\Psi _ m$ contains zero does not depend on that choice.
Proposition 11.29.4.07EC0797039C At each fixed $m$, exactly one of the following holds:
Some Hermitian matrix $B$ satisfies $\operatorname {tr}(B\Psi _ m)>0$ everywhere on $K$.
There is a probability measure $\mu $ on $K$, supported on at most $N_ m^2+1$ points, such that $\int _ K\Psi _ m\, \mathrm d\mu =0$.
The first alternative holds precisely when $0\notin \operatorname {conv}(\Psi _ m(K))$.
Proof. The real vector space of Hermitian $N_ m\times N_ m$ matrices has dimension $N_ m^2$. Carathéodory's theorem expresses every point of the convex hull as a convex combination of at most $N_ m^2+1$ points of the compact image $\Psi _ m(K)$, so the hull is compact. If it contains zero, the coefficients of such an expression define the required atomic probability measure $\mu $, and then $\int _ K\operatorname {tr}(B\Psi _ m)\, \mathrm d\mu =0$ excludes the first alternative.
If the hull misses zero, let $Z_0$ be its point of least Hilbert–Schmidt norm. Differentiating the squared norm along the segment from $Z_0$ to any $Z$ in the hull gives
Taking $B=Z_0$ proves the first alternative. $\square$
11.29.5 Approximation with two derivatives84DD621CC41A
Let $\mathcal H_ m=\{ h_{m,B}:B=B^*\} $. To pass from these finite-dimensional tests to all smooth metrics, we need approximation in $C^2$, since $D$ involves two derivatives.
Lemma 11.29.6.86B1F5EAC034 The matrix-valued map $Y=tt^*/\rho $ embeds $X$ as a compact smooth submanifold of the real vector space of Hermitian $d\times d$ matrices. Restrictions of real polynomials in its real coordinates are dense in $C^2(X,\mathbb R)$ for the $C^2$ norm.
Proof. The normalized rank-one projector $Y$ determines the point $[t]$ of projective space, and its differential vanishes only in directions in which $\mathrm dt$ is proportional to $t$. Since $A$ is very ample, $Y$ is therefore injective with injective differential, and compactness makes it an embedding. A $C^2$ function on this image extends to a $C^2$ function on an ambient cube: use local coordinates in which the image is a graph, extend constantly in the normal coordinates, and combine these extensions with a finite smooth partition of unity and a cutoff. Rescale the cube to $[0,1]^ q$.
For such an extension $f$, its multivariate Bernstein polynomial is
where $j=(j_1,\ldots ,j_ q)$. For a multi-index $\alpha $ with $|\alpha |\leq 2$, differentiation expresses $\partial ^\alpha B_ Nf$ as an average of $\Delta _{1/N}^\alpha f(j/N)$, with prefactor $\prod _\ell (N)_{\alpha _\ell }$ and Bernstein degrees $N-\alpha _\ell $. Here $(N)_ r=N(N-1)\cdots (N-r+1)$, and $\Delta ^\alpha _{1/N}$ is the corresponding iterated forward difference with step $1/N$. Repeated use of the fundamental theorem of calculus expresses $N^{|\alpha |}\Delta _{1/N}^\alpha f(j/N)$ as an average of $\partial ^\alpha f$ at points within $2/N$ in each affected coordinate.
The remaining Bernstein weights are the law of independent binomial variables of degrees $N-\alpha _\ell $, divided by $N$. Their means differ from $y_\ell $ by at most $2/N$ and their variances are at most $1/(4N)$. Uniform continuity of $\partial ^\alpha f$, splitting the average into a fixed small neighborhood of $y$ and its complement, and Chebyshev's inequality give uniform convergence to $\partial ^\alpha f$. The prefactor divided by $N^{|\alpha |}$ tends to one. This proves convergence for every derivative of order at most two, including the function itself. Restrict to the embedded image and use the chain rule in a finite collection of charts to obtain the asserted $C^2$ approximation. $\square$
Lemma 11.29.7.944943C60132 The spaces $\mathcal H_ m$ are nested, and their union is $C^2$ dense in the space of $C^2$ Hermitian forms on $F$.
Proof. For nesting, multiply the sections in $V_ m$ by each $t_\alpha $ and express the products in the basis of $V_{m+1}$. The identity $\sum _\alpha \mathrm e^{-\phi }|t_\alpha |^2=1$ writes $G_{m,B}$ as the sum of the corresponding weighted quadratic functions at level $m+1$, which combine into one Hermitian coefficient matrix at that level. Surjectivity of the multiplication map is not needed.
For density, write $p_{m_0}(v)=P_0v$ in a local frame of $F$, and put $L=\mathrm e^{-m_0\phi /2}P_0$. Global generation of $E\otimes A^{m_0}$ means that $L$ is injective, so $M=L^*L$ is positive definite. If a $C^2$ Hermitian form has matrix $H$, define
Then $L^*CL=H$. A change of frame of $F$ cancels in this expression; a change of frame of $A$ multiplies $L$ by a scalar of modulus one, which also cancels. Thus $C$ is a globally defined $C^2$ Hermitian matrix function on $X$.
Applying lemma 11.29.6 to its real coordinates in the vector space of Hermitian matrices, we obtain Hermitian matrix polynomials in the entries of $Y$ that converge to $C$ in $C^2$; their degrees may increase along the sequence. Let $C_ q(Y)$ be such a polynomial, of degree at most $q$. Every monomial of degree $k$ in $Y$ has the form $t^ I\bar t^ J/\rho ^ k$, with $|I|=|J|=k$. For a polynomial of degree at most $q$, multiply each numerator by $\rho ^{q-k}$ to use the common denominator $\rho ^ q$. Consequently
is a quadratic expression in the product sections $\sigma _ i t^ I$ of $E\otimes A^{m_0+q}$, with weight $\mathrm e^{-(m_0+q)\phi }$. Express these products in the basis of $V_{m_0+q}$ and replace the coefficient matrix by its Hermitian part, which does not change the real function. The result is an element of $\mathcal H_{m_0+q}$. Since $L^*CL=H$ and multiplication by the fixed smooth matrix $L$ is continuous in $C^2$, these elements converge to $H$. $\square$
11.29.8 The metric and measure alternativesB1D39F6DC8D4
A positive matrix-valued measure on $W\to S$ is a linear functional $T$ on the space of continuous Hermitian forms $\Phi $ on $W$ such that $\langle T,\Phi \rangle \geq 0$ whenever $\Phi \geq 0$. Its trace mass relative to $b$ is $\langle T,b\rangle $. We write $D^*T=0$ if $\langle T,Dh\rangle =0$ for every smooth Hermitian form $h$ on $F$.
Theorem 11.29.9.C334EF3CAF72 With the fixed projective data above, the following are equivalent:
$E$ admits a smooth strictly Griffiths-positive Hermitian metric.
For some $m\geq m_0$, a Hermitian matrix $B$ makes $G_{m,B}$ strictly plurisubharmonic off the zero section.
For some $m\geq m_0$, $0\notin \operatorname {conv}(\Psi _ m(K))$.
There is no nonzero positive matrix-valued measure $T$ on $W\to S$ satisfying $D^*T=0$.
Proof. Given a metric as in the first assertion, its dual metric $h$ on $F$ satisfies $Dh\geq \delta b$ on $S$ for some $\delta >0$, by lemma 11.29.1 and compactness. The $C^2$ density in lemma 11.29.7 and the continuity of $D$ give a form $h_{m,B}$ with $D h_{m,B}\geq \delta b/2$, which proves the second assertion. Conversely, if $G_{m,B}$ is strictly plurisubharmonic, its vertical Levi form shows that $h_{m,B}$ is positive, and lemma 11.29.1 shows that its dual metric is strictly Griffiths-positive. The second and third assertions are equivalent by proposition 11.29.4.
If the first assertion holds, every nonzero positive matrix-valued measure $T$ has positive trace mass, and
so such a measure cannot satisfy $D^*T=0$. Conversely, suppose that the third assertion fails at every level. By proposition 11.29.4, choose at each level $m$ an atomic probability measure $\mu _ m$ on $K$ with $\int _ K\Psi _ m\, \mathrm d\mu _ m=0$, so that $\mu _ m$ annihilates $Dh$ for every $h\in \mathcal H_ m$. A subsequence converges weakly to a probability measure $\mu $ on the compact set $K$. Since the spaces $\mathcal H_ m$ are nested, the limit annihilates $Dh$ for every $h$ in each fixed $\mathcal H_ j$, and the $C^2$ density and the continuity of $D$ extend this to every smooth Hermitian form.
Define
This is a positive matrix-valued measure on $W\to S$ with $\langle T,b\rangle =1$ and $D^*T=0$. Its entries in local frames are represented by scalar measures: positivity bounds the functional on Hermitian forms of uniform norm at most one by the trace mass, so the Riesz representation theorem applies to the coordinate functionals, and a finite partition of unity combines these local representations. This contradicts the fourth assertion and completes the proof. $\square$
The equation $D^*T=0$ concerns a bundle over the sphere bundle $S\subset F=E^*$; it should not be confused with the equation $\mathrm{dd}^{\mathrm{c}}T=0$ for currents on $X$. The theorem gives no bound for the first successful level, and no finite procedure for deciding that all levels fail.
Corollary 11.29.10.A86A6F25FF07 If $E$ is globally generated and $m_0=0$, the level-zero test succeeds exactly when the complete evaluation map $\Phi :F\to H^0(X,E)^*$ is an immersion off the zero section.
Proof. At $m=0$, one has $\Psi _0=\mathrm d\Phi (\xi )\mathrm d\Phi (\xi )^*$, where $\Phi $ is written in the coordinates given by the chosen basis. If $\mathrm d\Phi $ is injective off the zero section, $B=\operatorname {Id}$ gives a strictly positive Levi form. If $\mathrm d\Phi $ has a nonzero kernel vector there, fiber dilation and normalization of the tangent vector place such a vector in $K$, where $\Psi _0=0$, and no Hermitian matrix can separate the image from zero. $\square$
When this untwisted test fails, the higher twists remain available. The finite alternatives use all Hermitian coefficient matrices and the complete spaces of sections; testing only positive coefficient matrices or a smaller collection of sections gives a different criterion.
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