Can the unit complex 1-dimensional disc be embedded isometrically into complex euclidean space? - MathOverflow

Can the unit complex 1-dimensional disc be embedded isometrically into complex euclidean space?

2010-10-30T07:40:02Z

Let $D$ denote the unit complex 1-dimensional disc, together with the hyperbolic metric $h_D=\frac{4dz\wedge d\bar{z}}{(1-|z|^2)^2}$of curvature $-1$. By Nash's embedding theorem, we can always embed the disc $D$ real-analytically and isometrically into real Euclidean space ${\mathbb{R}}^n$ for some large $n$. (I think $n=5=2\dim_{\mathbb{R}}D+1$ is sufficient.)

I'm wondering whether we have a complex-analytic embeddeding of the disc $D$ into complex Euclidean space. Consider complex Euclidean space ${\mathbb{C}}^n$ with the standard Euclidean metric $h_{\mathbb{C}^n}=dz_1\wedge d\bar{z_1}+\cdots+dz_n\wedge d\bar{z_n}$. Does there exist a holomorphic map $f:D\to {\mathbb{C}}^n$ for some large $n$ that is at the same time an isometry, i.e. $f^*h_{\mathbb{C}^n}=h_D$?

If we write the map $f$ in terms of coordinates $f=(f_1,\ldots, f_n)$, this question has the an equivalent formulation solely in terms of holomorphic functions. This question is asking whether there are holomorphic functions $f_i:D\to {\mathbb{C}}$ on the disc, such that $\frac{4}{(1-|z|^2)^2}=f_1(z)\overline{f_1(z)}+\cdots + f_n(z)\overline{f_n(z)}$ for all $z\in D$.

Answer by David Roberts for Can the unit complex 1-dimensional disc be embedded isometrically into complex euclidean space?

2010-10-30T08:13:41Z

There is a body of work on embedding Riemann surfaces into $\mathbb{C}^n$ for small $n$ (it's possible for large $n$). If I recall correctly, Riemann surfaces can be holomorphically embedded in $\mathbb{C}^3$ easily, but the sharp lower bound that is known for complex manifolds of complex dimension $n > 1$ (which is believe is $2n$) is not known to be sharp for curves in general, although it is known in several cases constructed by hand. Forsternic is a name to look for, e.g.

Holomorphic curves in complex spaces
Barbara Drinovec Drnovšek and Franc Forstnerič
Duke Math. J. Volume 139, Number 2 (2007), 203-253.

(Edit: but this doesn't contain the answer).

I asked Finnur Larusson (one of Forstneric's collaborators) and I was told that $\mathbb{C}^2$ is impossible for the hyperbolic disk as you ask. However I didn't get a positive answer, with Larusson being a bit sceptical about the possibility, but he pointed to the following paper as indicating that an approximate embedding is possible for large $n$:

Bremermann, H. J.
Note on plurisubharmonic and Hartogs functions.
Proc. Amer. Math. Soc. 7 (1956), 771–775.

This may be not known.

Answer by Robert Bryant for Can the unit complex 1-dimensional disc be embedded isometrically into complex euclidean space?

2011-04-16T23:00:28Z

The answer is 'no', there is no holomorphic curve in $\mathbb{C}^n$ (for any $n$) such that the induced metric has constant negative curvature. To my knowledge, this was first proved by E. Calabi many many years ago, essentially using the structure equations for holomorphic curves in $\mathbb{C}^n$. The proof is easy, but it involves knowing something about the structure equations, and I don't want to try to explain that here. A reasonable source for this is Blaine Lawson's Lectures on Minimal Submanifolds, Chapter IV (see Theorem 11).

There is a more general fact, namely that the only minimal surface (let alone holomorphic curve) in $\mathbb{C}^n$ that has constant Gaussian curvature is a flat plane. This is due to M. Pinl, Minimalfl\"achen fester Gau{\ss}chen Kr\"ummung, Math. Ann. 136 (1958), 34-40.