Embedding of Kähler manifolds into $\Bbb C^n$ Consider $\Bbb C^n$ with its standard hermitian product. This space produces many example of Kähler manifolds simply by taking a smooth affine variety $X\subseteq\Bbb C^n$ with the induced metric. Now I am wondering about the converse:

Question: Suppose that $X$ is a Kähler manifold that is also a smooth affine variety over $\Bbb C$. Can we assume that $X$ has an embedding $X\subseteq\Bbb C^n$ such that the metric of $X$ is the one inherited from the hermitian product on $\Bbb C^n$?

Note that I assume that $X$ is algebraic to begin with. It would false for a general Kähler manifold.
 A: Such manifolds are called Stein.
They have a lot of very strong properties, starting with Cartan's A and B theorems. Obviously not all (non-compact) Kähler manifolds satisfy them. For example take out a point from $\mathbb{CP}^2$. Then any embedding of it into affine space would give a contradiction with the Theorem B.
A: The answer is 'no'.  For example, take $X=\mathbb{C}$, which is a smooth affine variety over $\mathbb{C}$.  If you take any embedding of $X$ into $\mathbb{C}^n$, the induced Kähler metric will be real-analytic and will have non-positive curvature.  As a result, if you fix any Kähler metric $h$ on $\mathbb{C}$ that is somewhere either not real-analytic or has positive Gauss curvature, then that $h$ cannot be induced by any embedding of $\mathbb{C}$ into any $\mathbb{C}^n$.
The condition of $h$ being real-analytic and having non-positive curvature is not enough to guarantee that $(\mathbb{C},h)$ can be induced by an embedding into some $\mathbb{C}^n$.  This was investigated by Eugenio Calabi in a famous paper: Isometric embeddings of complex manifolds, Ann. of Math. 58 (1953), 1–23, in which he derived necessary and sufficient conditions.
