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As Faisal says, there is no hope to have a general Nash-type theorem for all complex manifold, when the ambient space considered for the (isometric) embedding is some $\mathbb C^N$: no compact complex …

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 $X$ 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 …

If $(X,\omega)$ is Kähler, then it is always true that $$ \Delta'=\Delta''=\frac 12\Delta, $$ where these three Laplacians are with respect, in order, to $\partial$, $\bar\partial$ and $d$. This is va …

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?

Here is the answer. Let $(X,\omega)$ be a Kähler $n$-dimensional manifold. Fix a point $x_0\in X$ an choose local holomorphic coordinates $(z_1,\dots,z_n)$ centered at $x_0$ and such that $(\partial/ …

Call $C^{\infty}_{p,q}(M)$ the space of smooth complex sections of the bundle $\Lambda^{p,q}T^*_M$ and let $2n$ be the real dimension of $M$. The fact that $$ dC^{\infty}_{p,q}(M)\subset C^{\infty}_ …

Maybe what follows is not exactly what you are looking for, but it gives you an answer at least when you don't want to restrict yourself to the tangent bundle and work rather with general (complex) ve …

I will try to clarify a little bit your question and at the same time to give you an answer. All that I say, you can find on Demailly's book "Complex Analytic and Differential Geometry". GENERAL THEO …

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 …

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 …

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 …

The answer is yes. The holonomy principle states that a given a riemannian manifold $(M,g)$ and a point $x\in M$, the datum of a parallel tensor field of a given type is equivalent to the datum of a …

The whole point is that if the metric is Kähler, then the Chern connection coincides with the complexification of Levi-Civita connection of the riemanian metric on the underlying real manifold given b …