Given a tangent direction $X\in T_p S$, being $f:S\rightarrow \mathbb{R}^3$ an inmersed surface we have that for $p$ we can see the Weingarten map $dN_p$ sends $X$ to a tangent vector in $f(p)\in f(S)$
$$ dN_p: T_p S\rightarrow T_{f(p)} f(S) $$With this in mind, the shape operator is a linear map $\mathcal{S}:T_pS\rightarrow T_pS$ such that
$$ dN_p(X)=df_p(\mathcal{S}(X)) $$The idea is that we are studying the different curvatures of the surface, but in the abstract manifold, not in $\mathbb{R}^3$. That is, we are substituting the immersion by a $(1,1)$-tensor. We can use the Riemannian metric (called in this case the first fundamental form) of the surface to lower the index and construct the second fundamental form. We are connecting this way Gaussian curvature and extrinsic curvature.
In what follows, we are going to identify $S\equiv f(S)$, and will work inside $\mathbb{R}^3.$ Observe that for a unit tangent vector $V$ at $p$:
$$ \mathcal{S}(V)=dN_p(V) $$and therefore the shape operator is measuring the variation of the normal vector $N$ to the surface $S$ when we move along a normal section with tangent vector $V$ at $p$.
On the other hand, this normal section has a curvature, $\kappa(V)$, and it turns out that
Lemma
The curvature $\kappa(V)$ of a normal section along $V$ is given by
$$ \kappa(V)=-\langle V, \mathcal{S}(V)\rangle $$Proof
Think of $V$ as the velocity of a particle travelling at unit speed along the normal section $\alpha$, with $\alpha(0)=p, \alpha'(0)=V$, so
$$ \dfrac{d}{dt}|_{t=0}\alpha'(t)=\kappa(V) N_p $$(observe that at $p$, the normal to the curve $n$ coincides with the normal to the surface $N$).
But now observe that
$$ 0=\dfrac{d}{dt}|_{t=0}\langle \alpha'(t),N_{\alpha(t)} \rangle=\langle \kappa(V)N_p,N_p\rangle +\langle V,\dfrac{d}{dt}|_{t=0}N_{\alpha(t)}\rangle= $$ $$ =\kappa(V) +\langle V,\mathcal{S}(V)\rangle. $$$\blacksquare$
The shape operator is symmetric (I have to think this yet) and therefore it has eigenvectors and eigenvalues. The eigenvectors of $\mathcal{S}$ define the principal directions, and their eigenvalues define the principal curvatures. Since the extrinsic curvature is the local area magnification of the Gauss map, it corresponds to the determinant of the shape operator when expressed in a basis of tangent vectors. In case we take the basis of eigenvectors we recuperate the famous formula
$$ K=\kappa_1 \kappa_2, $$Observe that given $V\in T_pS$ represented by a curve $\alpha$ ($\alpha(0)=p$, $\alpha'(0)=V$) the shape operator $\mathcal{S}(V)$ is represented by the curve
$$ \beta(t)=N_{\alpha(t)}-N_p+p, $$in the sense that $\beta(0)=p$ and $\beta'(0)=dN_p(V)$, and interpreting $N$ as a map from $\mathbb{R}^3$ to $\mathbb{R}^3$.
If we now consider the standard Riemannian metric of $\mathbb{R}^3\equiv (x,y,z)$ and the corresponding Levi-Civita connection $\nabla$, we know that in the coordinate frame $\{\partial_x,\partial_y,\partial_z\}$ the connection form is the matrix $\Theta=0$. And we can wonder about $\nabla_V N$, provided we extend $N$ to a vector field on an open neighbourhood of $p$. Suppose $N=n_x \partial_x+n_y\partial_y+n_z\partial_z$, then
$$ \nabla_V N=V(n_x)\partial_x+V(n_y)\partial_y+V(n_z)\partial_z $$On the other hand, by computations we obtain
$$ dN_p(V)=\beta'(0)=V(n_x)\partial_x+V(n_y)\partial_y+V(n_z)\partial_z=(\nabla_VN)_p. $$Giving a particular orthonormal frame $\{e_1,e_2,N\}$, we can express $\nabla$ as another matrix $\tilde{\Theta}$, not necessarily null. The components $\Theta^i_j$ are 1-forms, and indeed the shape operator is given by the map:
$$ S:V\mapsto \nabla_V N=V\lrcorner\tilde{\Theta}^1_3 e_1+V\lrcorner\tilde{\Theta}^2_3 e_2 $$The determinant of $S$ is the Gaussian curvature, so $K=\tilde{\Theta}^1_3 \wedge \tilde{\Theta}^2_3(e_1,e_2)$.
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Author of the notes: Antonio J. Pan-Collantes
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