# Laplace-Beltrami of the mean curvature

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?

• 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 $H,K$. Aug 9, 2021 at 16:49
• 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. Aug 9, 2021 at 19:00
• 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 Aug 10, 2021 at 0:21
• 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 $F$. Aug 10, 2021 at 12:16
• Thank @RobertBryant. Are there some references or books on this topic? Aug 11, 2021 at 14:53

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).