# Why do the first Chern classes of these line bundles span the Dolbeault cohomology group $H^{1,1}(X;\mathbb{R})$?

Forgive me for what is probably a simple question, I am new to this field. I am studying the Hirzebruch surfaces and their higher dimensional analogues $M_{n,k}$, defined to be the projective line bundles

$$M_{n,k}=\mathbb{P}(\mathcal{O}(-k)\oplus\mathcal{O}(0))$$ over $\mathbb{CP}^{n-1}$. Here, $\mathcal{O}(-1)$ is the tautological line bundle, and $\mathcal{O}(0)$ the trivial line bundle, both over $\mathbb{CP}^{n-1}$. We can define two divisors on $M_{n,k}$, namely $D_0$ to be the section with zero $\mathcal{O}(-k)$ component and $D_\infty$ to be the section with zero $\mathcal{O}(0)$ component. These two divisors then determine holomorphic line bundles $[D_0]$ and $[D_\infty]$ in the usual way.

But to a line bundle $L$ on a complex manifold $X$, we can associate its first Chern class $c_1(L)$, which I understand to be the cohomology class in $H^{1,1}(X;\mathbb{R})$ determined by the curvature form $R_h$ of any Hermitian metric $h$ on $L$ (this is independent of $h$).

I am then told that the cohomology classes $c_1([D_0])$ and $c_1([D_\infty])$ span $H^{1,1}(M_{n,k};\mathbb{R})$, and that for any Kahler class $\alpha\in H^{1,1}(M_{n,k};\mathbb{R})$ (i.e. any class for which a Kahler metric/form can be chosen as a representative), one can find constants $0<a<b$ such that $$\alpha=\frac{b}{k}[D_\infty]-\frac{a}{k}[D_0].$$

I'm afraid that I have absolutely no idea as to how one can show this, or why it may be obvious. I'm aware that there are other ways of defining the 1st Chern class of a line bundle, and perhaps one of these may be more useful. Any help would be much appreciated!

Note that $c_1$ in fact lies in $H^{1,1}(X,\mathbb{Z}):=H^2(X,\mathbb{Z})\cap H^{1,1}(X)$. Then, this is precisely the content of Lefschetz' (1,1)-theorem. See the Wikipedia article on it: https://en.wikipedia.org/wiki/Lefschetz_theorem_on_(1,1)-classes
• Thanks for the answer. So I now understand that the Lefschetz theorem tells us we can write any element of $H^{1,1}(M_{n,k};\mathbb{R})$ as the linear combination of such classes of hypersurfaces, but is it clear in this case that $H^{1,1}$ is only two-dimensional? I am also unsure as to how the result for the Kahler class $\alpha$ follows. – jl2 Aug 15 '18 at 10:05
• @jl2 you're right. I didn't read your question carefully enough to notice its other parts. I wouldn't know how to go about the Kahler classes part. However, one could try to use the following two results for the other claim: 1) $H^1(X,\mathcal{O}_X^\times)\cong\operatorname{Pic}(X)$, and 2) $\operatorname{Pic}(M_{n,k})\cong\operatorname{Pic}(\mathbb{P}^{n-1})\times\mathbb{Z}(\mathcal{O}(-k)\oplus\mathcal{O})$. See mathoverflow.net/questions/61053/… . – user347489 Aug 15 '18 at 18:39