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For an embedded submanifold of a Riemannian manifold, use $R(v,w)y={\nabla }_v{\nabla }_{w}y-{\nabla }_w{\nabla }_{v}y-{\nabla }_{[v,w]}y$. The Gauss formula and tangential derivative of a normal field giveSubtract the expression with $v,w$ interchanged and the bracket derivative, whose normal part pairs to zero. This proves the displayed curvature identity. For flat Euclidean ambient space, its ambient curvature term vanishes and one obtains the ordinary Gauss equation. Declaring the four-slot convention $R(x,y,v,w)=⟨R(v,w)y,x⟩$ avoids sign ambiguities when permuting arguments.

Let $i:M\hookrightarrow N$ be the inclusion. The restriction of a connection to an embedded submanifold is the pullback connection on $i^{\ast }TN=TN{|}_M$. Concretely, for a tangent vector field $X$ on $M$ and a local section $S$ of $TN{|}_M$, extend $S$ smoothly off $M$ and define ${\nabla }_{X}S$ by differentiating the extension in direction $di(X)$. Two extensions differ by a field whose coefficients vanish on $M$. Their derivatives along every tangent curve in $M$ vanish too, and the connection's coefficient terms are multiplied by the zero field. The answer is therefore independent of the extension. Dependence only on the value of the first argument follows from the connection's linearity over smooth functions.

The induced Riemannian metric is positive definite, so the fibrewise orthogonal projection $\pi :TN{|}_M\to TM$ is smooth. For tangent fields define the projected ambient connection $D_{X}Y=\pi ({\nabla }_{X}Y)$. It is real-linear and satisfiesbecause $\pi Y=Y$. Thus $D$ is a Koszul connection on $M$.

The Levi-Civita connection is characterized by the torsion-free connection and metric connection conditionsThese determine it uniquely. For the ambient Levi-Civita connection, the Gauss formula is ${\nabla }_{X}Y=D_{X}Y+II(X,Y)$, withLinearity over smooth functions in $X$ is immediate. In $Y$, the additional $X(f)Y$ term is tangent and is killed by $I-\pi$. Thus tensoriality makes $II$ a bilinear map $TM{\times }_{M}TM\to \nu M$ into the normal bundle. Its antisymmetric part isbecause the Lie bracket of vector fields tangent to $M$ is tangent. This proves the symmetry of the second fundamental form in an arbitrary Riemannian ambient manifold.

Write the curvature operators as $R^N(v,w)y={\nabla }_v{\nabla }_{w}y-{\nabla }_w{\nabla }_{v}y-{\nabla }_{[v,w]}y$ and $R^M$ with $D$ in place of $\nabla$. To match the four-slot notation in the requested identity, useThis convention is stated because permuting the slots can change the displayed signs.

If $\eta$ is a normal field and $x$ is tangent, metric compatibility applied to $⟨\eta ,x⟩=0$ gives the tangential derivative of a normal field identityUsing the Gauss formula and this identity,Here the normal part of ${\nabla }_v(D_{w}y)$ pairs to zero with $x$. Interchange $v,w$ and subtract; the bracket term obeys $⟨{\nabla }_{[v,w]}y,x⟩=⟨D_{[v,w]}y,x⟩$. It follows thatRearranging proves the Gauss equation in a curved ambient manifold:All expressions are tensorial, so choosing local extensions of the four tangent vectors proves the pointwise identity at $P$. With flat Euclidean ambient space, $R^N=0$ and this reduces to the usual Gauss equation.