$\newcommand{\R}{\mathbb{R}}$ Everything is assumed to be $C^\infty$.

### Pullback of a vector bundle

Let $E$ be a rank $r$ vector bundle over an $m$-manifold $M$. Given an $n$-manifold $N$ and a smooth map $\Phi: N \rightarrow M$, the pullback bundle $E^\Phi$ over $N$ is defined as follows:

• The fiber at $x \in N$ is $E^\Phi_x = E_{\Phi(x)}$.
• Given any local frame $(e_1, \dots, e_r)$ of $E$, $(e_1\circ\Phi, \cdots, e_r\circ\Phi)$ is a local frame of $E^\Phi$. In particular, any section $f$ of $E^\Phi$ can be written locally as $f(x) = a^k(x)e_k(\Phi(x)),$ where $(e_1,\dots, e_r)$ is a local frame of $E$ and $(a^1, \dots, a^r): N \rightarrow \R^r$ is smooth.
In other words, the space of smooth sections of $E^\Phi$ is the $C^\infty(N)$-module generated by the space of smooth sections of $E$ pulled back by $\Phi$.

Given any section $f$ of $E$, $f\circ\Phi$ is a section of $E^\Phi$. For convenience, we will also denote $f\circ\Phi$ by simply $f$.

### Pushforward of the Lie bracket

Using the pullback of the tangent bundle of $M$, we can derive a formula for the pushforward of the Lie bracket of two vector fields on $N$.

Let - $(x^1, \dots, x^n)$ be local coordinates on $N$, and the corresponding coordinate vector fields by $(\partial_1, \dots, \partial_n)$. - $(y^1, \dots, y^m)$ be local coordinates on $M$ and $(Y_1, \dots, Y_m)$ be the corresponding coordinate vector fields. If $\Phi(x) = (y^1(x)), \dots, y^m(x))$, then, for any $1 \le p \le n$ and $1 \le j \le m$, \begin{align*} \Phi_*\partial_p &= \phi_p^jY_j\\ \end{align*} where \begin{align*} \phi_p^j &= \partial_py^j. \end{align*} Therefore, for any $1 \le p, q \le n$ and $1 \le j \le m$, $$\label{commute} \partial_p\phi_q^j = \partial_p\partial_qy^j = \partial_q\partial_py^j =\partial_q\phi_p^j.$$

Given $V \in T_xN$ and a section $\widehat{W} = b^kY_k$ of the pullback $T^\Phi_*N$ of the tangent bundle $T_*N$, we can define the directional derivative of $\widehat{W}$ at $x$ to be $V\widehat{W} = \langle V, db^k\rangle Y_k.$

The following lemma provides a formula for $\Phi_*[V,W]$ using the pullback bundle $T^\Phi_*M$.

Given tangent vector fields $V$ and $W$ on $N$, $$\label{LieBracket} \Phi_*[V,W] = V\Phi_*W - W\Phi_*V.$$
If $V = a^p\partial_p$ and $W = b^p\partial_p$, then \begin{align*} V\Phi_*W - W\Phi_*V &= a^p\partial_p(\phi^k_qb^q)Y_k - b^p\partial_p(\phi^k_qa^q)Y_k\\ &= (a^p\partial_pb^q - b^p\partial_pa^q)\phi^k_qY_k + a^pb^q(\partial_p\phi^k_q - \partial_q\phi^k_p))Y_k\\ &= (a^p\partial_pb^q - b^p\partial_pa^q)\Phi_*\partial_q\\ &= \Phi_*[V,W]. \end{align*}

If the pushforwards $\Phi_*V$ and $\Phi_*W$ can be extended smooth vector fields $\widetilde{V}$ and $\widetilde{W}$ on $M$, then they are said to be $\Phi$-related. In that case, the left side of the equation below is equal to the right side of \eqref{LieBracket}, and therefore $[\Phi_*V, \Phi_*W] = \Phi_*[V,W].$ follows from the lemma. In particular, this is the case, if $\Phi$ is an embedding.

Otherwise, the left side of this identity is ill-defined and therefore the identity does not hold in general.

### Pullback of a connection

Let $\nabla$ be a connection on $E$. Observe that, since each $e_k$ is a section on $M$, given any vector field $v$ on $M$, $(\nabla_ve_k)$ is a section of $E$.

The pullback connection $\nabla^\Phi$ is a connection on $E^\Phi$ that is defined as follows: Given any $V \in T_xN$, $\nabla^\Phi_Ve_k = \nabla_{\Phi_*V}e_k.$ Therefore, given a smooth section $f = a^ke_k$ of $E^\Phi$, \begin{align*} \nabla^\Phi_V f(x) &= \nabla^\Phi_V (a^ke_k)\\ &= \langle V, da^k(x)\rangle e_k + a^k\nabla_{\Phi_*V}e_k. \end{align*} Observe that if $V$ is a tangent vector field on $N$, then $\nabla^\Phi_Vf$ is a smooth section of $E^\Phi$.

### Pullback of curvature

Given a section $f = a^ke_k$ of $E^\Phi$, \begin{align*} \nabla^\Phi_{\partial_p}(\nabla^\Phi_{\partial_q}f) &= \nabla^\Phi_{\partial_p}(\partial_qa^k e_k + a^k\nabla_{\Phi_*\partial_q}e_k)\\ &= \nabla^\Phi_{\partial_p}(\partial_qa^k e_k + a^k\phi_q^j\nabla_{Y_j}e_k)\\ &= \partial_p(\partial_qa^k)e_k + \partial_qa^k\nabla_{\Phi_*\partial_p}e_k + \partial_p(a^k\phi_q^j)\nabla_{Y_j}e_k + a^k\phi_q^j\nabla_{\Phi_*\partial_p}(\nabla_{Y_j}e_k)\\ &= \partial_p(\partial_qa^k)e_k + \phi_p^j\partial_qa^k\nabla_{Y_j}e_k + \partial_p(a^k\phi_q^j)\nabla_{Y_j}e_k + a^k\phi_q^j\phi_p^i\nabla_{Y_i}(\nabla_{Y_j}e_k) \end{align*} Therefore, since $[\partial_p,\partial_q] = 0$ and $[Y_i,Y_j] = 0$, if $R$ is the curvature tensor of $\nabla$, \begin{align*} \nabla^\Phi_{\partial_p}(\nabla^\Phi_{\partial_q}f) - \nabla^\Phi_{\partial_q}(\nabla^\Phi_{\partial_p}f) &= a^k\phi_p^i\phi_q^j(\nabla_{Y_i}(\nabla_{Y_j}e_k) - \nabla_{Y_j}(\nabla_{Y_i}e_k)\\ &= a^k\phi_p^i\phi_q^jR(Y_i,Y_j)e_k\\ &= R(\phi_p^iY_i, \phi_q^jY_j)a^ke_k\\ &= R(\Phi_*\partial_p, \Phi_*\partial_q)f. \end{align*} Therefore, if $R^\Phi$ is the curvature tensor of $\nabla^\Phi$, then, for any $x \in N$ and $f \in E^\Phi_x = E_{\Phi(x)}$, $R^\Phi(\partial_p,\partial_q)f = R(\Phi_*\partial_p,\Phi_*\partial_q)f.$ From this, a straightforward calculation shows that given any vector fields $V, W$ on $N$ and section $f$ of $E^\Phi$, $$\label{curvature} R^\Phi(V,W)f = \nabla^\Phi_{V}(\nabla^\Phi_{W}f) - \nabla^\Phi_{W}(\nabla^\Phi_{V}f) - \nabla^\Phi_{[V,W]}f = R(\Phi_*V,\Phi_*W)f.$$

### Pullback of torsion-free connection

Assume that $\nabla$ is a torsion-free connection on $T_*M$, i.e., $$\label{torsion-free} \nabla_{Y_j}Y_k = \nabla_{Y_k}Y_j.$$ By the definition of a pullback connection, \begin{align*} \nabla^\Phi_{\partial_p} \Phi_*\partial_q &= \nabla^\Phi_{\partial_p}(\phi_q^jY_j)\\ &= (\partial_p \phi_q^j)Y_j + \phi_q^j\nabla_{\Phi_*\partial_p}Y_j\\ &= (\partial_p \phi_q^j)Y_j + \phi_q^j\phi_p^k\nabla_{Y_k} Y_j. \end{align*} Similarly, \begin{align*} \nabla^\Phi_{\partial_q} \Phi_*\partial_p &= (\partial_q \phi_p^j)Y_j + \phi_p^j\phi_q^k\nabla_{Y_k}Y_j. \end{align*} Therefore, by \eqref{commute} and \eqref{torsion-free}, $$\label{symmetric} \begin{split} \nabla^\Phi_{\partial_p} \Phi_*\partial_q -\nabla^\Phi_{\partial_q} \Phi_*\partial_p &= (\partial_p\phi_q^j - \partial_q\phi_p^j)Y_j + \phi_q^j\phi_p^k(\nabla_{Y_k}Y_j-\nabla_{Y_j}Y_k)\\ &= 0. \end{split}$$ This and the properties of a connection imply the following:

Given a smooth map $\Phi: N \rightarrow M$, and tangent vector fields $V, W$ on $N$, \begin{align*} \nabla^\Phi_{V} \Phi_*W -\nabla^\Phi_{W} \Phi_*V &= \Phi_*[V,W]. \end{align*}

### Riemann curvature

If $\nabla$ is the Levi-Civita connection and $R$ is the Riemannian curvature of a Riemannian metric on $M$, then for any smooth tangent vector fields $U, V, W$ on $N$, then by \eqref{curvature}, the curvature of $\nabla^\Phi$ is given by \begin{align*} R^\Phi(V,W)U &= \nabla^\Phi_{V}(\nabla^\Phi_{W}U) - \nabla^\Phi_{W}(\nabla^\Phi_{V}U) - \nabla^\Phi_{[V,W]}U = R(\Phi_*V, \Phi_*W)\Phi_*U. \end{align*} Observe that this identity holds for any smooth map $\Phi$, and there are no assumptions about the dimensions of $N$ and $M$. Also, no connection or Riemannian metric on $N$ is needed.

### Jacobi fields

A Jacobi field on a Riemannian manifold $M$ is the variation of a $1$-parameter family of geodesics, which is a map $\Phi: I_1\times I_2 \rightarrow M$, where $I_1, I_2$ are intervals. Since the vector fields $\partial_1, \partial_2$ are not necessarily $\Phi$-releated, the lemma above plays a crucial role in calculations involving Jacobi fields, including the proof that a Jacobi field satisfies the Jacobi equation. In most expositions of Jacobi fields, this is done in an ad hoc manner. This article shows that it fits within the more general framework of the pullback of the tangent bundle and Levi-Civita connection.