Definition of Willmore energy 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?
 A: 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.
A: 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 $v(M)$).
