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A Riemannian metric determines a unique torsion-free connection that is a metric connection. Metric compatibility and the torsion identity imply the Koszul formula, which proves uniqueness. Conversely the right-hand side of the Koszul formula is a smooth one-form in its last vector-field argument. The musical isomorphism defines $D_{X}Y$ from it; expanding the Lie bracket of vector fields verifies the connection rules, metric compatibility and zero torsion. Thus it constructs the Levi-Civita connection intrinsically.

A metric connection satisfies metric compatibility, ${\nabla }_{\alpha }{g}_{\mu \nu }=0$. This allows the metric tensor to commute with covariant derivatives and index raising or lowering.

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.

For a metric connection on a vector bundle, the covariant derivative along a curve satisfies $(d/dt)⟨u,v⟩=⟨D_{t}u,v⟩+⟨u,D_{t}v⟩$. If both sections are parallel, the right side vanishes. Thus parallel transport preserves the fibre inner product, not merely vector lengths, and is a linear isometry between the endpoint fibres.

For a scalar, the Lie derivative of a function is its directional derivative:The Lie derivative of a vector field is their Lie bracket of vector fields:For a torsion-free connection,The last term vanishes because the connection coefficients are symmetric in $b,c$. This proves ${\mathcal{L}}_{\xi }X={\nabla }_{\xi }X-{\nabla }_X\xi$.

For the metric tensor statements use the Levi-Civita connection, which also has metric compatibility. The tensor derivation property of the Lie derivative of a tensor field givesExpanding the first term using $\nabla g=0$ and using the vector identity cancels the derivatives of $X$ and $Y$. What remains isSince $X,Y$ are arbitrary, the Killing equation isMetric compatibility is needed here, even though symmetry alone sufficed for the vector identity. In general the Lie derivative of a metric with nonmetricity contains a term ${\xi }^c{\nabla }_c{g}_{ab}$. For example, with the metric tensor $g=e^{2x}d{x}^2$, zero connection coefficients and $\xi ={\partial }_x$, the Lie derivative of a tensor field is $2e^{2x}$, whereas $2{\nabla }_x{\xi }_x=4e^{2x}$. The usual metric connection interpretation of the question supplies the required compatibility.

For the Riemann curvature tensor calculation, keep the original PDF's convention. With $R_{abcd}=g_{ae}{R}^e{}_{bcd}$ and $R_{abc}{}^d=g^{de}{R}_{abce}$, it meansThus the commutator on a covector has the opposite sign. Write $K_{abc}={\xi }_{a;bc}={\nabla }_c{\nabla }_b{\xi }_a$. Taking a covariant derivative of the Killing equation makes $K$ antisymmetric in its first two indices, while the Ricci identity givesCombining this with the first-two-index antisymmetry gives the three relationsAdd the first and third and subtract the second. Pair symmetry and the first Bianchi identity imply $R_{adbc}+R_{bdca}-R_{cdab}=-2R_{abcd}$. ConsequentlyThis is the second covariant derivative of a Killing vector in the paper's convention; its sign is tied to the declared curvature-sign convention in Killing derivative identities, not to raising a particular index.

Set $L_{ab}={\xi }_{a;b}$. Along any smooth curve with tangent vector $V$, the chain rule and the just-proved identity give the Killing transport equationsChoose a parallel-propagated orthonormal frame along the curve. These become a homogeneous linear system of ordinary differential equations for the components of $(\xi ,L)$. Uniqueness of its initial value problem implies that zero initial data remain zero. Any point in the same connected component as $P$ can be joined to $P$ by a piecewise smooth curve. Therefore zero value and zero first derivative at $P$ force the Killing field and its derivative to vanish throughout that component, with no completeness assumption. On a manifold with one connected component this is the requested global result. Connectedness cannot be dropped: on two disjoint flat spacetimes, take the zero field on one and a nonzero translation on the other.

On a pseudo-Riemannian manifold of dimension $n$ with one connected component, the map $\xi \mapsto ({\xi }_a(P),L_{ab}(P))$ is therefore injective. There are $n$ independent components of the value and $n(n-1)/2$ of its antisymmetric derivative. The dimension bound for Killing vector fields isIt is attained in flat space by ${\xi }_a=A_a+B_{ab}{x}^b$ with constant antisymmetric $B$, giving translations and infinitesimal rotations or Lorentz transformations. The four-dimensional maximum is 10. For a disconnected manifold the bound applies componentwise; there is no uniform global bound depending only on $n$ if the number of components is unrestricted.

A vector field along a curve $\gamma :[a,b]\to M$ is a smooth section of the pullback tangent bundle ${\gamma }^{\ast }TM$: it assigns $V(t)\in T_{\gamma (t)}M$ smoothly. It need not be the restriction of one ambient vector field, since its values at different times with the same image point may differ.

Given a Koszul connection, use a local frame $e_j$ of $TM$ and write $V(t)={\sum }_j{v}^j(t)e_j(\gamma (t))$. Its covariant derivative along a curve isThe first term differentiates the time-dependent coefficients, and the second uses the connection on the local frame. Under a change of frame, differentiating the change matrix supplies exactly the extra term in the connection transformation, so the expression is independent of the frame. This definition also works when $\dot{\gamma }(t)=0$. In coordinate components it isA field is parallel when $D_{t}V=0$.

Thus a parallel field solves the linear system $\dot{v}=-A(t)v$, with smooth coefficient matrix $A(t)$. Given its value at one time, existence and uniqueness for a linear ordinary differential equation determine it on each coordinate segment. To obtain it on all of $[a,b]$, cover the compact parameter interval by finitely many frame neighborhoods and subdivide so that each closed subinterval lies in one such neighborhood. The coefficients are bounded on each subinterval, and the standard estimate $\Vert v(t)\Vert \le \Vert v(t_0)\Vert \exp (\int \Vert A(s)\Vert ds)$ prevents finite-time blowup. The solutions match uniquely at the subdivision points. This proves existence and uniqueness of the requested global parallel field.

Define parallel translation by ${\tau }_t(v)=V_v(t)$, where $V_v(a)=v$ and $D_t{V}_v=0$. Linearity of the equation and uniqueness make ${\tau }_t$ linear. Solving the same equation backwards gives its inverse; equivalently, a solution zero at time $t$ must be identically zero. Therefore

A metric connection for a Riemannian metric satisfies $\nabla g=0$, orPulling this identity along the curve givesFor parallel fields the right side vanishes, provingFor a piecewise smooth path compose the parallel transport maps on its smooth segments. On a closed path their domain and codomain are the same tangent space, so the resulting map is orthogonal. Its determinant need not be positive on a nonorientable surface.

An explicit reflection example is the reflection holonomy along a closed projective geodesic. Give ${\mathbb{RP}}^2$ the round metric induced from the sphere by the antipodal quotient. The Levi-Civita connection descends because the antipodal map is an isometry, so the quotient map is a local isometry carrying geodesics and parallel fields to their counterparts. The great-circle arcprojects to a smooth closed geodesic: $c(t+\pi )=-c(t)$, including all derivatives under the quotient identification. Along it the fieldsare parallel. The first is the velocity of a great-circle geodesic. The second is tangent, unit length, and orthogonal to the first. Metric compatibility makes its covariant derivative orthogonal to both frame vectors: differentiate its constant length and its zero inner product with the parallel first vector. Its derivative therefore vanishes in the two-dimensional tangent plane. At $t=\pi$, $E_1(\pi )=-E_1(0)$ and $E_2(\pi )=E_2(0)$. Identifying the endpoint with the start through the derivative of the antipodal map, namely $-I$, fixes the returned $E_1$ and negates the returned $E_2$. HenceThis is a reflection fixing the geodesic direction.

Finally orient the unit sphere outward and traverse a spherical triangle with its interior on the left. Let its angles be $\alpha ,\beta ,\gamma$. Its parallel transport preserves both the metric and orientation, so is a rotation by some angle $\delta$ in the initial tangent plane. On each geodesic side the unit tangent is parallel. At a vertex with angle $\alpha$, the actual forward tangent turns left by $\pi -\alpha$ relative to the transported incoming tangent. Tracking the tangent around all three sides and making those three corner turns returns it to its initial direction.

Rotations of oriented two-dimensional tangent planes commute with orientation-preserving parallel transport. Consequently the total angle in this tracking procedure is the transport angle plus the three corner turns:This determines parallel transport around a spherical triangle asBy the spherical excess formula, this is rotation by the enclosed area on the unit sphere. Reversing traversal negates the angle. For an octant triangle all three angles are $\pi /2$, giving a positive $\pi /2$ rotation.

The Levi-Civita connection is the unique affine connection which is both a metric connection and a torsion-free connection. Uniqueness is encoded by the Koszul formulaIt follows by adding the metric-compatibility identities with derivative fields $X,Y$, subtracting the one with derivative field $Z$, and replacing differences of covariant derivatives by Lie brackets. Nondegeneracy of $g$ determines ${\nabla }_{X}Y$ uniquely from all its pairings with $Z$. The existence and uniqueness of the Levi-Civita connection supplies existence; no existence proof is needed here.

A connection on $TM$ is a metric connection for $g$ if for all smooth vector fields $X,Y,Z$,Equivalently, its induced covariant derivative of the metric satisfies $\nabla g=0$. This is also called a metric-compatible connection.

Take $\mathbf{n}$ to be the unit normal, so $P(v)=v-⟨v,\mathbf{n}⟩\mathbf{n}$ is orthogonal projection onto $TM$. With $D$ the Euclidean Levi-Civita connection, the proposed derivative isIt is a smooth tangent field, and independence of the extensions is given. Linearity follows from linearity of $D$ and $P$. Moreover, for $f\in C^{\infty }(M)$, take local extensions of $f,X,Y$ to obtainsince $PY=Y$. Thus it is an affine connection on $M$.

For tangent fields $Y,Z$, the normal component of $DY$ pairs to zero with $Z$. Differentiating the ambient inner product along $X$ givesHence it is a metric connection for the induced Riemannian metric. The ambient torsion vanishes, soHere the restriction of the ambient bracket is the intrinsic bracket and is tangent to $M$: derivations of functions restricted to $M$ give the same bracket, or this follows in submanifold coordinates. Thus the projected derivative is also a torsion-free connection. Uniqueness now provesThis is the projected ambient connection. The projection formula requires a unit normal. If a nonunit normal is used, its normal term must instead be $⟨DY,\mathbf{n}⟩\mathbf{n}/⟨\mathbf{n},\mathbf{n}⟩$. For example, on the unit circle, taking $\mathbf{n}=2p$ and differentiating its unit tangent along itself gives $DY=-p$; the unnormalized printed expression would be $3p$, which is not tangent. Normalizing the normal removes that ambiguity.

Use the Levi-Civita connection interpretation of the curvature convention. Its Riemann curvature tensor satisfiesThe first antisymmetry comes from the derivative commutator; the second comes from compatibility with the metric tensor. The cyclic identity is the first Bianchi identity, and together these imply exchange symmetry of the two antisymmetric index pairs. Contracting gives the symmetric Ricci tensor, $\boxed{R_{ab}=R_{ba}}$. The differential Bianchi identity is ${\nabla }_{[a}{R}_{bc]de}=0$.

There is a hypothesis qualification. A metric connection need not be a torsion-free connection. For a general such connection, the tensorial derivative commutator contains $-T^c{}_{ab}{\nabla }_c$ as well as curvature; pair exchange, the stated cyclic identity and Ricci symmetry need not hold. The printed pure-curvature commutator, and the scalar identity requested later, use the torsion-free interpretation. Under that intended interpretation all the boxed symmetries follow. They are not assertions about arbitrary metric-compatible connections with torsion tensor.

For example, with a constant Minkowski metric a connection having only ${\Gamma }^1{}_{02}=1$ and ${\Gamma }^2{}_{01}=-1$ nonzero is metric compatible, since its connection matrix in direction zero is a spatial rotation. It has $T^1{}_{02}=1$. For the scalar $f=x^1$ the tensorial Hessian commutator is $[{\nabla }_0,{\nabla }_2]f=-1$, rather than zero. This explicitly demonstrates the missing zero-torsion condition used in the subsequent scalar proof.

A vector field along a map $\gamma$ assigns $V(t)\in T_{\gamma (t)}M$ smoothly to each parameter value. Equivalently, it is a section of a vector bundle ${\gamma }^{\ast }TM$. The pullback connection defines its covariant derivative along a curve, $D_{t}V$. To make the definition explicit, use connection coefficients with the conventionFor $V=V^k(t){\partial }_k{|}_{\gamma (t)}$ the derivative isThe change of frame of a vector-bundle connection ensures this expression is independent of the coordinate chart. It remains meaningful when $\dot{\gamma }=0$ and does not require extending $V$ to one ambient vector field.

The field is parallel exactly when $\boxed{D_{t}V=0}$. A geodesic for $\nabla$, with its specified affine parameter, is a smooth curve whose own tangent is parallel:This definition applies to any affine connection, whether or not it is a metric connection. A non-affine change of parameter generally gives a pregeodesic instead of this parametrized geodesic.

An affine connection $\nabla$ is a metric connection for the Riemannian metric $g$ whenEquivalently the induced covariant derivative of the metric tensor satisfies $\nabla g=0$. This is metric compatibility; it imposes no condition on the torsion tensor. A Levi-Civita connection has the additional property of being a torsion-free connection.

We use the following existence and uniqueness statement for parallel transport: for a smooth affine connection, a smooth curve $\gamma :[0,1]\to M$, an initial time $t_0$ and a vector $v\in T_{\gamma (t_0)}M$, there is a unique smooth vector field along a map $V$ on the whole curve satisfying $D_{t}V=0$ and $V(t_0)=v$. In a frame of a vector bundle this is the linear ordinary differential equation $\dot{V}=-\Gamma (\dot{\gamma })V$. Its smooth coefficient matrix is bounded on every compact time subinterval in a trivializing neighborhood, and the linear existence theorem gives a solution throughout that subinterval. A finite subdivision of $[0,1]$ into such neighborhoods and uniqueness patch the solutions. No completeness hypothesis on $M$ is needed.

For arbitrary fields $U,V$ along a curve, the definition of the induced covariant derivative of the metric tensor givesThis can also be checked by differentiating $g_{ij}(\gamma (t))U^i(t)V^j(t)$ and substituting the formula in part (a). If $\nabla$ is a metric connection and $V$ is parallel, it follows that $d(g(V,V))/dt=0$. Positivity of the Riemannian metric makes $|V|=\sqrt{g(V,V)}$ constant. Thus parallel transport preserves lengths.

Conversely assume every parallel field along every smooth curve has constant length. Fix $p\in M$ and arbitrary $x,v\in T_{p}M$. Choose a curve through $p$ with velocity $x$ at an interior time $t_0$. Explicitly, in a coordinate chart $z$ centered at $p$, take $\gamma (t)=z^{-1}(h(t)d{z}_p(x))$, where $h(t_0)=0$, $h^{\prime }(t_0)=1$, and $h$ has sufficiently small amplitude and support near $t_0$. A smooth bump function produces such $h$, so the curve is defined on all of $[0,1]$ and remains inside the chart. By the stated parallel transport theorem there is a parallel field with $V(t_0)=v$. Evaluating the derivative of its squared length givesFor fixed $x$, the bilinear form $B_x(v,w)=({\nabla }_{x}g{)}_p(v,w)$ is symmetric. Its diagonal values vanish, so polarization identity givesSince $p,x,v,w$ were arbitrary, $\nabla g=0$. A connection is metric-compatible if and only if all parallel fields have constant length. Polarization also shows that its parallel transport preserves all inner products, not just lengths.

A Levi-Civita connection on a Riemannian manifold is a covariant derivative $D$ that is a torsion-free connection, $D_{X}Y-D_{Y}X=[X,Y]$, and a metric connection, $Xg(Y,Z)=g(D_{X}Y,Z)+g(Y,D_{X}Z)$. A covariant derivative is $C^{\infty }(M)$-linear in $X$, real linear in $Y$, and obeys $D_X(fY)=X(f)Y+f{D}_{X}Y$.

Combining metric compatibility for the cyclic triples and eliminating reversed derivatives with the torsion identity gives the Koszul formulaSince $g$ is nondegenerate, this determines $D_{X}Y$ uniquely. For existence, use the right-hand side to define $D_{X}Y$: expansion of the Lie bracket of vector fields shows it is $C^{\infty }(M)$-linear in $Z$, hence defines a smooth one-form, which the musical isomorphism converts to a smooth vector field. The same expansion shows $D_{fX}Y=f{D}_{X}Y$ and $D_X(fY)=X(f)Y+f{D}_{X}Y$. Subtracting the formula with $X,Y$ exchanged gives torsion zero; adding its versions for $Y,Z$ exchanged gives metric compatibility. Thus it is a Levi-Civita connection. Equivalently its coefficients in a manifold chart are the Christoffel symbolsThe intrinsically defined Koszul formula ensures these local expressions fit together. This proves the fundamental theorem of Riemannian geometry.

The product Riemannian metric on $M\times N$ isThe two factor tangent spaces are orthogonal, and the sum is positive definite. Lift $X$ from $M$ and $Y$ from $N$. We check $D_{X}Y$ against lifted local frame fields from both factors, which together span each product tangent space.

If $Z$ is lifted from $M$, then $g(Y,Z)=g(X,Y)=0$, while $g(Z,X)$ depends only on $m$, so $Yg(Z,X)=0$. Also $[Y,Z]=[X,Y]=0$ and $[Z,X]$ is horizontal, hence orthogonal to $Y$. Every term in the Koszul formula is zero. If $W$ is lifted from $N$, $g(W,X)=g(X,Y)=0$ and $g(Y,W)$ depends only on $n$, so $Xg(Y,W)=0$. Now $[W,X]=[X,Y]=0$ and $[Y,W]$ is vertical, hence orthogonal to $X$. Again every term is zero. Thus $D_{X}Y$ is orthogonal to both spanning frame families, and positive definiteness yieldsThis calculation uses the lifts' independence of the other factor coordinates; a vector field with varying coefficients in those coordinates can have a nonzero mixed derivative.

Let $E\to M$ have rank $r$. A vector bundle covariant derivative is a real-linear operatorPairing its one-form slot with a vector field $X$ gives ${\nabla }_{X}s$, with ${\nabla }_{fX}s=f{\nabla }_{X}s$ and ${\nabla }_X(fs)=X(f)s+f{\nabla }_{X}s$.

In the horizontal description, a connection on a vector bundle $A$ is a smooth splitting $T_{a}E=V_{a}E\oplus H_{a}E$, where $V_{a}E=\ker (d\pi {)}_a$, and the horizontal choice is compatible with fibre addition and scalar multiplication. This linearity is essential; an arbitrary nonlinear horizontal splitting is not a vector-bundle connection. Identify a vertical vector at $a\in E_x$ with the element of $E_x$ obtained by differentiating the curve $a+tv$. Let $K_A:T_{a}E\to E_x$ be vertical projection followed by this identification. The induced vector bundle covariant derivative isIn a local frame of a vector bundle $e=(e_1,\dots ,e_r)$, a linear horizontal splitting has the formfor a matrix-valued connection one-form $\Gamma$. Thus for $s=ev$, ${\nabla }^{A}s=e(dv+\Gamma v)$, which directly verifies the Leibniz rule.

Conversely, the Leibniz rule makes $\nabla$ local: if a section vanishes near a point, multiply it by a smooth cutoff function supported in that neighborhood and equal to one near the point to see that its derivative vanishes there too. Local sections can therefore be differentiated by any extension agreeing near the point. Given $\nabla$, define its local connection one-form by $\nabla e_b={\sum }_a{e}_a{\Gamma }_b^a$. The Leibniz rule gives $\nabla (ev)=e(dv+\Gamma v)$. Under $e^{\prime }=eg$, $v^{\prime }=g^{-1}v$, one hasThe tangent fibre coordinate transforms by $w^{\prime }=g^{-1}w-g^{-1}(dg)(\xi )g^{-1}v$, so $w=-\Gamma (\xi )v$ transforms precisely to $w^{\prime }=-{\Gamma }^{\prime }(\xi )v^{\prime }$. Consequently these local horizontal subspaces of a vector bundle connection glue to a smooth linear splitting inducing the given $\nabla$. Its local formula also shows uniqueness. This proves the horizontal connection associated to a covariant derivative construction.

A horizontal lift $\tilde{\gamma }$ of $\gamma$ is a smooth curve in $E$ with $\pi \circ \tilde{\gamma }=\gamma$ and $\dot{\tilde{\gamma }}(t)\in H_{\tilde{\gamma }(t)}E$. In a local frame write $\tilde{\gamma }(t)=e(\gamma (t))v(t)$. Horizontality is exactly the linear ordinary differential equationWe use existence and uniqueness for an ODE with smooth coefficients, and the continuation theorem that a solution extends across a finite interval when it remains bounded inside the coordinate domain. For a linear system $\dot{v}=-C(t)v$ with continuous coefficients bounded by $L$ on a compact time interval, the Gronwall inequality bounds $|v(t)|$ by its initial norm times $e^{L|t-t_0|}$, preventing finite-time fibre escape. Cover a compact subinterval of $I$ by finitely many bundle charts and partition it so each piece lies in one chart. Solve successively, changing fibre frames at the partition points. Coordinate invariance of the horizontal equation and uniqueness make the pieces agree. Exhausting $I$ by compact intervals containing $0$ produces a unique horizontal lift on all of $I$ for every prescribed $a_0$. This is global continuation of linear parallel transport.

The lift equation is linear in its initial value. Hence the endpoint map $P_{\gamma }:E_{\gamma (0)}\to E_{\gamma (1)}$ is linear. Solving the same equation backwards, or transporting along $t\mapsto \gamma (1-t)$, gives its inverse, soThis is parallel transport. For two parallel lifts $s_1(t),s_2(t)$, the assumed metric connection identity extends along the embedded curve and becomesHere $D_t$ is the covariant derivative along a curve. The permitted extension of sections and the tangent vector along the embedded curve justifies applying the bundle identity; equivalently it follows in each local frame. Thereforeso parallel transport is an isometry. This is the principle that parallel transport preserves a fibre metric.

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.

The Levi-Civita connection is also a metric connection, soand similarly for the two cyclic permutations. Add the identities with leading derivatives $u$ and $v$, subtract the one with leading derivative $w$, and useCancellation and symmetry of $g$ giveThis is the Koszul formula written with the bracket terms on the right.

For a Riemannian metric on an ambient manifold and an embedded submanifold, let $\pi :TN{|}_M\to TM$ be orthogonal projection. Project the restriction of a connection to an embedded submanifold to obtain $D_{X}Y=\pi {\nabla }_{X}Y$. Since $\pi Y=Y$, the Leibniz rule and linearity over smooth functions hold, making $D$ an affine connection. If the ambient connection is the Levi-Civita connection, pairing with tangent vectors shows that $D$ is a metric connection; projecting the zero-torsion identity shows it is a torsion-free connection. By uniqueness, $D$ is the induced Levi-Civita connection.

Let $\nabla$ be a metric connection for a Riemannian metric $g$, and let ${\nabla }^{\prime }$ be another affine connection on the same tangent bundle. The difference of affine connections is a tensor $A=\nabla -{\nabla }^{\prime }$. Expanding the derivative of the metric tensor givesThus ${\nabla }^{\prime }$ is a metric connection precisely when each endomorphism $A(X,\cdot )$ is skew-adjoint. This imposes no antisymmetry in $X,Y$; that separate condition characterizes agreement of parametrized geodesics. If both conditions hold, $g(A(X,Y),Z)$ is alternating in all three arguments.