I'm reading Griffiths and Harris, Principles of algebraic geometry, and it's written that
"any effective nonzero divisor on $\mathbb{P}^n$ is positive"
on page 159. Is that obvious? There doesn't seem to be any explanation in the book.
Here, a divisor is called positive if the associated line bundle has a hermitian metric whose curvature form is positive.
I know that hyperplane divisors are positive since they induce the Fubini-Study metric, but I'm not sure how to deal with a general irreducible hypersurface.
Best Answer
Just to get this off the unanswered list.
As mentioned in the original post, an effective divisor $D$ on $\mathbb{P}^n$ is a finite sum $\displaystyle \sum_i a_i V_i$ where $a_i\in\mathbb{Z}^{\geqslant 0}$ and $V_i$ are irreducible subvarieties of $\mathbb{P}^n$. We call such a divisor positive if the associated line bundle $\mathcal{O}(D)$ has a Hermitian metric with positive curvature. It's easy to see that if $D_1,\ldots,D_m$ are positive then so is $D_1+\cdots +D_m$.
So, to see that every effective divisor on $\mathbb{P}^n$ is positive it suffices to show that every divisor of the form $V$ is positive where $V$ is a closed analytic submanifold. But, by Chow's theorem we know that $V=V(f)$ where $f$ is a homogenous polynomial in $n+1$-variables of degree $d$. But, one can quickly check by hand that $V(f)$ is then equivalent to $dH$ where
$$\mathbb{P}^{n-1}\cong H:=\{[0:z_1:\cdots:z_n]\}\subseteq \mathbb{P}^n$$
So, by our above observations it suffices to show that $H$ is positive. But, this is clear.
One can also verify the claim that every $V$ is linearly equivalent to a multiple of $H$ without appealing to Chow's theorem as follows. We know that the line bundles on $\mathbb{P}^n$ are classified by $H^1(\mathbb{P}^n,\mathcal{O}^\times)$. But, by the exponential sequence
$$0\to \underline{2\pi i\mathbb{Z}}\to \mathcal{O}\to \mathcal{O}^\times\to 0$$
we get the exact sequence
$$H^1(\mathbb{P}^n,\underline{2\pi i\mathbb{Z}})\to H^1(\mathbb{P}^n,\mathcal{O})\to H^1(\mathbb{P}^n,\mathcal{O}^\times)\to H^2(\mathbb{P}^n,\underline{2\pi i \mathbb{Z}})\to H^2(\mathbb{P}^n,\mathcal{O})$$
But,
$$H^i(\mathbb{P}^n,\underline{2\pi i \mathbb{Z}})\cong H^i_\text{sing}(\mathbb{P}^n,\mathbb{Z})=\begin{cases} \mathbb{Z} & \mbox{if}\quad i\in\{0,2,\ldots,2n\}\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$$
and
$$H^i(\mathbb{P}^n,\mathcal{O})=\begin{cases} \mathbb{C} & \mbox{if}\quad i=0\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$$
from where we see that
$$\mathrm{Pic}(\mathbb{P}^n)\cong H^1(\mathbb{P}^n,\mathcal{O}^\times)\cong H^1(\mathbb{P}^n,\underline{2\pi i \mathbb{Z}})\cong \mathbb{Z}$$
call the map
$$H^1(\mathbb{P}^n,\mathcal{O}^\times)\to H^2(\mathbb{P}^n,\underline{2\pi i\mathbb{Z}})\cong \mathbb{Z}$$
the 'Chern class map' and denote it by $c$. What the above shows is that a line bundle on $\mathbb{P}^n$ is entirely determined by its Chern class.
Now, it's not hard to directly check that $c(\mathcal{O}(H))=1$ and in general $c(\mathcal{O}(D))>0$ if $D$ is effective (just think about the cycle class map) from which case it follows that $\mathcal{O}(D)\cong \mathcal{O}(H)^{c(\mathcal{O}(D))}$ or, in other words, $D\sim c(\mathcal{O}(D))H$ (where $\sim$ denotes equivalence).