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With outward normal $N(x)=x$, the unit sphere has $A(X,Y)=-⟨X,Y⟩N$. The Gauss equation therefore gives sectional curvature one on every tangent two-plane.

The Gauss–Codazzi equations relate the intrinsic curvature and connection of an embedded submanifold to its second fundamental form and the ambient curvature. The Gauss equation controls tangential curvature; the Codazzi equation controls the covariant derivative of the second fundamental form. The Gauss formula is the tangential-normal splitting used to derive these relations.

For a hypersurface with unit normal vector $n$, put $\epsilon =g(n,n)=\pm 1$, $P=1-\epsilon n\otimes n^{♭}$, and $K(X,Y)=g({\nabla }_{X}n,Y)$ for tangent vectors. The induced metric is nondegenerate precisely when the normal is non-null. With $R(X,Y)Z=[{\nabla }_X,{\nabla }_Y]Z-{\nabla }_{[X,Y]}Z$ and $R_{abcd}=g(e_a,R(e_c,e_d)e_b)$, the Gauss formula is ${\nabla }_{X}Y=D_{X}Y-\epsilon K(X,Y)n$. Substituting twice into the ambient Riemann curvature tensor and projecting tangentially gives the displayed Gauss equation. Its normal projection gives $g(n,R(X,Y)Z)=-(D_{X}K)(Y,Z)+(D_{Y}K)(X,Z)$, the curved-ambient Codazzi equation. A null hypersurface needs additional geometric data because its induced metric is degenerate.

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.

Rolling a sheet into a circular cylinder bends it without changing its internal lengths and angles. Fitting that same sheet smoothly onto a sphere requires distortion of its first fundamental form. Gaussian curvature distinguishes these situations: it measures the intrinsic curvature of a smooth surface, even though its simplest definition uses how the surface bends in Euclidean space.

The Gaussian curvature at a point is the determinant of the shape operator, or the product of the two principal curvatures:Here $I=Ed{u}^2+2Fdudv+Gd{v}^2$ is the first fundamental form and $II=ed{u}^2+2fdudv+gd{v}^2$ is the second fundamental form, with $S=-dN$. Regularity makes $EG-F^2>0$. Reversing the unit normal negates $e,f,g$ and both principal curvatures, leaving Gaussian curvature unchanged. ^[ref]

The sign describes bending at the point: positive Gaussian curvature means the principal curvatures have the same nonzero sign, as at an elliptic point; negative Gaussian curvature means opposite signs, as on a saddle; zero Gaussian curvature means at least one principal curvature vanishes. Zero Gaussian curvature at a single point does not imply that a neighborhood is a plane. ^[ref]

Four worked models make the signs concrete:

The patches use a common length unit, with $R=1$ for the sphere and circular cylinder. The marked red and blue curves have principal directions at the black point; the arrow is the chosen unit normal. For physical lengths, principal curvature and mean curvature have units of inverse length, while Gaussian curvature has units of inverse squared length. Rescaling a surface by $\lambda >0$ divides its principal curvatures by $\lambda$ and its Gaussian curvature by ${\lambda }^2$. Writing the saddle as $h=(x^2-y^2)/(2L)$ restores an explicit length scale: $K(0)=-1/L^2$.

The Theorema Egregium says that Gaussian curvature depends only on the first fundamental form and its first two coordinate derivatives. A local isometry therefore preserves Gaussian curvature, even if the shape operator and the principal curvatures change. ^[ref]

Here is the outline, rather than a full proof. The Gauss formula splits second derivatives of a parametric surface into tangential Christoffel symbols and the second fundamental form. The Christoffel symbols are determined by the first fundamental form and its first derivatives. Commuting the next partial derivatives gives the Gauss equation, which identifies $\det S$ with the intrinsic sectional curvature computed from the Riemann curvature tensor. Thus the apparently extrinsic product $k_1{k}_2$ is determined by the Riemannian metric. ^[ref]

One can see the distinction directly by rolling a plane strip with coordinates $(u,v)$ into the circular cylinder $X(u,v)=(R\cos (u/R),R\sin (u/R),v)$. Both have $I=d{u}^2+d{v}^2$, so the map is a local isometry, and is one-to-one on an open strip of width less than $2\pi R$. Extending $u$ over all real numbers wraps around periodically; it does not give a global isometry between the entire plane and the entire circular cylinder. A nonempty open patch of a sphere cannot be flattened by a local isometry into a plane, because its Gaussian curvature $1/R^2$ cannot become zero. The obstruction concerns preservation of the first fundamental form, rather than a particular drawing or choice of seam.

For a tangential covector, the projected derivative is $D_{\nu }{W}_{\gamma }=P^{\beta }{}_{\nu }{P}^{\delta }{}_{\gamma }{\nabla }_{\beta }{W}_{\delta }$. Its second derivative isExpanding by the product rule, the derivative of the first inner projector gives $K_{\mu \nu }{n}^{\eta }{P}^{\kappa }{}_{\gamma }{\nabla }_{\eta }{W}_{\kappa }$. This contribution cancels on antisymmetrizing $\mu ,\nu$, because $K$ is symmetric.

The derivative of the second inner projector gives $K_{\mu \gamma }{n}^{\kappa }{P}^{\eta }{}_{\nu }{\nabla }_{\eta }{W}_{\kappa }$. Differentiating tangentiality, $n^{\kappa }{W}_{\kappa }=0$, rewrites it asThe remaining term is the projected four-dimensional curvature commutator. Using the curvature convention of the question,and replacing the tangential $W_{\xi }$ by $W_{\lambda }{P}^{\lambda }{}_{\xi }$, we obtainSince this holds for every tangential covector, the Gauss equation follows:The signs use a timelike unit normal and the specified commutator convention; reversing a curvature convention without changing the commutator would change the comparison.

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.

Double contraction of the projected Gauss equation, with the timelike-normal sign and curvature convention in the question, givesHere $q^{\mu \nu }=g^{\mu \nu }+n^{\mu }{n}^{\nu }$. Expanding the two projectors givesThe four-normal contraction vanishes by Riemann antisymmetry. Because $g_{\mu \nu }{n}^{\mu }{n}^{\nu }=-1$, the right side is $2G_{\mu \nu }^{(4)}{n}^{\mu }{n}^{\nu }$. Rearranging proves the Hamiltonian constraint identity:For flat FRW, $R^{(3)}=0$, $K^2=9H^2$ and $K_{\mu \nu }{K}^{\mu \nu }=3H^2$, so $G_{nn}=3H^2$. Einstein's equation is therefore the first Friedmann equation, $3H^2=8\pi G\rho$, or $3H^2=8\pi G\rho +\Lambda$ if a cosmological constant is placed separately in Einstein's equation.

The scalar $R^{(3)}$ measures intrinsic spatial curvature, $K^2$ measures squared volume expansion, and the norm of $K_{ij}$ measures directional expansion including shear. More generally $K_{ij}=K{q}_{ij}/3+s_{ij}$ with trace-free $s_{ij}$ gives $K_{ij}{K}^{ij}=K^2/3+s_{ij}{s}^{ij}$. The shear vanishes in FRW, leaving the positive $3H^2$ contribution. This is an instantaneous constraint on initial data, not the separate acceleration equation.

Assume the level set is regular and non-null. The latter hypothesis is necessary: a null hypersurface has a degenerate induced metric and hence no ordinary intrinsic Levi-Civita connection determined by that metric. Normalize the normal vector and define the non-null hypersurface projection byHere $h$ denotes the induced metric restricted to tangent vectors, while its ambient extension annihilates $n$. The intrinsic covariant derivative is $D_{X}Y=P{\nabla }_{X}Y$ for tangent vector fields. Projection of the ambient torsion-free connection and metric compatibility shows that $D$ is the Levi-Civita connection of $h$. Define the extrinsic curvature by $K(X,Y)=g({\nabla }_{X}n,Y)$, or $K_{ab}=P^c{}_a{P}^d{}_b{\nabla }_c{n}_d$. It is symmetric: differentiating the defining function twice and projecting removes the derivative of its normalization, leaving a multiple of its symmetric Hessian. This sign convention makes the scalar second fundamental form $g(n,{\nabla }_{X}Y)=-K(X,Y)$.

Let $S$ be the tangent linear map characterized by $h(SX,Y)=K(X,Y)$. Differentiating $g(n,n)=\epsilon$ and $g(n,Y)=0$ gives the Gauss formulaFix the Riemann curvature tensor conventionApplying the Gauss formula twice givesSubtracting the expression with $X,Y$ interchanged, and then subtracting the bracket derivative, provesThus the Gauss–Codazzi equations for a non-null hypersurface give the intrinsic Riemann curvature tensor asAll free indices here are tangent to the hypersurface. The normal component of the same calculation isAfter antisymmetrizing and using $D_{X}Y-D_{Y}X=[X,Y]$, this yields the curved-ambient Codazzi equationThe extrinsic curvature measures how tangent planes turn in the ambient manifold; the Gauss equation says that this turning contributes to intrinsic curvature even when the ambient Riemann curvature tensor is zero.

For the quadric in Minkowski spacetime, write its position vector as $X^a$. On the hypersurface, $g(X,X)=1$ and differentiating this identity in a tangent direction gives $g(X,Y)=0$. Hence $n=X$ is a unit normal and a spacelike vector, $\epsilon =1$, and ${\nabla }_{Y}n=Y$ in flat Cartesian coordinates. Therefore $K=h$. The ambient Riemann curvature tensor vanishes, soThis is unit-radius de Sitter spacetime. Reversing the normal changes $K$ to $-K$ but does not change this curvature. An opposite overall convention for the Riemann curvature tensor would reverse both curvature signs below.

To retain a four-dimensional Lorentzian metric while reversing the curvature, take a flat ambient metric with two negative directions and the quadricNow the position vector is a unit normal with negative squared norm, so $\epsilon =-1$ while $K=h$ remains true. The Gauss equation givesThis is unit-radius Anti-de Sitter spacetime. If a positive-definite induced metric is acceptable, an even simpler alternative is to keep the original one-time ambient Minkowski metric and change the quadric level to $g(X,X)=-1$. Each sheet then gives four-dimensional hyperbolic space with the same negative curvature, but its induced signature differs from that of de Sitter spacetime.

The Theorema Egregium states that the Gaussian curvature of a surface in ${\mathbb{R}}^3$ is determined entirely by its first fundamental form; in particular a local isometry preserves it.

Let $r(u^1,u^2)$ be a regular parametrization, $g_{ij}=r_i\cdot r_j$, $n$ a unit normal, and $b_{ij}=r_{ij}\cdot n$ the second fundamental form. Differentiating the metric and decomposing $r_{ij}$ into tangent and normal parts gives the Gauss formulaDifferentiating $n\cdot r_j=0$ gives $n_i=-b_i{}^k{r}_k$, the Weingarten formula. Compute two successive ambient derivatives of a tangent vector and subtract them. Ambient derivatives commute, so their tangential parts yield the Gauss equationwhere $S$ is the shape operator and $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$ is the intrinsic curvature of the connection with the displayed Christoffel symbols. Taking $X={\partial }_1,Y={\partial }_2,Z={\partial }_2$ and the inner product with ${\partial }_1$ givesBut $K=\det S=\det (b)/\det (g)$, soBoth the connection and its curvature on the right are computed from $g$ and its derivatives alone. This proves the theorem, rather than merely using its name as a substitute for the argument.

The Theorema Egregium states that Gaussian curvature is determined by the first fundamental form, and therefore preserved by local isometries. For a local surface parametrization $r$, write $g_{ij}=r_i\cdot r_j$, choose a unit normal $N$, and set $b_{ij}=r_{ij}\cdot N$. Decomposing the second derivatives into tangent and normal parts givesCommute the third partial derivatives of $r$ and take tangent components. The terms involving derivatives of $b$ are normal; the tangent terms give the Gauss equationwhere $R$ is computed from derivatives and products of the displayed Christoffel symbols, using $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. For example the tangent term from differentiating $b_{ij}N$ is $-b_{ij}{b}_k{}^{\ell }{r}_{\ell }$, which produces exactly the difference of the two products on the right. ConsequentlyThe last expression depends only on $g$ and its derivatives. Pulling back $g$ under a local isometry therefore preserves $K$, proving the theorem.

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.

For a local parametrization $\varphi (u,v)$, write the first fundamental form and second fundamental form asThe Gaussian curvature is

To prove the Theorema Egregium, set $g_{ij}=⟨{\varphi }_i,{\varphi }_j⟩$ and decomposeDifferentiating $g_{ij}$ and solving for the tangential coefficients givesso every Christoffel symbol depends only on $E,F,G$ and their first derivatives. Equality of mixed third derivatives yields the Gauss equationThe Riemann curvature tensor is formed from first derivatives and quadratic products of the Christoffel symbols, and hence from the first and second derivatives of $E,F,G$. Consequentlydepends only on the first fundamental form.

Contract the Gauss equation with $h^{ac}{h}^{bd}$. Since $h^{ab}=g^{ab}+n^a{n}^b$ and the antisymmetries of the Riemann curvature tensor eliminate the term containing four normals,The extrinsic terms contract to $K^2$ and $K_{ab}{K}^{ab}$. Hence the intrinsic Ricci scalar is

Let $D$ be the Levi-Civita connection of $h$. Apply $(D_c{D}_d-D_d{D}_c)$ to a tangent vector, replace each $D$ by the tangential projection of $\nabla$, and decompose $\nabla n$ using $K_{ab}$. Comparison with the intrinsic Riemann curvature tensor gives the Gauss equationChanging the sign convention for $K_{ab}$ does not alter these quadratic terms.

With the curvature convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Gauss equation for a Euclidean embedded submanifold isThe Codazzi equation iswhere the derivative uses the Levi-Civita connection on tangent arguments and the normal connection on the value of $A$.

On the unit sphere $S^n\subseteq {\mathbb{R}}^{n+1}$ choose the outward unit normal $N(x)=x$. For tangent vector fields $X,Y$,soIf $X,Y$ are orthonormal, the Gauss equation givesThus the round unit sphere has sectional curvature one. Tracing over an orthonormal basis gives its scalar curvature

For the convention $K_{\mu \nu }=-{\perp }^{\rho }{}_{\mu }{\nabla }_{\rho }{n}_{\nu }$, contraction of the Gauss equation gives

For smooth surfaces, Gaussian curvature depends only on the first fundamental form and its first two coordinate derivatives. Consequently any smooth local isometry $f:M\to N$ satisfies $K_N(f(p))=K_M(p)$. The surprise is that $K$ is defined using the shape operator, which describes bending in space, yet its value can be recovered from measurements made entirely within the surface.

In his 1827 paper, Carl Friedrich Gauss expressed Gaussian curvature through the metric coefficients and their partial derivatives in Article 11, then stated its invariance under development in Article 12. Article 13 explicitly compares a plane with surfaces developable onto it, including cylindrical surfaces. The paper was presented on 8 October 1827 and printed in 1828; the Morehead–Hiltebeitel English translation appeared in 1902. ^[ref]

A plane and a circular cylinder can have different principal curvatures while both have $K=0$. Their matching first fundamental forms exhibit the actual local isometry. In contrast, no nonempty open patch of a sphere of radius $R$ can be flattened smoothly into a plane while preserving its first fundamental form: $1/R^2$ cannot become zero. Making the patch small reduces practical distortion, but never produces an exact local isometry. This is a geometric obstruction to a perfectly distance-preserving flat map of a spherical world, independent of where one places a seam.

The following proof outline connects the two descriptions of Gaussian curvature. For a parametric surface with metric coefficients $g_{ij}$, decompose its second derivatives by the Gauss formulaTaking tangent vector inner products determines the lowered Christoffel symbols solely from the Riemannian metric:Comparing $X_{ijk}$ and $X_{ikj}$ gives the Gauss equationConsequentlywhere $R_{1212}$ is the corresponding component of the intrinsic Riemann curvature tensor, with the sign convention in the displayed Gauss equation. The final expression uses only $g_{ij}$ and its first two derivatives.