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Here $R(X,Y)=D_{[X,Y]}-[D_X,D_Y]$, the negative of the convention in the ordinary Gauss equation in a curved ambient manifold. Insert the Gauss formula and Weingarten formula into this definition and project tangentially to obtain the displayed identity. Positive sectional curvature uses $⟨R(X,Y)X,Y⟩/|X\wedge Y{|}^2$ with this convention. Thus a Euclidean surface still has Gaussian curvature $K=\det A_{\xi }$; reversing the operator sign without changing the slot order would give the wrong sign.

The ambient Levi-Civita connection induces this metric connection on the normal bundle of an embedded submanifold. Its tangential counterpart appears in the Weingarten formula. Projecting the derivative preserves the connection on a vector bundle Leibniz rule and the fiber inner product.

Let $D$ be the ambient Levi-Civita connection and $\nabla$ the one for the induced metric on $M$. The Gauss formula iswhere the second fundamental form $II(X,Y)=(D_{X}Y{)}^{\perp }$ is normal. The tangential part of $D$ is metric-compatible and torsion-free, so the existence and uniqueness of the Levi-Civita connection identifies it with $\nabla$.

For a normal field $\xi$, the Weingarten formula defines the shape operator in a normal direction $A_{\xi }$ and the normal connection byDifferentiate $⟨\xi ,Y⟩=0$ using metric compatibility to getThe symmetry of the second fundamental form follows from the ambient torsion-free condition:since the Lie bracket of vector fields tangent to $M$ is tangent. Thereforeso every normal-direction shape operator is self-adjoint.

The paper uses $R(X,Y)=D_{[X,Y]}-[D_X,D_Y]$, so use the Gauss equation with reversed curvature convention. Applying the Gauss formula and Weingarten formula twice givesSubtracting in the stated curvature order yieldsPair with a tangent field $W$ and use the defining relation for the shape operator in a normal direction:This is the Gauss equation in the supplied sign convention. If one defines curvature by $[D_X,D_Y]-D_{[X,Y]}$ instead, the two quadratic terms have the opposite signs.

For an orthonormal frame $X,Y$ on a surface, positive Gaussian curvature with the convention here is $K=⟨R^M(X,Y)X,Y⟩$. If the ambient space is Euclidean and $\xi$ is a unit normal, the equation becomesThe Koszul formula determines $\nabla$ from the induced metric, and hence determines its Riemann curvature tensor and $K$. Therefore the Gaussian curvature depends only on that metric, even though its expression as a product of principal curvatures appears extrinsic. This is Theorema Egregium.

The Weingarten formula decomposes the ambient derivative of a normal field into a tangential shape operator term and the normal connection. The minus sign corresponds to $⟨A_{\xi }X,Y⟩=⟨II(X,Y),\xi ⟩$. For the outward unit normal of a unit sphere, $A_{\xi }=-I$ with this convention.