PomeProposition 11.28.7.19F2C00A6056 Let $Y=\mathbb C^ m/\Lambda $ be a complex torus, with $m\geq 1$. A smooth Hermitian tangent metric $h$ on $Y$ has nonpositive Chern holomorphic sectional curvature if and only if it is translation invariant. Such a metric is Kähler and has zero Chern curvature.
Proposition 11.28.719F2C00A6056
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
\[ \partial _ i h_{k\bar l}+\partial _ k h_{i\bar l}=0 \qquad \text{for all }i,k,l. \]
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
\[ \sum _{i,k}\int _ Y\partial _ i a_ k\, \overline{\partial _ k a_ i}\, \mathrm dV_ g =\int _ Y\left|\sum _ k\partial _{\bar k}a_ k\right|^2\mathrm dV_ g. \]
Consequently,
\[ \begin{aligned} 0& =\sum _{i,k}\int _ Y|\partial _ i a_ k+\partial _ k a_ i|^2\, \mathrm dV_ g\\ & =2\sum _{i,k}\int _ Y|\partial _ i a_ k|^2\, \mathrm dV_ g +2\int _ Y\left|\sum _ k\partial _{\bar k}a_ k\right|^2\mathrm dV_ g. \end{aligned} \]
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. $\square$
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