Total mean curvature in $L^2$ and minimal surfaces in spaces with non-positive sectional curvature

Let's suppose we have a Riemannian $n$-manifold $(N,g)$ and an immersed surface $f:\Sigma\rightarrow N$, with genus zero, equipped with the induced metric. Let's further assume that the ambient space has non-positive sectional curvature.

If $f$ is minimal, i.e. $$ \vec{H} = 0 $$ where $\vec{H}$ is the mean curvature vector of $f$ (with respect to the induced metric $f^*g$), then $f$ is not closed. (The only proof I know of this fact uses the maximum principle. A geometric proof would be nice, if anybody has a reference. Perhaps using Cartan-Hadamard?)

Clearly then, for $f$ closed with genus zero, an inequality of the form $$ \int_\Sigma |\vec{H}|^2d\mu > c_g > 0 $$ should hold, for some $c_g$ depending only on $(N,g)$.

My question is: what are the known values of $c_g$?

Solution 1:

We can follow the derivation of the Willmore energy bound for genus 0 surfaces in $\mathbb{R}^3$.

By the Gauss equation we have that, letting $e_1, e_2$ denote an orthonormal frame of $T\Sigma$, $K$ be the sectional curvature operator of the ambient manifold, $S$ be the induced scalar curvature of $\Sigma$, and $II$ be the second fundamental form (using that $H = \frac{1}{n} \mathrm{tr} II$ for submanifolds of dimension $n$)

$$ 2 K(e_1,e_2) = S + \sum_{i,j = 1,2} |II(e_i,e_j)|^2 - 4|H|^2 $$

Now, we can write $II$ into its trace and trace-free parts:

$$ II(v,w) = H g_\Sigma(v,w) + \tilde{II}(v,w) $$

we see that the double trace $ ||II||^2 $ can be expanded as $2 |H|^2 + \sum |\tilde{II}(e_i,e_j)|^2 $. Hence from the Gauss equation we derive that

$$ 2 K(e_1,e_2) = S + \sum_{i,j = 1,2} |\tilde{II}(e_i,e_j)|^2 - 2|H|^2 $$

which implies, using non-positive sectional curvature that

$$ 2|H|^2 - S \geq 0 $$

This implies that the Willmore energy

$$ W[\Sigma; N,g] = \int_\Sigma (|H|^2 - S/2) \mathrm{d}\mu \geq 0 $$

So by Gauss-Bonnet (for genus 0) we have that

$$ \int_{\Sigma} |H|^2 \mathrm{d}\mu \geq 4\pi $$

in exactly the same way as in the case where the ambient space is Euclidean. Furthermore, we see that the Euclidean bound is "optimal" in the following sense (using that $K \leq 0$):

$$ W[\Sigma; N,g] = 0 \iff K(e_1,e_2) = 0 \text{ and } \tilde{II} = 0 $$

this implies that $\Sigma$ is umbilic and "locally" (in a very infinitesimal sense) the sphere looks like the round one and the ambient space looks like the Euclidean one.

Furthermore, I claim that this bound is the best possible! By using the "conformal invariance" (of course, if the ambient space does not have a scaling symmetry, this is a problem, but bear with me for a moment here), we see that in the Euclidean case all round spheres are equal for the purpose of Willmore energy. Now consider the non-positively curved case. The excess from the $4\pi$ bound is $$ \int_{\Sigma} |\tilde{II}|^2 - 2K ~\mathrm{d}\mu$$ Now consider a point $p\in N$ and let $\Sigma_{p,r}$ be the image under the exponential map of a round sphere in $T_pM$ of radius $r$. As $r\to 0$ we have that the total volume goes to $0$ and hence the sectional curvature integral also vanishes. A local computation (if I remember correctly) shows that as $r\to 0$, we have that the trace part of $II$ (that is $H$) scales like $O(r^{-1})$ while the trace free part of $II$ scales like $o(r^{-1})$ (one can check this in geodesic normal coordinates); this captures the fact that every Riemannian manifold, on a sufficiently small scale, looks like Euclidean space. Hence necessarily as $r\to 0$ we have that the excess goes to 0, so the bound $c_g = 4\pi$ is sharp.

Lastly, we note that from the above discussion, if you have a strict sectional curvature bound $K < c < 0$, then minimisers of Willmore energy cannot exist, because the "point" is the actual minimiser.