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For now I would be content with understanding why the eigenvalues of the shape operator of a surface are the principle curvatures, let's call them $k_1,k_2$.

Let $f: M \rightarrow S^2$ be the Gauss map of an oriented surface $M$ into the sphere. This map simply sends the unit normal vector at any point of our surface to it's point on the sphere, I like to think of this map sort of like a trippy compass.

The differential of this map is called the Shape Operator.

Given a point $x \in M$, the tangent plane at $x$ is denoted $T_xM$ is an inner product space. The shape operator can be defined as a linear operator on $T_xM$ by the equation:

$$ (S_x(v),w)=(df_x(v),w) \quad \text{for any $v,w \in T_x M.$} $$

Apparently, the equation is above is symmetric in $v$ and $w$, and thus the shape operator is a self-adjoint operator. Hm... So it's symmetric in $v$ and $w$, so $(df_x(v),w)=(df_x(w),v)$? Can somebody explain to me why that makes sense??

And then yeah, if somebody could help me understand why the eigen-values of this operator are the principle curvatures, i.e. the maximum and minimum values of possible curvatures as you depart from our given point, $x$, I'd really appreciate it.

Thanks!

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Let $\mathbf{x}(u,v)$ be a parametrisation of $M$ around $x$, then $\{\mathbf{x}_u,\mathbf{x}_v\}$ is a basis of $T_x M$. If $N$ is the normal vector on $M$, then $df_x(\mathbf{x_u}) = N_u$ and $df_x(\mathbf{x}_v)=N_v$. Derive $(N,\mathbf{x}_u) = 0$ w.r.t $v$ and $(N, \mathbf{x}_v)=0$ w.r.t. $u$: $$ \begin{align*} (N_v, \mathbf{x}_u) + (N_, \mathbf{x}_{uv}) &= 0 \\ (N_u, \mathbf{x}_v) + (N_, \mathbf{x}_{vu}) &= 0 \end{align*} $$ Hence $$ (N_u, \mathbf{x}_v) = -(N,\mathbf{x}_{uv}) = (N_v,\mathbf{x}_{u}). $$ By linearity, it follows that $(df_x(v),w)= (df_x(w),v)$ for all $v$, $w \in T_x M$.

For your second question, let $\mathbf{e}_1$, $\mathbf{e}_2 \in T_x M$ be the eigenvectors of $df_x$, i.e. the principal directions, and let $k_1$, $k_2$ be the corresponding eigenvalues. Let us say that $k_1 \geq k_2$. The normal curvature $k(v)$ in a direction $v=\cos \theta\, \mathbf{e}_1+\sin \theta\, \mathbf{e}_2$ is by definition $(df_x(v),v)$. We get $$ \begin{align*} k(v)= (df_x(v),v) &= (k_1 \cos \theta \,\mathbf{e}_1+ k _2 \sin \theta \,\mathbf{e}_2,\cos \theta \mathbf{e}_1+\sin \theta \mathbf{e}_2)\\ &= k_1\cos^2\theta + k_2\sin^2 \theta. \end{align*} $$ In some references this formula is called Eulers formula for the normal curvatures. Finally we note that. $$ k_1 = k_1 \cos^2\theta + k_1\sin^2\theta \geq k_1\cos^2\theta + k_2\sin^2\theta \geq k_2\cos^2\theta +k_2\sin^2\theta = k_2, $$ so $k_1\geq k(v) \geq k_2$. This shows that the eigenvalues of $df_x$ are the maximal en minimal normal curvatures.

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    $\begingroup$ Awesome, thanks man!! $\endgroup$
    – user637978
    Commented May 9, 2020 at 11:41
  • $\begingroup$ @Ernie060 Why can you assume that the two eigenvectors are linearly independent? In general, the shape operator matrix is not symmetric. $\endgroup$
    – gipouf
    Commented Dec 11, 2022 at 13:40
  • $\begingroup$ The shape operator matrix is always symmetric, that is what the line $(N_u, \mathbf{x}_v) = (N_v, \mathbf{x}_u)$ says. Since the shape operator is symmetric (self-adjoint), it has an orthonormal basis of eigenvectors by the spectral theorem. $\endgroup$
    – Ernie060
    Commented Dec 11, 2022 at 15:06
  • $\begingroup$ @Ernie060 I am a bit confused. see the answer for that post math.stackexchange.com/questions/859884/…, specifically, the first paragraph of the answer. It says that the shape operator matrix might not be symmetric. $\endgroup$
    – gipouf
    Commented Dec 12, 2022 at 21:30
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    $\begingroup$ Here is my mistake: the shape operator is indeed a self-adjoint operator, but this does not imply that its matrix is symmetric. My bad. The main difference is actually explained very well in the movie and the linked question you mentioned. In the clip, the vectors $X_i$ are vectors in the domain in $\mathbb{R}^2$. The images $df(X_i)$ are vectors in the tangent plane. (Also, as you probably have noticed, $f$ in this question stands for the Gauss map, while in the video, it stands for a parametrisation, so comparing the different sources is painful...) $\endgroup$
    – Ernie060
    Commented Dec 13, 2022 at 22:04

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