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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 $0$.

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 $0$.

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 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, $A^2 = A$ and $A$ is diagonalizable with eigenvalues $0$ and $1$. Let $X$ and $Y$ be any vector fields such that $AX = X$ and $AY = 0$. Then $A[X,Y] = [AX,AY] = 0$, which is absurd. Therefore, $A = 0$ or $A = 1$.

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 $0$.

Let $f: S \rightarrow \mathbb{R}^n$ be the embedding of $S$ in $\mathbb{R}^n$ and $A: T_*S \rightarrow T_*S$ its shape operator. Then given any local diffeomorphism $\Phi: S\rightarrow S$, the shape operator of $f\circ\Phi$ is $(\Phi^{-1})_*\circ A \circ (\Phi_*)$. Therefore, if a shape operator preserves the Lie algebra structure, then the bundle map $(\Phi^{-1})_*\circ A \circ (\Phi_*)$ is also a Lie algebra preserving shape operator. Therefore, the class of shape operators that preserve the Lie algebra structure is invariant under local diffeomorphisms. It is now straightforward to verify that the only possibilities for this class are, for each $0\le k \le n-1$, the class of endomorphisms of rank $k$, or unions of these classes. It’s easy now to check that only $k=n-1$ and $k=0$ work.

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 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, $A^2 = A$ and $A$ is diagonalizable with eigenvalues $0$ and $1$. Let $X$ and $Y$ be any vector fields such that $AX = X$ and $AY = 0$. Then $A[X,Y] = [AX,AY] = 0$, which is absurd. Therefore, $A = 0$ or $A = 1$.