= Solution
Let $i:M\hookrightarrow N$ be an embedded <submanifold>. In the general terminology, a <vector field along a map> $i$ is a smooth section $X\in\Gamma(i^*TN)$, so $X_p\in T_pN$. A <local extension of a vector field along a submanifold> near $p$ is a <vector field> $\widetilde X$ on an open neighbourhood in $N$ with $\widetilde X|_M=X$ on that neighbourhood. Choose adapted coordinates $(x^1,\ldots,x^m,y^{m+1},\ldots,y^n)$ in which $M$ is given by $y=0$. Writing $X=X^a(x)\partial_a|_{y=0}$, extend the coefficient functions independently of $y$:
$$
\widetilde X(x,y)=\sum_aX^a(x)\partial_a.
$$
This proves local existence, including for fields with normal components.
For the <Lie bracket> assertion, the fields must be tangent to $M$, namely sections of $TM$ viewed inside $i^*TN$. In this case the useful criterion for a <local extension of a vector field along a submanifold> is
$$
\boxed{(\widetilde XF)|_M=X(F|_M)\quad\text{for every local smooth function }F\text{ on }N.}
$$
If $\widetilde X$ extends $X$, the <chain rule> gives this identity because their tangent values agree. Conversely, applying it to the adapted coordinate functions forces the tangent components to equal those of $X$ and all normal components on $M$ to vanish, so $\widetilde X|_M=X$. The criterion also shows that functions vanishing on $M$ are differentiated to functions vanishing on $M$ by $\widetilde X$.
For tangent <vector fields> $X,Y$ and local extensions $\widetilde X,\widetilde Y$, apply the criterion twice:
$$
\begin{aligned}
([\widetilde X,\widetilde Y]F)|_M
&=X\bigl((\widetilde YF)|_M\bigr)-Y\bigl((\widetilde XF)|_M\bigr)\\
&=X\bigl(Y(F|_M)\bigr)-Y\bigl(X(F|_M)\bigr)\\
&=[X,Y](F|_M).
\end{aligned}
$$
Hence \b[the ambient bracket restricts to the intrinsic bracket, independently of the chosen extensions]. Tangency is essential. For example, on the $x$-axis in $\mathbb R^2$, let $X=\partial_y|_M$ and $Y=0|_M$. The two extensions $\widetilde Y=0$ and $\widetilde Y=y\partial_x$, with $\widetilde X=\partial_y$, give bracket restrictions $0$ and $\partial_x|_M$. Thus general sections with normal components do not possess the extension-independent <Lie bracket> asserted for tangent fields.
For the <hypersurface> $M^n\subset\mathbb R^{n+1}$, choose a smooth unit normal $\nu$ locally. A global choice is needed only if one wants a globally signed <shape operator>; an arbitrary <hypersurface> need not come equipped with such a choice. With $D$ the Euclidean <Levi-Civita connection>, define the <shape operator>, also called the <Weingarten map>, by
$$
\boxed{S_p:T_pM\to T_pM,\qquad S_p(v)=-D_v\nu.}
$$
The derivative is defined along $M$ and is linear in $v$. Since $v\langle\nu,\nu\rangle=2\langle D_v\nu,\nu\rangle=0$, its value is tangent to $M$, so this is an endomorphism of the <tangent space>. Differentiating $\langle\nu,Y\rangle=0$ gives, for tangent <vector fields> $X,Y$,
$$
h(X,Y):=\langle SX,Y\rangle=\langle\nu,D_XY\rangle.
$$
This is the scalar <second fundamental form> for the chosen normal. The Euclidean <connection on a vector bundle> has zero <torsion tensor>, giving $D_XY-D_YX=[X,Y]$. The bracket is tangent to $M$ by the previous argument; therefore
$$
h(X,Y)-h(Y,X)=\langle\nu,[X,Y]\rangle=0.
$$
Consequently $\boxed{\langle S_pv,w\rangle=\langle v,S_pw\rangle}$ for every $v,w\in T_pM$: the <shape operator> is <self-adjoint>. Reversing $\nu$ reverses $S$ and $h$ but preserves this self-adjointness.
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