# Trying to intuitively understand Christoffel symbols

I recently watched Sean Carroll's YouTube series on "The Biggest Ideas in the Universe". In his Geometry and Topology video, he says that the connection in Riemannian geometry describes how a vector gets parallel transported through the geometry.

I looked more into it and found that the Christoffel symbols are an array of numbers that describe the metric connection which itself describes how the basis varies from point to point.

The formula for the Christoffel symbols of the first kind is $$\Gamma_{kij} = \frac{\partial \vec{e_{i}}}{\partial x^j} \cdot \vec{e_k}$$. I'm trying to understand this formula intuitively, specifically how it describes how the covariant basis varies from point to point.

It seems to describe the projection of the rate of change of $$\vec{e_i}$$ with respect to the $$x^j$$ direction onto $$\vec{e_k}$$. As I understand it, this means that $$\frac{\partial \vec{e_{i}}}{\partial x^j}$$ is a vector that represents the change in the covariant basis vector $$\vec{e_i}$$ when you move in the $$x^j$$ direction. This resulting derivative vector then has components/projections along each covariant basis direction $$\vec{e_k}$$. So the Christoffel symbol describes the components/projections of each of these derivative vectors along each of the covariant basis directions at various points.

Is this the correct intuition behind the Christoffel symbols or do I have a misconception somewhere?

Also if this intuition is correct, I couldn't seem to get similar intuition for the Christoffel symbols of the second kind. Their definition is $$\Gamma ^k_{ij} = \frac{\partial \vec{e_i}}{\partial x^j} \cdot \vec{e^k}$$.

But I don't know how to intuitively grasp the projection onto $$\vec{e^k}$$ because the contravariant basis vectors $$\vec{e^k}$$ are vectors orthogonal to the covariant basis directions right? I guess the symbols of the second kind would be the projections of the derivatives onto the various directions orthogonal to the covariant basis?

Yes, that's about the size of it. If you want to test your intuition, consider polar coordinates on a Euclidean plane. At a point $$(r,\theta)$$, the orthonormal polar basis vectors $$\hat e_r$$ and $$\hat e_\theta$$ are given by

$$\hat e_r = \cos(\theta) \hat e_x + \sin(\theta) \hat e_y$$ $$\hat e_\theta = -\sin(\theta) \hat e_x + \cos(\theta) \hat e_y$$

where $$\hat e_x$$ and $$\hat e_y$$ are the standard Cartesian unit vectors. Clearly $$\hat e_r$$ and $$\hat e_\theta$$ change as you wander around the plane. One has that

$$\frac{\partial \hat e_r}{\partial r} = \frac{\partial \hat e_\theta}{\partial r} = 0$$ $$\frac{\partial \hat e_r}{\partial \theta} = \hat e_\theta$$ $$\frac{\partial \hat e_\theta}{\partial \theta} = -\hat e_r$$ As a result, in this coordinate system there are only two nonzero Christoffel symbols:

$$\Gamma_{\theta r \theta} = 1$$ $$\Gamma_{r\theta\theta} = -1$$

The remaining six are all zero. Because the metric is $$g_{ij}=\pmatrix{1&0\\0&1}$$ in this basis, these are trivially related to the Christoffel symbols of the second kind, $$\Gamma^i_{j k} = g^{im}\Gamma_{mjk}$$

The Christoffel symbols of the second kind $$\Gamma^i_{jk}$$ is the $$i^{th}$$ component of $$\frac{\partial \hat e_k}{\partial e_j}$$. That is, when we say $$\frac{\partial \hat e_r}{\partial \theta} = \hat e_\theta$$, it follows immediately that the $$\theta$$ component of the vector on the right hand side is equal to $$1$$, i.e. $$\Gamma^\theta_{\theta r} = 1$$.

The Euclidean plane is flat, of course, so if you calculate e.g. the Riemann tensor, you would find that all of the components vanish. This would be a nice, quick exercise.

• Thanks! This explains the Christoffel symbols of the first kind. But I edited my question cause I wanted to check if my intuition for those of the second kind was correct as well. But wait it seems that in polar coordinates the symbols of the second kind have factors of $r$ in them, how come your derivation does not have them? I think I just don't understand covariant vs. contravariant basis that intuitively. I also thought the metric in polar coordinates had an $r^2$ in it? Jul 23 '20 at 18:33
• Jul 23 '20 at 18:36
• @mihirb I have updated my answer with a small addendum about the Christoffel symbols of the second kind. As for the lost factors of $r$, I am working in the orthonormal polar basis. The polar coordinate basis, with basis vectors $\frac{\partial}{\partial r}$ and $\frac{\partial}{\partial \theta}$ is not orthonormal, so your results will be different. Jul 23 '20 at 18:41
• I see. I thought it was something to do with normalization. Thanks! Jul 23 '20 at 18:42