Ngô Quốc Anh

October 20, 2013

Rayleigh-Type Quotient For The Conformal Killing Operator on manifolds with boundary

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 22:50

Following the previous note, today we discuss a similar Rayleigh-type quotient for the conformal Killing operator \mathbb L on manifolds (M,g) with boundary. We also prove that

whenever M admits no non-zero conformal Killing vector fields, the following holds

\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.

Since 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,

is no longer available, we use a new approach in order to estimate \int_M |\mathbb L X|^2 dv_g from below. To this purpose, we make use of a Riemannian version for the Korn inequality recently proved by S. Dain [here].

First, in view of Corollary 1.2 in Dain’s paper, the following inequality holds

\displaystyle \int_M |\nabla X|^2 dv_g \leqslant C \left( \int_M |X|^2 dv_g + \int_M |\mathbb LX|^2 dv_g \right)

for some positive constant C independent of X. This helps us to conclude that

\displaystyle C\int_M |\mathbb LX|^2 dv_g \geqslant \|X\|_{H^1}^2 - (C+1)\|X\|_{L^2}^2

as in Dahl et al’ paper. Therefore, we can argue by contradiction by assuming 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 weakly in H^1(M) and
  • X_k \to X_\infty strongly in L^2(M).

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\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{s_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{s_g}} + \int_{\partial M} {\mathbb{L}{X_\infty }(\nu ,\xi )d{s_g}}.\end{array}

We further have

\begin{array}{lcl} \displaystyle\int_M {\left\langle {{\text{div}}(\mathbb{L}\xi ),{X_\infty }} \right\rangle d{v_g}} &=&\displaystyle \mathop {\lim }\limits_{k \to + \infty } \int_M {\left\langle {{\text{div}}(\mathbb{L}\xi ),{X_k}} \right\rangle d{v_g}} \hfill \\ &=&\displaystyle - \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}} + \mathop {\lim }\limits_{k \to + \infty } \int_{\partial M} {\mathbb{L}\xi (\nu ,{X_k})d{s_g}} \hfill \\ &=&\displaystyle - \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}\xi (\nu ,{X_\infty })d{s_g}} . \end{array}

Hence, we have proved that

\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{s_g}} .

Following the argument in Dahl et al.’ paper, one can prove that the limit in the preceding equality vanishes showing that

\displaystyle \int_M {\left\langle {{\text{div}}(\mathbb{L}{X_\infty }),\xi } \right\rangle d{v_g}} = \int_{\partial M} {\mathbb{L}{X_\infty }(\nu ,\xi )d{s_g}}

for any test vector field \xi. Hence, X_\infty solves

\left\{ \begin{gathered} {\text{div}}(\mathbb{L}{X_\infty }) = 0 \quad \text{ in } M, \hfill \\ \mathbb{L}{X_\infty }(\nu , \cdot ) = 0 \quad \text{ on }\partial M, \hfill \\ \end{gathered} \right.

in the sense of distributions. Finally, to show that X_\infty is a conformal Killing vector field, we again use the divergence theorem as follows

\displaystyle \frac{1}{2}\int_M {|\mathbb{L}{X_\infty }{|^2}d{v_g}} = - \int_M {\left\langle {{\text{div}}(\mathbb{L}{X_\infty }),{X_\infty }} \right\rangle d{v_g}} + \int_{\partial M} {\mathbb{L}{X_\infty }(\nu ,{X_\infty })d{s_g}} = 0

which obviously shows that \mathbb LX_\infty =0.

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