Minor technical question in diff geometry

Question

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.

Answer 1

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, $y_i$ 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.

Answer 2

The following comments are pedantic. Let $X=(X_{1},\ldots,X_{n+1})$ be the position vector of an embedded hypersurface $M^{n}$ in $\mathbb{R}^{n+1}$. Then $|\nabla X|^{2}=\sum_{i=1}^{n+1}|\nabla X_{i}|^{2}$ is indeed equal to $n$. Here $\nabla$ is the covariant derivative. Since we are acting on functions (or vector-valued ($\mathbb{R}^{n+1}$) functions), $\nabla$ is the same as the exterior derivative $d$. In local coordinates $\{x^{i}\}_{i=1} ^{n}$ on $M$, $|\nabla X|^{2}=g^{ij}\nabla_{i}X\cdot\nabla_{j}X$, where $(g^{ij})$ is the inverse matrix of $g_{ij}=\nabla_{i}X\cdot\nabla_{j} X=\frac{\partial X}{\partial x^{i}}\cdot\frac{\partial X}{\partial x^{j}}$; so we get $g^{ij}g_{ij}=n$. Alternatively, given an orthonormal frame $\{e_{i}\}_{i=1}^{n}$ at a point in $M$, $|\nabla X|^{2}=\sum_{i=1}^{n} |\nabla_{e_{i}}X|^{2}=\sum_{i=1}^{n}|e_{i}|^{2}=n$ ($\nabla_{Y}X=Y$ since $X$ is the position vector). A nice formula for the Laplacian of the square distance to the origin is $\frac{1}{2}\Delta|X|^{2}=|\nabla X|^{2} +\langle\Delta X,X\rangle=n+\langle H,X\rangle$, where $\Delta$ is the Laplacian on $M$ and where $H$ is the mean curvature vector. For example, assuming $M$ is closed and integrating yields $\int_{M}\langle H,X\rangle =-n\int_{M}1$.