# Is there a coordinate free proof for this result?

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.

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 $\gamma : N\to\gamma(N)$.)
• 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?
• @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. Feb 7 '18 at 3:04
• 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. Feb 7 '18 at 3:12