Consider the geodesic equation \begin{equation} 0=\frac{d^2 x^\lambda}{d\tau^2}+ \Gamma^\lambda_{\mu\nu} \frac{d x^\nu}{d\tau}\frac{d x^\mu}{d\tau} \end{equation} In Gravitation and Cosmology, on page 71 Weinberg derives this equation from the equivalence principle. In the process of doing so, he defines the Christoffel symbol as \begin{equation} \Gamma^\lambda_{\mu\nu} \equiv \frac{\partial x^\lambda}{\partial \xi^{\alpha}}\frac{\partial^2 \xi^\alpha}{\partial x^\mu \partial x^\nu} \end{equation} I am still confused as to what exactly the Christoffel symbol does mathematically. I gather it is an affine connection. And I have read many times the usual definition of an affine connection in the context of Riemannian geometry, but I still don't see how this fits the definition of an affine connection.

According to do Carmo, in Riemannian Geometry pages 49-50, he says let $\mathcal{X}(M)$ denote the set of all vector fields of class $C^{\infty}$ on $M$. Let $\mathcal{D}(M)$ denote the ring of all real-valued functions of class $C^{\infty}$ defined on $M$. An affine connection $\nabla$ on differential manifold $M$ is a mapping $\nabla : \mathcal{X}(M) \times \mathcal{X}(M) \rightarrow \mathcal{X}(M)$ which is denoted by $(X,Y) \xrightarrow{\nabla} \nabla_{X}Y$ and which satisfies the following properties:

  1. $\nabla_{fX+gY}Z = f\nabla_{X} Z+ g\nabla_{Y}Z$
  2. $\nabla_{X}(Y+Z) = \nabla_{X}Y + \nabla_{X}Z$
  3. $\nabla_{X}(fY) = f\nabla_{X}Y+ X(f)Y$

in which $X,Y,Z \in \mathcal{X}(M)$ and $f,g \in \mathcal{D}(M)$.

My Question:

Can someone explain how Weinberg's definition of an affine connection with simple partial derivatives matches do Carmo's complicated definition?

I can't seem to grasp do Carmo's definition but I know it's important. I feel like if I understood the relation between Weinberg's example and do Carmo's definition, I would be able to understand the definition of an affine connection much better. Then I could use Weinberg's explanations of this example as a start point for understanding the formal definition given by do Carmo.


2 Answers 2


When you consider geodesics, it might be easier to compute the Christoffel's symbols using their definition in terms of the metric, i.e.: $$\Gamma^{\rho}_{\ \mu\nu}=\dfrac{1}{2}g^{\rho\sigma}\left(g_{\sigma\mu,\nu}+g_{\sigma\nu,\mu}-g_{\mu\nu,\sigma}\right)$$ The expression you have arises when you consider how the differentiation operator, $\partial_\mu$, behaves under a general transformation of coordinates. In particular, you should observe that $\partial_\mu A^{\nu}$ does not transform as a $(1,1)$ tensor: there is some left-over term.

$$\begin{align} \partial_\mu A^\nu\to\partial_{\mu '}A^{\nu '}&=\left(\dfrac{\partial x^{\mu}}{\partial x^{\mu '}}\partial_\mu\right)\left(\dfrac{\partial x^{\nu '}}{\partial x^\nu}A^\nu\right)\\ &=\dfrac{\partial x^{\mu}}{\partial x^{\mu '}}\dfrac{\partial x^{\nu '}}{\partial x^{\nu}}\left(\partial_\mu A^\nu\right)+\dfrac{\partial x^\mu}{\partial x^{\mu '}}\dfrac{\partial^2 x^{\nu '}}{\partial x^\nu\partial x^\mu} A^\mu \end{align}$$

Now, the left term transforms as we expected, but the right term is the "undesirable left-over". That's where the affine connection comes up. We need to introduce an extra element in our derivative, that will "take care" of this left-over.

We need something that cancels the rightmost term above, but has also a tensor-like part (since we need a clean overall transformation). Therefore:

$$\Gamma^{\nu '}_{\ \mu ' \lambda '}=\dfrac{\partial x^{\mu}}{\partial x^{\mu'}}\dfrac{\partial x^{\nu'}}{\partial x^\nu}\dfrac{\partial x^\lambda}{\partial x^{\lambda'}}\Gamma^{\nu}_{\ \mu\lambda}-\dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial x^\lambda}{\partial x^{\lambda '}}\dfrac{\partial^2x^{\nu'}}{\partial x^\mu\partial x^\nu}$$

We can now define the new covariant derivative: $$\nabla_\mu A^\nu=\partial_\mu A^\nu+\Gamma^\nu_{\ \mu\lambda}A^\lambda$$ And you can see that the second part of the Christoffel symbol will indeed cancel out the unwanted term, whereas its first part transforms correctly. In effect, $\nabla_\mu A^\nu$ does now transform as a proper $(1,1)$ tensor.

  • $\begingroup$ Are both ${\nabla_\mu}A^\nu$ and $\Gamma^\mu_{\nu\lambda}$ affine connections? $\endgroup$ Mar 15, 2015 at 22:54
  • $\begingroup$ @StanShunpike I'm writing the details. $\endgroup$
    – Demosthene
    Mar 15, 2015 at 22:59
  • $\begingroup$ @StanShunpike $\Gamma^{\mu}_{\ \nu\lambda}$ is the affine connection (not a tensor), whereas $\nabla_{\mu}A^{\nu}$ is the covariant derivative of $A^\nu$ and transforms as a $(1,1)$ tensor. $\endgroup$
    – Demosthene
    Mar 15, 2015 at 23:12

The upshot is that Weinberg's definition (3.2.4) on p.71 of the Christoffel symbols $\Gamma^{\lambda}_{\mu\nu}$ can only describe a local flat space (in some local coordinate neighborhood). So Weinberg's definition does not apply to the generic curved case. See also e.g. this related Phys.SE post.


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