Ngô Quốc Anh

November 16, 2010

The Kazdan-Warner identity

Filed under: Uncategorized — Ngô Quốc Anh @ 22:52

In this short note, we shall discuss a very beautiful identity named after Jerry L. Kazdan and F. W. Warner published in the Ann. of Math. (2) long time ago [here].

To be precise, let us consider the following partial differential equation

\displaystyle \Delta u + h e^{2u}=c

over a 2-sphere \mathbb S^2 where h is a function and c is a constant. We shall prove the following

Theorem (Kazdan-Warner). It holds

\displaystyle \int_{{\mathbb{S}^2}} {\left\langle {\nabla h,\nabla {x_j}} \right\rangle {e^{2u}}d{v_g}} = 2(1 - c)\int_{{\mathbb{S}^2}} {h{x_j}{e^{2u}}d{v_g}} , \quad j = 1,2,3,

where x_j are coordinates.

Proof.

For any j we first multiple both sides of PDE with \left\langle {\nabla u,\nabla {x_j}} \right\rangle and then integrate over the sphere to arrive at

\displaystyle\begin{gathered} \int_{{\mathbb{S}^2}} {\left\langle {\nabla u,\nabla {x_j}} \right\rangle (\Delta u + h{e^{2u}})d{v_g}} \hfill \\ \qquad= c\int_{{\mathbb{S}^2}} {\left\langle {\nabla u,\nabla {x_j}} \right\rangle d{v_g}} \hfill \\ \qquad= c\int_{{\mathbb{S}^2}} {( - \Delta u){x_j}d{v_g}} \hfill \\\qquad = c\int_{{\mathbb{S}^2}} {(h{e^{2u}} - c){x_j}d{v_g}}. \hfill \\ \end{gathered}

On the other hand,

\displaystyle\begin{gathered} \int_{{\mathbb{S}^2}} {\left\langle {\nabla u,\nabla {x_j}} \right\rangle h{e^{2u}}d{v_g}} \hfill \\ \qquad= \frac{1}{2}\int_{{\mathbb{S}^2}} {\left\langle {\nabla ({e^{2u}}),h\nabla {x_j}} \right\rangle d{v_g}} \hfill \\ \qquad= - \frac{1}{2}\int_{{\mathbb{S}^2}} {{e^{2u}}\nabla (h\nabla {x_j})d{v_g}} \hfill \\ \qquad= - \frac{1}{2}\int_{{\mathbb{S}^2}} {{e^{2u}}\left[ {\left\langle {\nabla h,\nabla {x_j}} \right\rangle - h\Delta {x_j}} \right]d{v_g}} \hfill \\ \qquad= - \frac{1}{2}\int_{{\mathbb{S}^2}} {{e^{2u}}\left[ {\left\langle {\nabla h,\nabla {x_j}} \right\rangle - 2h{x_j}} \right]d{v_g}} \hfill \\ \qquad= - \frac{1}{2}\int_{{\mathbb{S}^2}} {{e^{2u}}\left\langle {\nabla h,\nabla {x_j}} \right\rangle d{v_g}} + \int_{{\mathbb{S}^2}} {h{x_j}{e^{2u}}d{v_g}} . \hfill \\ \end{gathered}

Similarly,

\displaystyle\begin{gathered} \int_{{\mathbb{S}^2}} {\left\langle {\nabla u,\nabla {x_j}} \right\rangle \Delta ud{v_g}} \hfill \\ \qquad= - \int_{{\mathbb{S}^2}} {\nabla \left\langle {\nabla u,\nabla {x_j}} \right\rangle \nabla ud{v_g}} \hfill \\ \qquad= - \int_{{\mathbb{S}^2}} {\left[ {\left\langle {\nabla (\nabla u),\nabla {x_j}} \right\rangle + \left\langle {\nabla u,\nabla (\nabla {x_j})} \right\rangle } \right]\nabla ud{v_g}} \hfill \\ \qquad= - \int_{{\mathbb{S}^2}} {\left[ {\left\langle {\nabla (\nabla u),\nabla {x_j}} \right\rangle - \left\langle {\nabla u,{x_j}} \right\rangle } \right]\nabla ud{v_g}} \hfill \\ \qquad= - \frac{1}{2}\int_{{\mathbb{S}^2}} {\left\langle {\nabla |\nabla u{|^2},\nabla {x_j}} \right\rangle d{v_g}} + \int_{{\mathbb{S}^2}} {{x_j}\left\langle {\nabla u,\nabla u} \right\rangle d{v_g}} \hfill \\ \qquad= \frac{1}{2}\int_{{\mathbb{S}^2}} {|\nabla u{|^2}\Delta {x_j}d{v_g}} + \int_{{\mathbb{S}^2}} {{x_j}|\nabla u{|^2}d{v_g}} \hfill \\ \qquad= 0. \hfill \\ \end{gathered}

Then

\displaystyle - \frac{1}{2}\int_{{\mathbb{S}^2}} {{e^{2u}}\left\langle {\nabla h,\nabla {x_j}} \right\rangle d{v_g}} + \int_{{\mathbb{S}^2}} {h{x_j}{e^{2u}}d{v_g}} = c\int_{{\mathbb{S}^2}} {(h{e^{2u}} - c){x_j}d{v_g}} = c\int_{{\mathbb{S}^2}} {h{x_j}{e^{2u}}d{v_g}}.

Thus

\displaystyle\int_{{\mathbb{S}^2}} {{e^{2u}}\left\langle {\nabla h,\nabla {x_j}} \right\rangle d{v_g}} = 2(1 - c)\int_{{\mathbb{S}^2}} {h{x_j}{e^{2u}}d{v_g}} .

The proof follows.

There are lots of variations and generalization in the literature so far. We are going to discuss later.

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