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Let $\Omega \subset \mathbf{R}^2$ be a domain, and let the cylinder $\Omega \times \mathbf{R}$ above it be endowed with a Riemannian metric $g$. (Note this is not assumed invariant in the vertical direction.) Let $u: \Omega \to \mathbf{R}$ be a $C^2$ function.

What is the expression for the quasilinear operator $F(x,z,p,X)$ expressing the mean curvature of $G$, i.e. so that $F(x,u,Du,D^2 u) = 0$ if $G$ is minimal?

Edit. I would be happy with an expression valid for example for diagonal metrics, that is of the form $g = g_{11} \mathrm{d} x^2 + g_{22} \mathrm{d} y^2 + g_{33} \mathrm{d} z^2$.

In the book of Colding and Minicozzi (pp. 235-236) the following formula is given: \begin{equation} F(x,z,p,X) = h^{ij}(X_{ij} + \Gamma_{ij}^3 + p_i \Gamma_{3j}^3 + p_j \Gamma_{i3}^3 + p_i p_j \Gamma_{33}^3) \\ - p_k h^{ij}(\Gamma_{ij}^k + p_i \Gamma_{3j}^k + p_j \Gamma_{i3}^k + p_i p_j \Gamma_{33}^k), \end{equation} where $h^{ij}$ is the inverse of $h_{ij} = \langle \partial_i + p_i \partial_3,\partial_j + p_j \partial_3 \rangle_g$.

Is this formula accurate? I can't follow their derivation, because it starts with an expression for the unit normal to $G$ that looks incorrect to me: \begin{equation} N = \frac{1}{W}(-u_1 \partial_1 - u_2 \partial_2 + \partial_3), \end{equation} with $W = (1 + g^{ij} u_i u_j)^{1/2}$.

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  • $\begingroup$ If you have an isometric immersion $\phi: M \rightarrow N$, then I'm pretty sure that the Hessian of $\phi$, $$\partial^2\phi: S^2T_*M \rightarrow T_*N$$ is the second fundamental form multiplied by the unit normal. Therefore, the embedding is minimal if $g^{ij}\nabla^2_{ij}\phi = 0$, where $g$ is the metric on $M$. Here, $\phi(x) = (x,u(x))$. So you just need to compute the Hessian of this map. $\endgroup$
    – Deane Yang
    Commented Jan 16, 2023 at 15:19
  • $\begingroup$ @DeaneYang Hmm, that might be a good idea. I mean, I guess $\phi: x \mapsto (x,u(x))$ would generally not be an isometry $\Omega \to \Omega \times \mathbf{R}$, but it ought to be harmonic at least. I don't know that much about harmonic maps, but perhaps for graphs one can ignore the point with conformality, and the harmonic map and minimal surface equation are just equivalent. Is that in essence what you're suggesting in your comment? $\endgroup$
    – Leo Moos
    Commented Jan 16, 2023 at 16:13
  • $\begingroup$ Leo, the map is an isometric embedding of the pullback metric. The differential of the map is the Jacobian which does not depend on the metric but is a bundle map from one tangent bundle to the other. A nice exercise is to prove the covariant derivative of this map is the second fundamental form. $\endgroup$
    – Deane Yang
    Commented Jan 16, 2023 at 18:34
  • $\begingroup$ Is $g$ really an arbitrary metric on $U\times\mathbb{R}$? Or have you omitted any assumptions about $g$/ $\endgroup$
    – Deane Yang
    Commented Jan 17, 2023 at 16:02
  • $\begingroup$ @DeaneYang Yup - the metric is arbitrary, there's no secrets. (I'd be delighted with a formula when $n = 3$, not sure if that counts.) I assumed that a formula was possible, seeing as Colding-Minicozzi give one in their book. $\endgroup$
    – Leo Moos
    Commented Jan 17, 2023 at 21:21

1 Answer 1

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There are no simpler expressions for your case.

Essentially you ask for a minimal-surface equation assuming that your surface is a graph in local coordinates, but any smooth-regular surface is a graph in some local coordinates.

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  • $\begingroup$ What do you mean, simpler? What you write in the second paragraph is of course true, but so what? $\endgroup$
    – Leo Moos
    Commented Jan 16, 2023 at 11:20
  • $\begingroup$ @LeoMoos simpler than the standard one. $\endgroup$
    – Anton Petrunin
    Commented Jan 16, 2023 at 11:21
  • $\begingroup$ And what would the "the standard one" be? $\endgroup$
    – Leo Moos
    Commented Jan 16, 2023 at 11:27
  • $\begingroup$ @LeoMoos Say vanishing mean curvature. $\endgroup$
    – Anton Petrunin
    Commented Jan 16, 2023 at 11:29
  • $\begingroup$ It looks like you've misunderstood the question. I've rephrased it; hopefully it's easier to understand now. $\endgroup$
    – Leo Moos
    Commented Jan 16, 2023 at 12:38

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