11.28 Holomorphic sectional curvatureAF4D6581275D

Proof. In a $g$-unitary frame, let $Z_1,\ldots ,Z_ n$ be independent standard complex Gaussian variables, with density $\pi ^{-1}\mathrm e^{-|z|^2}$ on $\mathbb C$. Their second and fourth moments are $\mathbb E|Z_ i|^2=1$ and $\mathbb E|Z_ i|^4=2$. Independence and invariance under coordinate phases give

Thus $\mathbb E R^ h(Z,\bar Z,Z,\bar Z)=Q_ g(h)+B_ g(h)$ and $\mathbb E|Z|_ h^4=u^2+q$. Taking expectations in the homogeneous inequality $R^ h(Z,\bar Z,Z,\bar Z)\leq -\kappa |Z|_ h^4$ proves the result. $\square$

Proof. We first compute the exact term that distinguishes the two curvature traces. Define

This is a tensor: putting holomorphic differentials first gives

For $D_ g=\sum _{i,k}D_{i\bar i k\bar k}$ in a $g$-unitary frame, expansion of the wedge product gives

Indeed, each pair $i<k$ contributes $D_{i\bar i k\bar k}\mathrm dV_ g$ to the wedge product and occurs twice in the contraction. Closedness of $\omega _ g$ and Stokes' theorem therefore give $\int _ XD_ g\, \mathrm dV_ g=0$.

The four-term antisymmetrization of the curvature formula is

The second derivatives give $-D$, and the quadratic terms give the squared connection difference in the two skew indices. Contracting gives $2(Q_ g(h)-B_ g(h))=|T|_{g,h}^2-D_ g$, hence

In $g$-normal holomorphic coordinates, the first derivatives of $g$ vanish. Differentiating $u=g^{k\bar l}h_{k\bar l}$ twice, the inverse-metric term is $I_ g(h)$ by the Kähler curvature symmetries of $g$, while the second derivatives of $h$ give $-Q_ g(h)+|A|_{g,h}^2$. Thus

where symmetric and skew tensors are orthogonal and $|A|_{g,h}^2=|S|_{g,h}^2+|T|_{g,h}^2/4$. Integrating removes both $P_ g u$ and $D_ g$, proving (11.28.4). $\square$

Proof. The Ricci assumption gives $I_ g(h)\geq 0$, and lemma 11.28.2 gives $-\mathcal R_ g(h)\geq 0$. All three continuous terms in (11.28.4) are nonnegative, so each vanishes everywhere. A positive definite $h$ pairs strictly positively with every nonzero semipositive Ricci matrix; hence $I_ g(h)=0$ forces $\operatorname {Ric}(g)=0$.

At each point, the Gaussian expectation in the averaging proof is now zero. Its integrand $R^ h(Z,\bar Z,Z,\bar Z)$ is nonpositive. If it were negative in one direction, continuity and the strictly positive Gaussian density would make its expectation negative. It is therefore zero in every direction, which gives $H_ h=0$ and excludes quasi-negativity. $\square$

Proof. Suppose first that $m\geq 2$, and take the flat Kähler metric $g$ induced by a constant Hermitian inner product on $\mathbb C^ m$. Choose linear coordinates in which this inner product is the identity. The coordinate vector fields descend to $Y$, even though the coordinates themselves need not. By corollary 11.28.5, $H_ h=0$; (11.28.4) also gives $S=0$. Since $g$ has zero connection coefficients, multiplying $S=0$ by $h$ gives

Fix $l$ and put $a_ k=h_{k\bar l}$. With respect to the flat volume, integration by parts and commutation of constant coordinate derivatives give

Consequently,

Thus every $\partial _ i h_{k\bar l}$ vanishes. Hermitian symmetry gives the vanishing of the conjugate derivatives as well, so all coefficients of $h$ are constant.

For $m=1$, write the lifted metric as $h(z)|\mathrm dz|^2$, with $h$ positive and $\Lambda $-periodic. Its curvature coefficient is $-h\, \partial _ z\partial _{\bar z}\log h$. Nonpositive sectional curvature therefore gives $\partial _ z\partial _{\bar z}\log h\geq 0$. The complex Laplacian has zero integral over $Y$, so it vanishes. Integration by parts then shows that $\log h$ has zero gradient, and $h$ is constant.

Conversely, a constant positive Hermitian matrix defines a closed form and has zero connection and curvature coefficients. This proves all the assertions. $\square$

Proof. The covering is locally biholomorphic, so the pullback metric has the same sectional-curvature sign. By proposition 11.28.7, its associated form is closed and its Chern curvature is zero. Both conclusions descend through the covering. $\square$