# What does $|h|_g$ even mean if $h$ is the scalar second fundamental form of a Riemannian hypersurface $(M,g)$?

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.

• 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 $A_{ij} A_{kj} = \delta_{ik}$. Apr 11, 2023 at 8:38
• @ArcticChar Thank you for your suggestion. I'm working on it.
– Boar
Apr 11, 2023 at 13:26

## 2 Answers

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.

• One word---brilliant.
– Boar
Apr 14, 2023 at 15:20

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.