I am perplexed by the following statement in the paper "A New Conformal Invariant and Its Applications to the Willmore Conjecture and the First Eigenvalue of Compact Surfaces" by Li and Yau found here <http://www.doctoryau.com/papers/NewConformalInvariant.pdf>


On page 275, in the derivation of (2.7) it seems that the following claim is made.
Let $X\_i$ be the coordinates on $\mathbb R^{n+1}$ and let $S\_n$ be the unit $n$-sphere.
Then $\sum\_{i=1}^{n+1} \vert \nabla X\_i\vert\,^2 = 2$ for the functions $X\_i$ viewed as functions on the sphere. I am getting $n$ rather than $2$ for this calculation, but it is possible that I either misinterpreted the statement or did the calculation wrong. 


Is anyone here familiar with this paper or is bored enough to have a quick look?


For what it's worth, my interest is guided by <http://arxiv.org/abs/alg-geom/9609012> "A linear lower bound on the gonality of modular curves" by Dan Abramovic.



Minor technical question in diff geometry
The notation is potentially confusing, but the end result is correct. 
Essentially this claim is the relationship between the area and the energy of of a conformal map. Here is a slightly expanded proof of their computation:


Let $X\_i:\mathbb{R}^{n+1}\to\mathbb{R}$ denote the $i$-th coordinate function.


Suppose that $\phi:M^2\to \mathbb{S}^n$ is a conformal map from a *surface*. Then, they argue that there is a new conformal map $F:=g\circ \phi:M\to \mathbb{S}^n$ which balances the $X\_i$ in the sense that
$$
\int\_M X\_i\circ F = 0
$$
for each $i$. Thus, $X\_i\circ F$ is an *acceptable test function* for the variational characterization of the first eigenfunction.


Choose [isothermal coordinates](http://en.wikipedia.org/wiki/Isothermal_coordinates), $y\_i$ near some point $p\in M$. That is, $g = \lambda^2 \delta\_{\mathbb{R}^2}$. Because $F$ is conformal and because we have chosen isothermal coordinates, we see that the vectors in $\mathbb{R}^{n+1}$
$$
F\_\*\left(\frac{\partial}{\partial y\_i}\right)
$$
for $i=1,2$ are orthogonal and of the same length $\lambda^{-1}$ (i.e. not necessarily unit vectors). Then, we may compute the area of the image of $F$ as (lets assume that the coordinates $y\_i$ cover $M$, otherwise this computation is valid locally and then must be patched together)
\begin{align\*}
\mathrm{Area}(F)&=\int\_M \Vert F\_\*(\partial/\partial y\_1)\wedge F\_\*(\partial/\partial y\_2)\Vert dy\_1\wedge dy\_2\\
& = \int\_M \sqrt{ \Vert F\_\*(\partial/\partial y\_1) \Vert^2 \Vert F\_\*(\partial/\partial y\_2) \Vert^2 - \langle F\_\*(\partial/\partial y\_1), F\_\*(\partial/\partial y\_2)\rangle} dy\_1 \wedge dy\_2\\
& = \int\_M { \Vert F\_\*(\partial/\partial y\_1) \Vert \Vert F\_\*(\partial/\partial y\_2) \Vert } dy\_1 \wedge dy\_2\\
& = \frac 12 \sum\_{i=1}^2\int\_M \Vert F\_\*(\partial/\partial y\_i) \Vert^2 dy\_1 \wedge dy\_2\\
& = \frac 12 \sum\_{i=1}^2 \sum\_{j=1}^{n+1} \int\_M\left( \frac{\partial F\_j}{\partial y\_i}\right)^2 dy\_1 \wedge dy\_2\\
& = \frac 12 \sum\_{j=1}^{n+1} \int\_M \lambda^{-2}\Vert \nabla (X\_j\circ F)\Vert^2 dy\_1 \wedge dy\_2\\
& = \frac 12 \sum\_{j=1}^{n+1} \int\_M \Vert \nabla (X\_j\circ F)\Vert^2 dV\_g\\
\end{align\*}


Let me explain these steps: The first equality is by definition, as is the second. Then, the third follows because the two vectors are orthogonal, so their inner product vanishes. Furthermore, they have the same length, so the fourth equality holds. The next is basically by definition of the Euclidean norm. Then, we change this into the $g$-gradient of $X\_j\circ F$ by definition of the gradient and by the choice of the isothermal coordinates. Finally, we use the expression for the volume form in coordinates.


