Ngô Quốc Anh

October 11, 2013

Bochner-type formula for the conformal Killing operator on manifolds without boundary

Filed under: Uncategorized — Tags: , — Ngô Quốc Anh @ 9:59

Assuming the Riemannian manifold (M,g) is compact without boundary. In the previous post, we showed that

\displaystyle \frac{1}{2}\int_M |\mathbb L_X g|^2 dv_g= \int_M |\nabla X|^2 dv_g +\int_M |{\rm div}X|^2 dv_g - \int_M {\rm Ric}(X,X)dv_g .

Today, we use the above formula to derive a Bochner-type formula for the conformal Killing operator \mathbb L given by

\displaystyle \mathbb L X = \mathbb L_Xg - \frac{2}{n}\text{div}(X)g.

Clearly, for any vector field X,

\begin{array}{lcl} |\mathbb{L}X{|^2} &=& \displaystyle \Big|{\mathbb{L}_X}g - \frac{2}{n}( \text{div}X)g \Big|^2 \hfill \\ &=&\displaystyle {\left( {{\mathbb{L}_X}g - \frac{2}{n}( \text{div}X)g} \right)_{im}}{\left( {{\mathbb{L}_X}g - \frac{2}{n}( \text{div}X)g} \right)_{jn}}{g^{ij}}{g^{mn}} \hfill \\ &=& \displaystyle |{\mathbb{L}_X}g{|^2} + \frac{4}{{{n^2}}}{( \text{div}X)^2}\underbrace {{g_{im}}{g_{jn}}{g^{ij}}{g^{mn}}}_{\delta _n^i\delta _i^n = n} - \frac{2}{n}( \text{div}X)\left[ {{g_{im}}{\mathbb{L}_X}{g_{jn}} + {g_{jn}}{\mathbb{L}_X}{g_{im}}} \right]{g^{ij}}{g^{mn}} \hfill \\ &=& \displaystyle |{\mathbb{L}_X}g{|^2} + \frac{4}{n}{( \text{div}X)^2} - \frac{2}{n}( \text{div}X)[({\mathbb{L}_X}{g_{jn}})\underbrace {{g_{im}}{g^{ij}}{g^{mn}}}_{\delta _m^j{g^{mn}} = {g^{jn}}} + ({\mathbb{L}_X}{g_{im}})\underbrace {{g_{jn}}{g^{ij}}{g^{mn}}}_{\delta _n^i{g^{mn}} = {g^{im}}}] \hfill \\ &=&\displaystyle |{\mathbb{L}_X}g{|^2} + \frac{4}{n}{( \text{div}X)^2} - \frac{2}{n}( \text{div}X)[({\nabla _j}{X_n}+\nabla_n X_j){g^{jn}} + ({\nabla _i}{X_m}+\nabla_m X_i){g^{im}}] \hfill \\ &=& \displaystyle |{\mathbb{L}_X}g{|^2} + \frac{4}{n}{( \text{div}X)^2} - \frac{2}{n}( \text{div}X)[\underbrace {2{\nabla ^n}{X_n} + 2{\nabla ^m}{X_m}}_{4 \text{div}X}] \hfill \\ &=& \displaystyle |{\mathbb{L}_X}g{|^2} - \frac{4}{n}{( \text{div}X)^2}. \end{array}

Hence, we eventually have

\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 .

Note that the identity

\displaystyle |\mathbb{L}X|^2=|{\mathbb{L}_X}g{|^2} - \frac{4}{n}{( \text{div}X)^2}

is independent of whether the manifold (M,g) has boundary or not. It is interesting to mention that in a book by K. Yano, he comes up with a slightly different formula for

\displaystyle \frac{1}{2}\int_M |\mathbb L_X g|^2 dv_g= \int_M |\nabla X|^2 dv_g +\int_M |{\rm div}X|^2 dv_g - \int_M {\rm Ric}(X,X)dv_g .

In fact, using

\displaystyle \int_M \Delta_g (|X|^2)dv_g =0

for any vector field X, we obtain by direct calculation

\begin{array}{lcl} \displaystyle\int_M {{\Delta _g}(|X{|^2})d{v_g}} &=&\displaystyle \int_M {{g^{ij}}{\nabla _i}{\nabla _j}({X_p}{X_q}{g^{pq}})d{v_g}} \hfill \\ &=&\displaystyle \int_M {{g^{ij}}[{\nabla _i}({X_q}{\nabla _j}{X_p} + {X_p}{\nabla _j}{X_q})){g^{pq}}]d{v_g}} \hfill \\ &=&\displaystyle \int_M {{g^{ij}}[({\nabla _i}{X_q}{\nabla _j}{X_p} + {X_q}{\nabla _i}{\nabla _j}{X_p}) + ({\nabla _i}{X_p}{\nabla _j}{X_q} + {X_p}{\nabla _i}{\nabla _j}{X_q})){g^{pq}}]d{v_g}} \hfill \\ &=&\displaystyle \int_M {{g^{ij}}({X_q}{\nabla _i}{\nabla _j}{X_p} + {X_p}{\nabla _i}{\nabla _j}{X_q}){g^{pq}} + {g^{ij}}({\nabla _i}{X_q}{\nabla _j}{X_p} + {\nabla _i}{X_p}{\nabla _j}{X_q}){g^{pq}}d{v_g}} \hfill \\ &=&\displaystyle 2\int_M {{g^{ij}}{\nabla _i}{\nabla _j}{X^h}{X_h} + {g^{ij}}{\nabla _i}{X_q}{\nabla _j}{X_p}{g^{pq}}d{v_g}} \hfill \\ &=&\displaystyle 2\int_M {({g^{ij}}{\nabla _i}{\nabla _j}{X^h}){X_h} + |\nabla X{|^2}d{v_g}} \end{array}

which implies that

\displaystyle - \int_M {{g^{ij}}{\nabla _i}{\nabla _j}{X^h}{X_h}d{v_g}} = \int_M {|\nabla X{|^2}d{v_g}}.

It is important to note that our long calculation above plays nothing with the integrals. Thus, we have shown that

\displaystyle \frac{1}{2}\int_M |\mathbb L_X g|^2 dv_g=- \int_M {{g^{ij}}{\nabla _i}{\nabla _j}{X^h}{X_h} - {\rm Ric}(X,X)d{v_g}} +\int_M |{\rm div}X|^2 dv_g.

Yano then defines

\displaystyle \square X := -( {({g^{ij}}{\nabla _i}{\nabla _j}{X^h})} + {\rm Rm}_i^hX^i )

which then helps him to rewrite the identity as follows

\displaystyle \frac{1}{2}\int_M |\mathbb L_X g|^2 dv_g= \int_M g(\square X, X) d{v_g} +\int_M |{\rm div}X|^2 dv_g.

Hence, we obtain the following

\displaystyle \frac{1}{2}\int_M |\mathbb L_X g|^2 dv_g= \int_M g(\square X, X) d{v_g} + \left( 1-\frac{2}{n}\right)\int_M |{\rm div}X|^2 dv_g.

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