# Parallel transport for infinitesimal displacement

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 $$pr_1 = ps + sr_1 = δ^μ + ε^μ − \Gamma^{\mu}_{\nu \lambda} \varepsilon^{\lambda} \delta^{\nu} .$$

• 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 ? – reuns May 24 '16 at 20:27
• The connection or the Christoffel is not a tensor – user156175 May 24 '16 at 20:40
• 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 ? – reuns May 24 '16 at 20:44
• 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). – user156175 May 24 '16 at 22:56
• I'm just saying that what you wrote is almost the definition of the Christoffel symbols – reuns May 24 '16 at 23:01

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.