# The minimal surface equation in a Riemannian metric

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: $$$$F(x,z,p,X) = h^{ij}(X_{ij} + \Gamma_{ij}^3 + p_i \Gamma_{3j}^3 + p_j \Gamma_{i3}^3 + p_i p_j \Gamma_{33}^3) \\ - p_k h^{ij}(\Gamma_{ij}^k + p_i \Gamma_{3j}^k + p_j \Gamma_{i3}^k + p_i p_j \Gamma_{33}^k),$$$$ where $$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: $$$$N = \frac{1}{W}(-u_1 \partial_1 - u_2 \partial_2 + \partial_3),$$$$ with $$W = (1 + g^{ij} u_i u_j)^{1/2}$$.

• 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. Jan 16 at 15:19
• @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? Jan 16 at 16:13
• 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. Jan 16 at 18:34
• Is $g$ really an arbitrary metric on $U\times\mathbb{R}$? Or have you omitted any assumptions about $g$/ Jan 17 at 16:02
• @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. Jan 17 at 21:21