[Tu Differential Geometry] 22. Connections and Curvature Again

 This article is one of Manifold, Differential Geometry, Fibre Bundle.

22.1 Connection and Curvature Matrices Under a Change of Frame

In Section 10.6, we showed that a connection on a vector bundle $E\rightarrow M$ may be restricted to a connection over any open subset $U$ of $M$: $$\nabla^U:\mathfrak{X}(U)\times \Gamma(U,E)\rightarrow \Gamma(U,E).$$ We will usually omit the superscript $U$ and write $\nabla$ for $\nabla^U$.

Suppose there is a frame $e_1,\cdots,e_r$ for $E$ over $U$. Then the connection matrix $\omega$ of $\nabla$ relative to the frame $e_1,\cdots,e_r$ over $U$ is the matrix $[\omega^i_j]$ of $1$-forms defined by $$\nabla_Xe_j=\sum \omega^i_j(X)e_i,$$ for $X\in \mathfrak{X}(U)$. If we write the frame $e_1,\cdots,e_r$ as a row vector $e=[e_1,\cdots,e_r]$, then in matrix notation $$\nabla_Xe=e\omega(X).$$ As a function of $X$, $$\nabla e=e\omega.$$

Similarly, for $C^\infty$ vector fields $X,Y$ on $U$, the curvature matrix $\Omega$ of the connection $\nabla$ is the matrix $[\Omega^i_j]$ of $2$-forms defined by $$R(X,Y)e_j=\sum \Omega^i_j(X,Y)e_i.$$

We next study how the connection and curvature matrices $\omega$ and $\Omega$ transform under a change of frame. Suppose $\bar{e}_1,\cdots,\bar{e}_r$ is another frame for $E$ over $U$. Let $\bar{\omega}=[\bar{\omega}^i_j]$ and $\bar{\Omega}=[\bar{\Omega}^i_j]$ be the connection and curvature matrices of the connection $\nabla$ relative to this new frame. At each point $p$, the basis vector $\bar{e}_l(p)$ is a linear combination of $e_1(p),\cdots,e_r(p)$: $$\bar{e}_l(p)=\sum a^k_l(p)e_k(p).$$ As sections on $U$, $$\bar{e}_l=\sum a^k_le_k.$$ So we get a matrix of functions $a=[a^k_l]$. As the coefficients of a smooth section with respect to a smooth frame, the $a^k_l$ are smooth functions on $U$. At each point $p$, the matrix $a(p)=[a^k_l(p)]$ is invertible because it is a change of basis matrix. Thus we can think of $a$ as a $C^\infty$ function $U\rightarrow GL(r,\mathbb{R})$. As row vectors, $$[\bar{e}_1\cdots\bar{e}_r]=[e_1\cdots e_r][a^k_l],$$ or in matrix notation, $$\begin{align} \bar{e}=ea.\end{align}$$

Since in matrix notation (1), the matrix of functions $a^i_j$ must be written on the right of the row vector of sections $e_i$, we will recast the Leiniz rule in this form. Suppose $s$ is a $C^\infty$ section of the vector bundle $E$ and $f$ is a $C^\infty$ function on $M$. By the Leibniz rule, for any $C^\infty$ vector field $X\in \mathfrak{X}(M)$, $$\begin{align}\nabla_X(sf)=(\nabla_Xs)f+sX(f)=(\nabla_Xs)f+sdf(X).\end{align}$$ We may view $\nabla_s$ as a function $\mathfrak{X}(M)\rightarrow \Gamma(E)$ with $(\nabla s)(X)=\nabla_Xs$. Then we may suppress the argument $X$ from (2) and rewrite the Leibniz rule as $$\begin{align}\nabla(sf)=(\nabla s)f+sdf\quad\mbox{for }s\in\Gamma(E),\ f\in C^\infty(M).\end{align}$$

Theorem 22.1. Suppose $e$ and $\bar{e}$ are two frames for the vector bundle $E$ over $U$ such that $\bar{e}=ea$ for some $a:U\rightarrow GL(r,\mathbb{R})$. If $\omega$ and $\bar{\omega}$ are the connection matrices and $\Omega$ and $\bar{\Omega}$ are the curvature matrices of a connection $\nabla$ relative to the two frames, then

(i) $\bar{\omega}=a^{-1}\omega a+a^{-1}da,$

(ii) $\bar{\Omega}=a^{-1}\Omega a$.

Proof. (i) We use the Leibniz rule (3) to derive the transformation rule for the connection matrix under a change of frame. Since $\bar{e}=ea$, we have $e=\bar{e}a^{-1}$. Recall that if $a=[a^i_j]$ is a matrix of $C^\infty$ functions, then $da$ is the matrix of $C^\infty$ $1$-forms $[da^i_j]$. Thus, $$\nabla \bar{e}=\nabla (ea)=(\nabla e)a+eda=(e\omega)a+eda=\bar{e}a^{-1}\omega a+\bar{e}a^{-1}da=\bar{e}(a^{-1}\omega a+a^{-1}da).$$ Therefore, $$\bar{\omega}=a^{-1}\omega a+a^{-1}da.$$

(ii) Since the structural equation (Theorem 11.1) $$\begin{align} \bar{\Omega}=d\bar{\Omega}+\bar{\omega}\wedge\bar{\omega}\end{align}$$ express the curvature $\bar{\Omega}$ in terms of the connection matrix $\omega$, the brute-force way to obtain the transformation rule for $\bar{\Omega}$ is to plug the formula in part (i) into (4), expand the terms, and try to write the answer in terms of $\Omega$. (Good!)

Another way is note that since for any $X_p,Y_p\in T_pM$, the curvature $R(X_p,Y_p)$ is the linear transformation of $E_p$ to $E_p$ with matrix $[\Omega^i_j(X_p,Y_p)]$ relative to the basis $e_1,\cdots,e_r$ at $p$, a change of basis should lead to a conjugate matrix. Indeed, for $X,Y\in \mathfrak{X}(U)$, $$R(X,Y)e_j=\sum_i \Omega^i_j (X,Y)e_i,$$ or in matrix notation, $$R(X,Y)e=R(X,Y)[e_1,\cdots,e_r]=[e_1,\cdots,e_r][\Omega^i_j(X,Y)]=e\Omega (X,Y).$$ Suppressing $X,Y$, we get $$R(e)=e\Omega.$$ Hence, $$R(\bar{e})=R(ea)=R(e)a=e\Omega(a)=\bar{e}a^{-1}\Omega a.$$ This proves that $\bar{\Omega}=a^{-1}\Omega a$.

22.2 Bianchi Identities

When $E\rightarrow M$ is the tangent bundle, a connection on $E$ is an affine connection as defined in Section 6.1. For an affine connection $\nabla$, the torsion $$T(X,Y)=\nabla_XY-\nabla_YX-[X,Y]$$ is defined and is a tensor. If $e_1,\cdots,e_n$ is a frame for the tangent bundle $TM$ over $U$, then the torsion forms $\tau^i$ are given by $$T(X,Y)=\sum \tau^i(X,Y)e_i.$$ We collect $\tau^1,\cdots,\tau^n$ into a column vector $\tau=[\tau^i]$ of $2$-forms. Similarly, if $\theta^1,\cdots,\theta^n$ are the $1$-forms dual to the basis $e_1,\cdots,e_n$, then $\theta$ is the column vector $[\theta^i]$. Recall the two structural equations (Theorem 11.7): $$(1)\quad \tau=d\theta+\omega\wedge\theta,$$ $$(2)\quad \Omega=d\omega+\omega\wedge\omega.$$

Proposition 22.2 (First Bianchi identity). Let $\nabla$ be a connection on the tangent bundle $TM$ of a manifold $M$. Suppose $\theta$ and $\tau$ are the column vectors of dual $1$-forms and torsion forms, respectively, on a framed open set, and $\omega$ and $\Omega$ are the connection and curvature matrices respectively. Then $$d\tau=\Omega\wedge\theta-\omega\wedge\tau.$$

Proof. Differentiation the first structural equation gives $$d\tau=d(d\theta)+d\omega\wedge\theta-\omega\wedge d\theta=0+(\Omega-\omega\wedge\omega)\wedge\theta-\omega\wedge(\tau-\omega\wedge\theta)=\Omega\wedge\theta-\omega\wedge\tau.$$

Like the second structural equation, the next two identities apply more generally to a connection $\nabla$ on any smooth vector bundle, not just the tangent bundle.

Proposition 22.3 (Second Bianchi identity). Let $\nabla$ be a connection on a smooth vector bundle $E$. Suppose $\omega$ and $\Omega$ are the connection and curvature matrices of $\nabla$ relative to a frame for $E$ over an open set. Then $$d\Omega=\Omega\wedge\omega-\omega\wedge\Omega.$$

Proof. Differentiating the second structural equation gives $$d\Omega=d(d\omega)+(d\omega)\wedge\omega-\omega\wedge d\omega=0+(\Omega-\omega\wedge\omega)\wedge\omega-\omega\wedge(\Omega-\omega\wedge\omega)=\Omega\wedge\omega-\omega\wedge\Omega.$$

We will use the notation $\Omega^k$ to mean the wedge product $\Omega\wedge\cdots\wedge\Omega$ of the curvature matrix $k$ times.

Propsotion 22.4 (Generalized second Bianchi identity). Under the same hypotheses as in Proposition 22.3, for any integer $k\ge 1$, $$d(\Omega^k)=\Omega^k\wedge\omega-\omega\wedge\Omega^k.$$

Proof.

22.3 The First Bianchi Identity in Vector Form

For a Riemannian connection, the torsion form $\tau$ is zero and the first Bianchi identity (Proposition 22.2) simplifies to $$\Omega\wedge\theta=0.$$ This identity translates into a symmetry property of the curvature tensor.

Theorem 22.5. If $X,Y,Z$ are $C^\infty$ vector fields on a Riemannian manifold $M$, then the curvature tensor of the Riemannian connection satisfies $$R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0.$$

Proof. Let $e_1,\cdots,e_n$ be a frame over an open set $U$ in $M$, and $\theta^1,\cdots,\theta^n$ the dual $1$-forms. On $U$, $Z=\sum_j\theta^j(Z)e_j$. In the notation of the preceding section, $$R(X,Y)Z=\sum_jR(X,Y)\theta^j(Z)e_j=\sum_{i,j}\Omega^i_j(X,Y)\theta^j(Z)e_i.$$

The $i$th component of $\Omega\wedge\theta$ is $\sum_j\Omega^i_j\wedge\theta^j$. Thus, for each $i$, the first Bianchi identity gives $$0=\sum_j(\Omega^i_j\wedge\theta^j)(X,Y,Z)=\sum_j\Omega^i_j(X,Y)\theta^j(Z)-\Omega^i_j(X,Z)\theta^j(Y)+\Omega^i_j(Y,Z)\theta^j(X).$$ Multiplying by $e_i$ and summing over $i$, we obtain $$0=R(X,Y)Z+R(Z,X)Y+R(Y,Z)X.$$

22.4 Symmetry Properties of the Curvature Tensor

With the first Bianchi identity, we have now proven three symmetry properties of the curvature tensor. Let $X,Y,Z,W$ be $C^\infty$ vector fields on a Riemannian manifold. Then the curvature tensor of the Riemannian connection satisfies

(i) (skew-symmetry in $X$ and $Y$) $R(X,Y)=-R(Y,X)$.

(ii) (skew-symmetry in $Z$ and $W$, Proposition 12.5) $$\langle R(X,Y)Z,W\rangle=-\langle R(X,Y)W,Z\rangle.$$

(iii) (first Bianchi identity) $$R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0.$$

As an algebraic consequence of these three properties, a fourth symmetry property follows.

Theorem 22.6. If $X,Y,Z,W$ are $C^\infty$ vector fields on a Riemannian manifold, then the curvature tensor of the Riemannian connection satisfies $$\langle R(X,Y)Z,W\rangle = \langle R(Z,W)X,Y\rangle.$$

Proof. By the first Bianchi identity, $$\begin{align}\langle R(X,Y)Z,W\rangle + \langle R(Y,Z)X,W\rangle+\langle R(Z,X)Y,W\rangle=0.\end{align}$$ To cancel out the term $\langle R(Y,Z)X,W\rangle$ in (5), we add to (5) the first Bianchi identity starting with the term $\langle R(Y,Z)W,X\rangle$: $$\begin{align}\langle R(Y,Z)W,X\rangle + \langle R(X,W)Y,X\rangle+\langle R(W,Y)Z,W\rangle=0.\end{align}$$ Similarly, to cancel out the term $\langle R(Z,X)Y,W\rangle$ in (5), we add $$\begin{align}\langle R(Z,X)W,Y\rangle+\langle R(X,W)Z,Y\rangle + \langle R(W,Z)X,Y\rangle=0.\end{align}$$ Finally, to cancel out $\langle R(W,Y)Z,X\rangle$ in (6), we add $$\begin{align}\langle R(W,Y)X,Z\rangle + \langle R(Y,X)W,Z\rangle+\langle R(X,W)Y,Z\rangle=0.\end{align}$$

Adding up the four equations (5), (6), (7), (8) and making use of the skew-symmetry properties, we get $$2\langle R(X,Y)Z,W\rangle -2\langle R(Z,W)X,Y\rangle =0,$$ which proves the theorem.

22.5 Covariant Derivative of Tensor Fields

A connection $\nabla$ on the tangent bundle $TM$ of a manifold $M$ induces a covariant derivative on all tensor fields. This is the content of the following proposition.

Proposition 22.7. Let $\nabla$ be a connection and $X,Y$ $C^\infty$ vector fields on the manifold $M$.

(i) If $\omega$ is a $C^\infty$ $1$-form on $M$, then $\nabla_X\omega$ defined by $$(\nabla_X\omega)(Y):=X(\omega (Y))-\omega(\nabla_XY)$$ is a $C^\infty$ $1$-form on $M$.

(ii)If $T$ is a $C^\infty$ $(a,b)$-tensor field on $M$, then $\nabla_XT$ defined by $$(\nabla_XT)(\omega_1,\cdots,\omega_a,Y_1,\cdots,Y_b):=X(T(\omega_1,\cdots,\omega_a,Y_1,\cdots,Y_b))\\ -\sum^a_{i=1}T(\omega_1,\cdots,\nabla_X\omega_i,\cdots,\omega_a,Y_1,\cdots,Y_b)-\sum^b_{j=1}T(\omega_1,\cdots,\omega_a,Y_1,\cdots,\nabla_XY_j,\cdots,Y_b),$$ for $\omega_i\in \Omega^1(M)$, $Y_j\in \mathfrak{X}(M)$, is a $C^\infty$ $(a,b)$-tensor field on $M$.

Proof. By the tensor criterion (Proposition 21.11), it suffices to check that $\nabla_X\omega$ and $\nabla_XT$ are $\mathcal{F}$-linear in its arguments, where $\mathcal{F}:=C^\infty(M)$ is the ring of $C^\infty$ functions on $M$. Let $f\in\mathcal{F}$.

(i) By definition and the Leibniz rule in $Y$ of a connection, $$(\nabla_X\omega)(fY)=X(f\omega(Y))-\omega(\nabla_XfY)=(Xf)\omega(Y)+fX(\omega(Y))-\omega((Xf)Y)-f\omega(\nabla_XY)=f(\nabla_X\omega)(Y).$$

(ii) Replace $\omega_i$ by $f\omega_i$ in the definition of $\nabla_XT$. On the right-hand side, we have $$\begin{align}X(T(\omega_1,\cdots,f\omega_i,\cdots,\omega_a,Y_1,\cdots,Y_b))=(Xf)T(\omega_1,\cdots,\omega_a,Y_1,\cdots,Y_b+fX(T(\omega_1,\cdots,\omega_a,Y_1,\cdots,Y_b))\end{align}$$ and $$\begin{align}T(\omega_1,\cdots,\nabla_X f\omega_i,\cdots,\omega_a,Y_1,\cdots,Y_b)=T(\omega_1,\cdots,(Xf)\omega_i,\cdots,\omega_a,Y_1,\cdots,Y_b)+fT(\omega_1,\cdots,\nabla_X\omega_i,\omega_a,Y_1,\cdots,Y_b).\end{align}$$ The first terms of these two expressions cancel out. All the other terms on the right of (9) and (10) are $\mathcal{F}$-linear in the $i$th argument.

Similarly, the right-hand side of the definition of $\nabla_XT$ is $\mathcal{F}$-linear in the argument $Y_j$.

On $C^\infty$ functions $f$, we defined $\nabla_Xf=Xf$.

Let $T^{a,b}(M)=\left( \bigotimes^aTM\right)\otimes \left( \bigotimes^b T^*M\right)$ be the bundle of $(a,b)$-tensors on $M$.

Theorem 22.8. Let $\nabla$ be a connection on a manifold and $X\in\mathfrak{X}(M)$. The covariant derivative $\nabla_X:\Gamma\left( \oplus^\infty_{a,b=0}T^{a,b}(M)\right)\rightarrow \Gamma\left( \oplus^\infty_{a,b=0}T^{a,b}(M)\right)$ satisfies the product rule for any two tensor fields $T_1,T_2$: $$\nabla_X(T_1\otimes T_2)=(\nabla_XT_1)\otimes T_2+T_1\otimes \nabla_XT_2.$$

Proof.

22.6 The Second Bianchi Identity in Vector Form

If $\nabla$ is a connection and $R(X,Y)$ is its curvature endomorphism on a manifold, then we define the Riemann curvature tensor or simply the curvature tensor to be $$ \mbox{Rm}(X,Y,Z,W)=\langle R(X,Y)Z,W\rangle,\quad X,Y,Z,W\in\mathfrak{X}(M).$$ The second Bianchi identity is obtained by differentiating the curvature. In vector form, this means taking the covariant derivative of the curvature tensor $\mbox{Rm}$.

Theorem 22.9 (The second Bianchi identity in vector form). On a Riemannian manifold, if a connection $\nabla$ is compatible with the metric, then for all $X,Y,Z,V,W\in\mathfrak{X}(M)$, $$\sum_{cyclic\ in\ X,Y,Z} (\nabla_X\mbox{Rm})(Y,Z,V,W)=0.$$ (This notation means we cyclically rotate $X,Y,Z$ to obtain two other terms.)

Proof. Since $(\nabla_X\mbox{Rm})(Y,Z,V,W)$ is $\mathcal{F}$-linear in all of its argument, it suffices to verify the identity at a point $p\in M$ and for $X,Y,Z,V,W$ a frame of $C^\infty$ vector fields near $p$. Such a frame is the coordinate frame $\partial_1,\cdots,\partial_n$ of a normal coordinate neighborhood $U$ of $p$. The advantage of using a normal coordinate system if that $\nabla_XY=0$ at $p$ and $[X,Y]=0$ on $U$ for any coordinate vector fields $X,Y$, i.e., $X=\partial_i$, $Y=\partial_j$ for some $i,j$ (Theorem 15.4). This greatly simplifies the expression for the covariant derivative of a tensor field.

Thus, if $X,Y,Z,V,W$ are coordinate vector fields on a normal neighborhood of $p$, then at the point $p$ $$(\nabla_X\mbox{Rm})(Y,Z,V,W)=X(\mbox{Rm}(Y,Z,V,W))-\mbox{Rm}(\nabla_XY,Z,V,W)-\cdots =X(\mbox{Rm}(Y,Z,V,W))\\ =X\langle R(Y,Z)V,W\rangle =\langle \nabla_XR(Y,Z)V,W\rangle+\langle R(Y,Z)V,\nabla_XW\rangle=\langle \nabla_X\nabla_Y\nabla_ZV,W\rangle -\langle \nabla_X\nabla_Z\nabla_Y V, W\rangle .$$ Cyclically permuting $X,Y,Z$, we get $$(\nabla_Y\mbox{Rm})(Z,X,V,W)=\langle \nabla_Y\nabla_Z\nabla_XV,W\rangle-\langle \nabla_Y\nabla_X\nabla_ZV,W\rangle$$ and $$(\nabla_Z\mbox{Rm})(X,Y,Z,V,W)=\langle \nabla_Z\nabla_X\nabla_YV,W\rangle-\langle \nabla_Z\nabla_Y\nabla_XV,W\rangle.$$ Summing the three equations gives $$(\nabla_X\mbox{Rm})(Y,Z,V,W+(\nabla_Y\mbox{Rm})(Z,X,V,W)+(\nabla_Z\mbox{Rm})(X,Y,V,W)\\ =\langle R(X,Y)\nabla_ZV,W\rangle+\langle R(Y,Z)\nabla_XV,W\rangle+\langle R(Z,X)\nabla_YV,W\rangle.$$ On the right-hand side, because $R(-,-)-$ is a tensor, we can evaluate the arguments at a single point $p$. Since $\nabla_XV$, $\nabla_YV$, $\nabla_ZV$ all vanish at $p$, all three terms are zero. This establishes the second Bianchi identity in vector form.

22.7 Ricci Curvature

The curvature tensor $\langle R(X,Y)Z,W\rangle$ is a complicated object, but from it one can construct other invariants of a Riemannian manifold $M$. One such is the sectional curvature introduced in Section 12.4.

Another is the Ricci curvature, which associates to two tangent vectors $n,v\in T_pM$ at $p$ the trace of the linear endomorphism $$w\mapsto R(w,u)v.$$ Thus, if $X,Y$ are $C^\infty$ vector fields on an open set $U$ in $M$ and $E_1,\cdots,E_n$ is an orthonormal frame on $U$, then $$\mbox{Ric}(X,Y)=\mbox{tr}(R(-,X)Y)=\sum^n_{i=1}\langle R(E_i,X)Y,E_i\rangle.$$ By the symmetry properties of $R$, for the Riemannian connection $$\mbox{Ric}(X,Y)=\sum \langle R(E_i,X)Y,E_i\rangle =\sum\langle R(Y,E_i)E_i,X\rangle =\sum \langle R(E_i,Y)X,E_i\rangle=\mbox{Ric}(Y,X).$$ Thus, the Ricci curvature $\mbox{Ric}(X,Y)$ is a symmetric tensor of type $(0,2)$.

22.8 Scalar Curvature

At each point $p$ of a Riemannian manifold $M$, the Ricci curvature is a symmetric bilinear form $$\mbox{Ric}:T_pM\times T_pM\rightarrow T_pM.$$ The scalar curvature at $p$ is the trace of the Ricci curvature at $p$, defined as follows.

By the nondegeneracy of the Riemannian metric $\langle\ ,\ \rangle$, there is a unique linear map $\rho:T_pM\rightarrow T_pM$ such that $$\mbox{Ric} (u,v)=\langle \rho(u),v\rangle\quad \mbox{for all } u,v\in T_pM.$$ The scalar curvature $S(p)$ at the point $p$ is defined to be the trace of $\rho$.

If $e_1,\cdots,e_n$ is an orthonormal basis for $T_pM$, then $$S(p)=\mbox{tr}\rho=\sum_j \langle \rho(e_j),e_j\rangle=\sum_j \mbox{Ric}(e_j,e_j)=\sum_{i,j}\langle R(e_i,e_j)e_j,e_i\rangle.$$

22.9 Defining a Connection Using Connection Matrices

In the next section we will describe how a connection on a vector bundle induces a connection on a pullback bundle. The easiest way to do this is to define a connection using connection matrices.

If $\nabla$ is a connection on a vector bundle $E\rightarrow M$, then relative to each framed open set $(U,e_1,\cdots,e_r)$ for $E$, there is a connection matrix $\omega_e$. By Theorem 22.1, the connection matrix relative to another frame $\bar{e}=ea$ is related to $\omega_e$ by formula $$\begin{align}\omega_\bar{e}=a^{-1}\omega_ea+a^{-1}da.\end{align}$$

For $s\in\Gamma(E)$ and $X\in\mathfrak{X}(M)$, to define $\nabla_Xs\in \Gamma(E)$, it suffices to define $(\nabla_Xs)|_U$ for the open set $U$ of an open cover of $M$. By the definition of the restriction of a connection, $$(\nabla_Xs)|_U=\nabla^U_X(s|_U).$$ Thus, a connection $\nabla$ on $E$ is completely determined by its restrictions $\nabla^U$ to the open sets of an open cover for $M$. Conversely, a collection of connection $\nabla^U$ on the opne sets $U$ of a trivializing open cover of $M$ that agree on pairwise intersections $U\cap U'$ defined a connection on $E$.

Equivalently, since a connection on a trivial bundle is completely specified by its connection matrix relative to a frame (see Section 11.2), a connection on a vector bundle can also be given by a collection of connection forms $\{\omega_e,(U,e)\}$ satisfying the compatibility conditions (22.11).

22.10 Induced Connection on a Pullback Bundle

In this section we show that under a $C^\infty$ map $f:N\rightarrow M$, a connection on a vector bundle $E\rightarrow M$ pulls back to a unique connection on $f^*E\rightarrow N$.

Theorem 22.10. Let $\nabla$ be a connection on a vector bundle $E\rightarrow M$ and $f:N\rightarrow M$ at $C^\infty$ map. Suppose $M$ is covered by framed open sets $(U,e_1,\cdots,e_r)$ for $E$ and the connection matrix relative to the frame $e$ is $\omega_e$. Then there is a unique connection on $f^*E$ whose connection matrix relative to the frame $f^*e_1,\cdots,f^*e_r$ on $f^{-1}(U)$ is $f^*(\omega_e)$.

Proof. The matrix $f^*(\omega_e)$ of $1$-forms defined a connection on the trivial bundle $(f^*E)|_{f^{-1}(U)}$ relative to the frame $f^*e$ (Section 11.2). If $\bar{e}=ea$ is another frame on $U$, then by Theorem 22.1 $$\omega_\bar{e}=a^{-1}\omega_ea+a^{-1}da\mbox{ on }U.$$ Taking the pullback under $f^*$ gives $$f^*(\omega_\bar{e})=(f^*a)^{-1}f^*(\omega_e)f^*a+(f^*a)^{-1}df^*a\mbox{ on }f^{-1}(U).$$ Since $f^*\bar{e}=(f^*e)(f^*a)$, the equation above shows that $\{f^*(\omega_e)\}$ satisfies the compatibility condition (11) and defines a unique connection on $f^*E$.