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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.

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    $\begingroup$ It's true and I suggest you to try to check it. Say if there is another o.n. basis $e_j$'s, then $e_j = A_{ij} E_i$ and $A = (A_{ij})$ is orthogonal, so $A_{ij} A_{kj} = \delta_{ik}$. $\endgroup$ Apr 11, 2023 at 8:38
  • $\begingroup$ @ArcticChar Thank you for your suggestion. I'm working on it. $\endgroup$
    – Boar
    Apr 11, 2023 at 13:26

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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.

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  • $\begingroup$ One word---brilliant. $\endgroup$
    – Boar
    Apr 14, 2023 at 15:20
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This is only a partial answer to my question. I still fail to directly handle it by assuming another orthonormal frame as suggested by @ArcticChar. If someone can finish what I fail to do at present, much will be appreciated. Thank you.

Alright, let's look at the question. We must keep in mind that $h$ is a (symmetric) $2$-tensor field on $M$. Then it makes sense to define $$|h|_g=\sqrt{\langle h,h\rangle_g},\tag{2}$$ where the inner product is described in Proposition 2.40. Now expand $h$ in the orthonormal frame $\beta$ to get $$h=h_{ij}\epsilon^i\otimes\epsilon^j,$$ where $\{\epsilon^i\}$ denotes the coframe dual to $\beta$. Then (2) gives $$|h|_g=\sqrt{h_{ij}h^{ij}}=\sqrt{\sum_{i,j}(h(E_i,E_j))^2}.$$ Independence of $\beta$ should be clear now.

Regretfully, what I really want to do is assume another orthonormal frame $\gamma:=\{F_1,\ldots,F_n\}$ and try to show that $(1)_\beta=(1)_\gamma$ with (1) denoting equation (1) in my question. Letting $F_ j=\sum_i Q_{ij}E_i$, we know the matrix $Q:=(Q_{ij})_{n\times n}$ is orthogonal, which gives $Q_{ij}Q_{ik}=\delta_{jk}$. But I don't know how to use this result to see $(1)_\beta=(1)_\gamma$. Specifically, what I have now is $$\begin{align} (1)_\gamma&=\sqrt{\sum_{\ell,j}\left(\sum_{m,k}Q_{m\ell}Q_{kj}h(E_m,E_k)\right)^2}\\ &\overset{\color{red}?}{=}(1)_\beta. \end{align}$$ I admit that I'm completely lost in those summations under the square. Could someone please tell me how to turn $Q_{ij}Q_{ik}=\delta_{jk}$ into effect? Thank you.

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