# Simple question regarding shape operator/minimal surface equation

Briefly, I'm asking why commuting the metric and the second fundamental form should make a difference.

Normally I wouldn't think much of this, but I came across the issue in the paper "Hypersurfaces of constant curvature in hyperbolic space. II." by Guan and Spruck. They in turn cite another paper, "Nonlinear second order elliptic equations IV. Starshaped compact Weingarten hypersurfaces." by Caffarelli, Nirenberg, and Spruck, which, sadly, is not easy to access. I found it strange that such a seemingly basic fact needed such a "high powered" reference to be addressed.

For example, consider a function $f:\mathbb{R}^2\to\mathbb{R}$ and let $S = (x,y,f(x,y))$ be the graph of $f$ in $\mathbb{R}^{3}$. Then the induced metric on the surface is $g_{ij} = \delta_{ij}+f_{i}f_{j}$ and the second fundamental form is $\frac{f_{ij}}{\sqrt{1+f_{1}^{2}+f_{2}^{2}}}$. The shape operator $s$ should then be $s_{j}^{i} = g^{ik}h_{kj}$. However, when I take the trace of $s$ (as a matrix) I obtain $\frac{f_{11}(1+f_{2}^{2})+f_{22}(1+f_{1}^{2})}{(1+f_{1}^{2}+f_{2}^2)^{3/2}}$. Note that this is not the expression that arises in the minimal surface equation ... but it should be, because it's the trace of the second fundamental form with respect to $g$! (At least, that's what I would believe by reading geometry texts, such as Lee, "Riemannian manifolds: an introduction to curvature").

However, let $\gamma_{ij} = \delta_{ij}+\frac{f_{i}f_{j}}{1+\sqrt{1+f_{1}^{2}+f_{2}^{2}}}$, so that $\gamma_{ij}$ is the square root of $g_{ij}$ via $\gamma_{ik}\gamma_{kj} = g_{ij}$. Consider the operator $t = \gamma^{ik}h_{kl}\gamma^{lj}$ ... in other words, $t$ is $s$ but after commuting a square root of $g_{ij}$. Then if I take the trace of $t$, as a matrix, I obtain $\frac{(1+f_{1}^{2}+f_{2}^{2})\Delta f - f_{ij}f_{i}f_{j}}{(1+f_{1}^{2}+f_{2}^{2})^{3/2}}$, which is exactly the expression arising in the minimal surface equation.

What is going on here? The idea of taking the trace after commuting a square root power seems strange. None of the geometry books I've looked at address this, and indeed in my time studying geometry I've never seen this issue arise before. Any help (even just a suggestion for a reference) would be appreciated.

• My solution to such questions: never do calculations because you(=I) never do it right. – Anton Petrunin Apr 19 '12 at 5:19
• I give a pretty clear description of this in Michigan Mathematical Journal, volume 38 (1991), pages 255-270. I had not seen explicit formulae for what I needed so I included some. – Will Jagy Apr 19 '12 at 5:46
• Are you sure that you have computed $s$ correctly? If $g_{ij}=\delta_{ij}+f_if_j$ then the inverse metric is $g^{ij}=\delta_{ij}-\tfrac{f_if_j}{1+|Df|^2}$ ... – Robert Haslhofer Apr 19 '12 at 7:30
• The two different ways of doing the calculation have to lead to the same answer, since the trace of $(\gamma^{-1} h) \gamma^{-1}$ equals the trace of $\gamma^{-1}(\gamma^{-1} h)$, which equals the trace of $g^{-1}h$. – Deane Yang Apr 19 '12 at 8:29
• The difficulty arose from the software I was using for the matrix product, that's why I kept running into this problem. The claim in Prof. Yang's comment is correct, I had thought that this claim was true but kept obtaining a different result. Thank you for helping to clear this up. – Nick Apr 20 '12 at 20:56

## 1 Answer

Posting this so I can close the problem. Question has been answered above.