# Lower bounds on the Ricci curvature of Kähler submanifolds of $\mathbb{C}^n$

Say that $M$ is a smooth complex algebraic variety inside $\mathbb{C}^n$, and that $M$ has Ricci curvature bounded from below when endowed with the Kähler metric induced by the Euclidean metric of the ambient space. Is there something meaningful that can be inferred from this hypothesis? Is there a simple way to produce examples of such varieties?

I agree that what Dmitri said should work. Just add some comments that could help:

In your setting a lower bound on Ricci is equivalent to a lower bound on holomorphic sectional curvature.

Every smooth complex algebraic curve in $\mathbb{C}^2$ has ends asymptotic to flat cones of angle $2\pi m$ where $m$ is the multiplicity of the corresponding asymptotic line.

In higher dimensions a generic hypersurface is asymptotic to a cone with smooth cross section and has quadratic curvature decay.

It is known that around an ordinary multiple singularity of a plane complex curve in $\mathbb{CP}^2$ the metric is modeled by a union of infinitely negative cusps (see Ness, Curvature oon algebraic plane curves I). As we let $z \to \infty$ in Dmitri's example we aproach an ordinary double point on the complex link, the negative curvature should compensate the quadratic decay.

This is not a real answer, but rather a suggestion (with details missing).

There is definitely a simple way to produce such varieties. For example, any hypersurface whose completion (at infinity) is smooth in $\mathbb CP^n$ and intersects the infinity ($\mathbb CP^n$) transversely should have such a property. By scaling such a hypersuface one can make its Ricci curvature as close to zero as you want.

However, in the case the variety has singularities at infinity, Ricci curvature could be unbounded from below as (I believe should happen) for the following hypersurface: $\{z^2(x^2+y^2)=1\}\subset \mathbb C^3$, if we take $z\to \infty$. (I don't have a proof of this fact but rather confident that such example or an example of similar kind should work out.)