The differential of the Gauss normal map from a Lie algebraic view point Let  $S\subset \mathbb{R}^3$  be  a  smooth surface  with the Gauss normal map $N:S\to  S^2$.
Then for  every $x\in S$, the  differential $(dN)_x:T_xS\to T_{N(x)}S^2$ can be  considered as  an  endomorphism of the  tangent space $T_xS$ since $T_xS$ is  parallel to  $T_{N(x)}S^2$. So  from  now  on, without any  ambiguity and in a  unique way,   we  count $dN$ as an  endomorphism of  the  tangent  bundle $TS$ of  the  surface $S$. So  $dN$  defines  a  linear  operator  $$dN:\chi^{\infty}(S)\to \chi^{\infty}(S)$$  where $\chi^{\infty}(S)$ is the  space  of  all smooth vector  fields on $S$. 

Under which geometric conditions on $S$ this operator preserve the  Lie  bracket  of  $\chi^{\infty}(S)$? Under which  conditions on $S$, the  range $dN(\chi^{\infty}(S))$ of this operator is  a  Lie  algebra?

Of course we  can ask the  same  question for every codimension $1$ submanifold $S$ of  $\mathbb{R}^n$.
 A: ADDED: I checked my calculation of the first displayed equation, and it appears to be correct. If so, it looks to me that it already implies that $A$ is either the identity or zero and therefore the second paragraph isn't even needed.
The shape operator is an example of a bundle map $A: S\rightarrow S$, where $S$ is a $d$-dimensional smooth manifold. If it preserves the Lie algebra structure of vector fields, then $[AX,AY] = A[X,Y]$ for any smooth vector fields $X$ and $Y$. For any $p \in S$ choose local coordinates $x = (x^1, \dots, x^d)$ such that $p$ is at the origin. For any $1 \le i,j,k \le d$, let $X = \partial_i$ and $Y= x^j\partial_k$. A straightforward calculation shows that at the origin,
$$ [AX,AY] =A^i_jA_k^p\partial_p\text{ and }A[X,Y] = \delta_i^jA_k^p\partial_p. $$
Therefore, at each point in $S$, $A^2 = A$ and therefore $A$ is diagonalizable with eigenvalues $0$ and $1$.
Now suppose $A$ is neither the identity nor $0$ at a point $p \in S$. Let $X$ and $Y$ be smooth vector fields that are nonzero at $p$ and satisfy $AX = 0$ and $AY = Y$ at $p$. Then $A[X,Y] = [AX,AY] = 0$ at $p$. Let $f$ be a smooth function such that $Xf(p) \ne 0$. Then on one hand,
$$ [AX,A(fY)] = 0. $$
On the other hand, at $p$,
$$
[AX,A(fY)] = A[X,fY] = A((Xf)Y + f[X,Y]) = (Xf)AY + fA[X,Y] = (Xf)Y \ne 0.
$$
This is a contradiction and therefore $A$ is either the identity or $0$.
