Gaussian curvature from isometric embedding

Let $M$ be a two-dimensional Riemannian manifold, and assume that $M$ can be realized as a surface $S$ in $\mathbb{R}^{3}$.

I know that one can compute the Gaussian curvature $K$ of $M$ as the determinant of the shape operator of $S$.

Question: What if you are given an isometric embedding of $M$ into some Riemannian three-manifold $N \neq \mathbb{R}^{3}$? Can you still compute $K$ as in the classical case?

• To supplement the existing answers, it's easy to see the product of the principal curvatures is not generally the Gaussian curvature: If $S$ is a great sphere (or "equator") in a round $3$-sphere, itself sitting in Euclidean $4$-space, a continuous unit normal field is constant, so the shape operator of $S$ in $S^{3}$ vanishes identically. – Andrew D. Hwang Aug 2 '17 at 10:55

Denote by $\mathbb{M}^3(c)$ a simply-connected Riemannian $3$-space with constant sectional curvature $c=-1,0,1$. That is, $\mathbb{M}^3(c)$ denotes the hyperbolic space $\mathbb{H}^3$ if $c=-1$, the Euclidean space $\mathbb{R}^3$ when $c=0$, or the sphere $\mathbb{S}^3$ if $c=1$.

Now, consider an oriented Riemannian surface $\Sigma$ with induced metric $I=\langle,\rangle$ and $f:\Sigma\to\mathbb{M}^3(c)$ an isometric immersion. Let $N$ be a unit normal along $\Sigma$ which is compatible with the orientation, and $II$ its associated second fundamental form. That is, $II(X,Y)=\langle-\nabla_X N,Y\rangle$, where $X,Y$ are smooth vector fields in $\Sigma$ and $\nabla$ the Levi-Civita connection of the ambient space.

If $k_1$, $k_2$ are the principal curvatures of the immersion, namely, the eigenvalues of the shape operator $S$, given by $SX=-\nabla_X N$, then we shall denote by $$K=k_1k_2,\quad H={1\over 2}(k_1+k_2)$$ the extrinsic curvature and the mean curvature of the immersion, respectively. These two quantities are extrinsic and depend on how $S$ is immersed into $\mathbb{M}^3(c)$.

On the other hand, we denote by $K(I)$ the curvature of the metric $I$, namely, the Gauss curvature, which is an intrinsic quantity. For the previous immersion $f:\Sigma\to\mathbb{M}^3(c)$ the following relations must be satisfied $$K(I)=K+c,\quad\text{(Gauss equation)}$$ $$\nabla_X SY-\nabla_Y SX-S[X,Y] = 0,\quad\text{(Mainardi-Codazzi equation)},$$ where $\nabla$ is the Levi-Civita connection of the metric $I$.

You can see with these examples, the space forms, that the Gauss curvature is the product of the eigenvalues of the shape operator when your ambient space is $\mathbb{R}^3$, but there is a term (the sectional curvature $c$) that modifies it in general.

You'll have a correction term given by the sectional curvature of the ambient $N$ itself. If $(v,w)$ is an orthonormal basis for the tangent plane of $M$ at some point $p$, then: $$K_M(v,w)= K_N(v,w) + \langle \alpha(v,v),\alpha(w,w)\rangle - \langle \alpha(v,w),\alpha(v,w)\rangle,$$where $\alpha$ is the second fundamental form of $M$ in $N$. This is true for all dimensions and codimensions.

In the present case we simplify it: since $M$ has dimension $2$, you can write $K_M(p)$ for the Gaussian Curvature, and if $\nu$ is an unit normal field, and $X$ is a parametrization, you can write $\alpha(X_i,X_j)=h_{ij}\nu$ for $1\leq i,j\leq 2$, apply Gram-Schmidt to $\{X_1,X_2\}$ and plug all in the Gauss Formula above. For $\Bbb R^3$, $K_N =0$ so you'll get the usual formula.