Let $(\widetilde{M},\widetilde{g})$ be a Lorentz $(n+1)$-manifold and let $(M,g)$ be a Riemannian hypersurface in $\widetilde{M}$. Suppose $h$ is the scalar second fundamental form of $(M,g)$ defined by $\langle sX,Y\rangle=h(X,Y)$ for vector fields $X,Y$ on $M$ with $s$ being the shape operator. Then one of the Einstein constraint equations on $M$ states that
$$S-2\Lambda-|h|_g+(\mathrm{tr}_g h)^2=2\rho.$$
I was wondering the meaning of the symbol $|h|_g$ used by John M. Lee in his IRM book (Problem 8-20). Thank you.
According to the "Mean curvature" part of https://en.wikipedia.org/wiki/Gauss%E2%80%93Codazzi_equations,
it looks reasonable to interpret $|h|_g$ as
$$\sqrt{\sum_{i,j}(h(E_i,E_j))^2}\tag{1}$$
with $\beta:=\{E_1,\ldots,E_n\}$ being an orthonormal frame on $M$, but does the symbol without reference to $\beta$ suggest that $|h|_g$ is independent of the orthonormal frame chosen? Thank you.
Edit 1. I know (1) is simply the Frobenius norm of the matrix of the bilinear form $h$ in $\beta$. Is this norm independent of the basis $\beta$ chosen? Is any $\beta$ going to give the same value to (1)? Thank you.
Edit 2. I'm sorry. Maybe I asked the question in a bad way. Actually, I'm not going to study Professor Lee's introduction to the Einstein constraint equation. My question is really that I came across $|h|_g$, $||h||$, or something like that a lot, I know they all probably refer to (1), but I don't know why those statements never clearly specify an orthonormal frame like our $\beta$ here. This is the very question I wanted to ask about. Thank you.
Best Answer
If we work in the basis $\beta$ and use matrix notation (to avoid needing a zillion different indices), then the quantity being squared is $A_{lj} = (Q^T h Q)_{lj};$ so the squared norm is $$\sum_{lj} A_{lj} A_{lj} = \mathrm{tr} \left( A^T A \right ) = \mathrm{tr} \left( Q^Th^TQQ^ThQ.\right)$$
Since $Q$ is orthogonal we know $QQ^T$ is the identity. Combining this with the cyclic permutation symmetry of $\mathrm{tr}$, we arrive at $$\mathrm{tr}(A^T A) = \mathrm{tr}(h^TQQ^ThQQ^T) = \mathrm{tr}(h^Th) = \sum_{ij} h_{ij} h_{ij}$$ as desired.