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I have a question about the following calculation about the parallel transport of an infinitesimal vector. I read the following text but I do not understand where the expression for the components of vector $sr_1$ comes from, especially not the minus sign. If I understood it correctly a vector is parallel transported when its covariant derivative along a parametrized curve vanishes, but I do not understand how I get the expression for $sr_1$ by parallel transport of $X$ along $ps$. Could someone explain to me how to derive this expression? Especially I don't know where the minus sign comes from.

Thanks a lot

Let $p \in M$ be a point whose coordinates are $\{x^μ\}$. Let $X = \varepsilon^μ e_μ$ and $Y = \delta^μ e_μ$ be infinitesimal vectors in $T_pM$. If these vectors are regarded as small displacements, they define two points $q$ and $s$ near $p$, whose coordinates are $\{x^μ + ε^μ\}$ and $\{x^μ + δ^μ\}$ respectively . If we parallel transport $X$ along the line $ps$, we obtain a vector $sr_1$ whose component is $\varepsilon^μ − \varepsilon^{\lambda} \Gamma^{\mu}_{\nu \lambda} \delta^{\nu}$ . The displacement vector connecting $p$ and $r_1$ is $$pr_1 = ps + sr_1 = δ^μ + ε^μ − \Gamma^{\mu}_{\nu \lambda} \varepsilon^{\lambda} \delta^{\nu} .$$

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  • $\begingroup$ your notation (naming) is awful... and everything comes from the definition of $\Gamma_{\mu \lambda}^\mu$. do you understand why it is a order $3$ tensor ? $\endgroup$ – reuns May 24 '16 at 20:27
  • $\begingroup$ The connection or the Christoffel is not a tensor $\endgroup$ – user156175 May 24 '16 at 20:40
  • $\begingroup$ let's say it is a linear operator $\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n$ where $\mathbb{R}^n$ is $T_pM$ in local coordinates. my question was : do you see why it is $\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n$ ? how do you explain it ? $\endgroup$ – reuns May 24 '16 at 20:44
  • $\begingroup$ I don't really know what you are trying to say. We motivated the definition of the symbol for the affine connection so that the covariant derivative of a vector field transforms as a tensor which it would not do by just taking the partial derivative. However, as far as I understood this example should be more "technical" and was calculated just by using the definition of parallel transport and a affine connection ( not the christoffel symbol). $\endgroup$ – user156175 May 24 '16 at 22:56
  • $\begingroup$ I'm just saying that what you wrote is almost the definition of the Christoffel symbols $\endgroup$ – reuns May 24 '16 at 23:01
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If I understand correctly, you are essentially asking how to derive the formula for the infinitesimal parallel transport of a vector $v\in T_pM$ in some direction $a\in T_pM$.

One usually defines parallel transport of vectors along a curve. So let's do this first. Let $v\in T_xM$ be some vector and $\gamma:[0,1]\to M$ a curve starting at $p$, i.e., $\gamma(0)=p$. We say that a vector field $\bar v:I\to TM$ along $\gamma$ (that is, $\bar v(t)\in T_{\gamma(y)}M$) parallel transports $v$ if $\bar v(0)=v$ and $\nabla_{\dot\gamma}\bar v=0$. Employing index notation, the latter condition can be written in terms of the connection coefficients $\Gamma^\rho_{\mu\nu}$ as

$$ 0 = \dot{\gamma}^\mu\nabla_\mu\bar v^\rho = \dot\gamma^\mu\left(\partial_\mu\bar v^\rho + \Gamma^\rho_{\mu\nu}\bar v^\nu\right) = \dot\gamma^\mu\partial_\mu\bar v^\rho + \Gamma^\rho_{\mu\nu}\dot\gamma^\mu\bar v^\nu = \left(\frac{\text{d}\bar v}{\text{d} t}\right)^\rho + \Gamma^\rho_{\mu\nu}\dot\gamma^\mu\bar v^\nu.$$

This differential equation has a unique solution $\bar v(t)$ given any initial vector $\bar v(0)$, meaning that there is a unique way to parallel transport our vector $v$ along $\gamma$.

Now let's try to understand infinitesimal parallel transport in a direction $a\in T_pM$. That means we transport $v$ over an infinitesimal time interval $\Delta t$ along a curve $\gamma$ that goes in the direction $a$ at $p$, that is, $\dot\gamma(0)=a$. (I'm not writing d$t$ because people might interpret this as a 1-form.) Infinitesimally, we may write

$$ \bar v^\rho(t) = \bar v^\rho(0) + \left(\frac{\text{d}\bar v}{\text{d} t}\bigg|_{t=0}\right)^\rho\Delta t.$$

Using the formula we found for parallel transport, in which we substituting $t=0$, we obtain

$$\bar v^\rho(\Delta t) = v^\rho-\Gamma^\rho_{\mu\nu}\dot\gamma^\mu(0)\bar v^\nu(0)\Delta t = v^\rho-\Gamma^\rho_{\mu\nu} (a^\mu \Delta t) v^\nu = v^\rho-\Gamma^\rho_{\mu\nu} \delta^\mu v^\nu, $$ where $\delta$ is the infinitesimal displacement in the direction $a$.

Thus we see that if we infinitesimally parallel transport $v$ from $x^\mu$ to $x^\mu + \delta^\mu$, the resulting vector is given by $v^\rho-\Gamma^\rho_{\mu\nu} \delta^\mu v^\nu$. Hope that helps.

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