An identity related to generalized divergence theorem

I am trying to verify one proposition proved in Reilly’s paper. For the notation of this, please consult the paper.

Propsotion 2.4 Let {{D}} be a domain in {\mathbb{R}^n}. {f} is a smooth function on {{D}}. Then

\displaystyle \int_{{D}}(q+1)S_{q+1}(f)dx_1\cdots dx_m=\int_{\partial D}\left(\tilde S_q(z)z_n-\sum_{\alpha\beta}\tilde T_{q-1}(z)^{\alpha\beta}z_{\alpha}z_{n,\beta}\right)dA

Proof: Take an orthonormal frame field {\{e_\alpha,e_n\}} such that {\{e_\alpha\}} is tangent to {\partial D}. Notice

\displaystyle D^2(f)(X,Y)=X(Yf)-(\nabla_{X}Y)f

\displaystyle D^2(f)=\left(\begin{matrix} z_{,\alpha\beta}-z_nA_{\alpha\beta} & z_{n,\alpha}\\ z_{n,\beta} & f_{nn} \end{matrix} \right)

where {f_{nn}=D^2(f)(e_n,e_n)}. It follows from Remark 2.3 that

\displaystyle \int_{D}(q+1)S_{q+1}(f)dx_1\cdots dx_m=\int_{\partial D}\sum_{i,j}T_q(f)^{ij}f_jt_idA

where {t=(t_1,\cdots,t_n)} is the outward unit normal to {\partial D}. Changing the coordinates to {e_\alpha} and {e_n}, we can get

\displaystyle \int_{\partial D}\sum_{i,j}T_q(f)^{ij}f_jt_idA=\int_{\partial D}T_q(f)^{\alpha n}z_{\alpha}+T_{q}(f)^{nn}z_ndA

It is easy to see

\displaystyle T_q(f)^{nn}=\tilde S_{q}(z)


\displaystyle T_q(f)^{\alpha n}z_\alpha=\sum\delta\binom{i_1,i_2\cdots,n,\alpha}{j_1,j_2\cdots,\beta, n}z_\alpha

\displaystyle =-\sum\delta\binom{\alpha_1,\alpha_2\cdots,\alpha}{\beta_1,\beta_2\cdots,\beta}f_{n\beta}z_\alpha=-\tilde T_{q-1}(z)^{\alpha\beta}z_\alpha z_{n,\beta}

Therefore the proposition is established.

Remark: Robert Reilly, On the hessian of a function and the curvature of its graph

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: