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Let $i:M\hookrightarrow N$ be an embedded submanifold. In the general terminology, a vector field along a map $i$ is a smooth section $X\in \Gamma (i^{\ast }TN)$, so $X_p\in T_{p}N$. A local extension of a vector field along a submanifold near $p$ is a vector field $\tilde{X}$ on an open neighbourhood in $N$ with $\tilde{X}{|}_M=X$ on that neighbourhood. Choose adapted coordinates $(x^1,\dots ,x^m,y^{m+1},\dots ,y^n)$ in which $M$ is given by $y=0$. Writing $X=X^a(x){\partial }_a{|}_{y=0}$, extend the coefficient functions independently of $y$: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 $i^{\ast }TN$. In this case the useful criterion for a local extension of a vector field along a submanifold isIf $\tilde{X}$ 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 $\tilde{X}{|}_M=X$. The criterion also shows that functions vanishing on $M$ are differentiated to functions vanishing on $M$ by $\tilde{X}$.

For tangent vector fields $X,Y$ and local extensions $\tilde{X},\tilde{Y}$, apply the criterion twice:Hence 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 $\tilde{Y}=0$ and $\tilde{Y}=y{\partial }_x$, with $\tilde{X}={\partial }_y$, 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, byThe derivative is defined along $M$ and is linear in $v$. Since $v⟨\nu ,\nu ⟩=2⟨D_v\nu ,\nu ⟩=0$, its value is tangent to $M$, so this is an endomorphism of the tangent space. Differentiating $⟨\nu ,Y⟩=0$ gives, for tangent vector fields $X,Y$,This is the scalar second fundamental form for the chosen normal. The Euclidean connection on a vector bundle has zero torsion tensor, giving $D_{X}Y-D_{Y}X=[X,Y]$. The bracket is tangent to $M$ by the previous argument; thereforeConsequently $\boxed{⟨S_{p}v,w⟩=⟨v,S_{p}w⟩}$ for every $v,w\in T_{p}M$: the shape operator is self-adjoint. Reversing $\nu$ reverses $S$ and $h$ but preserves this self-adjointness.