One way to state Gauss's theorema egregium is as follows.
Theorema Egregium. Suppose $(M,g)$ is a $2$-dimensional Riemannian submanifold of $\Bbb{R}^3$. For every $p\in M$, the Gaussian curvature $K$ of $M$ at $p$ is equal to one-half the scalar curvature $R_g$ of $g$ at $p$; thus, the Gaussian curvature is a local isometry invariant of $(M,g)$.
I'm wondering if the theorem still holds with $\Bbb{R}^3$ replaced by an abstract Riemannian $3$-manifold $(N,\bar{g})$. I asked this question because I don't know why
$$R_g=g^{ij}R_{ij}=K g^{ij}g_{ij}=2K$$
has anything to do with $\Bbb{R}^3$. Here $R_{ij}$ is the local expression of the Ricci tensor.
Does anyone have an idea? Thank you.
Edit. I grabbed a proof of this theorem. It seems to exploit flatness of $\Bbb{R}^3$, which prohibits our freedom to replace the Euclidean space with a general Riemannian manifold. But then I began to wonder why $R_g=2K$ still holds for a general ambient manifold. Maybe the point is that we have to show that Gaussian curvature is in fact an intrinsic property.
Best Answer
Let’s go general to special. Suppose $(M,g)$ is a Riemannian submanifold of $(\widetilde{M},\widetilde{g})$, and let $\alpha:TM\oplus TM\to (TM)^{\perp}$ be the (vector) second fundamental form. Then, for each point $p\in M$ and linearly independent tangent vectors $x,y\in T_pM$, Gauss’ equation tells us \begin{align} A_p(x,y)&=\widetilde{A}_p(x,y)+\frac{\langle\alpha(x,x),\alpha(y,y)\rangle-\|\alpha(x,y)\|^2}{\|x\|^2\|y\|^2-\langle x,y\rangle^2}, \end{align} where $A_p(x,y)$ is the sectional curvature of the tangent plane in $T_pM$ spanned by $x,y$ as measured in $(M,g)$, and $\widetilde{A}_p(x,y)$ is the sectional curvature measured by $(\widetilde{M},\widetilde{g})$. So, this equation tells you that the sectional curvature of the submanifold (which is intrinsic to the submanifold) equals the sectional curvature of the ambient manifold (intrinsic to the ambient manifold) + some stuff involving the second fundamental form (i.e extrinsic to the submanifold).
Now, suppose $\dim \widetilde{M}=n\geq 3$ and $M$ is a hypersurface. Fixing a unit normal $\nu$ for $T_pM$, we have that the shape operator/Weingarten map $S_{\nu}: T_pM\to T_pM$ is related to the second fundamental form $\alpha$ by $\alpha(x,y)=\langle S_{\nu}(x),y\rangle\nu$. With this, the above equation becomes \begin{align} A_p(x,y)&=\widetilde{A}_p(x,y)+\frac{\langle S_{\nu}(x),x\rangle\langle S_{\nu}(y),y\rangle-\langle S_{\nu}(x),y\rangle^2}{\|x\|^2\|y\|^2-\langle x,y\rangle^2}, \end{align} Symmetry of $\alpha$ implies $S_{\nu}$ is self-adjoint, so by the spectral theorem, we can find an orthonormal basis of eigenvectors $\{e_1,\dots, e_{n-1}\}$ of $S_{\nu}$ for $T_pM$, say with corresponding eigenvalues $\lambda_1,\dots,\lambda_{n-1}$. Then, for each $1\leq i<j\leq n-1$, we have that \begin{align} A_p(e_i,e_j)&=\widetilde{A}_p(e_i,e_j)+\lambda_i\lambda_j, \end{align} or rearranging, $\lambda_i\lambda_j=A_p(e_i,e_j)-\widetilde{A}_p(e_i,e_j)$ . Now, multiply over all possible pairs $i,j$ such that $1\leq i<j\leq n$. Then, we get \begin{align} (\lambda_1\cdots\lambda_{n-1})^{n-2}&=\prod_{1\leq i<j\leq n-1}\left[A_p(e_i,e_j)-\widetilde{A}_p(e_i,e_j)\right]. \end{align} On the left this is precisely $(\det S_{\nu})^{n-2}$, and on the right, we have a bunch of products of the difference in sectional curvatures. Therefore, depending on the parity of the dimension $n$ of the ambient manifold, we can solve for $\det S_{\nu}$: \begin{align} \begin{cases} \det S_{\nu}=\left(\prod_{1\leq i<j\leq n-1}\left[A_p(e_i,e_j)-\widetilde{A}_p(e_i,e_j)\right]\right)^{\frac{1}{n-2}}&,\quad\text{if $n$ odd}\\ |\det S_{\nu}|= \left(\prod_{1\leq i<j\leq n-1}\left[A_p(e_i,e_j)-\widetilde{A}_p(e_i,e_j)\right]\right)^{\frac{1}{n-2}}&,\quad\text{if $n$ even} \end{cases} \end{align} In particular,
That this sign ambiguity must be present is obvious because the choice of normal $\nu$ is only determined up to sign, so replacing $\nu$ by $-\nu$ changes $\det S_{\nu}$ to $\det S_{-\nu}=\det(-S_{\nu})=(-1)^{n-1}\det S_{\nu}$, i.e it stays the same if $n$ is odd and flips sign if $n$ is even.
Since $\det S_{\nu}$ is originally how Gauss defined his curvature, we get the result that the Gaussian curvature is intrinsic to the submanifold. In particular, specializing to $n=3$ with flat ambient space, we see that there is only one term on the right: \begin{align} \det S_{\nu}=A_p(e_1,e_2)\equiv A_p,\tag{$*$} \end{align} where $A_p\in\Bbb{R}$ denotes the sectional curvature of the plane $T_pM$ (there is only one since $M$ is 2-dimensional).
A completely separate result is the following: for any $(n-1)$-dimensional Riemannian manifold $(M,g)$, the scalar curvature of $(M,g)$ is equal to the sum of the distinct sectional curvatures in any orthonormal basis $\{e_1,\dots, e_{n-1}\}$ for $T_pM$: \begin{align} R_p&=2\sum_{1\leq i<j\leq n-1}A_p(e_i,e_j). \end{align} This follows just by definition of the various things, and using orthonormality. In particular, specializing to $n-1=2$-dimensional manifold $M$, there is only one term on the right, so \begin{align} R_p&=2A_p.\tag{$**$} \end{align}
So, if you now combine the two situations $(*)$ and $(**)$, we get that for a Riemannian 2-manifold $M$ embedded in a flat 3-dimensional space, \begin{align} \det S_{\nu}&=A_p=\frac{R_p}{2}.\tag{$***$} \end{align} It is this equation, $(***)$, which motivates the abstract intrinsic definition of Gaussian curvature, $K$, of a Riemannian 2-manifold $(M,g)$: we define $K:=\frac{R}{2}$, where $R$ is the scalar curvature. So, with this definition, the Gaussian curvature of a Riemannian 2-manifold coincides with the single sectional curvature $A$ of $M$, and when embedded into $\Bbb{R}^3$ it coincides with $\det S_{\nu}$, the product of the principal curvatures. So, I think you’re having an issue with what’s a definition vs what’s being proved.