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The MAA has posted to its facebook page a link to an article about a recent proposed proof of what is called the Willmore conjecture, after Thomas Willmore.

Wikipedia's article titled Wilmore conjecture includes the following:

Let $v:M\to\mathbb{R}^3$ be a smooth immersion of a compact, orientable surface (of dimension two). Giving $M$ the Riemannian metric induced by $v$, let $H:M\to\mathbb{R}$ be the mean curvature (the arithmetic mean of the principal curvatures $\kappa_1$ and $\kappa_2$ at each point). Let $K$ be the Gaussian curvature. In this notation, the Willmore energy $W(M)$ of $M$ is given by $$W(M) = \int_S H^2 \, dA - \int_S K \, dA.$$ In the case of the torus, the second integral above is zero.

A little bit of this came from editing by me within the past hour.

Knowing very little of differential geometry, I hesitate to do much more with this paragraph before clarifying some things. It seems $M$ is a particular parametrization of the surface, but the integrals look like things that should not depend on which suitably well-behaved parametrization is chosen. Yet the definition seems to attribute the Willmore energy to the parametrization $M$, rather than to the surface, which might be parametrized in any of many different ways. Notice the use of the capital letter $S$ in the expression $\displaystyle\int_S$, when nothing called $S$ was defined! Presumably $S$ means the image of $M$.

Ought one to write $W(S)$ instead of $W(M)$, to be clear about a lack of dependence on a choice of parametrization?

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The key is the line "Giving $M$ the Riemannian metric induced by $\nu$." This means to put on $M$ the pullback metric $\nu^*\delta$, where $\delta$ is the euclidean metric. Thus $M$ is not a parametrization of a manifold, but the $2$-manifold itself. This allows us to define the mean curvature $H$ (an extrinsic quantity) on $M$. Mean curvature is usually thought of as being defined on the parametrization $\nu(M)$ however you can define it on $M$ by just pulling back, and this also avoids the issue of the image $\nu(M)$ possibly having self-intersections ($H$ may not be well-defined on $\nu(M)$ but it is well defined on $M$). Then the integrals are appropriately defined over $M$ (instead of $S$ - looks like a typo there). Thus the Willmore energy is the difference of an extrinsic (mean curvature) and intrinsic (Gauss curvature) quantity.

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The article had $v$ rather than $\nu$, but after reading this answer, I've changed it to $\nu$. Alright, so $\nu$ is an immersion. Might the mean curvature, and hence the Willmore energy, depend on which immersion it is? If not, then why not state the definition without speaking of an immersion at all? –  Michael Hardy Jun 2 '12 at 1:53
    
Oh, I only used $\nu$ because I thought that was the character you were using in your question (they look very similar on my browser). Yes, the mean curvature depends on the immersion. For instance, a sheet of notebook paper sitting flat on a table has zero mean curvature as an immersion in $3$-space, but if you fold this sheet of notebook paper into a cylinder, you have made an immersion with nonzero mean curvature (but the gauss curvature is still zero). So we say that the bending of the notebook paper in this case is extrinsic. –  treble Jun 2 '12 at 1:59
    
Actually on the browser I'm using, although I can tell a $v$ from a $\nu$ when it's set in $\TeX$, I'm not sure I can tell the difference in the inline non-$\TeX$ notation used at that point in the Wikipedia article, but I can tell the difference when I click on "edit" and look at it then. It appears from your example of the cylinder that the Willmore energy, which is an integral over the whole surface of a certain function of the curvatures, will also depend on which immersion it is. Next there's the question of whether that's.... –  Michael Hardy Jun 2 '12 at 2:34
    
....also true of the torus. Could a torus, with its intrinsic metric, be immersed not just in ways that make the mean curvature different, but also make the integral of the square of the mean curvature over the whole surface different? I'm guessing probably yes, since the statement that that integral would not change if you change the immersion seems pretty strong. –  Michael Hardy Jun 2 '12 at 2:36
    
I've now changed it back to $v$, and split the section on statement of the conjecture into two sections, and done a bit of formatting, in particular putting the actual statement in an indented "displayed" setting. –  Michael Hardy Jun 2 '12 at 2:44

The principal curvatures of $v(M)$ are quantities which do not depend on the parametrization, but only on the geometry of the image of the immersion with respect to the surrounding manifold.

For this reason you are right, the notation for the Willmore energy should reflect this fact and has been poorly chosen in your example. The notation $W(M)$, or even $W(M, g)$ does not reflect this, a metric on $M$ can be defined w/o having an immersion into some target manifold. In this case the principal curvatures are not even defined. The metric on $M$ alone obtained by pulling back the metric of the target space does not allow to define this, only if you know the normal to the image you can define principal curvatures (they are basically the eigenvalues of the derivative of the normal or, without being too specific, the second derivative of $v$, the mean curvature is the sum of them (trace of the Weingarten map) and the Gauss curvature their product (determinant of the Weingarten map)).

(As an off topic side remark: The fact that the integral over $K $ vanishes is a consequence of the Gauss- Bonnet Theorem which relates this integral to the Euler characteristic of $v(M)$).

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"the notation for the Willmore energy should reflect this fact and has been poorly chosen in your example." So what notation would you use instead? –  Michael Hardy Jun 2 '12 at 19:37
    
@MichaelHardy This depends on the purpose you are aiming at. It makes perfect sense to define $W(i)$ for an immersion $i: M\rightarrow \mathbb{R}^3$. You have to be aware then, that this this is invariant under the group of diffeomorphisms $\Phi: M \rightarrow M$, i.e. $W(i) = W(i\circ \Phi)$. For the sake of the calculus of variation this is not very pleasant, since you don't have any kind of compactness for minimizing sequences -- the group of diffeomorphims is too large. For the wikipedia article you were citing this might be the best option, nonetheless. Does this answer your question? –  user20266 Jun 2 '12 at 19:50
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Maybe.... Of course the "purpose" in the present case was to make the Wikipedia article as good as it can be. –  Michael Hardy Jun 3 '12 at 3:48

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