Calculation of traceless second fundamental form

Question

The traceless part of second fundamental form is $$ \mathring A = A -\frac{H}{n}g $$ where $A$ is second fundamental form, $H$ is mean curvature, $g$ is metric. The norm square is $$ |\mathring A|^2 = |A|^2-\frac{1}{n}H^2. $$ I want to verify it. What I do: \begin{align} |\mathring A|^2 &=g^{ij}g^{kl}\mathring A_{ik}\mathring A_{jl} \\ &=g^{ij}g^{kl}(A_{ik} -\frac{H}{n}g_{ik})(A_{jl} -\frac{H}{n}g_{jl}) \\ &=g^{ij}g^{kl} (A_{ik}A_{jl}-\frac{H}{n}g_{ik}A_{jl} -\frac{H}{n}g_{jl}A_{ik} +(\frac{H}{n})^2g_{ik}g_{jl}) \\ &=|A|^2 -\frac{2H}{n}g^{ij}A_{ij} + (\frac{H}{n})^2\delta_k^j\delta_j^k \end{align} Then, I don't know how to deal it. I can't think though why $\delta_k^j\delta_j^k =n$ and what is $g^{ij}A_{ij}$.

Answer 1

$H$ is by definition the trace of $A$ with respect to $g$, i.e. $g^{ij} A_{ij}.$ The contraction $\delta^j_k \delta^k_j$ is equal to $n$ because it reduces to $\delta^j_j,$ which is just the trace of the $n \times n$ identity matrix.