# Kähler form on the Blow up of a Kähler manifold

Given a Kähler manifold $X$, the blow up is a Kähler manifold aswell (see Hodge theory and CAG by Voision Prop. 3.24). The idea is of course to use the pull-back the Kähler form $\pi^*\omega_X$. It says that this form is not positive, but only semi-positive and I fail to see why exactly.

Positivity of $\omega_x$ means that locally we can write $\omega_x= \sum_i \alpha_i dz_i\wedge d\bar z_i$ with $\alpha_i$ real and non-negative. Now $\pi^*\omega_X= \sum_i (\alpha_i\circ\pi) d(z_i\circ\pi)\wedge d(\bar z_i\circ\pi)$. Obviously the coefficients $(\alpha_i\circ\pi)$ still remain positive, so somehow the differential forms $d(z_i\circ\pi)\wedge d(\bar z_i\circ\pi)$ must vanish on some special tangent vectors in a neighborhood around the blow up?

I am also not sure, if my notion of positivity is correct or appropriate here.

• Did you understand the proof in the end? I am currently trying to figure it out, but I have my problems. see math.stackexchange.com/questions/3276030/…. I am under the impression that what is proved in the consecutive lemma 3.25 is noth enough. Could you help me? – Michael Jun 30 '19 at 11:48

Let $\widetilde X$ be the blowup of $X$, and let $\pi\colon \widetilde X\to X$ denote the blow-down map. If the local coordinates $(z_i)$ are chosen so that $z=0$ is the point being blown up, then $D = \pi^{-1}(0)\subseteq \widetilde X$ is a complex hypersurface called the exceptional divisor. Because $z_i\circ\pi\equiv 0$, it follows that $d(z_i\circ \pi)$ annihilates every vector tangent to $D$, and the same goes for $d(\bar z_i\circ\pi)$.

• This makes sense, but what do you mean by "every vector tangent to D". Just any $v\in T_x \widetilde X$ for $x\in D$? – Notone Nov 26 '17 at 19:24
• @Notone: No, that's not what it means. For each $x\in D$, there is a natural way to identify $T_xD$ with the subspace $di_x(T_xD) \subseteq T_x\widetilde X$, where $i\colon D\to\widetilde X$ is the inclusion map. A vector in $T_x\widetilde X$ is said to be tangent to $D$ if it is in the image of this map. – Jack Lee Nov 26 '17 at 19:35
• In your last sentence do you mean "it follows that $dz_i\circ\pi$ annihilates every vector tangent to D"? But other than that it's clear now, thanks alot – Notone Nov 26 '17 at 20:25
• @Notone: Sorry, what I means was $d(z_i\circ\pi)$. Fixed now. – Jack Lee Nov 26 '17 at 20:28

If $M$ and $N$ be complex manifolds of the same dimension and $π:M→N$ is a holomorphic mapping, then for a volume form (as measure )$Ψ$ on $N$ the pull-back $π^∗Ψ$ is positive outside aramification divisor of $M$ and may not be a positive on the whole of $M$. There is a classical paper of P. Griffiths http://publications.ias.edu/sites/default/files/nevanlinna.pdf

From Principles of Algebraic Geometry by Phillip Griffiths and Joseph Harris, we have:

Let $Y\subset X$. If $Y$ is compact then blow up is Kahler but to construct the metric on $Bl_YX$substantially you use $π^∗ω+εc_1(\mathcal O(−E))$ where $E$ is the exceptional divisor and $π:Bl_YX\to X$ is the canonical surjection

Definition of positivity of pull back of kahler form as current is different with what you wrote , see this paper. See also the definition of pull back of current https://arxiv.org/pdf/math/0606248.pdf

See Lemma 34. of this paper