The differential of the Gauss normal map from a Lie algebraic view point

Question

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 $\mathbb{R}^n$.

Answer 1

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 $0$.