# Can someone explain how Weinberg's definition of the affine connection for the geodesic equation matches the definition of an affine connection?

Consider the geodesic equation $$0=\frac{d^2 x^\lambda}{d\tau^2}+ \Gamma^\lambda_{\mu\nu} \frac{d x^\nu}{d\tau}\frac{d x^\mu}{d\tau}$$ 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 $$\Gamma^\lambda_{\mu\nu} \equiv \frac{\partial x^\lambda}{\partial \xi^{\alpha}}\frac{\partial^2 \xi^\alpha}{\partial x^\mu \partial x^\nu}$$ 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.

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.

• Are both ${\nabla_\mu}A^\nu$ and $\Gamma^\mu_{\nu\lambda}$ affine connections? Mar 15 '15 at 22:54
• @StanShunpike I'm writing the details. Mar 15 '15 at 22:59
• @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. Mar 15 '15 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.