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Let $(M,g)$ be a smooth manifold with metric tensor $g$. For this problem, the signature of $g$ seems to be of no importance. Consider a vector bundle $\pi : E\to M$ and a smooth embedding $\phi : N\to M$. We can form the pullback bundle $\phi^\ast E$ and introduce on it the induced connexion, characterized by

$$(\phi^\ast\nabla)_X (\phi^\ast f)=\phi^\ast(\nabla_{\phi_\ast X}f).$$

First of all it is worth to remark that if $f : N\to \phi^\ast E$ is a section, we know it to be of the form $f(x)=(x,\bar{f}(x))$ where $\bar{f} : N\to E$ with $\pi\circ \bar{f}=\phi$. In other words, we extract $\bar{f}$ from $f$ by composing with the projection on the second factor $\operatorname{pr}_2: N\times E\to E$.

If $f : N\to \phi^\ast E$ is a section, we shall denote

$$(\phi^\ast \nabla)_{X}\bar{f}=\operatorname{pr}_2\circ (\phi^\ast\nabla)_X f.$$

I have first of all shown the following result:

Let $\{e_a\}$ be a local frame of $E$ so that $f = f^a \phi^\ast e_a$. Suppose further that $(y,V)$ is a coordinate system on $N$ and $(x,U)$ a coordinate system on the image of $\phi$ inside $M$. Define $\Gamma_{\kappa a}^b$ by $$\nabla_{\frac{\partial}{\partial x^\kappa}}e_a=\Gamma_{\kappa a}^b e_b$$ then we have

$$(\phi^\ast \nabla)_X \bar{f}=\left[X(f^b)+f^a(\phi_\ast \circ X)^\kappa \Gamma_{\kappa a}^b\circ \phi\right]e_b\circ \phi.$$

With this we can prove the following result: let $E=TM$ be the tangent bundle and $\nabla$ some connexion on it (not necessarily the Levi-Civita). Let $N=(-\epsilon,\epsilon)\times [a,b]$ equiped with the natural coordinate functions $(s,u)$ which are just the components of the identity. Let $\gamma : N\to M$ be a one-parameter family of geodesics. Introduce the tangent and deviation vector fields on $\gamma^\ast(TM)$ by

$$\bar{T}=\gamma_\ast \circ \dfrac{\partial}{\partial u},\quad \bar{S}=\gamma_\ast \circ \dfrac{\partial}{\partial s}.$$

It is clear that the components of these vector fields over $\gamma$ are just

$$\bar{T}^\beta = \dfrac{\partial \gamma^\beta}{\partial u},\quad \bar{S}^\beta=\dfrac{\partial \gamma^\beta}{\partial s}.$$

In that case I have shown in local coordinates on the image of $\gamma$ the following

\begin{align}(\gamma^\ast\nabla)_{\frac{\partial}{\partial s}}\bar{T}^\beta&=\dfrac{\partial T^\beta}{\partial s}+T^\alpha S^\kappa \Gamma_{\kappa \alpha}^\beta\circ \gamma\\ &= \dfrac{\partial^2 \gamma^\beta}{\partial u\partial s}+T^\alpha S^\kappa \Gamma_{\kappa \alpha}^\beta\circ \gamma\\&=\dfrac{\partial^2 \gamma^\beta}{\partial s\partial u}+T^\alpha S^\kappa \Gamma_{\kappa \alpha}^\beta\circ \gamma\\&=\dfrac{\partial S^\beta}{\partial u}+T^\alpha S^\kappa \Gamma_{\kappa \alpha}^\beta\circ \gamma\\&=\dfrac{\partial S^\beta}{\partial u}+T^\alpha S^\kappa \Gamma_{\alpha\kappa }^\beta\circ \gamma+2 S^\kappa T^\alpha \Gamma_{[\kappa\alpha]}^\beta\circ\gamma\\&=(\gamma^\ast\nabla)_{\frac{\partial}{\partial u}}S^\beta+2S^\kappa T^\alpha\Gamma_{[\kappa\alpha]}^\beta\circ\gamma.\end{align}

In particular if the connexion is torsion-free (the Levi-Civita for example) we get the nice-looking result:

\begin{align}(\gamma^\ast\nabla)_{\frac{\partial}{\partial s}}\bar{T}^\beta&=(\gamma^\ast\nabla)_{\frac{\partial}{\partial u}}S^\beta.\end{align}

This proof seems fine to me (of course I could have made some mistake and not noticed), but I'm wondering: is there some coordinate free proof of this result?

The only way I've found to prove this result was going through the coordinate expression of the pullback connexion, but I believe some quicker and cleaner way without coordinates there should be.

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Every book I've seen this result in proves it using coordinates. It's hard to avoid when dealing with a pullback connection - since the defining formula only defines the connection on pullback sections, in general you necessarily have to do some kind of decomposition, so you might as well work in coordinates.

If you're assuming $\gamma$ is an embedding (as you did initially with $\phi$) then things are much neater: in this case the fields $\gamma_* \partial_u, \gamma_* \partial_s$ can be extended to genuine sections $V,W$ of $TM$ defined on some neighbourhood of $\gamma(N)$, so that we have \begin{align} \bar T &= \gamma_* \partial_u = \gamma^*V, \\ \bar S &= \gamma_* \partial_s = \gamma^*W. \end{align} Applying the pullback connection formula then gives $$(\gamma^* \nabla)_{\partial_u} (\gamma_* \partial_s) = (\gamma^* \nabla)_{\partial_u} (\gamma^* W)= \gamma^*(\nabla_{\gamma_* \partial_u} W)=\gamma^*(\nabla_VW).$$

Swapping $u,s$ here just swaps $V,W$, which has no effect thanks to the symmetry of $\nabla$. (We have $[V,W]=0$ along $\gamma(N)$ since the restrictions $V,W\in \Gamma(T \gamma(N))$ are related to $\partial_s,\partial_u$ by the diffeomorphism $\gamma : N\to\gamma(N)$.)

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  • $\begingroup$ thanks for the answer! Indeed, what you point out was exactly the reason I felt the need to work in coordinates. By the way, what you say about $\gamma_\ast \partial_u$ and $\gamma_\ast \partial_s$ admiting global extensions to genuine sections of $TM$ when $\gamma$ is an embedding, this seems quite one standard result. Can you point me out where I could read a proof of it? $\endgroup$ – user1620696 Feb 7 '18 at 2:35
  • $\begingroup$ @user1620696: I don't know a reference for a full proof - e.g. Lee's book on smooth manifolds leaves it as an exercise. The local version is very easy: in submanifold slice coordinates where $N$ is locally described by $x^1=0,\ldots,x^k=0$ we can simply extend the vector field to be independent of $x^1,\ldots,x^k$. To make this global you need to use a partition of unity. $\endgroup$ – Anthony Carapetis Feb 7 '18 at 3:04
  • $\begingroup$ I edited my answer - what I said was not quite true, in general you cannot extend globally but only to some open neighbourhood of $N$. (Think about the standard embedding $(0,1) \to \mathbb R^2$ with a vector field that does not continuously extend to $[0,1]$. This can be extended to $(0,1)\times \mathbb R$ but not to all of $\mathbb R^2.$) Thankfully, a neighbourhood is good enough for our purposes. $\endgroup$ – Anthony Carapetis Feb 7 '18 at 3:12

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