I try to understand the following version of the Kodaira embedding theorem:
Let $X$ be a compact Kähler manifold. A line bundle $L$ is positiv if and only if it is ample.

I have a problem with the 'easy part' of the proof: An ample line bundle $L$ is positive.

I know (per definition), there exist a positive integer $m > 0$, such the map $\iota_{L^{\otimes m}}: X \rightarrow \mathbb{P}^{n}$ is an embedding. Also I know that I look at the pullback $\iota_{L^{\otimes m}}^{*}\mathcal{O}_{\mathbb{P}^{n}}(1) \cong L^{\otimes m}$ and when I calculate the curvature I use the Fubini-Study-metric $\omega_{\text{FS}}$. But here, I have my problem(s):

  1. What happens with the positive integer $m$ on the left side of this identification? Why vanish this in the sense, that I don't look at $m$-fold of $\iota_{L^{\otimes m}}^{*}\mathcal{O}_{\mathbb{P}^{n}}(1)$? Why is the following identification not right?: $\iota_{L^{\otimes m}}^{*}\mathcal{O}_{\mathbb{P}^{n}}(m) \cong L^{\otimes m}$ Or is it right but the calculation is too difficult? In my opinion it makes more sense and I try to use this and I know, that the transition functions of $\mathcal{O}_{\mathbb{P}^{n}}(m)$ are the $m$-fold of $\mathcal{O}_{\mathbb{P}^{n}}(1)$. But I didn't know a hermitian metric for $\mathcal{O}_{\mathbb{P}^{n}}(m)$, so I couldn't go on with the calculation of the curvature.

  2. Initially I accept 1. and I search a little bit in the internet and I found the same idea to calculate the curvature $\Theta(L^{\otimes m})$. For example the book 'An Introduction to the Kähler-Ricci Flow' by Boucksom, Eyssidieux and Guedj (page 104) or the book 'Proceedings of the International Congress of mathematicians' (page 820) by Demailly: $\Theta(L^{\otimes m}) = \iota_{L^{\otimes m}}^{*}(\omega_{\text{FS}})$ and extracting the $m$-th root of the metric solve the problem. But my problem is the same: What happens with the $m$ in this equation. Isn't it possible to say ' if $h^{m}$ is any hermitian metric on $L^{\otimes m}$ ' and then you calculate the curvature? Perhaps it will be clearer for me, when someone show me this calculation?!

I hope, someone can help me with my dilemma ;-)


1 Answer 1


Consider the case where $L=\mathcal{O}(1)$ on $X=\mathbb{P}^1$, but using $L^{\otimes 2}$ to make the map to projective space $\mathbb{P}^2$. Then you immediately see that the pullback of $\mathcal{O}(1)$ is $\mathcal{O}(2)$, looking at where the sections vanish geometrically.

  • $\begingroup$ But then if you find a Hermitian metric for $L^{\otimes m}$, you check that it defines a Hermitian metric for $L$, with curvature divided by $m$, so still positive. $\endgroup$
    – Ben McKay
    Jul 27, 2016 at 11:50
  • $\begingroup$ First of all, thank you for your help! I try this what you said: $L^{\otimes 2} = L \otimes L = \mathcal{O}(1) \otimes \mathcal{O}(1) = \mathcal{O}(2)$; then I have the embedding $\iota_{L^{\otimes m}}: \mathbb{P} \rightarrow \mathbb{P}^{2}$; For $\mathcal{O}(2) \rightarrow \mathbb{P}^{2}$ I have the transition functions $g_{ij} = (\frac{z_{j}}{z_{i}})^{2}$ and the sections of $\mathcal{O}(2)$ are given by $z_{k}$. For the metric the only possibility is, that I choose the one, which I use for $\mathcal{O}(1): h_{i} = \frac{|z_{i}|^{2}}{\sum_{j}|z_{j}|^{2}}$. So far is that right? $\endgroup$
    – danielg
    Jul 27, 2016 at 17:36
  • $\begingroup$ Sorry, I don't understand, what you mean by 'looking at where the sections vanish geometrically'. Normally, the sections $s_{i}$ must be $\neq 0$ so that $\iota_{\mathcal{O}(2)}$ is an embedding or is that wrong? $\endgroup$
    – danielg
    Jul 27, 2016 at 17:39
  • $\begingroup$ Every holomorphic section of $\mathcal{O}(k)$ on $\mathbb{P}^1$ has $k$ zeroes counting multiplicities. Every section of $\mathcal{O}(1)$ on $\mathbb{P}^2$ has zeroes along a projective line. This is getting a little too detailed, and is not really a research problem, so perhaps ask on math.stackexchange. $\endgroup$
    – Ben McKay
    Jul 27, 2016 at 19:53

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