Ngô Quốc Anh

October 13, 2013

Rayleigh-type quotient for the conformal Killing operator

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 10:05

Suppose $(M,g)$ is a compact Riemannian manifold without boundary of dimension $n \geqslant 3$. We further assume that $M$ admits no conformal Killing vector fields.

In this entry, we discuss a beautiful result due to Dahl-Gicquaud-Humbert recently published in Duke Math. J. They proved that $\displaystyle\inf \frac{{{{\left( {\int_M {|\mathbb LX|^2 d{v_g}} } \right)}^{1/2}}}}{{{{\left( {\int_M {|X|^{2n/(n - 2)}d{v_g}} } \right)}^{(n - 2)/(2n)}}}} > 0$

where the infimum is taken over all smooth vector fields $X$ on $M$ with $X \not\equiv 0$.

Their proof goes as follows: First by the compactness of $M$, ${\rm Ric} \leqslant \lambda g$ for some constant $\lambda$. We now use the the Bochner-type formula for the conformal Killing operator on manifolds without boundary, i.e., $\displaystyle \frac{1}{2}\int_M |\mathbb L X|^2 dv_g= \int_M |\nabla X|^2 dv_g + \left( 1-\frac{2}{n}\right)\int_M |{\rm div}X|^2 dv_g - \int_M {\rm Ric}(X,X)dv_g ,$

which now yields $\displaystyle \frac{1}{2}\int_M |\mathbb L X|^2 dv_g \geqslant \int_M |\nabla X|^2 dv_g - \lambda\int_M |X|^2 dv_g,$

thanks to $1-\frac{2}{n}>0$. Using the standard norm for $H^1(M)$, we rewrite the preceeding inequality as follows $\displaystyle \frac{1}{2}\int_M |\mathbb L X|^2 dv_g \geqslant \|X\|_{H^1}^2 - (\lambda+1)\|X\|_{L^2}^2.$

We now argue by contradiction. Assume that there exists a sequence of vector fields $\{X_k\}_k \in H^1(M)$ such that

• $\|X_k\|_{L^{2n/(n-2)}}=1$ and
• $\int_M |\mathbb LX_k|^2 dv_g \leqslant 2/k$.

Since $M$ is compact, it follows that $X_k$ is uniformly bounded in $H^1(M)$. Therefore, it is standard to conclude that there exists some $X_\infty \in H^1(M)$ such that

• $X_k \hookrightarrow X_\infty$ in $H^1(M)$ weakly and
• $X_k \to X_\infty$ in $L^2(M)$ strongly.

Consequently, $\|X_\infty\|_{L^2} = 1$. In particular, $X_\infty \not\equiv 0$. For any smooth vector field $\xi$, using the divergence theorem, we obtain $\begin{array}{lcl} \displaystyle\frac{1}{2}\left| \int_M {\left\langle {\text{div}}(\mathbb LX_\infty ),\xi \right\rangle dv_g} \right|&=&\displaystyle\frac{1}{2}\left| {\int_M {\left\langle {{\text{div}}(\mathbb{L}\xi ),{X_\infty }} \right\rangle d{v_g}} } \right| \\&=&\displaystyle \frac{1}{2}\mathop {\lim }\limits_{k \to + \infty } \left| {\int_M {\left\langle {{\text{div}}(\mathbb{L}\xi ),{X_k}} \right\rangle d{v_g}} } \right| \hfill \\ &=&\displaystyle \frac{1}{2}\mathop {\lim }\limits_{k \to + \infty } \left| {\int_M {\left\langle {\mathbb{L}\xi ,\mathbb{L}{X_k}} \right\rangle d{v_g}} } \right| \hfill \\ &\leqslant&\displaystyle \mathop {\lim }\limits_{k \to + \infty } {\left( {\int_M {|\mathbb{L}\xi {|^2}d{v_g}} } \right)^{1/2}}{\left( {\int_M {|\mathbb{L}{X_k}{|^2}d{v_g}} } \right)^{1/2}} \hfill \\ &=& 0.\end{array}$

Hence $X_\infty$ solves the equation $\text{div}(\mathbb L X_\infty)=0$ in the sense of distributions. In particular, $X_\infty$ is a nonzero conformall Killing vector fields, (see this topic for an argument). This is a contradiction.

When the manifold $M$ has boundary $\partial M$, the situation is now complicated since $\begin{array}{lcl}\displaystyle\frac{1}{2}\int_M |\mathbb LX |^2 dv_g -\int_{\partial M} (\mathbb LX)(\nu ,X) d\sigma_g=&-&\displaystyle\int_M (\Delta_g X^h){X_h} dv_g+ \left(1-\frac{2}{n}\right) \int_M {|\text{div} X|^2d{v_g}}\\ &-&\displaystyle \int_M {\rm Ric}(X,X) dv_g-\left( 1 - \frac{2}{n} \right) \int_{\partial M} {\text{div} X{{\left\langle {\nu ,X} \right\rangle } }d{\sigma _g}}\end{array}$

and $\begin{array}{lcl} \displaystyle\int_M {\left\langle {{\text{div}}(\mathbb{L}{X_\infty }),\xi } \right\rangle d{v_g}} &=&\displaystyle - \frac{1}{2}\int_M {\left\langle {\mathbb{L}\xi ,\mathbb{L}{X_\infty }} \right\rangle d{v_g}} + \int_{\partial M} {\mathbb{L}{X_\infty }(\nu ,\xi )d{v_g}} \hfill \\ &=&\displaystyle \int_M {\left\langle {{\text{div}}(\mathbb{L}\xi ),{X_\infty }} \right\rangle d{v_g}} - \int_{\partial M} {\mathbb{L}\xi (\nu ,{X_\infty })d{v_g}} + \int_{\partial M} {\mathbb{L}{X_\infty }(\nu ,\xi )d{v_g}}\end{array}$

which eventually implies $\displaystyle\int_M {\left\langle {{\text{div}}(\mathbb{L}{X_\infty }),\xi } \right\rangle d{v_g}} = - \frac{1}{2}\mathop {\lim }\limits_{k \to + \infty } \int_M {\left\langle {\mathbb{L}\xi ,\mathbb{L}{X_k}} \right\rangle d{v_g}} + \int_{\partial M} {\mathbb{L}{X_\infty }(\nu ,\xi )d{v_g}} .$

We shall come back to this later.