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For a smooth map $\varphi :M^{\prime }\to M$ and a connection on a vector bundle $E\to M$, the curvature form of a connection on the pullback vector bundle satisfiesas an endomorphism of $E_{\varphi (p)}$. In a frame of a vector bundle written as a row, with section components written as columns, the connection one-form is $\Omega$ and its curvature form of a connection is $d\Omega +\Omega \wedge \Omega$. The pullback connection has matrix ${\varphi }^{\ast }\Omega$; commutation of the pullback of a differential form with the exterior derivative and wedge product of differential forms proves the identity. No pushforward to an ambient vector field is required: $d\varphi (X)$ is a vector field along a map.

The expression $d\varphi (X)$ is in general a vector field along a map, rather than a vector field on $M$. Its intended meaning in the required formula is pointwise evaluation of the vector-bundle-valued differential form $\nabla s$:using the canonical identification $({\varphi }^{\ast }E{)}_p=E_{\varphi (p)}$. This interpretation works for every smooth map, including a constant map or a map with self-intersections.

Choose a frame of a vector bundle over $U$, written as a row $e=(e_1,\dots ,e_r)$. Define its connection one-form $\Omega$ byfor component columns $u$. The pulled-back frame $e^{\varphi }=({\varphi }^{\ast }{e}_1,\dots ,{\varphi }^{\ast }{e}_r)$ is a frame of a vector bundle over ${\varphi }^{-1}U$. For any component column $v$ of smooth functions on that open set, defineHere the matrix entries of ${\varphi }^{\ast }\Omega$ are ordinary pullbacks of a differential form. The displayed operator is linear and satisfies the Leibniz rule, hence is locally a connection on a vector bundle.

To prove these local operators glue, change the original frame of a vector bundle to $\tilde{e}=eh$. Expanding $\nabla (ehu)$ gives the change of frame of a vector-bundle connectionThe pullback of a differential form and the chain rule giveThis is precisely the required transformation law for the pulled-back frame ${\tilde{e}}^{\varphi }=e^{\varphi }(h\circ \varphi )$. Thus the definitions agree and give a global pullback connection.

For $s=eu$, substitute $v=u\circ \varphi$ in the defining formula and use $d(u\circ \varphi )(X)=du(d\varphi (X))$. This proves the required identity. Conversely that identity fixes ${\nabla }_X^{\prime }({\varphi }^{\ast }{e}_a)$; the Leibniz rule then fixes the derivative of every ${\sum }_a{v}^a{\varphi }^{\ast }{e}_a$, proving uniqueness. If the identity is initially imposed only on global sections of a vector bundle, multiply each local frame section by a smooth bump function equal to one near the point in question and supported in $U$, and extend by zero. These global sections agree with the frame locally, so the same uniqueness argument applies. The pullback connection exists and is unique.

The curvature form of a connection is the endomorphism-valued differential two-form defined byIt is alternating in $X,Y$. To see its tensoriality, replacing $X$ by $fX$ produces an extra $-Y(f){\nabla }_{X}s$ from the second term and an extra $+Y(f){\nabla }_{X}s$ from $[fX,Y]=f[X,Y]-Y(f)X$, so these cancel. Alternation gives linearity over smooth functions in $Y$ too. Replacing $s$ by $fs$ produces the additional coefficient $(X(Yf)-Y(Xf)-[X,Y]f)s=0$; all terms involving one derivative of $s$ also cancel. Thus $R^{\nabla }$ belongs to ${\Omega }^2(M;\mathrm{End}E)$.

In the coefficient-column convention of part (b), expanding the definition givesThe first three terms are $d\Omega (X,Y)$, and the last two are $(\Omega \wedge \Omega )(X,Y)$. This proves the Cartan curvature matrix equation $F=d\Omega +\Omega \wedge \Omega$, with matrix multiplication combined with the wedge product of differential forms.

The pullback connection has matrix ${\Omega }^{\prime }={\varphi }^{\ast }\Omega$. Since the pullback of a differential form commutes with the exterior derivative and the wedge product of differential forms, its curvature form of a connection has matrixChanges of frame of a vector bundle conjugate both sides by the same pulled-back transition matrix, so this identity is intrinsic. In fibre notation the answer isBoth sides act on the same fibre $E_{\varphi (p)}$. No claim that $d\varphi$ preserves Lie brackets of vector fields for an arbitrary map is needed.

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.

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

An affine connection, also called a Koszul connection, induces a pullback connection on this bundle. In coordinates, write $V=V^i(t){\partial }_i$ and ${\nabla }_{{\partial }_j}{\partial }_k={\Gamma }^i{}_{jk}{\partial }_i$. Its covariant derivative along a curve isThe connection transformation law, or the Leibniz rule applied to a change of frame, makes this expression independent of the frame. In particular it is defined even when $\dot{\gamma }=0$; it differentiates the section's coefficients as well as the frame.

The field is parallel precisely when $D_{t}V=0$. In a frame along a coordinate segment this is the linear ordinary differential equationThe Picard-Lindelöf theorem gives a unique solution with prescribed initial value. Smooth coefficients on compact subintervals are bounded, and the usual norm estimate for a linear ordinary differential equation prevents finite-time blowup there. A finite sequence of coordinate segments covers the image of the compact interval $[a,b]$; solve successively and use uniqueness on overlaps. This gives a unique parallel field on the whole interval for every $V_a\in T_{\gamma (a)}M$.

Define parallel transport by ${\tau }_t(V_a)=V(t)$. Linearity of the equation and uniqueness give ${\tau }_t(c{V}_a+d{W}_a)=c{\tau }_t(V_a)+d{\tau }_t(W_a)$. Solving along the reversed curve supplies its inverse. Therefore each ${\tau }_t:T_{\gamma (a)}M\to T_{\gamma (t)}M$ is a linear isomorphism.

For two affine connections, put $\Delta (X,Y)={\nabla }_{X}Y-{\tilde{\nabla }}_{X}Y$. Their rules givesince the two $X(f)Y$ terms cancel. The tensoriality of this operation proves that the difference of affine connections is a tensor, of type $(1,2)$, with smooth coordinate coefficients ${\Gamma }^i{}_{jk}-{\tilde{\Gamma }}^i{}_{jk}$.

A parametrized geodesic satisfies $D_t\dot{\gamma }=0$, or the geodesic equationWith $v^i={\dot{\gamma }}^i$, this becomes the first-order system ${\dot{\gamma }}^i=v^i$, ${\dot{v}}^i=-{\Gamma }^i{}_{jk}(\gamma )v^j{v}^k$. Its right side is smooth, so the Picard-Lindelöf theorem gives a unique local solution for each initial point and tangent vector. This establishes uniqueness with the parametrization fixed.

The two accelerations differ byIf $\Delta (v,v)=0$ for every tangent vector, either acceleration vanishes exactly when the other does. Conversely, start the $\nabla$-geodesic with arbitrary initial vector $v$ at an arbitrary point. If it is also a $\tilde{\nabla }$-geodesic with the same parameter, evaluating this identity initially gives $\Delta (v,v)=0$. HencePolarization makes the latter condition equivalent to $\Delta (u,v)+\Delta (v,u)=0$: parametrized geodesics determine the symmetric part of an affine connection. In particular torsion-free connections are determined by their parametrized geodesics, since their difference is also symmetric and must then vanish.

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.

Let $p_i:M_1\times M_2\to M_i$ be the projections. The derivative of the two projections identifies the tangent bundle withTake the direct sum of the two pullback connections. This is the product affine connection. In product coordinates $(x^a,y^{\alpha })$, its only nonzero Christoffel symbols are the two blocks inherited from the factors; all mixed coefficients are zero.

To make the connection on arbitrary fields explicit, write $V=V^a{\partial }_{x^a}+V^{\alpha }{\partial }_{y^{\alpha }}$ and $W=W^a{\partial }_{x^a}+W^{\alpha }{\partial }_{y^{\alpha }}$, allowing all coefficients to depend on both variables. ThenThe terms $V(W^a)$ and $V(W^{\alpha })$ include derivatives in both factors. Omitting these cross derivatives for fields with variable coefficients would not define a connection.

The formula is real-bilinear, linear over smooth functions in $V$, and obeys the Leibniz rule in $W$. Under a change of product coordinates each block has the factor's connection transformation law, so the local formulas agree; equivalently the pullback connection construction already guarantees this gluing.

A field $X_i$ on one factor has a canonical lift to the product. Opposite-factor lifts commute, and their mixed covariant derivatives vanish. For such lifted fields the coefficient derivatives reduce to their factor derivatives, givingThese values determine the connection uniquely: lifted coordinate frames span locally, and the connection's product rules determine its action on all their smooth linear combinations.

A covariant derivative is an $\mathbb{R}$-linear mapsatisfying $d_A(fs)=df\otimes s+f{d}_{A}s$. It is local: its value over an open set depends only on the restriction of the section there. A horizontal connection differentiates a section, projects its derivative vertically, and identifies the vertical tangent with the fibre.

In a local frame write $d_{A}s=ds+A_{i}sd{y}^i$. If $x^a$ are coordinates on $M$, the defining identity in the question forceswithThus $A^{\prime }=f^{\ast }A$, the pullback connection.

Connections ${\nabla }^1,{\nabla }^2$ on $T{M}_1,T{M}_2$ give a connection onby taking the direct sum of their pullback connections. In product coordinates its Christoffel symbols have the two factor blocks and zero mixed blocks. On arbitrary fields, the ordinary derivatives of their coefficients are still taken in both factors. On fields lifted separately from the factors it satisfies ${\nabla }_{Y_1+Y_2}(X_1+X_2)={\nabla }_{Y_1}^1{X}_1+{\nabla }_{Y_2}^2{X}_2$.

For an embedded submanifold $i:M\hookrightarrow N$, the ambient affine connection induces a pullback connection on $i^{\ast }TN=TN{|}_M$. Extend a local section off $M$ and differentiate in a direction tangent to $M$. If two extensions agree on $M$, their difference has zero coefficients on $M$, whose tangential derivatives are zero. Thus the restricted derivative is independent of the extension. Its value may have a normal component, so obtaining a connection on $TM$ requires a further projection.