For a surface $S$ defined in 3D space, denote its mean curvature as $H$, and the Laplace-Beltrami operator as $\Delta_S$. I know that there is a result for the Laplace-Beltrami of coordinate functions: $$ \Delta_S \vec{r} = 2 H \vec{n}. $$

My question is that: how to calculate the Laplace-Beltrami of the mean curvature? $$ \Delta_S H = ? $$

If there isn't a simple expression, is it possible to expand it like this way: $$ \Delta_S H = a_1 H + a_2 H^2 + a_3 K + ...$$ where $a_i$ are some constant numbers?

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    $\begingroup$ What sort of quantities do you allow in the dots? Obviously not the Laplace Beltrami of $H$, so maybe not derivatives. Perhaps polynomials in $H,K$? Clearly it is not a function of $H,K$. $\endgroup$
    – Ben McKay
    Aug 9, 2021 at 16:49
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    $\begingroup$ A reasonable analogy of your question would be the following: Suppose $f$ and $g$ are scalar functions (on $\mathbb{R}^2$ for simplicity) such that $\Delta f = g$. What's the formula for $\Delta g$? There's not much you can say beyond the fact that $\Delta g = \Delta^2f$. The answer here would be similar. $\endgroup$
    – Deane Yang
    Aug 9, 2021 at 19:00
  • $\begingroup$ Thanks Ben McKay and Deane Yang. My aim is to approximate $\Delta_S H$ by some polynomials of $H, K$. I don't know how to derive it. Is there some reference on this direction? Why the result is not a function of $H,K$? @BenMcKay $\endgroup$
    – Lightmann
    Aug 10, 2021 at 0:21
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    $\begingroup$ The Laplacian of $H$ with respect to the induced metric is 4-th order invariant of the surface, while $H$ and $K$ are only 2-nd order invariants. It's easy to show that there cannot be any generally valid formula of the form $\Delta_SH = F(H,K)$, though, of course, there are plenty of examples of local surfaces that satisfy this equation for a given function $F$. $\endgroup$ Aug 10, 2021 at 12:16
  • $\begingroup$ Thank @RobertBryant. Are there some references or books on this topic? $\endgroup$
    – Lightmann
    Aug 11, 2021 at 14:53

1 Answer 1


Here is a bit more detail on the answer to the following question, which is my interpretation of what the OP is asking:

Suppose that one knows the induced metric $g$ on a surface $S\subset\mathbb{R}^3$. What equations involving the metric does the mean curvature function $H$ satisfy?

There have to be some nontrivial equations, since one can't freely specify the mean curvature function $H$ once the induced metric is fixed. At the very least, $H$ has to satisfy the inequality $H^2-K\ge0$, where $K$ is the Gauss curvature of $g$, since $H^2-K = \tfrac14(\kappa_1{-}\kappa_2)^2$. This simple inequality is not sufficient though, since, as was classically observed, if a surface $S\subset \mathbb{R}^3$ has $H=0$ (i.e., is minimal), then the (singular) metric $(-K)\,g$ is the pullback of the metric on $S^2$ via the Gauss map $\nu:S\to S^2$ of the surface $S$ and hence must have Gauss curvature $+1$, which turns out to be an equation of order $4$ on the metric $g$.

Setting aside the trivial all-umbilic case where $H^2-K=0$, one can assume that $H^2-K = r^2 > 0$ for some function $r$ on $S$. In this case, it turns out that pursuing the structure equations, one derives a further necessary inequality of the form $F_g(H,\nabla H,\nabla^2H)\ge 0$, i.e., a polynomial inequality that is second-order in $H$, but whose coefficients involve 4 derivatives of the metric $g$.

In some cases, when $F_g(H,\nabla H,\nabla^2H)\equiv 0$, one can actually show that there is an entire circle of isometric immersions of $(S,g)$ into $\mathbb{R}^3$ with mean curvature $H$. Such data $(S,g,H)$ is said to constitute a Bonnet surface, after the work of Ossian Bonnet in the 19th century.

Assuming that $H$ satisfies the strict inequality $F_g(H,\nabla H,\nabla^2H)> 0$, it then turns out that there is a pair of equations $$ M_g(H,\nabla H,\nabla^2H,\nabla^3H) = N_g(H,\nabla H,\nabla^2H,\nabla^3H) = 0, $$ which are third-order in $H$ and fifth-order in $g$ that $H$ must satisfy in order to be the mean curvature of an isometric immersion of $(S,g)$ into $\mathbb{R}^3$ When $H$ satisfies these equations, there are at most two non-congruent isometric immersions of $(S,g)$ into $\mathbb{R}^3$ up to isometry. (When two exist, this pair of immersions is nowadays said to be a Bonnet pair.)

In particular, it follows from the structure equations that there is no universal identity of the form $\Delta_g H = E_g(H,\nabla H)$ where $E_g$ is a (possibly nonlinear) operator constructed from the metric $g$ and its derivatives (and hence, for example, might contain $K$ and some of its derivatives).


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