I was struggling with the motivation for connection and found it more comfortable to think from parallel transport. In Wiki, parallel transport is a map between tangent spaces denoted by $\Gamma(\gamma)_s^t:T_{\gamma(s)}M \rightarrow T_{\gamma(t)}M$, where $\gamma$ is the flow of the vector field $X$. This map satisfies the necessary smoothness properties and also (*) $\Gamma(\gamma)_u^t \circ \Gamma(\gamma)_s^u = \Gamma(\gamma)_s^t$. This is the intuition in preserving parallelism along the curve $\gamma$. Finally, the connection is defined by \begin{align*} \nabla_X Y = \lim_{s \rightarrow 0} \frac{\Gamma(\gamma)^0_s Y-Y}{s}. \end{align*} All of this makes sense to me.

However, I tried to prove $\nabla_{fX+gZ} = f\nabla_X +g\nabla_Z$ and I don't know if I can do it without assuming that $\Gamma(\gamma)^0_s$ must be linear. My feeling is that the linearity of the map can be proven using condition (*), but I can't do it. Any help and hints are greatly appreciated.

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    $\begingroup$ surely you have to assume linearity, I would think $\endgroup$ Jan 29, 2018 at 16:03

1 Answer 1


Assume that, if $\gamma:[a,b]\to M$, and take a diffeomorphism $\phi:[\alpha,\beta]\to [a,b]$ such that $a=\phi(\alpha),\ b=\phi(\beta)$, then $\Gamma(\gamma)_a^b=\Gamma(\gamma\circ\phi)_\alpha^\beta$.

We will use a lemma: if $f:\mathbb R^n\to\mathbb R^n$ is $C^\infty$, and satisfy $\forall_{\lambda\in \mathbb R \\ x\in \mathbb R^n}: f(\lambda x)=\lambda f(x)$, then $\underbrace{\frac{d}{d\lambda} f(\lambda x)}_{=df(\lambda x)\cdot x}= f(x)$ (note that $df(\lambda x)\in \operatorname{Lin}(\mathbb R^n,\mathbb R^n)$ is differential), and set $\lambda=0$ we get $df(0)\cdot x = f(x)$, that is, $f$ is linear map. (Euler homogeneous function)

Then, w.l.o.g. let $\gamma:[0,1]\to M$, $\phi(t)=\lambda t$ we have $\dot\gamma(0)=\lambda \frac{d}{dt}(\gamma\circ\phi)(0)$. so, denote $\dot\gamma(0)=X$,

$$\begin{align}\nabla_{\lambda X} Y&=\lim_{t\to 0}\frac{\Gamma(\gamma\circ\phi)_{t}^0 Y_{\gamma(\lambda t)} - Y_{\gamma(0)}}{\frac{1}{\lambda }\cdot\lambda t}\\ &=\lambda\cdot \lim_{\tau=\lambda t\to 0} \frac{\Gamma(\gamma)_{\tau}^0 Y_{\gamma(\tau)}-Y_{\gamma(0)}}{\tau}=\lambda\cdot\nabla_X Y\end{align}$$

If we take some extra assumption, to ensure the map $$\begin{align} T_p M &\longrightarrow T_p M \\ X&\longmapsto \nabla_X Y \end{align}$$

to be $C^\infty$ (or just $C^1$), then by the lemma, we get linearity.

The linearity of $\Gamma(\gamma)_0^1:T_{\gamma(0)}M\to T_{\gamma(1)}M​​$ is often an assumption. But when you let the parallel transport always preserve metric or distance, the map will be force to be linear. See Are non-bijective isometries necessarily affine functions?. This is also true in semi-Riemannian case.

(from a textbook written in Chinese, by Hung-Hsi Wu.)


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