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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 $c_g$?

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Incidentally, the argument below should also give you a proof that there are no genus zero closed minimal surfaces in a Hadamard manifold which does not use the maximum principle. –  Willie Wong Oct 3 '12 at 8:57

1 Answer 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.

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A slightly more interesting question to ask if for the case where $\Sigma$ has non-trivial genus, or higher dimensions. Considering all the efforts gone into proving the Willmore conjecture, I suspect that these other numbers are not known in general. –  Willie Wong Oct 3 '12 at 8:55
    
Thanks for the answer. I think you'll find that the mean curvature scales like $r^{-1}$; this is easy to remember from the scale invariance of the Willmore energy in Euclidean space: the measure scales like $r^2$ and so the integrand $H^2$ must scale like $r^{-2}$. Or just from thinking about the prinicipal curvatures, which must scale like $r^{-1}$. –  Glen Wheeler Oct 5 '12 at 10:51
    
I don't understand the application of G-B here. I'm probably being dense, but I thought we had $\int K = 4\pi$ and not $\int S = 8\pi$. (The $S$ is the scalar curvature of the background space, and should be negative, right?) Perhaps you could add a bit more detail there for the slowpokes like me. –  Glen Wheeler Oct 5 '12 at 10:57
    
$S$ is the induced scalar curvature on the surface (as I wrote in paragraph 2); just to clarify, that means the scalar curvature of the induced metric on the surface. So $S$ is equal to twice the Gauss curvature (I avoid the letter $K$ since I use it for sectional curvature). Oh, and thanks for the correction: I meant that $H^2$ scales like $r^{-2}$ but mistyped. –  Willie Wong Oct 5 '12 at 12:39
    
To further clarify, for the trace of the Gauss equation, I am taking the double trace with respect to the induced metric on the surface (which I guess I denoted as $g_{\Sigma}$ in one of the expressions), hence the left hand side we are left with the (background) sectional curvature of the tangent plane to $\Sigma$ and not the scalar curvature of the background. –  Willie Wong Oct 5 '12 at 12:42

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