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Define h : ℂ → ℂ2 via

    h(z) = (f(z), g(z)),

where f, g : ℂ → ℂ are analytic functions such that the complex velocity h'(z) = (f'(z), g'(z)) ∈ ℂ2 is never equal to (0,0). Suppose further that h is an embedding of ℂ into ℂ2.

Then the image h(ℂ) is a smooth (C∞) real surface at each of its points h(z), deriving its metric in the standard way as a submanifold of ℂ2 = ℝ4. As such it has a well-defined gaussian curvature K(z) at each point h(z) ∈ ℂ2.

After finding a formula for K(z) in no book or paper that I was able to access, I calculated such a formula as

    K(z) = -2 |h''(z) ^ h'(z)|2 / |h'(z)|6,

where h''(z) ^ h'(z) is shorthand for f''(z) g'(z) - g''(z) f'(z).

Question: Is this formula well known? Can anyone give a reference for where it can be found?

(Note: I am not looking for formulas from which this one can be derived.)

Remark: We could have replaced the domain by a nonempty open subset of ℂ and asked only that h be an immersion, but those refinements seemed to obscure the main point.

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  • $\begingroup$ I think it is well-known, as are the cases $\mathbb{C}^2$ (or $\mathbb{CP}^2$) equipped with the Fubini-Study metric. I can't give you a reference though. They are almost immediate from the chern connection matrix. $\endgroup$
    – user10354138
    Commented Oct 14, 2023 at 4:09
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    $\begingroup$ Perhaps try papers of Linda Ness in the late 70‘s. $\endgroup$
    – Ted Shifrin
    Commented Oct 14, 2023 at 6:08
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    $\begingroup$ Any chance the math could be rewritten in LaTeX? $\endgroup$
    – Deane
    Commented Oct 14, 2023 at 13:23
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    $\begingroup$ This is a formula that is too specialized to be mentioned in most textbooks. I’m sure it’s been used and appears in at least one paper somewhere, but I wouldn’t know how to find them. And since the derivation is straightforward, the proof is probably left to the reader. $\endgroup$
    – Deane
    Commented Oct 14, 2023 at 16:06
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    $\begingroup$ How did you prove this exactly? The general formula for Gaussian curvature in terms of the metric is complicated but in isothermal coordinates $f^2(x,y)(dx^2+dy^2)$ it simplifies to $K=-\Delta_{LB}\ln f$ where $\Delta_{LB}$ is the Laplace-Beltrami operator. Since your map is holomorphic, it provides isothermal coordinates and therefore the simpler formula applies. I haven't checked the details but the formula you provide does not seem to be that surprising. Interestingly, it is the square of the formula for the geodesic curvature of a curve. I am not sure I have a ready explanation othis. $\endgroup$
    – Mikhail Katz
    Commented Oct 23, 2023 at 14:04

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