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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 the range $dN(\chi^{\infty}(S))$ of this operator is a Lie algebra?

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 $\mathbb{R}^n$.

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 bundle $TS$ 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 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 the image of this operator is a Lie algebra?

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 the range $dN(\chi^{\infty}(S))$ of this operator is a Lie algebra?