Search Results

Let X be a complex hermitian manifold with hermitian form $\omega$. How can you prove that if $\omega$ has negative holomorphic sectional curvature, then its scalar curvature is negative, too?

I assume that for $\operatorname{Hom}(X,Y)$ you mean $\operatorname{Hol}(X,Y)$, that is the family of all holomorphic maps from $X$ to $Y$, endowed with its universal complex structure (which exists s …

Let me write this too long comment as an answer. As abx says, what we do know is Theorem 1. If a Kähler manifold $X$ is homeomorphic to $\mathbb{CP}^n$, then $X$ is biholomorphic to it. This is due to …

Let $(X,\omega)$ be a complex hermitian manifold, and call $\Theta$ its Chern curvature tensor. Out of this we can consider different notions of curvature, namely the holomorphic bisectional curvature …

Let me expand a bit Henri's (hi Henri!) answer, even if this is completely standard. In general, given a compact Kähler manifold $X$ of any dimension, given a holomorphic line bundle $L\to X$, and giv …

Examples of compact Kähler manifolds with non positive holomorphic bisectional curvature are given by: Closed submanifolds of complex tori. Smooth compact quotients of bounded symmetric domains. M …

As soon as the base manifold has dimension greater than one, the existence of Griffiths positive metrics on an ample vector bundle is not known (and even in the one dimensional case, it is not obvious …

Just for fun, here is an answer in the purely algebraic setting. So, suppose that $X$ is irreducible projective algebraic of dimension $n$, $\alpha=c_1(\mathcal O_X(D))$ is the class of a nef diviso …

Kobayashi hyperbolicity (or Brody hyperbolicity, the two notions coincide for compact complex spaces), is an open condition with respect to the analytic topology. Thus, for instance, once you find an …

Maybe you already were aware of that, or maybe it really doesn't answer to your question, but I'll try anyhow... Take a look at this, Subsection 7.5 on page 39. The construction you talk about in you …

I think the answers you are looking for are in this paper by V. Tosatti, see in particular Proposition 1.1, point (4) and Proposition 1.3. Warning (in view of the comment below by S.S.): the holonom …

The answer to your question is exactly the same of the answer to this older question of mine. Enjoy!

With your notations, the hermitian form $\theta_E$ on $T_X\otimes E$ defined by $\Theta_E$ is given in a somewhat more extrinsic way by $$ \theta_E(v\otimes\sigma,v\otimes\sigma):=h(\Theta_E(v,\bar v) …

My answer will consist mainly of a collection of trivial facts but which nonetheless often generate some confusion. I begin by fixing some notation. Let $V$ be a complex vector space of complex dimen …

Let $X$ be a compact complex manifold of complex dimension $n$. The Hodge-Frölicher spectral sequence starts with $$ E_1^{p,q}=H^{p,q}(X,\mathbb C) $$ and the limit term $E^{p,q}_\infty$ is the grade …