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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.

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  • $\begingroup$ How do you define the Gaussian curvature? $\endgroup$ Commented Dec 11, 2023 at 13:00
  • $\begingroup$ @ArcticChar Hello, I would define the Gaussian curvature as the determinant of the Weingarten map, also known as the shape operator. That way, I would obtain $K$ as the product of the two principal curvatures. $\endgroup$
    – Boar
    Commented Dec 11, 2023 at 14:19
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    $\begingroup$ If $M$ is a submanifold of $N$ $(e_1,e_2)$ is an orthonormal basis of $T_pM$,, then the generalized Theorem Egregium states that $$ g(e_1,R_M(e_1,e_2)e_2) = \det W + g(e_1,R_N(e_1,e_2)e_2), $$ where $W$ is the Weingarten map, $R_M$ and $R_N$ are the Riemann curvature tensors of $M$ and $N$ respectively. $\endgroup$
    – Deane
    Commented Dec 11, 2023 at 16:37
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    $\begingroup$ You are using an extrinsic definition of the Gaussian curvature that is misleading you. A good example to think through is a great $S^2\subset S^3$. It is totally geodesic, so the second fundamental form is zero. Yet its curvature is surely not $0$. $\endgroup$ Commented Dec 11, 2023 at 18:38

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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,

  • if $n\geq 3$ is odd, and the ambient space $\widetilde{M}$ is flat then its sectional curvatures all vanish, and so we have expressed $\det S_{\nu}$ completely in terms of the sectional curvatures $A$ of the submanifold (which is completely intrinsic to the submanifold), thereby proving that $\det S_{\nu}$ (which is calculated in an extrinsic fashion) actually depends only on the geometry of the submanifold. Also, reading the formula right-to-left, we see that it does not matter which orthonormal basis $\{e_1,\dots, e_{n-1}\}$ one chooses to compute the sectional curvatures; the resulting product is independent of this choice.
  • if $n\geq 4$ is even, and the ambient space is flat, then it is $|\det S_{\nu}|$ which is intrinsic to the

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.

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  • $\begingroup$ Hello, in the last paragraph where you define $K$, are you telling me that the Riemannian $2$-manifold $(M,g)$ is not necessarily a submanifold? Thank you. $\endgroup$
    – Boar
    Commented Dec 12, 2023 at 5:27
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    $\begingroup$ @Boar yes exactly. The equation $(***)$ which we proved for $M$ embedded in $\Bbb{R}^3$ is now used as motivation to define $K$ for abstract 2-dimensional $M$. $\endgroup$
    – peek-a-boo
    Commented Dec 12, 2023 at 5:30
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    $\begingroup$ @Boar yes, $g$ is defined in terms of $\widetilde{g}$, but intrinsic here means roughly ‘depends only on $g$’. This is not the same as ‘depends only on $\widetilde{g}$’. Anyway, I really don’t see what the confusion is. Once you have $(M,g)$, then forget about everything else. Don’t ask ‘where did it come from’ or anything else. No more distractions. Define everything using $(M,g)$ only and you have ‘intrinsic’ stuff. $\endgroup$
    – peek-a-boo
    Commented Dec 12, 2023 at 5:38
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    $\begingroup$ yes, it doesn’t matter how $g$ came to be. All we care about is we have a smooth manifold $M$ and a Riemannian metric $g$ on it, and all subsequent quantities/properties are being discussed relative to $g$ only. $\endgroup$
    – peek-a-boo
    Commented Dec 12, 2023 at 6:30
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    $\begingroup$ It is possible for a Riemannian manifold to be isometrically embedded into the same higher dimensional Riemannian manifold in more than one way. Intrinsic means the invariant depends on the Riemannian metric and not on the embedding. The second fundamental form is an extrinsic invariant. The theorem egregium and its generalizations identify invariants of the second fundamental form that are intrinsic. $\endgroup$
    – Deane
    Commented Dec 12, 2023 at 12:59

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