David Mumford remarks in his book Algebraic Geometry I, Complex Projective Varieties on page 109 that the fact that the canonical ring $\oplus_{k=0}^{\infty} \Omega_{k, \mathbb{P}^n}$ of projective space $\mathbb{P}^n$ has a very poor structure ($\Omega_{k, \mathbb{P}^n}=0$ for all $k \ge 1$) is closely related to the fact from differential-geometry that $\mathbb{P}^n$ 'stands at one end of the curvature spectrum having the most positive curvature of any algebraic variety'.
Moreover it is also stated that for a smooth variety the canonical class $K_{\mathbb{P}^n}$ turns out to represent minus the cohomology class of the Ricci curvature of $X$ (source Cornalba-Griffiths: Algebraic Geometry Arcata 1974 AMS).
Unfortunately my basic knowledge of differential geometry is not sufficient to understand the meaning of these two remarks. At first, what does it mean that a smooth variety or say a manifold has the most positive curvature? In case of surfaces the Gaussian curvature associates to every point on the surface a real number and therefore it is natural to compare the curvatures of surfaces by the relation "$\le$" and it's reasonable to call one curvature 'more positive than the other' or vice versa. On the other hand if we deal with manifolds of dimension $n >2$ then the curvature (presumbly Mumford means here the Riemann curvature tensor ) the curvature is not a number any more, so that's incomprehensible to me how in higher dimensions one can call a curvature 'most positive'. Any idea what Mumford here wants to say.
The second point is how generally for divisor $D$ on manifold $M$ or say it's associated bundle $L(D) \to M$ one can associate a certain cohomology class in the way Mumford here did in order to compare it with the class associated to Ricci curvature? As 1st Chern class of $L(D)$?
